from binomial model to risk measures unitext · 2020. 1. 17. · emanuelarosazzagianin dipartimento...

Emanuela Rosazza Gianin • Carlo Sgarra

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Mathematical Finance: Theory Review and Exercises

From Binomial Model to Risk Measures

ABC

UNITEXT – La Matematica per il 3+2

Volume 70

For further volumes:http://www.springer.com/series/5418

Emanuela Rosazza Gianin · Carlo Sgarra

Mathematical Finance:Theory Review andExercisesFrom Binomial Model to Risk Measures

Emanuela Rosazza GianinDipartimento di Statistica e Metodi QuantitativiMilano-BicoccaMilanItaly

Carlo SgarraDipartimento di MatematicaPolitecnico di MilanoMilanItaly

UNITEXT – LaMatematica per il 3+2ISSN 2038-5722 ISSN 2038-5757 (electronic)ISBN 978-3-319-01356-5 ISBN 978-3-319-01357-2 (eBook)DOI 10.1007/978-3-319-01357-2Springer Cham Heidelberg New York Dordrecht London

Translated and extended version of the original Italian edition:Emanuela Rosazza Gianin, Carlo Sgarra: Esercizi di finanza matematica,© Springer-Verlag Italia 2007

Library of Congress Control Number: 2013942577

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To my family

To Francesco and Teresa

Preface

This exercise textbook is the result of several years of experience in teachingcourses on Quantitative Finance at our Universities. Although part of thecontent was already published in an Italian edition a few years ago (2007),this new version has been substantially modified: the exercise collection ismuch more consistent and three new chapters have been included: one chap-ter on Arbitrage Theory for discrete-time models, which is a relevant issue inmany courses on Mathematical Finance, another on Risk Measures, and oneon option pricing in models beyond the classical Black-Scholes framework.The subjects covered by the textbook include option-pricing methods basedboth on Stochastic Calculus and Partial Differential Equations, and somebasic notions about portfolio optimization and risk theory. Many textbookson option pricing focus either on stochastic methods or on PDE methods.The aim of the present collection of exercises is to provide examples of bothapproaches and try to illustrate their mutual interplay with the aid of thefundamental notions. Some of the exercises can be solved by students with arather basic mathematical background (essentially, we mean an introductorycourse on Calculus and a basic course in Probability), while others require amore developed mathematical knowledge (a course on Differential Equations,and an introductory course on stochastic processes). The book can be usedin courses with mathematical finance contents aimed at graduate students ofEngineering, Economics and Mathematics.The material proposed has been organized in 12 chapters. The first chap-

ter reviews the basic notions of probability and stochastic processes throughspecific examples, while the second chapter illustrates the fundamental re-sults of (static) portfolio optimization. Chapter 3 presents applications of thebasic option-pricing techniques in a discrete-time framework, mainly in thebinomial setting. Chapter 4 provides examples of the fundamental resultsof arbitrage theory in a discrete-time setting. Chapter 5 reviews basic no-tions on Itô’s integration and on SDEs, while Chapter 6 presents exerciseson some standard methods in partial differential equations, which turn outto be helpful in solving valuation problems for derivatives in continuous-time

viii Preface

models. Chapter 7 focuses on the main results of the Black-Sholes model re-lated to European option-pricing and hedging; examples involving both staticand dynamic hedging strategies are illustrated. Chapter 8 deals with morecomplex derivative products, namely American options: valuation and hedg-ing problems for these options do not admit explicit solutions, except in veryfew (somehow trivial) cases; approximate solutions can be found by applyingsuitable numerical techniques. We present a few examples in a discrete-timesetting and simple applications of the basic notions. Chapter 9 deals withvaluation and hedging of Exotic options; a huge number of different kind ofderivative contracts belong to this class, but we focused on the most commontype of contracts, in particular those for which an explicit solution for thevaluation problem exists in a diffusion setting: Barrier, Lookback, geometricAsian options; moreover, we provide several examples of valuation in a bi-nomial setting. Chapters 10 provides applications of the valuation results forinterest rate derivatives: the concern is mainly on short-rate models, but afew examples on the so-called change of numéraire technique are included.Chapter 11 attempts to illustrate how the derivatives valuation problem canbe attacked in models that drop some of the main assumptions underlyingthe Black-Scholes model: a few examples, mainly involving affine stochas-tic volatility models, are provided together with simple examples of jump-diffusion models; all these are typically incomplete market models, and somespecific assumptions about the risk-neutral measure adopted by the marketin order to assign an arbitrage-free price to contingent claim must be made;the theoretical issues arising in this framework go far beyond the purpose ofthe present textbook, so we decided to limit our description to the most basic(and popular) models, in which these problems can be avoided by makingsimple, but reasonable assumptions. Chapter 12 presents some applicationsof the most important notions related to risk measures; these seem to playan increasingly relevant role in many introductory courses in mathematicalfinance, so we decided to include some examples in our exercise collection.

We have to thank several people for reading the manuscript and providinguseful comments on the material included: we thank Fabio Bellini, MarcoFrittelli, Massimo Morini, Paolo Verzella for reading the first draft of thetextbook, and Andrea Cosso, Daniele Marazzina, Lorenzo Mercuri for readingthe final version and suggesting important modifications. We thank our col-league Giovanni Cutolo for the unvaluable help and assistance on LaTeX andthe graphical packages necessary to edit the manuscript. We thank FrancescaFerrari and Francesca Bonadei of Springer for providing highly qualified edi-torial support. Finally we want to thank all our colleagues and students whooffered any kind of help or comment in support and encouragement to thepresent work.

Milano, May 2013 Emanuela Rosazza GianinCarlo Sgarra

Contents

1 Short review of Probability and of Stochastic Processes . . . . 11.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Portfolio Optimization in Discrete-Time Models . . . . . . . . . . . 172.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3 Binomial Model for Option Pricing . . . . . . . . . . . . . . . . . . . . . . . . 313.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4 Absence of Arbitrage and Completeness of Market Models 614.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

5 Itô’s Formula and Stochastic Differential Equations . . . . . . . . 855.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

6 Partial Differential Equations in Finance . . . . . . . . . . . . . . . . . . . 1016.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1046.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

x Contents

7 Black-Scholes Model for Option Pricing and HedgingStrategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1237.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1237.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1287.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

8 American Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1538.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1538.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1548.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

9 Exotic Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1739.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1739.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1779.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

10 Interest Rate Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20110.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20110.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20710.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230

11 Pricing Models beyond Black-Scholes . . . . . . . . . . . . . . . . . . . . . . 23311.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23311.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23611.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

12 Risk Measures: Value at Risk and beyond . . . . . . . . . . . . . . . . . 24712.1 Review of Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24712.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25012.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275

1

Short review of Probability and ofStochastic Processes

1.1 Review of Theory

Given a probability space (Ω,F , P ), where Ω denotes a non-empty set, F aσ-algebra and P a probability measure on Ω:

• A random variable (r.v.) is a function X : Ω → R such that {ω ∈ Ω :X (ω) ∈ A} ∈ F for any Borel set A in R.

• A stochastic process is a family (Xt)t≥0 of random variables defined on(Ω,F , P ).The stochastic process (Xt)t≥0 is said to be a discrete-time stochastic pro-cess if t takes values in N and a continuous-time stochastic process if t takesvalues in R+.

• A filtration on Ω is a family (Ft)t≥0 of σ-algebras on Ω such that Fu ⊆ Fv

for any u ≤ v.

A stochastic process (Xt)t≥0 is called adapted to the filtration (Ft)t≥0 if,for any s ≥ 0, Xs is Fs-measurable.Given a stochastic process (Xt)t≥0, a filtration (Ft)t≥0 is said to be gener-

ated by (Xt)t≥0 if, for any s ≥ 0, Fs is the smallest σ-algebra that makes Xs

measurable.A filtration can be interpreted as the evolution of the information available

up to a given time.

For a more detailed and exhaustive treatment of the notions recalled here,we refer to the books of Mikosch [28] and Ross [34].

The following families of random variables and stochastic processes arewidely used in Mathematical Finance.

Bernoulli random variableA random variable X has a Bernoulli distribution if it assumes only two

values (typically 1 and 0) with probability p and (1− p) respectively. In sucha case, we write X ∼ B(p).

EmanuelaRosazzaGianin, Carlo Sgarra:Mathematical Finance:Theory Review and Exercises.UNITEXT – La Matematica per il 3+2 70DOI 10.1007/978-3-319-01357-2_1, © Springer International Publishing Switzerland 2013

2 1 Short review of Probability and of Stochastic Processes

As a consequence, the expected value and the variance of X are equal,respectively, to E[X] = p and V (X) = p(1 − p).

Binomial random variable and Binomial processA binomial random variable Yn counts the number of successes in a series

of n independent trials, where p is the probability of success in any one trial.In such a case, Yn ∼ Bin(n; p).

Yn can be written as a sum Yn =∑n

i=1 Xi where (Xi)i=1,...,n are inde-pendent and identically distributed (i.i.d.), Xi ∼ B(p) and Xi = 1 denotes asuccess in trial i. The expected value and the variance of a binomial randomvariable Yn ∼ Bin(n; p) are, respectively, E[Yn] = np and V (Yn) = np(1−p).The sequence (Yn)n∈N is called a binomial process.

Poisson random variable and Poisson processA random variable Z has a Poisson distribution with parameter λ > 0 (in

symbols, Z ∼ Poi(λ)) if it takes values in N and its probability mass functionis given by

P (Z = k) =e−λλk

k!, ∀k = 0, 1, 2, ..., n, ....

for any n ∈ N. Consequently, the expected value and the variance of Z areequal, respectively, to E[Z] = λ and to V (Z) = λ.Furthermore, recall that a Poisson random variable can be obtained by tak-

ing the limit of a sequence of binomial random variables as p → 0, n → +∞and with pn = λ.

Setting λ = νt (for t ≥ 0), ν can be understood as the rate or averagenumber of arrivals per unit of time.A process (Zt)t≥0 is said to be a Poisson process of rate ν if Z0 = 0 and

if all the increments Zt − Zs (for 0 ≤ s ≤ t) are independent and identicallydistributed as a Poisson with parameter ν(t − s).If (Zt)t≥0 is a Poisson process of rate ν counting the number of arrivals

and T denotes the time between two arrivals, then T has an exponential dis-tribution with parameter ν > 0 (T ∼ Exp (ν)). The density function of T isgiven by

fT (t) = νe−νt, ∀t > 0.

The expected value and the variance of T are then equal to E [T ] = 1ν and

V (T ) = 1ν2 , respectively.

Pareto random variableA random variable X has a Pareto distribution with parameters x0 > 0

and a > 0 if its cumulative distribution function is given by

P (X ≤ x) =(1 −(x0

x

)a)1[x0,+∞) (x) .

For a > 1, X has finite expected value E [X] = aa−1x0.

1.1 Review of Theory 3

Normal (or Gaussian) random variableA random variable X taking values in R has a Gaussian or normal distri-

bution with parameters μ ∈ R and σ2 > 0 (X ∼ N(μ, σ2)) if its cumulativedistribution function (denoted by N) is given by

N(x) � P (X ≤ x) =1√

2πσ2

∫ x

−∞e−

(y−μ)2

2σ2 dy.

As a consequence, its expected value and variance are equal, respectively,to E[X] = μ and V (X) = σ2. When μ = 0 and σ2 = 1, the random variableX is called standard normal. Tables providing the values of N are widelyavailable in the literature.Furthermore, a normal random variable X ∼ N(μ, σ2) can be transformed

into a standard normal Z ∼ N(0, 1) by taking Z = (X − μ)/σ.

The Central Limit Theorem guarantees that the sum of n random variables,that are independent and identically distributed (i.i.d.) with finite expectedvalues and variances, converges in distribution to a standard normal. Moreprecisely: given a sequence (Xn)n∈N of i.i.d. random variables with E (Xn) = μand V (Xn) = σ2 for any n ∈ N, where μ ∈ R and σ2 > 0, it follows that∑n

i=1 Xi−nμ

σ√

n

d→n N (0, 1).In particular, if Xn is a binomial random variable with parameters n, p

(i.e. the sum of n independent Bernoulli random variables with parameter p),then Xn−np√

np(1−p)can be approximated by a standard normal.

By the Central Limit Theorem the following stochastic processes can beobtained as limits of processes discussed previously.

Brownian MotionA Brownian Motion (Xt)t≥0 is a continuous-time stochastic process with

continuous trajectories such that:

• X0 = 0;• its increments are stationary and independent;• for any t ≥ 0, Xt ∼ N(μt, σ2t).

μ and σ are called drift and diffusion, respectively. When μ = 0 and σ2 = 1,the Brownian motion is called standard and denoted by (Wt)t≥0.By definition of Brownian motion it follows that, for any 0 ≤ s ≤ t,

Xt − Xs ∼ N(μ (t − s) , σ2 (t − s)).

For a standard Brownian motion we have

Wt − Ws ∼ N(0, t− s).


A standard Brownian motion can be obtained as limit of a suitable bino-mial process. Suppose that, at any interval Δt of time, the random variable atthe previous time may have a positive or negative increment Δx with the sameprobability p. After n = [t/Δt] periods, we have Yn =

∑ni=1 Yi(Δx), where

Yi takes values in {−1, 1}. Such a process is called binomial random walk.By taking the limit of Yn (under the assumption that Δx → 0, Δt → 0 andlim[(Δx)2/Δt] = σ2 as n → +∞) and applying the Central Limit Theorem,one obtains a Brownian motion with mean 0 and variance σ2t. In order toobtain a standard Brownian motion, it is sufficient to have lim(Δx)2/Δt = 1.

Log-normal stochastic processA stochastic process (St)t≥0 such that

St = exp (Xt) ,

where (Xt)t≥0 is a Brownian motion, is said to be log-normal. It follows,therefore, that for any 0 ≤ s ≤ t

ln(

St

S0

)∼ N(μt, σ2t)

ln(

St

Ss

)∼ N(μ (t − s) , σ2 (t − s)).

As for the Brownian motion, also the log-normal process can be seen asthe limit of a suitable stochastic process. It can be obtained, indeed, as thelimit (for tk − tk−1 = Δt → 0) of a discrete process of the form Sk = Sk−1X,where X is a Bernoulli random variable taking values u = exp(σ

√Δt) and

d = exp(−σ√

Δt) with probabilities pu = (1 + μ√

Δt/σ)/2 and pd = (1 −μ√

Δt/σ)/2, respectively. By passing to the limit, the process (St)t≥0 satisfies

ln(

St

S0

)∼ N(μt, σ2t), (1.1)

so that (ln (St) / ln (S0))t≥0 is a Brownian motion with drift μ and diffusionσ. Such a process (St)t≥0 is also called geometric Brownian motion with driftμ and diffusion σ.

Among the different families of stochastic processes, a remarkable one isthe family of martingales.

Discrete-time MartingalesA stochastic process (Xn)n≥0 is said to be a discrete-time martingale with

respect to the filtration (Fn)n≥0 if (Xn)n≥0 is (Fn)n≥0-adapted and if, forany n ≥ 0, we have E [|Xn|] < +∞ and

E [Xn+1| Fn] = Xn.

1.2 Solved Exercises 5

Continuous-time MartingalesA stochastic process (Xt)t≥0 is said to be a continuous-time martingale

with respect to the filtration (Ft)t≥0 if (Xt)t≥0 is (Ft)t≥0-adapted and, forany 0 ≤ s ≤ t, we have E [|Xt|] < +∞ and

E [Xt| Fs] = Xs.

1.2 Solved Exercises

Exercise 1.1. Let (Wt)t≥0 be a standard Brownian motion.

1. Calculate P (W6 − W2 < 0; W1 > 0) and V ((W6 − W2)W1).2. Consider the stochastic process (Xt)t≥0 defined as Xt = μt+Wt. Establishif there exists a drift μ > 0 such that the probability P (X6−X2 < 0; X1 >0) is at least equal to 20%.

3. Set (Yt)t≥0 to be the process Yt = μt + σWt. Find the distribution of Yt

for μ = 0 and for μ �= 0.4. Compute P (Y6 − Y2 < 0; Y1 > 0), P (Y6 − Y2 < 4σ) andE [(Y6 − Y2)Y1] for

μ = 0.1 and σ = 0.4.

Solution

1. Since (W6 − W2) and W1 are independent random variables (by definitionof Brownian motion), we deduce that

P (W6 − W2 < 0; W1 > 0) = P (W6 − W2 < 0) · P (W1 > 0) .

By (W6 − W2) ∼ N (0; 4) (hence, W6−W2√4

∼ N(0, 1)) and W1 ∼ N (0; 1), itfollows that

P (W6 − W2 < 0; W1 > 0) = P (W6 − W2 < 0) · P (W1 > 0)

= P

(W6 − W2√

4< 0)· P (W1 > 0)

= N (0) · [1 − N (0)] =12· 12

= 0.25,

where N denotes the cumulative distribution function of a standard nor-mal.Furthermore,

V ((W6 − W2)W1) = E[(W6 − W2)

2W 2

1

]− (E [(W6 − W2) W1])

2, (1.2)

and because of the properties of standard Brownian motions we get

V ((W6 − W2)W1) = E[(W6 − W2)

2]E[W 2

1

]− (E [W6 − W2])2 (E [W1])

2

= V (W6 − W2)V (W1) = 4 · 1 = 4.


2. We have to establish if there exists μ > 0 such that

P (X6 − X2 < 0; X1 > 0) ≥ 0.2.

To this end we rewrite the left-hand side of the inequality above in termsof μ and of the Brownian motion. This gives

P (X6 − X2 < 0; X1 > 0) = P (6μ + W6 − (2μ + W2) < 0; μ + W1 > 0)= P (W6 − W2 < −4μ; W1 > −μ)= P (W6 − W2 < −4μ) · P (W1 > −μ)

= P

(W6 − W2

2< −2μ

)· P (W1 > −μ)

= N (−2μ) · [1 − N (−μ)] = N (−2μ) · N (μ) ,

where the last inequality is due to the symmetry of the normal distribution.The initial problem is, therefore, equivalent to establish if there exists adrift μ > 0 so that N (−2μ) · N (μ) ≥ 0.2. The answer is yes, becausetaking μ = 0.1, for instance, gives N (−2μ) · N (μ) = 0.227 ≥ 0.2.

3. Consider, first, the case where Yt = μt + σWt with μ = 0.Recall that Wt ∼ N (0; t) for any t > 0. It follows that σWt is distributedas a normal with mean

E [σWt] = σE [Wt] = 0

and varianceV (σWt) = σ2V (Wt) = σ2t.

Hence, σWt ∼ N(0; σ2t

).

In general, for an arbitrary μ ∈ R and for any t > 0, Yt = μt + σWt isdistributed as a normal with mean

E [Yt] = E [μt + σWt] = μt + σE [Wt] = μt

and variance

V (Yt) = V (μt + σWt) = V(σ2Wt

)= σ2t.

Hence, for any t > 0, Yt ∼ N(μt; σ2t

)either for μ = 0 or μ �= 0.

4. Proceeding as above, we get

P (Y6 − Y2 < 0; Y1 > 0) = P (σ (W6 − W2) < −4μ; σW1 > −μ)

= P

(W6 − W2

2<

−2μ

σ

)· P(W1 > −μ

σ

)

= N

(−2μ

σ

)[1 − N

(−μ

σ

)]= N (−0.5) [1 − N (−0.25)] = 0.185,


and also

P (Y6 − Y2 < 4σ) = P (σ (W6 − W2) < −4μ + 4σ)

= P

(W6 − W2

2<

−2μ

σ+ 2)

= N

(32

)= 0.933.

By the properties of Brownian motions, we also find

E [(Y6 − Y2)Y1]

= E [(4μ + σ (W6 − W2)) (μ + σW1)]

= E[4μ2 + 4μσW1 + μσ (W6 − W2) + σ2 (W6 − W2)W1

]= 4μ2 + 4μσE [W1] + μσE [W6 − W2] + σ2E [(W6 − W2)W1]

= 4μ2 + σ2E [(W6 − W2)W1] = 4μ2,

where the last inequality can be deduced in two different ways. The first isbased on the independence of (W6−W2) andW1. Then E[(W6−W2)W1] =E [W6 − W2] E [W1] = 0. The second comes from E [WtWs] = min{s; t}; soE [(W6 − W2)W1] = E [W6W1] − E [W2W1] = 1 − 1 = 0.

Exercise 1.2. Consider two geometric Brownian motions(S1

t

)t≥0

and(S2

t

)t≥0, representing the prices of two stocks, with drifts μ1, μ2 and diffu-

sions σ1 = σ > 0, σ2 = 2σ respectively, and with the same initial priceS1

0 = S20 = S0 > 0.

1. Compute P (S1t ≥ S2

t ).2. For μ1 = 4μ2 and t = 4 years, find under which conditions on μ2/σ onehas P (S1

4 ≥ S24) ≥ 1

4 .

3. Compute E[S1t − S2

t ] and V(

S1t

S2t

).

SolutionRecall that for a geometric Brownian motion (Xt)t≥0 with

Xt = X0eμt+σWt ,

one has

ln(

Xt

Xs

)∼ N

(μ (t − s) ; σ2 (t − s)

)(1.3)

for any 0 ≤ s ≤ t.


1. By (1.3) and S10 = S2

0 = S0 > 0 we deduce that

P(S1

t ≥ S2t

)= P

(S0e

μ1t+σWt ≥ S0eμ2t+2σWt

)= P (μ1t + σWt ≥ μ2t + 2σWt)= P (σWt ≤ (μ1 − μ2) t)

= P

(Wt√

t≤ (μ1 − μ2)

√t

σ

)

= N

((μ1 − μ2)

√t

σ

).

2. By substituting μ1 = 4μ2 and t = 4 in the expression of P(S1

t ≥ S2t

)just

obtained, we get

P(S1

4 ≥ S24

)= N

((μ1 − μ2)

√4

σ

)= N

(6μ2

σ

).

In order to have P(S1

4 ≥ S24

)= N

(6μ2σ

) ≥ 0.25, the inequality 6μ2σ

≥N−1(0.25) ∼= −0.67 should hold (see the tables of the cumulative distribu-tion function of the standard normal). This forces μ2

σ ≥ −0.11.

3. First of all, we remind that E[eY ] = em+ s22 for Y ∼ N(m, s2). Hence

E[S1

t − S2t

]= E

[S1

t

S0· S0

]− E

[S2

t

S0· S0

]= S0 · E

[eμ1t+σWt

]− S0 · E[eμ2t+2σWt

]= S0

[eμ1t+ σ2t

2 − eμ2t+2σ2t],

since μ1t + σWt ∼ N(μ1t; σ2t

)and μ2t + 2σWt ∼ N

(μ2t; 4σ2t

).

As for the variance, we obtain that

V

(S1

t

S2t

)= V

(S1

t

S0· S0

S2t

)= V

(eμ1t+σWt · e−(μ2t+2σWt)

)

= V(e(μ1−μ2)t−σWt

)= E

[(e(μ1−μ2)t−σWt

)2]

−[E(e(μ1−μ2)t−σWt

)]2= E

[e2(μ1−μ2)t−2σWt

]−[e(μ1−μ2)t+

σ2t2

]2= e2(μ1−μ2)t+2σ2t − e2(μ1−μ2)t+σ2t = e2(μ1−μ2)t+σ2t

[eσ2t − 1

],

where the previous equalities are based on the fact that Wt and (−Wt)have the same distribution.


Exercise 1.3. Suppose that the number of shares of a given stock boughtover time (measured in minutes) follows a Poisson process of rate λ = 12 perminute.

1. Establish how many minutes are needed so that more than 36 shares arebought with probability of at least 94.8%. Denote by n∗ this minimumnumber of minutes.

2. Compute the average waiting time for the purchase of 40 shares.3. Assume that the same stock is also sold on a different market and thatthe number of its shares bought in time (measured in minutes) in such amarket follows a Poisson process of rate μ = 8 per minute, independent ofthe first one.Compute the probability that the total number of shares (bought in thetwo markets combined) in n∗ minutes is greater than 72.

Solution

1. Denote by Xt the number of shares bought in t minutes on the first mar-ket. We have to find how many minutes t are needed so that P (Xt > 36) ≥0.948. Denote by n∗ the smallest following integer, namely n∗ = [t]+1. Weknow that

Xt ∼ Poi (λt) .

Hence,

P (Xt > 36) = 1 − P (Xt ≤ 36) = 1 −36∑

κ=0

e−λt · (λt)k

k!

= 1 −36∑

κ=0

e−12t · (12t)k

k!.

Since one has P (Xt > 36) = 1.44 · 10−7 for t = 1, P (Xt > 36) = 0.008 fort = 2, P (Xt > 36) = 0.456 for t = 3 and P (Xt > 36) = 0.956 for t = 4, wededuce that the minimum number of minutes to wait in order for P (Xt >36) ≥ 0.948 is n∗ = 4.

2. Let Ti be the waiting time (in minutes) of the i-th purchase. It is wellknown that

Ti+1 − Ti ∼ Exp (λ) .

Hence, the average time to wait for the purchase of 40 shares (measured inminutes) is given by

E [T40 − T0] = E [T40 − T39] + E [T39 − T38] + ... + E [T1 − T0]

=1λ

+1λ

+ ... +1λ

=40λ

=4012

= 3.33.


3. Denote now by Yt the number of shares bought in t minutes on the secondmarket.Since Yt is assumed to be a Poisson process of intensity μ (so, Yt ∼ Poi (μt))and Xt and Yt are assumed to be independent,

Xt + Yt ∼ Poi ((λ + μ) t)

for any t ≥ 0.Since n∗ = 4 (by item 1.), it follows that

P (Xn∗ + Yn∗ > 72) = 1 − P (X4 + Y4 ≤ 72)

= 1 −72∑

k=0

e−4(λ+μ) · (4 (λ + μ))k

k!

= 1 −72∑

k=0

e−80 · (80)k

k!∼= 0.80.

Exercise 1.4. Consider a stock with current price of 8 euros. In each of thefollowing two years the stock price may increase by 20% (with probability40%) or decrease by 20% (with probability 60%).Denote with (Sn)n=0,1,2 the process representing the evolution of the stock

price in time. S1 may then take the values Su1 = S0u and Sd

1 = S0d, while S2

the values Suu2 = S0u

2, Sud2 = S0ud and Sdd

2 = S0d2, where u is the growth

factor and d the decreasing factor.

1. Is (Sn)n=0,1,2 a martingale with respect to the above probability and withrespect to the filtration generated by (Sn)n=0,1,2?

2. Consider the stochastic process (Sn)n≥0 defined as

S0 � S0

Su1 � Su

1 − k; Sd1 � Sd

1 + k

S2 � S2.

Establish whether (Sn)n=0,1,2 can be a martingale with respect to the prob-ability and the filtration of the previous item for a suitable k > 0.

3. Discuss if there exists a probability measure Q such that the probabilitythat the stock price increases (respectively, decreases) in the first year isequal to the probability that the stock price increases (respectively, de-creases) in the second year, and such that (Sn)n=0,1,2 is a martingale withrespect to Q.

4. Establish if there exist u > 1 and d > 0 such that the new stock priceprocess is a martingale with respect to the probability measure of item 1.


SolutionSince the growth factor is u = 1.2 and the decreasing factor is d = 0.8, the

stock prices evolves as follows.

11.52 = S0u2

↗S0u = 9.6

↗ ↘S0 = 8 7.68 = S0ud

↘ ↗S0d = 6.4

↘5.12 = S0d

2

− −−− − −− −−− −− −− −−0 1 year (S1) 2 years (S2)

It is also easy to check that

P (S1 = 9.6) = 0.4; P (S1 = 6.4) = 0.6P (S2 = 11.52) = 0.16; P (S2 = 7.68) = 0.48; P (S2 = 5.12) = 0.36P (S2 = 11.52|S1 = 9.6) = P (S2 = 7.68|S1 = 6.4) = 0.4P (S2 = 7.68|S1 = 9.6) = P (S2 = 5.12|S1 = 6.4) = 0.6,

and so on.

1. Let us verify if (Sn)n=0,1,2 is a martingale with respect to P .From

E [S1|S0] = Su1 p + Sd

1 (1 − p) = 9.6 · 0.4 + 6.4 · 0.6= 7.68 �= S0

it follows immediately that (Sn)n=0,1,2 is not a martingale with respectto P .

2. We need to verify if the following equalities hold for some k > 0:{E[S1

∣∣S0

]= S0

E[S2

∣∣S1

]= S1

.

The first equality is equivalent to

E[S1

∣∣S0

]= S0

(9.6− k) · 0.4 + (6.4 + k) · 0.6 = 87.68 + 0.2 · k = 8k = 1.6,


hence Su1 = Sd

1 = 8. This implies that

E[S2

∣∣S1 = Su1

]= 11.52 · 0.4 + 7.68 · 0.6 = 9.216 �= Su

1 ,

so (Sn)n=0,1,2 is not a martingale with respect to P .3. Define now a probability measure Q as follows:

Q (S1 = 9.6) = q; Q (S1 = 6.4) = 1 − q

Q (S2 = 11.52) = q2; Q (S2 = 7.68) = 2q (1 − q)Q (S2 = 5.12) = (1 − q)2

Q (S2 = 11.52|S1 = 9.6) = Q (S2 = 7.68|S1 = 6.4) = q

Q (S2 = 7.68|S1 = 9.6) = Q (S2 = 5.12|S1 = 6.4) = 1 − q.

We look now for q ∈ (0, 1) such that

{EQ

[S1

∣∣S0

]= S0

EQ

[S2

∣∣S1

]= S1

,

or, equivalently,

EQ [S1 |S0] = S0 (1.4)

EQ [S2|S1 = Su1 ] = Su

1 (1.5)

EQ

[S2|S1 = Sd

1

]= Sd

1 . (1.6)

By equation (1.4) it follows that

S0uq + S0d (1 − q) = S0

q = 1−du−d = 1

2 .

There remains therefore to check if such a q also satisfies equations (1.5)and (1.6) or not.Since uq + d (1 − q) = 1, we obtain that

EQ [S2|S1 = Su1 ] = S0u

2q + S0ud (1 − q)

= S0u [uq + d (1 − q)]= S0u = Su

1 ,

hence (1.5) is satisfied. In a similar way, it is easy to check that also (1.6)is true.Consequently, (Sn)n=0,1,2 is a martingale with respect to the probabilitymeasure Q defined above with q = 0.5.

4. We need to establish if there exist u > 1 and d > 0 such that the process(Sn)n=0,1,2 defined below is a martingale with respect to P (and to the


natural filtration):

S0u2

↗S0u

↗ ↘S0 = S0 = 8 S0ud

↘ ↗S0d

↘S0d

2

−− −− −− −− −− −− − −− −−−0 1 year (S1) 2 years (S2)

Proceeding as above, we need to check if there exist u > 1 and d > 0 sat-isfying E[S1

∣∣S0] = S0 and E[S2

∣∣S1] = S1. Or, equivalently, if the followingsystem in the unknowns (u, d) admits solutions:⎧⎪⎨

⎪⎩S0up + S0d (1 − p) = S0

S0u2p + S0ud (1 − p) = S0u

S0udp + S0d2 (1− p) = S0d

Since the previous system is equivalent to up + d (1 − p) = 1, we find

up + d− dp = 1

u = 1−d+dpp

.

Taking, for instance, d = 0.8, we obtain u = 1.3 > 1. The pair (u = 1.3; d =0.8) is therefore one among the (infinitely many) pairs of factors for whichthe corresponding process

(Sn

)n=0,1,2

is a martingale with respect to P .

Exercise 1.5. Consider the stock of Exercise 1.4 and assume that in each pe-riod the price may increase with probability p and decrease with probability(1 − p).Let X1 and X2 be two random variables defined as

X1 ={

1; if S1 = Su1

0; if S1 = Sd1

X2 =

⎧⎨⎩

2; if S2 = Suu2

1; if S2 = Sud2

0; if S2 = Sdd2


1. Write P (S2 = Suu2 ) as a function of P (X2 = 2).

2. Establish if (Xn)n=1,2 is a binomial process.

Solution

1. By the definition of X2 it follows that

P (X2 = 2) = P (S2 = Suu2 ) = p2

P (X2 = 1) = P(S2 = Sud

2

)= 2p (1 − p)

P (X2 = 0) = P(S2 = Sdd

2

)= (1 − p)2 .

2. We have to check if (Xn)n=1,2 is a binomial process, i.e. if X1 ∼ Bin (1; p)and X2 ∼ Bin (2; p).It is immediate to see thatX1 ∼ Bin (1; p). By definition of a binomial vari-able with parameters n = 2 and p and using the probabilities computedabove, it follows that X2 ∼ Bin (2; p) .Consequently, (Xn)n=1,2 is a binomial process.

1.3 Proposed Exercises

Exercise 1.6. Let (Wt)t≥0 be a standard Brownian motion.Consider two stocks whose associated gains (difference between current

stock price and buying price) evolve as

Xt = μt + σWt

Yt = μ1t + σ1Wt,

with X0 = 1, Y0 = 10, μ = 20, μ1 = 10, σ = 10 and σ1 = 20 euros per year.Taking into account the stocks above, consider also a derivative of value

Zt = X2t Yt

at time t.

1. Suppose we buy the derivative only if in two years its expected value willexceed 10X2

0Y0 (that is, 10 times its current price). Decide whether weeventually buy the derivative or not.

2. Compute the probability of having a positive net gain (Z1 − Z0) in oneyear or the probability of having a net gain (Z2 − Z1) between the firstand the second years greater than 100 euros.

3. Compute the probability of both events occurring (Z1 − Z0 > 0 and Z2 −Z1 > 100). Are such events independent?

1.3 Proposed Exercises 15

Exercise 1.7. Let (Wt)t≥0 be a standard Brownian motion.Consider a stock whose price evolves as the following stochastic process

(Yt)t≥0:Yt = Y0e

μt+σWt .

1. Compute P(Y 2

t ≥ Yt

), when Y0 = 4, t = 2, μ = 0.1 and σ = 0.4.

2. Establish if Y2t and Y 2t have the same distribution when Y0 = 1.

3. Compute E[ln(

Y 28

Y16

)]and V

(ln(

Y 28

Y16

))when Y0 = 1.

[Hint: remember that E [WtWs] = min (s; t).]

2

Portfolio Optimization in Discrete-TimeModels


We restrict our interest to one-period models, and in this case the “portfolio”,which we are going to define formally below, is characterized through its com-position at the initial time, and the returns of different assets are assumedto be random variables. A systematic exposition of the notions briefly sum-marized below can be found in the textbooks by Barucci [3], Capiński andZastawniak [9] and Luenberger [26].

Let us consider a market model where n assets are traded, and whose val-ues at a prescribed date are represented by n random variables S1, S2, ..., Sn.We shall denote by S the random vector S = (S1, S2, ..., Sn) with componentsS1, S2, ..., Sn.We define portfolio a vector v = (v1,v2, ...vn) with each component vi iden-

tifying the number of units of the asset with value Si. The portfolio value Vis given by the scalar product of the two vectors v and S, i.e.:

V = v · ST =n∑

j=1

vjSj .

More often a portfolio is characterized by the vector of its relative weightsw = (w1, w2, ...wn), where the components wi are defined by:

wi �viSi

V=

viSi∑nj=1 vjSj

. (2.1)

One can immediately verify that the weights wi sum up to 1. A negative valueof the component wi denotes a short position in the asset with value Si. Theset of portfolios satisfying the normalization condition on the weights wi iscalled the set of admissible portfolios.


18 2 Portfolio Optimization in Discrete-Time Models

The portfolio optimization problem consists, briefly, in finding a portfoliowhose expected utility is the maximum possible among all admissible port-folios. The utility function whose expectation must be maximized should bechosen so to describe the investors preferences, and it should take into accounttheir risk-aversion attitude.A utility function is a nondecreasing, concave function U : R→ [−∞, +∞)

of class C2. Because of their simplicity, the most commonly adopted utilityfunctions are the following:

• exponential utility:

U(x) = 1 − exp(−αx), α > 0; (2.2)

• logarithmic utility:

U(x) ={ −∞, x ≤ 0

ln(x), x > 0; (2.3)

• power utility:

U(x) ={−∞, x ≤ 0

xα, x > 0, 0 < α < 1. (2.4)

In the simplest formulation, the portfolio optimization problem can be re-duced to the problem of finding a portfolio with the maximum expected returnand the minimal variance, assuming the latter is adopted as a risk measureassociated to the portfolio under consideration. A portfolio with minimumvariance with respect to all other portfolios with the same expected return, orwith maximum expected return with respect to all portfolios with the samevariance, is called efficient .If we denote by rK the return of portfolio K, ri (i = 1, ..., n) the return of

asset Si, μi its expected value and σij (i, j = 1, ..., n) the covariance betweenri and rj , then the expected return and the variance1 of portfolio K can beeasily proved to be given by:

E(rK) =n∑

i=1

wiE(ri) =n∑

i=1

wiμi (2.5)

V ar(rK) =n∑

i=1

w2i V ar(ri) +

∑i=1,...,n

i �=j

n∑j=1

wiwjCov(ri, rj)

=n∑

i=1

n∑j=1

wiwjσij. (2.6)

1 In contrast to the other chapters, to avoid confusion with the notation adoptedfor the portfolio value we shall denote the variance of a random variable by V ar (·)and not by V (·).


The portfolio with minimum variance (with no assigned expected return)can be determined via the following formula:

w =uC−1

uC−1uT, (2.7)

where u is the n-dimensional vector with components equal to 1 and C is thecovariance matrix of returns.The portfolio with minimum variance among those with expected return

μV can be determined via the following formula:

w =det(

1 uC−1mT

μV mC−1mT

)uC−1 + det

(uC−1uT 1mC−1uT μV

)mC−1

det(

uC−1uT uC−1mT

mC−1uT mC−1mT

) , (2.8)

where m = (μ1, μ2, ...μn) is the vector of the expected returns of the n assets.The set of efficient portfolios with expected return μV can also be charac-

terized as the solution of the following:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

n∑j=1

σijwj − λ1μi − λ2 = 0, i = 1, ..., n

n∑i=1

wiμi = μV

n∑i=1

wi = 1

or, equivalently, ⎧⎪⎨⎪⎩

C · wT − λ1m − λ2 = 0w · mT = μV

w · 1T = 1,

where λ1, λ2 are the Lagrange multipliers relative to the last two constraintequations. The set of efficient portfolios obtained by varying expected returnsis called the efficient frontier .

Sometimes the portfolio optimization problem includes some constraintson the portfolio weights; the most typical example of this situation is theno-short-selling constraint that requires all the portfolio weights to be non-negative, i.e. wi ≥ 0 for i = 1, ..., n.When the portfolio optimization problemconsists in variance minimization with a prescribed expected return, and theno-short-selling constraints are imposed, the problem can be identified as aquadratic programming problem. This kind of problem, in general, cannot besolved explicitly without the support of a computer program (there are severalavailable on the market), unless the dimension of the problem is small enoughto allow explicit calculations.


The Mutual Funds Theorem is another fundamental result of mathemati-cal portfolio theory: it states that given a market with n assets, it is alwayspossible to determine two funds (i.e. two portfolios) such that every efficientportfolio (as far as the expected return and variance are concerned) can beexpressed as a linear combination of them. In order to compute these twoportfolios, it is sufficient to find two solutions of the optimization problem fortwo arbitrary values of the Lagrange multipliers λ1, λ2. In general the two so-lutions determined by this procedure will violate the normalization conditionon the weights, but this can be easily amended by simply dividing the weightsobtained by their sum. We shall provide an example in order to illustrate themethod just outlined.

The Capital Asset Pricing Model (CAPM) assumes as a starting point thelinear dependence of the return of a generic portfolioK (the model postulatesthe same relation for all different returns of each asset) on the return rM of abenchmark portfolio, called market portfolio:

rK = rf + βK(rM − rf) + εi, (2.9)

where rf is the risk-free interest rate and εi is a zero-mean Gaussian ran-dom variable. By simply taking the expectation of each side of the previousequality, one obtains:

μK = E(rK) = rf + βK(μM − rf). (2.10)

The (straight) regression line for the present model is identified by the coef-ficients βK , αK, where βK and αK can be obtained by the following formulas:

βK =Cov(rM , rK)

σ2M

(2.11)

αK = μK − βKμM . (2.12)


Exercise 2.1. Determine the expected return and the variance of the portfo-lio formed by the two assets S1, S2 with weights w1 = 0.6 and w2 = 0.4. Theassets’ returns are described by the following scheme:

scenario probability r1 r2

ω1 0.1 −20% −10%ω2 0.4 0% 20%ω3 0.5 20% 40%


SolutionSince the expected returns of the two assets are given by:

E(r1) = 0.1 · (−0.2) + 0.4 · 0 + 0.5 · 0.2 = 0.08

E(r2) = 0.1 · (−0.1) + 0.4 · 0.2 + 0.5 · 0.4 = 0.27,

the expected return of the portfolio K with weights w1 = 0.6 and w2 = 0.4for the assets S1 and S2 is then:

E(rK) = w1E(r1) + w2E(r2) = 0.6 · 0.08 + 0.4 · 0.27 = 0.156.

In order to calculate the portfolio variance it is necessary to know thecovariance matrix of the two returns. In the present case, we obtain:

V ar(r1) = E(r21

)− (E (r1))2

= 0.1 · (−0.2)2 + 0.4 · 02 + 0.5 · (0.2)2 − (0.08)2 = 0.0176

V ar(r2) = E(r22

)− (E (r2))2

= 0.1 · (−0.1)2 + 0.4 · (0.2)2 + 0.5 · (0.4)2 − (0.27)2 = 0.0241

Cov(r1, r2) = E (r1r2) − E (r1)E (r2)= 0.1 · (−0.2) (−0.1) + 0.4 · 0 · 0.2 + 0.5 · 0.2 · 0.4− 0.08 · 0.27= 0.0204

Hence

V ar(rK) = w21V ar(r1) + w2

2V ar(r2) + 2w1w2Cov(r1r2)

= (0.6)2 · 0.0176 + (0.4)2 · 0.0241 + 2 · 0.6 · 0.4 · 0.0204 = 0.019984.

Exercise 2.2. Suppose that the returns r1, r2 of the two assets S1, S2 are asin the following scheme:

scenario probability r1 r2

ω1 0.2 −10% 10%ω2 0.3 5% −2%ω3 0.5 20% 15%

1. Compare the risk (assume the variance as a risk measure) of a portfolio Kcomposed by the two assets above with weights w1 = 0.3, w2 = 0.7, withthe risk of the two single assets considered separately.


2. By using the same table, determine the composition of a portfolio K∗ stillconsisting of the two previous assets, but with expected return equal to9%. Calculate the variance of the portfolio K∗.

Solution

1. In strict analogy with Exercise 2.1, one obtains easily

E (r1) = 0.095; E (r2) = 0.089V ar(r1) = 0.0137; V ar(r2) = 0.0054Cov (r1, r2) = 0.0042.

Then

ρ12 =Cov(r1, r2)√

V ar(r1)V ar(r2)= 0.49.

The variance of the portfolio K is then given by:

V ar(rK) = w21V ar(r1) + w2

2V ar(r2) + 2ρ12w1w2

√V ar(r1)V ar(r2)

= (0.3)2 · 0.0137 + (0.7)2 · 0.0054

+ 2 · 0.49 · 0.3 · 0.7 ·√

0.0137 · 0.0054

= 0.0056.

2. We have to find a new portfolio K∗ with expected return equal to 9%.More precisely, we have to compute the weights corresponding to the twoassets in the new portfolio K∗.Reminding that E(r1) = 0.095 and E(r2) = 0.089, the two required weightsw∗

1, w∗2 must satisfy the following system of equations:{

w∗1 · 0.095 + w∗

2 · 0.089 = 0.09w∗

1 + w∗2 = 1

.

The solution is then given by

w∗1 =

16, w∗

2 =56.

The weights for the two assets S1 and S2 in the new portfolio K∗ are thenw∗

1 = 16and w∗

2 = 56.

The variance of the return of K∗ is then given by:

V ar(rK∗) = (w∗1)

2V ar(r1) + (w∗

2)2V ar(r2) + 2w∗

1w∗2Cov(r1r2)

= (1/6)2 · 0.0137 + (5/6)2 · 0.0054 + 2 · 536

· 0.0042 = 0.0053.


Exercise 2.3. Let three risky assets be given. Their returns have expectationsand covariance matrix described in the following scheme:

μ1 = 0.20 σ1 = 0.25 ρ12 = −0.20

μ2 = 0.12 σ2 = 0.30 ρ23 = 0.50μ3 = 0.15 σ3 = 0.22 ρ13 = 0.30

Find the minimum variance portfolio and compute its expected return andvariance.

SolutionLet us recall the formula providing the weight vector w of the minimum

variance portfolio:

w =uC−1

uC−1uT,

where C is the returns covariance matrix and u is the vector with all compo-nents equal to 1. In the present case, we have:

C =

⎛⎜⎝ 0.0625 −0.015 0.0165

−0.015 0.09 0.0330.0165 0.033 0.048

⎞⎟⎠ .

Since the inverse matrix C−1 of C is:

C−1 =

⎛⎜⎝

21.497 8.4132 −13.1748.4132 18.1487 −15.369

−13.174 −15.369 35.928

⎞⎟⎠ ,

we can easily obtain the weights of the minimum variance portfolio:

w1 = 0.474, w2 = 0.317, w3 = 0.209.

The expected return and the variance of this portfolio are then

μV = w1μ1 + w2μ2 + w3μ3 = 0.164

σ2V = w2

1σ21 + w2

2σ22 + w2

3σ23 + 2w1w2Cov (r1, r2)

+ 2w1w3Cov (r1, r3) + 2w2w3Cov (r2, r3)

= 0.0283.


Exercise 2.4. Consider a portfolio K with return rK and the market port-folio KM with return rM , where rK and rM take the following values:

scenario probability rK rM

ω1 0.3 3% 8%ω2 0.2 2% 7%ω3 0.3 4% 10%ω4 0.2 1% 9%

Find the coefficients βK and αK of the regression line.

SolutionFrom the table above we obtain

μK = E(rK) = 0.027; μM = E(rM ) = 0.086;σ2

M = 0.000124; Cov(rK , rM) = 0.000058.

By applying directly the formulas providing the coefficients βV and αV , weobtain

βV =Cov(rK , rM)

σ2M

=0.0000580.000124

= 0.467

αK = μK − βKμM = 0.027− 0.467 · 0.086 = −0.01316.

Exercise 2.5. 1. Consider two assets S1, S2 with jointly normally distributedreturns of expected value and standard deviation E(r1) = μ1 = 0.20,E(r2) = μ2 = 0.16, σ1 =

√V ar(r1) = 0.30, σ2 =

√V ar(r2) = 0.36,

respectively, and with correlation coefficient ρ12 = −0.5.Determine the portfolio maximizing the expected utility, assumed to be theexponential utility function:

U(x) = 1 − exp(−αx)

with parameter α = 0.04.2. Consider now a new portfolio J made by the same asset S1 of the previ-ous item and another asset B (riskless) with return rB = 0.06. Find theoptimal composition of the new portfolio with respect to the same utilityfunction as in item 1. (exponential utility).

Solution

1. Denote by K the portfolio with weights w1 = x (in the asset S1) andw2 = 1 − x (in the asset S2). The expectation and the variance of the


return of K are given by

E[rK ] = E [w1r1 + w2r2] = 0.20 · x + 0.16 · (1 − x) = 0.16− 0.04x

V ar(rK) = V ar (w1r1 + w2r2)

= (0.3)2 x2 + (0.36)2 (1 − x)2 − 2 · 0.5 · 0.30 · 0.36 · x(1 − x)

= 0.3276 · x2 − 0.828 · x + 0.1296.

If a random variable Z is normally distributed, then exp(Z) is distributedaccording to a log-normal and the following relation holds:

E [exp(Z)] = exp{

E(Z) +12V ar(Z)

}.

Since maximizing the expectation of the exponential utility is equivalentto maximizing the quantity:

E[U(Z)] = 1 − E [exp(−αZ)] = 1 − exp{−αE(Z) +

12α2V ar(Z)

},

under the assumption of jointly normally distributed returns of both assets,it is enough to maximize the following expression:

E[Z] − 12αV ar(Z).

In the present case Z = rK , so we need to find the value x maximizing thesecond-order polynomial:

g(x) = 0.16− 0.04x− 0.042 (0.3276x2 − 0.828x + 0.1296)

= 0.1574− 0.0234x− 0.0066x2.

It is easy to obtain that the optimal point is x∗ = −1.776.2. The expected return of the new portfolio J , identified via its weights w1 = yand w2 = 1 − y, is then:

E[rJ ] = E [w1r1] + w2rB = 0.20 · y + (1 − y) · 0.06, (2.13)

and its variance is:V ar(rJ) = (0.3)2 · y2. (2.14)

The new function to be maximized is now the following:

f(y) = 0.14 ·y+0.06− 0.042

(0.3 · y)2 = −0.0018 ·y2+0.14 ·y+0.06. (2.15)


It is immediate to verify that the maximum point is y∗ = 38.89. The corre-sponding weight for the riskless asset is 1− y∗ = −37.89, that correspondsto a short position in the non-risky asset B.

Exercise 2.6. Consider the market consisting of three assets S1, S2, S3 withreturns r1, r2, r3, uncorrelated and all with unit variance, with expectationsE[r1] = 1/4, E[r2] = 1/2, E[r3] = 3/4, respectively.Determine the efficient portfolio with expected return r, under the con-

straint of no short selling (i.e. wi ≥ 0 for i = 1, 2, 3).

SolutionSince unilateral constraints are imposed, the problem cannot be solved by

means of a system of equations. It is a quadratic optimization problem withlinear constraints. Since the number of assets involved is small, the problemcan be solved by considering all different pairs of assets.Denote by K the portfolio with weight wi in the asset Si (for i = 1, 2, 3)

and by rK its return. Since assets’ returns are uncorrelated and all with unitvariance, the expectation and the variance of rK are given by

E[rK] = w1E[r1] + w2E[r2] + w3E[r3]

= w1 · 14

+ w2 · 12

+ (1 − w1 − w2) · 34

=34− w1

2− w2

4;

V ar(rK) = V ar (w1r1 + w2r2 + w3r3)= w2

1V (r1) + w22V (r2) + w2

3V (r3)= w2

1 + w22 + (1 − w1 − w2)2

= 2w21 + 2w2

2 + 1 − 2w1 − 2w2 + 2w1w2.

Set x � w1 and y � w2.Our goal is to find a portfolio with return r and with minimum variance.

Notice, however, that (because of the no-short-selling constraints) only targetreturns r ∈ [ 1

4; 3

4] are admissible.

By the condition on the target return, we get

34− x

2− y

4= r,

so y = 3 − 2x − 4r. We have thus to minimize (under the no-short-sellingconstraints) the following

V ar(rK) = 2x2 + 2y2 + 1 − 2x − 2y + 2xy

= x2 + 2(3 − 2x − 4r)2 + 1 − 2x − 2(3− 2x − 4r) + 2x(3− 2x− 4r)= 6x2 − 16x + 24xr + 12 + 32r2 − 40r � f(x).


Since f ′(x) = 12x− 16 + 24r, the weights

⎧⎨⎩

x∗ = 43 − 2r

y∗ = 3 − 2x∗ − 4r = 13

z∗ = 1 − x∗ − y∗ = 2r − 23

(2.16)

are optimal if admissible. It is easy to check that the optimal weights abovesatisfy all the constraints if and only if 1

3≤ r ≤ 2

3. In such a case, the efficient

portfolio weights are the following:

w1 =43− 2r, w2 =

13, w3 = 2r − 2

3,

and the portfolio mean square deviation turns out to be σ =√

31/9− 8r + 8r2.

Take 23 < r ≤ 3

4 . Since f ′(x) = 12x − 16 + 24r ≥ 0 for any x ∈ [0, 1], it iseasy to check that the efficient portfolio weights are the following:

w1 = 0, w2 = 3 − 4r, w3 = 4r − 2


13 − 40r + 32r2.Take now 2

3 < r ≤ 34 . Since for any x ∈ [0, 1], it is easy to check that the

efficient portfolio weights are the following:

w1 = 0, w2 = 3 − 4r, w3 = 4r − 2,


13 − 40r + 32r2.Finally, take 1

4 ≤ r < 13 . In order to have admissible weights one should

have ⎧⎨⎩

x ∈ [0, 1]y = 3 − 2x − 4r ∈ [0, 1]z = 1 − x − y = x + 4r − 2 ∈ [0, 1]

(2.17)

so 2 − 4r ≤ x ≤ 32 − 2r. Since 4

3 − 2r ≤ 2 − 4r for r ≤ 13 , we deduce that the

efficient portfolio weights are the following:

w1 = 2 − 4r, w2 = 4r − 1, w3 = 0


5 − 24r + 32r2.

Exercise 2.7. Consider a portfolio composed by five assets S1, S2, S3, S4, S5

with expected returns E[r1] = 0.183, E[r2] = 0.085, E[r3] = 0.121, E[r4] =


0.112, E[r5] = 0.096, respectively, and with return covariance matrix

C =

⎛⎜⎜⎜⎜⎝

0.0690 0.0279 0.0186 0.0222 −0.00690.0279 0.042 0.0066 0.0168 0.00780.0186 0.0066 0.054 0.0234 −0.00810.0222 0.0168 0.0234 0.102 −0.168

−0.0069 0.0078 −0.0081 −0.168 0.078

⎞⎟⎟⎟⎟⎠ . (2.18)

Determine two funds whose linear combinations generate any efficient port-folio.

SolutionThe two funds we are looking for must belong to the set of portfolios with

minimum variance. We briefly recall that the efficient portfolio (with expectedreturn μK) weights must satisfy the following linear system:{

C · wT − λ1m − λ2 = 0,w ·mT = μK ,

(2.19)

together with the normalization condition:

n∑j=1

wj = 1. (2.20)

Here wi are the portfolio weights, w in vector notation, and μi = E[ri] the ex-pected returns of assets Si, m in vector notation. C is the returns’ covariancematrix. The Two Funds theorem implies that, in order to compute an efficientportfolio for every value of μK , it is sufficient to determine two solutions of(2.19) and (2.20), since every other solution is a linear combination of them.A simple method to determine two solutions is that of specifying two possiblevalues for the Lagrange multipliers λ1 and λ2, for example we can pick thesolution with λ1 = 0 and λ2 = 1, and the solution with λ1 = 1 and λ2 = 0. Ifwe look for the solution with Lagrange multipliers λ1 = 0, λ2 = 1, we have tosolve the following linear system:

C · v1T = 1. (2.21)

The solution can be easily found to be the following:

v1 = (2.1191; 26.1615; 17.6417;−8.5490;−5.9895).

In order to compute the relative weights w1,i we must impose the normal-ization condition on the components of the vector v1, i.e. we must divide eachcomponent by their sum:

w1,i =vi∑n

j=1 vj.


We obtain then:

w1 = (0.0721; 0.8233; 0.6004;−0.2909;−0.2038).

In order to determine the solution with λ1 = 1 and λ2 = 0 we must solvethe second linear system:

C · v2T = m. (2.22)

We obtain:

v2 = (2.0736; 0.9003; 1.7540;−1.0116;−0.6725),

that, after normalization, provides the second fund expressed by its relativeweights:

w2 = (0.6813; 0.2958; 0.5763;−0.3324;−0.2210).


Exercise 2.8. Given two risky assets whose values are normally distributedwith S1 ∼ N(3; 0.5), S2 ∼ N(4; 0.7) and with correlation ρ12 = 0.5, determinethe efficient frontier and the optimal portfolio with respect to the exponentialutility function with parameter α = 1.5.

Exercise 2.9. In the set of all admissible portfolios composed by three assetsof expected returns μ1 = 0.30, μ2 = 0.15, μ3 = 0.18, with standard deviationσ1 = 0.22, σ2 = 0.30, σ3 = 0.26, and correlations ρ12 = 0.34, ρ23 = 0.02,ρ13 = 0.25, find the minimum variance portfolio and compute its expectedreturn and variance.

Exercise 2.10. In the set of all admissible portfolios composed by the threeassets of the previous exercise and with expected return μV = 20%, determinethe minimum variance portfolio and compute its variance.

Exercise 2.11. Assume that the risk-free interest rate is rf = 6%, the expec-tation and the variance of the market portfolio are 10% and 20%, respectively,and the correlation between a given asset and the market portfolio return is0.5. Compute the expected return of the asset considered in a CAPM frame-work.

Exercise 2.12. Consider a market with three assets S1, S2 , S3 with expectedreturns E[r1] = 1/8, E[r2] = 1/4, E[r3] = 3/8, respectively. Suppose the re-turns are uncorrelated and all with unit variance. Find the efficient portfoliowith expected return r = 1

3 when short-selling is allowed and also when it isnot.


Exercise 2.13. Consider a portfolio composed by four assets S1, S2, S3 , S4

with expected returns E[r1] = 0.112, E[r2] = 0.076, E[r3] = 0.191, E[r4] =0.125, and with return covariance matrix

C =

⎛⎜⎜⎝

0.1024 0.05376 −0.0144 0.034560.05376 0.0441 0.0189 0.00756−0.0144 0.0189 0.2025 −0.12960.03456 0.00756 −0.1296 0.1296

⎞⎟⎟⎠ . (2.23)

Determine two funds able to generate every efficient portfolio by linearcombination.

3

Binomial Model for Option Pricing


In the following, we consider a market model where a non-risky asset (calledbond) and a risky asset (called stock) are available. The bond price is denotedby B, while the stock price is denoted by S.

Let us focus, first, on a one-period model. This means that any asset onthe market has to be evaluated just at the beginning and at the end of thegiven time interval. By convention, let t = 0 and t = 1 be the correspondingdates.The initial prices of the (non-risky and risky) assets are known: B0 for the

bond, S0 for the stock. The bond price is deterministic, equal to

B1 = B0(1 + r)

at t = 1, where r is the risk-free interest rate and the rate is compoundedonce at the end of the period. The stock price at t = 1 is given by

S1 = S0X,

where X indicates a Bernoulli random variable that may assume u and d,with probability p and (1 − p), respectively.The market model described above (known as one-period binomial model)

can be summarized as follows:

Bond B0 B1 = B0(1 + r)

Stock S0 S1 =

{S0u = Su

1 ; with prob. p

S0d = Sd1 ; with prob. (1 − p)

t = 0 t = 1


32 3 Binomial Model for Option Pricing

For d < 1+ r < u, the market above is free of arbitrage. Roughly speaking,it is not possible to have a profit for free (see Chap. 4 for details on arbitrageon more general market models). As the binomial model is arbitrage-free, theprice of any financial derivative (or contingent claim) on the underlying S isuniquely determined. This fact can be seen in two different ways. The first isbased on the construction of a portfolio composed by one derivative and bya suitable number of stock shares rendering the portfolio riskless (hence itsdynamics are deterministic). Such a strategy is called Delta-Hedging strategy.The second is based on the construction of a portfolio (composed by a suitablenumber of bonds and stock shares) “replicating” the value of the derivative,that is assuming at maturity T the same value of the derivative in any stateof the world. This last strategy is called replicating strategy (see Chap. 4 formore details).Under the no-arbitrage assumption, the initial cost F (S0) of the derivative

written on the stock S is given by the following formula:

F (S0) =1

1 + r[qu · F (S0u) + qd · F (S0d)] , (3.1)

where qu = (1+r)−du−d , qd = u−(1+r)

u−d = 1 − qu, and F (S0u) and F (S0d) denotethe payoff of the derivative when the price of the underlying is equal to S0uand S0d, respectively. In the following, we will write φu = F (S0u) and φd =F (S0d).As we could expect, the initial price of the derivative depends on its payoff

F (S1) at maturity.Since qu, qd verify qu, qd ≥ 0 and qu + qd = 1, they can be seen as the

new probabilities associated to the values u, d assumed by the random vari-able X. The derivative price in formula (3.1) can be thus interpreted as theexpected value of the discounted payoff of the derivative with respect to thenew probability measure induced by qu, qd. In other words

F (S0) =1

1 + rEQ [F (S1)] . (3.2)

The new probability measure Q is called risk-neutral measure or equivalentmartingale measure. By definition of qu and qd, the expected value (with re-spect to Q) of the discounted final value of the stock is indeed equal to itscurrent value:

11 + r

[quS0u + qdS0d] = S0, (3.3)

which means that the discounted price process of the stock is a martingaleunder Q.

The above market model can be generalized to a multi-period setting asfollows. Consider a discrete-time model where a bond and a stock are traded.In each period (from t to t + 1, for t = 0, 1, ..., T − 1), the bond price and the


stock price are assumed to evolve as in a one-period binomial model. In otherwords:

Sn up;0 downn

↗p

↘1−p

. . . S(n−1) up;1 downn

↗p

S0u = Su1

......

↗p...

...S0

↘1−p

......

S0d = Sd1

↘1−p S1 up;(n−1) downn

. . . ↗p

↘1−p

S0 up;n downn

The market described above is known as multi-period binomial model.By the Markov property fulfilled by the model, it is easy to check that

under the risk-neutral measure Q

Sk =1

(1 + r)l−kEQ [Sl|Sk] =

1(1 + r)l−k

EQ [Sl|Fk] , 0 ≤ k ≤ l ≤ n. (3.4)

Once discounted, the process (Sk)k=0,1,...,n is then a martingale under Q andwith respect to the filtration (Fk)k=0,1,...,n generated by (Sk)k=0,1,...,n.The pricing method for the derivative F (S) is based on the assumption

that, in the binomial model, any derivative is attainable, namely there alwaysexists a strategy consisting in a suitable number of bonds and of stocks repli-cating exactly the derivative value at any time and in any state of the world.Or, equivalently, there always exists a portfolio strategy composed by a suit-able amount of the derivative and the underlying, so to “completely hedge”the risk. Such a property is called completeness of the market (see Chap. 4for more details on completeness). More precisely, a market model is said tobe complete if any derivative can be attained (or replicated).Both the one-period and the multi-period binomialmarket models described

above are complete. In the multi-period binomial model, the initial price F (S0)of the derivative is given by the expected value of the discounted payoff underthe risk-neutral measure Q, i.e.

F (S0) =1

(1 + r)nEQ [F (Sn)] =

1(1 + r)n

n∑k=0

(nk) qk

uqn−kd F (S0u

kdn−k). (3.5)


Necessary and sufficient conditions for a (quite general) market model tobe free of arbitrage and/or complete are given by the Fundamental Theoremsof the Asset Pricing . What just recalled in a binomial model can be seen asa particular case of these conditions.The First Fundamental Theorem of the Asset Pricing establishes that, in

“suitable” market models, no-arbitrage and existence of an equivalent martin-gale measure are equivalent conditions.For a market that is free of arbitrage, the Second Fundamental Theorem

of the Asset Pricing guarantees the uniqueness of the equivalent martingalemeasure when the market model is complete.

As already underlined, the binomial model is complete. Nevertheless, if atany time the risky asset could assume not only two but three (or more) dif-ferent values, such new market model would not any longer be complete. Aswe will see in Chapter 4, in such a case, indeed, an arbitrary derivative couldnot be attainable only by means of the bond B and of the risky asset S.

European Put and Call options (sometimes called vanilla options) andAmerican options, i.e. options that can be exercised at any time betweenthe initial 0 and the maturity T , can be easily evaluated in a binomial model.Concerning European options, it is sufficient to apply the arguments aboveto the payoff of a European Call (respectively, Put) option with strike K,maturity T and written on the stock S:

FCall(ST ) = (ST − K)+, resp. FPut(ST ) = (K − ST )+.

It is also worth to mention that the price (at time t = 0, 1, ..., T ) of a Eu-ropean Call option (Ct) and of a European Put option (Pt) with the samecharacteristics are related by the so called Put-Call Parity :

Ct − Pt = St − K

(1 + r)T−t. (3.6)

The pricing of some other options whose value depends not only on theunderlying value at maturity, but also on its value at intermediate times be-tween 0 and T , will be illustrated in some exercises of the present chapter andof the next ones.Based on the arguments above, it is also possible to obtain the strategy

able to make the portfolio riskless between two consecutive times: it would becomposed by a long position in the derivative and by Δ = (Fu − Fd)/(u − d)short positions in the underlying.

For a detailed treatment and further details on the subject, we refer toBjörk [6], Hull [23], Pliska [32] and Ross [34], among others.



Exercise 3.1. Different options having the same stock as underlying are avail-able on the market. Suppose that the risk-free interest rate is 4% per year,that the current stock price is 20 euros and that such price may go up or downby 25% in each of the next 2 semesters.

1. Assume that the stock price may go up or down in each semester with (ob-jective) probability of 50%. Verify whether the exercise of a European Calloption with strike of 18 euros and with maturity of 6 months is more likelythan the exercise of a Call option as the one above, but with maturity ofone year.

2. Compare the prices of the Call options of item 1.3. Discuss whether the options prices of items 1. and 2. would change if:

(a) the objective probability of an increase in the stock price was of 80%(consequently, 20% of a decrease);

(b) if the strike was of 20 euros. If yes, compute the price of a EuropeanCall option with maturity 6 of months and with strike of 20 euros.

SolutionWe deduce that the stock price evolves as follows:

31.25 = S0u2

↗S0u = 25

↗ ↘S0 = 20 18.75 = S0ud

↘ ↗S0d = 15

↘11.25 = S0d

2

− −−− − −− −−− −− −− −−0 6 months T = 1 year

where u = 1.25 (growth factor) and d = 0.75.

1. Remember that a European Call option is exercised when the underlyingprice is greater (or equal) than the strike. With this fact in mind, it is easyto compute the probability to exercise the options above.Before doing that, it is worth to underline that the behavior of the stockprice under the “real” probability is


25 p↗

20↘

15 1 − p−− −− −− −− −− −− −− − −− −− −− −− −−−

0 T ′ = 6 months with probability

31.25 p2

↗20 → 18.75 2p(1− p)

↘11.25 (1 − p)2

− −− −− −− −− −− −− − −− −− −− −− −−−0 T = 1 year with probability

where p = 1 − p = 0.5.It follows that the option with maturity (T ′) of 6 months and with strikeof 18 euros is exercised with probability

P (ST ′ ≥ K) = P (ST ′ ≥ 18) = P (ST ′ = 25) = 0.5,

while the option with maturity (T ) of one year and with strike of 18 eurosis exercised with probability

P (ST ≥ K) = P (ST ≥ 18)= P ({ST = 18.75} ∪ {ST = 31.25})= p2 + 2p (1 − p) = (0.5)2 + 2 · 0.5 · 0.5 = 0.75.

The exercise of a European Call option with strike of 18 euros and withmaturity of one year is then more likely than the exercise of a Call optionas the one above but with maturity of 6 months.

2. In order to compare the prices of the options taken into account, we look forthe “risk-neutral” probability Q that will be useful to evaluate the options.Such a probability Q corresponds to:

qu =(1 + r)1/2 − d

u − d=

√1.04− 0.75

1.25− 0.75= 0.54

qd = 1 − qu = 0.46,

because r = 0.04 is the annual interest rate and the time intervals are of 6months.


For the European Call (A) with maturity T ′ = 6 months we get:

S0u = 25 φuA = 7

↗qu

S0 = 20↘qd

S0d = 15 φdA = 0

− −−− −− −− −− − −− −− −−−0 T ′ = 6 months Payoff of option A

since the payoff of a European Call is given by

(ST ′ − K)+ ={

ST ′ − K; if ST ′ ≥ K0; if ST ′ < K

.

It follows that the initial price of the European Call option A is given by

CA0 =

1

(1 + r)1/2

[qu · φu

A + qd · φdA

]=

1√1.04

[0.54 · 7 + 0] = 3.71 euros.

For the European Call (B) with maturity T = 1 year, we have

S0u2 = 31.25 φuu

B = 13.25↗qu

S0u = 25↗qu ↘qd

S0 = 20 S0ud = 18.75 φudB = 0.75

↘qd ↗qu

S0d = 15↘qd

S0d2 = 11.25 φdd

B = 0− −−− − −− −−− − −− −− −−− − −− −− − −− −−

0 6 months 1 year Payoff of option B

To compute the initial cost of such an option we can proceed in two differentways:

• translate the problem in a one-period setting by considering only theinitial date (0) and the maturity T . Consequently,

0 T = 1 year with probability payoff− −− − −− −−− − −− −− −− −−− −− −− −−

31.25 (qu)2 13.25 = φuuB

↗20 −→ 18.75 2quqd 0.75 = φud

B

↘11.25 (qd)

2 0 = φddB


It follows that the initial price of the European Call B is given by

CB0 =

11 + r

[q2uφuu

B + 2quqdφudB + q2

dφddB

]=

11.04

[(0.54)2 · 13.25 + 2 · 0.54 · 0.46 · 0.75 + 0

]= 4.07 euros;

• proceed backwards in time as illustrated below.We first consider the sub-tree in bold and compute the “price” Cu

6m atthe node “Su

6m” at time t = 6 months.

S0u2= 31.25 φuuB = 13.25

↗S0u = 25

↗ ↘S0 = 20 S0ud = 18.75 φud

B = 0.75↘ ↗

S0d = 15↘

S0d2 = 11.25 φdd

B = 0−− −− −− −− −− −− −− −− −− −− −− −− −−


This reduces to option pricing in a one-period model. Hence

Cu6m =

1

(1 + r)1/2

[qu · φuu

B + qd · φudB

]=

1√1.04

[0.54 · 13.25 + 0.46 · 0.75] = 7.3545.

The same procedure can be applied to the sub-tree below:

S0u2 = 31.25 φuu

B = 13.25↗

S0u = 25↗ ↘

S0 = 20 S0ud = 18.75 φudB = 0.75

↘ ↗S0d = 15

↘S0d2= 11.25 φdd

B = 0−− −− −− −− −− −− −− −− −− − −− −−−



It follows that:

Cd6m =

1

(1 + r)1/2

[qu · φud

B + qd · φddB

]=

1√1.04

[0.54 · 0.75 + 0] = 0.397.

Finally, there remains to deduce the initial price of the Call from theprices Cu

6m and Cd6m just computed. In other words,

Cu6m = 7.3545 qu

↙CB

0 =???↖

Cd6m = 0.397 qd

− −− −−− − −− −− − −− −− −− −− −−0 t = 6 months with probability

Also in the present case the option pricing reduces to the pricing in aone-period setting, hence:

CB0 =

1

(1 + r)1/2

[qu · Cu

6m + qd · Cd6m

]=

1√1.04

[0.54 · 7.3545 + 0.46 · 0.397] = 4.07 euros.

3. (a) The prices of options A and B studied in items 1.-2. do not change onvarying the real probability. Such a probability, indeed, does not influ-ence the option pricing.

(b)The price of the European Call option with maturity of 6 months de-creases as the strike increases.Recall that the initial price of the European Call option A with maturityT ′ = 6 months and with strike of K = 18 euro is CA

0 = 3.85 euros.Consider now the same option as before except for the strike, now ofK = 20 euros. We get

S0u = 25 φu = 5↗qu

S0 = 20↘qd

S0d = 15 φd = 0− −−− −− −− −− −− − −− −− −− −− −− −−−

0 T ′ = 6 months Payoff of option with strike K = 20


Consequently, the price of the European Call (as A) but with strike Kis given by

C0 =1

(1 + r)1/2

[qu · φu + qd · φd

]=

1√1.04

[0.54 · 5 + 0] = 2.65 < CA0 .

Exercise 3.2. Different options having the same stock as underlying are avail-able on the market. Suppose that the risk-free interest rate is 4% per year,that the current stock price is 20 euros and that such price may go up or downby 25% in each of the next 3 semesters.

1. Evaluate a European Call option with strike of 18 euros and with maturityof 18 months.

2. Evaluate the corresponding Put option.3. Consider the Call and Put options above.

(a) How many shares of the stock would you be able to buy, and how muchmoney could you invest in a bank account at the initial time, if yousold eight Call options and bought one Put option (supposing it is notpossible to buy fractions of shares)?

(b)Compute the profit (or the loss) of the investment above if the stockprice grows in the next 3 semesters. What about the profit (or the loss)if the stock price grows in 2 semesters and decreases in one semester?

4. Evaluate the Call option of item 1. when the length of the periods is oneyear instead one semester.

Solution

1. Recall that, when the model consists of n periods (months, years, ...) andrp denotes for the interest rate per period1 (monthly, annual, ...), we have:

F (S0) = initial option price

=1

(1 + rp)n

n∑k=0

(n

k

)qku · qn−k

d · φ(k) up;(n−k) down,

where φ(k) up;(n−k) down stands for the payoff of the option when the priceof the underlying has moved k times up and (n − k) times down.

1 If time and interest rate rp refer to a different unit of time, it is enough to transformthe interest rate according to compounding. For instance, if ryear is the annual in-terest rate and time intervals are months, then the monthly interest rate (equivalentto ryear) is given by

rmonth = (1 + ryear)1/12 − 1.


In the present case, we have that time periods are of 6 months and thatthe price of the underlying evolves as follows

39.0625 φuuu = 21.0625↗

31.25↗ ↘

25 23.4375 φuud = 5.4375↗ ↘ ↗

20 18.75↘ ↗ ↘

15 14.0625 φudd = 0↘ ↗

11.25↘

8.4375 φddd = 0−− − −−− − −−− − −− −−− − −− −− −−−0 6 months 1 year 18 months Payoff of the Call

The monthly interest rate (equivalent to r = 0.04 per year) is rm =(1 + r)1/12 − 1 = (1.04)1/12 − 1, so (1 + rsem)3 = (1 + r)3/2. From thearguments above it follows that

C0 = price of the European Call with maturity T (18 months )

=1

(1 + r)3/2

[(qu)3 · φuuu + 3 (qu)2 qd · φuud + 3qu (qd)

2 · φudd

+(qd)3 · φddd

]=

1

(1.04)3/2

[(0.54)3 · 21.0625 + 3 · (0.54)2 · 0.46 · 5.4375 + 0 + 0

]=5.19,

where qu = 0.54 and qd = 0.46 are the same as in Exercise 3.1.2. We can evaluate the European Put option in two different ways.First, we can proceed as above. Since the payoff of the European Put optionis

(K − ST )+ = max{K − ST ; 0} ={

K − ST ; if ST ≤ K0; if ST > K

,

it follows that

39.0625 (qu)3 φuuuPut = 0

↗↗ 23.4375 3 (qu)2 qd φuud

Put = 020

↘ 14.0625 3qu (qd)2

φuddPut = 3.9375

↘8.4375 (qd)

3φddd

Put = 9.5625−− −− −− −− −− −− −− −− −− − −− −− −−−

0 T = 18 months with probability Payoff of the Put


Hence

P0 = price of the European Put with maturity T

=1

(1 + r)3/2

[(qu)3 · φuuu

Put + 3 (qu)2 qd · φuudPut + 3qu (qd)

2 · φuddPut

+(qd)3 · φddd

Put

]=

1

(1.04)3/2

[0 + 0 + 3 · 0.54 · (0.46)2 · 3.9375 + (0.46)3 · 9.5625

]= 2.15 euros.

An alternative way to evaluate the Put option above is to apply the Put-Call Parity true for European options with the same maturity and strikeand written on the same underlying, that is

C0 − P0 = S0 − K

(1 + r)T.

Due to the equality above, we obtain immediately

P0 = C0 − S0 +K

(1 + r)T∼= 2.15 euros.

3. By the previous items, C0 = 5.19 and P0 = 2.15 euros.

(a) If we sell eight Call options and buy one Put option as above, we disposeof

π = 8 ·C0 − P0 = 8 · 5.19− 2.15 = 39.37 euros.

Under the restrictions imposed, we can buy only one share of the under-lying (its initial price is indeed S0 = 20 euros) and invest the remainingbudget of 19.37 euros at risk-free interest rate.

(b) If the stock price grows in each of the next three semesters (hence,ST = S0u

3 = 39.0625 euros), the profit (or loss) of our strategy is

8 Call sold −8 (ST − K)+ = −8 · φuuu = −168.51 Put bought (K − ST )+ = 0profit/loss by the stock (ST − S0) = 19.0625cash invested in a bank account 19.37 · (1.04)3/2 = 20.54375Total Profit / Loss −128.894 euro

If the stock price grows in two of the next three semesters and decreasesin one (hence ST = S0u

2d = 23.4375 euros), the profit (or loss) of our


strategy is

8 Calls sold −8 (ST − K)+ = −8 · φuud = −43.51 Put bought (K − ST )+ = 0profit/loss by the stock (ST − S0) = 3.4375cash invested in a bank account 19.37 · (1.04)3/2 = 20.54375Total Profit / Loss −19.5188 euros

4. With annual periods instead of semestral ones, the stock price would evolveas before:

39.0625 φuuu = 21.0625↗

31.25↗ ↘

25 23.4375 φuud = 5.4375↗ ↘ ↗

20 18.75↘ ↗ ↘

15 14.0625 φudd = 0↘ ↗

11.25↘

8.4375 φddd = 0−− − − −−− − − −− − − −− −− −− −− −− −−

0 1 year 2 years 3 years Payoff of the Call

while the maturity and the equivalent martingale measure Q would change.For annual periods, indeed, Q would correspond to

q∗u =(1 + r) − d

u − d=

1.04− 0.751.25− 0.75

= 0.58

q∗d = 1 − q∗u = 0.42,

hence the price of the European Call with maturity 3 years would be givenby

C(3y)0 =

1(1 + r)3

[(q∗u)3 · φuuu + 3 (q∗u)2 q∗d · φuud + 3q∗u (q∗d)2 · φudd

+(q∗d)3 · φddd]

=1

(1.04)3[(0.58)3 · 21.0625 + 3 · (0.58)2 ·0.42 · 5.4375 + 0 + 0

]=5.70.

Exercise 3.3. Consider a stock whose current price is S0 = 40 euros. In eachof the next two four-month periods, the stock price may move up by 4% ordown by 2%. The risk-free rate available on the market is of 4% per year.


Compute the initial price of an option with maturity of T = 8 months andwith payoff

φ =

(ST − (K − S0)

2

10

)+

=

{0; ST <

(K−S0)2

10

ST − (K−S0)2

10; ST ≥ (K−S0)2

10

when the strike K is of 60 euros.

SolutionWe know that on each of the next four-month periods the stock price may

increase by a growth factor u = 1.04 or decrease by a factor d = 0.98. Accord-ingly, the stock price evolves as below:

43.264 = S0u2

↗41.6

↗ ↘40 40.768 = S0ud

↘ ↗39.2

↘38.416 = S0d

2

−− − − −−− − −− −− −−−0 4 months 8 months

The payoff of the option may thus take the following values:

φuu =

(Suu

T − (60 − 40)2

10

)+

=(

43.264− 202

10

)+

= 3.264

φud =

(Sud

T − (60− 40)2

10

)+

=(

40.768− 202

10

)+

= 0.768

φdd =

(Sdd

T − (60 − 40)2

10

)+

=(

38.416− 202

10

)+

= 0.

Consequently,

43.264 = S0u2 3.264 = φuu

↗41.6

↗ ↘40 40.768 = S0ud 0.768 = φud

↘ ↗39.2

↘38.416 = S0d

2 0 = φdd

−− − −− −− − −− − −− −− −− −− −− −−0 4 months 8 months Payoff of the option


The risk-neutral probability Q to be used for option pricing correspondsto:

qu =(1 + r)4/12 − d

u − d= 0.55

qd = 1 − qu = 0.45,

since periods are of 4 months while r is the annual interest rate.From the arguments above, the option with payoff φ now costs

F (S0) =1

(1 + r)T

[(qu)2 φuu + 2quqdφ

ud + (qd)2φdd]

=1

(1 + r)8/12

[(0.55)2 · 3.264 + 2 · 0.55 · 0.45 · 0.768 + 0

]= 1.33 euros.

Exercise 3.4.We take a short position in a European Call option with ma-turity 4 months and with strike of 20 euros, having a stock with current priceof 20 euros as underlying. In the next 4 months, the stock price may increaseby a growth factor u = 1.2 or decrease by a factor d = 0.8. The risk-freeinterest rate available on the market is of 4% per year.

1. Find the replicating strategy of the option.2. Compute the initial price of the option.3. Suppose now that the stock price does not follow a binomial model anymore, but that in 4 months it may either increase by a growth factoru = 1.2, decrease by a factor d = 0.8 or remain unchanged. Discuss whetherit is still possible to replicate the Call option above only with the underlyingand with cash (to be deposited or borrowed).

Solution

1. The stock price and the option payoff are the following:

S0u = 24 φu = 4↗

S0 = 20↘

S0d = 16 φd = 0−− −− −− −− −− − − −− −−−

0 T = 4 months Payoff

In order to find the replicating strategy of the option we need to find anumber Δ of shares of the underlying and an amount x of cash so that theseller of the option holding such a portfolio can be “covered” against losses(due to the option) at maturity and in any state of the world. We are then


looking for Δ and x such that

{ΔS0u + x (1 + r)T = φu

ΔS0d + x (1 + r)T = φd.

The previous system of linear equations is equivalent to

{ΔS0u − ΔS0d = φu − φd

ΔS0d + x (1 + r)T = φd ,

whose solution is given by:

{Δ = φu−φd

S0(u−d)

x = 1(1+r)T

[φd − ΔS0d

] .

The replicating strategy is then composed as below:

{Δ = φu−φd

S0(u−d)= 4−0

20(1.2−0.8)= 0.5

x = 1(1.04)4/12 [0 − 0.5 · 20 · 0.8] = −7.9

To be covered against losses deriving from the exercise of the option, theseller has to buy 0.5 shares of the stock and to borrow 7.9 euros.Since the cost of such a strategy is equal to

ΔS0 + x = 0.5 · 20 − 7.9 = 2.1,

the No Arbitrage principle implies that the Call price should be equal tothe cost of the replicating strategy, i.e. C0 = ΔS0 + x = 2.1.

2. We have already computed the option price above by means of the repli-cating strategy.An alternative way to compute C0 is to proceed as in Exercises 3.1 and3.2. By the no arbitrage principle, the result has to coincide with the oneabove.Since the equivalent martingale measure corresponds to qu = (1+r)4/12−d

u−d =0.53 and qd = 1 − qu = 0.47, we obtain indeed

C0 =1

(1 + r)T

[quφu + qdφ

d]

=1

(1.04)4/12[0.53 · 4 + 0] = 2.10 euros.


3. The stock price is now assumed to evolve as follows:

S0u = 24 φu = 4↗

S0 = 20 → S0 · 1 = 20 φm = 0↘

S0d = 16 φd = 0− −−− − −− −− −−− −− −− −−

0 T = 4 months Payoff

Verifying the existence of a replicating strategy for the option is equivalentto verifying the existence of a number Δ of shares of the underlying and ofan amount x of cash such that⎧⎪⎨

⎪⎩ΔS0u + x (1 + r)T = φu

ΔS0 + x (1 + r)T = φm

ΔS0d + x (1 + r)T = φd

.

Since the previous system (of 3 linear equations in 2 variables) is equivalentto the following ones

⎧⎨⎩

ΔS0u + x (1 + r)T = φu

ΔS0 + x (1 + r)T = φm

ΔS0 (u − d) = φu − φd

⎧⎪⎨⎪⎩

ΔS0u + x (1 + r)T = φu

ΔS0 + x (1 + r)T = φm

Δ = φu−φd

S0(u−d)⎧⎨⎩

x (1 + r)T = φu − ΔS0u

x (1 + r)T = φm − ΔS0

Δ = 12

⎧⎨⎩

x (1 + r)T = −8x (1 + r)T = −10

Δ = 12

,

it is evident that there is no solution. Consequently, it is not possible toreplicate the option only by means of the underlying and of cash investedat risk-free rate.From the arguments above it follows that the market model is incomplete(i.e. not any option is attainable) and that the option price is not uniquelydetermined, in general.

Exercise 3.5.We take a short position in a European Call option with ma-turity 2 years and with strike of 25 euros, having a stock with current priceof 40 euros as underlying. In each of the next 2 years, the stock price mayincrease by a growth factor u = 1.5 or decrease by a factor d = 0.5, while therisk-free interest rate is of 4% per year.

1. Find the replicating trading strategy of the option above and deduce theoption price.

2. Find the replicating strategy of a portfolio formed by two short positions ina Call with maturity of 1 year, in one long position in a Put with maturityof 1 year (where both Call and Put options have strike of 25 euros and are


written on the underlying described above) and in one long position in theunderlying.

3. Discuss whether the replicating strategy and the option price change if ev-erything (underlying dynamics, strike, maturity, ...) stays the same exceptfor the risk-free rate that increases to 6% per year.

Solution

1. The stock price evolves as follows:

90 = S0u2

↗S0u = 60

↗ ↘S0 = 40 30 = S0ud

↘ ↗S0d = 20

↘10 = S0d

2

− −−− − −− −− −− − −− −−0 1 year 2 years

where u = 1.5 (growth factor) and d = 0.5.First of all, let us find the risk-neutral measure to be used for option pricing.Such a measure Q corresponds to:

qu =(1 + r) − d

u − d=

1.04− 0.51.5− 0.5

= 0.54

qd = 1 − qu = 0.46,

because both the risk-free rate r and the periods are annual.We will find simultaneously the option price and the replicating strategy.For a European Call option with strike of 25 euros:

90 = S0u2 φuu = 65

↗S0u = 60

↗ ↘S0 = 40 30 = S0ud φud = 5

↘ ↗S0d = 20

↘10 = S0d

2 φdd = 0− −−− − −− − − − −− −− −− − −− −− −−−

0 1 year 2 years Payoff of the option

To find the replicating strategy of the option, we consider separately anyperiod and we proceed backwards in time.


We first consider the sub-tree in bold so to compute the option “price” Cu1

at the node “Su1 ” in t = 1 year.

90 = S0u2 φuu = 65↗

S0u = 60↗ ↘

S0 = 40 30 = S0ud φud = 5↘ ↗

S0d = 30↘

10 = S0d2 φdd = 0

−− −− −− −− −− −− −− −− −− −− −−0 1 year 2 years Payoff of the option

We have reduced to evaluate an option and/or find its replicating tradingstrategy in a one-period model. We obtain that:

Cu1 =

11 + r

[qu · φuu + qd · φud

]=

11.04

[0.54 · 65 + 0.46 · 5] = 35.96

and that the replicating strategy (Δu1 , xu

1) at time t = 1 and at the node“Su

1 ” is given by:⎧⎪⎪⎨⎪⎪⎩

Δu1 =

φuu − φud

Su1 (u − d)

=65 − 5

60 (1.5− 0.5)= 1

xu1 =

11 + r

[φud − Δu

1Su1 d]

=1

1.04[5 − 30] = −24.04

As expected,

Cu1 = Δu

1Su1 + xu

1 = Δu1S0u + xu

1 = 35.96.

Proceeding analogously for the sub-tree in bold below:

90 = S0u2 φuu = 65

↗S0u = 60

↗ ↘S0 = 40 30 = S0ud φud = 5

↘ ↗S0d = 20

↘10 = S0d2 φdd = 0



we get

Cd1 =

11 + r

[qu · φud + qd · φdd

]=

11.04

[0.54 · 5 + 0] = 2.6

and ⎧⎪⎪⎨⎪⎪⎩

Δd1 =

φud − φdd

Sd1 (u − d)

=5 − 0

20 (1.5 − 0.5)= 0.25

xd1 =

11 + r

[φdd − Δd

1Sd1d]

= −2.4

Also in the present case, we obtain the same result as previously, that is

Cd1 = Δd

1Sd1 + xd

1 = Δd1S0d + xd

1 = 2.6.

Finally, we consider the first period:

Cu1 = 35.96 qu

↙C0 =???

↖Cd

1 = 2.6 qd

−− −− −− −− −−0 t = 1 year

It is a one-period case, hence the option price can be computed immediatelyas

C0 =1

1 + r

[qu ·Cu

1 + qd · Cd1

]=

11.04

[0.54 · 35.96 + 0.46 · 2.6] = 19.8 euros,

while the replicating strategy is given by⎧⎪⎨⎪⎩

Δ0 =Cu

1 − Cd1

S0 (u − d)=

35.96− 2.640 (1.5− 0.5)

= 0.834

x0 = (1 + r)−1 [Cd

1 − Δ0S0d]

= −13.54

Consequently, we find again C0 = Δ0S0 + x0 = 19.8 euros.

Summing up: the initial price of the option is 19.8 euros and its replicatingtrading strategy consists in the following. The seller of the option buys0.834 stocks and borrows 13.54 euros at t = 0. Then, at t = 1: (a) if thestock price is increased to S0u = 60, he buys (1 − 0.834) = 0.166 morestocks and borrows (24.04 − 13.54) = 10.50 euros more. So, his portfolioconsists in 1 stock and in 24.04 euros borrowed. (b) If the stock price is


decreased to S0d = 20, he sells (0.834−0.25) = 0.584 stocks and gives back(13.54 − 2.4) = 11.14 euros of what he had borrowed. Thus, his portfolioconsists in 0.25 stocks and in 2.4 euros borrowed.

2. We consider now a portfolio formed by two short positions in a Call withmaturity of 1 year, by one long position in a Put with maturity of 1 year(where both Call and Put options have strike of 25 euros and are written onthe underlying described above) and by one long position in the underlying.Since the underlying and the Call and Put payoffs read:

S0u = 60 φuCall = 35 φu

Put = 0↗

S0 = 40↘

S0d = 20 φdCall = 0 φd

Put = 5−− −− −− −− −− − −− −−− − −− −−−

0 T ∗ = 1 year Call payoff Put payoff

the portfolio π to be replicated is given by

φπ = −2φCall + φPut + ST∗

=

{φu

π ; if S1 = Su1

φdπ ; if S1 = Sd

1

=

{−2φuCall + φu

Put + Su1 = −10; if S1 = Su

1

−2φdCall + φd

Put + Sd1 = 25; if S1 = Sd

1

In order to find a replicating strategy of the above portfolio of options andstocks, we have to look for a number Δ of shares of the underlying and foran amount x of cash such that{

ΔS0u + x (1 + r)T∗= φu

π

ΔS0d + x (1 + r)T∗= φd

π

.

In the present case, we obtain

{ΔS0u − ΔS0d = φu

π − φdπ

ΔS0d = φdπ − x (1 + r)T∗

⎧⎪⎪⎨⎪⎪⎩

Δ =φu

π − φdπ

S0 (u− d)=

−10 − 2540(1.5− 0.5)

= −0.875

x =1

1 + r

[φd

π − ΔS0d]

= 40.865

Consequently, the replicating strategy we are looking for consists in sell-ing 0.875 shares of the underlying and in investing 40.865 euros in a bankaccount. Such a strategy costs then ΔS0 +x = −35+40.865 = 5.865 euros.


3. To find the replicating trading strategy of the option (that exists for sure!),we recall that the dynamics of the stock, as well as the option payoff, donot depend on the risk-free rate and are the following:

90 = S0u2 φuu = 65

↗S0u = 60↗ ↘

S0 = 40 30 = S0ud φud = 5↘ ↗

S0d = 20 ↘10 = S0d

2 φdd = 0−− −− −− −− −− −− −− −− −− −− −−

0 1 year 2 years Payoff of the option

In order to find the replicating strategy we analyze separately the twoperiods and we proceed backwards in time as in the previous item.First, let us consider the sub-tree in bold so to compute the option priceCu

1 at the node “Su1 ” and at time t = 1:

90 = S0u2 φuu = 65↗

S0u = 60↗ ↘S0 = 40 30 = S0ud φud = 5

↘ ↗S0d = 20 ↘

10 = S0d2 φdd = 0


Under the new risk-free rate r, the risk-neutral measure corresponds nowto

qu =(1 + r) − d

u − d=

1.06− 0.51.5− 0.5

= 0.56

qd = 1 − qu = 0.44.

Accordingly, the replicating strategy (Δu1 , xu

1) to be held at t = 1 and atthe node “Su

1 ” is now⎧⎪⎨⎪⎩

Δu1 =

φuu − φud

Su1 (u − d)

=65 − 5

60 (1.5− 0.5)= 1

xu1 = (1 + r)−1 [

φud − Δu1Su

1 d]

= (1 + 0.06)−1 [5 − 30] = −23.58

Therefore, Cu1 = Δu

1Su1 + xu

1 = Δu1S0u + xu

1 = 36.42 euros.


Analogously for the sub-tree in bold below:

90 = S0u2 φuu = 65

↗S0u = 60

↗ ↘S0 = 40 30 = S0ud φud = 5

↘ ↗S0d = 20

↘10 = S0d2 φdd = 0


In the present case, we get⎧⎪⎨⎪⎩

Δd1 =

φud − φdd

Sd1 (u − d)

=5 − 0

20 (1.5 − 0.5)= 0.25

xd1 = (1 + r)−1 [

φdd − Δd1S

d1d]

= −2.36

and, consequently, Cd1 = Δd

1Sd1 + xd

1 = Δd1S0d + xd

1 = 2.64 euros.

Finally, we take into account the first period:

Cu1 = 36.42 qu

↙C0 =???

↖Cd

1 = 2.64 qd

−− −− −− −− −−0 1 year

The replicating trading strategy is given by

⎧⎪⎪⎨⎪⎪⎩

Δ0 =Cu

1 − Cd1

S0 (u − d)=

36.42− 2.6440 (1.5− 0.5)

= 0.8445

x0 =1

1 + r

[Cd

1 − Δ0S0d]

= −13.44

,

so we deduce that C0 = Δ0S0 + x0 = 20.34 euros.

Exercise 3.6. Consider a market where a bond (corresponding to a risk-freerate of 4% per year) and a stock are traded. The current price of the stock isS1

0 = 8 euros that, in one year, may move up to 16 euros or to 12 euros, ordown to 2 euros.


1. Consider a European Call option on the stock above, with strike of 8 eurosand with maturity of one year. Verify if it is possible to replicate such anoption only by means of the underlying and of cash invested (or borrowed)at risk-free rate.

2. Consider another stock of current price S20 = 10 euros and that, in one year,

may cost 20, 8 or 6 euros whenever the other stock costs, respectively, 16,12 or 2 euros.A new option is also available, with maturity T of one year, with the samestrike K as the Call above and with payoff

φ =(ST − K

)+

,

where ST = S1T +S2

T

2 −∣∣∣S1

T −S2T

4

∣∣∣.Verify if it is possible to replicate the option above by means of the twostocks and of cash invested (or borrowed). If yes, find the cost of the repli-cating strategy.

Solution

1. The price of the first stock and the payoff of the European Call on thestock, with maturity of one year and strike of 8 euros are the following:

S1,u1 = 16 φu

Call = 8↗

S10 = 8 → S1,m

1 = 12 φmCall = 4

↘S1,d

1 = 2 φdCall = 0

− −−− − −− −−− − − −− −− −−0 1 year Payoff of the Call

In order to verify if the Call option above is attainable by means of theunderlying stock and of cash, we need to check if the following system hassolutions or not: ⎧⎪⎪⎨

⎪⎪⎩ΔS1,u

1 + x (1 + r)T = φuCall

ΔS1,m1 + x (1 + r)T = φm

Call

ΔS1,d1 + x (1 + r)T = φd

Call

,

where Δ represents the number of shares of the underlying to buy or selland x the amount of cash to be invested or borrowed.


Replacing T with 1 (year), the previous system (of 3 equations in 2 vari-ables) is equivalent to the following ones:

⎧⎪⎪⎪⎨⎪⎪⎪⎩

ΔS1,u1 + x (1 + r) = φu

Call

Δ(S1,u

1 − S1,m1

)= φu

Call − φmCall

ΔS1,d1 + x (1 + r) = φd

Call

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

ΔS1,u1 + x (1 + r) = φu

Call

Δ =φu

Call − φmCall

S1,u1 − S1,m

1

ΔS1,d1 + x (1 + r) = φd

Call

⎧⎪⎪⎨⎪⎪⎩

x (1 + r) = φuCall − ΔS1,u

1

Δ = 1

x (1 + r) = φdCall − ΔS1,d

1

⎧⎪⎨⎪⎩

x (1 + r) = −8

Δ = 1

x (1 + r) = −2

Since the last system has no solution, the Call option cannot be replicatedonly by means of the underlying and of cash.

2. We display below the dynamics of the two stocks and the payoff of theoption written on them:

S1,u1 = 16; S2,u

1 = 20 φu = 9↗

S10 = 8; S2

0 = 10 → S1,m1 = 12; S2,m

1 = 8 φm = 1↘

S1,d1 = 2; S2,d

1 = 6 φd = 0−− −− −− −− − −− −− −− −−− − −− −− −− −−−

0 1 year Payoff of the option

Verifying the existence of a replicating strategy for the option is equiva-lent to verifying the existence of a number Δ1 of shares of one stock, of anumber Δ2 of shares of the other stock and of an amount x of cash suchthat ⎧⎪⎪⎨

⎪⎪⎩Δ1S1,u

1 + Δ2S2,u1 + x (1 + r) = φu

Δ1S1,m1 + Δ2S2,m

1 + x (1 + r) = φm

Δ1S1,d1 + Δ2S2,d

1 + x (1 + r) = φd

.

In the present case, the previous system (of 3 equations in 3 variables) isequivalent to

⎧⎪⎨⎪⎩

16Δ1 + 20Δ2 + x (1 + r) = 9

12Δ1 + 8Δ2 + x (1 + r) = 1

2Δ1 + 6Δ2 + x (1 + r) = 0

⎧⎪⎨⎪⎩

16Δ1 + 20Δ2 + x (1 + r) = 9

10Δ1 + 2Δ2 = 1

2Δ1 + 6Δ2 + x (1 + r) = 0

56 3 Binomial Model for Option Pricing⎧⎪⎨⎪⎩

16Δ1 + 20Δ2 + x (1 + r) = 9

Δ2 = 1/2 − 5Δ1

14Δ1 + 14Δ2 = 9

⎧⎪⎨⎪⎩

16Δ1 + 20Δ2 + x (1 + r) = 9

Δ2 = 1/2− 5Δ1

−56Δ1 = 2⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

x =9 − 16Δ1 − 20Δ2

1 + r

Δ2 =1928

Δ1 = − 128

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x = −3.846

Δ2 =1928

Δ1 = − 128

It is then possible to replicate the option by selling −Δ1 = 128shares of

the first stock, buying Δ2 = 1928shares of the second stock and borrowing

3.846 euros. So, the cost of the replicating trading strategy is equal to

Δ1S10 + Δ2S2

0 + x = − 128

· 8 +1928

· 10 − 3.846 = 2.654 euros.

Exercise 3.7. A stock (indexed by A) is available on the market at the cur-rent price SA

0 = 8 euros. In one year, the price may increase by 25% or decreaseby 25%. Another stock (indexed by B) is also available. Its current price isSB

0 = 12 euros that, in one year, may increase by 25% (when also the price ofstock A is increased) or decreased by 25% (when also the price of stock A isdecreased). The risk-free interest rate on the market is 4% per year.

1. Consider a European Call option with maturity of one year, with strike of8 euros and written on stock A. Verify if it is possible to replicate such anoption only by means of stock A and of cash invested or borrowed at risk-free rate.

2. Verify if it is possible to replicate the European Call option above by tak-ing long positions in stock A, short positions in stock B and cash. If yes,compute the cost of the replicating strategy.

3. Verify if it is possible to replicate the European Call option above only bymeans of stock A, of cash and of one (and only one) short position in stockB. If yes, compute the cost of the replicating strategy.

Solution

1. The prices of stocks A and B may increase by a factor u = 1.25 (growthfactor) or decrease by a factor d = 0.75. These prices and the payoff of theEuropean Call option on stock A, with maturity of one year and strike of


8 euros, can be summarized as follows:

SA,u1 = SA

0 u = 10SB,u

1 = SB0 u = 15

φuCall = 2

↗SA

0 = 8

SB0 = 12

↘SA,d

1 = SA0 d = 6

SB,d1 = SB

0 u = 9φd

Call = 0

−− −− −− −− −− −− −− −− −− − −− −− −− −− −0 1 year Payoff of Call on A

In order to verify if the Call option above is attainable by means of stockA and of cash, we need to check if the following system has solutions ornot: ⎧⎨

⎩ΔASA,u

1 + x (1 + r)T = φuCall

ΔASA,d1 + x (1 + r)T = φd

Call

,

where ΔA represents the number of shares of stock A to buy or sell and xthe amount of cash to be invested or borrowed.Substituting the price of A, the previous system (of 2 equations in 2 vari-ables) becomes

{10ΔA + x (1 + r) = 2

6ΔA + x (1 + r) = 0

{10ΔA + x (1 + r) = 2

4ΔA = 2⎧⎪⎨⎪⎩

x = − 31.04

ΔA =12

⎧⎨⎩

x = −2.88

ΔA =12

Consequently, the Call option can be replicated by buying ΔA = 12 shares

of stock A and by borrowing 2.88 euros at risk-free rate. The replicatingstrategy then costs

ΔASA0 + x = 4 − 2.88 = 1.12 euros.

2. Verifying the existence of a replicating strategy for the Call option aboveby means of long positions in stock A, of short positions in stock B and ofcash is equivalent to verifying if there exist a number ΔA ≥ 0 of shares ofstock A, a number ΔB ≤ 0 of shares of stock B and an amount x of cash


such that ⎧⎪⎨⎪⎩

ΔASA,u1 + ΔBSB,u

1 + x (1 + r) = φu

ΔASA,d1 + ΔBSB,d

1 + x (1 + r) = φd

ΔA ≥ 0; ΔB ≤ 0

.

Substituting our data, the previous system becomes⎧⎪⎨⎪⎩

10ΔA + 15ΔB + x (1 + r) = 2

6ΔA + 9ΔB + x (1 + r) = 0

ΔA ≥ 0; ΔB ≤ 0

⎧⎪⎨⎪⎩

10ΔA + 15ΔB + x (1 + r) = 2

4ΔA + 6ΔB = 2

ΔA ≥ 0; ΔB ≤ 0

⎧⎪⎪⎨⎪⎪⎩

10ΔA + 15ΔB + x (1 + r) = 2

ΔA =12− 3

2ΔB

ΔA ≥ 0; ΔB ≤ 0

⎧⎪⎪⎨⎪⎪⎩

5 − 15ΔB + 15ΔB + x (1 + r) = 2

ΔA =12− 3

2ΔB

ΔA ≥ 0; ΔB ≤ 0

⎧⎪⎪⎪⎨⎪⎪⎪⎩

x = − 31.04

= −2.88

ΔA =12− 3

2ΔB

ΔB ≤ 0

Consequently, the Call option can be replicated with ΔB ≤ 0 stocks B,ΔA = 1

2 − 32ΔB stocks A and by borrowing 2.88 euros. The replication cost

is then equal to

ΔASA0 + ΔBSB

0 + x =(

12− 3

2ΔB

)· 8 + 12ΔB − 2.88 = 1.12 euros.

3. In the present case, we should verify if it is possible to replicate the Calloption only by means of stock A, of cash and by selling one (and only one)short position in stock B. We just need to impose ΔB = −1 and apply theresult obtained in the previous item.The required replicating strategy of the Call reduces to⎧⎪⎪⎪⎨

⎪⎪⎪⎩x = − 3

1.04= −2.88

ΔA =12− 3

2(−1)

ΔB = −1

⎧⎪⎪⎨⎪⎪⎩

x = − 31.04

= −2.88

ΔA = 2

ΔB = −1

In the present case, then, the Call option can be replicated by buying2 shares of stock A, selling one share of stock B and borrowing 2.88 eu-ros. It is immediate to check that such a replicating strategy, as well, costs1.12 euros.


Exercise 3.8. The current price of a stock is S0 = 8 euros. In one year, suchprice may move up to 16 euros or to 12 euros, or move down to 8.32 euros.Each event may happen with probability 1/3.On the market, the risk-free interest rate is 4% per year and it is possible

to sell (at the price of one euro) a European Call option written on the stockabove, with maturity of 1 year and with strike of 11 euros.Consider the following strategy:

• buy one share of the stock;• sell the Call;• borrow 7 euros at the risk-free rate.

Verify that such a strategy represents an arbitrage opportunity.

SolutionConsider strategy (A):

• buy one share of the stock;• sell the Call;• borrow 7 euros at the risk-free rate.

The initial value V0 (A) and the final value V1 (A) of strategy A are thefollowing:

t = 0 t = 1 year

buy one stock stock price⇒ −S0 = −8 ST

sell the Call Call payoff⇒ +C0 = 1 − (ST − K)+

borrow 7 euros give back the loan⇒ +c = 7 −c (1 + r)1 = −7 · 1.04 = −7.28

V0 (A) = C0 − S0 + c = 0 V1 (A) = ST − (ST − K)+ − c (1 + r)

Recall that ST − (ST − K)+ = min{ST ; K}.Since V0 (A) = 0, in order to verify that the strategy above is an arbitrage

opportunity we just need to check that V1 (A) = ST −(ST − K)+−c (1 + r) =min{ST ; K} − c (1 + r) ≥ 0 with P (V1 (A) > 0) > 0.Since P (ST = 16) = P (ST = 12) = P (ST = 8.32) = 1

3, the final value

V1 (A) of the strategy A is given by

V1 (A) =

⎧⎪⎨⎪⎩

min{16; 11} − 7.28 = 3.72; with prob. 1/3

min{12; 11} − 7.28 = 3.72; with prob. 1/3

min{8.32; 11}− 7.28 = 1.04; with prob. 1/3

Therefore, V1 (A) ≥ 0 and P (V1 (A) > 0) = 1 > 0.


As V0 (A) = 0, and because of the previous arguments, strategy A is anarbitrage opportunity.


Exercise 3.9. A stock A is available on the market at the current price of12 euros. In 4 months, such price may move up to 16 or down to 10 euros. Ineach of the next four-month periods, the stock price may increase or decreaseby the same percentage as in the previous period. The risk-free interest rateon the market is 4% per year.Consider also a stock B whose price is the square of the price of stock A

at any time and at any node of the tree.

1. Denote by St and St the prices (at time t) of stocks A and B, and by uand d (respectively, u and d) the increase factor and the decrease factor oneach period.Verify that the price of stock B follows a binomial model with S0 = S2

0 ,u = u2 and d = d2.

2. Compute the current price of a European Put option on stock B, withmaturity of 8 months and strike of 144 euros.

3. Find the replicating strategy of the European Put option. Verify if theresult found in item 2. is in line with the replicating strategy cost.

4. Compare the above Put with a similar Put option on stock A with strikeof 12 euros. Which one is more expensive?

5. Using the Put-Call Parity, deduce the prices of the corresponding EuropeanCall options.

Exercise 3.10. The current price of a stock is S0 = 8 euros. In one year, suchprice (S1) may move up to 16 euros, to 12 euros or down to 8.32 euros, eachwith probability 1/3.The risk-free interest rate available on the market is 4% per year.

1. Consider the following strategy: buy one share of the stock and borrow 8euro at the risk-free rate.

Verify if the strategy is an arbitrage opportunity or not.

2. Discuss what happens if the stock price (in 1 year) assumes the values 16,12 and 8.32 euros with probability 1/2, 1/4 and 1/4, respectively.

3. Discuss whether the interest rate can be chosen so that the correspondingmarket model is free of arbitrage.

4. Verify if the strategy in item 1. remains an arbitrage opportunity even ifS1 = S1 − 4 represents the value of another stock in one year and theinterest rate is 4% per year.

4

Absence of Arbitrage and Completenessof Market Models


In the following, we recall the notions of arbitrage, completeness and optionpricing in quite general one-period (or multi-period) market models, but al-ways based on a finite sample space.For a detailed treatment and further details on the subject, we refer to

Pliska [32], among others.

Consider a market model consisting in a non-risky asset, called bond, whosevalue will be denoted by B, and in n risky assets, called stocks, whose valueswill be denoted by S1, S2, ..., Sn.

Let us focus, first, on a one-period model. This means that any asset on themarket can be exchanged just at the beginning and at the end of the giveninterval of time. By convention, let t = 0 and t = 1 be the correspondingdates.The initial prices of the (non-risky and risky) assets are known: B0 for the

bond, S10 , S2

0 , ..., Sn0 for the stocks. The bond price is deterministic, equal to

B1 = B0(1 + r)

at t = 1, where r is the risk-free interest rate on the period. On the otherhand, the prices of the stocks at t = 1 will be random, namely

Si1 = Si

1 (ω) , for i = 1, 2, ..., n,

with ω ∈ Ω = {ω1, ω2, ..., ωm} and m ∈ N (finite sample space). Any stateof the world will be assumed to be possible, that is P (ωk) > 0 for any k =1, 2, ...,m (under the real probability P on Ω).

It is then possible to form different trading portfolios by means of the riskyand non-risky assets above. A trading strategy or portfolio is characterized by


62 4 Absence of Arbitrage and Completeness of Market Models

a vectorα =

(α0, α1, ..., αn

) ∈ Rn+1,

where α0 and αi (i = 1, 2, ..., n) stand, respectively, for the number of non-risky assets and the number of shares of stock Si in the portfolio. In particular,negative values of αj (j = 0, 1, 2, ..., n) correspond to short positions, whilepositive values correspond to long positions in asset j.

Given a strategy α, its value at time t ∈ {0, 1} is given by

Vt (α) = α0Bt +n∑

i=1

αiSit. (4.1)

A strategy α is said to be an arbitrage opportunity if V0 (α) = 0 and V1 (α) ≥ 0with P (V1 (α) > 0) > 0. In other words, an arbitrage opportunity is a way tomake a profit at zero cost (hence the name of free lunch).It is therefore reasonable to exclude arbitrage opportunities or, better, to

focus on market models with no arbitrage opportunities. As already recalledin Chapter 3, the First Fundamental Theorem of Asset Pricing guaranteesthat the one-period market model above has no arbitrage opportunities if andonly if there exists (at least) one equivalent martingale measure (EMM, forshort), where with equivalent martingale measure (or risk-neutral measure)we mean a probability measure Q such that Q (ωk) > 0 for any k = 1, 2, ..., m(hence, equivalent) and such that

EQ

[Si

1

B1

]= Si

0 (4.2)

for any i = 1, 2, ..., n (hence, martingale measure). The set of all equivalentmartingale measures will be denoted by M.

There remains to establish how to evaluate a derivative. First of all, we willdistinguish between attainable derivatives and non-attainable derivatives. Werecall that a derivative with payoff Φ is said to be attainable if there existsa trading strategy α such that V1 (α) = Φ. If all the derivatives available onthe market are attainable (that is, they can be replicated), then the marketis said to be complete; otherwise, it is said to be incomplete.The following result provides a necessary and sufficient condition for a

one-period model to be complete.

Theorem 4.1. Suppose that the market model considered is free of arbitrage.It is also complete if and only if the matrix⎡

⎢⎢⎢⎣B1 (ω1) S1

1 (ω1) · · · Sn1 (ω1)

B1 (ω2) S11 (ω2) · · · Sn

1 (ω2)...

......

...B1 (ωm) S1

1 (ωm) · · · Sn1 (ωm)

⎤⎥⎥⎥⎦

has rank equal to the cardinality of Ω, that is m.


The Second Fundamental Theorem of Asset Pricing guarantees that if aone-period model as above is Free of Arbitrage, then it is complete if and onlyif there exists a unique equivalent martingale measure (i.e. card(M) = 1).

If the market model is free of arbitrage, it is well known that:

• a derivative (with payoff Φ) is attainable if and only if EQ

[ΦB1

]is constant

for any Q ∈ M;• if a derivative (with payoff Φ) is attainable, then its initial price coincideswith the cost of its replicating strategy:

EQ

[Φ

B1

], with Q ∈ M; (4.3)

• if a derivative (with payoff Φ) is not attainable, then its initial price belongsto the no-arbitrage interval (V− (Φ) , V+ (Φ)), where

V− (Φ) = infQ∈M

EQ

[Φ

B1

]; V+ (Φ) = sup

Q∈MEQ

[Φ

B1

]. (4.4)

If the infimum and supremum V−(Φ) and V+(Φ) are attained, then theno-arbitrage interval is closed.

Consider now a multi-period model where transactions may occur at times

0, 1, 2, ..., T

and consisting in a non-risky asset (called bond), whose value will be denotedby B, and in n risky assets (called stocks), whose values will be denoted byS1, S2, ..., Sn. Assume (as in the one-period model) that the sample space isfinite

Ω = {ω1, ω2, ..., ωm}and that any elementary event is possible with respect to a probability Pgiven a priori, that is P (ωk) > 0 for any k = 1, ..., m. The information avail-able in time is represented by a filtration (Ft)t=0,1,...,T with F0 = {∅, Ω} andFT = P (Ω), where P (Ω) denotes the power set of Ω.The bond price at t ∈ {1, 2, ..., T} is assumed to be equal to

Bt = B0(1 + r)t, (4.5)

where B0 = 1 and r stands as usual for the risk-free interest rate. The stocksprices

(Si

t

)t=0,1,...,T

(for i = 1, 2, ..., n) are assumed to be stochastic processes,adapted to (Ft)t=0,1,...,T and such that Si

0 > 0 for any i = 1, 2, ..., n.In the multi-period model, a trading strategy is a vector α =(

α0t , α

1t , ..., α

nt

)t=1,...,T

of processes, where α0t and αi

t (i = 1, 2, ..., n) repre-sent, respectively, the number of bonds and stock Si in the portfolio fromtime (t − 1) to t.


The value of a trading strategy α is then given by

V0 (α) = α01B0 +

n∑i=1

αi1S

i0

Vt (α) = α0tBt +

n∑i=1

αitS

it , for t = 1, ..., T.

A strategy α is said to be self-financing if Vt (α) = α0t+1Bt +

∑ni=1 αi

t+1Sit

for any t = 1, ..., T . In other words, a strategy is self-financing when at anytime t what one obtains by selling the strategy αt is enough (and not toomuch) to finance the purchase of the strategy αt+1; hence nothing has to beadded and nothing can be withdrawn.Furthermore, a strategy α is called an arbitrage opportunity if it is self-

financing and if it satisfies V0 (α) = 0 and VT (α) ≥ 0, with P (VT (α) > 0) > 0.Similarly to the one-period case, the First Fundamental Theorem of As-

set Pricing guarantees that the multi-period model described above is freeof arbitrage opportunities if and only if there exists (at least) one equivalentmartingale measure, that is, a probability measure Q such that Q (ωk) > 0for any k = 1, 2, ...,m (equivalent) and such that

EQ

[Si

t

Bt

∣∣∣∣Fs

]=

Sis

Bs, (4.6)

for any s ≤ t and i = 1, 2, ..., n (a martingale measure). The set of all equiva-lent martingale measures will be denoted by M.

It is also well known that if the multi-period model is free of arbitrage,then all one-period sub-markets are free of arbitrage.

Furthermore, a derivative with payoff Φ is said to be attainable if there ex-ists a self-financing trading strategy α such that VT (α) = Φ. If all the deriva-tives available on the market are attainable (that is, they can be replicated),then the market is said to be complete; otherwise, it is called incomplete.The Second Fundamental Theorem of Asset Pricing guarantees that if a

multi-period model as above is free of arbitrage, then it is complete if and onlyif there exists a unique equivalent martingale measure (i.e. card(M) = 1).Moreover, if the multi-period model is free of arbitrage, then it is complete

if and only if all its one-period sub-markets are complete.

As in the one-period case, if the market model is free of arbitrage it is wellknown that:

• a derivative (with payoff Φ) is attainable if and only if EQ

[Φ

BT

]is constant

for any Q ∈ M;


• if a derivative (with payoff Φ) is attainable, then its price at t ∈ {0, 1, ..., T}coincides with the cost of its replicating strategy, which equals

Vt (Φ) = Bt · EQ

[Φ

BT

∣∣∣∣Ft

], with Q ∈ M. (4.7)


Exercise 4.2. Consider a one-period (annual) market model consisting in anon-risky asset (paying a risk-free rate of 5% per year) and in two stocks withprices S1 and S2 :

S10 = 10; S1

1 (ω) =

⎧⎨⎩

12; if ω = ω1

8; if ω = ω2

6; if ω = ω3

S20 = 5; S2

1 (ω) =

⎧⎨⎩

10; if ω = ω1

4; if ω = ω2

5; if ω = ω3

,

where P (ω1) = P (ω2) = P (ω3) = 13 .

Establish if there exist arbitrage opportunities on the market. If yes, givean example.

SolutionIn the present example a one-period market model is considered, where

time is measured in years. The bond price evolves as follows:

B0 = 1B1 = 1 + r

with r = 0.05 per year, while the prices of the stocks can be summarized as(S1,u

1 ; S2,u1

)= (12; 10) (on ω1)

↗(S1

0 ; S20

)= (10; 5) −→

(S1,m

1 ; S2,m1

)= (8; 4) (on ω2)

↘ (S1,d

1 ; S2,d1

)= (6; 5) (on ω3)

− −− −− −−− −− −− −− −− −− −−t = 0 T = 1 year

By the First Fundamental Theorem of Asset Pricing, a market is free ofarbitrage if and only if there exists an equivalent martingale measure. So,we only need to establish if there exists a probability measure Q satisfying


qk = Q (ωk) > 0 for any k = 1, 2, 3 and⎧⎪⎪⎨⎪⎪⎩

S10 = EQ

[S1

1

B1

]

S20 = EQ

[S2

1

B1

] .

The previous system can be rewritten as⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

121 + r

q1 +8

1 + rq2 +

61 + r

q3 = 10

101 + r

q1 +4

1 + rq2 +

51 + r

q3 = 5

q1 + q2 + q3 = 1⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

12q1 + 8q2 + 6q3 =212

10q1 + 4q2 + 5q3 =214

q1 + q2 + q3 = 1

,

so its unique solution is ⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

q1 =516

q2 =2116

> 1

q3 = −58

< 0

Accordingly, there do not exist any equivalent martingale measures. Hence,the market above is not free of arbitrage.An example of arbitrage opportunity is given by the strategy α =(

α0, α1, α2)

= (0;−1; 2), consisting in zero positions in the bond, a shortposition in stock S1 and two long positions in stock S2. For such a strategy,indeed,

V0 (α) = 0 ·B0 − 1 · S10 + 2 · S2

0 = −10 + 10 = 0,

whileV1 (α) = −1 · S1

1 + 2 · S21

=

⎧⎨⎩

8; if ω = ω1

0; if ω = ω2

4; if ω = ω3

.

Consequently, V0 (α) = 0 and V1 (α) ≥ 0 with P (V1 (α) > 0) = 23.

Exercise 4.3. Consider a one-period (annual) model formed by a bond (pay-ing a risk-free rate of 5% per year) and by two stocks with prices S1 and S2


evolving as follows:

S10 = 10; S1

1 (ω) =

⎧⎨⎩

12; if ω = ω1

10; if ω = ω2

6; if ω = ω3

S20 = 10; S2

1 (ω) ={

15; if ω = ω1

8; if ω ∈ {ω2, ω3} ,

with P (ω1) , P (ω2) , P (ω3) > 0 and P (ω1) + P (ω2) + P (ω3) = 1.

1. Establish if the market is free of arbitrage and complete.2. Consider two derivatives (A and B) with maturity T of one year and with

payoffs ΦA =(

S1T +S2

T

2 − 8)+

and ΦB =(S1

T − S2T

)+, respectively. Find

their initial prices and their replicating strategies.3. Discuss whether the market formed only by the bond and by stock S1

would remain free of arbitrage and complete.4. In the market formed by the bond and by the stock S1, find the no-arbitrageintervals for the prices of derivatives A and B.

Solution

1. The market model we are dealing with is a one-period model with initialtime 0 and final time 1 year. The bond price evolves as

B0 = 1B1 = 1 + r

with r = 0.05 per year, while the prices of the stocks can be summarizedas follows: (

S1,u1 ; S2,u

1

)= (12; 15) (on ω1)

↗(S1

0 ; S20

)= (10; 10) −→

(S1,m

1 ; S2,m1

)= (10; 8) (on ω2)

↘ (S1,d

1 ; S2,d1

)= ( 6; 8) (on ω3)

− −− −− −− −−− −− −− −− −− −− −− −−t = 0 T = 1 year

In order to establish if the market above is free of arbitrage (or not), we needto verify if there exists (at least) one equivalent martingale measure, thatis a probability measure Q satisfying qk = Q (ωk) > 0 for any k = 1, 2, 3


and ⎧⎪⎪⎨⎪⎪⎩

S10 = EQ

[S1

1

B1

]

S20 = EQ

[S2

1

B1

] .

The previous system is equivalent to

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

121 + r

q1 +10

1 + rq2 +

61 + r

q3 = 10

151 + r

q1 +8

1 + rq2 +

81 + r

q3 = 10

q1 + q2 + q3 = 1⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

12q1 + 10q2 + 6q3 =212

15q1 + 8q2 + 8q3 =212

q1 + q2 + q3 = 1

.

The systems above have a unique solution (q1, q2, q3) satisfying q1, q2, q3 ∈(0, 1), namely (q1, q2, q3) =

(514 , 33

56 , 356

). Hence, the market is free of ar-

bitrage and, by the Second Fundamental Theorem of Asset Pricing, alsocomplete. The unique equivalent martingalemeasureQ is given byQ (ω1) =514

, Q (ω2) = 3356and Q (ω3) = 3

56.

An alternative way to deduce the completeness of the market consists inobserving

rk

⎡⎣B1(ω1) S1

1 (ω1) S21 (ω1)

B1(ω2) S11 (ω2) S2

1 (ω2)B1(ω3) S1

1 (ω3) S21 (ω3)

⎤⎦ = rk

⎡⎣1.05 12 15

1.05 10 81.05 6 8

⎤⎦ = 3,

i.e. the rank is equal to the cardinality of the sample space.

2. Since the market formed by the bond and stocks S1 and S2 is free ofarbitrage and complete, derivatives A and B (with maturity of one yearand with payoffs ΦA and ΦB) are attainable. Their initial prices are given,respectively, by

V0

(ΦA)

=1

1 + rEQ

[ΦA]; V0

(ΦB)

=1

1 + rEQ

[ΦB]. (4.8)

It is immediate to check that

ΦA (ω) =

⎧⎨⎩

5.5; if ω = ω1

1; if ω = ω2

0; if ω = ω3

; ΦB (ω) =

⎧⎨⎩

0; if ω = ω1

2; if ω = ω2

0; if ω = ω3

,


hence, by (4.8), we deduce that

V0

(ΦA)

=1

1 + rEQ

[ΦA]

=1

1.05[5.5 · Q (ω1) + 1 · Q (ω2) + 0] = 2.432

V0

(ΦB)

=1

1 + rEQ

[ΦB]

=1

1.05[0 + 2 ·Q (ω2) + 0] = 1.1224.

There remains to find the replicating strategies of the two derivatives above.A replicating trading strategy for the derivative A with payoff ΦA is avector αA =

(α0

A, α1A, α2

A

)satisfying V1 (α) = ΦA; equivalently, a vector

solving the following system of equations

⎧⎪⎪⎨⎪⎪⎩

α0A · B1 + α1

A · S1,u1 + α2

A · S2,u1 = 5.5

α0A · B1 + α1

A · S1,m1 + α2

A · S2,m1 = 1

α0A · B1 + α1

A · S1,d1 + α2

A · S2,d1 = 0⎧⎪⎨

⎪⎩α0

A · 1.05 + α1A · 12 + α2

A · 15 = 5.5

α0A · 1.05 + α1

A · 10 + α2A · 8 = 1

α0A · 1.05 + α1

A · 6 + α2A · 8 = 0

The unique solution of the systems is αA = (−5.7823, 0.25, 0.5714), givingthe required replicating strategy of A. Its cost V0 (αA) = −5.7823 + 0.25 ·10 + 0.5714 · 10 = 2.432 coincides with the initial price of the derivative A(as we should expect, and indeed happens).Similarly, one obtains that the replicating strategy of option B is αB =(α0

B, α1B, α2

B

)= (0.40816, 0.5,−0.42857). Consequently, its cost V0 (αB) =

0.40816 + 5 − 4.2857 = 1.1225 coincides with the initial price of derivativeB (again, as it should be).

3. Proceeding as above, it is easy to check that the one-period model formedonly by the bond and by stock S1 is free of arbitrage but incomplete.There exist, indeed, infinitely many equivalent martingale measures, i.e.probability measures Q satisfying qk = Q (ωk) > 0 (for any k = 1, 2, 3)and S1

0 = EQ

[S1

1B1

]. In the present case, the set of equivalent martingale

measures is given by

M ={

Q

∣∣∣∣Q (ω1) = q ∈(

14;34

); Q (ω2) =

98− 3

2q; Q (ω3) =

12q − 1

8

}.

4. According to the previous item, the no-arbitrage interval for the price ofthe option with payoff ΦA is given by

(1

1 + rinf

Q∈MEQ

[ΦA];

11 + r

supQ∈M

EQ

[ΦA])

.


Since EQ

[ΦA]

= 5.5 ·Q (ω1) +1 ·Q (ω2) +0 = 5.5q + 98 − 1.5q = 4q + 9

8 forQ ∈ M, we get

11 + r

infQ∈M

EQ

[ΦA]

=1

1.05inf

14 <q< 3

4

(4q +

98

)=

11.05

(1 +

98

)= 2.0238

11 + r

supQ∈M

EQ

[ΦA]

=1

1.05sup

14 <q< 3

4

(4q +

98

)=

11.05

(3 +

98

)= 3.9286.

It follows that the no-arbitrage interval for the price of option A is(2.0238; 3.9286).Similarly, one can obtain that the no-arbitrage price of option B belongsto the interval (0; 1.4286).

Exercise 4.4. Consider a bond (paying a risk-free rate of 3% per year) andtwo stocks with prices S1 and S2 evolving as follows:

↗ S1,C2 = 14; S2,C

2 = 8

↗ S1,D2 = 12; S2,D

2 = 4

S1,A1 = 12; S2,A

1 = 6 → S1,E2 = 9; S2,E

2 = 2

↗ ↘ S1,F2 = 7; S2,F

2 = 9S1

0 = 10; S20 = 5

↘ ↗ S1,G2 = 8; S2,G

2 = 4S1,B

1 = 6; S2,B1 = 3 ↘

S1,H2 = 4; S2,H

2 = 5−− −− −− −− −− − −− −− −− −− −−− − −− −− −− −− −−−

0 1 year 2 years

Establish if the two-period market model formed by the bond and twostocks above is free of arbitrage or not.

SolutionDenote by ω1 the state of the world corresponding to node C where at

T = 2 the price of the first stock has reached 14 and the price of the secondstock has reached 8, by ω2 the state of the world corresponding to node D,and so on.By the two-period model described above, we deduce that

Ω = {ω1, ω2, ..., ω6}

and the filtration corresponding to the information available in time is thefollowing one:

F0 = {∅, Ω}F1 = {∅, {ω1, ω2, ω3, ω4}, {ω5, ω6}, Ω}F2 = P (Ω) .

We shall write U = {ω1, ω2, ω3, ω4} and D = {ω5, ω6}.


Recall that if a multi-period model is free of arbitrage then all its one-periodsub-markets are free of arbitrage.Let us consider the first one-period sub-market, written in bold:

↗ S1,C2 = 14; S2,C

2 = 8

↗ S1,D2 = 12; S2,D

2 = 4

S1,A1 = 12; S2,A

1 = 6 → S1,E2 = 9; S2,E

2 = 2

↗ ↘ S1,F2 = 7; S2,F

2 = 9S1

0= 10; S20= 5

↘ ↗ S1,G2 = 8; S2,G

2 = 4S1,B

1 = 6; S2,B1 = 3 ↘

S1,H2 = 4; S2,H

2 = 5−− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −−

0 1 year 2 years

The sub-market above is free of arbitrage if and only if there exists atleast an equivalent martingale measure, i.e. a probability measure Q suchthat Q (U) , Q (D) ∈ (0, 1) and

⎧⎨⎩

EQ

[S1

1B1

]= S1

0

EQ

[S2

1B1

]= S2

0

.

Calling q = Q (U) and observing that Q (D) = 1−Q (U), the previous condi-tion becomes ⎧⎪⎪⎪⎨

⎪⎪⎪⎩S1,A

1

1 + r· Q (U) +

S1,B1

1 + r· Q (D) = S1

0

S2,A1

1 + r· Q (U) +

S2,B1

1 + r· Q (D) = S2

0

⎧⎪⎪⎨⎪⎪⎩

121 + r

· q +6

1 + r· (1 − q) = 10

61 + r

· q +3

1 + r· (1 − q) = 5

.

Trivially, the systems above reduce to 12q + 6 (1 − q) = 10 (1 + r). Hence

q =23

+53r ∼= 0.72

andQ (U) = q = 0.72; Q (D) = 1 − q = 0.28.

It follows that the first sub-market is free of arbitrage and, because of theuniqueness of the equivalent martingale measure, complete.


We consider now the one-period sub-market in bold:

↗ S1,C2 = 14; S2,C

2 = 8

↗ S1,D2 = 12; S2,D

2 = 4

S1,A1 = 12; S2,A

1 = 6 → S1,E2 = 9; S2,E

2 = 2

↗ ↘ S1,F2 = 7; S2,F

2 = 9S1

0 = 10; S20 = 5

↘ ↗ S1,G2 = 8; S2,G

2 = 4S1,B

1 = 6; S2,B1 = 3

↘S1,H

2 = 4; S2,H2 = 5

−− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −−0 1 year 2 years

As previously, the sub-market is free of arbitrage if and only if there existsat least one probability measure Q ( ·|U) satisfying qk = Q (ωk|U) > 0 (fork = 1, 2, 3, 4) and ⎧⎪⎪⎪⎨

⎪⎪⎪⎩EQ

[S1

2

B2

∣∣∣∣U]

=S1,A

1

B1

EQ

[S2

2

B2

∣∣∣∣U]

=S2,A

1

B1

.

The condition above becomes⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

14(1 + r)2

· q1 +12

(1 + r)2· q2 +

9(1 + r)2

· q3 +7

(1 + r)2· q4 =

121 + r

8(1 + r)2

· q1 +4

(1 + r)2· q2 +

2(1 + r)2

· q3 +9

(1 + r)2· q4 =

61 + r

q1 + q2 + q3 + q4 = 1⎧⎪⎨⎪⎩

7q1 + 5q2 + 2q3 = 5 + 12r

−q1 − 5q2 − 7q3 = −3 + 6r

q4 = 1 − q1 − q2 − q3

Consequently, ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

q1 =56q3 +

13

+ 3r

q2 =8 − 27r

15− 47

30q3

q3 ∈ (0, 1)

q4 =215

− 65r − 4

15q3

.


By imposing qk > 0, we obtain⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

q1 =56q3 +

13

+ 3r

q2 =8 − 27r

15− 47

30q3

0 < q3 < min{

2 − 18r

4;4 − 18r

5;16 − 54r

47

}= 0.306

q4 =215

− 65r − 4

15q3

.

It follows that also the second sub-market is free of arbitrage (but incomplete).Furthermore,

Q (ωk) = Q (ωk|U) · Q (U) , for k = 1, 2, 3, 4.

Finally, we consider the last one-period sub-market:

↗ S1,C2 = 14; S2,C

2 = 8

↗ S1,D2 = 12; S2,D

2 = 4

S1,A1 = 12; S2,A

1 = 6 → S1,E2 = 9; S2,E

2 = 2

↗ ↘ S1,F2 = 7; S2,F

2 = 9S1

0 = 10; S20 = 5

↘ ↗ S1,G2 = 8; S2,G

2 = 4S1,B

1 = 6; S2,B1 = 3 ↘

S1,H2 = 4; S2,H

2 = 5−− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −−

0 1 year 2 years

As previously, such a sub-market is free of arbitrage if and only if thereexists at least one probability measure Q ( ·|D) satisfying qk = Q (ωk|D) > 0(for k = 5, 6) and ⎧⎪⎪⎪⎨

⎪⎪⎪⎩EQ

[S1

2

B2

∣∣∣∣D]

=S1,B

1

B1

EQ

[S2

2

B2

∣∣∣∣D]

=S2,B

1

B1

.

The condition above can be rewritten as⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

8(1 + r)2

· q5 +4

(1 + r)2· q6 =

61 + r

4(1 + r)2

· q5 +5

(1 + r)2· q6 =

31 + r

q5 + q6 = 1

74 4 Absence of Arbitrage and Completeness of Market Models⎧⎨⎩

8q5 + 4q6 = 6 + 6r4q5 + 5q6 = 3 + 3rq5 + q6 = 1

.

It is easy to check that the system above is incompatible, hence there doesnot exist any equivalent martingale measure for the sub-market taken intoaccount. This implies that such sub-market is not free of arbitrage and, con-sequently, the same holds for the overall two-period market model.

Exercise 4.5. Consider a two-period model (with annual periods) formed bya non-risky asset (B, paying a risk-free rate of 2% per year) and by a stock(S) whose price evolves as follows:

↗ SC2 = 18

SA1 = 15 → SD

2 = 15↗ ↘ SE

2 = 10S0 = 10

↘ ↗ SF2 = 10

SB1 = 5

↘SG

2 = 4−− −− −− −− −− −− − − −− −− −− −−

0 1 year T = 2 years

Denote by ω1 (respectively, ω2, ..., ω5) the state of the world correspondingto the node C (respectively, D, E, F, G). The probability measure P given apriori satisfies P (ωi) = 1

5 for any i = 1, 2, ..., 5.

1. Check whether the two-period model above is free of arbitrage and com-plete.

2. Consider a European Call option with maturity of two years and with strikeof 10, written on stock S. Establish if such option is attainable and computeits price of no arbitrage.

3. Find the no-arbitrage interval for the price of a derivative with payoff

ΦC = max{(ST − 10)+ ; CT},

where CT (ω1) = CT (ω3) = 4 and CT (ω2) = CT (ω4) = CT (ω5) = 0.4. Compute the price of the derivative with payoff ΦC by means of the mar-tingale measure Q∗ ∈ M1 = {Q ∈ M| 0.1 ≤ Q (ω1) ≤ 0.2} minimizing therelative entropy between all martingale measures inM1 and the probabilitymeasure P .

SolutionBy the two-period model specified above, we deduce that

Ω = {ω1, ω2, ..., ω5}


and the filtration corresponding to the information available is

F0 = {∅, Ω}F1 = {∅, {ω1, ω2, ω3}, {ω4, ω5}, Ω}F2 = P (Ω) .

Denote by U = {ω1, ω2, ω3} and by D = {ω4, ω5}.1. Recall that if a multi-period model is free of arbitrage then all its one-period sub-markets are free of arbitrage. Furthermore, it is complete if andonly if all its one-period sub-markets are complete.Consider now the first one-period sub-market model in bold:

↗ SC2 = 18

SA1 = 15 → SD

2 = 15↗ ↘ SE

2 = 10S0= 10

↘ ↗ SF2 = 10

SB1 = 5

↘SG

2 = 4− −− −− −− −− − −− −− −− −− −−


Such a sub-market is free of arbitrage if and only if there exists at leastone probability measure Q satisfying Q (U) , Q (D) ∈ (0, 1) and

EQ

[S1

B1

]= S0.

Denoting by q = Q (U) and observing that Q (D) = 1−Q (U), the conditionabove becomes

151 + r

q +5

1 + r(1 − q) = 10.

Hence

Q (U) = q =12

+ r = 0.52

Q (D) = 1 − q =12− r = 0.48.

By the arguments above, we conclude that the first one-period sub-marketis free of arbitrage and complete (because of the existence and uniquenessof the martingale measure).


We focus now on the one-period sub-market in bold:

↗ SC2 = 18

SA1 = 15 → SD

2 = 15↗ ↘ SE

2 = 10S0 = 10

↘ ↗ SF2 = 10

SB1 = 5

↘SG

2 = 4−− −− − −− −− −− − − −− −− −− −−


As previously, such a sub-market is free of arbitrage if and only if there ex-ists at least one probability measure Q ( ·|U) (conditional probability, giventhe event U) satisfying qk = Q (ωk|U) > 0 (for k = 1, 2, 3), q1 +q2 +q3 = 1and

EQ

[S2

B2

∣∣∣∣U]

=SA

1

B1.

The condition above becomes⎧⎨⎩

18(1 + r)2

· q1 +15

(1 + r)2· q2 +

10(1 + r)2

· q3 =15

1 + r

q1 + q2 + q3 = 1{8q1 + 5q2 = 5 + 15rq3 = 1 − q1 − q2

Hence ⎧⎪⎪⎨⎪⎪⎩

q1 ∈ (0, 1)

q2 = 1 + 3r − 85q1

q3 =35q1 − 3r

.

Nevertheless, we have to restrict our attention only to those solutions thatcorrespond to equivalent martingale measures. We need, therefore, to im-pose qk > 0 for any k = 1, 2, 3, obtaining that⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

0.1 = 5r < q1 <58

(1 + 3r) ∼= 0.66

q2 = 1.06− 85q1

q3 =35q1 − 0.06

.

By the arguments above, the sub-market taken into account is free of ar-bitrage but incomplete.


Furthermore, Q (ωk) = Q (ωk|U) · Q (U) for k = 1, 2, 3, so

Q (ω1) =(

12

+ r

)q1

Q (ω2) =(

12

+ r

)(1.06− 8

5q1

)

Q (ω3) =(

12

+ r

)(35q1 − 0.06

).

We finally focus on the last one-period sub-market in bold:

↗ SC2 = 18

SA1 = 15 → SD

2 = 15↗ ↘ SE

2 = 10S0 = 10

↘ ↗ SF2 = 10

SB1 = 5 ↘

SG2 = 4

−− −− − −− −− −− − − −− −− −− −−0 1 year T = 2 years

As previously, such a sub-market is free of arbitrage if and only if there ex-ists at least one probability measure Q ( ·|D) (conditional probability, giventhe event D) satisfying qk = Q (ωk|D) > 0 (for k = 4, 5), q4 + q5 = 1 and

EQ

[S2

B2

∣∣∣∣D]

=SB

1

B1.

The previous condition becomes⎧⎨⎩

10(1 + r)2

q4 +4

(1 + r)2q5 =

51 + r

q4 + q5 = 1,

that admits as unique solution⎧⎪⎨⎪⎩

q4 =16

(1 + 5r) ∈ (0, 1)

q5 =56

(1 − r) ∈ (0, 1).

Accordingly, such a sub-market is free of arbitrage and complete. Further-more, Q (ωk) = Q (ωk|D) · Q (D) for k = 4, 5, so

Q (ω4) =16

(12− r

)(1 + 5r)

Q (ω5) =56

(12− r

)(1 − r) .


To conclude, the two-period model taken into account is free of arbitrageand incomplete. The set of all equivalent martingale measures for the two-period model is

M =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Q

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

Q (ω1) =(

12

+ r

)q1

Q (ω2) =(

12

+ r

)(1 + 3r − 8

5q1

)

Q (ω3) =(

12

+ r

)(35q1 − 3r

)

Q (ω4) =16

(12− r

)(1 + 5r)

Q (ω5) =56

(12− r

)(1 − r)

; for 5r < q1 <58

(1 + 3r)

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

.

(4.9)2. Due to the incompleteness of the above two-period model, there is no guar-antee that the European Call option with maturity of two years, with strikeof 10 and written on S is attainable. The payoff of the Call option is given by

Φ (ω) = (ST (ω) − 10)+ =

⎧⎨⎩

8; ω = ω1

5; ω = ω2

0; ω ∈ {ω3, ω4, ω5}.

Looking for a self-financing replicating strategy for the option with payoffΦ is the same as looking for a self-financing replicating strategy (αt)t=1,2

with αt =(αB

t , αSt

)(where αB, resp. αS , represents the number of bonds,

resp. of stocks, to be held in portfolio).At maturity (T = 2), such a strategy (if it exists) has to satisfy V2 (α) = Φ,i.e. ⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

αB,u2 · B2 + αS,u

2 · 18 = 8αB,u

2 · B2 + αS,u2 · 15 = 5

αB,u2 · B2 + αS,u

2 · 10 = 0αB,d

2 · B2 + αS,d2 · 10 = 0

αB,d2 · B2 + αS,d

2 · 4 = 0

where α·,u2 and α·,d

2 denote, respectively, the number of shares to be heldin portfolio from time 1 to time 2 in the node “up” or “down” at t = 1. Theunique α2 satisfying the system above is given by⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

αB,u2 = − 10

(1 + r)2

αS,u2 = 1

αB,d2 = 0

αS,d2 = 0

.


It follows that V u1 (α) = αB,u

2 · B1 + αS,u2 · SA

1 = 5+15r1+r

and V d1 (α) =

αB,d2 · B1 + αS,d

2 · SB1 = 0.

At time t = 1, a replicating self-financing strategy has to satisfy

{αB

1 ·B1 + αS1 · SA

1 = V u1 (α)

αB1 ·B1 + αS

1 · SB1 = V d

1 (α),

hence

⎧⎨⎩αB

1 (1 + r) + αS1 · 15 =

5 + 15r

1 + rαB

1 (1 + r) + αS1 · 5 = 0

⎧⎪⎪⎨⎪⎪⎩

αB1 = −5 (1 + 3r)

2 (1 + r)2

αS1 =

1 + 3r

2 (1 + r)

.

The European Call option of concern is then attainable, so the price ofno arbitrage has to coincide with the cost of the self-financing replicatingstrategy, that is

V0 = αB1 ·B0 + αS

1 · S0 = −5 (1 + 3r)

2 (1 + r)2+

1 + 3r

2 (1 + r)· 10 = 2.6490. (4.10)

An alternative way to compute the option price is to find the no-arbitrageinterval and observe that it reduces to a point, namely the value V0 in(4.10).

3. The payoff of the claim C is equal to

ΦC (ω) = max{(ST (ω) − 10)+ ; CT (ω)} =

⎧⎪⎪⎨⎪⎪⎩

8; ω = ω1

5; ω = ω2

4; ω = ω3

0; ω ∈ {ω4, ω5}.

It is easy to check that such a claim is not attainable in the market we aredealing with. We will then look for the no-arbitrage interval for its price.Let Q be an equivalent martingale measure, i.e. Q ∈ M. From (4.9) itfollows that

EQ

[ΦC

B2

]=

1(1 + r)2

[8 · Q (ω1) + 5 · Q (ω2) + 4 · Q (ω3)]


=1

(1 + r)2

[8(

12

+ r

)q1 + 5

(12

+ r

)(1 + 3r − 8

5q1

)

+4(

12

+ r

)(35q1 − 3r

)]

=1 + 2r

2 (1 + r)2

[5 + 3r +

125

q1

].

Hence

infQ∈M

EQ

[ΦC

B2

]= inf

5r<q1< 58 (1+3r)

{1 + 2r

2 (1 + r)2

[5 + 3r +

125

q1

]}

=5 (1 + 2r) (1 + 3r)

2 (1 + r)2= 2.65

and

supQ∈M

EQ

[ΦC

B2

]= sup

5r<q1< 58 (1+3r)

{1 + 2r

2 (1 + r)2

[5 + 3r +

125

q1

]}

=(13 + 15r) (1 + 2r)

4 (1 + r)2= 3.32.

The price of option C belongs then to the no-arbitrage interval (2.65; 3.32).4. As verified in the previous items, the market considered is free of arbitrageand incomplete (there exist indeed infinitely many equivalent martingalemeasures) and the derivative with payoff ΦC is not attainable. A priori,therefore, we are only able to find the no-arbitrage interval for the price ofsuch a claim.A criterion to “choose” one among the infinitely many equivalent martin-gale measures (using which we can evaluate the derivative C) is the Min-imal Relative Entropy Criterion. More precisely, such a criterion suggeststo select the equivalent martingale measure Q that minimizes the relativeentropy of Q with respect to P :

H (Q, P ) � EP

[dQ

dPln(

dQ

dP

)].

In the present case, H (Q, P ) (with Q ∈ M1) becomes

H (Q, P ) = EP

[dQ

dPln(

dQ

dP

)]

=5∑

i=1

[Q (ωi)P (ωi)

ln(

Q (ωi)P (ωi)

)]P (ωi)


=5∑

i=1

Q (ωi) [ln (Q (ωi)) − ln (P (ωi))]

=5∑

i=1

Q (ωi) ln (Q (ωi)) + ln 5

=(

12

+ r

)·[q1 ln (q1) +

(1 + 3r − 8

5q1

)ln(

1 + 3r − 85q1

)

+(

35q1 − 3r

)ln(

35q1 − 3r

)]

+(

12− r

)[16

(1 + 5r) ln(

16

(1 + 5r))

+56

(1 − r) ln(

56

(1 − r))]

+(

12

+ r

)ln(

12

+ r

)+(

12− r

)ln((

12− r

))+ ln5.

We deduce that

arg minQ∈M1

H (Q, P )

= arg minQ∈M1

⎡⎢⎢⎣q1 ln (q1) +

(1 + 3r − 8

5q1

)ln(

1 + 3r − 85q1

)

+(

35q1 − 3r

)ln(

35q1 − 3r

)⎤⎥⎥⎦. (4.11)

Since Q ∈ M1 if and only if Q ∈ M and 0.1 ≤ Q (ω1) ≤ 0.2 (or, equiva-lently, 0.1 ≤ ( 1

2+ r)q1 ≤ 0.2), we deduce that Q ∈ M1 if and only if

⎧⎪⎪⎨⎪⎪⎩

0.1 ≤(

12

+ r

)q1 ≤ 0.2

5r < q1 <58

(1 + 3r)

⎧⎪⎪⎨⎪⎪⎩

15 + 10r

≤ q1 ≤ 25 + 10r

5r < q1 <58

(1 + 3r)

{0.199 ≤ q1 ≤ 0.3980.1 < q1 < 0.6625

,

hence q1 ∈ [0.199; 0.398]. It follows that (4.11) can be rewritten as

arg minQ∈M1

H (Q, P ) = arg minq1∈[0.199;0.398]

g (q1) ,

where g (q1) = q1 ln (q1) +(1 + 3r − 8

5q1

)ln(1 + 3r − 8

5q1

)+(

35q1 − 3r

) ·


ln(

35q1 − 3r

). Since g′ (q1) = ln (q1)−8

5 ln(1 + 3r − 8

5q1

)+3

5 ln(

35q1 − 3r

) ≤0 on the interval [0.199; 0.398], minQ∈M1 H (Q, P ) is attained at q∗1 =

25+10r = 0.398. Accordingly, the equivalent martingale measure chosen bythe criterion is given by

Q∗ (ω1) =(

12

+ r

)q∗1 = 0.2

Q∗ (ω2) =(

12

+ r

)(1 + 3r − 8

5q∗1

)= 0.22

Q∗ (ω3) =(

12

+ r

)(35q∗1 − 3r

)= 0.09

Q∗ (ω4) =16

(12− r

)(1 + 5r) = 0.09

Q∗ (ω5) =56

(12− r

)(1 − r) = 0.4.

It follows that the price of the derivative C (evaluated by means of Q∗) isequal to

EQ∗

[ΦC

B2

]=

1(1 + r)2

[8 · Q∗ (ω1) + 5 · Q∗ (ω2) + 4 · Q∗ (ω3)]

=1

(1.02)2[8 · 0.2 + 5 · 0.22 + 4 · 0.09] = 2.94

which, as expected, belongs to the no-arbitrage interval of the price of Cfound previously.


Exercise 4.6. Consider a one-period (annual) model formed by a non-riskyasset (B, paying a risk-free interest rate of 4% per year) and by a stock (S)whose price is

S0 = 5; S1 (ω) =

⎧⎨⎩

8; if ω = ω1

5; if ω = ω2

3; if ω = ω3

,

with P (ω1) , P (ω2) , P (ω3) > 0.

1. Establish if the market above is free of arbitrage and complete.2. Find the no-arbitrage price (or the no-arbitrage interval for the price) of aEuropean Put option on the underlying S, with strike of 5 euros and withmaturity of one year.

3. Suppose now that an additional asset is introduced on the market. Suchan asset is a derivative (denoted by C) with initial price of 1 euro and with


payoff

ΦC (ω) =

⎧⎨⎩

0; if ω = ω1

2; if ω = ω2

1; if ω = ω3

.

Discuss if the new market (formed by B, S and the derivative C) is free ofarbitrage and complete.

Exercise 4.7. Consider a two-period model (with annual periods) formed bya non-risky asset (B, paying a risk-free interest rate of 4% per year) and by astock (S) whose price evolves as follows

↗ SD2 = SA

1 · uSA

1 = 8 → SE2 = SA

1 · d↗

S0 = 5 −→ SB1 = 5 −→ SF

2 = SB1 · u

↘ ↘ SG2 = SB

1 · dSC

1 = 3 −→ SH2 = SC

1 · u↘ SL

2 = SC1 · d

−− −− − −− −−− − −− −− − −−0 1 year T = 2 years

with u = 1.5 and d = 0.5.

1. Establish if the two-period market above is free of arbitrage and complete.2. Find the no-arbitrage price (or the no-arbitrage interval for the price) of aEuropean Call option on the underlying S, with strike of 7 euros and withmaturity of two years.

Exercise 4.8. Consider a one-period (annual) model formed by a non-riskyasset (B, paying a risk-free interest rate of 4% per year) and by two stocks(S1 and S2) whose prices evolve as follows

↗ S1,A1 = 12; S2,A

1 = 15

↗ S1,B1 = 10; S2,B

1 = 8S1

0 = 10; S20 = 10

↘ S1,C1 = 8; S2,C

1 = 4

↘ S1,D1 = 4; S2,D

1 = 8− −− −− −− −−− − −− −− −− −− −−−

0 1 year

Establish if the one-period market above is free of arbitrage and complete.

5

Itô’s Formula and Stochastic DifferentialEquations


Given a stochastic process (Vt)t≥0 having trajectories with bounded variationand a sufficiently regular function f , it is possible to define the integral ofZt = f(Vt) with respect to dVt as follows

∫ t

0

ZsdVs =∫ t

0

f(Vs)dVs, (5.1)

i.e. in the sense of a Riemann-Stieltjes integral.On the other hand, if the trajectories of the process (Vt)t≥0 are not of

bounded variation and if the function f is not regular enough, the integralabove may not make sense as a Riemann-Stieltjes integral. This is the case ofthe Brownian motion, for which (in general) one cannot define the integral

∫f(Ws)dWs

as a Riemann-Stieltjes integral.A way out is to introduce Itô’s stochastic integral

∫ t

0 HsdWs for processes(Ht)t≥0 that satisfy a “suitable measurability assumption” (for instance, pro-

gressive measurability) together with P(∫ t

0H2

sds < +∞)

= 1. Under thesehypotheses, one can write the following equation

Xt = X0 +∫ t

0

g(Xs, s)ds +∫ t

0

σ(Xs, s)dWs, (5.2)

where (under some suitable regularity hypothesis on g and σ) the first integralcan be understood as a Riemann-Stieltjes integral, while the second as an Itô


86 5 Itô’s Formula and Stochastic Differential Equations

integral. The previous expression can also be written as:{dXt = g(Xt, t)dt + σ(Xt, t)dWt

X0 = x,

which is referred to as a Stochastic Differential Equation (SDE). Note thatthe two formulations are equivalent: the latter is just a different way to writethe former.

A fundamental result in stochastic analysis is the so-called Itô’s Lemma.In the following, we will denote with f(x, t) ∈ C2,1(R,R+) a function

that is continuously differentiable twice in x and once in t. Similarly forf(x1, ..., xn, t) ∈ C2,1(Rn,R+).

Lemma 5.1 (Itô’s Lemma for functions of one variable plus time).Let f (x, t) ∈ C2,1(R,R+) be a given function.If the process (Xt)t≥0 satisfies the following equation:

dXt = μ (t) dt + σ (t) dWt, (5.3)

then the process (Zt)t≥0, defined by Zt � f (Xt, t), satisfies

dZt =∂f

∂t(Xt, t) dt +

∂f

∂x(Xt, t) dXt +

12

∂2f

∂x2(Xt, t)σ2 (t) dt

=(

∂f

∂t+ μ

∂f

∂x+

σ2

2∂2f

∂x2

)(Xt, t) dt + σ (t)

∂f

∂x(Xt, t)dWt. (5.4)

For functions of two or more variables plus time, Itô’s Lemma can be for-mulated as follows.

Lemma 5.2 (Itô’s Lemma for functions of two or more variables plustime).Let f (x1, x2, ..., xn, t) be a function in C2,1(Rn,R+) depending on n vari-

ables x1, ..., xn and on t.If, for any i = 1, 2, ..., n, Xi =

(Xi

t

)t≥0

satisfies the following equation:

dXit = μi (t) dt + σi (t) dW i

t , (5.5)

then the process (Zt)t≥0, defined by Zt � f(X1

t , X2t , ..., Xn

t , t), satisfies

dZt =∂f

∂t

(X1

t , ..., Xnt , t)dt +

n∑i=1

∂f

∂xi

(X1

t , ..., Xnt , t)dXi

t

+12

n∑i,j=1

σiσjρij∂2f

∂xi∂xj

(X1

t , ..., Xnt , t)dt


=

⎡⎣∂f

∂t+

n∑i=1

μi ∂f

∂xi+

12

n∑i,j=1

σiσjρij∂2f

∂xi∂xj

⎤⎦ (X1

t , ..., Xnt , t)dt

+n∑

i=1

σi ∂f

∂xi

(X1

t , ..., Xnt , t)dW i

t , (5.6)

where ρij = E[dW idW j

]/dt is the correlation between the (standard) Brow-

nian motions W i and W j .

When f does not depend on time t, equation (5.4) simplifies to

dZt =(

μf ′ (Xt) +σ2

2f ′′ (Xt)

)dt + σf ′ (Xt) dWt, (5.7)

while equation (5.6) reduces to

dZt =

⎡⎣ n∑

i=1

μi ∂f

∂xi+

12

n∑i,j=1

σiσjρij∂2f

∂xi∂xj

⎤⎦(X1

t , ..., Xnt

)dt

+n∑

i=1

σi ∂f

∂xi

(X1

t , ..., Xnt

)dW i

t . (5.8)

We finally recall the notion of Brownian local time and the Tanaka formula.The occupation time of a set A by time t of a Brownian motion is defined byY A

t �∫ t

0 1A (Wu) du, where 1A(x) is the indicator function of the set A takingvalue 1 if x ∈ A, taking value 0 if x /∈ A. The Brownian local time Lt is theRadon-Nikodym derivative of Yt with respect to the Lebesgue measure on thereal line (they can be proved to be equivalent), i.e.

Y At =

∫A

Lt (u) du. (5.9)

The Tanaka formula provides a representation of the function of a Brow-nian motion f(x) = (x − K)+ = max {x − K; 0} in terms of the Brownianlocal time of K. Namely,

(Wt − K)+ = (W0 − K)+ +∫ t

0

1(K,+∞)(Wu)dWu +12Lt(K), (5.10)

where Lt(K) can be proved to be given by

Lt(K) =∫ t

0

δK(Wu)du (5.11)


and δK (x) is the Dirac Delta distribution defined by:

∫ t

0

δK(Wu)du = limn→+∞

(n

∫ t

0

1[K− 12n ,K+ 1

2n )(Wu)dWu

). (5.12)

For an exhaustive treatment we recommend the texts by Björk [6], Mikosch[28], Øksendal [30] and Pascucci [31] among many others.


Exercise 5.3. Let (Wt)t≥0 be a standard Brownian motion, and assume(S1

t

)t≥0

and(S2

t

)t≥0

satisfy the following SDEs:

dS1t = μ1S

1t dt + σ1S

1t dWt; S1

0 = s10 > 0

dS2t = μ2S

2t dt + σ2S

2t dWt; S2

0 = s20 > 0

with μ1, μ2 ∈ R and σ2 > σ1 > 0.

1. For f (x) = lnx, find the stochastic differential equation satisfied by theprocess

(f(S1

t

))t≥0

and the SDE satisfied by(f(S2

t

))t≥0.

2. Find the stochastic differential equation satisfied by Yt = g(S1

t , S2t

)=

ln(

S1t

S2t

)when μ = μ1 = μ2. Establish whether (Yt)t≥0 can be a geometric

Brownian motion and/or a Brownian motion and/or a Brownian motionwith drift.

3. Do the same as above but this time for Zt = h(S1

t , S2t

)= ln

(S1

t · S2t

).

Solution

1. Since f ′ (x) = 1x and f ′′ (x) = − 1

x2 , Itô’s formula implies that

df(S1

t

)= d

(lnS1

t

)= f ′ (S1

t

)dS1

t +12f ′′ (S1

t

) [σ1S

1t

]2dt

=1S1

t

dS1t − 1

21

(S1t )2[σ1S

1t

]2dt

=1S1

t

[μ1S

1t dt + σ1S

1t dWt

]− 12σ2

1dt

=(

μ1 − 12σ2

1

)dt + σ1dWt.

It follows that(f(S1

t

))t≥0

satisfies the following stochastic differential equa-tion: {

df(S1

t

)=(μ1 − 1

2σ21

)dt + σ1dWt

f(S1

0

)= ln

(S1

0

) .


Note that the previous SDE implies

ln(S1

t

)= ln

(S1

0

)+(

μ1 − 12σ2

1

)t + σ1Wt.

Hence

S1t = S1

0 · exp((

μ1 − 12σ2

1

)t + σ1Wt

). (5.13)

Proceeding as above, we get{df(S2

t

)=(μ2 − 1

2σ22

)dt + σ2dWt

f(S2

0

)= ln

(S2

0

) .

2. In order to find the stochastic differential equation satisfied by g(S1

t , S2t

),

we can proceed (at least) in two ways.One (longer) is to apply Itô’s formula to the function g.

Another is based on the fact that Yt = g(S1

t , S2t

)= ln

(S1

t

S2t

)= ln

(S1

t

) −ln(S2

t

)(using the properties of logarithms, as S1

0 , S20 > 0). From item 1.

and from μ = μ1 = μ2 it follows that

dYt = d(lnS1

t

)− d(ln S2

t

)= df

(S1

t

)− df(S2

t

)=

12(σ2

2 − σ21

)dt + (σ2 − σ1) dWt.

Hence (Yt)t≥0 satisfies the following SDE:{dYt = 1

2

(σ2

2 − σ21

)dt + (σ2 − σ1) dWt

Y0 = ln(

S10

S20

).

We deduce that (Yt)t≥0 is neither a Brownian motion, nor a Brownianmotion with drift, nor a geometric Brownian motion, yet it is a Brownianmotion with drift μ∗ = 1

2

(σ2

2 − σ21

)and diffusion σ∗ = σ2 − σ1.

3. Proceeding as above, we obtain

dZt = d(lnS1

t

)+ d(ln S2

t

)= df

(S1

t

)+ df

(S2

t

)=(

2μ − 12(σ2

1 + σ22

))dt + (σ1 + σ2) dWt.

Hence, (Zt)t≥0 is a Brownian motion with drift μ = 2μ − 12

(σ2

1 + σ22

)and

diffusion σ = σ1 + σ2.


Exercise 5.4. Assume that the price dynamics of two stocks is representedby the processes

(S1

t

)t≥0

and(S2

t

)t≥0, solutions of the following SDEs:

dS1t = μ1S1

t dt + σ1S1t dW 1

t , S10 = s1,

dS2t = μ2S2

t dt + σ2S2t dW 2

t , S20 = s2,

where(W 1

t

)t≥0,(W 2

t

)t≥0

are independent standard Brownian motions,μ1, μ2 ∈ R and σ1, σ2 > 0.

1. Find the SDE satisfied by

f(S1

t , S2t

)=(S1

t

)2 − S1t S2

t − K,

with K > 0, representing the value of a derivative on S1 and S2.2. Find the SDE satisfied by

g(t, S1

t , S2t

)=(S1

t

)2 − S1t S2

t − Kt.

Solution

1. By Itô’s formula in several variables (see (5.8)),

df(S1

t , S2t

)=

∂f

∂S1

(S1

t , S2t

)dS1

t +∂f

∂S2

(S1

t , S2t

)dS2

t

+12

[∂2f

∂ (S1)2(S1

t , S2t

) (σ1

t S1t

)2dt

+∂2f

∂ (S2)2(S1

t , S2t

) (σ2

t S2t

)2dt

]

+12

[2

∂2f

∂S1∂S2

(S1

t , S2t

)σ1S1

t σ2S2t ρ12dt

]=(2S1

t − S2t

)dS1

t − S1t dS2

t

+12

[2(σ1

t S1t

)2dt + 0 − 2σ1S1

t σ2S2t ρ12dt

]=(2S1

t − S2t

)dS1

t − S1t dS2

t +(σ1

t S1t

)2dt,

because, in our case, ∂f∂x1

(x1, x2) = 2x1 − x2,∂f∂x2

(x1, x2) = −x1,∂2f∂x2

1(x1, x2) = 2, ∂2f

∂x22

(x1, x2) = 0, ∂2f∂x1∂x2

(x1, x2) = −1 and ρ12 = 0 (bythe independence of the two Brownian motions).Furthermore, f

(S1

0 , S20

)=(S1

0

)2 − S10S2

0 − K.


2. Using Itô’s formula in several variables (see (5.6)) and proceeding as above,we obtain

dg(t, S1

t , S2t

)=

∂g

∂t

(t, S1

t , S2t

)dt +

∂g

∂S1

(t, S1

t , S2t

)dS1

t +∂g

∂S2

(t, S1

t , S2t

)dS2

t

+12

[∂2g

∂ (S1)2(t, S1

t , S2t

) (σ1

t S1t

)2dt

+∂2g

∂ (S2)2(t, S1

t , S2t

) (σ2

t S2t

)2dt

]

+12

[2

∂2g

∂S1∂S2

(t, S1

t , S2t

)σ1S1

t σ2S2t ρ12dt

]

= −Kdt +(2S1

t − S2t

)dS1

t − S1t dS2

t +(σ1

t S1t

)2dt,

with g(0, S1

0 , S20

)=(S1

0

)2 − S10S2

0 .

Exercise 5.5. Consider a stock price evolving as the following geometricBrownian motion

dSt = μStdt + σStdWt, (5.14)

with current stock price S0 = 40 euros, drift μ = 0.1 and volatility σ = 0.4(per year).

1. Establish whether the probability to exercise a European Call option isgreater than the probability to exercise a European Put option, both withstrike of 24 euros, maturity of 2 years and written on the stock above.

2. Suppose now that the price ST of another stock at time T = 2 (years) isa multiple of the price of the first stock, i.e. ST = cST for some c > 0.Establish if there exist c > 0 such that the probability to exercise the Eu-ropean Put option having the second stock as underlying is at least twicethe probability to exercise the corresponding Call option.

3. Find the greatest strikeK∗ making the Call payoff φ = (ST − K∗)+ greaterthan 6 euros (or equal) with probability at least of 20%.

SolutionAs already pointed out in Exercise 5.3 (see equation (5.13)), the stock price

is given by:

St = S0 exp{(

μ − 12σ2

)t + σWt

}. (5.15)

1. In order to exercise the European Call option, the stock price at maturityshould be greater than or equal to the strike. Hence,

P ({ Call is exercised}) = P (ST ≥ K)= P (ln(ST ) ≥ ln(K))


= P

(ln (S0) +

(μ − 1

2σ2

)T + σWT ≥ ln (K)

)

= P

(WT ≥ ln(K) − ln(S0) −

(μ − 1

2σ2)T

σ

).

Since WT ∼ N (0; T ) and, consequently, WT√T

∼ N (0; 1), we get

P ({Call is exercised}) = P

(WT√

T≥ ln(K) − ln(S0) −

(μ − 1

2σ2)T

σ√

T

)

= 1 − N

(ln(K) − ln(S0) −

(μ − 1

2σ2)T

σ√

T

)

= 1 − N

⎛⎝ ln(24) − ln(40) −

(0.1 − 1

2(0.4)2

)· 2

0.4 · √2

⎞⎠

= 1 − N (−0.974) = N (0.974) = 0.835,

where, as usual, N(·) denotes the cumulative distribution function of astandard normal.In order to exercise the European Put option, at maturity the stock priceshould not exceed the strike. Since St is a continuous random variablefor any fixed t > 0 (by equation (5.15)), we deduce that P (ST ≤ K) =P (ST < K), so

P ({Put is exercised}) = P (ST ≤ K) = P (ST < K)= 1 − P (ST ≥ K)= 1 − P ({Call is exercised}) = 0.165.

The probability of exercising the Call option is therefore greater than theone of exercising the Put option.

2. Suppose now that ST = cST for some c > 0. We have to establish if thereexist some suitable c > 0 so that

P ({new Put is exercised}) ≥ 2P ({new Call is exercised}) . (5.16)

In order to solve this problem, we start with computing the probabilitiesabove. On the one hand,

P ({new Call is exercised}) = P(ST ≥ K

)= P (cST ≥ K) = P

(ST ≥ K

c

),


which corresponds to the probability to exercise a Call with strike K/c andwith the first stock as underlying. By the arguments of item 1., we get

P ({new Call is exercised}) = P

(ST ≥ K

c

)

= 1 − N

(ln(K/c) − ln(S0) −

(μ − 1

2σ2)T

σ√

T

)

= 1 − N

(ln(K) − ln (c) − ln(S0) −

(μ − 1

2σ2)T

σ√

T

).

On the other hand,

P ({new Put is exercised}) = 1 − P(ST ≥ K

)

= N

(ln(K) − ln (c) − ln(S0) −

(μ − 1

2σ2)T

σ√

T

).

Equation (5.16) becomes, therefore,

3N

(ln(K) − ln (c) − ln(S0) −

(μ − 1

2σ2)T

σ√

T

)≥ 2

ln(K) − ln(S0) −(

μ − 12σ2

)T − q2/3 · σ

√T ≥ ln (c) ,

where q2/3 denotes the quantile at level α = 2/3 of the standard Normal,defined as the solution of N

(q2/3

)= 2/3. Since q2/3 = 0.4307, it follows

0 < c ≤ 0.452.3. First, notice that

P((ST − K)+ ≥ 6

)= P (ST ≥ K + 6)

= P (lnST ≥ ln (K + 6))

= P

⎛⎝WT ≥

ln(

K+6S0

)− (μ − 1

2σ2)T

σ

⎞⎠

= P

⎛⎝WT√

T≥

ln(

K+6S0

)− (μ − 1

2σ2)T

σ√

T

⎞⎠

= 1 − N

⎛⎝ ln

(K+6S0

)−(0.1 − 1

2 (0.4)2)· 2

0.4√

2

⎞⎠ .


Since such a probability has to be greater than (or equal to) 20%, we deducethat

1 − N

⎛⎝ ln

(K+6S0

)−(0.1− 1

2 (0.4)2)· 2

0.4 · √2

⎞⎠ ≥ 0.2

N

⎛⎝ ln

(K+6S0

)−(0.1 − 1

2 (0.4)2)· 2

0.4 · √2

⎞⎠ ≤ 0.8

ln(

K+6S0

)−(0.1− 1

2 (0.4)2)· 2

0.4 · √2≤ 0.8416

ln(

K + 640

)≤ 0.516

K ≤ 61.01.

The largest ‘admissible’ strike is, then, K∗ = 61.01 euros.

Exercise 5.6. Consider a standard Brownian motion (Wt)t≥0 and a stochas-tic process (St)t≥0 satisfying the following stochastic differential equation:

dSt = μtdt + σtdWt, (5.17)

with μt and σt not constant everywhere but time-dependent.

1. When σt = 0.1 for t ∈ [0, 4] and

μt ={

0.04t; for 0 ≤ t ≤ 20.02 (10 − t) ; for 2 < t ≤ 4

,

compute: (a) the probability of having (at time t = 4) a profit greater than(or equal to) 0.4; (b) the average profit at time t = 4.

2. When μt is the same as above, while

σt ={

0.3; for 0 ≤ t ≤ 20.4; for 2 < t ≤ 4

,

compute: (a) the probability of having a profit at least equal to 0.4 at timet = 4; (b) the average profit at time t = 4.

3. Set Xt � St − S0 for any t ≥ 0. Discuss whether there exist (μt)t≥0 and(σt)t≥0 such that (Xt)t≥0 is a martingale (with respect to the filtered spaceconsidered at the beginning).


Solution

1. By integrating equation (5.17), we obtain

St = S0 +∫ t

0

μsds +∫ t

0

σsdWs.

Hence

S4 − S0 =∫ 4

0

μsds +∫ 4

0

σsdWs

=∫ 2

0

0.04 s ds +∫ 4

2

0.02 (10 − s) ds +∫ 4

0

0.1 dWs

=[0.02 s2

]∣∣20+[0.02

(10s− s2

2

)]∣∣∣∣4

2

+ 0.1 · (W4 − W0)

= 0.36 + 0.1 · (W4 − W0)= 0.36 + 0.1 W4.

(b) It is straightforward to obtain that the average profit at time t = 4 isequal to E [S4 − S0] = 0.36 +0.1 ·E [W4] = 0.36, where the last equality isdue to a well-known property of standard Brownian motions.For item (a), we deduce that

P ({profit at time t = 4 is at least equal to 0.4})= P (0.36 + 0.1 W4 ≥ 0.4)

= P (0.1 W4 ≥ 0.04) = P

(W4√

4≥ 0.04

0.1√

4

)= 1 − N (0.2) = 0.42.

2. Under the hypothesis on μt and on σt, we get

S4 − S0 =∫ 4

0

μsds +∫ 4

0

σsdWs

=∫ 2

0

0.04 s ds +∫ 4

2

0.02 (10 − s) ds +∫ 2

0

0.3 dWs +∫ 4

2

0.4 dWs

= 0.36 + 0.3 (W2 − W0) + 0.4 (W4 − W2)= 0.36 + 0.3 W2 + 0.4 (W4 − W2) .

By the properties of standard Brownian motions we know, indeed, thatW2

and (W4 − W2) are independent Normal random variables. We remind alsothat:

X ∼ N(mX ; s2

X

)and Y ∼ N

(mY ; s2

Y

)are independent

⇒ aX + bY ∼ N(amX + bmY ; a2s2

X + b2s2Y

)for real numbers a and b.


Consequently, we deduce that S4 − S0 ∼ N (0.36; 0.09 · 2 + 0.16 · 2), soS4 − S0 ∼ N (0.36; 0.5). The average profit at time t = 4 is then equal toE [S4 − S0] = 0.36, while

P (profit at time t = 4 is at least equal to 0.4)= P (S4 − S0 ≥ 0.4)

= P

(S4 − S0 − E [S4 − S0 ]√

V (S4 − S0)≥ 0.4 − E [S4 − S0 ]√

V (S4 − S0)

)

= 1 − N

(0.04√

0.5

)= 0.48.

3. Taking μt = 0 and σt = σ (with σ ∈ R) for any t ≥ 0, we get

Xt = St − S0 =∫ t

0

σdWs = σWt

for any t ≥ 0. By the properties of Brownian motions and by the argumentsabove, it follows that (Xt)t≥0 is a martingale (when μt ≡ 0, σt ≡ σ).

Exercise 5.7. By applying the Tanaka formula and the definition of Brow-nian local time, prove that the payoff of a European Call option, on a stockwhose dynamics is described by a geometric Brownian motion with null driftand diffusion coefficient σ, admits the following representation:

(ST − K)+ = (S0 − K)+ +∫ T

0

1(K,+∞)dSt +12

∫ T

0

σ2S2t δK(St)dt.

SolutionA naive application of Itô’s formula would provide:

f(ST ) = f(S0) +∫ T

0

f ′(St)dSt +12σ2

∫ t

0

f ′′(St)dt,

but in order to apply correctly the Itô’s formula we must verify the conditionthat both the first and the second derivative of the function f exist and arecontinuous. Actually, Itô’s formula holds also if there is a finite number ofpoints where f ′′ is not defined, provided that f ′ is continuous on the wholedomain of f (and |f ′′| is bounded wherever it is defined). This is not the caseof the function under consideration. We can nevertheless apply an alterna-tive procedure which could provide the desired result. Define the followingsequence of functions:

fn(x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

0; x ≤ K − 12n

n

2(x − K)2 +

12

(x − K) +18n

; K − 12n

< x < K +12n

x − K; x ≥ K +12n

.


Hence

f ′n(x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

0; x ≤ K − 12n

n (x − K) +12; K − 1

2n< x < K +

12n

1; x ≥ K +12n

and it can be immediately verified that all the functions fn are continuous onthe whole domain of f . Moreover, their second derivatives

f ′′n (x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

0; x < K − 12n

,

n; K − 12n

< x < K +12n

0; x > K +12n

are not defined at the points x = K ± 12n (even if their absolute values are

bounded everywhere else). Itô’s formula can therefore be applied to the func-tions fn, so that

fn (ST ) = fn (S0) +∫ T

0

f ′n(St)dSt +

12σ2

∫ T

0

f ′′n (St)S2

t dt (5.18)

since the stock price St is assumed to solve dSt = σStdWt.We can also remark that the sequence (fn)n∈N converges to the function

f , and the sequence of the first derivatives converges to

f ′(x) = limn→∞ f ′

n(x) =

⎧⎪⎪⎨⎪⎪⎩

0; x < K12; x = K

0; x > K

.

By passing to the limit in the first integral appearing in equation (5.18),we notice that the value of f ′ at the single point x = K will not affect itsvalue, hence that integral can be expressed as follows:

∫ T

0

1(K,+∞)(St)dSt. (5.19)

By collecting the results and passing to the limit n → +∞ (this is possibleby the dominated convergence theorem), we finally obtain:

f(ST ) = f(S0)+∫ T

0

1(K,+∞)(St)dSt+12σ2 lim

n→∞n

∫ T

0

S2t 1(K− 1

2n ,K+ 12n)(St)dt


which, by (5.12), can be written as:

(ST − K)+ = (S0 − K)+ +∫ T

0

1(K,∞)(St)dSt +12

∫ T

0

σ2S2t δK(St)dt.


Exercise 5.8. Consider a stock price (St)t≥0 evolving as the following geo-metric Brownian motion

dSt = μStdt + σStdWt.

Find the stochastic differential equation satisfied by f (St, t) = ert ·ln (S4t − K

)and compare it with the one satisfied by g (St) = ln

(S4

t − K)(where K is a

given positive real number).

Exercise 5.9. The financial position Xt (at time t) evolves as

dXt = μdt + σtdWt,

with X0 = 80000 euros, μ = 400 (per year) and with σt (per year) given by

σt =

⎧⎪⎨⎪⎩

8∑k=0

100 (k + 1)1[k;k+1) (t) ; 0 ≤ t < 9

σ∗; t ≥ 9

(5.20)

where σ∗ > 0. (We remind that the indicator function 1A (t) equals 1 whent ∈ A, and 0 when t /∈ A.)

1. Find the distribution of Xt (as a function of t).2. Make Xt explicit as function of X0, μ, σt, t and Wt.3. Find σ∗ such that P (X10 < 10000) < 1

100 . Is such a σ∗ reasonable from afinancial point of view?

4. Compute E [X100 − X10], V (X100 − X10), E [X200 − X10] and V(X200 −

X10

)for (σt)t≥0 as in (5.20).

Due to the properties of Brownian motions, the quantities above are easyto compute. Is the same true for E [X200/X10] and V (X200/X10)?

Exercise 5.10. Let S1t , S2

t and S3t be the prices of three different stocks at

time t. Suppose that(S1

t

)t≥0,(S2

t

)t≥0,(S3

t

)t≥0

satisfy

dS1t = μ1S1

t dt + σ1S1t dW 1

t

dS2t = μ2S2

t dt + σ2S2t dW 2

t

dS3t = μ3S3

t dt + σ3S3t dW 3

t ,

respectively, where W 1, W 2 and W 3 are standard Brownian motions.


1. Find the dynamics of the value f(S1

t , S2t , S3

t

)of the portfolio formed by

four shares of the first stock, two shares of the second and one share of thethird.

2. Recall that cash invested (or borrowed) in a bank account at a risk-freerate r evolves as follows:

Xt = X0ert.

Applying the fact above, find the dynamics of the value g(S1

t , S2t , S3

t , t)

of the portfolio formed by four shares of the first stock, two shares of thesecond and one of the third and with a bank loan equal to the quantity ofcash necessary (at the initial time) to buy the stocks.

6

Partial Differential Equations in Finance


Let u be a function of several real variables (t, x1, x2, ..., xn):

u : R+ ×Rn → R(t, x1, x2, ..., xn) �→ u (t, x1, x2, ..., xn)

An equation involving one or more partial derivatives of the function u withrespect to its variables (sometimes also u itself) is called a Partial DifferentialEquation (PDE). Every function u satisfying the equation is called a solution.The notion of PDE generalizes in a natural way the notion of Ordinary

Differential Equation (ODE) to the case where the unknown map depends onmore than one variable. While the most general solution of an ODE dependsin general on some arbitrary constants, the set of solutions satisfying a PDE isparameterized by arbitrary functions. Sometimes it is necessary to determinea solution satisfying extra conditions, for example some initial and/or bound-ary conditions. These conditions are also necessary in order to have existence,uniqueness and stability results for the corresponding solutions.PDEs are one of the richest and most fruitful subjects of Mathematical

Analysis. A detailed discussion, even of the most basic results of the theory,would require a much deeper treatment than that allowed in a summary likethe present one. Below we shall just recall in a very concise way the conceptsnecessary to understand and solve the exercises we are going to propose, andwe prompt the interested reader to consult one of the several textbooks writ-ten on the subject. We recommend, for instance, Salsa [35] and, concerningthe most important applications of derivatives valuation, Wilmott et al. [37].In our applications we shall focus on PDEs for functions of two real vari-

ables of semilinear type, where the partial derivatives of highest order appearalways with power 1.Semilinear PDEs of second order can be classified according to some basic

properties. In the particular framework we have chosen to work, i.e. real func-


102 6 Partial Differential Equations in Finance

tions of two real variables, the classification can be described in the followingway.Once mixed partial derivatives of second order have been eliminated (pos-

sibly by a linear transformation of the independent variables), one faces thefollowing possibilities:

• if both second-order derivatives appear with the same sign, the equation issaid to be of elliptic type;

• if both second-order derivatives appear with opposite sign, the equation issaid to be of hyperbolic type;

• if the equation contains only one second-order derivative, and the first-orderderivative in the other variable, the equation is of parabolic type.

For parabolic equations a further distinction is introduced: if the two deriva-tives with highest order with respect to the two variables have the same sign,the equation is called backward parabolic, otherwise forward parabolic.The classification of a PDE provides some useful information about the

kind of conditions to assign in order to ensure existence of solutions. It is pos-sible to prove, for example, that the initial-value problem for elliptic equationsis ill-posed: in this case in fact, the existence of a solution is not guaranteedand, if a solution exists, it can be not unique and/or it can depend on theinitial data in a discontinuous manner.

The Black-Scholes equation

∂u

∂t+

12σ2S2 ∂2u

∂S2+ rS

∂u

∂S− ru = 0 (6.1)

(see Chap. 7) is the most popular PDE in financial applications. The unknownfunction depends on two independent variables S, t and the equation involvesfirst derivatives in both, but only the second derivative in S. Moreover, thehighest derivatives in S and t (the first derivative with respect to time t andthe second with respect to the underlying S) appear with the same sign. TheBlack-Scholes equation is thus a backward parabolic PDE.

For a backward parabolic equation it is possible to prove that, under quitegeneral regularity conditions on both data and coefficients, given a final con-dition (in t) and (possibly) two boundary conditions (in S), a solution exists,is unique and depends continuously on the data.For forward parabolic equations existence and uniqueness of a solution is

guaranteed once an initial (instead of final) datum is assigned.The interest in parabolic equations, in particular in the backward parabolic

ones, for financial applications is motivated by a deep relationship betweenthis type of equations and diffusion stochastic processes, i.e. processes de-scribed by stochastic differential equations driven by a standard Brownian


motion. This relationship is clarified by the following very important result,known in the literature as the Feynman-Kac Representation Theorem.

Theorem 6.1. Let u(x, t) be a solution of the PDE:

∂u

∂t+

12σ2(x, t)

∂2u

∂x2+ μ(x, t)

∂u

∂x− ru = 0 (6.2)

with final condition:u(x, T ) = Φ(x). (6.3)

Then this solution can be represented as follows:

u(x, t) = e−r(T−t)Et,x[Φ(XT )], (6.4)

where (Xs)s≥t satisfies the stochastic differential equation:{dXs = μ(Xs, s)ds + σ(Xs, s)dWs

Xt = x(6.5)

and Et,x denotes the dependence on the variables t and x. This dependence isinherited via the condition (6.5).

The result holds under the further hypothesis that the processσ(s, Xs)∂u

∂x (s, Xs) is square-integrable, i.e.∫ t

0

[σ(s, Xs)∂u

∂x (s, Xs)]2

ds < +∞,P -almost surely (a.s.).The relevance of this result is primary: it allows to represent the solution

of a backward parabolic PDE as the expectation of its final condition withrespect to a probability measure determined by a suitable stochastic differ-ential equation. The reader eager to study in depth the relationship existingbetween backward parabolic PDEs and stochastic differential equations of dif-fusion type can find a more detailed treatment in the textbook by Björk [6].

Since only for a very limited number of PDEs explicit solutions are avail-able, numerical procedures are needed to obtain approximate solutions. Sev-eral fast and accurate numerical methods can be used to compute solutionswhen these are not available in explicit form.Now let us concentrate on one specific class of methods allowing to get so-

lutions of PDEs in explicit form: the so-called similarity methods, sometimessuccessful in providing parabolic PDE solutions, in particular for diffusionequations:

∂u

∂t(x, t) =

∂2u

∂x2(x, t) (6.6)

and some variations thereof.The similarity method is based on the invariance properties of the PDE

and its initial (final) and boundary data under a class of transformations ofthe independent variables, and it can be applied if this invariance property


holds and can be identified. It is easy to see, for example, that the classicaldiffusion equation (6.6) does not change if t is replaced by λ2t and at thesame time x is replaced by λx. This invariance property, known as scalinginvariance (in this particular case the scaling is called parabolic for obviousreasons), suggests to look for a solution of a special form, where the two in-dependent variables appear always in the same combination, depending onthe ratio x/

√t. It is necessary that also the data exhibit the same invariance

property and, as we shall see immediately in the applications, sometimes thedata themselves suggest the particular form to look for.Once this special combination has been detected, both the unknown and

the data of the PDE can be expressed as functions of this new variable, whichbecomes the only independent variable for the problem, and the PDE is re-duced to an ordinary differential equation.Here is a rule of thumb to find similarity solutions for second order parabolic

PDEs with two independent variables. The rule consists in finding, by trials,a solution of the form u = tαf(x/tβ) by looking for α and β such that theequation admits a similarity reduction, i.e. it becomes an ordinary differentialequation for the unknown U(y) in the new independent variable y = x/tβ.The applications we are going to provide should explain in a more satisfactoryway the concepts just exposed.


Exercise 6.2. Show that, for functions a, b : R→ R of class C1,

V (St, t) = a(t)St + b(t)

satisfies the Black-Scholes equation:

∂V

∂t+

12σ2S2

t

∂2V

∂S2+ rSt

∂V

∂S− rV = 0 (6.7)

if and only if a(t) = a and b(t) = bert with a and b real constants.

SolutionFirst of all, we should remark that in the problem considered no bound-

ary condition has been assigned. If (6.7) is furnished with the final conditionV (ST , T ) = option payoff, then the problem admits one and only one solution:the one provided by the Black-Scholes formula for Call or Put options.

Let us consider V (St, t) = a(t)St + b(t) and verify that if V satisfies equa-tion (6.7) then a and b are constant. We get

∂V

∂t= a′(t)St + b′(t)

∂V

∂S= a(t)

∂2V

∂S2= 0,


and, by substituting these partial derivatives in (6.7), we can verify that, if Vsatisfies equation (6.7), then

∂V

∂t+

12σ2S2

t

∂2V

∂S2+rSt

∂V

∂S−rV = a′(t)St+b′(t)+rSta(t)−ra(t)St−rb(t) = 0

or, equivalently,a′ (t)St + b′ (t) − rb (t) = 0.

From the previous equation and by imposing the coefficients of both thefirst- and the second-degree terms in S to be equal to zero, we get

a′(t) = 0b′(t) − rb(t) = 0.

Consequently, if V satisfies (6.7) then a(t) = a and b(t) = bert with a, b realconstants.The inverse implication is immediate to prove: if a and b are constant, then

V (St, t) = aSt + b satisfies equation (6.7).

Exercise 6.3. Suppose that the real function u (x, t) satisfies the followingproblem on the real, positive half-line:

∂u

∂t=

∂2u

∂x2, for x > 0, t > 0

withu (x, 0) = u0 (x) , for x > 0,u (0, t) = 0, for t > 0.

Find a function h(x, s, t) allowing to write the solution of the problem in theform:

u (x, t) =∫ +∞

0

u0 (s) h(x, s, t)ds. (6.8)

SolutionLet us start by pointing out that the Green function for the problem con-

sidered allows to write the solution of the problem in the following form:

u (x, t) =∫ +∞

−∞u0 (s) g(x − s, t)ds, (6.9)

i.e. as a convolution integral extended to the whole real line. The functiong playing the role of the integral kernel in the previous expression is thefundamental solution of the PDE under examination, i.e. the solution withDirac’s function δ(x) as initial datum. It is well known that the Green functiong(x, s, t) for the present problem is the following:

g(x, s, t) =1

2√

πtexp{−(x − s)2

4t

}. (6.10)


Notice that the function we are looking for is not the Green function, which isalready known, since the integral providing the solution is not necessarily ofconvolution type and the integration domain is just the positive real half-line.With this in mind, let us consider the following auxiliary problem (AP).

Let the function v (x, t) be defined by reflection with respect to x = 0, i.e.

v (x, t) ={−u (−x, t) ; for x < 0

u (x, t) ; for x > 0

with v0 (x) = v (x, 0) for x ∈ R.We will verify that v (0, t) = 0 and prove that the solution of the problem:

∂v

∂t=

∂2v

∂x2, x ∈ R, t > 0

with

v (x, 0) = v0 (x) , for x > 0v (0, t) = 0, for t > 0

is

v (x, t) =1

2√

πt

∫ +∞

0

u0 (s)[e−

(x−s)2

4t − e−(x+s)2

4t

]ds.

By definition of v and by the simple remark that u (0, t) = 0, one obtainssimultaneously the following conditions:

v(0+, t

)= lim

x→0+u (x, t) = 0

v(0−, t

)= lim

x→0−[−u (−x, t)] = 0.

Since v must be continuous, then v (0, t) = 0.From the preliminary remark, if v(x, t) is regular enough and if

limx→±∞ v(x, t) = 0, we know that the following equality holds:

v (x, t) =1

2√

πt

∫ +∞

−∞v0 (s) e−

(x−s)2

4t ds. (6.11)

By definition, however,

v0 (x) = v (x, 0) ={−u (−x, 0) = −u0 (−x) ; for x < 0

u (x, 0) = u0 (x) ; for x > 0(6.12)


By plugging (6.12) into (6.11), we obtain:

v (x, t) =1

2√

πt

∫ +∞

−∞v0 (s) e−

(x−s)2

4t ds

=1

2√

πt

[∫ 0

−∞−u0 (−s) e−

(x−s)2

4t ds +∫ +∞

0

u0 (s) e−(x−s)2

4t ds

](6.13)

=1

2√

πt

[∫ 0

+∞u0 (y) e−

(x+y)2

4t dy +∫ +∞

0

u0 (s) e−(x−s)2

4t ds

](6.14)

=1

2√

πt

[−∫ +∞

0

u0 (y) e−(x+y)2

4t dy +∫ +∞

0

u0 (s) e−(x−s)2

4t ds

]

=1

2√

πt

[−∫ +∞

0

u0 (z) e−(x+z)2

4t dz +∫ +∞

0

u0 (z) e−(x−z)2

4t dz

]

=1

2√

πt

∫ +∞

0

u0 (z)[e−

(x−z)2

4t − e−(x+z)2

4t

]dz,

where equality (6.14) follows from the change of variable y = −s (dy = −ds).

From the last equality and the relationship between the initial problem andproblem (AP), one obtains that the function h(x, s, t) we are looking for is

h (x, s, t) = 12√

πt

[e−

(x−s)2

4t − e−(x+s)2

4t

].

Remark. The result just obtained can be applied to barrier option valuationand consists in a slightly modified version of the “image method”.

Exercise 6.4. Find the similarity solution of the following mixed problem:

∂u

∂t=

∂2u

∂x2+ 3x2, for x > 0, t > 0 (6.15)

with

u (0, t) = 0, for t > 0 (6.16)

limx→∞

u(x, t)x4

= 0, for t > 0 (6.17)

u (x, 0) = 0, for x > 0. (6.18)

SolutionOur goal is to find a transformation of x and t of the kind:

ξ =x

tα,


generating a solution u (x, t) of the following type:

u (x, t) = tβU( x

tα

)= tβU (ξ) ,

and to write (6.15) in terms of ξ and U . The resulting second-order ODE inU will be much simpler to solve.

Let us begin by determining suitable coefficients α and β. First of all, interms of U and ξ, we get

∂u

∂t= βtβ−1U

( x

tα

)+ tβU ′

( x

tα

)(−α

x

tα+1

)= βtβ−1U (ξ) − αxtβ−α−1U ′ (ξ) = βtβ−1U (ξ) − αξtβ−1U ′ (ξ) ,

∂u

∂x= tβU ′

( x

tα

) 1tα

= tβ−αU ′( x

tα

)= tβ−αU ′ (ξ) ,

∂2u

∂x2= tβ−αU ′′

( x

tα

) 1tα

= tβ−2αU ′′( x

tα

)= tβ−2αU ′′ (ξ) .

Equation (6.15) becomes then:

βtβ−1U (ξ) − αξtβ−1U ′ (ξ) = tβ−2αU ′′ (ξ) + 3x2

βtβ−1U (ξ) − αξtβ−1U ′ (ξ) = tβ−2αU ′′ (ξ) + 3ξ2t2α

and, dividing both sides by t2α,

tβ−2α−1 [βU (ξ) − αξU ′ (ξ)] = tβ−4αU ′′ (ξ) + 3ξ2. (6.19)

In order to eliminate the explicit dependence on t in the previous equation,let us choose α and β such that:

{β − 2α − 1 = 0

β − 4α = 0.

This implies α = 1/2 and β = 2.The transformation induced on x and t is then given by

ξ =x√t

and we are looking for a solution u (x, t) of the following type:

u (x, t) = t2U (ξ) .


By writing again (6.15) in terms of ξ and U or, in other words, equation (6.19)with α = 1/2 and β = 2, one obtains

βU (ξ) − αξU ′ (ξ) = U ′′ (ξ) + 3ξ2

U ′′ (ξ) +12ξU ′ (ξ) − 2U (ξ) = −3ξ2. (6.20)

Condition (6.16) on u becomes U (0) = 0, while condition (6.17) becomeslimξ→+∞

U(ξ)ξ4 = 0. That is: as t → 0 we have ξ → +∞ and

0 ← u (x, t)x4

=t2U (ξ)

x4=

U (ξ)ξ4

.

We must then find the general solution of equation (6.20). Since this is anon-homogeneous ordinary differential equation, its general solution will beprovided by the general solution of the associated homogeneous equation:

U ′′ (ξ) +12ξU ′ (ξ) − 2U (ξ) = 0 (6.21)

plus a particular solution of (6.20).Let us start from the particular solution. We can try with a polynomial,

and we immediately observe that Up (ξ) = −14ξ4 is a particular solution of

(6.20).

Now we look for the general solution of equation (6.21). If we knew two“independent” solutions UH

1 and UH2 of equation (6.21), the general solution

would be provided by a linear combination of UH1 and UH

2 .By looking again for a polynomial solution, we find that UH

1 (ξ) = ξ4 +12ξ2 + 12 satisfies equation (6.21).

Another solution of equation (6.21) can be obtained using the method ofvariation of constants:

UH2 (ξ) = a (ξ)UH

1 (ξ) ,

with a being a suitable function, to be determined. To this aim we requirethat UH

2 satisfies equation (6.21).Since

(UH

2

)′(ξ) = a′ (ξ)UH

1 (ξ) + a (ξ)(UH

1

)′(ξ)(

UH2

)′′(ξ) = a′′ (ξ)UH

1 (ξ) + 2a′ (ξ)(UH

1

)′(ξ) + a (ξ)

(UH

1

)′′(ξ) ,

the function a must satisfy

a′′U1 + 2a′U ′1 + aU ′′

1 +12ξaU ′

1 +12ξa′U1 − 2aU1 = 0, (6.22)


where for notational simplicity the dependence on ξ has been omitted and U1

denotes UH1 . Since UH

1 satisfies (6.21), equation (6.22) can be written as:

a′′U1 + 2a′U ′1 +

12ξa′U1 = 0,

i.e.a′′

a′ = −ξ

2− 2

U ′1

U1. (6.23)

By denoting by h the function a′, the differential equation (6.23) can bewritten as a first-order ordinary differential equation whose solution can beobtained via separation of variables.Excluding h (ξ) ≡ 0, we obtain:

ln (h (ξ)) = ln(h(0)) − ξ2

4− 2 ln(ξ4 + 12ξ2 + 12), (6.24)

hence

h (ξ) = a′ (ξ) = a′(0)1

(ξ4 + 12ξ2 + 12)2e−ξ2/4.

Now, in order to get UH2 we have two possibilities. First, we can integrate

a′ to get a and consequently UH2 . Second (more simply), looking at a′ we

notice that a good candidate for UH2 is

f (ξ) e−ξ2/4 + g (ξ)∫ ξ

−∞e−

14 s2

ds,

where f(ξ) and g(ξ) are polynomials in ξ, the former of degree not higher than3 and the latter of degree not higher than 4. The coefficients of these polyno-mials can be easily computed by substituting f(ξ) and g(ξ) in the equationthat UH

2 must satisfy.

Proceeding as described above we obtain another solution of equation(6.21), independent of UH

1 :

UH2 (ξ) = 2ξ

(ξ2 + 10

)e−

14 ξ2

+(ξ4 + 12ξ2 + 12

)∫ ξ

0

e−14 s2

ds.

The general solution of the homogeneous equation (6.21) is then

UHG (ξ) = aUH

1 (ξ) + bUH2 (ξ)

and, consequently, the general solution of the non-homogeneous (6.20) is

UG (ξ) = Up (ξ) + aUH1 (ξ) + bUH

2 (ξ) .


By imposing U (0) = 0 and limξ→+∞U(ξ)ξ4 = 0, it is possible to determine the

constants a and b. By the first condition we get a = 0. Since limξ→∞U(ξ)ξ4 =

−14 + b

√π, from the second condition we get b = 1

4√

π.

The similarity solution required is then:

U (ξ) = −14ξ4 +

12√

π(ξ3 + 10ξ)e−

14 ξ2

+1

4√

π(ξ4 + 12ξ2 + 12ξ)

∫ ξ

0

e−s24 ds.

Finally, the solution of the initial problem (6.15)-(6.18) is

u (x, t) = t2U

(x√t

)= −1

4x4 +

12√

π

[√tx3 + 10xt3/2

]e−

14 ξ2

+1

4√

π

(x4 + 12x2t + 6xt3/2

) ∫ x/√

t

0

e−s24 ds.

It is easy to check that this expression solves our problem by direct substitu-tion into (6.15).

Exercise 6.5. By applying the Feynman-Kac representation formula, find thesolution of the following problem on the domain [0, T ]× R:

∂V

∂t+ μx

∂V

∂x+

12σ2x2 ∂2V

∂x2= 0 (6.25)

V (T, x) = ln(x4)

+ k

with given constants μ, σ and k.

SolutionBy the Feynman-Kac representation formula, we know that the solution of

the problem:

∂V

∂t+ m (t, x)

∂V

∂x+

12s2 (t, x)

∂2V

∂x2= 0

V (T, x) = Φ (x)

isV (t, x) = E [Φ (XT )] ,

where, for u ≥ t {dXu = m (u, Xu) du + s (u, Xu) dWu

Xt = x.


In the present case, m (t, x) = μx, s (t, x) = σx and Φ (x) = ln(x4)

+ k.The solution of the problem proposed is then given by:

V (t, x) = E [Φ (XT )] ,

where for u ≥ t {dXu = μXudu + σXudWu

Xt = x(6.26)

Since the solution of (6.26) is given by

Xu = Xt exp{(

μ − 12σ2

)(u − t) + σ (Wu − Wt)

}

= x exp{(

μ − 12σ2

)(u − t) + σ (Wu − Wt)

},

we obtain that

XT = x exp{(

μ − 12σ2

)(T − t) + σ (WT − Wt)

}

and

ln(X4

T

)+ k = ln

(x4 exp

{4(

μ − 12σ2

)(T − t) + 4σ (WT − Wt)

})+ k

= ln(x4)

+ 4(

μ − 12σ2

)(T − t) + 4σ (WT − Wt) + k.

We finally get the solution of the problem proposed:

V (t, x) = E [Φ (XT )]

= E

[ln(x4)

+ 4(

μ − 12σ2

)(T − t) + 4σ (WT − Wt) + k

]

= ln(x4)

+ 4(

μ − 12σ2

)(T − t) + k,

once we observe E [WT − Wt] = 0 (W is a standard Brownian motion).

Exercise 6.6. In the Black-Scholes setting, consider an option written on anunderlying with the following dynamics:

dSt = μStdt + σStdWt

with S0 = 20 Euro, μ = 0.16 and σ = 0.36 (per year). The risk-free interestrate is 0.04 per year.


By applying the Feynman-Kac formula, compute the initial value of theoption with underlying S and payoff

Φ (ST ) =(ln(S2

T

)− K)+

with strike K = 6 euros and with maturity T = 1 year.

SolutionIn the Black-Scholes model the value of the option above at time t is as-

sumed to be a function of St and t (denoted by F (St, t)) satisfying the fol-lowing PDE:

∂F

∂t+ rx

∂F

∂x+

12σ2x2 ∂2F

∂x2− rF = 0

F (T, ST ) =(ln(S2

T

) − K)+ = Φ (ST ) .

By the Feynman-Kac formula, we know that the solution of

∂F

∂t+ m (t, x)

∂F

∂x+

12s2 (t, x)

∂2F

∂x2− rF = 0

F (T, x) = Φ (x)

is given byF (t, x) = e−r(T−t)E [Φ (XT )] ,

where, for u ≥ t {dXu = m (u, Xu) du + s (u, Xu) dWu

Xt = x.

Since, in our case, m (t, x) = rx, s (t, x) = σx and Φ (x) =(ln(x2)− K

)+,

the solution of the initial problem is given by

F (t, x) = e−r(T−t)E [Φ (XT )] ,

where, for u ≥ t {dXu = rXudu + σXudWu

Xt = x. (6.27)

Since the solution of equation (6.27) is

Xu = x exp{(

r − 12σ2

)(u − t) + σ (Wu − Wt)

},

we obtain that

XT = x exp{(

r − 12σ2

)(T − t) + σ (WT − Wt)

}


and

ln(X2

T

)= ln

(x2 exp

{2(

r − 12σ2

)(T − t) + 2σ (WT − Wt)

})

= ln(x2)

+ 2(

r − 12σ2

)(T − t) + 2σ (WT − Wt) .

We finally get the initial value of the option considered:

F (S0, 0) = e−rT E [Φ (XT )]

= e−rT E

[(ln(X2

0

)+ 2(

r − 12σ2

)T + 2σWT − K

)+]

= e−rT E[(

ln(S2

0

)+(2r − σ2

)T + 2σWT − K

)+].

There remains then to compute

E[(

ln(S2

0

)+(2r − σ2

)T + 2σWT − K

)+].

Since (Wt)t≥0 is a standard Brownian motion and T = 1,

WT = W1 ∼ N (0; 1) .

Accordingly, if we denote by Z = W1 ∼ N (0; 1), we get:

F (S0, 0) = e−rE[(

2 ln (S0) + 2r − σ2 + 2σZ − K)+]

= e−r

∫R

(2 ln (S0) + 2r − σ2 + 2σz − K

)+fZ (z) dz

= e−r

∫R

(2 ln (S0) + 2r − σ2 + 2σz − K

)+ 1√2π

e−z22 dz

= e−r

∫ +∞

K−2 ln(S0)−2r+σ22σ

(2 ln (S0) + 2r − σ2 + 2σz − K

) 1√2π

e−z22 dz

= e−r

∫ +∞

K−2 ln(S0)−2r+σ2

2σ

(2 ln (S0) + 2r − σ2 − K

) 1√2π

e−z22 dz

+e−r

∫ +∞

K−2 ln(S0)−2r+σ22σ

2σz1√2π

e−z22 dz

= e−r(2 ln (S0) + 2r − σ2 − K

) [1 − N

(K − 2 ln (S0) − 2r + σ2

2σ

)]

+e−r

[−2σ

1√2π

e−z22

]∣∣∣∣+∞

K−2 ln(S0)−2r+σ22σ


= e−r(2 ln (S0) + 2r − σ2 − K

) [1 − N

(K − 2 ln (S0) − 2r + σ2

2σ

)]

+e−r2σ1√2π

e−[K−2 ln(S0)−2r+σ2]2

8σ2

= 0.25,

where N (·) denotes the cumulative function of a standard normal randomvariable. The initial value of the option is then 0.25 euros.

Exercise 6.7. Compute the solution φ(x, t) of the diffusion equation:

∂φ

∂t=

∂2φ

∂x2, x ∈ R, t > 0, (6.28)

satisfying the following conditions:

φ(x, 0) = δ(x), limx→±∞ φ(x, t) = 0,

limx→±∞ ∂φ∂x

= 0,∫ +∞−∞ φ(x, t)dx = 1.

SolutionThe solution to be determined is called the fundamental solution of the

diffusion equation; it can be used to find solutions of the diffusion equationsatisfying more general data, as we shall see later.A first method to solve this initial/boundary-value problem is based on the

Fourier transform. The Fourier transform of a function f(x, t) is defined as

ϕ(k, t) �∫ +∞

−∞f(x, t)eikxdx. (6.29)

Notice that, by integrating twice by parts, the following relations hold forthe Fourier transforms of the derivatives of f (under the hypothesis that bothf and its first derivative in x vanish at infinity):

∫ +∞

−∞

∂f

∂xeikxdx =

[f(x, t)eikx

]+∞−∞ − ik

∫ +∞

−∞f(x, t)eikxdx = −ikϕ,

∫ +∞

−∞

∂2f

∂x2eikxdx =

[∂f

∂xeikx

]+∞

−∞− ik

∫ +∞

−∞

∂f

∂xeikxdx = −k2ϕ.

It is then immediate to verify that the function φ must satisfy

∂ϕ

∂t= −k2ϕ, (6.30)

with the initial conditionϕ(k, 0) = 1. (6.31)


The solution satisfying the required initial condition of the previous ODE canbe found immediately

ϕ(k, t) = e−k2t. (6.32)

Now, in order to find the solution φ(x, t) of the original problem we haveto invert the Fourier transform:

φ(x, t) =12π

∫ +∞

−∞ϕ(k, t)e−ikxdx. (6.33)

The explicit computation of the integral (easily performed by completing thesquare appearing in the argument of the exponential function) provides thefollowing result

φ(x, t) =1

2√

πtexp(−x2

4t

). (6.34)

Another method for solving the problem is based on a similarity technique.We look for a solution f(x, t) of the following form: f(x, t) = tαU(ξ), withξ = x/tβ. The form of the PDE suggests for β the value β = 1/2, while thecondition on the integral of the unknown extended to the entire real line sug-gests for α the value α = 1/2. The similarity solution we are looking for isthen of the following kind:

f(x, t) =1√tU(ξ). (6.35)

The similarity reduction provides the following ODE for the new unknownU(ξ):

U ′′(ξ) +ξ

2U ′(ξ) +

12U(ξ) = 0, (6.36)

together with boundary conditions:

limξ→±∞

U(ξ) = 0∫ +∞

−∞U(ξ)dξ = 1.

A straightforward computation provides the solution U(ξ):

U(ξ) = Ae−ξ2

4 + B. (6.37)

While the first boundary condition implies B = 0, the second implies A =1/(2

√π). Hence the solution of our original problem φ(x, t) becomes:

φ(x, t) =1

2√

πtexp(−x2

4t

). (6.38)


Exercise 6.8. Let φ be the fundamental solution of the diffusion equationobtained in the previous exercise. Verify that the function

f(x, t) �∫ +∞

−∞φ(ξ − x, t)u(ξ)dξ (6.39)

is a solution of the diffusion equation

∂f

∂t=

∂2f

∂x2, x ∈ R, t > 0, (6.40)

satisfying the initial condition f(x, 0) = u(x).

SolutionWe already know that the function φ(x, t) is a solution of the diffusion

equation satisfying the initial condition φ(x, 0) = δ(x). It is immediate toverify that the diffusion equation is invariant under the change of variablex → ξ −x, i.e. if φ(x, t) is a solution, φ(ξ −x, t) is also a solution, but satisfy-ing the initial datum φ(ξ − x, 0) = δ(ξ − x). Moreover the diffusion equationis linear, so that multiplying the equation:

∂φ

∂t=

∂2φ

∂x2(6.41)

by u(ξ) and integrating in ξ, the function

f(x, t) �∫ +∞

−∞φ(x − ξ, t)u(ξ)dξ (6.42)

satisfies the diffusion equation with initial datum

f(x, 0) =∫ +∞

−∞φ(x − ξ, 0)u(ξ)dξ =

∫ +∞

−∞δ(x − ξ)u(ξ)dξ = u(x). (6.43)

Exercise 6.9. Find a suitable change of variables turning the Black-Scholesequation into a PDE with constant coefficients.

SolutionThe Black-Scholes PDE, holding for the value F (St, t) of every derivative,

written on an underlying asset whose price dynamics is described by a geo-metric Brownian motion, and consistent with the no-arbitrage requirement,is the following:

∂F

∂t+

12σ2S2 ∂2F

∂S2+ rS

∂F

∂S− rF = 0. (6.44)

This PDE exhibits an explicit dependence on the variable S, denoting theunderlying value, in both the coefficients of the terms involving the first andthe second derivatives in S.


As a geometric Brownian motion is the exponential of a Brownian motion(with drift), an “educated guess” suggests the following change of variable:

x = ln(S),S = ex,

F (S, t) = h(x(S), t),h(x, t) = F (S(x), t).

By computing the partial derivatives of the new unknown with respect to thenew independent variables, we get

∂F

∂S=

∂h

∂x

∂x

∂S=

1S

∂h

∂x,

∂2F

∂S2=

∂2h

∂x2

(∂x

∂S

)2

+∂h

∂x

∂2x

∂S2=

1S2

∂2h

∂x2− 1

S2

∂h

∂x.

By plugging these expressions into the original equation, one obtains

∂h

∂t+

12σ2 ∂2h

∂x2+(

r − σ2

2

)∂h

∂x− rh = 0, (6.45)

where the coefficients’ dependence on the independent variables has beendropped.

Exercise 6.10. Find a suitable change of variables turning the Black-Scholesequation for a European Call option into the diffusion equation with properinitial-boundary conditions.

SolutionSolving the problem requires two steps. In the first, we shall remove the

dependence on the independent variable S in the PDE coefficients. To thisend, inspired both by Exercise 6.9 and by the functional form of the EuropeanCall option payoff, we exploit a change of variables of the following kind:

x = ln(S/K),S = Kex,

τ =σ2

2(T − t),

t = T − 2σ2

τ

F (S, t) = Kh(x(S), τ (t)),h(x, t) = K−1F (S(x), t(τ )),

where K is the strike of the European option considered. We introduce thenew time variable τ , which is the dimensionless version of the “time to matu-rity” variable (T − t); this change of sign in the time variable allows to turn


the parabolic type of the Black-Scholes equation (parabolic backward) intothe parabolic type of the diffusion equation (parabolic forward). Moreover, weintroduce the new parameter q = 2r/σ2 to simplify the notation. After thesechanges, the Black-Scholes equation becomes

∂h

∂τ− ∂2h

∂x2− (q − 1)

∂h

∂x+ qh = 0. (6.46)

The final condition of the Black-Scholes equation for the European Call op-tion is its payoff F (S, T ) = max(S − K; 0); expressed in the new variables, itcan be written as h(x, 0) = max(ex − 1; 0).As far as the second step is concerned, we try to remove the terms involving

h and its first derivative in x. To this end we introduce a further change ofvariables of the following kind:

h(x, τ) = exp{αx + βτ}g(x, τ), (6.47)

where α, β have to be determined so that the required terms can be removed,and we look for the PDE that the new function g must satisfy. After a directcomputation of all the derivatives of the old function in terms of the newvariables, we obtain

βg +∂g

∂τ=

∂2g

∂x2+ (q − 1)

[αg +

∂g

∂x

]+ 2α

∂g

∂x+ (α2 − q)g. (6.48)

To remove the terms involving g and its first derivative in x, the coefficientα, β must fulfill the following conditions:{

2α + (q − 1) = 0β = α2 + (q − 1)α − q

.

The required values are: α = −(q − 1)/2, β = (q + 1)2/4. By choosing then αand β as above, the function g appearing in

h(x, τ) = exp{−q − 1

2x − (q + 1)2

4τ

}g(x, τ), (6.49)

must satisfy the diffusion equation for x ∈ (−∞, +∞) and τ > 0:

∂g

∂τ=

∂2g

∂x2, (6.50)

withg(x, 0) = max

{e

q+12 x − e

q−12 x; 0

}. (6.51)

Exercise 6.11. Find the explicit solution of the Black-Scholes equation for aEuropean Call option by using the results obtained in the previous exercises.


SolutionBy the previous exercise we know that the Black-Scholes equation for a Eu-

ropean Call option can be reduced, by a suitable change of variables (whichwe performed in two separate steps), to the diffusion equation:

∂g

∂τ=

∂2g,

∂x2(6.52)

with initial datum:

g(x, 0) = max{

eq+12 x − e

q−12 x; 0

}. (6.53)

Moreover, we know from Exercise 6.8 that the solution of the diffusion equa-tion satisfying a general initial datum can be expressed, via an integral rep-resentation formula, as a convolution of the fundamental solution:

g(x, τ) �∫ +∞

−∞φ(ξ − x, τ)g(ξ, 0)dξ, (6.54)

where

φ(ξ − x, τ) =1√2πτ

exp

{−(ξ − x)2

2τ

}. (6.55)

Hence, in order to find explicitly this solution, we must compute the followingintegral:

g(x, τ) �∫ +∞

−∞

1√2πτ

exp

{−(ξ − x)2

2τ

}max

(e

q+12 ξ − e

q−12 ξ ; 0

)dξ. (6.56)

We remark that the support of the integrand is simply ξ > 0. Due tomonotonicity of the exponential function ( q > 1), we have indeed

eq+12 ξ − e

q−12 ξ > 0 ⇐⇒ ξ > 0. (6.57)

So, by changing the integration domain, the integral can be written in thefollowing way:

g(x, τ) �∫ +∞

0

1√2πτ

exp

{−(ξ − x)2

2τ

}[e

q+12 ξ − e

q−12 ξ]dξ. (6.58)

Let us consider separately the two contributions:

J1 =∫ +∞

0

1√2πτ

exp

{−(ξ − x)2

2τ

}e

q+12 ξdξ


J2 =∫ +∞

0

1√2πτ

exp

{−(ξ − x)2

2τ

}e

q−12 ξdξ.

By introducing the new variable z = (ξ−x)/√

2τ (dz = dξ/√

2τ), the integralJ1 can be written as follows:

J1 =1√2π

∫ +∞

−x/√

2τ

exp(−z2

2

)e

q+12 (x+z

√2τ)dz, (6.59)

by completing the square in the argument of the exponential function, it canbe computed in the following way:

J1 =1√2π

eq+12 x

∫ +∞

−x/√

2τ

e(q+1)2

4 τ exp

{−(z − q+1

2

√2τ)2

2

}dz

=1√2π

[e

q+12 xe

(q+1)2

4 τ

] ∫ +∞

− x√2τ

− q+12

√2τ

exp{−y2

2

}dy,

where we introduced a further integration variable y = z − (q + 1)√

2τ/2(dy = dz). The last integral can be immediately expressed in terms of thestandard normal distribution function, simply observing that:

∫ +∞

−x

exp(−u2

2

)du =

∫ x

−∞exp(−u2

2

)du = N(x). (6.60)

Hence we finally get:

J1 = exp

{q + 1

2x +

(q + 1)2

4τ

}N

(x√2τ

+q + 1

2

√2τ

). (6.61)

The integral J2 can be computed similarly:

J2 = exp

{q − 1

2x +

(q − 1)2

4τ

}N

(x√2τ

+q − 1

2

√2τ

). (6.62)

In order to find the solution to our original problem (the Black-Scholesequation for a European Call option) we need to write our solution in termsof the original variables. From

x = ln(S/K), τ =σ2

2(T − t),

C(S, t) = K · h(x(S), τ (t)),

h(x, τ) = exp{−q − 1

2x − (q + 1)2

4τ

}g(x, τ),


we obtain the well-known formula:

C(S, t) = S · N(d1) − Ke−r(T−t)N(d2), (6.63)

where d1, d2 are defined, as usual, by

d1 =ln(

SK

)+(r + σ2

2

)(T − t)

√T − t

,

d2 =ln(

SK

)+(r − σ2

2

)(T − t)

√T − t

.


Exercise 6.12. In a Black-Scholes setting, consider an option with underly-ing evolving according to the following dynamics:

dSt = μStdt + σStdWt,

with S0 = 16 euros, μ = 0.12 and σ = 0.3. The risk-free interest rate ravailable on the market is 0.04. The time unit is one year.

1. If the payoff of the option considered is

Φ (ST ) =(ln (ST /K)3

)+

=(ln(S3

T

)− 3 lnK)+

with K = 20 euros and maturity T = 1, compute the price of the optiontoday (t = 0) by using the Feynman-Kac representation formula.

2. What is the event with higher probability: to have a strictly positive payofffor the option of item 1. or for a European Call option with the same Kand T? Or are their probabilities the same?

3. Is the probability to get a payoff at least of 10 euros for the option of item1. higher than for a European Call option with the same K and T? Or aretheir probabilities the same?

Exercise 6.13. By solving the Black-Scholes equation, equipped with theproper final condition, find the formula for the European Put option price.

7

Black-Scholes Model for Option Pricingand Hedging Strategies


As in the previous chapters, we consider a market model consisting in twoassets: one non-risky (bond), the other risky (stock). While before we focusedon discrete-time market models, here we introduce the so-called Black-Scholesmodel : a well-known example of continuous-time market model.The key assumption behind the Black-Scholes model is that the stock price

S evolves as a geometric Brownian motion, i.e.

St = S0 exp{(

μ − 12σ2

)t + σWt

}, (7.1)

where S0 is the initial stock price, μ the drift, σ the volatility (or diffusionparameter) and (Wt)t≥0 is a standard Brownian motion. By Itô’s Lemma, theprice of S is described by the following Stochastic Differential Equation:{

dSt = μStdt + σStdWt

S0 = s0.

The bond price B is evaluated by continuous compounding with a risk-freerate r. Namely, if B0 is the initial price of the bond:

Bt = B0 exp(rt) (7.2)

or, equivalently, {dBt = rBtdtB0 = b0

.

Furthermore, the model is based on the following assumptions: perfect liq-uidity (one can buy or sell any real number of an asset), short selling is allowed,transaction costs are never applied, bid and ask prices of any asset coincide.Moreover, the market model is also assumed to be free of arbitrage.


124 7 Black-Scholes Model for Option Pricing and Hedging Strategies

In the Black-Scholes model, it has been proved (see [6], [19], [29] for de-tails) that there exists an equivalent martingale measure useful for the pricingof a derivative and consistent with the no-arbitrage assumption. The equiv-alent martingale measure is also called “risk-neutral”; roughly speaking, infact, under such a measure the underlying evolves (more or less) as it wouldwith a non-risky asset. Under the risk-neutral measure, the underlying priceis described by the following Stochastic Differential Equation:

{dSt = rStdt + σStdWt

S0 = s0.

It was also shown that the Black-Scholes model, as well as the binomialmodel, is complete. The Second Fundamental Theorem of Asset Pricing im-plies therefore the uniqueness of the risk-neutral measure for such a marketmodel. For complete market models, the price of any derivative is uniquelydetermined by the no-arbitrage assumption.Denote by F (St, t) the price (at time t) of a derivative having S as underly-

ing. Such a price can be obtained by solving the following Partial DifferentialEquation (called Black-Scholes equation):

∂F

∂t+

σ2

2∂2F

∂S2+ rS

∂F

∂S− rF = 0, (7.3)

endowed with some suitable final and boundary conditions. Equivalently,F (St, t) can be obtained by the discounted expected value of the payoff ofthe derivative, under the equivalent martingale measure and conditioned byFt (where (Ft)t≥0 is the filtration generated by (St)t≥0). In other words,

F (St, t) = e−r(T−t)EQ[F (ST , T )|Ft]. (7.4)

Black-Scholes Formula for European Call/Put options on stockswithout dividends

Consider a European Call option, whose payoff is given by

C(ST , T ) = max{ST − K; 0} ,

where K stands for the strike of the option and ST for the underlying priceat maturity. The solution of the Black-Scholes PDE (as well as the discountedexpected value of the payoff under the equivalent martingale measure) givesthe following price:

C(St, t) = StN(d1) − Ke−r(T−t)N(d2), 0 ≤ t < T (7.5)


where N(·) denotes the cumulative distribution function of a standard normaland d1, d2 are defined as:

d1 �ln(St/K) + (r + σ2

2)(T − t)

σ√

T − t(7.6)

d2 �ln(St/K) + (r − σ2

2)(T − t)

σ√

T − t= d1 − σ

√T − t (7.7)

For the pricing of a Put option one can proceed similarly by taking into ac-count that the final value (equal to the payoff) of a Put is given by

P (ST , T ) = max {K − ST ; 0} .

An alternative way to evaluate the Put option is based on the relation be-tween the prices of Call and a Put options written on the same underlying,with same strike and maturity. The prices of Call and Put options as above,indeed, satisfy the so-called Put-Call parity :

C(St, t) − P (St, t) = St − Ke−r(T−t). (7.8)

Proceeding as explained above, the price (at time t) of a European Put optionwith maturity T , with strike K and written on the underlying S is given by

P (St, t) = Ke−r(T−t)N(−d2) − StN(−d1), 0 ≤ t < T. (7.9)

Black-Scholes Formula for European Call/Put options on stocks pay-ing “continuous” dividends

The stock S is said to pay “continuous” dividends when the dividend rateD0 is constant in time, that is, the sum D0Sdt is distributed in the infinites-imal interval dt to the owners of the stock.The Black-Scholes formula for a European Call option written on an un-

derlying paying continuous dividends can be obtained by (7.5), by replacingr with the risk-free rate (r − D0) in the expressions of d1 and d2 (d′

1, d′2 will

denote the new values) and by replacing St with Ste−D0(T−t) in the first term

of the formula. This gives

C(St, t) = Ste−D0(T−t)N(d′

1) − Ke−r(T−t)N(d′2), (7.10)

where

d′1 �

ln(St/K) + (r − D0 + σ2

2 )(T − t)σ√

T − t(7.11)

d′2 �

ln(St/K) + (r − D0 − σ2

2 )(T − t)σ√

T − t(7.12)

A similar formula holds for European Put options.


Black-Scholes Formula for European Call/Put options on stocks pay-ing “discrete” dividends

When dividends are not paid “continuously” (as in the previous case wherea dividend rate was applied) but just at fixed times (known a priori), dividendsare said to be “discrete”.The Black-Scholes formula for a European Call option written on a stock

paying discrete dividends is the following:

C (S0, 0) = (S0 − D)N(d1

)− Ke−rT N

(d2

), (7.13)

where D denotes the discounted value of all dividends paid until the optionmaturity T and

d1 �ln ((S0 − D)/K) +

(r + 1

2σ2)T

σ√

T

d2 �ln ((S0 − D)/K) +

(r − 1

2σ2)T

σ√

T= d1 − σ

√T

A similar formula holds for European Put options written on stocks payingdiscrete dividends.

As for options on stocks without dividends, a Put-Call Parity holds forEuropean Call/Put options with the same maturity and strike and written onthe same stock paying discrete dividends:

C (S0, 0)− P (S0, 0) = S0 − D − Ke−rT .

In order to simplify notations, in the followingwe will often write Ct insteadof C (St, t) and Pt instead of P (St, t).

The Black-Scholes model allows not only to compute explicitly the price ofa European Call/Put option on a given underlying S, but also to find the so-called hedging strategy that the seller of a derivative should have to be hedgedagainst the risk assumed. An hedging strategy can be static (sometimes called“buy and hold” strategy) or dynamic. In the first case, the portfolio composi-tion, once built, will be unchanged for the whole duration of the strategy; inthe dynamic case, continuous (or periodic) changes in the composition of theportfolio would be needed.The simplest dynamic hedging strategy is the one called Delta-hedging ,

where an “instantaneously riskless” portfolio is composed by one derivativeand by −Δ (Delta) shares of the underlying (the negative sign denotes a“short” position). The ‘Delta’ of a derivative of value F (St, t) is defined as:

ΔF �∂F

∂S(7.14)


and, obviously, it changes with time. By the Black-Scholes formula, one candeduce that the Delta of a European Call option on a stock without dividendsis given by:

ΔCall = N(d1),

while the Delta of the corresponding Put option is given by

ΔPut = N (d1) − 1.

Notice that ΔCall > 0, while ΔPut < 0.From a theoretical point of view, in order to keep the portfolio riskless

one should be able to modify its composition at any instant of time. Obvi-ously, this is unreasonable and too expensive in practice. Nevertheless, theportfolio should be re-balanced at intervals of time close enough to keep theriskiness under control. When the value of the derivative varies so quicklythat it becomes difficult to control the risk associated to such a variation,Gamma-hedging strategies are used. The ‘Gamma’ Γ of a derivative is de-fined as follows:

ΓF �∂2F

∂S2. (7.15)

In the Black-Scholes model, the Gamma of a European Call/Put option writ-ten on a stock without dividends is given by:

ΓCall = ΓPut =N ′ (d1)

σSt

√T − t

, (7.16)

where N ′(·) denotes the density function of a standard normal.The Gamma-hedging strategy consists in composing a portfolio whose

Gamma is equal to zero. In order to cancel both Delta and Gamma of a port-folio (in such a case the portfolio is said to be Delta and Gamma-neutral),the portfolio is required to be composed by more than two assets (differentlyfrom the Delta-neutral case). We will explain how to apply this idea in theexercises.

Besides Delta and Gamma, that quantify the riskiness of a derivative asso-ciated to changes in the underlying price, the Greek letters ρ (Rho), ν (Vega)and θ (Theta) are used to index the risk associated to changes in the interestrate r, in the volatility σ and in time t, respectively. Because of their symbols,these quantities are collectively known as “Greeks”. They are used to introduceportfolios that are either Rho-neutral, Vega-neutral or Theta-neutral.For a European Call option on a stock without dividends, ρ, ν and θ are

given by

ρ � ∂F

∂r= K(T − t)e−r(T−t)N(d2) (7.17)

ν � ∂F

∂σ= StN

′(d1)√

T − t (7.18)


θ � ∂F

∂t= −σStN

′(d1)2√

T − t− rKe−r(T−t)N(d2). (7.19)

For hedging strategies we will adopt the convention (and approximation)not to discount their future cash-flows. This approximation is supported bythe short duration of such strategies.By composing portfolios formed by suitable shares of the underlyings and

of the derivatives one can achieve different risk profiles, which fit differentgoals and risk attitudes of investors. Some of these tools will be considered inthe exercises below.

For a detailed treatment and further details on the subject, we send theinterested reader to Björk [6], Hull [23] and Wilmott et al. [37], [38], amongmany others.


Exercise 7.1. It is well known that, in market models that are free of arbi-trage, the prices of a European Call and a European Put option written onthe same stock (without dividends) satisfy

C0 ≤ S0 and P0 ≤ Ke−rT , (7.20)

where K stands for the strike of the options, T for the maturity of the optionsand r for the risk-free rate per year.

1. Verify that−Ke−rT ≤ C0 − P0 ≤ S0

and find an arbitrage opportunity in the following cases:

(a) C0 − P0 > S0;(b) C0 − P0 < −Ke−rT .

2. It is well known that, in market models that are free of arbitrage, the fol-lowing Put-Call Parity holds for European options written on the sameunderlying paying discrete dividends:

C0 − P0 = S0 − D − Ke−rT ,

where D stands for the discounted value of all the dividends paid until thematurity T of the options.

(a) Find an arbitrage opportunity when C0 = P0 + S0 − Ke−rT and thediscounted value of dividends is strictly positive (D > 0).

(b) Show that, under the no-arbitrage assumption, C0 > S0 − D can neveroccur.


Solution

1. From inequalities (7.20) it follows immediately that

C0 − Ke−rT ≤ C0 − P0 ≤ S0 − P0.

Since C0, P0 ≥ 0 (see below), we deduce

−Ke−rT ≤ C0 − Ke−rT ≤ C0 − P0 ≤ S0 − P0 ≤ S0,

that is, the claim.Let us verify that C0, P0 ≥ 0. Suppose by contradiction that C0 < 0 andconsider the following strategy:

t = 0 t = T

buy the Call⇒ −C0 > 0 (ST − K)+

invest (−C0) at rate r⇒ +C0 −C0e

rT > 0

−C0 + C0 = 0 (ST − K)+ − C0erT > 0

Consequently, if C0 were strictly negative then there would exist arbitrageopportunities (for instance, the one built above). This would contradict theno-arbitrage assumption, hence it follows that C0 ≥ 0.Similarly, one can check P0 ≥ 0.

(a) Assume now that C0 > P0 + S0.As shown below, the following strategy represents an arbitrage oppor-tunity.

t = 0 t = T

sell the Call⇒ +C0 − (ST − K)+

buy one stock and one Put⇒ −S0 − P0 ST + (K − ST )+

invest (C0 − S0 − P0)⇒ − (C0 − S0 − P0) (C0 − S0 − P0) erT

C0 − S0 − P0 − (C0 − S0 − P0) = 0 K + (C0 − S0 − P0) erT > 0

(b) Suppose now that C0 − P0 < −Ke−rT , from which C0 − P0 < 0.As checked below, the following strategy represents an arbitrage oppor-tunity.


t = 0 t = T

sell the Put⇒ +P0 − (K − ST )+

buy the Call⇒ −C0 (ST − K)+

invest (P0 − C0)⇒ − (P0 − C0) (P0 − C0) erT

P0 − C0 − (P0 − C0) = 0 ST − K + (P0 − C0) erT > ST > 0

2. Note, first, that the Put-Call Parity (in the presence of dividends) can berewritten as

C0 + D = S0 + P0 − Ke−rT .

(a) It is clear that if we had C0 = P0 + S0 − Ke−rT for D > 0, then therewould exist arbitrage opportunities.Suppose indeed that C0 = P0 + S0 − Ke−rT and consider the followingportfolio:• one long position in the stock;• one long position in the Put;• one short position in the Call;• bank loan of Ke−rT .

The value V0 of such a portfolio at t = 0 is then equal to

V0 = −S0 − P0 + C0 + Ke−rT = 0,

while its value VT at maturity T of the two options is equal to

VT = ST + DerT + (K − ST )+ − (ST − K)+ − K = DerT > 0.

The portfolio above represents then an arbitrage opportunity.

(b) Suppose now that C0 > S0 − D.The following strategy is an arbitrage opportunity.

t = 0 t = T

sell the Call⇒ +C0 − (ST − K)+

buy one share of the underlying⇒ −S0 ST + DerT

invest/borrow (C0 − S0)⇒ − (C0 − S0) (C0 − S0) erT

C0 − S0 − (C0 − S0) = 0 − (ST − K)+ + ST

+DerT + (C0 − S0) erT


In fact, the initial value of such a strategy is zero, while its final valueequals

VT = − (ST − K)+ + ST + DerT + (C0 − S0) erT

= min{ST ; K} + (C0 − S0 + D) erT

≥ (C0 − S0 + D) erT > 0,

where the last inequality is due to the initial assumption C0 > S0 − D.

Exercise 7.2. Consider a stock (without dividends) whose price evolves as inthe Black-Scholes model with drift μ of 10% per year, volatility σ of 40% peryear and current price of S0 = 16 euros. The risk-free interest rate r on themarket is 4% per year.

1.(a)Compute the initial price of a European Call option on the stock above,with strike of 18 euros and maturity T of one year.

(b)Deduce the price of the corresponding Put option.2. Suppose now that in 6 months’ time the stock will cost 16.4 euros. Discusswhether it is convenient to wait 6 months before buying the Call optionabove and investing (at risk-free rate) what we would have paid for buyingthe Call at the initial time.What if the stock costed 19.2 euros in 6 months from now?

3. Discuss whether it is possible to establish a priori some bounds on the ini-tial prices of the Call and the Put option. If yes, verify that such boundshold in the present case.

4. Consider the same Call option as above but written on a stock with S0 = 16euros, paying a dividend of 2 euros in 4 months and of 4 euros in 8 months.

(a) Compute the initial price of such an option and compare it with theprice of the Call option (of item 1.) on the stock without dividends.Determine whether the option price on the stock paying dividends isgreater than the one not paying dividends.

(b) Discuss whether, in general, the price of a Call option on a stock withdividends could be greater than the same without dividends. Justify theresult by means of no-arbitrage arguments.

Solution

1.(a) Recall that the Black-Scholes formula for European Call options on stockwithout dividends is the following:

C0 = S0N (d1) − Ke−rT N (d2) ,

where N denotes the cumulative distribution function of a standard nor-mal and

d1 =ln (S0/K) +

(r + 1

2σ2)T

σ√

T; d2 = d1 − σ

√T .


In the present case, we get

d1 =ln (16/18) +

(0.04 + 1

2 (0.4)2)

0.4= 0.0055

d2 = 0.0055− 0.4 = −0.3945,

since T = 1 year.The initial price of the European Call option above is then equal to

C0 = 16 · N (0.0055)− 18 · e−0.04 ·N (−0.3945) = 2.04 euros.

(b) Applying the Put-Call Parity (for European options with same matu-rity and strike, written on the same stock without dividends), we deducethat

P0 = C0 − S0 + Ke−rT = 2.04− 16 + 18 · e−0.04 = 3.34 euros.

We would obtain the same result by applying again the Black-Scholesformula in the case of European Put options on stocks without dividends.In such a case, indeed,

P0 = Ke−rT N (−d2) − S0N (−d1) .

2. We compute now the price of the Call option at t = 6 months when weassume that S

(1)6m = 16.4 euros. Under this assumption,

d(1)1 =

ln(S

(1)6m/K

)+(r + 1

2σ2)(T − t)

σ√

T − t

=ln (16.4/18)+

(0.04 + 1

2 (0.4)2)

12

0.4 ·√

12

= −0.117

d(1)2 = d

(1)1 − 0.4 ·

√12

= −0.40,

hence

C(1)6m = 16.4 · N

(d(1)1

)− 18 · e−0.04·0.5N

(d(1)2

)= 1.356 euros.

It would be convenient to wait 6 months before buying the Call option ofitem 1. and investing (at risk-free rate) what we would have paid for buyingthe Call at the initial time depending on whether

C0 · er/2 > C(1)6m.


Since C0 · er/2 = 2.04 · e0.04·0.5 = 2.08 > C(1)6m = 1.356, we conclude that it

would be convenient to wait before buying the Call option.

Proceeding as above, we obtain that for S(2)6m = 19.2 euros

d(2)1 =

ln (19.2/18)+(0.04 + 1

2 (0.4)2)

12

0.4 ·√

12

= 0.440

d(2)2 = d

(2)1 − 0.4 ·

√12

= 0.157

and

C(2)6m = 19.2 · N

(d(2)1

)− 18 · e−0.04·0.5N

(d(2)2

)= 2.94 euros.

Since C0 · er/2 = 2.04 · e0.04·0.5 = 2.08 < C(2)6m = 2.94, we conclude that in

such a case it would not be convenient to wait 6 months to buy the option.

3. As well known, prices of European options written on stocks without divi-dends fulfill the following inequalities:

S0 − Ke−rT ≤ C0 ≤ S0

Ke−rT − S0 ≤ P0 ≤ Ke−rT

In the present case, we deduce immediately that

0 = max{−1.29; 0} = max{S0 − Ke−rT ; 0} ≤ C0 ≤ S0 = 161.29 = max{1.29; 0} = max{Ke−rT − S0; 0} ≤ P0 ≤ Ke−rT = 17.29.

The previous bounds are then fulfilled by the initial price of the Call andthe Put.

4. Consider now a stock paying a dividend of 2 euros in 4 months and of 4euros in 8 months.

(a) We recall that the Black-Scholes formula for European options writtenon stocks paying discrete dividends is the following:

C0 = (S0 − D) N(d1

)− Ke−rT N

(d2

),

where N denotes as usual the cumulative distribution function of a stan-dard normal and

d1 =ln ((S0 − D)/K) +

(r + 1

2σ2)T

σ√

T

d2 = d1 − σ√

T .


Since the discounted dividends amount to

D = 2 · e−0.04· 412 + 4 · e−0.04· 8

12 = 5.87

and

d1 =ln ((16 − 5.87)/18)+

(0.04 + 1

2(0.4)2

)0.4

= −1.137

d2 = −1.137− 0.4 = −1.537,

it follows that the initial price of the new Call option equals

C0 = (16 − 5.87) · N (−1.137)− 18 · e−0.04N (−1.537) = 0.22 euros.

Consequently, the price of the Call “without dividends” is greater thanthe other one. More precisely, the difference between the two pricesamounts to C0 − C0 = 2.04− 0.22 = 1.82 euros.

(b) In general, we cannot have C0 > C0. Suppose by contradiction thatC0 > C0 and consider the following strategy, where ST stands for theprice (at time T ) of the stock that does not pay any dividend:

t = 0 t = T

buy the Call without dividends⇒ −C0 (ST − K)+

sell the Call paying dividends⇒ +C0 − (ST − DerT − K

)+invest

(C0 − C0

)⇒ −

(C0 − C0

) (C0 − C0

)erT

C0 − C0 −(C0 − C0

)= 0 VT

Since(ST − DerT − K

)+ ≤ (ST − K)+, we deduce that the final valueVT of the strategy above is

VT = − (ST − DerT − K)+

+ (ST − K)+ +(C0 − C0

)erT

≥(C0 − C0

)erT > 0.

If C0 > C0 at any time, there would exist an arbitrage opportunity (e.g.the one just built). Under the no-arbitrage assumption, then, necessarilyC0 ≤ C0.


Exercise 7.3. Consider the same stock A – paying discrete dividends– andthe same European Call option of Exercise 7.2, item 4.The risk-free interest rate r available on the market is of 4% per year.

1. Consider a Call option with strike of 18 euros, maturity of one year andwritten on a stock B with the same features as stock A but paying contin-uous dividends (with dividend rate d).Find the dividend rate d such that the price of the Call option written onthe stock A paying discrete dividends coincides with the price of the Calloption above written on stock B.

2. Compute the discounted value of the dividends paid by stock B.

Solution

1. From Exercise 7.2 we already know that the price of the Call option writtenon stock A (paying discrete dividends), with strike of 18 euros and maturityof one year is equal to C0 = 0.22 euros.Furthermore, we remind that the price of a Call with strike of 18 euros,maturity of one year and written on stock B (paying continuous dividends)is given by

Cd0 = S0e

−dT N(d′1) − Ke−rT N(d′

2),

where

d′1 =

ln(S0/K) + (r − d + σ2

2 )T

σ√

T

d′2 =

ln(S0/K) + (r − d − σ2

2)T

σ√

T.

Hence

Cd0 = 16 · e−d · N

(ln(16/18) + 0.12− d

0.4

)

−18 · e−0.04 · N(

ln(16/18)− 0.04− d

0.4

).

We look then for a dividend rate d so that

Cd0 = C0 = 0.22 euros.


Numerically, we get

Cd0

d = 0.1 1.3505d = 0.2 0.8586d = 0.3 0.5232d = 0.4 0.3050

d = 0.45 0.2289d = 0.454 0.2236

d= 0.4567 0.2200d = 0.457 0.2196

d = 0.46 0.2158d = 0.5 0.1697

We can conclude that the “implicit” dividend rate we are looking for isd ∼= 0.4567.

2. Since the price of stock B evolves as

SBt = SB

0 · exp{(

r − d− 12σ2

)t + σWt

}

with SB0 = S0 = 16, we deduce that the discounted value of all dividends

paid by stock B is equal to

D∗ = S0

(1 − e−dT

)= 16 · (1− e−0.4567

)= 5.866 euros.

As expected, D∗ is approximately equal to the discounted value D of alldividends paid by stock A (where D = 5.87, see Exercise 7.2).

Exercise 7.4. Consider a stock whose price evolves as in the Black-Scholesmodel, with volatility σ = 0.32 (per year) and initial price S0 = 30 euros.Such a stock pays a dividend of one euro in 3 months and of one euro in 9months. The risk-free interest rate available on the market is of 4% per year.

1. Compute the price of a European Call option with maturity of one year,strike of 25 euros and written on the stock above.

2. What is the price of the corresponding Put option?3. Discuss whether the price would change if the risk-free interest rate de-creased from 4% to 2% per year. What would such a change affect?

4. Suppose now that the risk-free interest rate is equal to 4% per year. Com-pute the new price of the stock of item 1. “adjusted” by the discounted valueof dividends, i.e. obtained by detracting the discounted value of dividends.

5. Compare the price of the Call of item 1. to the price of the Call with strikeof 25 euros, maturity of one year and written on the stock without divi-dends, with volatility σ = 0.32 (per year) and with the current price foundin item 4.


Solution

1. The discounted value D of all dividends of the stock amounts to

D = 1 · e−0.04· 312 + 1 · e−0.04· 9

12 = 1.96 euros.

Recall that, by the Black-Scholes formula, the price of a European Calloption written on a stock paying discrete dividends is given by

C0 = S0 · N(d1

)− Ke−rT · N

(d2

),

where S0 = S0 − D and

d1 =ln(S0/K

)+(r + 1

2σ2)T

σ√

T

d2 = d1 − σ√

T .

In the present case, we get

d1 =0.2060.32

= 0.644

d2 = d1 − 0.32 = 0.324

and, accordingly,

C0 = 28.04 · N (0.644)− 25 · e−0.04 ·N (0.324)= 28.04 · 0.74− 24.02 · 0.627 = 5.69 euros.

2. By the Put-Call parity on stocks paying dividends, we obtain immediatelythe Put price:

P0 = C0 − S0 + D + Ke−rT

= 5.69− 30 + 1.96 + 24.02 = 1.67 euros.

3. Changes in the interest rate affect the discounted value of the dividendspaid and the strike, as well as d1 and d2 in the Black-Scholes formula.If the risk-free interest rate was r∗ = 0.02, the new discounted value D∗ ofthe dividends paid by the stock would amount to

D∗ = 1 · e−0.02· 312 + 1 · e−0.02· 9

12 = 1.98.

Hence, we would get

d∗1 =

ln ((S0 − D∗) /K) +(r∗ + 1

2σ2)T

σ√

T= 0.579

d∗2 = d∗

1 − σ√

T = 0.259


and, accordingly,

C∗0 = (S0 − D∗)N (d∗

1) − Ke−r∗T N (d∗2)

= 28.02 ·N (0.579)− 25 · e−0.02 · N (0.259)= 5.38 euros.

4. In case r∗ = 0.04, the current stock price adjusted by the discounted valueof the dividends would be

S0 = S0 − D = 28.04 euros.

5. The price of the Call written on the stock paying the dividends of item 1.is equal to 5.69 euros.If we computed the price of a Call option with maturity of one year, strikeof 25 euros and written on a stock with volatility σ = 0.32 (per year) andcurrent price S0 = 28.04 euros, we would find exactly the same price as theone of the Call of item 1.

Exercise 7.5. Consider a market model where it is possible to trade on astock (with current price S0 = 20 euros) and on different European Call andPut options written on such a stock.The price of the stock follows a geometric Brownian motion, namely St =

S0 exp{(μ − 12σ2)t + σWt} with S0 = 20, μ ∈ R and σ > 0.

Our goal is an investment with one year as horizon of time.

1. Consider two European Call options (both with maturity T = 1 year andwritten on the stock above), with strikes of

K1 = 20 eurosK2 = 40 euros

and prices

C10 = 6 euros

C20 = 2 euros,

respectively.Establish under which conditions on μ and σ > 0 the profit (in one year)due to a bear spread1 formed by the previous options is non-negative withprobability at least of 50%.

1 A spread is a trading strategy composed by two or more options of the same kind,that is all options are either European Calls or European Puts written on the sameunderlying.Among the different spreads, the so-called “bear spread” is used when a decrease

in the stock price is expected. For Call options, a bear spread is obtained by sellingthe Call option with lower strike and by buying the Call with higher strike.


2. Compare the profit due to the bear spread with profit arising from thefollowing strategy:

• short position in a Call with strike K1 (the lower strike);• short selling a share of the underlying at the beginning and giving itback at maturity.

Establish which strategy is most convenient as the price of the underlyingdecreases to 16 euros. What if the price increases to 32 euros?

3. Suppose that on the market it is also possible to buy/sell Call options onthe same underlying as before but with different strikes (in particular, wecould think that options of any strike between K1 and K2 are available).Assume that on this market (where arbitrage opportunities might exist)the prices of the options above decrease linearly from the price of the op-tion with strike K1 = 20 to the price C∗ = 4 euros for the option withstrike K∗ = 28. Furthermore, the prices decrease linearly from the price ofthe option with strike K∗ to that of the option with strike K2.

(a) Build a “butterfly spread”2 using the options above, which necessarilyrelies on the Call options with strikes K1 and K2.What is the profit when we use a butterfly spread instead of the previousstrategy (see item 2.)?

(b) Suppose now that the stock price in one year is distributed as in item 1.with μ = 0.24 and σ = 0.48. Compute the probability with which it ismore convenient to use the butterfly spread instead of the bear spread.

Solution

1. By assumption, the price (at time T ) of the underlying is given by

ST = S0 exp{(

μ − 12σ2

)T + σWT

}

with S0 = 20 and T = 1 year.We need to find under which conditions on μ and σ the profit (in one year)due to the bear spread is non-negative with probability at least of 50%. Inother words, μ and σ have to satisfy the following inequality

P ({profit of the bear spread ≥ 0}) ≥ 0.5. (7.21)

Then we have to analyze the random variable representing the profit dueto the bear spread.

2 The so-called “butterfly spread” is used when we expect that the stock price re-mains more or less stable. A butterfly spread for Call options can be obtained bybuying one Call option with lower strike, buying one Call option with higher strikeand selling two Call options with intermediate strike.


Recall that a bear spread can be built – by means of Call options– asfollows:

(A) sell the Call option with lower strike (K1, in the present case)(B) buy the Call options with higher strike (K2)

The payoff of a bear spread is then given by:

(A) (B) Total Payoff

if ST ≤ K1 0 0 0

if K1 < ST ≤ K2 − (ST − K1) 0 K1 − ST

if ST > K2 − (ST − K1) ST − K2 K1 − K2

since the payoff of a Call option is (ST − K)+ for the buyer and− (ST − K)+ for the seller. Consequently, the profit (due to the payoffand to the Call prices) is given by:

Total Profit

if ST ≤ 20 C10 − C2

0 = 4

if 20 < ST ≤ 40 K1 − ST + C10 − C2

0 = 24 − ST

if ST > 40 K1 − K2 + C10 − C2

0 = −16

Figure 7.1 shows the profit due to the bear spread on varying the underlyingprice.

-20

-15

-10

-5

0

5

0 10 20 30 40 50 60 70 80

underlying price

prof

it o

f th

e be

ar s

prea

d

Fig. 7.1. Profit of the bear spread


Condition (7.21) reduces then to

0.5 ≤ P ({profit ≥ 0})= P ({0 ≤ ST ≤ 20} ∪ {20 < ST ≤ 40; 24− ST ≥ 0}) .

By the arguments above and by the assumptions on ST , we deduce that

P ({profit ≥ 0}) = P (0 ≤ ST ≤ 24)

= P(20e(μ− 1

2 σ2)T+σWT ≤ 24)

= P

((μ − 1

2σ2

)+ σW1 ≤ ln

(2420

))

= P

(W1 ≤ ln (1.2)− (μ − 1

2σ2)

σ

)

= N

(ln (1.2) − (μ − 1

2σ2)

σ

),

because W1 is distributed as a standard normal.

Since μ and σ have to verify N

(ln(1.2)−(μ− 1

2 σ2)σ

)≥ 0.5, we obtain

ln (1.2) − (μ − 12σ2)

σ≥ 0.

The condition on μ and σ (> 0) becomes then

μ ≤ 12σ2 + ln (1.2) .

Let us compare the profit due to the bear spread with the profit due to thefollowing strategy:

• short position in the Call with strike K1 (the lower strike);• short selling a share of the underlying at the beginning and giving itback at maturity.

The profit (at maturity) of the strategy above is then given by:

Profit due to the Call Profit due to the stock Total Profit

if ST ≤ K1 C10 S0 − ST 26− ST

if ST > K1 C10 − (ST − K1) S0 − ST 46 − 2ST

The comparison between the strategy just considered and the bear spreadis manifest from Figure 7.2.


-140

-120

-100

-80

-60

-40

-20

0

20

40

0 10 20 30 40 50 60 70 80

underlying price

pro

fit

bear spread

alternative strategy

Fig. 7.2. Comparison between the profit of a bear spread and the profit of ourstrategy

Notice that “our strategy” is very convenient when the stock price de-creases, while it is very risky and could cause big losses when the stockprice increases steeply.If the stock price decreased to 16 euros, we would have a total profit of10 euros with our strategy, while a profit of 4 euros with the bear spread,making our strategy more convenient.If the stock price increased to 32 euros, we would have a total loss of 18euros with our strategy, while a loss of 8 euros with the bear spread. Thebear spread would be thus more favorable.

2. First, remind that a butterfly spread can be built – by means of Call op-tions – as follows:

(A) buy the Call with lower strike (K1);(B) buy the Call with higher strike (K2);(C) sell 2 Calls with intermediate strike (namely, with strike K3 = K1+K2

2 ).

(a) By the initial assumptions, we obtain that the prices of the Call optionswith strike between K1 and K2 are given by

Ci0 =

{6 − Ki−20

4 ; if 20 ≤ Ki ≤ 28

4 − Ki−286 ; if 28 ≤ Ki ≤ 40

.

In the present case, then, the butterfly spread would be built as explainedbefore where the Call options in (C) have strike K3 = K1+K2

2 = 30 andprice C3

0 = 4 − 30−286 = 11

3 .


The payoff of the butterfly spread is then given by

(A) (B) (C) Total Payoff

if ST ≤ K1 0 0 0 0

if K1 < ST ≤ K3 ST − K1 0 0 ST − K1

if K3 < ST ≤ K2 ST − K1 0 − 2 (ST − K3) K2 − ST

if ST > K2 ST − K1 ST − K2 − 2 (ST − K3) 0

Consequently, the profit (due to the payoff and the Call prices) is givenby:

Total Profit

if ST ≤ 20 2C30 − C1

0 − C20 = −2

3

if 20 < ST ≤ 30 ST − K1 + 2C30 − C1

0 − C20 = ST − 62

3

if 30 < ST ≤ 40 K2 − ST + 2C30 − C1

0 − C20 = 118

3 − ST

if ST > 40 2C30 − C1

0 − C20 = −2

3

By comparing the profit of the strategy of item 2. to the profit of the but-terfly spread, we deduce that the latter is more advantageous when thestock price is close to the intermediate strike, while it is less advantageouswhen the stock price goes under 20 euros. When the underlying price in-creases, however, the butterfly spread becomes more advantageous than“our strategy”, since loss is bounded with the butterfly spread.

(b) In order to compute the probability for which the butterfly spread ismore convenient than the bear spread, we need to compare the butterflyprofit (see Fig. 7.3) with the bear profit.It is evident that for ST ≤ 20 the profit due to the bear is greater thanthe profit of the butterfly; vice versa for ST ≥ 40. For 20 < ST < 40 theprofit of the bear is equal to 24− ST while the profit of the butterfly isequal to ST − 62

3for 20 < ST < 30 and 118

3− ST for 30 ≤ ST < 40.

So, we deduce that

for ST ≤ 20: the bear is the more advantageous

for 20 < ST < 673 : the bear is the more advantageous

for ST = 673 : indifferent

for 673 < ST < 40: the butterfly is the more advantageous

for ST ≥ 40: the butterfly is the more advantageous


-2

0

2

4

6

8

10

0 10 20 30 40 50 60 70 80

underlying price

pro

fit

of

the

bu

tter

fly

spre

ad

Fig. 7.3. Profit of the butterfly spread

Consequently, the probability for which it is more advantageous to usea butterfly spread than a bear spread is equal to

P

(ST >

673

)= P

(S0 exp

{(μ − 1

2σ2

)T + σWT

}>

673

)

= P

((μ − 1

2σ2

)T + σWT > ln

(673S0

))

= P

⎛⎝WT >

ln (67/60)−(0.24− 1

2 (0.48)2)

0.48

⎞⎠

= 1 − N

⎛⎝ ln (67/60) −

(0.24− 1

2 (0.48)2)

0.48

⎞⎠

= 0.512.

Notice that μ = 0.24 and σ = 0.48 fulfill the condition found in item 1.,hence the profit due to the bear spread is non-negative with probabilityat least of 50%.

Exercise 7.6.We take a short position in a European Call option on a stock“XYZ” with strike K1 = 8 euros and maturity T = 1 year. The price of theunderlying follows a geometric Brownian motion with S0 = 8 euros, μ = 20%and σ = 40% per year. The risk-free interest rate available on the market isr = 4% per year.

1.(a)Compute the Delta (Δ1) of the Call option and, accordingly, establishhow many shares of stock “XYZ” we need to buy/sell in order to makeour short position Delta-neutral.


(b)What if our short position was in a Put option (instead of a Call option)?(c)Find the probability for which the seller of the Call has a (net) lossby the Delta-neutral portfolio built in item (a), given that the Call isexercised.

2. Discuss whether we can make a Delta-neutral short position in a Call op-tion (as the one above) but with maturity T ∗ = 2 years instead of T = 1year.

3. Suppose that another Call option on the same underlying “XYZ” but withstrike K2 = 12 euros is available on the market.

(a) Compute Delta and Gamma (Δ2 and Γ2) of the new option.(b) Establish how one could make Delta- and Gamma-neutral the portfolio

with a short position in the first Call option. Is it possible for the Delta-and Gamma-neutral portfolio to contain a short position in the secondCall?

(c) We take a short position in the Call option of item 1. Explain how tomake such a portfolio Gamma-neutral and with a total Delta smaller orequal (in absolute value) to 1. How many shares of the underlying couldwe buy at most?

4. Consider the portfolio composed by a short position in the Call (of item 1.)and by two long positions in the Put (of item 1.). Verify whether it is pos-sible to make it Delta-, Gamma- and Vega-neutral if we dispose of a thirdCall option on the same underlying but with strike K3 = 6 euros. If yes,explain how.

SolutionRemind that:

• once a time t is fixed, the Delta of an option is defined by Δ = Δt �∂F∂S

(St, t), where F (St, t) stands for the price at time t of the option withunderlying S;

• the Delta of the underlying (ΔS) is equal to 1;• a portfolio is said to be Delta-neutral if the Delta of the portfolio is zero;• for European options written on a stock without dividends:

ΔCall = N (d1) > 0ΔPut = N (d1) − 1 < 0.

1. (a) By the arguments above, we obtain immediately that

Δ1 = N (d1) = N

(ln (S0/K1) +

(r + 1

2σ2)T

σ√

T

)

= N

⎛⎝ ln (1) +

(0.04 + 1

2 (0.4)2)

0.4

⎞⎠ = N (0.3) = 0.618.


Since our goal is to make a short position in the Call option Delta-neutral, we need to find the number x of shares of the underlying to bebought (if x > 0) or sold (if x < 0) in order to make the portfolio ofstocks-option Delta-neutral. We are then looking for x verifying

−1 · Δ1 + x ·ΔS = 0,

where the sign “-” is due to the short position in the option.Since ΔS = 1, x has to solve

−0.618 + x = 0.

So, x = 0.618. We conclude that to make a short position in the Calloption Delta-neutral we need to buy 0.618 shares of the underlying.

(b) Suppose now that the short position to be “hedged” is in a Put optioninstead of a Call option. In order to make it Delta-neutral we look for anumber x of shares of the underlying to be bought (if x > 0) or sold (ifx < 0) satisfying

−1 · Δ1,Put + x ·ΔS = 0−1 · Δ1,Put + x = 0.

Since Δ1,Put = Δ1,Call − 1 = 0.618− 1 = −0.382, to make the portfolioabove Delta-neutral we need to sell 0.382 shares of the underlying.

(c)We have to compute the probability that the seller of the Call has a netloss by the Delta-neutral portfolio built in item (a), given that the Calloption is exercised. This reduces to compute the following conditionalprobability

P(

C0 − (ST − K1)+ + x (ST − S0) ≤ 0

∣∣∣ST ≥ K1

).

First, let us compute the initial price C0 of the Call option. By item (a),we already know that d1 = 0.3, hence d2 = d1 − σ

√T = −0.1 and

C0 = S0 · N (d1) − K1e−rT · N (d2)

= 8 · N (0.3)− 8 · e−0.04 · N (−0.1) = 1.407 euros.

It follows that the probability we are looking for equals

P(

C0 − (ST − K1)+ + x (ST − S0) ≤ 0

∣∣∣ST ≥ K1

)

=P(ST ≥ K1; ST ≥ C0+K1−xS0

1−x

)P (ST ≥ K1)


=P(ST ≥ C0+K1−xS0

1−x

)P (ST ≥ K1)

=P(WT ≥ 1

σ

[ln(

C0+K1−xS0(1−x)S0

)− (μ − 1

2σ2)T])

P(WT ≥ 1

σ

[ln(

K1S0

)− (μ − 1

2σ2)T])

=1 − N

(1σ

[ln(

C0+K1−xS0(1−x)S0

)− (μ − 1

2σ2)])

1 − N(

1σ

[ln(

K1S0

)− (μ − 1

2σ2)])

= 0.419.

2. For a Call option as in item 1.(a) but with maturity T ∗ = 2 years (insteadof T = 1) the Delta would be equal to

Δ(T∗=2)1 = N

⎛⎝ ln (S0/K1) +

(r + σ2

2

)T ∗

σ√

T ∗

⎞⎠

= N

⎛⎝ ln (1) +

(0.04 + 1

2 (0.4)2)· 2

0.4√

2

⎞⎠ = 0.664.

It is then easy to check that for Delta-neutrality we need to buy 0.664shares of the underlying.

3. Remember that:

• the Gamma of the underlying (ΓS) is zero;• a portfolio is said to be Gamma-neutral if the Gamma of the portfoliois zero.

Since the Gamma of the underlying is zero, to make the initial portfolioGamma-neutral we should buy or sell other options.

(a) We start computing Δ2 and Γ2 (Delta and Gamma of the new option)as well as Γ1 (Gamma of the Call option we are selling).

Proceeding as in item 1., we get

Δ2 = N (d1) = N

(ln (S0/K2) +

(r + 1

2σ2)T

σ√

T

)

= N

⎛⎝ ln (8/12) +

(0.04 + 1

2 (0.4)2)

0.4

⎞⎠ = N (−0.71) = 0.238.


To compute Gamma, we recall that for European Call or Put options inthe Black-Scholes model we have

ΓCall = ΓPut =N ′ (d1)S0σ

√T

=1√

2πS0σ√

Te−

d212 .

Hence

Γ1 = =1√

2π · 8 · 0.4e−

(0.3)2

2 = 0.12

Γ2 =1√

2π · 8 · 0.4e−

(−0.71)2

2 = 0.097

(b) First of all, let us make our portfolio Gamma-neutral. We will subse-quently make it Delta-neutral by buying or selling a suitable number ofshares of the underlying. Notice that it is necessary to respect this order.Indeed, making Gamma-neutral a portfolio that is already Delta-neutralmight generate a portfolio that, in the end, is no more Delta-neutral.The idea is then to find a number y of options of the second kind to bebought/sold, that makes the portfolio Gamma-neutral. Namely, we arelooking for y solving

−1 · Γ1 + y · Γ2 = 0, (7.22)

where the sign “-” is due to the short position in the Call option. Theequation above becomes then

−0.12 + y · 0.097 = 0.

Hence y = 1.237. In order to make the portfolio Gamma-neutral we haveto buy 1.237 options with strike K2.Since Γ1, Γ2 > 0 and y has to solve (7.22), it is evident that to make ourportfolio Gamma-neutral it is necessary to take a long position in thesecond Call.In order to make the portfolio Delta-neutral as well, we need to find anumber x of shares of the underlying to be bought or sold that makesthe new stocks-options portfolio Delta-neutral, i.e. such that

−1 · Δ1 + 1.237 ·Δ2 + x = 0−0.618 + 1.237 · 0.238 + x = 0.

It follows that x = 0.324. To summarize: in order to make our portfolioDelta- and Gamma-neutral we have to buy 1.237 Call options with strikeK2 and 0.324 shares of the underlying.

(c) Our aim is to make our portfolio Gamma-neutral and with Delta smalleror equal (in absolute value) to 1. We have therefore to look for a numbery of Call options of the second kind and for a number x of shares of the


underlying satisfying the following conditions{ −1 · Γ1 + y · Γ2 = 0−1 ≤ −1 · Δ1 + y · Δ2 + x ≤ 1

Consequently, {y = 1.237

−0.676 ≤ x ≤ 1.324As we would expect, the number of shares of the underlying to be boughtto make the portfolio Delta-neutral (equal to x = 0.618, see item 1.(a))belongs to the interval above. Furthermore, the number of shares thatwe are allowed to buy so to respect the constraints above is 1 (and nomore).

4. In the following we will make our portfolio Delta-, Gamma- and Vega-neutral.Remember that:

• the Vega of the underlying is zero;• a portfolio is said to be Vega-neutral if the Vega of the portfolio is null.

First of all, let us make our portfolio Gamma- and Vega-neutral. Onlyafterwards we will make it Delta-neutral by buying or selling a suitablenumber of shares of the underlying.The idea is then to find a number y of options of the second kind and zof the third kind to be bought or sold to make the portfolio Gamma- andVega-neutral.Let us compute Δ, Γ and ν of the three options available:

Δ3 = N (d1) = N

(ln (S0/K3) +

(r + 1

2σ2)T

σ√

T

)

= N

⎛⎝ ln (8/6) +

(0.04 + 1

2(0.4)2

)0.4

⎞⎠ = N (1.02) = 0.846

Γ3 =1√

2π · 8 · 0.4e−

(1.02)2

2 = 0.07

It is well known that the Vega of European Call or Put options in theBlack-Scholes model is given by

νCall = νPut = S0

√TN ′ (d1) = S2

0σΓT.

Hence, in our setting we get

ν1 = 82 · 0.4 · Γ1 = 3.072ν2 = 82 · 0.4 · Γ2 = 2.483ν3 = 82 · 0.4 · Γ3 = 1.792


In order to make our portfolio Gamma-neutral and Vega-neutral y and zhave to solve the following system of linear equations{−Γ1,Call + 2Γ1,Put + y · Γ2 + z · Γ3 = 0

−ν1,Call + 2ν1,Put + y · ν2 + z · ν3 = 0{Γ1 + y · Γ2 + z · Γ3 = 0ν1 + y · ν2 + z · ν3 = 0{y = −1.24z = −0.000069 ,

because Γ1,Call = Γ1,Put = Γ1 and ν1,Call = ν1,Put = ν1.Finally, to make such a portfolio Delta-neutral we have to find a number xof shares of the underlying to be bought or sold so that the new portfolioof stocks-options becomes Delta-neutral. This means that x, y and z haveto solve

−Δ1,Call + 2Δ1,Put + y · Δ2 + z · Δ3 + x = 0.

It follows that x = 1.68.To conclude: in order to make our position Delta-, Gamma- and Vega-neutral we have to: sell 1.24 options with strike K2; sell 0.000069 optionswith strike K3; and buy 1.68 shares of the underlying.

Exercise 7.7. Suppose we want to hedge a European Call option written onan underlying stock with dynamics described by a geometric Brownian mo-tion with null drift by means of the following strategy (known as stop-lossstrategy): whenever the stock price is greater than the strike K we buy oneunit of the stock, and we sell it whenever the stock price becomes smaller thanthe strike. By applying the Tanaka formula verify that this strategy cannotreplicate the European Call option payoff.

SolutionIn order to replicate the Call option, we should determine an adapted pro-

cess (ht)t∈[0,T ] such that:

(ST − K)+ = (S0 − K)+ +∫ T

0

htdSt. (7.23)

The stop-loss strategy can be expressed as follows:

ht � 1(K,+∞)(St).

By applying the Tanaka formula to the Call option payoff, as in Chap. 5,we get:

(ST − K)+ = (S0 − K)+ +∫ T

0

1(K,+∞)(St)dSt +12

∫ T

0

σ2S2t δK(St)dt.


The last term in this expression does not vanish, unless the stock never as-sumes the value K (in which case the strategy becomes trivial) and, althoughthe process (ht)t∈[0,T ] is adapted to the natural filtration generated by theBrownian motion, it does not allow to write the option payoff in the desiredform (7.23). We can conclude that the stop-loss strategy cannot be a replicat-ing strategy for a European Call option.


Exercise 7.8. Consider a stock paying a dividend of 2 euros in 3 months andof 3 euros in 4 months. The stock price follows a geometric Brownian motionwith current price of 36 euros and volatility σ = 0.28 (per year).

1. Compute the discounted value of all dividends paid when the risk-free in-terest rate is r = 6% per year.

2. Find the initial price of a Call option with strike of 32 euros and maturityof one year, written on the stock above. Using the Put-Call Parity, deducethe initial price of the corresponding Put option.

3. Discuss whether the probability that the Call of item 1. is exercised isgreater, smaller than or equal to the probability to exercise the Call initem 2. written on the same stock but without dividends.

4. Establish if it is possible to choose suitable drift, volatility and currentprice of the stock (without dividends and distributed as a log-normal) sothat the price of the Call option with strike of 32 euros, maturity of oneyear and written on such a stock can equal the Call price found in item 1.

5. Discuss what would happen to the prices of the Call and Put of item 1. ifthe dividends were paid in 4 and 8 months and the maturity of the optionswere unchanged.

6. What if the amount of dividends stayed unchanged while their dates ofpayment were swapped?

Exercise 7.9. A spread is a strategy consisting in investing in two or moreoptions of the same kind. Whereas a butterfly spread is used when a smallchange in the underlying price is expected, a bull spread is considered whenan increase in the underlying price is expected.

1. Build a butterfly spread based on the following three Call options with thesame maturity and with strikes and prices as below:

• Call A: strike of 20 euros; price of 4 euros;• Call B: strike of 24 euros; price of 2 euros;• Call C: strike of 28 euros; price of 1 euro.

2. Build a bull spread using only two of the above options. [Hint: you shouldbuy the Call with lower strike (between the two options chosen) and sellthe Call of greater strike (same two).]

3. Suppose now that at maturity of the options nothing of what is expectedactually happens and the stock is, instead, ST = 16 euros.


(a) Compare the Profit and Loss due to the butterfly spread (respectively,bull spread) with the profit of a strategy consisting in a long position ina Put option with strike of 20 euros and price of 2 euros.

(b) Now compare with the Profit and Loss due to a bear spread built withthe options of item 1.

4. Suppose that the price (at T = 1 year) of the underlying is given byST = S0 exp

{(μ − 1

2σ2)T + σWT

}, with μ = 0.1 (per year), σ = 0.4 (per

year) and that the risk-free interest rate available on the market is r = 0.02(per year).

(a) Compute the probability to lose at most one euro with the bear, withthe bull and with the butterfly spread, respectively.

(b) Compute the probability that the profit due to the bear spread is notsmaller than the profit of the butterfly, and the probability that the lossdue to the bear spread is not greater than the loss of the butterfly.Is there any relationship between the probabilities just computed?

5. Say if one can obtain a more satisfactory result by means of long/shortpositions in the stock and in the options.

Exercise 7.10.We take a short position in a European Call option with strikeK1 = 8 euros and maturity T = 1 year, written on a stock “XYZ” whose priceis distributed as a log-normal with S0 = 8 euros, σ = 40% per year. We alsotake a long position in two European Put options with strike K2 = 10 eurosand maturity of one year, written on the same stock.The risk-free interest rate available on the market is r = 4% per year.

1. After having computed the Delta of the Call option (denoted by Δ1) andof the Put option (Δ2), find the number of shares of the stock “XYZ” to bebought/sold to make Delta-neutral a portfolio PORT1 consisting in fourshort positions in the Call option and in a long position in the Put option.

2. Suppose that a Call option with strike K3 = 12 euros written on the samestock as before is also available on the market.

(a) Compute Delta and Gamma (Δ3 and Γ3) of such a new option.(b) Explain how to make Delta- and Gamma-neutral a short position in the

first Call option.(c) Discuss what would change in items (a) and (b) if we had a Put option

with strike K3 (instead of the Call option with strike K3).

3. Establish whether it is possible to make the portfolio (PORT1) Delta-,Gamma- and Vega-neutral when a third Put option with strike K4 = 6euro and written on stock “XYZ” is also available. If yes, explain how.

(a) Would this be still possible if we were forced to buy/sell at least oneoption with strike K3 and at least one option with strike K4?

(b) And if we had to buy/sell at most one option with strike K3 and at mostone option with strike K4?

8

American Options


An American Option is a contract giving the buyer the right to buy (Call)or sell (Put) a financial underlying asset for a strike price K at every instantbetween the agreement date and the maturity. The main difference betweenAmerican and European options consists thus in the early exercise feature.If the underlying does not pay any dividend, it is possible to prove that

the early exercise of an American Call option is never optimal. The Amer-ican Call option value – when written on dividend-free underlying– is thenthe same as that of the European Call option with the same parameters. Thesame conclusion does not hold for an American Call option on an underlyingpaying dividends, nor for an American Put option. This means that Put-Callparity cannot be formulated in the same way as for European options.Since American options provide the holder with “at least” the same rights of

their European counterpart (actually some more), their value cannot be lessthan the corresponding European options value. The initial values of bothkind of options are therefore related by the following inequalities:

CAm0 ≥ CEur

0

P Am0 ≥ P Eur

0 .

During the options’ lifetime it may happen that early exercise becomesmore remunerative than exercise at maturity, and the option holder will ex-ploit this opportunity, which is available for American options. The exercisetime will be then chosen so to optimize the income. A problem of this type iswell known in Stochastic Analysis as Optimal Stopping problem.The American options valuation problem in a Black-Scholes setting can be

formulated as a free boundary problem for the Black-Scholes PDE, and thisformulation can be proved to be equivalent to the optimal stopping problemfor the stochastic process describing the underlying dynamics. The Ameri-can option fair value (i.e. the option price consistent with the no-arbitrage


154 8 American Options

requirement) can be equivalently characterized as the free boundary problemsolution for the Black-Scholes PDE, or as the supremum over all possible risk-neutral expectations of the option payoff with respect to all exercise times.Both formulations do not enjoy a closed-form solution and the methods tosolve them require some numerical techniques that give approximate solu-tions. A systematic presentation of these techniques is beyond the purpose ofthis textbook.We invite the reader interested in a deeper understanding of both the ana-

lytical and the stochastic aspects of free boundary problems, and in particularAmerican options valuation problems, to consult some classical textbooks onthe subject, like Wilmott et al. [37], and Musiela and Rutkowski [29]. We aregoing to propose a number of exercises on American options valuation in thesimple context where the detailed knowledge of more complex techniques isnot required, and we mainly focus on the binomial model setting. We willshow how to evaluate an American option by assigning to each node of the bi-nomial tree an option value that takes the early exercise feature into account,and by proceeding with a backward recursion in time starting from maturity.


Exercise 8.1. A Put option (with strike of 8 euros) is written on an underly-ing stock with current price of 8 euros and without dividends. At each ensuingtwo semesters the stock price can move up by 40% with probability p = 0.8 ordown by 60% with probability 1−p = 0.2. Denote by (St)t=0,1,2 the stochasticprocess representing the underlying stock price, where t = 1 stands for onesemester and t = 2 for two semesters.The risk-free interest rate available on the market is 4% (per year).

1.(a) Compute the discounted expectation of the random variable S2 at thepresent date.Is the probability measure P obtained by the previous data an equivalentmartingale measure?Is the stochastic process (St)t=0,1,2 describing the underlying stock dy-namics a martingale with respect to P ?

(b) By letting St denote the underlying value at time t discounted atthe present date, find a probability measure Q with respect to which(St

)t=0,1,2

is a martingale.2. What is the fair initial value of a European Call option written on theunderlying considered before with maturity T = 1 year? And what is thefair value of the Put option on the same underlying and with the sameparameters?

3. Establish if it is optimal to exercise the American Call on the same un-derlying and with the same parameters of the options of the previous itembefore maturity. What would then be its fair value?


4. Verify if there exist arbitrage opportunities in the market where only the(risky) stock and the (risk-free) bond are traded if, for each one of the nexttwo semesters, the stock price can rise by 4% with probability 80% or fallby 4% with probability of 20% and the risk-free interest rate is 15% (peryear).

5. In the market model of item 4., establish if it is possible to evaluate boththe European and the American options as in items 2. and in 3..

SolutionFrom the data we see that we are working in a binomial model with factors

u = 1.4 and d = 0.4 in each time period (in the present case one semester).The dynamics of the stock price can then be described as follows:

Suu2 = 15.68

p ↗Su

1 = 11.2p=0.8 ↗ 1−p ↘

S0 = 8 Sud2 = 4.48

1−p=0.2 ↘ p ↗Sd

1 = 3.21−p ↘

Sdd2 = 1.28

− −−− −− −− −− −− − −− −− −−−t = 0 t = 1 (semester) T = 2 (semesters)

1. (a) By the binomial tree just described, we immediately get:

P (S2 = Suu2 ) = p2 = 0.64

P(S2 = Sud

2

)= 2p (1 − p) = 0.32

P(S2 = Sdd

2

)= (1 − p)2 = 0.04

and

EP [S2] = 15.68 · 0.64 + 4.48 · 0.32 + 1.28 · 0.04 = 11.52.

Since EP [S2] = 11.52 �= S0, the process (St)t=0,1,2 is not a martingalewith respect to the probability measure P .

Let us check if the stochastic process(St

)t=0,1,2

, describing the dis-counted (at the present date) underlying value, is a martingale withrespect to P , i.e. if P is a martingale measure.We immediately verify that P is not a martingale measure because:

EP

[S2

]=

EP [S2]1 + r

=11.521.04

= 11.076 �= 8 = S0 = S0.


(b) In order to determine the martingale measure Q, it is necessary to checkif there exists a probability measure Q such that

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

EQ [S2]1 + r

= S0

EQ [S1]

(1 + r)1/2= S0

EQ [S2|S1 = Su1 ]

(1 + r)1/2= Su

1

EQ

[S2|S1 = Sd

1

](1 + r)1/2

= Sd1

or, equivalently, if there exists qu ∈ [0, 1] (the “new” growth probabilityof the underlying for each time period) solving the following system ofequations:

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

Suu2 · q2

u + Sud2 · 2qu (1 − qu) + Sdd

2 · (1 − qu)2 = S0 (1 + r)

Su1 · qu + Sd

1 · (1 − qu) = S0 (1 + r)1/2

Suu2 · qu + Sud

2 · (1 − qu) = Su1 (1 + r)1/2

Sud2 · qu + Sdd

2 · (1 − qu) = Sd1 (1 + r)1/2

.

Since Suu2 = Su

1 · u, Sud2 = Su

1 · d = Sd1 · u, Sdd

2 = Sd1 · d, Su

1 = S0 · u andSd

1 = S0 · d, the system can be reduced to the following form:

{S0u

2q2u + S0ud · 2qu (1 − qu) + S0d

2 (1 − qu)2 = S0 (1 + r)

S0uqu + S0d (1 − qu) = S0 (1 + r)1/2

{u2q2

u + ud · 2qu (1 − qu) + d2 (1 − qu)2 = 1 + r

uqu + d (1 − qu) = (1 + r)1/2

Since the first equation is essentially the second one squared, we immedi-ately recognize that the solution of the previous system is the following:

qu =(1 + r)1/2 − d

u − d=

√1.04− 0.41.4− 0.4

= 0.62.

Moreover, we conclude that the martingale measure (required in orderto compute the options prices) corresponds to the following rise/fall


probabilities for each time step:

qu = 0.62

1 − qu = 0.38.

2. We remark that the European Call option payoff is 7.68, 0 and 0 whenST = 15.68, ST = 4.48 and ST = 1.28, respectively. The initial value of theEuropean Call is then:

CEur0 =

11 + r

[q2u · 7.68 + 2qu (1 − qu) · 0 + (1 − qu)2 · 0

]= 2.84 euros.

By the Put-Call parity we immediately get the European Put price:

P Eur0 = CEur

0 − S0 +K

1 + r= 2.84− 8 +

81.04

= 2.53 euros.

3. We know that for an American Call option, written on an underlying with-out dividends, early exercise is never optimal; so the initial value of suchan option coincides with that of a European Call option on the same un-derlying and with the same parameters:

CAm0 = CEur

0 = 2.84 euros.

4. In the present case, the framework for valuation is a binomial model withfactors u = 1.04 and d = 0.96 for each time step. The stock price dynamicsis then the following:

Suu2 = 8.6528

p ↗Su

1 = 8.32p=0.8 ↗ 1−p ↘

S0 = 8 Sud2 = 7.9872

1−p=0.2 ↘ p ↗Sd

1 = 7.681−p ↘

Sdd2 = 7.3728

−− −− −− −− −− −− −− −− −− −−t = 0 t = 1 (semester) T = 2 (semesters)

Since the risk-free interest rate for each semester is equivalent to the yearlyrate ryear of 15%, we obtain rsem = (1 + ryear)

1/2 − 1 = 0.072, and 1 +rsem = 1.072 > u = 1.04. So in the market model considered, where a stockand a bond are traded, there exist arbitrage opportunities. We are goingto present one of them:


t = 0 t = 2 = T

short-sell the stock give back the stock⇒ +S0 ⇒ −ST

invest S0 at the risk-free rate r

⇒ −S0 S0 (1 + rsem)2

S0 − S0 = 0 S0 (1 + rsem)2 − ST > S0u2 − ST ≥ 0

Moreover,

P(S0u

2 − ST > 0)

= P({S2 = S0ud} ∪ {S2 = S0d

2})

= 2p (1 − p) + (1 − p)2 > 0.

This shows that the strategy illustrated is an arbitrage opportunity.

5. By the previous item and from the First Fundamental Theorem of AssetPricing we can immediately conclude that an equivalent martingale mea-sure for the market model considered does not exist. This implies that theoptions previously mentioned cannot be evaluated as in items 2. and 3.The non-existence of an equivalent martingale measure can be proved alsodirectly. If such a measure (let us denote it by Q) existed, the followingcondition should be satisfied:

EQ [S1]1 + rsem

= S0,

i.e. there should exist qu ∈ [0, 1] such that

S0uqu + S0d (1 − qu) = S0 (1 + rsem) .

Since the last equation is equivalent to

uqu + d (1 − qu) = 1 + rsem

qu =1 + rsem − d

u − d>

u − d

u − d= 1,

it follows immediately that an equivalent martingale measure does not ex-ist.

Exercise 8.2. An option is written on a stock without dividends and withcurrent price of 10 euros. At the end of each one of next two semesters thestock price can rise by 25% or fall by 20%, the risk-free interest rate is 4%(per year) and the strike of the option is 11 euros.

1. Find the current price of a European Put option written on this underlyingand with maturity T = 1 year.


2. Is it optimal to exercise the corresponding (i.e. with the same parameters)American option before maturity? What is its fair value?

3. Discuss whether there exists a probability measure P (different from therisk-neutral measure) under which the event that the European Put optionexpires out-of-the money (OTM) or the American Put is exercised earlyhappens with probability that is twice the probability that the EuropeanPut expires out-of-the money and the American Put is not exercised early.

4. Assume now that the underlying stock pays a dividend of 1.5 euros in 6months.

(a) Compute the fair value of both the American and European Put optionswith strike of 11 euros.

(b) Find the fair value of the American Call option with the same parame-ters of the previous one.

5. Can the difference between the initial values of the American Put and theAmerican Call (both with strike K = 11 euros) be more than (or equal to)0.5 and less than (or equal to) 3 euros?

SolutionThe market model described above corresponds to a binomial model with

factors u = 1.25 and d = 0.8 for each time step (in the present case onesemester). The stock price dynamics is then the following:

Suu1y = 15.625

↗Su

6m = 12.5↗ ↘S0 = 10 Sud

1y = 10↘ ↗

Sd6m = 8 ↘

Sdd1y = 6.4

−− −− −− −− −− −− −− −− −− −−t = 0 t = 6 months T = 1 year

The risk-neutral probability measure, necessary to compute the options’value, can be easily obtained and is defined by the following rise/fall proba-bilities for each time step:

qu =(1 + r)1/2 − d

u − d=

√1.04− 0.8

1.25− 0.8= 0.488

1 − qu = 0.512

1. Since the European Put option payoff is 0, 1 and 4.6 (when ST = 15.625,ST = 10 and ST = 6.4, respectively), by the previous remark we can con-


clude that the initial fair value of the European Put option is given by:

P Eur0 =

11 + r

[q2u · 0 + 2qu (1 − qu) · 1 + (1 − qu)2 · 4.6

]= 1.64 euros.

2. We want to verify now if the early exercise of the American Put is optimalor not. To check this we are going to proceed backwards along the tree,starting from maturity and considering all subtrees corresponding to onetime-step separately.We shall denote by a capital letter each node of the tree:

15.625 (D)↗

12.5 (B)↗ ↘10 (A) 10 (E)

↘ ↗8 (C) ↘

6.4 (F)

−− −− − −− − −− − −− −−−t = 0 t = 6 months T = 1 year

Moreover, for each node we write the payoff of the American Put consid-ered:

15.625 (D)payoff = 0

↗12.5 (B)payoff = 0↗ ↘

10 (A)payoff = 1

10 (E)payoff = 1↘ ↗

8 (C)payoff = 3 ↘

6.4 (F)payoff = 4.6

Let us start by considering the subtree including nodes B, D and E. TheAmerican Put, when restricted to this subtree is “European-like”. So we cancalculate the corresponding (i.e. with the same parameters) European Putvalue at the nodes B, D and E. To avoid confusion between the EuropeanPut of item 1. (we mean the European Put written at t = 0 and maturingin one year) and the “artificial” put options just introduced, we shall callthe latter “European”(*).


At the nodes D and E the Put price coincides with its payoff, so

P Eur∗T,D = payoff at node D = 0

P Eur∗T,E = payoff at node E = 1

P Eur∗6m,B =

1

(1 + r)1/2[qu · 0 + (1 − qu) · 1] = 0.50.

Since the payoff of the American Put in B is 0, therefore less than the Eu-ropean* Put price in t = 6 months with maturity 1 year (P Eur∗

6m,B = 0.50),early exercise at node B is not optimal.The American Put value at each node is the maximum between the pay-off and the value of the corresponding European* Put. In other words, itcoincides with the European* counterpart when the early exercise is notoptimal, while it equals the payoff at the node considered when the earlyexercise is optimal.At the nodes B, D and E we have then:

P AmT,D = P Eur

T,D = payoff in D = 0

P AmT,E = P Eur

T,E = payoff in E = 1

P Am6m,B = max{P Eur

6m,B; payoff in B} = max{0.50; 0} = 0.50.

Let us now consider the subtree with nodes C, E and F. The AmericanPut option, restricted to this subtree is again “European-like”. We can thencompute the corresponding European* Put value at nodes C, E and F.At nodes E and F, the option value coincides with the payoff. Hence,

P Eur∗T,E = payoff in E = 1

P Eur∗T,F = payoff in F = 4.6

P Eur∗6m,C =

1

(1 + r)1/2[qu · 1 + (1 − qu) · 4.6] = 2.79.

Since the American Put payoff at node C is equal to 3 (hence greater thanthe European* Put value at t = 6 months and with maturity T = 1 year(P Eur∗

6m,C = 2.79)), the early exercise at node C is optimal.The American Put price at nodes C, E, F is then given by:

P AmT,E = P Eur∗

T,E = payoff at node E = 1

P AmT,F = P Eur∗

T,F = payoff at node F = 4.6

P Am6m,C = max{P Eur∗

6m,C ; payoff at node C} = max{2.79; 3} = 3.

Let us finally consider the subtree corresponding to nodes A, B and C. TheAmerican Put option restricted to this subtree is again “European-like”.


We can compute the corresponding European* Put value at nodes A, Band C.In B and C the value coincides with that of the American Put computedbefore, so

P Eur∗6m,B = P Am

6m,B = 0.50

P Eur∗6m,C < P Am

6m,C = 3

P Eur∗0,A =

1

(1 + r)1/2[qu · 0.50 + (1 − qu) · 3] = 1.75.

Since the payoff of the American Put at node A is 1, so less than the valueof the European* Put in t = 0 with maturity t = 6 months (P Eur∗

0,A = 1.75),the early exercise in A is not optimal.Eventually, we can conclude that the American Put value in A is given by:

P Am0,A = max{P Eur∗

0,A ; payoff at node A} = max{1.75; 1} = 1.75 euros.

We remark that, as one could expect, the initial value of the American Putis greater than that of the corresponding European (as computed in 1. andequal to 1.64 euros).

3. We will determine if there exists a probability measure P (not coincidingwith the risk-neutral) under which the event that the European Put optionexpires out-of-the money (OTM) or the American Put is exercised earlyhappens with probability that is twice the probability that the EuropeanPut expires out-of-the money and the American Put is not exercised early.We first denote by ω1 the event of the underlying price moving up at bothtime steps (up-up), by ω2 a rise in the first time step and a fall in thesecond (up-down), by ω3 a down-up movement and by ω4 the down-down.We also introduce the following events:

A1 = {European Put OTM or early exercise}A2 = {European Put OTM and no early exercise} .

We get then:

P (A1) = P ({ω1} ∪ {ω3 ∪ ω4}) = p2 + p (1 − p) + (1 − p)2

P (A2) = P ({ω1} ∩ {ω1 ∪ ω2}) = P (ω1) = p2.

We have now to look for p ∈ (0, 1) such that

p2 + p (1 − p) + (1 − p)2 = 2p2

p + 1 + p2 − 2p = 2p2

p2 + p − 1 = 0.


We conclude that the only admissible solution is given by p = −1+√

52 .

Finally, we obtain:

P (ω1) =

(−1 +

√5

2

)2

P (ω2) = P (ω3) =

(−1 +√

5) (

3 −√5)

4= 2 −√

5

P (ω4) =

(3 −√

52

)2

.

4. If the underlying stock pays a dividend of 1.5 euros in 6 months, its dy-namics can be described as follows:

SDiv,uu1 year = SDiv,u

6m · u = 13.75↗

SDiv,u6m = S0u − 1.5 = 11

↗ ↘ SDiv,ud1y = SDiv,u

6m · d = 8.8SDiv

0 = 10↘ ↗ SDiv,du

1y = SDiv,d6m · u = 8.125

SDiv,d6m = S0d − 1.5 = 6.5 ↘

SDiv,dd1y = SDiv,d

6m · d = 5.2

− −−− − −− −− −− −− − −− −− −− −− −− −− −−t = 0 t = 6 months T = 1 year

(a) We describe, with the help of the following diagram, the American Putand the European Put payoffs, both with strike K = 11 euros:

13.75 (node D)payoffAm = payoffEur = 0

↗11 (node B)payoffAm = 0

↗ ↘ 8.8 (node E)payoffAm = payoffEur = 2.2

10 (node A)payoffAm = 1

↘ ↗ 8.125 (node F)payoffAm = payoffEur = 2.875

6.5 (node C)payoffAm = 4.5 ↘

5.2 (node G)payoffAm = payoffEur = 5.8


Since

qu =(1.04)1/2 − d

u − d= 0.488,

we have that

P Eur6m,B =

1√1.04

[0 + (1 − qu) · 2.2] = 1.105

P Eur6m,C =

1√1.04

[qu · 2.875 + (1 − qu) · 5.8] = 4.288.

Consequently,

P Am6m,B = max{P Eur

6m,B; payoffAm; node B} = max{1.105; 0} = 1.105

P Am6m,C = max{P Eur

6m,C; payoffAm; node C} = max{4.288; 4.5}= 4.5.

Finally we obtain:

P Eur0 =

1√1.04

[qu · 1.105 + (1 − qu) · 4.288] = 2.682 euro

P Am0 = max

{1√1.04

[qu · 1.105 + (1 − qu) · 4.5] ; payoffAm; node A

}= max{2.788; 1} = 2.788 euros.

(b) If the American option considered in the previous point was a Call in-stead of a Put, the procedure would change as follows:

13.75 (node D)payoffCall Am = 2.75

↗11 (node B)

payoffCall Am = 0↗ ↘ 8.8 (node E)

payoffCall Am = 010 (node A)

payoffCall Am = 0

↘ ↗ 8.125 (node F)payoffCall Am = 0

6.5 (node C)payoffCall Am = 0 ↘

5.2 (node G)payoffCall Am = 0


By proceeding in the same way outlined before, we have that:

CEur6m,B =

1√1.04

[qu · 2.75 + 0] = 1.316

CEur6m,C = 0

and, consequently,

CAm6m,B = max{CEur

6m,B; payoffCall Am; node B}= max{1.316; 0} = 1.316

CAm6m,C = max{CEur

6m,C; payoffCall Am; node C} = 0.

Finally, we obtain

CAm0 = max

{1√1.04

[qu · 1.316 + 0] ; payoffAm; node A

}= max{0.629; 0} = 0.629.

5. From the previous items, we immediately get

P Am0 − CAm

0 = 2.788− 0.629 = 2.159 ∈ [0.5; 3] .

Consequently, the difference between the initial American Put and Callvalues (both with strike K = 11 euros) is included in the interval [a, b]where a = 0.5 and b = 3 euros (endpoints included).We could obtain the same conclusion without having to compute the twooptions prices directly. We know, indeed, that

K

1 + r− S0 < P Am

0 − CAm0 < D + K − S0,

where D is the discounted value of the dividends distributed between theinitial date 0 and the maturity T . In the present case we have D = 1.5√

1.04=

1.471 and

0.577 =K

1 + r− S0 < P Am

0 − CAm0 < D + K − S0 = 2.471.

Exercise 8.3. Consider the American Put option of item 2. in Exercise 8.2.This option expires in one year, its strike is K = 11 euros and it is written onan underlying stock without dividends with current price of 10 euros. Duringeach one of next two semesters the stock price can rise by 25% or fall by 20%.The risk-free interest rate is 4% (per year).Find the hedging strategy of the American option considered.


SolutionFrom Exercise 8.2, we recall that the dynamics of the underlying stock and

the dynamics of the American Put can be described as follows:

Suu1y = 15.625 (node D)valuePut Am = 0

↗Su

6m = 12.5 (node B)valuePut Am = 0.50↗ ↘

S0 = 10 (node A)valuePut Am = 1.75

Sud1y = 10 (node E)valuePut Am = 1

↘ ↗Sd

6m = 8 (node C)valuePut Am = 3 ↘

Sdd1y = 6.4 (node F)valuePut Am = 4.6

and that at node C early exercise is optimal.

In order to compute the hedging strategy, it is necessary to determine thestrategy at time t = 0 and at node B, but not in C since at this node earlyexercise is viable.To find the hedging strategy at time t = 0 means to find a quantity Δ0 of

the underlying stock and an amount of money x0 such that{Δ0S0u + x0 (1 + r)1/2 = P Am

6m,B

Δ0S0d + x0 (1 + r)1/2 = P Am6m,C

,

i.e. ⎧⎪⎪⎪⎨⎪⎪⎪⎩

Δ0 =P Am

6m,B − P Am6m,C

S0 (u − d)= −0.556

x0 =1

(1 + r)1/2

[P Am

6m,B − Δ0S0u]

= 7.31.

This means that the strategy consists in selling 0.556 stock units and investing7.3 euros. As one could expect, its initial cost is given by:

Δ0S0 + x0 = −0.556 · 10 + 7.31 = 1.75 = P Am0 .

The replicating strategy(ΔB

6m, xB6m

)to be performed at t = 6 months in

the node B must solve the following equations:{ΔB

6mSu6mu + xB

6m (1 + r)1/2 = P Am6 months, D

ΔB6mSu

6md + xB6m (1 + r)1/2 = P Am

6m,E

.


Then ⎧⎪⎪⎨⎪⎪⎩

ΔB6m =

P Am6m,D − P Am

6m,E

Su6m (u − d)

= −0.1778

xB6m =

1(1 + r)1/2

[P Am

6m,D − ΔB6mSu

6mu]

= 2.724.

As expected, then, at node B the cost of the strategy just outlined is givenby:

ΔB6mSu

6m + xB6m = ΔB

6mS0u + xB6m = 0.50 = P Am

6m,B.

Exercise 8.4. Compute the price at time t = 0 of a perpetual American Putoption with strike K and written on an underlying whose price dynamics isdescribed by a geometric Brownian motion with diffusion coefficient σ. Therisk-free interest rate available on the market is r.

SolutionSince the option is perpetual, i.e. its maturity is T = +∞, the valuation

problem does not depend explicitly on t. By removing the t-dependence inthe PDE providing the American Put value, we obtain the following ODE:

σ2

2S2 d2P

dS2+ rS

dP

dS− rP = 0. (8.1)

This ordinary differential equation must be solved taking into account theproper boundary conditions that, for the American Put option, turn out tobe the following

P (+∞) = limS→+∞

P (S) = 0,

P (Sf) = K − Sf ,

P ′(Sf ) =dP

dS

∣∣∣∣S=Sf

= −1.

We should remark that we need three conditions in order to determine thesolution of the ODE, since the free-boundary value Sf is unknown, i.e. we needone more condition than for a “fixed” boundary. Actually, the free-boundaryvalue Sf has to be computed as part of the solution of the problem considered,and becomes an extra unknown. The second and third conditions are calledthe “value matching” condition and the “smooth pasting” condition, respec-tively. Together with the condition at infinity they can be proved to be bothnecessary and sufficient in order to guarantee the existence and uniqueness ofthe solution for the problem under examination.To compute the general solution of the ODE above, we can perform a

simple variable change:S = ex. (8.2)


Hence, x = ln S and the ODE for the new unknown P (x) becomes

σ2

2d2P

dx2+(

r − σ2

2

)dP

dx− rP = 0. (8.3)

This is a linear homogeneous ordinary differential equation, whose solutioncan be immediately written as

P (x) = Aeλ1x + Beλ2x, (8.4)

where A, B are constants determined by imposing the boundary conditionsand λ1, λ2 (λ1 < λ2) are the two distinct solutions of the quadratic equation:

σ2

2λ2 +

(r − σ2

2

)λ − r = 0. (8.5)

Hence,

λ1,2 = σ−2

[−(r − σ2

2) ±√

(r − σ2

2)2 + 2rσ2

]

= σ−2

[−(r − σ2

2) ± (r +

σ2

2)]

,

λ1 = − 2r

σ2, λ2 = 1. (8.6)

Written in terms of the old independent variable S , the solution is then

P (S) = ASλ1 + BSλ2 = AS− 2r

σ2 + BS. (8.7)

The first boundary condition, i.e. P (∞) = 0, implies that the constant Bmust vanish, while the second and third conditions allow to determine simul-taneously the constant A and the free-boundary value Sf :

P (Sf ) = AS− 2r

σ2

f = K − Sf ,

P ′(Sf ) = − 2r

σ2AS

− 2rσ2 −1

f = −1.

From these conditions, we immediately obtain

A = (K − Sf )S2r

σ2

f

Sf =K

1 + σ2

2r

.

Exercise 8.5. The early exercise premium of an American option p(S, t) isdefined as p(S, t) � P Am(S, t) − P Eur(S, t), i.e. as the difference at time tbetween the values of an American and a European Put option on the same


underlying and with the same parameters. Assume as ansatz that the depen-dence of P on the time to maturity τ = T − t can be factorized through thefunction H(τ ) = 1 − exp(−rτ ) as follows:

p(S, τ ) = H(τ )f(S, H(τ )), (8.8)

for a suitable function f .

1. Find the PDE satisfied by the function f in the Black-Scholes framework(i.e. by assuming that the underlying price dynamics is described by ageometric Brownian motion).

2. Assume that the nonlinear terms appearing in the PDE determined inthe previous step are much smaller than the remaining terms. Solve theODE obtained by neglecting those terms by taking into account the properboundary conditions.

3. By using the solution obtained in the previous step, provide an approximateformula for the American Put option value.

SolutionWe know already that the function p(S, τ ) is non-negative. Since p is the

difference between two functions satisfying the Black-Scholes equation, it mustalso satisfy the same equation (due to the linearity of the Black-Scholes equa-tion):

−∂p

∂τ+

σ2

2S2 ∂2p

∂S2+ rS

∂p

∂S− rp = 0. (8.9)

The initial condition in the variable τ for the function p must be p(S, 0) = 0,since the American and the European options satisfy the same initial condi-tion (they have the same payoff).

1. By assuming that the dependence of p on τ can be factorized in the sug-gested form, it is easy to compute the partial derivatives with respect tothe new independent variables:

∂p

∂S= H

∂f

∂S,

∂2p

∂S2= H

∂2f

∂S2,

∂p

∂τ= H

∂f

∂H

dH

dτ+ f

dH

dτ,

dH

dτ= r(1 − H).

Hence, we get the following PDE for f :

S2 ∂2f

∂S2+ kS

∂f

∂S− k

Hf

[1 + H(1 − H)

1f

∂f

∂H

]= 0, (8.10)

where we introduced the parameter k = 2r/σ2.


2. By neglecting the quadratic term in H , and thus eliminating the explicitdependence of f on H , we finally obtain the approximate ODE for f :

S2 ∂2f

∂S2+ kS

∂f

∂S− k

Hf = 0. (8.11)

This ODE, in which H plays the role of a parameter, can be solved explic-itly.By solving the ODE exactly as in the previous exercise, we obtain thesolution:

f(S) = ASλ, (8.12)

where λ is defined as follows:

λ � −12

[(k − 1) +

√(k − 1)2 +

4k

H

]. (8.13)

The constant A must be determined by imposing the proper conditions.

3. Combining the results obtained, we can write an approximation for theAmerican Put option value for S > Sf :

P Am(S, τ ) ≈ P Eur(S, τ ) + AH(τ )Sλ, (8.14)

where the constant A must be such that the option value matches its payofffor S = Sf :

P Eur(S, τ ) + AH(τ )Sλf = K − Sf . (8.15)

There is still the unknown Sf to be determined, and this can be done byimposing the “smooth pasting” condition

∂P Am

∂S(Sf , τ ) =

∂P Eur

∂S(Sf , τ ) + AλHSλ−1

f

= N(d1(Sf )) − 1 + AλHSλ−1f = −1, (8.16)

where N(·) stands, as usual, for the cumulative distribution function of astandard normal. Solving the last equation, we get an expression for A interms of Sf :

A = −N(d1(Sf ))λHSλ−1

f

. (8.17)

By substituting into the equation (8.15), we obtain an equation for theremaining unknown Sf :

P Eur(Sf , τ ) − N(d1(Sf ))Sf

λ= K − Sf , (8.18)


which, taking into account the Put-Call parity and the trivial identityN(−d) = 1 − N(d), can be written as

SfN(d1(Sf ))−Ke−rτ N(d2(Sf ))−Sf +Ke−rτ −N(d1(Sf ))Sf

λ−K+Sf = 0.

Collecting terms, we write it as

SfN(d1(Sf ))(1 − 1λ

) + Ke−rτ [1 − N(d2(Sf ))] − K = 0. (8.19)

Since d1 and d2 depend on Sf , this is an implicit equation in the unknownSf and must be solved by an approximate iterative method, like Newton’smethod, where the initial seed S0 = K may be used for starting the itera-tive procedure.


Exercise 8.6. The current price of a stock is 40 euros. At the end of this yearthe stock price will be either 42 or 36 euros. The growth and fall factors ofthe stock in the following years will remain the same and the risk-free interestrate is 4% per year.

1. Compute the price of the American Put with maturity T = 2 years andwith strike of 40 euros. Is it optimal to exercise this option before maturity?

2. Suppose a first dividend of 1 euro is paid after one year and a second oneof 3 euros after two years: what changes in the valuation procedure? Whichoption is more expensive: the American Put or the American Call?

3. For which American option is the early exercise more likely to take place:a Call or a Put option?

4. Compute the price of the American Put and Call options with the sameparameters as in the previous point, but with maturity delayed by twoyears.

5. Is it possible to find a different value of the growth factor in a binomialmodel, so that the value of a Call option with maturity T = 2 years andstrike K = 40 euros written on a stock without dividends coincides withthe value of an American Put option with the same parameters written ona stock with the same dividends as item 2.?

Exercise 8.7. Compute the price (at time t = 0) of a perpetual AmericanCall option, written on an underlying paying dividends at a constant rate q,and whose price dynamics is described by a geometric Brownian motion withdiffusion coefficient σ. The risk-free interest rate is r, the strike K.

9

Exotic Options


An Exotic option is an option that is neither a European Call or Put optionnor an American Call or Put. A path-dependent option is an option whose pay-off does not depend only on the underlying value at maturity, but also on oneor more values that it can assume during its lifetime. Although the two optionclasses do not coincide, many exotic options exhibit path-dependence features.Among the most popular Exotic options that are not path-dependent we justrecall binary (or digital) options, compound options and chooser options.

Binary Options. The most traded kind of binary option is the so called“cash or nothing” Call. This is a contract guaranteeing the owner a fixedamount of money B if the underlying value at maturity exceeds a thresholdL, and nothing if the underlying value is less than L.The payoff of this option can be expressed in the following form:

B1{ST ≥L} = B ·H(ST − L),

where the function H is the so-called Heaviside step-function:

H(x) ={

0, x < 01, x ≥ 0 . (9.1)

Another kind of binary option is the “asset or nothing” Call. This option ex-pires out-of-the-money if the underlying value at maturity is less than thethreshold L, while it guarantees a unit of the underlying asset if the thresholdvalue is exceeded. The expression of this second kind of binary option payoffis given by:

ST 1{ST ≥L} = ST · H(ST − L). (9.2)


174 9 Exotic Options

“Chooser” Options are options with other options as underlying assets, giv-ing the owner the right to decide, at maturity T1 < T2, whether the underlyingoption must be a Call or a Put. The chosen option will be (possibly) exercisedat maturity T2 and its strike will be K2.The valuation of these options can be performed in a way that is strictly

similar to the European options, but with the following final condition:

F (ST2 , T2) = max{C(ST2 , T2) − K2; P (ST2, T2) − K2; 0} .

As far as path-dependent options are concerned, we are going to propose afew exercises on Asian options, Barrier options and Lookback options.

Asian Options are options depending on the average of the values assumedby the underlying during the option lifetime.Asian options can be of different kinds according to the average considered:

this can be the arithmetic mean or the geometric mean, and the underlyingvalues contributing to the average can be “observed” in continuous or discretetime.A further distinction among Asian options arises according to the func-

tional dependence of the payoff on the average considered; this can in factsubstitute the underlying final value or the strike in the payoff involved. Inthe former case we shall refer to average rate options, in the latter to averagestrike options.The valuation problem for these kind of options can be cast into the Black-

Scholes framework and, for some particular cases, closed-form solutions pro-viding options prices exist. Except for the very simple case of the geometricAsian options (with continuous time monitoring), these solutions do not ad-mit a representation that is easy to handle for applications. We shall thereforefocus our attention mainly on examples of valuation problems in a binomialsetting.

Barrier Options are options with the same payoff at maturity of the Euro-pean options, but with one extra feature: they expire valueless (or not) if, atany instant during the option lifetime, the underlying reaches some thresholdvalue from above or from below. In the first case they will be called up-and-outoptions, in the second down-and-out (or up-and-in and down-and-in respec-tively).For Barrier options (with continuous monitoring), explicit valuation formu-

las are available in a Black-Scholes setting. We recall these formulas below.

Down-and-out Call pricing formula

CDO(St, t) = C(St, t) −(

L

St

) 2rσ2 −1

C

(L2

St, t

), (9.3)


where C(St, t) is the value of the European Call option written on the sameunderlying and with the same parameters (strike, maturity, volatility andrisk-free interest rate) of the Barrier option, and L is the barrier value. Theprevious formula holds when K > L. When K < L, the down-and-out Callpricing formula reduces to the following:

CDO(St, t) = St [N(d3) − b(1 − N(d6))]

− Ke−r(T−t) [N(d4) − a(1 − N(d5))] , (9.4)

where a, b, d3, d4, d5, d8 are defined as follows:

a =(

L

St

) 2rσ2 −1

, b =(

L

St

) 2rσ2 +1

, (9.5)

d3 =ln(

St

L

)+(r + σ2

2

)(T − t)

σ√

T − t, (9.6)

d4 =ln(St

L) + (r − σ2

2)(T − t)

σ√

T − t, (9.7)

d5 =ln(

St

L

)− (r − σ2

2

)(T − t)

σ√

T − t, (9.8)

d6 =ln(

St

L

)− (r + σ2

2

)(T − t)

σ√

T − t. (9.9)

Up-and-out Call pricing formulaFor up-and-out Call, in case U > K, the expression providing the price attime t is the following:

CUO(St, t) = C(St, t) − St [N(d3) + b(N(d6) − N(d8))]

+ Ke−r(T−t) [N(d4) + a(N(d5) − N(d7))] (9.10)

= St [N(d1) − N(d3) − b (N(d6) − N(d8))]

− Ke−r(T−t) [N(d2) − N(d4) − a(N(d5) − N(d7)] , (9.11)

where

d7 =ln(

StKU2

) − (r − σ2

2

)(T − t)

σ√

T − t, (9.12)

d8 =ln(

StKU2

) − (r + σ2

2

)(T − t)

σ√

T − t(9.13)

and the quantities d3, d4, d5, d6 are defined as before, with U instead of L. It iseasy to observe that the value of an up-and-out Call option with U < K is zero.


For Barrier options of “in” type it is possible to obtain immediately the pric-ing formulas from those just presented, by observing that a portfolio composedby two Barrier options of in and out type respectively, written on the sameunderlying, with the same parameters and with the same threshold values,is equivalent to a European option, still written on the same underlying andwith the same parameters. As an example, we provide the following pricingformula for a down-and-in Call option, holding if K > L:

CDI(St, t) =(

L

St

)2r/σ2−1

C

(L2

St, t

). (9.14)

Lookback Options are options with payoff assuming one of the followingforms:

FPM (ST , Smax, T ) = max{Smax − ST ; 0} = max{M − ST ; 0} (9.15)

for the Put,

FCm(ST , Smin, T ) = max {ST − Smin; 0} = max {ST − m; 0} (9.16)

for the Call, where m = Smin, M = Smax are, respectively, the minimum andthe maximum values assumed by the underlying during the option lifetime.We remark thatM andm “substitute the strike” in the payoff of the EuropeanPut and Call options respectively.Explicit analytical expressions are available for the Lookback options just

mentioned in the Black-Scholes setting; they can be obtained both by solvingthe relative PDE with similarity methods or by computing the discountedexpectation of their payoff, since the joint probability distributions of therunning maximum and minimum of a Brownian motion and of the Brownianmotion itself is known. We briefly recall these pricing formulas for the reader’sconvenience:

FCm(St, m, t) = StN(d1) − me−r(T−t)N(d2)

+ Ste−r(T−t) σ2

2r ·[(St

m )−2r/σ2N(−d1 + 2r

σ

√T − t)

− er(T−t)N(−d1)],

FPM (St, M, t) = Me−r(T−t)N(−d2) − StN(−d1)

+ Ste−r(T−t) σ2

2r·[− (St

M)−2r/σ2

N(d1 − 2rσ

√T − t)

+ er(T−t)N(d1)].

It is worth mentioning a few issues related to Exotic options hedging. Whilesome of these options are easy to hedge – like Asian options (since the averagebecomes more and more stable as time goes on, and the number of observa-


tions grows), other options – like barrier options– can be extremely difficult tohedge: a small variation of the underlying price, close to the barrier value, cangive rise to a huge variation of the option value, i.e. the Delta can be discon-tinuous across the barrier. In these cases, a static replication approach can beused. The static replication methodology is based on a standard result of func-tional analysis that can be roughly resumed as follows: every continuous realfunction with compact support can be approximated (with the desired accu-racy) by piecewise linear functions; this means, in practice, that every payoff(with compact support) can be approximated by the payoff of a portfolio con-sisting of a suitable number of units of the underlying and a suitable numberof Put and/or Call European options. We shall illustrate this procedure byan example in the exercises.

The reader interested in a more detailed presentation of pricing methodsfor Exotic options can find many results in El Karoui [16], Hull [23], Musielaand Rutkowski [29] and Wilmott et al. [37].


Exercise 9.1. Consider a stock with value at time t denoted by (St)t≥0 andwith current value S0 = 8 euros. Compute the price at time t = 0 of a “cashor nothing” Call option, written on the stock considered, with maturity T = 3months, strike L = 10 euros and rebate B = 20 euros. The stock price dynam-ics is assumed to be described by a geometric Brownian motion with volatilityσ = 0.25 (per year). The risk-free interest rate on the market is r = 0.04 (peryear).We recall that, for a “cash or nothing” Call the strike is the threshold value

L such that, if the underlying at maturity is greater than or equal to L, theoption buyer gets the cash amount B (the rebate).

SolutionIn order to compute the value at time t = 0 of a binary option of “cash or

nothing” Call type, we can apply the usual risk-neutral valuation procedure.Hence, we need to determine the discounted expectation of the option payoffwith respect to the equivalent martingale measure:

CD(St, t) = e−r(T−t)EQ [Φ(ST )] , (9.17)

where we denote by CD the binary option’s value at time t, by Φ its payoffand by EQ the expectation with respect to Q (the risk-neutral measure). Werecall that the payoff of a binary option of “cash or nothing” Call type is givenby the following expression: Φ(ST ) = B ·H(ST −L), where L is the “threshold”value and B the rebate.


In the present case, since we are working in a Black-Scholes framework, wecan write:

CD(St, t) = e−r(T−t)

∫ +∞

−∞B · H

(Ste

(r−σ2/2)(T−t)+σx − L)

1√2π(T − t)

e−x2

2(T−t) dx

= e−r(T−t)

∫ +∞

−∞

B · H(Stez − L)

σ√

2π(T − t)

exp

{−(z − (r − σ2/2)(T − t)

)22σ2(T − t)

}dz.

The integral appearing in the last formula can be explicitly computed:

∫ +∞

−∞

B · H(Stez − L)

σ√

2π(T − t)exp

{−(z − (r − σ2/2)(T − t)

)22σ2(T − t)

}dz

= B

∫ +∞

ln L−ln St

exp

{−(z − (r − σ2/2)(T − t)

)22σ2(T − t)

}dz

σ√

2π(T − t)

= B

∫ +∞

ln L−ln St−(r−σ2/2)(T−t)

σ√

T−t

1√2π

e−u22 du

= B

[1 − N

(ln (L/St) −

(r − σ2/2

)(T − t)

σ√

T − t

)]

= B · N(d2).

In the last expression we have denoted by N the cumulative distribution func-tion of a standard normal and set

d2 =ln(St/L) + (r − σ2/2)(T − t)

σ√

T − t

(in strict analogy with the term appearing in the Black-Scholes formula forthe European Call option price).

The “cash or nothing” Call option value is then given by

CD(St, t) = Be−r(T−t)N(d2).

In our case, we have t = 0, T = 1/4, S0 = 8, B = 20, L = 10, r = 0.04, σ2 =0.0625, so

d2 =ln(8/10) + (0.04− 0.0625/2)1

4

0.25 ·√1/4= −1.767


andCD(0, 8) = 20 · e−0.04·14 N(−1.767) = 0.764 euros.

Exercise 9.2. Consider a chooser option with maturity T2 and strike K, i.e.an option written at time t = 0 with maturity T2 for which at time T1 (withT1 < T2) the option buyer can choose if the option will be exercised as a Callor a Put on the same underlying asset.Compute the initial price of the chooser option with maturity T1 = 3

months, T2 = 6 months and strike K = 30 euros, if the underlying dynamicsis described by a geometric Brownian motion with initial value S0 = 20 euros,volatility σ = 0.25 (per year), and the risk-free interest rate is r = 0.06 (peryear).

SolutionA chooser option can be easily valuated by showing that it is equivalent

to a portfolio composed by a European Call with maturity T1 and strikeKe−r(T2−T1) and by a European Put with maturity T2 and strike K (see Hull[23] for a detailed proof). We denote by C0

(Ke−r(T2−T1); T1

)and P0 (K; T2)

the prices at t = 0 of these options.The initial price Ch0 of the chooser option is then given by:

Ch0 = P0 (K; T2) + C0

(Ke−r(T2−T1); T1

), (9.18)

i.e. the initial prices of the Put and Call options mentioned before. By sub-stituting the data assigned, we get:

Ch0 = P0(30, 1/2) + C0(30e−0.0614 ; 1/4).

By applying the Black-Scholes formulas for European options directly, oneobtains:

C0(30e−0.06 14 ; 1/4) = 0.038

P0(30; 1/2) = 9.172,

by which we can compute the initial value of the chooser option:

Ch0 = 0.038 + 9.172 = 9.21 euros.

Exercise 9.3. Consider a stock with initial value S0 = 8 euros, with dynam-ics described by a geometric Brownian motion with volatility σ = 0.36 (peryear) in a market model with risk-free interest rate r = 0.04 (per year).

1. Compute the value at time t = 0 of an up-and-in and of an up-and-out Calloption with barrier U = 12 euros, strike K = 9 euros and maturity T = 4months.

2. What is the probability that, at t = 2 months, the underlying value willreach the barrier?


Solution

1. First of all, we recall the valuation formula for a Barrier Call option ofup-and-out type. By denoting by CUO the value of this Barrier option andby C0 the value the corresponding vanilla option (both at time t = 0), from(9.10) we get:

CUO,0 = C0 − S0 [N(d3) + b(N(d6) − N(d8))] (9.19)

+ Ke−rT [N(d4) + a(N(d5) − N(d7))] (9.20)

where a, b, d3, d4, d5, d6, d7 and d8 are defined by (9.5)-(9.13).

By the formulas just mentioned, with the assigned data, we have that:

a =(

128

) 0.08(0.36)2

−1

= 0.856; b =(

128

) 0.08(0.36)2

+1

= 1.927

d3 =ln(

812

)+(0.04 + (0.36)2

2

)412

0.36 ·√

412

= −1.783

d4 =ln(

812

)+(0.04− (0.36)2

2

)412

0.36 ·√

412

= −1.991

d5 =ln(

812

)− (0.04− (0.36)2

2

)412

0.36 ·√

412

= −1.911

d6 =ln(

812

)− (0.04 + (0.36)2

2

)412

0.36 ·√

412

= −2.119

d7 =ln(

8·9(12)2

)−(0.04− (0.36)2

2

)412

0.36 ·√

412

= −3.295

d8 =ln(

8·9(12)2

)−(0.04 + (0.36)2

2

)412

0.36 ·√

412

= −3.50

Since

d1 =ln(S0/K) + (r + σ2

2 )T

σ√

T= −0.399

d2 = d1 − σ√

T = −0.606,


we conclude that the initial (at time t = 0) value of the European Calloption is

C0 = S0 ·N(d1) − Ke−rT · N(d2)

= 8 · N (−0.399)− 9 · e−0.04·4/12 · N (−0.606)= 0.342 euros.

Finally, we obtain the initial value of the up-and-out Call:

CUO,0 = 0.20 euros.

By the relation holding between up-and-out, up-and-in and vanilla optionvalues (strictly analogous to that between down-and-out, down-and-in andvanilla), we have that:

C0 = CUI,0 + CUO,0, (9.21)

where CUI,0 denotes the initial value of the up-and-in Call.From (9.21) we finally obtain:

CUI,0 = C0 − CUO,0 = 0.342− 0.20 = 0.142 euros.

We remark that both the up-and-in Call and the up-and-out Call valuesare smaller that the value of the corresponding European Call. That is be-cause, in general, there is a non-vanishing probability that the barrier willbe reached. In general, the following relations hold:

C0 ≥ CDI,0

C0 ≥ CDO,0.

2. We now want to compute the probability that, at t = 2 months, the un-derlying value will reach the barrier, i.e. we want to compute P (St ≥ U)at t = 2/12 (2 months).Since the dynamics of the underlying’s value is described by a geometricBrownian motion,

St = S0 exp{(

μ − σ2/2)t + σWt

}and the probability that its value at t = 2/12 will be greater than U is thefollowing:

P (St ≥ U) = P (ln (St) ≥ ln (U))

= P

(ln (S0) +

(μ − σ2

2

)t + σWt ≥ ln (U)

)


= P

⎛⎝Wt√

t≥

ln (U) − ln (S0) −(μ − σ2

2

)t

σ√

t

⎞⎠

= 1 − N

⎛⎝ ln (U) − ln (S0) −

(μ − σ2

2

)t

σ√

t

⎞⎠ .

With the assigned data we obtain:

P (S2months ≥ U) = 1 − N

⎛⎝ ln (12) − ln (8) −

(0.2− (0.36)2

2

)212

0.36√

2/12

⎞⎠

= 1 − N (2.606) = 0.0046.

Exercise 9.4. In a binomial setting, consider a Lookback Call option withmaturity T = 8 months, written on an underlying stock (without dividends)whose price can rise or fall by 50% during each two-months period of the next8 months. The underlying’s initial value is S0 = 10 euros, the risk-free interestrate is 8% (per year).Compute the initial value of the Lookback option considered. What is the

probability (risk-neutral) that its payoff will be strictly positive?

SolutionThe data provide u = 1.5 and d = 0.5 as parameters of the binomial model

and S0 = 10 as initial stock price. In the present setting, the underlying’sdynamics can be described as follows:

50.625↗

33.75↗ ↘

22.5 16.875↗ ↘ ↗

15 11.25↗ ↘ ↗ ↘

S0 = 10 7.5 5.625↘ ↗ ↘ ↗

5 3.75↘ ↗ ↘

2.5 1.875↘ ↗

1.25↘

0.625

− −−− − −−− −− −− − −−− −− −−t = 0 t = 1

(2 months)t = 2

(4 months)t = 3

(6 months)T = 4

(8 months)


In order to simplify the notation, we shall denote as follows the nodes ofthe tree just described:

M↗

G↗ ↘

D N↗ ↘ ↗

B H↗ ↘ ↗ ↘

A E O↘ ↗ ↘ ↗

C I↘ ↗ ↘

F P↘ ↗

L↘

Q

We recall that the payoff of a Lookback Call option is (ST − Smin) , whereSmin denotes the minimum value assumed by the underlying during the op-tion lifetime. The initial value of this option (written on an underlying witha binomial dynamics) is then given by:

FCm (0) =1

(1 + r)TEQ [ST − Smin ] , (9.22)

where Q is the risk-neutral probability measure.Moreover, since Smin depends on the “trajectory” of the underlying value

process, we must examine the contribution of each single path followed by theunderlying along the binomial tree.Before considering all possible final nodes, let us compute the risk-neutral

probability measure with respect to which the expectation must be calculated.As usual, in the present binomial setting, qu is given by:

qu =(1 + r)2/12 − d

u − d=

(1.08)2/12 − 0.51.5− 0.5

= 0.513

and qd = 1 − qu = 0.487.Now we must consider all possible paths arriving at the final nodes M, N,

O, P and Q.

Final node: M If the underlying value at maturity was 50.625 (node M),then the only possible path would be that described by a growth of the


underlying value at each time step. In this case we should have

Smin = 10ST = 50.625

}=⇒ ST − Smin = 40.625.

The probability that this occurs is q4u = 0.0693.

Final node: Q If the underlying value at maturity was 0.625 (node Q),then the only possible path would be that described by a fall of the under-lying value at each time step. In this case we should have

Smin = 0.625ST = 0.625

}=⇒ ST − Smin = 0.

The probability that this occurs is q4d = (1 − qu)4 = 0.0562.

Final node: N If the underlying value at maturity was 16.875 (node N),then there would be 4 possible paths to consider.

First path. The price dynamics is the following:

G↗ ↘

D N↗

B↗

A

and the corresponding probability is q3u (1 − qu) = 0.0654. In this case we

have:Smin = 10

ST = 16.875

}=⇒ ST − Smin = 6.875.

Second path. The price dynamics is the following:

D N↗ ↘ ↗

B H↗

A

with probability q3u (1 − qu) = 0.0654. In this case we have:

Smin = 10ST = 16.875

}=⇒ ST − Smin = 6.875.


Third path. The price dynamics is the following:

N↗

B H↗ ↘ ↗

A E

and the probability of this path is q3u (1 − qu) = 0.0654. In this case we

have:Smin = 7.5

ST = 16.875

}=⇒ ST − Smin = 9.375.

Fourth path. The price dynamics is the following:

N↗

H↗

A E↘ ↗

C

and the probability is q3u (1 − qu) = 0.0654. In this case we have:

Smin = 5ST = 16.875

}=⇒ ST − Smin = 11.875.

Final node: O If the underlying value at maturity was 5.625 (node O),there would be 6 possible paths to consider.


D↗ ↘

B H↗ ↘

A O

and the probability is q2u (1 − qu)2 = 0.0560. In this case we have:

Smin = 5.625ST = 5.625

}=⇒ ST − Smin = 0.



B H↗ ↘ ↗ ↘

A E O


Smin = 5.625ST = 5.625

}=⇒ ST − Smin = 0.


A E O↘ ↗ ↘ ↗

C I


Smin = 3.75ST = 5.625

}=⇒ ST − Smin = 1.875.


A O↘ ↗

C I↘ ↗

F


Smin = 2.5ST = 5.625

}=⇒ ST − Smin = 3.125.

Fifth path. The price dynamics is the following:

B↗ ↘

A E O↘ ↗

I



Smin = 3.75ST = 5.625

}=⇒ ST − Smin = 1.875.

Sixth path. The price dynamics is the following:

H↗ ↘

A E O↘ ↗

C

and the probability that such occurs is q2u (1 − qu)2 = 0.0560. In this case

we have:Smin = 5

ST = 5.625

}=⇒ ST − Smin = 0.625.

Final node: P If the underlying value at maturity was 1.875 (node P),there would be 4 possible paths to consider.


B↗ ↘

A E↘

I↘

P

and the probability is qu (1 − qu)3 = 0.0593. In this case we have:

Smin = 1.875ST = 1.875

}=⇒ ST − Smin = 0.


A↘

C↘

F P↘ ↗

L



Smin = 1.25ST = 1.875

}=⇒ ST − Smin = 0.625.


A↘

C I↘ ↗ ↘

F P


Smin = 1.875ST = 1.875

}=⇒ ST − Smin = 0.


A E↘ ↗ ↘

C I↘

P


Smin = 1.875ST = 1.875

}=⇒ ST − Smin = 0.

By summing up the contributions of all cases, we conclude that the Look-back Call option value at time t = 0 is given by:

FCm (0) =1

(1 + r)8/12[q4

u · 40.625+q3u (1−qu) (6.875+6.875+9.375+11.875)

+ q2u (1 − qu)2 (0 + 0 + 1.875 + 3.125 + 1.875 + 0.625)

+ qu (1 − qu)3 (0 + 0.625 + 0 + 0) + (1 − qu)4 · 0]= 5.561 euros.

Moreover, we can easily find that the (risk-neutral) probability that thepayoff of the Lookback Call will be strictly positive is given by:

Q ({payoff > 0}) = q4u + 4q3

u (1 − qu) + 4q2u (1 − qu)2 + qu (1 − qu)3 = 0.6142.


Exercise 9.5.With the same data of the previous exercise and again in abinomial setting, compute the value at time t = 0 of an average strike Calland the (risk-neutral) probability that its payoff will be strictly positive.

SolutionAn “average strike” option is an Asian option with payoff at maturity ex-

plicitly depending on the average of the values assumed by the underlyingasset during the option lifetime, according to the following expression:

FAS(ST , Smed, T ) = (ST − Smed)+ = max{ST − Smed; 0} .

We shall assume that the average considered is the arithmetic mean, whichin our binomial setting with 4 time steps is:

Smed =∑4

t=1 St

4. (9.23)

By recalling the scheme provided in the previous exercise we have to con-sider the following possible paths.

Final node: M If the underlying value at maturity was 50.625 (node M),then the only possible path for the underlying price would be that of agrowth at each time step. In this case we have:

Smed = 30.469ST = 50.625

}=⇒ (ST − Smed)+ = 20.156.

The probability that this occurs is q4u = 0.06935.

Final node: Q If the underlying value at maturity was 0.625 (node Q),then the only possible path for the underlying price would be that of a fallat each time step. In this case we have:

Smed = 2.344ST = 0.625

}=⇒ (ST − Smed)+ = 0.

The probability that this occurs is q4d = (1 − qu)4 = 0.0562.

Final node: N If the underlying value at maturity was 16.875 (node N),there would be 4 possible paths for the underlying price.



G↗ ↘

D N↗

B↗

A

and the probability of this is q3u (1 − qu) = 0.0654. In this case we have:

Smed = 22.031ST = 16.875

}=⇒ (ST − Smed)+ = 0.


D N↗ ↘ ↗

B H↗

A


Smed = 16.406ST = 16.875

}=⇒ (ST − Smed)+ = 0.469.


N↗

B H↗ ↘ ↗

A E


Smed = 12.656ST = 16.875

}=⇒ (ST − Smed)+ = 4.219.



N↗

H↗

A E↘ ↗

C


Smed = 10.156ST = 16.875

}=⇒ (ST − Smed)+ = 6.719.

Final node: O If the underlying value at maturity was 5.625 (node O),then there would be 6 possible paths for the underlying price.


D↗ ↘

B H↗ ↘

A O


Smed = 13.594ST = 5.625

}=⇒ (ST − Smed)+ = 0.


B H↗ ↘ ↗ ↘

A E O


Smed = 9.844ST = 5.625

}=⇒ (ST − Smed)+ = 0.



A E O↘ ↗ ↘ ↗

C I


Smed = 5.469ST = 5.625

}=⇒ (ST − Smed)+ = 0.156.


A O↘ ↗

C I↘ ↗

F


Smed = 4.219ST = 5.625

}=⇒ (ST − Smed)+ = 1.406.

Fifth path. The price dynamics is the following:

B↗ ↘

A E O↘ ↗

I


Smed = 7.969ST = 5.625

}=⇒ (ST − Smed)+ = 0.

Sixth path. The price dynamics is the following:

H↗ ↘

A E O↘ ↗

C



Smed = 7.344ST = 5.625

}=⇒ (ST − Smed)+ = 0.

Final node: P If the underlying value at maturity was 1.875 (node P),there would be 4 possible paths for the underlying price.


B↗ ↘

A E↘

I↘

P


Smed = 7.031ST = 1.875

}=⇒ (ST − Smed)+ = 0.


A↘

C↘

F P↘ ↗

L


Smed = 2.656ST = 1.875

}=⇒ (ST − Smed)+ = 0.


A↘

C I↘ ↗ ↘

F P



Smed = 3.281ST = 1.875

}=⇒ (ST − Smed)+ = 0.


A E↘ ↗ ↘

C I↘

P


Smed = 4.531ST = 1.875

}=⇒ (ST − Smed)+ = 0.

The initial value of the option considered is then given by:

FAS (ST ; Smed; 0)= 1

(1.08)8/12 [0.0693 · 20.156 + 0.0654 · (0.469 + 4.219 + 6.719)+0.0560 · (0.156 + 1.406) + 0.0593 · 0 + 0.0562 · 0]

= 2.26.

Moreover, we can easily find that the (risk-neutral) probability that thepayoff of the average strike Call will be strictly positive is given by:

Q ({payoff > 0}) = q4u + 3q3

u (1 − qu) + 2q2u (1 − qu)2

= 0.0693 + 3 · 0.0654 + 2 · 0.0560 = 0.3775.

Exercise 9.6. Compute the value at time t = 0 of a geometric Asian option,of the average-rate kind, written on an underlying stock whose price dynamicsis described by a geometric Brownian motion with diffusion coefficient σ. Therisk-free interest rate is r, the strike K.

SolutionWe first remark that the payoff of a geometric Asian option of average-rate

type can be written as follows:

CAR(T, ST ) = max{ST − K; 0} = max{S0 exp(XT ) − K; 0}, (9.24)

where ST denotes the geometric average of the values assumed by the under-lying during the time interval [0, T ], while XT denotes the arithmetic meanof the values assumed by the log-returns during the same time interval. The


arithmetic average of the log-returns can be easily derived when the under-lying price dynamics is described by a geometric Brownian motion. If theunderlying price evolves as follows:

dSt = rStdt + σStdWt,

i.e. St = S0 exp(Xt) where

dSt = (r − σ2/2)dt + σdWt,

then, by integrating by parts, we can write immediately:

∫ t

0

Xudu = tXt −∫ t

0

udX(u). (9.25)

By inserting the dynamics of X and dividing by t, (9.25) becomes:

Xt =(

r − σ2

2

)∫ t

0

(1 − u

t

)du +

∫ t

0

(1 − u

t

)σdWt.

The last equation implies that XT is distributed as a normal random vari-able with mean:

E[XT ] =(

r − σ2

2

)∫ T

0

(1 − u

T

)du =

(r − σ2

2

)T

2, (9.26)

and variance:

V (XT ) =∫ T

0

(1 − u

T

)2

σ2du =σ2

3T. (9.27)

Now, since the value at time t = 0 of an average rate option is just thediscounted value of its expected payoff (computed with respect to the risk-neutral measure):

CAR(0, S0) = e−rT E[max{S0 exp(XT ) − K; 0}] , (9.28)

we can obtain this value by applying the Black-Scholes formula for EuropeanCall options, simply by comparing the payoffs and substituting the new valuesof the mean and the variance in the arguments. A direct substitution providesthe following formula for the (geometric) average rate Call option:

CAR(0, S0) = S0e−(

r+ σ26

)T2 N

(d∗ + σ

√T

3

)− Ke−rT N(d∗), (9.29)


where d∗ is defined as follows:

d∗ �ln(S0

K ) +(r − σ2

2

)T2

σ√

T3

. (9.30)

Exercise 9.7. Consider the following down-and-out Call option with matu-rity T = 1 year, strike K = 30 euros and barrier threshold L = 25 euros. Theunderlying price dynamics is described by a geometric Brownian motion withvolatility σ = 0.40 (per year). The current price of the underlying is S0 = 30euros, while the risk-free interest rate is r = 0.05 (per year).Construct a portfolio consisting of European options (approximately and

statically) replicating the Barrier option considered.

SolutionThere is a large amount of arbitrariness in constructing a portfolio repli-

cating the Barrier option. We start the process with a simple remark: if twoportfolios have the same value on the contour of a region in the (t, S)-space,then they must have the same value inside the region. The idea is then toapproximate the Barrier option on the contour of such region by means ofEuropean options with suitable maturities and strikes.In the present case, we consider as a contour the rectangle (in the (t, S)

space) delimited by the straight lines S = 25, S = 30 (corresponding to S = Land S = K) and t = 0, t = 1 (the last corresponding to t = T ), where time ismeasured in years. The down-and-out Call takes the following values on theselines:

C(S, 1) = max{S − 30; 0} (for S > 25) (9.31)

andC(25, t) = 0 (for 0 ≤ t ≤ 1). (9.32)

0

5

10

15

20

25

30

35

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 1,1

time

un

der

lyin

g p

rice

Fig. 9.1. Static replication of a down-and-out Call


The simplest way to replicate the upper contour (S = 30) is to buy aEuropean Call option with strike KA = 30. We denote by A such an option.

To replicate the lower contour we can divide the time interval [0, 1] in threesub-intervals of 4 months (i.e. 1/3 of a year) and choose three different Calloptions satisfying condition (9.32) for t = 0, 1

3 , 23 , respectively.

A second option (denoted by B) to be considered for replication on thelower contour can be a Call with strike KB = 25 and maturity TB = 1. Bythe Black-Scholes formula for European Call options (see (7.5)), the valuesof options A and B at the point (t = 2/3, S = 25) of the lower contour areCA(25, 2/3) = 0.86 and CB(25, 2/3) = 2.48.To satisfy the condition (9.32) for t = 1/3 we need another option (vanish-

ing at all the points previously considered): this third option (denoted by C)can be a Call option with maturity TC = 2/3 and strike KC = 25.Finally, we need to satisfy the condition (9.32) at time t = 0. To this end

we choose a fourth European Call (denoted by D) with maturity TD = 1/3and strike KD = 25.The number of options of each kind to be held in the replicating portfolio is

determined by the condition that the replicating portfolio assumes value zeroat each of the points considered on the contour. So, by assuming that one unitof the option A is held in the replicating portfolio, the number β of optionsB to buy (or sell) is obtained by imposing that at point (t = 2/3, S = 25)

CA(25, 2/3) + βCB(25, 2/3) = 0.

Since the values of options A and B at (t = 2/3, S = 25) are 0.86 and 2.48,respectively, this condition provides for β the value β = −0.35.As far as the number of options C is concerned, we can repeat the same

argument and impose the condition at (t = 1/3, S = 25):

CA(25, 1/3)+ βCB(25, 1/3) + γCC (25, 1/3) = 0.

Since A, B, and C at (S = 25, t = 1/3) are valued 1.85, 3.64 and 2.48, respec-tively, this gives γ = −0.232.Finally, we can determine the number of options of type D by imposing the

analogous condition at (t = 0, S = 25):

CA(25, 0) + βCB(25, 0) + γCC(25, 0) + δCD(25, 0) = 0.

Since the options’ values are 2.70, 4.51, 3.64 and 2.48 for A, B, C, and D,respectively, the value of δ = −0.113 follows.The replicating portfolio has then the following composition:

(1,−0.35,−0.232,−0.113)

in options A, B, C, D, respectively.



Exercise 9.8. Compute the value at time t = 0 of a Chooser option writ-ten on an underlying stock whose price dynamics is described by a geometricBrownian motion, with strike K = 20 euros and maturities T1 = 6 monthsand T2 = 1 year. The initial value of the underlying is S0 = 20 euros, therisk-free interest rate r = 10% (per year) and the volatility σ = 20% (peryear).

Exercise 9.9. Compute the value at time t = 0 of a Barrier option of “up-and-in” Call type, written on an underlying stock whose price dynamics isdescribed by a geometric Brownian motion, with initial value S0 = 10 euros,σ2 = 0.016, r = 0.04, K = 6 euros and barrier value U = 8 euros. By usingthe relationship between Barrier and European Call options, find the initialprice of the corresponding “up-and-out” Call.

Exercise 9.10. Compute the value at time t = 0 of a Binary option of “as-set or nothing” Call type, with the same parameters (except the strike K) ofExercise 9.1.

Exercise 9.11.With the same data of Exercises 9.4 and 9.5, and again in abinomial setting, compute the initial price of a Lookback Call option (on theminimum) with maturity T = 10 months. Find, moreover, the initial value ofthe corresponding Put option.

Exercise 9.12.With the same data of Exercises 9.4 and 9.5, and again in abinomial setting, compute the initial price of an average strike Call option,with a payoff at maturity expressed by (ST − SMG)+ = max{ST − SMG; 0},where SMG is the geometric mean of the values assumed by the underlyingduring the option lifetime, i.e. SMG = n

√Πn

i=1Si.

Exercise 9.13.With the same data of Exercises 9.4 and 9.5, and again in a bi-nomial setting, compute the price at time t = 0 of an average rate Call option,with a payoff at maturity expressed by (Smed − K)+ = max{Smed − K; 0},where Smed denotes the arithmetic mean of the values assumed by the under-lying during the option lifetime and K = 5 euros.

Exercise 9.14. Determine the payoff at maturity of a portfolio composed bythe Lookback Call of Exercises 9.4 and by the corresponding Lookback Putoption. Compute the value of this portfolio at time t = 0.

Exercise 9.15. Compute the initial value of a geometric Asian option, of theaverage-rate type, with maturity T = 1 year, strike K = 30 euros, written onan underlying stock with dynamics described by a geometric Brownian motionand with initial value S0 = 36 euros. The risk-free interest rate is r = 0.04(per year) and the volatility σ = 0.4 (per year).


Exercise 9.16. Compute the value at t = 0 of a (floating strike) LookbackCall option on the minimum, with maturity T = 1 year, written on an un-derlying stock with dynamics described by a geometric Brownian motion andwith initial value S0 = 36 euros. The risk-free interest rate is r = 0.04 (peryear) and the volatility σ = 0.4 (per year).Consider the binomial model with four time steps (each of three months)

approximating the geometric Brownian motion above. After computing theLookback option value at time t = 0 in this binomial setting, evaluate thebinomial approximation’s quality by comparing the two results obtained.

Exercise 9.17. Compute the initial value of a down-and-out Call option withthe same data of Exercise 9.7, except the barrier value (which is now a lowerthreshold), here assumed to be L = 40 euros, and construct a static replicatingportfolio composed by 4 European options.Do Call options give a better or worse replication than Put options?

10

Interest Rate Models


In the valuation problems presented in the previous chapters, the interest ratewas assumed to be a deterministic parameter, constant in most cases. This isof course a crude approximation of the real situation where interest rates varyover time in an unpredictable way. This assumption can be reasonable onlywhen dealing with financial contracts with short maturities (like options),while for financial derivatives with longer lifetime (like bonds) or any otherfixed-income products, it can be quite misleading. It is necessary, then, toadopt stochastic models for interest-rate dynamics.

Let us start by defining the relevant quantities that we wish to describe.A zero-coupon bond (ZCB) is a contract that guarantees the buyer one

unit (by convention) of the reference currency at maturity T . Since the valueof this contract depends both on the date at which it is written and on itsmaturity T , we will denote by Z(t, T ) its value at time t. If the zero-couponbond guarantees the buyer the cash amount N at maturity T , we will specifythat the zero-coupon bond’s face value is N . In this case, its value at time twill be expressed by N · Z (t, T ).A coupon bond with face valueN , maturity T and periodically paid coupons

(at the end of each semester, for example) at the annual rate rc is a contractthat guarantees the buyer the cash amount N at time T and the couponswith value c at the end of each period (in the case of one semester: c = rc

2·N)

until maturity T .We define instantaneous forward rate with respect to maturity T the fol-

lowing quantity:

f(t, T ) � −∂ lnZ(t, T )∂T

(10.1)

and short rate the quantity rt � f(t, t).


202 10 Interest Rate Models

The bond valuation is then strictly related to the interest-rate dynamics.Without any further specification, in the following we shall refer to shortrates. When dealing with the valuation problem for zero-coupon bonds thefirst difficulty is the incompleteness of a short-rate stochastic model, so theno-arbitrage requirement does not provide a unique price for a generic deriva-tive (in our case the zero-coupon bond). Nevertheless, no arbitrage impliessome consistency condition on the zero-coupon bonds’ values (with differentmaturities) which allows to reduce the valuation problem to the final-valueproblem for a partial differential equation; the coefficient of this PDE mustbe determined according to some “exogenous” criterion.If the short-rate dynamics is described by a model of the following kind:

drt = u(rt, t)dt + w(rt, t)dWt, (10.2)

then the zero-coupon value Z(t, T ) must satisfy the following PDE:

∂Z

∂t(t, T ) +

12w2 (t, T )

∂2Z

∂r2(t, T ) + (u − λw) (t, T )

∂Z

∂r(t, T )− (rZ) (t, T ) = 0

together with the final condition Z(T, T ) = 1. The term λ in the previousequation is called the Market Price of Risk and, by the consistency relationimplied by the no-arbitrage condition mentioned above, it can be shown to bethe same for all bonds available in the market considered. The computationof the market price of risk turns out to be equivalent to the determination ofthe risk-neutral measure, which is not unique any more due to market incom-pleteness, chosen by the market in order to assign prices to traded securities.Several different approaches can be followed in order to determine the coef-ficient λ: among these we just mention the general equilibrium framework.The most popular in practice, anyway, is the one based on a “calibration”procedure. In the following, we will assume that the coefficient λ is known.An explicit solution of the bond pricing PDE for short-rate models is avail-

able only in a few relevant cases. Among these the most relevant are thoseassuming the following general form:

Z(t, T ) = exp {A(t, T ) − rtB(t, T )} , (10.3)

where A and B are deterministic functions of t and T . These solutions arecalled affine term structures. It is possible to prove that a sufficient conditionfor the zero-coupon bond equation to admit a solution of the form (10.3) isthat the coefficients u, w of the stochastic differential equation describing theshort-rate dynamics in (10.2) have the following form:

u(r, t) = α(t)r + β(t) (10.4)

w(r, t) =√

γ(t)r + δ(t). (10.5)


Moreover, the functions A, B are related to the coefficients α, β, γ, δ by thefollowing ordinary differential equations (of Riccati type):

∂B(t, T )∂t

+ α(t)B(t, T ) − 12γ(t)B2(t, T ) = −1, B(T, T ) = 0

∂A(t, T )∂t

− β(t)B(t, T ) +12δ(t)B2(t, T ) = 0, A(T, T ) = 0.

(10.6)

Among these models, we briefly recall the most popular (and easiest tohandle) on which we will focus in view of their applications. In particular, weshall deal with the models proposed by Vasiček, Ho and Lee, Hull and White.

Vasiček Modeldrt = a (b − rt) dt + σdWt, (10.7)

i.e. α(t) = −a, β(t) = ab, γ(t) = 0 and δ(t) = σ2.In this case, we have:

B(t, T ) =1 − e−a(T−t)

a(10.8)

A(t, T ) =[B(t, T ) − (T − t)] (a2b − σ2/2)

a2− σ2B2(t, T )

4a. (10.9)

The explicit solution of the stochastic differential equation (10.7) describingthe short-rate dynamics is the following:

rt = r0e−at + b

(1 − e−at

)+ σ

∫ t

0

e−a(t−s)dWs. (10.10)

Moreover:E [rt] = r0e

−at + b(1 − e−at

). (10.11)

Ho-Lee Modeldrt = ϑtdt + σdWt, (10.12)

i.e. α(t) = 0, β(t) = ϑt, γ(t) = 0 and δ(t) = σ2.In this case, we have that:

B(t, T ) = T − t (10.13)

A(t, T ) =∫ T

t

ϑs(s − T )ds +σ2(T − t)3

6. (10.14)

Hull-White Model

drt = (ϑt − art) dt + σdWt, (10.15)

i.e. α(t) = −a, β(t) = ϑt, γ(t) = 0 and δ(t) = σ2.


In this case, we have that:

B(t, T ) =1 − e−a(T−t)

a(10.16)

A(t, T ) =∫ T

t

[12σ2B2(s, T ) − ϑsB(s, T )

]ds. (10.17)

By a suitable calibration procedure it is possible to obtain the functions ϑt

appearing in the Ho-Lee and Hull-White valuation formulas:

ϑHLt =

∂f(0, t)∂t

+ σ2t (10.18)

ϑHWt =

∂f(0, t)∂t

+ af(0, t) +σ2

2a(1 − e−2at). (10.19)

In both expressions f(0, t) denotes the instantaneous forward rate at timet = 0.When the above models are taken as pricing framework, the zero-coupon

bonds can be easily valuated, and so are the coupon bonds and the bondoptions. We shall examine in a more detailed way these valuation problemsin the exercises.The formulas providing the value of a Call option written on a zero-coupon

bond with face value N are, for the three models presented, the following.

Price of a Call option on a ZCB in the Vasiček Model

C(t, T, K, T ∗) = N · Z(t, T ∗) ·N(d) − K · Z(t, T ) · N(d − σp), (10.20)

where K, T denote the strike and the maturity of the Call option, respectively,while T ∗ is the bond maturity, N (·) the cumulative distribution function ofa standard normal random variable and the quantities d and σp are definedas follows:

d =1σp

ln[N · Z(t, T ∗)K · Z(t, T )

]+

12σp (10.21)

σp =1 − e−a(T∗−T )

a

√σ2

2a

[1 − e−2a(T−t)

]. (10.22)

Price of a Call option on a ZCB in the Ho-Lee Model



where

d =1σp

ln[N · Z(t, T ∗)K ·Z(t, T )

]+

12σp (10.24)

σp = σ(T ∗ − T )√

T . (10.25)

Price of a Call option on a ZCB in the Hull-White Model


where

d =1σp

ln[N · Z(t, T ∗)K · Z(t, T )

]+

12σp (10.27)

σp =1 − e−a(T∗−T )

a

√σ2

2a

[1 − e−2a(T−t)

]. (10.28)

Strictly analogous formulas exist for European Put options. These formu-las can be proved quite easily with the help of a useful technique, called theChange of Numéraire. The latter provides a solution to several valuation prob-lems arising in several, different contexts, like interest-rate models, currencyderivatives, and some exotic options. We briefly recall the main ideas underly-ing this powerful method. Roughly speaking, a Numéraire is the unit in whichall assets in a market model are expressed. This can be a currency unit, ora money market account, which is simply given by Bt = exp

(∫ t

0rudu

)in

market models with a deterministic short rate. Changing numéraire meansthat a new unit is assumed to express financial values, and the new numérairedoes not have to be constant, nor deterministic, but it can be described bya stochastic dynamics; the only requirement it must fulfill is to be strictlypositive for all t. The fundamental theorems of asset pricing are usually ex-pressed with respect to the numéraire Bt (bond) in such a way that, in orderto avoid arbitrage opportunities, all asset price processes in the market modelsconsidered (including derivatives), divided by Bt, must be local martingaleswith respect to the risk-neutral measure. The same results can be expressedin terms of a different numéraire, but the risk-neutral measure for the marketmodel will be different as well. It can be proved that, under suitable integra-bility conditions, the relationship between the two risk-neutral measures isprovided by the following formula:

Lt =dQ1

dQ0

∣∣∣∣Ft

=S1

t

S0t

S00

S10

, (10.29)


where Lt denotes the Radon-Nikodym derivative of the risk-neutral measureQ1 (with S1 as numéraire) with respect to the risk-neutral measure Q0 (withS0 as numéraire) at time t.A special role in interest rate models is played by the T-forward measure,

that is the risk-neutral measure adopting the value p(t, T ) (= Z(t, T )) of azero-coupon bond with maturity T as numéraire. We shall illustrate in theexercises how this method can be used to obtain pricing formulas.

Up to now we have focused our attention on short-rate models. In mar-ket practice, however, other relevant models play a role; among these, a veryimportant one is the LIBOR (London InterBank Offer Rate) market modeldescribing the dynamics of rates. The simple forward LIBOR rate at time tfor the time interval [S, T ] is defined by:

L(t, S, T ) � p(t, S) − p(t, T )(T − S)p(t, T )

, (10.30)

where 0 ≤ S ≤ T and p(t, T ) (= Z(t, T )) denotes the value at time t of azero-coupon bond with maturity T (supposed to be quoted on the market).The LIBOR spot rate for the time interval [S, T ] is defined by:

L(S, T ) � 1 − p(S, T )(T − S)p(S, T )

, (10.31)

and it is simply the forward rate for t = S.If the dynamics of the LIBOR rate for each interval [Ti−1, Ti] (where Ti are

the revision dates for the interest rate) is assumed to be described by a geo-metric Brownian motion, the well-known Black Formula provides an explicitvalue for the Caplets. These are financial contracts with the following payoff:

Xi � αi · max{L(Ti−1, Ti) − R; 0} ,

where αi = Ti − Ti−1 is the tenor and R is the Cap rate. The Black Formulafor Caplets is the following:

Capli(t) = αi · p(t, Ti) [L(t, Ti−1, Ti) ·N(d1) − R ·N(d2)] , i = 1, ..., N,(10.32)

where:

d1 =1

σi

√Ti − t

[ln(

L(t, Ti−1, Ti)R

)+

σ2i

2(Ti − t)

]

d2 =1

σi

√Ti − t

[ln(

L(t, Ti−1, Ti)R

)− σ2

i

2(Ti − t)

]

and σi denotes the Black volatility for the i-th Caplet. The Black volatilitiesmust be calibrated on the market values and, in order to simplify the treat-


ment, in the exercises we shall assume that σi take a constant value (we shallassume a flat structure for the Black volatilities).

The reader interested in a rigorous and systematic treatment of the valua-tion problems for interest rate derivatives should consult the books by Björk[6], Brigo and Mercurio [8], El Karoui [16] and Mikosch [28].


Exercise 10.1. Consider a short rate model of Vasiček type, that is (rt)t≥0

satisfyingdrt = a (b − rt) dt + σdWt,

with a = 0.4, b = 0.01, σ = 0.2 and initial value r0 = 4% on an annual basis.

1.(a) Compute the value at time t = 0 of a zero-coupon bond with face value40 euros and maturity T ∗ = 2.5 years.

(b) Discuss whether there exists a constant interest rate value r∗ such thatthe value of a zero-coupon bond with face value 40 euros and maturityT ∗ = 2.5 coincides with the bond value considered in item 1.(a).

2. Consider a European Call option with maturity T of one year and 2 months,strike K = 36 euros and written on a zero-coupon bond with face value 40euros.

(a) Compute the initial value of the European Call considered.(b) Compute the value of the same Call option of the previous item under

the assumption that the short rate has constant value re = E [rT ] , whereT denotes the maturity of the Call option.

3. Consider now a coupon bond with face value N∗, maturity T ∗ = 2.5 years,with coupon paid at the end of each semester with annual rate of 8%.

(a) Find the value of N∗ such that the zero-coupon bond of item 1.(a) coin-cides with the present value of the coupon bond, if this value is computedunder the assumption of a constant short rate with value r0.

(b) Compute the value (at time t = 0) of a European Call option with ma-turity T of one year and 2 months, strike K = 36 euros, written on thecoupon bond with face value N∗ and maturity T ∗ = 2.5.

(c) Compute the value (at time t = 0) of a European Call option with ma-turity T of one year and 2 months, strike K = 36 euros and written ona zero-coupon bond with face value N∗.

4. Compute the value of the European Put options with the same parametersof the Call options considered in items 2.(a) and 3.(b).


Solution

1. (a) The price of the zero-coupon bond (with unit face value and maturityT ∗) at time t in the Vasiček model is given by:

Z (t, T ∗) = eA(t,T∗)−rtB(t,T∗),

where

B (t, T ∗) =1 − e−a(T∗−t)

a

and

A (t; T ∗) =[B (t, T ∗) − (T ∗ − t)]

(a2b − σ2

2

)a2

− σ2B2 (t, T ∗)4a

.

To find the initial value of a zero-coupon bond with face value 40 eu-ros and maturity T ∗ = 2.5 years, we need to compute the quantitiesA (0; 2.5) and B (0; 2.5).Since

B (0; 2.5) =1 − e−0.4·2.5

0.4= 1.58

A (0; 2.5) =[B (0; 2.5)− 2.5]

((0.4)2 · 0.01− (0.2)2

2

)(0.4)2

− (0.2)2 B2 (0; 2.5)4 · 0.4

= 0.043,

the initial value of the zero-coupon bond with face value 40 euros is thengiven by:

N ·Z (0; 2.5) = 40 · eA(0;2.5)−r0·B(0;2.5) = 40 · e0.043−0.04·1.58 = 39.2.

This means that we have to invest 39.2 euros today in order to get 40euros in two-and-a-half years.By the previous results, it follows that the initial value of a zero-couponbond (with unit face value) is Z (0; 2.5) = 0.980 euros.

(b)We have to find a constant short rate value r∗ such that the price of azero-coupon bond with face value 40 euros and maturity T ∗ = 2.5 yearscoincides with the value found in item (a).This short rate r∗ must then satisfy the following equation:

N · Z (0; T ∗) = N · e−r∗T∗

or, equivalently, Z (0; 2.5) = e−r∗·2.5. We get then:

r∗ = − ln Z (0; 2.5)2.5

= 0.008.


2. We focus now on a European Call option with maturity T of one yearand 2 months (that is T = 7/6 years) and strike K = 36, written on azero-coupon bond with face value N = 40 euros and maturity T ∗ = 2.5years.

(a)We recall that the initial value of a European Call option written on azero-coupon bond with face value N is given by:

C0 = N · Z (0; T ∗) · N (d) − K ·Z (0; T ) · N (d − σp) , (10.33)

where T ∗ and T stand, respectively, for the maturity of the ZCB and ofthe option, and

d =1σp

ln[N · Z(0, T ∗)K · Z(0, T )

]+

12σp (10.34)

σp =1 − e−a(T∗−T )

a

√σ2

2a[1 − e−2aT ]. (10.35)

First of all, we remark that the previous formula extends the “classical”Black-Scholes formula and that the factor N ·Z (0; T ∗) simply representsthe current value of the underlying security, while K ·Z (0; T ) representsthe current value (at t = 0) of the strike. This happens just becauseZ (0; s) represents the present value of a unit of currency paid at time s.Now, we turn to the Call option valuation problem. Looking at formula(10.33), we can immediately observe that only Z (0; T ) must be com-puted, since the value of Z (0; T ∗) has been already computed in 1.(a).By proceeding in the same way as in 1.(a), we obtain that, for maturityT = 7/6 years (i.e. one year and two months),

B (0; 7/6) =1 − e−0.4·76

0.4= 0.9323

A (0; 7/6) =

[B(0; 7

6

)− 76

] ((0.4)2 · 0.01− (0.2)2

2

)(0.4)2

− (0.2)2 B2(0; 7

6

)4 · 0.4

= 0.0052.

The zero-coupon bond (with unit face value) price is then given by:

Z (0; 7/6) = eA(0;7/6)−r0·B(0;7/6) = e0.0052−0.04·0.9323 = 0.9684.

Hence

σp =1 − e−0.4(2.5−7/6)

0.4

√(0.2)2

2 · 0.4[1 − e−2·0.4·7/6

]= 0.18

d =ln[

40·Z(0;2.5)36·Z(0;7/6)

]0.18

+0.182

= 0.741.


As a direct consequence, the initial value of the European Call with ma-turity T = 7/6, strike K = 36, written on the zero-coupon bond withface value N = 40 and maturity T ∗ = 2.5 years is

C0 = 40 · 0.980 ·N (0.741)− 36 · 0.9684 ·N (0.741− 0.18) = 5.365 euros.

(b) Assume that the short rate is constant with value re = E [rT ] and con-sider a European Call option with maturity T = 7/6 years, strikeK = 36euros, written on a zero-coupon bond with face value N = 40 euros andmaturity T ∗ = 2.5 years.We recall that for the Vasiček model the explicit solution of the stochas-tic differential equation describing the short-rate dynamics is the follow-ing:

rt = r0e−at + b

(1 − e−at

)+ σ

∫ t

0

e−a(t−s)dWs.

Moreover, E [rt] = r0e−at + b (1 − e−at) = r0 +(b − r0)

(1 − e−aT

). This

implies that:

re = E [rT ] = r0 + (b − r0)(1 − e−aT

)= 0.04 + (0.01− 0.04)

(1− e−0.4·76

)= 0.029.

At maturity, the Call option payoff is then the following:

(40 · Zre (T ; T ∗) − 36)+ =(40 · e−re(T∗−T ) − 36

)+

=(40 · e−0.029·(2.5−7/6) − 36

)+

= 2.483

and the value of the Call option with this payoff is its present value(computed by using the short rate re), i.e.

Cre0 = Zre (0; T ) · (40 · Zre (T ; T ∗) − 36)+

= e−reT · (40 · Zre (T ; T ∗) − 36)+ = 2.40 euros.

3. We focus now on a coupon bond with face value N∗, maturity T = 2.5 andcoupons paid at the end of each semester with annual rate 8%.

(a)We need to compute the value N∗ that makes the price of the zero-coupon bond of item 1.(a) equal to the present value of the couponbond, when the present value is computed assuming a constant shortrate r0. Since this value is:

4N∗

100

[e−r0/2 + e−r0 + e−3r0/2 + e−2r0

]+

104N∗

100e−5r0/2,


we have to find N∗ as a solution of the following equation:

4N∗

100

[e−r0/2 + e−r0 + e−3r0/2 + e−2r0

]+

104N∗

100e−5r0/2 = 40·Z (0; 2.5) .

We obtain then:

N∗ = 100 · 40 · Z (0; 2.5)4[e−r0/2 + e−r0 + e−3r0/2 + e−2r0

]+ 104e−5r0/2

= 35.86 euros.

(b) Let us compute now the initial value of a European Call option withmaturity T = 7/6, strike 36 euros, written on the coupon bond withface value N∗ = 35.86, with maturity T ∗ = 2.5 and coupons paid at theend of each semester at annual rate 8%.A few preliminary remarks are in order. After the optionmaturity (whichdoes not coincide with any coupon payment date), the payment flow ofthe coupons will take place as follows: the cash amount c = N∗ · 0.04 =1.43 euros) will be paid at times (always expressed in yearly units) T ∗

1 =1.5, T ∗

2 = 2 and T ∗3 = T ∗ = 2.5 (this is the bond maturity). Since at this

date the bond cash value is paid back, the total cash amount receivedby the bond buyer is N∗ + c = 37.29 euros.

The basic idea in computing the price of a European option written ona coupon bond is the following:• “decompose” the underlying coupon bond as a linear combination of

n zero-coupon bonds of suitable face value, where n is the number ofcoupons paid after the option maturity;

• “decompose” the option considered as a sum of n options, each onewritten on the zero-coupon bonds of the previous step, and with suit-able strike;

• compute the value of the relevant option as the sum of the optionsconstructed in the previous step.

According to the previous remarks, let us try to decompose the Calloption strike into three contributions (K1, K2 and K3) representing:the strike (K1) of the “virtual” option with maturity T and written onthe zero-coupon bond (ZCB1), with face value N1 = c = 1.43 andmaturity T ∗

1 = 1.5 years; the strike (K2) of a further “virtual” optionwith maturity T and written the zero-coupon bond (ZCB2) with facevalue N2 = c = 1.43 and maturity T ∗

2 = 2 years; and the strike (K3)of another “virtual” option with maturity T and written on the zero-coupon bond (ZCB3) with face valueN3 = N∗+c = 37.29 and maturityT ∗

3 = T ∗ = 2.5 years.


We impose the following requirement:

N1 · Z (T ; T ∗1 ) + N2 · Z (T ; T ∗

2 ) + N3 · Z (T ; T ∗3 ) = K (10.36)

and define

K1 � N1 ·Z (T ; T ∗1 )

K2 � N2 ·Z (T ; T ∗2 )

K3 � N3 ·Z (T ; T ∗3 ) .

We remark that condition (10.36) requires just the decomposition of Kinto the three strikes K1, K2 and K3 taking into account the respectiveface values and the times to maturity of the three zero-coupon bonds.In order to compute K1,K2 and K3, we just need to calculate Z (T ; T ∗

1 ),Z (T ; T ∗

2 ) and Z (T ; T ∗3 ).

By proceeding in a strictly similar way to that outlined in item 1.(a),we get:

B (T ; T ∗1 ) =

1 − e−a(T∗1 −T )

a=

1 − e−0.4·(1.5−7/6)

0.4= 0.312

A (T ; T ∗1 ) = 0.00001

B (T ; T ∗2 ) =

1 − e−a(T∗2 −T )

a=

1 − e−0.4·(2−7/6)

0.4= 0.709

A (T ; T ∗2 ) = 0.0018

B (T ; T ∗3 ) =

1 − e−a(T∗3 −T )

a=

1 − e−0.4·(2.5−7/6)

0.4= 1.033

A (T ; T ∗3 ) = 0.0078

and

Z (T ; T ∗1 ) = eA(T ;T∗

1 )−r∗T ·B(T ;T∗

1 ) = e0.00001−0.312·r∗T

Z (T ; T ∗2 ) = eA(T ;T∗

2 )−r∗T ·B(T ;T∗

2 ) = e0.0018−0.709·r∗T

Z (T ; T ∗3 ) = eA(T ;T∗

3 )−r∗T ·B(T ;T∗

3 ) = e0.0078−1.033·r∗T ,

where r∗T is unknown and can be determined via (10.36). By the previousremarks and using N1 = N2 = 1.43 and N3 = 37.29, indeed, equality(10.36) can be written as follows:

1.43 ·e0.00001−0.312·r∗T +1.43 ·e0.0018−0.709·r∗

T +37.29 ·e0.0078−1.033·r∗T = 36.

The solution of this equation is r∗T = 0.1168. The required decompositionis then given by:

K1 = N1 · Z (T ; T ∗1 ) = 1.38

K2 = N2 · Z (T ; T ∗2 ) = 1.32

K3 = N3 · Z (T ; T ∗3 ) = 33.31.


By collecting the results available at this stage:• the underlying coupon bond has been written as a linear combinationof three zero-coupon bonds with face values N1 = 1.43, N2 = 1.43and N3 = 37.29, respectively;

• the relevant option has been decomposed into three options with ma-turity T = 7/6 years in the following way:

one option (Call 1) with strike K1 and written on ZCB1 with facevalue N1 and maturity T ∗

1 = 1.5 years;

a second option (Call 2) with strike K2 and written on ZCB2 withface value N2 and maturity T ∗

2 = 2 years;

a third option (Call 3) with strike K3 and written on ZCB3 with facevalue N3 and maturity T ∗

3 = 2.5 years;

• the initial value of the option we want to evaluate CCB0 is then:

CCB0 = C

(1)0 + C

(2)0 + C

(3)0 ,

where C(i)0 , i = 1, 2, 3, are the initial values of Call i.

So, we need to compute C(1)0 , C

(2)0 and C

(3)0 . Since Call 1, Call 2 and

Call 3 are options written on zero-coupon bonds, the following relationholds:

C(i)0 = Ni · Z (0; T ∗

i ) ·N(d(i))− Ki · Z (0; T ) · N

(d(i) − σ(i)

p

),

where, for i = 1, 2, 3,

d(i) =ln[

Ni·Z(0,T∗i )

Ki·Z(0,T )

]σ

(i)p

+12σ(i)

p

σ(i)p =

1 − e−a(T∗i −T )

a

√σ2

2a[1 − e−2aT ].

In our case, we have:

σ(1)p =

1 − e−a(T∗1 −T )

a

√σ2

2a[1 − e−2aT ] = 0.054

d(1) =ln[

N1·Z(0,T∗1 )

K1·Z(0,T )

]σ

(1)p

+12σ(1)

p = 0.6446.


Moreover, it is immediate to verify that Z (0; T ∗1 ) = 0.9664 and Z (0; T ) =

0.9684. Consequently,

C(1)0 = 1.43 ·0.9664 ·N (0.664)−1.38 ·0.9684 ·N (0.664− 0.054) = 0.058.

In the same way we can obtain:

C(2)0 = 0.1335

C(3)0 = 5.1871.

We can then conclude that the price of the European Call option con-sidered is:

CCB0 = C

(1)0 + C

(2)0 + C

(3)0 = 0.058 + 0.1335 + 5.1871 = 5.3786.

(c) If we have a European Call written on a zero-coupon bond with facevalue N∗ = 35.86 euros, by the same procedure outlined before we ob-tain:

σp =1 − e−0.4(2.5−7/6)

0.4

√(0.2)2

2 · 0.4[1 − e−2·0.4·7/6

]= 0.18

d =ln[

35.86·Z(0;2.5)36·Z(0;7/6)

]0.18

+0.182

= 0.135.

Finally,

CZCB0 = N∗ · Z (0; T ∗) · N (d) − K ·Z (0; T ) · N (d − σp)

= 35.86 · 0.980 · N (0.135)− 36 · 0.9684 · N (0.135− 0.18)

= 2.65 euros.

4. In order to compute the prices of the Put options with the same parame-ters of the Call examined in items 2.(a) and 3.(b) we can adopt two differ-ent approaches. The first consists in applying the European Put valuationformula, the second in applying the Put-Call Parity relation for optionswritten on zero-coupon bonds.

Let us compute the price of the Put option with the same parameters ofthe Call in item 2.(a).We recall that the initial value of a Put with maturity T and strike K, writ-ten on a zero-coupon bond with face value N and maturity T ∗, is given by:

P0 = K · Z (0; T ) · N (σp − d) − N ·Z (0; T ∗) ·N (−d) , (10.37)

where d and σp are defined as in (10.34) and in (10.35).


Since all the quantities appearing in (10.37) have been already computedin item 2.(a), we get immediately that:

P0 = 36 · 0.9684 ·N (0.18− 0.741)−40 · 0.98025 ·N (−0.741) = 1.029 euros.

Alternatively, and by directly applying the Put-Call parity holding for aEuropean Call and European Put with maturity T and strike K, writtenon the same zero-coupon bond with face value N and maturity T ∗:

C0 − P0 = N · Z (0; T ∗) − K · Z (0; T ) ,

we obtain the same result as before, i.e. P0 = 1.0274 ∼= 1.029 euros.

Let us compute now the price of the Put option with the same parame-ters of the Call considered in item 3.(b). As in the previous case, we havetwo possibilities: one consists in repeating step-by-step the computationsperformed for the Call option, the other consists in applying the Put-CallParity to each of the Call options appearing in the decomposition proposed(Call 1, Call 2 and Call 3, in the present case) and finally sum up the valuesof the Put options in the “decomposition”.We are going to compute the Put value relying on the second approach,while the calculations with the former strategy can be performed as anexercise by the motivated student.We have that:

P CB0 = P

(1)0 + P

(2)0 + P

(3)0 ,

where Put i is a European Put with maturity T ∗i , strike Ki and is writ-

ten on a zero-coupon bond with face value Ni. By the Put-Call Parity weobtain that

P(1)0 = C

(1)0 − N1 · Z (0; T ∗

1 ) + K1 ·Z (0; T ) = 0.012

P(2)0 = C

(2)0 − N2 · Z (0; T ∗

2 ) + K2 ·Z (0; T ) = 0.025

P(3)0 = C

(3)0 − N3 · Z (0; T ∗

3 ) + K3 ·Z (0; T ) = 0.889,

since Z (0; T ∗1 ) = 0.9664, Z (0; T ∗

2 ) = 0.9697, Z (0; T ∗3 ) = 0.9803 and

Z (0; T ) = 0.9684. Finally,

P CB0 = P

(1)0 + P

(2)0 + P

(3)0 = 0.926 euros.

Exercise 10.2. Consider the instantaneous forward rate with the followingaffine dynamics:

f (0, t) = c + mt

for t ≥ 0.


1. Determine ϑt and the price of a zero-coupon bond with maturity t, asa function of the parameters c and m when the short rate dynamics isdescribed by one of the following models: (a) the Ho-Lee model; (b) theHull-White model.

2. Assume that the short rate (rt)t≥0 dynamics is described by the Ho-Leemodel:

drt = ϑtdt + σdWt,

with r0 = c = 0.04, σ = 0.2 and with the ϑt obtained in the previous item.

(a) Find m such that the current value of a zero-coupon bond with maturityT = 7/6 and face value 40 euros is 36 euros.

(b) By using the parameters r0 and σ above and the value of m found inthe previous item, compute the value of a European Call option withmaturity T = 7/6, strike 36 euros and written on a zero-coupon bondwith face value 40 euros and maturity of two years and a half.

3. Suppose now that the short rate (rt)t≥0 is described by the Hull-Whitemodel:

drt = (ϑt − art) dt + σdWt,

with r0 = c = 0.04, σ = 0.2 and with the value of m found in item 2.(a).

(a) Since the parameter a value is not known beforehand, but we haveϑ1/(2a) − ϑ0 = 0.084, determine a.

(b) Compute the value of a European Call option with maturity T = 7/6,strike 36 euros and written on a zero-coupon bond with face value 40euros and maturity of two years and a half.

4. Determine ϑt and the value of a zero-coupon bond with maturity t, as afunction of c and m, when the instantaneous forward rate is of quadratictype, i.e.

f (0, t) = c + mt +m

2t2,

and when the short rate dynamics is described by the Ho-Lee model withthe parameters of item 2.(a).Compute the value of a zero-coupon bond with maturity one year and ahalf and face value 40 euros.

Solution

1. Let us start by considering case (a) of a short rate following a dynamics ofHo-Lee type. In this case we have:

ϑHLt = ϑt =

∂f

∂t(0, t) + σ2t = m + σ2t. (10.38)

As far as the price ZHL (0, t) of a zero-coupon bond with maturity t isconcerned, we need to find the quantities A (0, t) and B (0, t). Recall that


for the Ho-Lee model they are given by:

B(t, T ) = T − t

A(t, T ) =∫ T

t

ϑs(s − T )ds +σ2(T − t)3

6,

so we immediately get B (0, t) = t. It is then enough to determine A (0, t)in order to find ZHL (0, t).Since r0 = f (0, 0) = c and

A (0, t) =∫ t

0

ϑs(s− t)ds +σ2t3

6

=∫ t

0

(m + σ2s

)(s − t)ds +

σ2t3

6

=[ms2

2+

σ2s3

3− mst − σ2s2t

2

]∣∣∣∣t

0

+σ2t3

6

=mt2

2+

σ2t3

3− mt2 − σ2t3

2+

σ2t3

6

= −mt2

2,

we obtain that, in a Ho-Lee model with the assigned parameters, the initialvalue of a zero-coupon bond with maturity t is given by:

ZHL (0, t) = eA(0,t)−r0B(0,t) = e−mt22 −r0t = e−

mt22 −ct.

In case (b) of a Hull-White type dynamics, we have that:

ϑHWt = ϑt =

∂f

∂t(0, t) + a · f (0, t) +

σ2

2a

(1 − e−2at

)= m + a (c + mt) +

σ2

2a

(1 − e−2at

)= m + ac + amt +

σ2

2a

(1 − e−2at

).

As far as the price of a zero-coupon bond with maturity t is concerned, wejust recall that ZHW (0, t) = eA(0,t)−r0B(0,t) with

B(0, t) =1 − e−at

a

A(0, t) =∫ t

0

[12σ2B2(s, t) − ϑsB(s, t)

]ds.


Hence

A(0, t) =∫ t

0

[12σ2

(1− e−a(t−s)

a

)2

− ϑs1 − e−a(t−s)

a

]ds.

In order to simplify the notation, we denote by Cs the integrand:

Cs �12σ2

(1 − e−a(t−s)

a

)2

− ϑs1 − e−a(t−s)

a.

Since

Cs =σ2

2a2

[1 + e−2a(t−s) − 2e−a(t−s)

]− m

a− c − ms − σ2

2a2

(1 − e−2as

)+

m

ae−a(t−s) + c · e−a(t−s) + ms · e−a(t−s)

+σ2

2a2e−a(t−s) − σ2

2a2e−at−as

=σ2

2a2e−2a(t−s) −

(m

a+ c)− ms +

σ2

2a2e−2as

+(

m

a+ c − σ2

2a2

)e−a(t−s)

+ ms · e−a(t−s) − σ2

2a2e−at−as,

we get:

A(0, t) =[

σ2

4a3e−2a(t−s) −

(m

a+ c)

s − ms2

2− σ2

4a3e−2as

]∣∣∣∣t

0

+[(

m

a2+

c

a− σ2

2a3

)e−a(t−s) +

σ2

2a3e−at−as

]∣∣∣∣t

0

+∫ t

0

ms · e−a(t−s)ds

=σ2

4a3− σ2

4a3e−2at −

(m

a+ c)

t − mt2

2− σ2

4a3e−2at

+σ2

4a3+

m

a2+

c

a− σ2

2a3

−(

m

a2+

c

a− σ2

2a3

)e−at +

σ2

2a3e−2at − σ2

2a3e−at

+[ms

ae−a(t−s) − m

a2e−a(t−s)

]∣∣∣t0


= −(m

a+ c)

t − mt2

2+

m

a2+

c

a−(m

a2+

c

a

)e−at

+mt

a− m

a2+

m

a2e−at

= − ct − mt2

2+

c

a− c

ae−at

=c

a

(1 − e−at

)− ct − mt2

2.

Since r0 = f (0, 0) = c, in the Hull-White model considered the initial valueof a zero-coupon with maturity t is given by:

ZHW (0, t) = eA(0,t)−r0B(0,t)

= exp{

r0

a

(1 − e−at

)− r0t − mt2

2− r0

1 − e−at

a

}

= exp{−r0t − mt2

2

}.

2. Assume now that the short rate (rt)t≥0 is described by the following Ho-Leemodel:

drt = ϑtdt + σdWt,

with r0 = c = 0.04, σ = 0.2 and ϑt as obtained in (10.38), i.e.

ϑHLt = ϑt = m + σ2t.

(a) Let us start by determining the value m such that the current price of azero-coupon bond with face value 40 euros and maturity T = 7/6 yearsis equal to 36 euros.We have already obtained (see item 1.) the current price of a zero-couponbond as a function of m, that is:

ZHL (0, t) = eA(0,t)−r0B(0,t) = e−mt2

2 −r0t.

We need then to find the value m such that 40 ·Z (0; T ) = 36 or, equiv-alently,

e−mT2

2 −r0T =3640

.

It follows that:

m = −2ln (36/40) + r0T

T 2= −2

ln (36/40) + 0.04 · 7/6(7/6)2

= 0.0862.

The valuem required is then m = 0.0862. Consequently, ϑHLt = 0.0862+

0.04 · t and f (0, t) = 0.04 + 0.0862 · t.


(b) We have to find the price of a European Call with maturity of one yearand two months, strike 36 euros and written on a zero-coupon bond withface value 40 euros and maturity of two years and a half years.Since, in a Ho-Lee model, the initial value of such a Call option is givenby

C0 = 40 · Z(0, T ∗) · N(d) − 36 · Z(0, T ) · N(d− σp),

with

d = σ−1p ln

[40 · Z(0, T ∗)36 ·Z(0, T )

]+

12σp

σp = σ(T ∗ − T )√

T ,

it is necessary to compute Z (0, T ∗), Z (0, T ), d and σp.The value Z (0; T ) is immediate to obtain: m has been actually chosenin order to get Z (0; T ) = 36/40 = 0.9 euros.Moreover: since T = 7/6, T ∗ = 2.5 and σ = 0.2, we obtain also

Z (0, T ∗) = e−m(T∗)2

2 −r0T∗= e−

0.0862·(2.5)2

2 −0.04·2.5 = 0.691

σp = 0.2 (2.5− 7/6)√

7/6 = 0.288

d =ln[40·0.69136·0.9

]0.288

+0.288

2= −0.408.

Finally, the initial value of a European Call with maturity T = 7/6 (i.e.one year and two months), strike 36 euros and written on a zero-couponbond with face value 40 euros and maturity of two years and a half, isgiven by:

C0 = 40 · 0.691 ·N(−0.408)− 36 · 0.9 ·N(−0.408− 0.288) = 1.563 euros.

3. Let us consider now a short rate (rt)t≥0 with dynamics described by theHull-White model with ϑt as in item 1., a to be determined and with pa-rameters r0 = 0.04 and σ = 0.2.

(a)We need to find the value of a first. We know that a must satisfy thefollowing condition:

ϑ1/(2a) − ϑ0 = α, (10.39)

with α = 0.084.By condition (10.39) we can obtain a. More precisely, (10.39) is equiva-lent to:

am · 12a

+σ2

2a

(1 − e−2a· 1

2a

)= α

m

2+

σ2

2a

(1 − e−1

)= α,


because (see item 1.)

ϑt = m + ac + amt +σ2

2a

(1 − e−2at

).

Then:

a =σ2(1 − e−1

)2(α − m

2

) =σ2 (e − 1)e (2α − m)

=(0.2)2 (e − 1)

e (2 · 0.084− 0.0862)= 0.309.

(b)We compute now the current price of a European Call with maturity ofone year and two months, strike 36 euros and written on a zero-couponbond with face value 40 euros and maturity of two years and a half.Since, in the Hull-White model, the initial value of such a Call option isgiven by:

C0 = 40 · Z(0, T ∗) · N(d) − 36 · Z(0, T ) · N(d− σp),

with

d = σ−1p ln

[40 · Z(0, T ∗)36 · Z(0, T )

]+

12σp

σp =1 − e−a(T∗−T )

a

√σ2

2a[1 − e−2aT ],

we need then to compute Z (0, T ∗), Z (0, T ), d and σp.Since T = 7/6, T ∗ = 2.5, σ = 0.2,m = 0.0862 and a = 0.309, we obtain:

Z (0, T ∗) = e−m(T∗)2

2 −r0T∗= e−

0.0862·(2.5)2

2 −0.04·2.5 = 0.691

Z (0; T ) = e−mT2

2 −r0T = e−0.0862·(7/6)2

2 −0.04·7/6 = 0.9

σp =1 − e−0.309·(2.5−7/6)

0.309

√(0.2)2

2 · 0.309[1 − e−2·0.309·7/6

]= 0.199

d =ln[40·0.69136·0.9

]0.199

+0.199

2= −0.699.

Finally, the initial price of the European Call option (with maturityof one year and two months, strike of 36 euros and written on a zero-coupon bond with face value 40 euros and maturity of two years and ahalf) equals

C0 = 40 · 0.691 ·N(−0.699)− 36 · 0.9 ·N(−0.699− 0.199) = 0.716 euros.


4. If the short rate dynamics is described by the Ho-Lee model and the in-stantaneous forward rate is f (0, t) = c + mt + m

2 t2, we have that:

ϑt =∂f

∂t(0, t) + σ2t = m + mt + σ2t = m +

(m + σ2

)t.

As far as the price Z (0, t) of a zero-coupon bond with maturity t is con-cerned, the unique quantity we still have to compute is A (0, t), sinceB (0, t) = t.As r0 = f (0, 0) = c and

A (0, t) =∫ t

0

ϑs(s − t)ds +σ2t3

6

=∫ t

0

[m +

(m + σ2

)s](s − t)ds +

σ2t3

6

=

[ms2

2+

(m + σ2

)s3

3− mst −

(m + σ2

)s2t

2

]∣∣∣∣∣t

0

+σ2t3

6

=mt2

2+

mt3

3+

σ2t3

3− mt2 − mt3

2− σ2t3

2+

σ2t3

6

= −mt2

2− mt3

6,

we obtain that

Z (0, t) = eA(0,t)−r0B(0,t) = e−mt2

2 −mt36 −r0t = e−

mt22 −mt3

6 −ct.

The value (at time t = 0) of a zero-coupon bond with maturity of one yearand two months and face value 40 euros is then equal to:

40 · Z (0; 7/6) = 40 · e− 0.0862·(7/6)2

2 − 0.0862·(7/6)3

6 −0.04·7/6 = 35.189 euros.

Exercise 10.3. Assume the short rate (rt)t≥0 dynamics is described by theHull-White model:

drt = (ϑt − art) dt + σdWt, (10.40)

with r0 = 0.04, σ = 0.2 and ϑt associated to an instantaneous forward ratewith the following affine dependence on t:

f (0, t) = c + mt,

with m = 0.0862 and a = 0.309 - as in Exercise 10.2, items 2.(a), 3.(a).

1. Compute the current price of a zero-coupon bond with maturity T of oneyear and two months and face value 32 euros, in the case of a short rate dy-


namics described by the Vasiček model (considered as a particular case ofHull-White model with ϑt = ab) with parameter b satisfying the followingcondition: ab = ϑ4

T /ϑ0.2. Consider now the Hull-White dynamics for the short rate as in (10.40), withflat term structure, i.e. with constant instantaneous forward rate f (0, t) =r0 for any t ≥ 0.Compute the value (at time t = 0) of a European Call option with ma-turity of one year and eight months, strike of 38 euros and written on acoupon bond with face value 40 euros, maturity of two years and a half andcoupons paid at the end of each semester at the annual coupon rate of 8%.

Solution

1. We already know from Exercise 10.2 that

ϑt =∂f

∂t(0, t) + a · f (0, t) +

σ2

2a

(1 − e−2at

)= m + a (c + mt) +

σ2

2a

(1 − e−2at

)= m + ar0 + amt +

σ2

2a

(1 − e−2at

).

Let us now consider the Vasiček model corresponding to the Hull-Whitemodel considered, i.e.

drt = a (b − rt) dt + σdWt, (10.41)

where the parameter b satisfies the condition ab = ϑ4T /ϑ0. The value of b

is then:

b =ϑ4

T

aϑ0=

(m + ar0 + amT + σ2

2a

(1 − e−2aT

))4

a (m + ar0)= 0.228,

because a = 0.309, m = 0.0862, r0 = 0.04, σ = 0.2 and T = 20/12 (1 yearand 8 months).We need to compute the current value of a zero-coupon bond with facevalue 32 euros and maturity of one year and eight months. In a Vasičekmodel, we obtain the following results:

B(0, T ) =1 − e−aT

a=

1 − e−0.309·20120.309

= 1.303

A(0, T ) =[B(0, T ) − T ] (a2b − σ2/2)

a2− σ2B2(0, T )

4a= −0.06166

Z (0, T ) = eA(0,T )−r0B(0,T ) = e−0.06166−0.04·1.303 = 0.8925.


The value (at t = 0) of a zero-coupon bond with face value 32 euros andmaturity of one year and eight months is then equal to:

32 · Z (0; 20/12) = 28.56 euros.

2. Let us determine first ϑt in the case f (0, t) = r0 for any t ≥ 0. Since weare working now with the Hull-White model, we can write

ϑt =∂f

∂t(0, t) + a · f (0, t) +

σ2

2a

(1 − e−2at

)= ar0 +

σ2

2a

(1 − e−2at

).

We just recall that for any affine model the price of zero-coupon bond withmaturity t can be written as Z (0, t) = eA(0,t)−r0B(0,t) with

B(0, t) =1 − e−at

a

A(0, t) =∫ t

0

[12σ2B2(s, t) − ϑsB(s, t)

]ds.

We get then A (0, t):

A(0, t) =∫ t

0

[12σ2

(1− e−a(t−s)

a

)2

− ϑs1 − e−a(t−s)

a

]ds.

In order to simplify the notation, denote by Cs the integrand, that is

Cs �12σ2

(1 − e−a(t−s)

a

)2

− ϑs1 − e−a(t−s)

a.

Since

Cs =σ2

2a2

[1 + e−2a(t−s) − 2e−a(t−s)

]− r0

(1 − e−a(t−s)

)− σ2

2a2

(1 − e−2as

)+

σ2

2a2e−a(t−s) − σ2

2a2e−at−as

=σ2

2a2e−2a(t−s) − σ2

2a2e−a(t−s) − r0

(1− e−a(t−s)

)+

σ2

2a2e−2as − σ2

2a2e−at−as,

the following relation holds:

A (0, t) =[

σ2

4a3e−2a(t−s) − σ2

2a3e−a(t−s) − r0s +

r0

ae−a(t−s)

]∣∣∣∣t

0

+[− σ2

4a3e−2as +

σ2

2a3e−at−as

]∣∣∣∣t

0

= −r0t +r0

a− r0

ae−at.


Consequently,Z (0, t) = eA(0,t)−r0B(0,t) = e−r0t,

as we could actually have guessed from the beginning.In a strictly analogous way we can prove that:

A (t, T ) = −r0 (T − t) +r0

a[1 − e−a(T−t)] (10.42)

+σ2

4a3

[2e−a(T−t) − 1 − e−2a(T−t) + e−2aT − 2e−a(T+t) + e−2at

]

Z (t, T ) = exp{−r0(T − t) +

r0

a(1 − e−a(T−t)) − rt

1 − e−a(T−t)

a

}(10.43)

· exp{

σ2

4a3

[2e−a(T−t) − 1 − e−2a(T−t)

+ e−2aT − 2e−a(T+t) + e−2at

]}.

Let us now turn our attention to the initial problem, that is compute theinitial value of a European Call option with maturity of one year and eightmonths, strike of 38 euros and written on a coupon bond with face value40 euros, maturity of two years and a half and with coupons paid at theend of each semester with coupon rate of 8%.We remark that, after the option maturity (which does not coincide withany of the coupon payment dates), the coupons are paid at dates T ∗

1 = 2years and T ∗

2 = T ∗ = 2.5 years (the bond maturity). At this last date, more-over, the face value is paid back, the total amount paid then is N +c = 41.6euros since the coupons (paid in each semester) are equal to c = N · 0.04 =1.6 euros.

In strict analogy with Exercise 10.2, we try to decompose the Call op-tion strike in two components (K1 and K2) representing, respectively, thestrike of the “virtual” option with maturity T and written on the zero-coupon bond (ZCB1) with face value N1 = c = 1.6 and maturity T ∗

1 = 2years; the strike of the “virtual” option with maturity T and written onthe zero-coupon bond (ZCB2) with face value N2 = N + c = 41.6 andmaturity T ∗

2 = T ∗ = 2.5 years.We then impose the condition:

N1 · Z (T ; T ∗1 ) + N2 · Z (T ; T ∗

2 ) = K (10.44)

and define:

K1 � N1 · Z (T ; T ∗1 )

K2 � N2 · Z (T ; T ∗2 ) .


We remark that condition (10.44) only requires to decompose K into twostrikes K1 and K2 as a function of the face values and the time to maturityof the two zero-coupon bonds.In order to determine K1 and K2 as mentioned above, we need Z (T ; T ∗

1 )and Z (T ; T ∗

2 ).From (10.43) we deduce

Z (T ; T ∗1 ) = exp

{−r0 (T ∗

1 − T ) +r0

a

(1 − e−a(T∗

1 −T ))− r∗T

1 − e−a(T∗1 −T )

a

}

+σ2

4a3

[2e−a(T∗

1 −T ) − 1 − e−2a(T∗1 −T )

+ e−2aT∗1 − 2e−a(T∗

1 +T ) + e−2aT]

= e−0.0021−0.3167·r∗T

Z (T ; T ∗2 ) = e−0.0112−0.7347·r∗

T ,

where r∗T is unknown but can be obtained by (10.44). In fact, as a conse-quence of our previous remarks and since N1 = 1.6 and N2 = 41.6, equality(10.44) can be written as follows:

1.6 · e−0.0021−0.3167·r∗T + 41.6 · e−0.0112−0.7347·r∗

T = 38.

By solving explicitly this equation, we find the solution r∗T ∼= 0.16337. Weget then:

K1 � N1 · Z (T ; T ∗1 ) = 1.516

K2 � N2 · Z (T ; T ∗2 ) = 36.484.

At this stage we have:

• the decomposition of the underlying coupon bond as a linear combina-tion of two zero-coupon bonds with face values N1 = 1.6 and N2 = 41.6respectively;

• the option considered has been then decomposed into two options (bothwith maturity T = 20/12): the first one (Call 1) with strike K1, writtenon ZCB1 with face value N1 and maturity T ∗

1 = 2 years; the secondone (Call 2) with strike K2, written on ZCB2 with face value N2 andmaturity T ∗

2 = 2.5 years;• the initial price of our option CCB

0 is then given by:

CCB0 = C

(1)0 + C

(2)0 ,

where C(i)0 , i = 1, 2 denotes the initial price of the European Call i.


The remaining quantities to compute are then C(1)0 and C

(2)0 . Since Call

1 and Call 2 are options written on zero-coupon bonds, in the Hull-Whitemodel, the following relation holds:

C(i)0 = Ni · Z (0; T ∗

i ) · N(d(i))− Ki · Z (0; T ) · N

(d(i) − σ(i)

p

),

where, for i = 1, 2,

d(i) =ln[

Ni·Z(0,T∗i )

Ki·Z(0,T )

]σ

(i)p

+12σ(i)

p

σ(i)p =

1 − e−a(T∗i −T )

a

√σ2

2a[1 − e−2aT ].

In this case we have:

σ(1)p =

1 − e−a(T∗1 −T )

a

√σ2

2a[1 − e−2aT ] = 0.0646

d(1) =ln[

N1·Z(0,T∗1 )

K1·Z(0,T )

]σ

(1)p

+12σ(1)

p = 0.6606.

It is immediate to verify that Z (0; T ∗1 ) = 0.9231 and Z (0; T ) = 0.9355.

Consequently:

C(1)0 = 1.6 · 0.9231 ·N (0.6606)− 1.516 · 0.9355 ·N (0.6606− 0.0646)

= 0.0739 euros.

By proceeding in a strictly similar way we obtain

C(2)0 = 4.3384.

We conclude then that the value at t = 0 of the European Call optionconsidered at the beginning is the following:

CCB0 = C

(1)0 + C

(2)0 = 0.0739 + 4.3384 = 4.4123 euros.

Exercise 10.4. By using the Change of Numéraire technique, prove that thepricing formula for a European Call option with strike K, maturity T , writtenon a zero-coupon bond (with unit face value and maturity T ∗) when the shortrate dynamics is described by the Ho-Lee model, is the following:

C(t, T, K, T ∗) = p(t, T ∗)N(d) − Kp(t, T )N(d − σp), (10.45)


where N denotes the cumulative density of a standard normal and d and σp

are defined as follows:

d =

[ln(

p(t,T∗)Kp(t,T )

)+

σ2p

2

]σp

σp = σ(T ∗ − T )√

T − t.

SolutionThe payoff of a Call option is the following: Φ = max{p(T, T ∗) − K; 0}. A

risk-neutral valuation argument provides the following expression for the Calloption price at time t:

C(t, T, K, T ∗) = EQ

[exp

(−∫ T

t

rsds

)max{p(T, T ∗) − K; 0} | Ft

]. (10.46)

Using the zero-coupon bond with maturity T as a numéraire, this formula canbe rewritten as:

C(t, T, K, T ∗) = p(t, T ) · EQT

[max

{p(T, T ∗)p(T, T )

− K; 0}∣∣∣∣Ft

], (10.47)

where EQT denotes expectation computed with respect to the T-forward mea-sure QT , defined as the probability measure with respect to which the valueof every asset divided by p(t, T ) is a martingale.Set now

Zt �p(t, T ∗)p(t, T )

. (10.48)

Zt is a QT -martingale and its QT -dynamics can be easily obtained by thatof the bond prices. By the zero-coupon bond pricing formula in the Ho-Leemodel, the bond dynamics with respect to Q is in fact:

dp(t, T ) = rtp(t, T )dt− σB(t, T )p(t, T )dWt. (10.49)

By applying Itô’s formula, we obtain the dynamics of Z:

dZt = σ2ZtB(t, T )(B(t, T ) − B(t, T ∗))dt + σZt[B(t, T ) − B(t, T ∗)]dWt

= σZt(B(t, T ) − B(t, T ∗))[σB(t, T )dt + dWt]= Ztσ [B(t, T ) − B(t, T ∗)] dWT

t

= Ztσ [(T − t) − (T ∗ − t)] dWTt

= Ztσ (T − T ∗) dWTt ,

where WTt = Wt +σ

∫ t

0B(s, T )ds and, by Girsanov’s Theorem, QT is the new

probability measure for which (WTt )t≥0 is a Brownian motion.


By comparing the present case with the usual one in the Black-Scholesmodel, we deduce that

C(t, T, K, T ∗) = p(t, T )EQT [max{Z(T ) − K; 0} | Ft]

= p(t, T ∗)N(d) − Kp(t, T )N(d − σp).

Exercise 10.5. By applying the Black Formula, find the value (at time t = 0)of a Caplet with maturity of two years and tenor of one semester in a LIBORmarket model. Assume a flat term structure for the Black (forward) impliedvolatility σi = 0.6 (on annual basis) and for the spot LIBOR at level 2% anda CAP rate R = 0.06 (i.e. the forward implied volatility and the LIBOR rateare assumed to be constant over the period considered). The face value isM = 100.000 euros.

SolutionWe recall the Black formula for Caplets:

Capl(0) = M · αi · p(0, Ti) [L(0, Ti−1, Ti)N(d1) − R · N(d2)] , (10.50)

where αi = Ti − Ti−1 and

d1 �

[ln(

L(0,Ti−1,Ti)R

)+ 1

2Σ2

i (0, Ti−1)]

Σ−1i (0, Ti−1)

d2 �

[ln(

L(0,Ti−1,Ti)R

)− 1

2Σ2

i (0, Ti−1)]

Σ−1i (0, Ti−1)

with

Σi(0, Ti−1) �

√∫ Ti−1

0

σ2i (u)du. (10.51)

In the present case, αi = Ti − Ti−1 = 0.5 years, σi = 0.6 , Ti = 2 years and

Σi(0, Ti−1) �√∫ Ti−1

0σ2

i (u)du =√

σ2i Ti−1 = σi

√Ti−1 = 0.7348.

From the definition of LIBOR spot rate we get:

1 = p(0, T ) [1 + L(0, T )T ] , (10.52)

hencep(0, T ) =

11 + L(0, T )T

=1

1 + 0.02 · T . (10.53)


In particular,

p(0, Ti) =1

1 + 0.02 · Ti=

11 + 0.02 · 2 = 0.96

p(0, Ti−1) =1

1 + 0.02 · Ti−1=

11 + 0.02 · 1.5

= 0.97.

It follows that the forward LIBOR rate at time t = 0 and for the interval[Ti−1, Ti] equals

L(0, Ti−1, Ti) =p(0, Ti−1) − p(0, Ti)

αi · p(0, Ti)= 0.0208. (10.54)

Consequently,

N(d1) = N

([ln(

0.02080.06

)+ 1

2 (0.7348)2]

0.7348

)

= N(−1.07) = 1 − N(1.07) = 1 − 0.8577 = 0.1423

N(d2) = N

([ln(

0.02080.06

)− 12(0.7348)2

]0.7348

)

= N(−1.81) = 1 − N(1.81) = 1 − 0.9649 = 0.0351.

Finally, Capl(0) = 40.98 euros.


Exercise 10.6. 1. Compute the current value of a zero-coupon bond withmaturity of one year and two months and face value of 24 euros, by as-suming that the short rate dynamics is described by the Hull-White modelwith the same value of a as in item 3.(a) of Exercise 10.2.

2. Compare the price obtained in the previous item with the one obtainedby assuming for the short rate dynamics of Vasiček type with parameter bsatisfying the condition ab = ϑ0, where a and ϑ0 are the same as for theHull-White model of item 1.

Exercise 10.7. Assume that the short rate dynamics is described by theVasiček model:

drt = a (b − rt) dt + σdWt,

where the values of the parameters a, b and σ are: a = 0.2, b = 0.1 and σ =0.2. Assume that, at time t = 0, the short rate is r0 = 4% (per year).

1. Compute the value (at time t = 0) of a zero-coupon bond with maturityT ∗ = 4 years.


2. Keeping the results of item 1. in mind, compute the price (at time t = 0) ofa zero-coupon bond with maturity T ∗ = 4 years and face value of 36 euros.

3. Establish if there exists a meaningful value of r0 (i.e. r0 ∈ (0, 1)) such thatthe current value of the bond of item 2. coincides with the current value ofthe same bond under the hypothesis of a constant short rate equal to theexpected value of rT∗ .

4. Consider a European Call option with maturity T = 11/3 years and strikeK = 32, written on a coupon bond with maturity T ∗ = 4 years, face valueof 36 euros and coupon rate of 8% per year (coupon rate paid at the endof each semester).

(a) Compute the value (at t = 0) of this European Call option.(b) How would the price of the option considered change if its maturity was

three years and two months?(c) Compute the price of the Put option with the same parameters of theCall of item 4.(b).

Exercise 10.8. Consider three short rates(r(1)t

)t≥0,(r(2)t

)t≥0

and(r(3)t

)t≥0,

with the following dynamics:

dr(1)t = a1

(b − r

(1)t

)dt + σdWt

dr(2)t = ϑtdt + σdWt

dr(3)t =

(ϑt − a2r

(3)t

)dt + σdWt,

with real parameters a1, a2, b ∈ R, σ > 0 and a2 �= 0.

1. Establish what is the model describing the dynamics of the short ratesdefined as follows:

rt � r(2)t − r

(3)t − a1

a2r(1)t

r∗t � r(2)t − r

(3)t + r

(1)t .

2. For a1 = a2 = a, find the probability distribution of r(2)T with T = 1 year.

Discuss whether P(r(2)T ≥ 0

)= 0.9 holds for some value of the parameters

b and σ.

Exercise 10.9. Assume that the short rate dynamics is described by the Hull-White model:

drt = (ϑt − art) dt + σdWt,

where ϑt is related to f (0, t) of Exercise 10.2, item 4., r0 = 0.04, σ = 0.2 andwith the same parameters a and m of Exercise 10.2.


1. Compute the initial value of a European Put option of strike 36 euros,maturity of one year and two months and written on a zero-coupon bondwith face value of 40 euros and maturity of two years and a half.

2. In the corresponding Ho-Lee model (obtained by assigning the values a =0 and σ), compute the price of the European Put of item 1.

3. Compute the value of the Put option of items 1.-2. in a Vasiček model withparameters a, σ and r0 as before, ϑt = c for every t and with c constantand arbitrary (but reasonable).

Exercise 10.10. By applying the Change of Numéraire technique, prove thatthe initial value of a European Call option with maturity T and strike K,written on a zero-coupon bond (with unit face value) with maturity S, whenthe short rate dynamics is described by the Vasiček model, is given by thefollowing formula:

C(0, T, K, S) = p(0, S) ·N(d) − p(0, T ) ·K ·N(d − σp),

where:

σp = σ(S − T )√

T

d =1σp

ln(

p(0, S)p(0, T )K

)+

12σp.

Exercise 10.11. By the Black Formula, compute the value (at time t = 0)of a Caplet with maturity of two years and tenor of one semester in a LIBORmarket model, by assuming a flat term structure for the Black (forward) im-plied volatility σi = 0.4 (on annual basis), and for the spot LIBOR at level 4%and a CAP rate R = 0.04 (i.e. the forward implied volatility and the LIBORrate are assumed to be constant over the period considered). The face valueis M = 400.000 euros.

11

Pricing Models beyond Black-Scholes


In the previous chapters we presented several pricing and hedging problemsboth in a discrete- and in a continuous-time setting. The basic model assumedin the first case was the binomial model, while for the continuous-time casethe Black-Scholes model was assumed to be the framework, and in this lastcase the dynamics of the risky assets was described by a geometric Brownianmotion.Empirical evidence suggests that these models provide a very rough de-

scription of the financial markets’ behaviour, and they can be used just as afirst approximation of real markets’ modelling.These models are nevertheless still somehow popular since they give several

explicit results that are not so easily available in more sophisticated models,and they provide a simple and consistent approach to valuation and hedg-ing problems. In spite of this remark, some empirical features exhibited byasset prices that cannot be explained by the Black-Scholes model are impos-sible to ignore; we briefly mention the so-called Aggregational Gaussianity ,Volatility Clustering , Fat Tails and Leverage Effects. The first indicates thatprices tend to be more normally distributed when observed on a bigger time-scale, the second that volatility exhibits some kind of persistence behaviour;moreover, the log-returns seem to assume extreme values more likely thanexpected by a Gaussian distribution (fat tails), and prices look negativelycorrelated with volatility (when volatility grows, prices go down on average).Finally, a well-known effect exhibited by option prices is completely outsidethe reach of the Black-Scholes description, and this is the so-called VolatilitySmile. This can be shortly summarized as follows. The coefficient σ enteringin the Black-Scholes formula for European Call options can, in principle, beestimated by the asset prices from historical data, but in practice it can beobtained by inverting the same formula if some of these options are activelytraded on the market. The volatility value obtained by this second procedureis called Black-Scholes implied volatility . Empirical evidence shows that im-


234 11 Pricing Models beyond Black-Scholes

plied volatility exhibits a strong dependence both on the strike price and ontime to maturity, while, if its behaviour is consistent with the Black-Scholesdescription, it should be independent from both.In order to obtain a description of financial markets incorporating the pre-

vious features, more complex models must be introduced. Models proposedin the literature in the attempt to capture at least some of the previous fea-tures can be collected into two main groups: the so-called Stochastic Volatilitymodels and models with jumps.The Stochastic Volatility models propose to model the dynamics of volatil-

ity (which is assumed simply to be constant in the Black-Scholes model) by astochastic differential equation driven by a Brownian motion correlated withthe one driving the asset’s dynamics. The most popular models of this kindare the following: {

dSt = μStdt +√

YtStdW(1)t

dYt = ξ(η − Yt)dt + dW(2)t

, (11.1)

{dSt = μStdt +

√YtStdW

(1)t

dYt = Ytdt + YtdW(2)t

, (11.2)

{dSt = μStdt +

√YtStdW

(1)t

dYt = ξ(η − Yt)dt + θ√

YtdW(2)t ,

, (11.3)

where Y is the square of the volatility process and(W

(1)t

)t≥0

and(W

(2)t

)t≥0

are two correlated Brownian motions with dW(1)t dW

(2)t = ρdt. The first of

the models just written was proposed by Stein and Stein [36], the second byHull and White [24], and the third by Heston [22]. All three models allowto obtain, under suitable simplifying assumptions, some kind of semi-closedform expression for European option prices. Typically, it is possible to obtainan explicit expression for the characteristic function of the log-return distri-bution, and hence – through a technique pioneered by Carr and Madan [14]and now available in several versions– get an explicit formula for the Fouriertransform of the option prices. Once this is available, a standard algorithmfor inverting the Fourier transform can provide the option prices for differentstrikes and maturities. Only the last step of this procedure requires a numer-ical treatment, and this is the reason why these pricing methods are said toprovide solutions in semi-closed form. We shall illustrate by an example howthe log-return characteristic function can be computed.Another direction explored in the attempt to explain some of the relevant

features of the market behaviour is the one introducing jumps in the stochasticprocess describing the asset price dynamics.The first model of this kind was proposed by Merton [27], and is called

the Jump-Diffusion model since it combines a geometric Brownian motiondynamics with a Compound Poisson process.


The asset price dynamics is then assumed to be:

St = S0 exp

{μt + σWt +

Nt∑i=1

Zi

}, (11.4)

where the term μt represents the usual deterministic drift part, σWt representsthe diffusion term and the term

∑Nt

i=1 Zi consists of a Compound Poisson pro-cess, i.e. it is the sum on Nt independent and identically distributed randomvariables Zi, where Nt is a random variable (independent of Wt and all the Yi)distributed as a Poisson with intensity λt (that is, P (Nt = k) = e−λt(λt)k/k!).The original choice proposed by Merton for the jump-size distribution wasGaussian with mean γ and variance δ2, i.e. Zi ∼ N(γ, δ2).Also in the jump-diffusion model described above, under suitable simplify-

ing assumptions, some explicit formulas for option prices are available. Merton[27] proved, assuming that the risk-neutral dynamics of the underlying pricediffers from the same dynamics under the historical measure only in the driftcoefficient, that the initial price of a European Call option with strike K andmaturity T is provided by the following formula:

CJD(S0, T, K, r, σ, λ, γ, δ2) =+∞∑n=0

e−λkT (λkT )n

n!CBS(S0, T, K, σ(n), r(n)),

(11.5)where CBS(S0, T, K, σ(n), r(n)) denotes the value expressed by the Black-Scholes formula of a European Call option with volatility parameter σ(n) andrisk-free interest rate parameter r(n) with

k = exp(

γ +δ2

2

),

σ(n) = σ2 + nδ2

T,

r(n) = r + λ(1 − k) +n

T

(γ +

δ2

2

).

Still in the attempt to capture some of the features exhibited by asset pricebehaviour, some stochastic volatility models in discrete-time have been pro-posed. Among them we just mention the class of GARCH (Generalized Auto-Regressive Conditional Heteroskedasticity) models introduced by Bollerslev[7], generalizing the class of ARCH models previously proposed by Engle [17].The asset price dynamics described by the simplest version of the GARCHmodel, called GARCH(1,1), is the following:

Sn = Sn−1 exp (Xn) = S0 exp {X1 + ... + Xn} ,

Xn = σnεn,

σ2n = α0 + α1X

2n−1 + βσ2

n−1 = α0 + (α1ε2n−1 + β)σ2

n−1,


where α0 > 0, α1 ≥ 0 and β ≥ 0 are real parameters, S0 and σ0 are the initialvalues of the asset price and the volatility, respectively, and (εi)i are i.i.d. (in-dependent and identically distributed) random variables, often assumed to bestandard normal. This model can explain the volatility persistence mentionedbefore and can provide fatter tails for the log-returns distributions than thosepredicted by the log-normal model. The option pricing issues arising in thismodel have been addressed by some authors [13], but their application goesfar beyond the purpose of the present textbook.Although both the stochastic volatility models and the models with jumps

improve the description of financial markets with respect to the Black-Scholesmodel, they still provide a description of the volatility term structure thatis not completely realistic: while stochastic volatility models usually performquite well in describing volatility smiles for long maturities, models with jumpsperform better in capturing volatility smiles for short maturities. Some mod-els have been proposed that incorporate both features, stochastic volatilityand jumps, in order to remove the inconsistencies exhibited by both classesof the previous models. Some of them allow to obtain an explicit expressionfor the characteristic function of log-returns distribution, in such a way thatsemi-explicit valuation formulas for European options are available. We shallprovide a single example of how to compute the log-returns characteristicfunction for a model of this class. An exhaustive list of examples related tothese topics is again far beyond the purpose of this textbook.


Exercise 11.1. Compute the value (at time t = 0) of a European Call optionwith maturity T = 1 year, strike K = 30 euros, written on an underlyingasset whose dynamics is of jump-diffusion type with σ = 0.4, r = 0.05, andS0 = 30 euros; the jump process is a compound Poisson with intensity λ = 1(this means an average of one jump each year) and jump-size distributed asa Gaussian with mean γ = 1 and variance δ2 = 0.49. Assume that the changeof measure only affects the drift coefficient in the SDE describing the assetprice dynamics.

SolutionAssume (as suggested by Merton) that, when the historical probability is

replaced by the risk-neutral one, only the drift part of the dynamics is af-fected. The explicit valuation formula for European Call options obtained byMerton can therefore be applied. Hence we need to compute:

CJD(S0, T, K, r, σ, λ, γ, δ2) =+∞∑n=0

e−λkT (λkT )n

n!CBS(S0, T, K, σ(n), r(n)),

(11.6)


where

k = exp(

γ +δ2

2

),

σ(n) = σ2 + nδ2

T,

r(n) = r + λ(1 − k) +n

T

(γ +

δ2

2

),

and CBS denotes the Black-Scholes value of a European Call option computedby assigning the values σ(n), r(n) to the volatility and the risk-free interestrate, respectively.The valuation formula for the Call option in a jump-diffusion setting is ac-

tually provided by a series, and only an approximate value can be computedby truncating the series.We remark that the general term contains a negative exponential-like fac-

tor (with n! in the denominator) which makes the convergence rate of thisseries fast enough; for this reason we decide to consider only the first 5 terms.By adopting the following notation, C

(n)BS � CBS(S0, T, K, σ(n), r(n)), we

obtain:

C(0)BS � 0, C

(1)BS = 0.47, C

(2)BS = 13.51, C

(3)BS = 24.41, C

(4)BS = 28.35.

We remark that the first term is similar to the value of a European Call op-tion valuated according to the Black-Scholes formula, since for n = 0 no jumpsoccur in the underlying dynamics. Note that with more than two jumps theprobability that the Call option will expire out of the money is negligible. Onthe other hand, for n ≥ 4, the factor 1/n! makes the contribution very small,allowing us to consider only the contribution of the first 5 terms of the seriesrelevant for the result.By summing up all these contributions, we finally obtain:

CJD � e−λkT

[C

(0)BS + λkTC

(1)BS +

(λkT )2

2C

(2)BS +

(λkT )3

6C

(3)BS +

(λkT )4

24C

(4)BS

]= 13.2.

Exercise 11.2. Compute the conditional characteristic function of the log-returns for the stochastic volatility model proposed by Heston:

{dSt = μStdt +

√YtStdW

(1)t

dYt = ξ(η − Yt)dt + θ√

YtdW(2)t

.


SolutionThe market model proposed by Heston [22] is incomplete. There is one

Brownian motion appearing in the dynamics of the asset price and a secondBrownian motion appearing in the square of the diffusion coefficient’s dynam-ics: these are two different (although correlated) sources of randomness, andonly one risky asset is traded on the market. We must specify then whichone, among the infinitely many equivalent measures turning the discountedasset price into a martingale, we are going to assume describes the market be-haviour. For simplicity reasons we are going to assume that the market priceof risk associated to the volatility vanishes, i.e. the risk-neutral volatility dy-namics is the same under both the historical and the risk-neutral measures.Applying Itô’s formula and a change of measure (from historical to risk-

neutral), it is easy to check that the dynamics of the log-returnXt � ln (St/S0)is described by the following SDE:

{dXt =

(r − Yt

2

)dt +

√YtdW

(1)t

X0 = 0. (11.7)

The characteristic function (conditionally to the log-return and volatilityvalues at time t) is defined as follows:

f(x, y, t) � E [exp (iuXT ) |Xt = x, Yt = y] , u ∈ R. (11.8)

The function f , depending on t, x, y, must satisfy a stochastic differentialequation which can be easily obtained by a straightforward application ofItô’s Lemma:

df =

[12Yt

∂2f

∂x2+ ρθYt

∂2f

∂x∂y+

12θ2Yt

∂2f

∂y2

+(

r − Yt

2

)∂f

∂x+ ξ(η − Yt)

∂f

∂y+

∂f

∂t

]dt

+√

Yt∂f

∂xdW

(1)t + θ

√Yt

∂f

∂ydW

(2)t . (11.9)

By looking at the definition of the function f we should remark that thisis a stochastic process obtained by conditioning an FT -adapted random vari-able with respect to the information available at time t (i.e. with respect tothe filtration Ft). This means that it must be an Ft-martingale. In a diffusionframework, like the present one, this means that the drift coefficient appearingin the SDE describing the dynamics of f must vanish identically, so that:

12Yt

∂2f

∂x2+ ρθYt

∂2f

∂x∂y+

12θ2Yt

∂2f

∂y2+(

r − Yt

2

)∂f

∂x+ ξ(η − Yt)

∂f

∂y+

∂f

∂t= 0.

(11.10)


In order to compute the solution of this PDE we need to impose a finalcondition (this is a backward parabolic PDE). This turns out to be:

f(x, y, T ) = eiux. (11.11)

Inspired by PDEs of similar kind encountered in the previous chapters, wemake an educated guess of the following type:

f(x, y, t) = exp {A(τ ) + yB(τ ) + iux} , (11.12)

where τ = T − t is the time to maturity. By plugging this expression for f inPDE (11.10), we get

fYt

[−1

2u2 + ρθiuB(τ ) +

12B2(τ ) − iu

2− ξB(τ )

]+ f [riu + ξηB(τ ) − A′(τ ) − yB′(τ )] = 0.

By imposing that both the sum of terms multiplying y and the sum of theterms in which y does not appear vanish separately, we obtain the followingsystem of ordinary differential equations together with their initial conditions:

⎧⎨⎩

A′(τ ) = ξηB(τ ) + iur, A(0) = 0

B′(τ ) = (iuρθ − ξ)B(τ ) +θ2

2B2(τ ) − u2 + iu

2, B(0) = 0

.

Although non-linear, the previous system exhibits some nice features: once thedifferential equations satisfied by B is solved, the first equation can be imme-diately reduced to quadratures and the unknown function A can be computedas an integral. The second equation does not contain the unknown A and isa differential equation which admits a closed-form solution; it is actually aRiccati equation with constant coefficients, and the explicit solution can befound by separating the variables as follows:

dB

aB2 + bB + c= dτ, (11.13)

where a = θ2/2, b = iuρθ − ξ , c = −(u2 + iu)/2. The right-hand side of theprevious equation can be written as follows:

[a (y2 − y1)]−1

dB

B − y2− [a (y2 − y1)]

−1dB

B − y1, (11.14)


where y1,2 are the roots of the second-order algebraic equation aB2 +bB+c =0:

y1 =−b −√

b2 − 4ac

2a=

1θ2

[ξ − iuρθ −

√(iuρθ − ξ)2 + θ2 (u2 + iu)

]

y2 =−b +

√b2 − 4ac

2a=

1θ2

[ξ − iuρθ +

√(iuρθ − ξ)2 + θ2 (u2 + iu)

].

By integrating the two rational functions and inverting the solution obtained,we finally get:⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

A(τ ) = iruτ +ξητ (ξ − iρθu)

θ2− 2ξη

θ2ln[cosh

(βτ

2

)+

ξ − iρθu

βsinh

(βτ

2

)],

B(τ ) = − u2 + iu

β coth(

βτ2

)+ ξ − iρθu

,(11.15)

where β =√

(ξ − iρθu)2 + θ2 (u2 + iu). By substituting A(τ ) and B(τ ) in(11.12), we obtain the characteristic function we were looking for.

Exercise 11.3. Compute the characteristic function of the log-returns in thejump-diffusion model proposed by Merton:

St = S0 exp

{μt + σWt +

Nt∑i=1

Yi

}. (11.16)

SolutionThe log-return process in the jump-diffusion model proposed by Merton is

the following:

Xt = ln(

St

S0

)μt + σWt +

Nt∑i=1

Yi. (11.17)

It is the sum of a linear deterministic component (μt), a diffusion component(a Brownian motion with diffusion coefficient σ), and a jump part representedby a compound Poisson process with intensity λ and normal jump-size distri-bution.The characteristic function of Xt is given by:

E [exp (iuXt)] = E

[exp

{iu

(μt + σWt +

Nt∑i=1

Zi

)}], u ∈ R. (11.18)

We remark that in the Merton model the jump and the diffusion processes areassumed to be independent, so that the expectation above can be computedas the product of the expectations of each contribution. The first contribu-tion is the characteristic function of a Brownian motion with drift, i.e. it is


the characteristic function of a normal random variable with mean μt andvariance σ2t:

E [exp {iu (μt + σWt)}] = exp{

iuμt − iu2 σ2

2t

}. (11.19)

The characteristic function of the compound Poisson process can be com-puted as follows, by conditioning with respect to the variable Nt:

E

[exp

(iu

Nt∑i=1

Zi

)]= E

[E

[exp

(iu

Nt∑i=1

Zi

)∣∣∣∣∣Nt

]]

=+∞∑n=0

E

[exp

(iu

Nt∑i=1

Zi

)∣∣∣∣∣Nt = n

]P (Nt = n).

Since the Zi are independent and identically distributed, the last conditionalexpectation in the previous expression becomes:

E

[exp

(iu

Nt∑i=1

Zi

)∣∣∣∣∣Nt = n

]= E

[n∏

i=0

exp(iuZi)

]

=n∏

i=0

E[exp(iuZi)]

= [φ(u)]n,

where φ(u) is the characteristic function of the random variable Zi. For anormal jump-size distribution with mean γ and variance δ2, we have:

φ(u) = exp{

iγu − δ2

2u2

}. (11.20)

Finally, we obtain:

E

[exp

(iu

Nt∑i=1

Zi

)]=

+∞∑n=0

E

[exp

(iu

Nt∑i=1

Zi

)∣∣∣∣∣Nt = n

]P (Nt = n)

=+∞∑n=0

[φ(u)]ne−λt (λt)n

n!

= exp {λt (φ(u) − 1)} .

Putting together the results in (11.19) and (11.21), we conclude that

E [exp (iuXt)] = exp{

iuμt − iu2 σ2

2t

}· exp {λt(φ(u) − 1)} . (11.21)

Exercise 11.4. By combining the results obtained in Exercises 11.2-11.3,compute the characteristic function of the log-returns for the following model


proposed by Bates [5]:{St = S0 exp

{μt +

√YtW

(1)t +

∑Nt

i=1 Zi

}dYt = ξ(η − Yt)dt + θ

√YtdW

(2)t

.

Specify by which equivalent martingale measure you are going to describe therisk-neutral dynamics of the asset price.

SolutionAs far as the jump-diffusion dynamics of the asset price is concerned, we

shall work under the usual assumption that the change of measure from thehistorical to the risk-neutral probability affects only the drift component, insuch a way that in order to get a local martingale for the discounted assetprice dynamics we can write:

St = S0 exp{(

r − Yt

2− λκ

)t +√

YtW(1)t +

Nt∑i=1

Zi

}, (11.22)

where κ is defined by:

κ = λ

{exp(γ +

δ2

2

)− 1}

. (11.23)

Moreover, we shall assume a vanishing risk-premium for the volatility compo-nent, so that the variance dynamics under the risk-neutral measure remainsunchanged.Under these assumptions, the characteristic function can be computed by

observing that in the model under consideration the jump process (the com-pound Poisson) appearing in the asset price dynamics is independent of thediffusion part. Consequently, the characteristic function of the log-returns issimply the product of the two characteristic functions of the diffusion and thejump components, which can be computed separately. The diffusion compo-nent is just the same as the one appearing in the Heston model (computedexplicitly in Exercise 11.2), while the characteristic function of the compoundPoisson has been computed in Exercise 11.3. The characteristic function ofthe log-return Xt = ln(St/S0) is the given by:

E [exp (iuXt)]

= exp{i (r − λκ)ut +

ξηt (ξ − iρθu)θ2

}[cosh

(βt

2

)+

ξ − iρθu

βsinh

(βt

2

)]− 2ξη

θ2

·

· exp

⎧⎨⎩− u2 + iu

β coth(

βt2

)+ ξ − iρθu

Y0

⎫⎬⎭ · exp {λt(φ(u) − 1)} ,


where φ(u) is the characteristic function of a normal random variable withmean γ and variance δ2, that is,

φ(u) = exp{

iγu − δ2

2u2

}. (11.24)

Exercise 11.5. Prove that the solution of the recurrence equation describingthe GARCH(1,1) model is the following:

⎧⎪⎨⎪⎩

σ2n = α0

[1 +∑n

i=1

∏ij=1(α1ε

2n−j + β)

]Xn = εn

√α0

[1 +∑n

i=1

∏ij=1(α1ε

2n−j + β)

] . (11.25)

SolutionThe equation characterizing the GARCH(1,1) model is a particular case of

a class of recurrence equations of the following form:

Mn = AnMn−1 + Bn, (11.26)

where (An)n∈N and (Bn)n∈N are sequences of independent and identically dis-tributed (i.i.d.) random variables. The solution of this equation can be easilyobtained by induction

Mn = An (An−1Mn−2 + Bn−1) + Bn

= Bn +k∑

i=1

⎛⎝Bn−i

i−1∏j=0

An−j

⎞⎠+ Mn−k−1

k∏i=0

An−i.

In order to ensure the convergence of this expression, as k → +∞ the thirdterm must vanish and the second must converge. By the Strong Law of LargeNumbers we know that:

1k + 1

k∑i=0

ln |An−i| → E[ln |An|], P -almost surely. (11.27)

So, under the assumption E [ln |An|] < 0, we have:

k∏i=0

|An−i| = exp

{k∑

i=0

ln |An−i|}

=

(exp

{1

k + 1

k∑i=0

ln |An−i|})k+1

→ 0, P -a.s.


By proceeding in a similar way we can prove that the conditionE [ln |An|] <0 ensures that the term:

k∑i=1

⎛⎝Bn−i

i−1∏j=0

An−j

⎞⎠ (11.28)

converges as k → ∞, in such a way that the unique solution is

Mn = Bn ++∞∑i=1

⎛⎝Bn−i

i−1∏j=0

An−j

⎞⎠ . (11.29)

In the particular case of the GARCH(1,1) models,Mn = σ2n, An = α1ε

2n−1+

β and Bn = α0 (with Bn−i = 0 for i > n). Equation (11.29) implies:

σ2n = α0

⎡⎣1 +

n∑i=1

i∏j=1

(α1ε2n−j + β)

⎤⎦ . (11.30)

Since Xn = σnεn, (11.25) follows.

Exercise 11.6. Recall that a time series (Xn)n∈Z is called stationary (in strictsense) if the random vector (Xt1 , ..., Xtn) has the same distribution of(Xt1+k, ..., Xtn+k) for all t1, ..., tn, k ∈ Z and for all n ∈ N.Prove that for the GARCH(1,1) model the condition α1+β < 1 is sufficient

to ensure that the sequence (σn)n∈N is stationary.

SolutionThe unique solution of the recurrence relationMn = AnMn−1+Bn (n ∈ Z),

under the assumption E [ln |An|] < 0, has been shown (see Exercise 11.5) tobe the following:

Mn = Bn ++∞∑i=1

⎛⎝Bn−i

i−1∏j=0

An−j

⎞⎠ .

For a GARCH(1,1) model, An = α1ε2n−1 +β and Bn = α0 (with Bn−i = 0

for i > n). The condition E [ln |An|] < 0, ensuring (σn)n∈Z is stationary,becomes E

[ln(α1ε

2n + β

)]< 0, which in turn translates into the condition

α1 + β < 1, since the innovations (εi)i are i.i.d. standard normal randomvariables. The solution obtained is a function of the i.i.d. random variablesAn, Bn, and this implies that the process (Mn)n∈Z is strictly stationary.



Exercise 11.7. Compute the value (at time t = 0) of a European Call optionwith maturity T = 1 year, strike K = 50 euros, written on an underlying assetwhose dynamics is of jump-diffusion type with σ = 0.8, r = 0.04, and S0 = 45euros; the jump process is a compound Poisson with intensity λ = 0.5 andjump-size distributed as a normal with mean γ = 2 and variance δ2 = 0.64.Assume that the change of measure only affects the drift coefficient in theSDE describing the asset price dynamics.

Exercise 11.8. Compute the conditional characteristic function of the log-returns for the stochastic volatility model proposed by Stein and Stein andunder the same assumptions of Exercise 11.2:{

dSt = μStdt +√

YtStdW(1)t

dYt = ξ(η − Yt)dt + dW(2)t

.

Exercise 11.9. Consider a GARCH(1,1) model with parameters α0 = 0.7,α1 = 0.5 and β = 0.4. Verify that the sequence σ2

n is stationary and writeexplicitly σ2

10.

Exercise 11.10. An ARCH(1) model describes the log-returns (squared) X2n

of a financial asset through the following recurrence equation:

X2n = α0ε

2n + α1ε

2nX2

n−1.

Assume that (εn)n is a sequence of i.i.d. standard normal random vari-ables, and prove that the condition α1 < 3.562 is sufficient to ensure

(X2

n

)n

is strictly stationary, and also that the solution of the recurrence equation isthe following:

X2n = α0

+∞∑i=0

⎛⎝αi

1

i∏j=0

ε2n−j

⎞⎠ .

12

Risk Measures: Value at Risk andbeyond


In the late 90s an increasing interest has been developing towards risk mea-sures, in particular the Value at Risk (VaR) and the Conditional Value atRisk (CVaR). The use of such risk measures is due, on the one hand, to therules imposed by the Basel Accord on the deposit of margins by banks andfinancial institutions because of the financial risks they are exposed to. Onthe other hand, these tools are important to quantify the riskiness assumedby an investor or an intermediary because of his financial transactions.

We recall briefly some notions on risk measures. For a detailed treatmentand further details, we refer to Föllmer and Schied [19], Hull [23], Jorion [25]and Barucci et al. [4], among many others.

Consider the Profit & Loss (P&L) or the return of a financial position at afuture date T , and denote with X the family of random variables represent-ing P&Ls or returns of the financial positions taken into account. The spaceX = L∞ (Ω,F , P ) is often considered, where L∞ (Ω,F , P ) (or, simply, L∞)is formed by essentially bounded random variables, i.e. by random variablesX satisfying P (|X| ≤ K) = 1 for some K > 0.

Given any random variableX ∈ X , the Value at Risk at the level α ∈ (0, 1)of X is defined as

V aRα (X) � −q+α (X) = − inf{x ∈ R : FX(x) > α}, (12.1)

i.e. as the opposite of the greatest α-quantile of X. Notice that, for continuousrandom variables X, V aRα (X) is nothing but the opposite of the quantileqα (X) solution of FX (x) = α.The financial interpretation of the V aR is the following. If V aRα (X) > 0,

then V aRα (X) represents the minimal amount to be deposited as a marginfor X.


248 12 Risk Measures: Value at Risk and beyond

We list now some results on the V aR for stock returns of the followingform:

ΔS

S0

∼= μ ·Δt + σ · ε ·√

Δt,

where Δt = T is the length of the period of time considered, S0 > 0 is theinitial stock price, μ ∈ R its drift, σ > 0 its volatility and ε ∼ N (0; 1).The V aR of the Profit & Loss of a stock of current price S0 and daily

return distributed as a normal with drift μ ∈ R and volatility σ > 0, is:

V aRα (ΔS) = S0

[σ√

Δt · N−1 (1 − α) − μ · Δt].

The V aR of the Profit & Loss of a portfolio formed by xi (for i = 1, 2, ..., n)stocks i with current prices Si

0, where the daily returns are jointly normaldistributed with drift μi ∈ R, volatility σi > 0 and correlation ρij (i, j =1, 2, ...n), is:

V aRα (ΔP ) = N−1 (1 − α)·√

Δt

√√√√ n∑i=1

n∑j=1

xixjSi0S

j0σiσjρij−

n∑j=1

xjSj0μj(Δt).

(12.2)In particular: if all drifts are zero, then

V aRα (ΔP ) = N−1 (1 − α) ·√

Δt

√√√√ n∑i=1

n∑j=1

xixjSi0S

j0σiσjρij. (12.3)

The first-order approximation (Delta approximation) of the V aR for theProfit & Loss of a portfolio formed by xi (i = 1, 2, ..., n) options having DeltaΔi, underlying stocks of current prices Si

0 and daily returns that are jointlynormal with null drifts, volatility σi > 0 and correlation ρij (i, j = 1, 2, ...n),is given by:

V aRΔ approxα (ΔP ) ∼= N−1 (1 − α) ·

√Δt

√√√√ n∑i=1

n∑j=1

xixjΔi0Δ

j0S

i0S

j0σiσjρij .

(12.4)

Although the V aR satisfies monotonicity (and, more in general, consistencywith respect to first-order stochastic dominance), cash-invariance and posi-tive homogeneity (recalled below), many authors (see, among others, Artzneret al. [1]) have underlined its several drawbacks. First of all, the V aR doesnot encourage diversification of risk (namely, it does not satisfies subaddi-tivity); furthermore, it does not distinguish tails that are even very differentat the left of −V aR (i.e. for losses greater than V aR). The need to over-


come the latter problem led to the introduction of the so-called Tail Con-ditional Expectation. More precisely, if X is a random variable having fi-nite expected value E [X], the Tail Conditional Expectation (TCE) of X atlevel α ∈ (0, 1) is defined as:

TCEα(X) � EP [−X|X ≤ −V aRα(X)] =EP

[−X1{X≤−V aRα(X)}]

P (X ≤ −V aRα(X)).

(12.5)Unfortunately, neither the TCE encourages diversification in general. For fur-ther details, see Artzner et al. [1] and Delbaen [12].

In order to amend the drawbacks of V aR and TCE but at the same timepreserve the main properties, Artzner et al. [1] and Delbaen [12] introducedthe so-called coherent risk measures. More details on coherent risk measuresand, in particular, on the CV aR can be found in Artzner et al. [1], Delbaen[12], Föllmer and Schied [19], Rockafellar and Uryasev [33], Barucci et al. [4]and Fiori et al. [18], among many others.A coherent risk measure ρ is defined (see [1], [12]) as a risk measure ρ :

X → R satisfying the following properties:

• monotonicity: if X ≤ Y , P -a.s., with X, Y ∈ X , then ρ(X) ≥ ρ(Y );• cash-invariance: ρ(X + c) = ρ(X) − c for any X ∈ X , c ∈ R;• subadditivity: ρ(X + Y ) ≤ ρ(X) + ρ(Y ) for any X, Y ∈ X ;• positive homogeneity: ρ(λX) = λ · ρ(X) for any X ∈ X , λ ≥ 0.

When X = L∞, the most relevant examples of coherent risk measures arethe Conditional Value at Risk and the risk measures generated by a set ofprobability measures. The Conditional Value at Risk at level α ∈ (0, 1) isdefined by:

CV aRα(X) � infx∈R

{E[(x− X)+]

α− x

}; (12.6)

while a risk measure associated to a set Q of probability measures (known asset of generalized scenarios) is given by:

ρQ (X) � supQ∈Q

EQ [−X] . (12.7)

Any coherent risk measure defined on a finite sample space or satisfying afurther hypothesis on continuity can be represented as in (12.7) (see Artzneret al. [1]), and conversely.It is also well known (see Föllmer and Schied [19]) that

CV aRα(X) =E[(qα − X)+]

α− qα, (12.8)

where qα is an arbitrary α-quantile of X. Moreover, if X is a continuousrandom variable, then CV aRα(X) = TCEα(X).


Since ρ (X) > 0 can be understood as the amount to deposit as a marginfor X, the acceptance set A of ρ is defined as the set

Aρ � {X ∈ X : ρ (X) ≤ 0} .

Coherent risk measures and acceptance sets are in one-to-one correspondence.Consider, for simplicity, a finite sample space. Then:

• if A satisfies the following properties (A):

A ⊇ {X ∈ X : X ≥ 0}A ∩ {X ∈ X : X ≤ 0 with X (ω) < 0 for some ω ∈ Ω} = ∅A is a convex cone,

then ρA (X) � inf {m ∈ R : m + X ∈ A} is a coherent risk measure;• if ρ is a coherent risk measure, then Aρ satisfies properties (A) and ρAρ = ρ.

Several extensions of coherent risk measures are studied in the literatureboth in a static and in a dynamic setting. Problems concerning the choiceof portfolios that minimize risk and dynamic risk measures and their time-consistency will be addressed in some of the exercises below. For portfoliooptimization we refer to Gaivoronski and Pflug [21], while for (dynamic time-consistent) risk measures see Artzner et al. [2], Cheridito and Stadje [10],Föllmer and Schied [19], Frittelli and Rosazza Gianin [20], amongmany others.


Exercise 12.1. Consider two stocks whose Profit & Loss (per year) are rep-resented, respectively, by the following random variables:

X =

⎧⎪⎪⎨⎪⎪⎩

−20; on ω1

−8; on ω2

0; on ω3

12; on ω4

; Y =

⎧⎨⎩

6; on ω1

0; on {ω2, ω3}−2; on ω4

with P (ω1) = 0.01, P (ω2) = 0.09, P (ω3) = 0.8 and P (ω4) = 0.1.

1. Consider the portfolio composed by 3 shares of the first stock (correspond-ing to X) and by 4 shares of the other (corresponding to Y ). Establishwhether risk diversification is encouraged or not by V aR at 1%.

2. Consider the portfolio composed by one share of the first stock and by 8shares of the second. Establish whether risk diversification is encouragedor not by V aR at 10%.

3. Consider the same portfolio as in item 2. Establish whether diversificationof risk is encouraged by CV aR at 10% and compute the profit due todiversification.


Solution

1. By the Profit & Loss of the two stocks we get the cumulative distributionfunctions of X and Y :

FX(x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

0; if x < −200.01; if − 20 ≤ x < −80.1; if − 8 ≤ x < 00.9; if 0 ≤ x < 12

1; if x ≥ 12

FY (x) =

⎧⎪⎪⎨⎪⎪⎩

0; if x < −20.1; if − 2 ≤ x < 0

0.99; if 0 ≤ x < 61; if x ≥ 6

As a consequence, we deduce that

V aR0.01(X) = −q+0.01(X) = − inf{x ∈ R : FX(x) > 0.01} = 8

V aR0.01(Y ) = −q+0.01(Y ) = − inf{x ∈ R : FY (x) > 0.01} = 2.

Let us consider a portfolio composed by 3 shares of X and by 4 shares ofY . The Profit & Loss of such a portfolio is then given by

3X + 4Y =

⎧⎪⎪⎨⎪⎪⎩

−60 + 24; on ω1

−24; on ω2

0; on ω3

36 − 8; on ω4

=

⎧⎪⎪⎨⎪⎪⎩

−36; on ω1

−24; on ω2

0; on ω3

28; on ω4

whose cumulative distribution function is

F3X+4Y (x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

0; if x < −360.01; if − 36 ≤ x < −240.1; if − 24 ≤ x < 00.9; if 0 ≤ x < 28

1; if x ≥ 28

.

Since

V aR0.01(3X + 4Y ) = 24 < 3 · V aR0.01(X) + 4 · V aR0.01(Y ) = 32,

we can conclude that, for such a portfolio, V aR at 1% encourages diver-sification of risk. The profit due to diversification (in terms of cash-savingon the margin to be deposited) is then 8 euros.

2. By the cumulative distribution functions of X and of Y found above, wededuce that

V aR0.1(X) = −q+0.1(X) = − inf{x ∈ R : FX(x) > 0.1} = 0

V aR0.1(Y ) = −q+0.1(Y ) = − inf{x ∈ R : FY (x) > 0.1} = 0.


Consider now a portfolio composed by one share of X and by 8 shares ofY . The Profit & Loss of such a portfolio is then given by

X + 8Y =

⎧⎪⎪⎨⎪⎪⎩

28; on ω1

−8; on ω2

0; on ω3

−4; on ω4

.

The cumulative distribution function of the Profit & Loss of the portfolioabove is then

FX+8Y (x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

0; if x < −80.09; if − 8 ≤ x < −40.19; if − 4 ≤ x < 00.99; if 0 ≤ x < 28

1; if x ≥ 28

,

hence we can conclude that risk diversification is not encouraged by V aRat 10%. In fact,

V aR0.1(X + 8Y ) = 4 > V aR0.1(X) + 8V aR0.1(Y ) = 0.

3. By formula (12.8) for the computation of CV aR and by item 2., we obtain

CV aR0.1(X) = E[(q0.1(X)−X)+]α

− q0.1(X) = 20·0.01+8·0.090.1

= 9.2

CV aR0.1(Y ) = 2·0.10.1

= 2

CV aR0.1(X + 8Y ) = 4·0.090.1 + 4 = 7.6,

by considering the quantiles q+0.1(X) = q+

0.1(Y ) = 0 and q+0.1(X + 8Y ) =

−4 = q−0.1(X + 8Y ). Since

CV aR0.1(X + 8Y ) = 7.6 < CV aR0.1(X) + 8 ·CV aR0.1(Y ) = 17.2,

we conclude that there is diversification of risk with respect to CV aR forthe portfolio considered above (as well as for any other portfolio, becauseof the properties of the CV aR) with a profit (in terms of cash-saving onthe margin to be deposited) of 9.6 euros.


Exercise 12.2. Consider a portfolio whose Profit & Loss is represented bythe following random variable:

X =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−400; on ω1

−160; on ω2

−40; on ω3

0; on ω4

80; on ω5

1200; on ω6

with P (ω1) = 0.01, P (ω2) = p, P (ω3) = q, P (ω4) = 0.8, P (ω5) = 0.05 andP (ω6) = 0.05, and p, q > 0 so that P is a probability measure.

1. Compute the Value at Risk at 1% and at 10% of X. Would we find adifferent result with a maximum loss of 4000 (instead of 400)?

2. Find (if possible) p and q such that the V aR at 5% of X is equal to 40 andthe Tail Conditional Expectation at 5% of X does not exceed 100.

SolutionThe condition p+q = 0.09 has to be satisfied in order for P to be a probabil-

ity measure. Moreover, the cumulative distribution function of X (dependingon p, q) is

FX(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

0; if x < −4000.01; if − 400 ≤ x < −160

0.01 + p; if − 160 ≤ x < −400.1; if − 40 ≤ x < 00.9; if 0 ≤ x < 80

0.95; if 80 ≤ x < 12001 if x ≥ 1200

,

because 0.01 + p + q = 0.1 (by the arguments above).

1. By the results above, it is straightforward to deduce that V aR0.01(X) = 160and V aR0.1(X) = 0. It is easy to check (also using FX) that such resultswould remain unchanged even if the maximum loss was 4000 instead of 400.

2. The conditions to be imposed on p and q (so that the V aR at 5% of Xis equal to 40 and the Tail Conditional Expectation at 5% of X does notexceed 100) are the following:⎧⎨

⎩p + q = 0.09 (P prob. measure)0.01 + p ≤ 0.05 (condition on V aR)40 + 1600p + 400q ≤ 100 (condition on CV aR)

(12.9)

where the last inequality is due to

TCE0.05 (X) = EP [−X|X ≤ −V aR0.05(X)] =EP

[−X1{X≤−40}]

P (X ≤ −40)

=400 · 0.01 + 160p + 40q

0.1= 40 + 1600p + 400q.


Solving system (12.9) we obtain⎧⎨⎩

q = 0.09− pp ≤ 0.04p ≤ 0.02

{0 < p ≤ 0.02q = 0.09− p

.

A pair of admissible p and q is, for instance, p = 0.02 and q = 0.07.

Exercise 12.3. Consider two stocks (A and B) whose daily returns are jointlynormally distributed. The correlation of the stock returns is 0.25, the currentprice of stock A is 4.2 euros, its daily drift 0.0002 and its daily volatility 1.2%,while the current price of the stock B is 3.6 euros, its daily drift 0.0008 andits daily volatility 2.8%.

1. Find the V aR (at 1% and on a time interval of 10 days) of the portfolio(PORT1) composed by 20 shares of A and by 15 shares of B.

2. Consider two other stocks (C and D) having the same features as stocks Aand B, respectively, but both having daily drift equal to zero.Compute the V aR (at 1% and on a time interval of 10 days) of the portfolio(PORT2) composed by 20 shares of C and 15 shares of D.

3. Consider now a portfolio (PORT3) composed by 20 options (with DeltaΔC = 0.4) on the underlying C and by 15 options (with Delta ΔD = −0.1)on the underlying D. Give the Delta-approximation of the V aR (at 1% andon a time interval of 10 days) of portfolio PORT3.

Solution

1. Applying formula (12.2) for the computation of the V aR of a portfolio ofstocks whose returns are jointly normally distributed, we get immediatelythat the V aR (at 1% and on a time interval of 10 days) of portfolio PORT1is

V aR10gg0.01

(ΔP port1

)= N−1 (1 − α) · √Δt ·

√∑ni=1

∑nj=1 xixjSi

0Sj0σiσjρij

−∑nj=1 xjS

j0μj(Δt)

= N−1 (0.99) · √10 ·[(20 · 4.2 · 0.012)2 + (15 · 3.6 · 0.028)2

+ 2 · 0.25 · (20 · 4.2 · 0.012 · 15 · 3.6 · 0.028)]1/2

− [20 · 4.2 · 0.0002 + 15 · 3.6 · 0.0008] · 10= 14.831− 0.6 = 14.231.

In other words, the amount of cash to be deposited as a margin for PORT1is equal to 14.231 euros.

2. Since the features of stocks C and D are the same as those of A and B,except for the drifts that are zero for C and D, the V aR of portfolio PORT2can be obtained by taking only the first term in the computation above for


the V aR of PORT1, i.e.

V aR10gg0.01

(ΔP port2

)=N−1 (1 − α) ·

√Δt

√√√√ n∑i=1

n∑j=1

xixjSi0S

j0σiσjρij = 14.831.

In other words, the amount of money to be deposited as a margin forPORT2 is 14.831 euros.

3. Applying formula (12.4) for the Delta approximation of the V aR of a port-folio of options having underlyings with jointly normal returns, we obtainimmediately that the V aR (at 1% and on a time interval of 10 days) ofportfolio PORT3 is

V aR10gg0.01

(ΔP port3

)= N−1 (1 − α) · √Δt ·

√∑ni=1

∑nj=1 xixjΔ

i0Δ

j0S

i0S

j0σiσjρij

= . . . = 2.896.

In other words, the amount of cash to be deposited as a margin for PORT3is 2.896 euros. Note that this margin is much smaller than the one ofPORT2. This means that some options in PORT3 have positive Δ, othersnegative Δ. Hence, PORT3 is more diversified than PORT2.

Exercise 12.4. Consider the daily log-return of a stock represented by therandom variable X of mean 0.0002 and standard deviation 0.02.

1. Compute the V aR of X at 5% and on a period of one day, once X isassumed to be normally distributed.

2. Applying the Cornish-Fisher expansion, compute the V aR of X at 5% andon a period of one day when the skewness of X is supposed to be equal to(−0.3).

SolutionDenote by m, s and ξ the mean, the standard deviation and the skewness

of X, respectively.

1. If X is distributed as a normal with mean m = 0.0002 and standard de-viation s = 0.02, then the V aR of X at 5% and on a period of one day isgiven by

V aR0.05 (X) = s · N−1 (0.95) − m = 0.0327.

2. Applying the Cornish-Fisher expansion, the V aR of X at α = 0.05 and ona period of one day can be approximated as follows:

V aR0.05 (X) ∼= −m− s ·[N−1 (0.05) +

16ξ((

N−1 (0.05))2 − 1

)]= 0.0344.


Exercise 12.5. Consider a financial position for which a margin of 400 euroshas been already deposited, and suppose that the new position (initial posi-tion plus margin) is distributed as a Pareto with parameters x0 = 100 anda = 2.

1. Compute the V aR at 10% of the new position. Is the deposited marginsufficient?

2. What about the V aR at 1%?3. Compute the CV aR at 10% of the new position.

SolutionDenote by X the random variable representing the P&L of the initial po-

sition and by Y the one representing the P&L of the new position, henceY = X + 400. The cumulative distribution function and the density functionof Y are, respectively,

FY (y) =

{0; if y < x0

1 −(

x0y

)a

; if y ≥ x0

fY (y) ={

0; if y < x0

axa0y

−a−1; if y ≥ x0

First of all, note that E [Y ] = aa−1x0 = 200 and that Y /∈ L∞ since Y may

take any value greater or equal to x0.

1. Because Y is a continuous random variable, the α0-quantile of Y solves

FY (y) = α0.

It follows, therefore, that (x0y

)a

= 1 − α0

y = x0

(1−α0)1/a .

For α0 = 0.10, we obtain that q0.10 (Y ) = 100√0.9

= 105.41 and thatV aR0.10 (Y ) = −105.41. Consequently, the margin of 400 euros added toX is enough and it is not necessary to deposit an extra amount of money.

2. The V aR at α1 = 0.01 of Y can be computed in (at least) two ways. Thefirst way consists in proceeding as before, so to obtain

V aR0.01 (Y ) = − x0

(1 − α1)1/a

= −100.50.

The second consists in recalling (see Barucci et al. [4], Fiori et al. [18]) thatfor Pareto distributed random variables one has

V aRα1 (Y ) = V aRα0 (Y ) ·(

1 − α0

1 − α1

)1/a

,


hence

V aRα1 (Y ) = −105.41 ·(

0.90.99

)1/2

= −100.50.

3. Since Y does not belong to L∞ but to L1, it is easy to check thatCV aRα0 (Y ) = TCEα0 (Y ) = EP [−Y |Y ≤ −V aRα0(Y )] is still true.Hence:

CV aRα0 (Y ) = EP [−Y |Y ≤ −V aRα0(Y )] = −EP

[Y 1{Y ≤−V aRα0(Y )}

]P (Y ≤ −V aRα0(Y ))

.

(12.10)By item 1., we deduce that V aRα0 (Y ) = − x0

(1−α0)1/a and that P (Y ≤

−V aRα0(Y )) = α0. Set now y∗ = x0

(1−α0)1/a = −V aRα0 (Y ). We obtain

that

EP

[Y 1{Y ≤−V aRα0(Y )}

]=∫ y∗

x0

axa0y

−ady =[

a

1 − axa

0y1−a

]∣∣∣∣∣y∗

x0

=a

a − 1x0

[1 − 1

(1 − α0)1/a−1

]

= E [Y ] +a

a − 1(1 − α0) V aRα0 (Y ) .

From the argument above and from (12.10) it follows that

CV aRα0 (Y ) = EP [−Y |Y ≤ −V aRα0(Y )]

= −E [Y ] + aa−1 (1 − α0)V aRα0 (Y )

α0

= −200 + 2 · 0.9 · (−105.41)0.1

= −102.62.


X =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−1200; on ω1

−400; on ω2

−80; on ω3

0; on ω4

16; on ω5

x; on ω6

,

where 16 < x < 4000, P (ω1) = 0.01, P (ω2) = 0.04, P (ω3) = 0.05, P (ω4) =0.7, P (ω5) = 0.1 and P (ω6) = 0.1.Find the smallest x such that, according to the coherent risk measure gen-

erated by the set Q = {P, Q1} (with Q1(ω1) = Q1(ω2) = Q1(ω3) = Q1(ω6) =


1/8 and Q1(ω4) = Q1(ω5) = 1/4), the margin to be deposited for X is smallerthan or equal to 200.

SolutionFirst of all, recall that the coherent risk measure generated by the set of

generalized scenarios Q = {P, Q1} is given by ρQ (X) � supQ∈Q EQ [−X].Let us start by computing EP [−X] and EQ1 [−X]:

EP [−X] = 1200 · 0.01 + 400 · 0.04 + 80 · 0.05 + 0 − 16 · 0.1− x · 0.1

= 30.4− x

10

EQ1 [−X] =1200

8+

4008

+808

+ 0 − 164

− x

8= 206 − x

8.

Since 30.4− x10 ≥ 206− x

8 holds if and only if x ≥ 7024, in the present case(16 < x < 4000) we deduce that

ρQ (X) = supQ∈Q

EQ [−X] = sup {EP [−X] ; EQ1 [−X]} = 206− x

8.

In order to fulfill the constraint on the margin for X (according to ρQ), oneshould have 206− x

8 ≤ 200, hence x ≥ 48.Consequently, 48 is the smallest x ∈ (16, 4000) verifying the conditions

required in terms of margin for X.

Exercise 12.7. Consider a sample space with only two elementary events,i.e. Ω = {ω1, ω2}.Find the coherent risk measure ρA associated to the acceptance set

A ={

(x, y) ∈ R2 : y ≥ −2x; y ≥ −x

2

}.

Establish which of the following positions is preferred according to the riskmeasure ρA:

X1 ={−30; on ω1

60; on ω2; X2 = 0.

SolutionIt is easy to check that the acceptance set A (see Fig. 12.1) satisfies all the

properties (A), hence the risk measure

ρA (X) = inf {m ∈ R : m + X ∈ A}

associated to A is coherent.Since Ω has only two elements, any random variable X on Ω can be iden-

tified with (X(ω1), X(ω2)) ∈ R2. Note that m + X ∈ A if and only if{m + X (ω2) ≥ −2 (m + X (ω1))m + X (ω2) ≥ −m+X(ω1)

2

{m ≥ −2

3X (ω1) − 13X (ω2)

m ≥ −13X (ω1) − 2

3X (ω2).


−4 −2 0 2 4

−2

2

4

Fig. 12.1. Acceptance set A

It follows that

m ≥ sup{−2

3X (ω1) − 1

3X (ω2) ;−1

3X (ω1) − 2

3X (ω2)

},

hence

ρA (X) = sup{−2

3X (ω1) − 1

3X (ω2) ;−1

3X (ω1) − 2

3X (ω2)

}.

In terms of sets of generalized scenarios, ρA can also be written as

ρA (X) = supQ∈{Q1,Q2}

EQ [−X]

where Q1 (ω1) = 23 , Q1 (ω2) = 1

3 and Q2 (ω1) = 13 , Q2 (ω2) = 2

3 .

For what concerns positions X1 and X2, we obtain that

ρA (X1) = sup{−2

3· (−30) − 1

3· 60;−1

3· (−30) − 2

3· 60}

= −30

ρA (X2) = 0,

hence both positions are acceptable even if X1 is preferable over X2 (withrespect to ρA).

Exercise 12.8. Consider a sample space with only two elementary events,i.e. Ω = {ω1, ω2}.Find the risk measure ρA associated to the acceptance set

A ={(x, y) ∈ R2 : y ≥ x

2; y ≤ 4x

}and verify that ρA does not satisfy monotonicity.


−4 −2 0 2 4

−2

2

4

Fig. 12.2. Acceptance set A

SolutionIt is easy to check that the acceptance set A (see Fig. 12.2) does not satisfy

properties (A) (in particular A � R2+). Consequently, the risk measure ρA

associated to such a set is not coherent.Since Ω has only two elements, any random variable X on Ω can be iden-

tified with (X (ω1) , X (ω2)) ∈ R2. Note that m + X ∈ A if and only if{m + X (ω2) ≥ m+X(ω1)

2m + X (ω2) ≤ 4 (m + X (ω1))

{m ≥ X (ω1) − 2X (ω2)m ≥ −4

3X (ω1) + 1

3X (ω2)

.

It follows that

m ≥ sup{

X (ω1) − 2X (ω2) ;−43X (ω1) +

13X (ω2)

},

hence

ρA (X) = sup{

X (ω1) − 2X (ω2) ;−43X (ω1) +

13X (ω2)

}.

It is immediate to verify that ρA satisfies subadditivity, positive homogene-ity and cash-invariance.The following counterexamples shows that monotonicity is not guaranteed

by ρA.Taking, for instance,

X1 ={

0; on ω1

30; on ω2,

we obtain that X1 ≥ 0 but ρA (X1) = 10 > 0 = ρA (0), so X1 /∈ A. In otherwords, even if X1 will never give a loss, an amount of 10 euros has to bedeposited as a margin.


Furthermore, take the two positions:

Y1 ={−30, on ω1

60, on ω2; Y2 =

{ −6, on ω1

180, on ω2.

We obtain that ρA (Y2) > ρA (X1) even if Y2 (ω) ≥ Y1 (ω) for any ω ∈ Ω. Infact,

ρA (Y1) = sup{−30 − 120;

43· 30 +

13· 60}

= 60

ρA (Y2) = sup{−6 − 360;

43· 6 +

13· 180

}= 68.

It is also straightforward to check that

Y2 − Y1 ={

24, on ω1

120, on ω2≥ 0

and ρA (Y2 − Y1) = 8 > 0. Consequently, Y2 − Y1 /∈ A.

Exercise 12.9. Consider a portfolio formed by two stocks whose returns arejointly distributed as bivariate normal. Suppose that the mean returns of thestocks are, respectively, 10% and 6% per year, their standard deviations 40%and 30% per year and their correlation amounts to 0.5.

1. Supposing short-selling is not allowed, find the optimal portfolio so to min-imize the Conditional Value at Risk of the portfolio return [minus its mean]at level α = 0.01 and so to have an average return of 6.5% at least. Whatabout the optimal portfolio for an average return of at least 8%?

2. Show that the optimal portfolio of the previous item is the same whenthe riskiness is measured by Value at Risk, Conditional Value at Risk orstandard deviation.

SolutionOur goal is to construct a portfolio by investing a percentage x1 in one

stock (with return R1) and a percentage x2 in the other stock (with returnR2). Since short-selling is not allowed, x1 and x2 have to satisfy:{

x1, x2 ≥ 0,x1 + x2 = 1,

hence x1 = x ∈ [0, 1] and x2 = 1 − x.Let μi = E[Ri], σi =

√V (Ri) (for i = 1, 2) and denote with ρ the correla-

tion between R1 and R2. Since R1 and R2 are jointly normal, then also theportfolio return

Rp = x1R1 + x2R2 = xR1 + (1 − x)R2


is distributed as a normal with mean E[Rp] = xE[R1] + (1 − x)E[R2] =x(μ1 − μ2) + μ2 and with variance

V (Rp) = V (xR1 +(1−x)R2) = x2σ21 +(1−x)2σ2

2 +2ρσ1σ2x(1−x). (12.11)

1. When Conditional Value at Risk is used as a criterion for risk minimization,the optimization problem can be written as

minx ∈ [0, 1];E[Rp] ≥ l

CV aRα (Rp − E[Rp]) , (12.12)

where l stands for the target on the portfolio return.For μ1 > μ2, the constraint in the minimization becomes x ∈ [0, 1] suchthat x ≥ l−μ2

μ1−μ2. Hence, the minimization problem above reduces to

minx ∈ [0, 1];x ≥ l−μ2

μ1−μ2

CV aRα (Rp − E[Rp]) . (12.13)

Since Rp − E[Rp] ∼ N(0; V (Rp) and CV aRα(Y ) = −m + sN′(N−1(α))α

forY ∼ N(m; s2) (see, for instance, Barucci et al. [3]) with N ′(·) denoting thedensity function of a standard normal, problem (12.13) becomes

minx ∈ [0, 1];x ≥ l−μ2

μ1−μ2

{N ′(N−1(α))

α

√x2(σ2

1 + σ22 − 2ρσ1σ2)−2x(σ2

2 − ρσ1σ2) + σ22

}.

(12.14)It is then easy to check that the unconstrained minimum is attained atx = σ2

2−ρσ1σ2

σ21+σ2

2−2ρσ1σ2. In our case, x = 3

13.

Consider now l1 = 0.065. Since the constraints on x become x ∈ [0, 1] andx ≥ l1−μ2

μ1−μ2= 1

8 , the constrained minimization is obtained for x∗ = 313 .

The optimal portfolio consists thus in investing in 313 shares of the first

stock and in 1013 shares of the second stock. The corresponding value of the

Conditional Value at Risk is CV aR0.1(Rp − E[Rp]) = 2.614.

When l2 = 0.08, instead, the constraints on x become x ∈ [0, 1] and x ≥l2−μ2μ1−μ2

= 0.5 and the constrained minimization is obtained for x∗ = 0.5.The optimal portfolio consists thus in investing in 0.5 shares of the firststock and in 0.5 shares of the second stock. The corresponding value of theConditional Value at Risk is CV aR0.1(Rp − E[Rp]) = 2.758. As one couldexpect, the minimal riskiness of this last portfolio is greater than the other(we have indeed required an average portfolio return greater than before).


2. Consider now the optimization problem studied above, where the criterionof risk minimization of Conditional Value at Risk is replaced by that ofValue at Risk or of standard deviation.Because of the normality assumption on the joint distribution of (R1, R2)(hence, normality of Rp), the V aR minimization reduces to

minx ∈ [0, 1];x ≥ l−μ2

μ1−μ2

{N−1(1 − α)

√x2(σ2

1 + σ22 − 2ρσ1σ2) − 2x(σ2

2 − ρσ1σ2) + σ22

},

(12.15)while the minimization with respect to the standard deviation becomes

minx ∈ [0, 1];x ≥ l−μ2

μ1−μ2

√x2(σ2

1 + σ22 − 2ρσ1σ2) − 2x(σ2

2 − ρσ1σ2) + σ22. (12.16)

By the arguments above and since N−1(1 − α) > 0 for α = 0.01, in thepresent case the optimal composition of the portfolio is the same with re-spect to the CV aR criterion, to the V aR criterion and to the standarddeviation criterion. Indeed, the minimum in problems (12.13), (12.15) and(12.16) is always attained at the same x∗. The minimal riskiness of theportfolio is, respectively, equal to

CV aR0.1(Rp) = 2.614

V aR0.1(Rp) = 0.369

σ(Rp) = 0.288

for l1 = 0.065, while for l2 = 0.08

CV aR0.1(Rp) = 2.758

V aR0.1(Rp) = 0.38989

σ(Rp) = 0.304.

Exercise 12.10. Consider a financial investment whose Profit and Loss in 2years is represented by the random variable:

X =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

100, on ω1

20, on ω2

0, on ω3

−10, on ω4

−40, on ω5

,


where the filtrated probability space (Ω,F , (Ft)t=0,1,2, P ) is given by

Ω = {ω1, ω2, ω3, ω4, ω5}U = {ω1, ω2}D = {ω3, ω4, ω5}F0 = {∅, Ω}F1 = {∅, U, D, Ω}F2 = P(Ω)

andP (U) = 0.8; P (D) = 0.2

P (ω1|U) = 0.4; P (ω2|U) = 0.6P (ω3|D) = 0.6; P (ω4|D) = 0.3; P (ω5|D) = 0.1

1. Verify whether V aR(0)0.05(X) = V aR

(0)0.05(−V aR

(1)0.05(X)), where V aR

(t)α (X)

stands for the Value at Risk of X at time t 1.2. Verify whether CV aR

(0)0.05(X) = CV aR

(0)0.05(−CV aR

(1)0.05(X)), where

CV aR(t)α (X) stands for the Conditional Value at Risk of X at time t.

3. Consider now the dynamic risk measure (ρt)t=0,1,2 defined as

ρt(X) � ess supQ∈Q

EQ[−X| Ft], (12.17)

where Q = {P, Q1, Q2, Q3} and

Q1(U) = Q2(U) = Q3(U) = 0.8; Q1(D) = Q2(D) = Q3(D) = 0.2

Q1(ω1|U) = Q2(ω1|U) = 0.5; Q1(ω2|U) = Q2(ω2|U) = 0.5

Q3(ω1|U) = 0.4; Q3(ω2|U) = 0.6; Q1(ω3|D) = Q3(ω3|D) = 0.4

Q1(ω4|D) = Q3(ω4|D) = 0.4; Q1(ω5|D) = Q3(ω5|D) = 0.2

Q2(ω3|D) = 0.6; Q2(ω4|D) = 0.3; Q2(ω5|D) = 0.1

Verify whether ρ0(X) = ρ0(−ρ1(X)) or not.

1 We recall (see, among many others, [2]) that a dynamic risk measure is a fam-ily (ρt)t=0,1,2 defined on a space X of random variables such that ρt(X) is Ft-measurable, i.e. it takes into account all the information available until time t.In particular, ρ0(X) ∈ R. A risk measure is then said to be time-consistent ifρ0(−ρt(X)) = ρ0(X) for any X ∈ X and t ∈ [0, T ].


SolutionBy the evolution of information in time, we obtain the following tree:

ω1↗U = {ω1, ω2}↗ ↘

ω2·ω3↘ ↗

D = {ω3, ω4, ω5} → ω4↘ω5−− − − −− −− −−− − −−−

0 1 year 2 years

Furthermore, it also follows that

P (ω1) = 0.32; P (ω2) = 0.48;P (ω3) = 0.12; P (ω4) = 0.06;P (ω5) = 0.02.

1. In order to compute the Value at Risk of X at the initial time 0 we canjust consider the last nodes of the tree and the corresponding probabilities(evaluated by means of P ). We deduce that

V aR(0)0.05(X) = 10. (12.18)

Let us compute now V aR(1)0.05(X), a random variable taking two values: one

at the node U (denoted by V aR(1,U)0.05 (X)), the other at the node D (denoted

by V aR(1,D)0.05 (X)).

Note that V aR(1,U)0.05 (X) is the V aR of the random variable X on the sub-

tree starting from time t = 1 and node U . More precisely:

ω1↗U = {ω1, ω2} ↘

ω2−− −− −− − −−−1 year 2 years

or, in terms of X and P ,

X(ω1) = 100↗prob.0.4

U ↘prob.0.6

X(ω2) = 20− −−− −− −− −− −−1 year 2 years

It follows that V aR(1,U)0.05 (X) = −20.


In order to compute V aR(1,D)0.05 (X) we can proceed as previously by consid-

ering the sub-tree starting from time t = 1 and node D. More precisely:

ω3↗D = {ω3, ω4, ω5} → ω4↘

ω5− −− −− − −− −− −−1 year 2 years


X(ω3) = 0↗prob.0.6

D →prob.0.3 X(ω4) = −10↘prob.0.1

X(ω5) = −40−− −− − −− −− −−−1 year 2 years

It follows that V aR(1,D)0.05 (X) = 40.

Finally, to find V aR(0)0.05(−V aR

(1)0.05(X)) we proceed backwards once again

by considering the sub-tree starting from 0 and ending at nodes U, D:

U = {ω1, ω2}↗·

↘D = {ω3, ω4, ω5}−− − − −− −− −−−

0 1 year

or, in terms of V aR(1)0.05(X) and P ,

−V aR(1,U)0.05 (X) = 20 (with prob. 0.8)

↗·

↘−V aR

(1,D)0.05 (X) = −40 (with prob. 0.2)

−− − − −− −− −− − −−0 1 year

HenceV aR

(0)0.05(−V aR

(1)0.05(X)) = 40 �= 10 = V aR

(0)0.05(X),

i.e. the dynamic version of V aR is not time-consistent.


2. Proceeding as above, we deduce that

CV aR(0)0.05(X) =

E[(−10 − X)+ ]0.05

+ 10 = 22. (12.19)

CV aR(1)0.05(X) is then a random variable taking two values: CV aR

(1,U)0.05 (X)

at node U, CV aR(1,D)0.05 (X) at node D.

Note that CV aR(1,U)0.05 (X) is the CV aR of the random variable X on the

sub-tree starting from time t = 1 and node U . More precisely:

ω1

↗U = {ω1, ω2}

↘ω2

−− −− −− − −−−1 year 2 years


X(ω1) = 100↗prob.0.4

U↘prob.0.6

X(ω2) = 20−− −− −− −− −−1 year 2 years

It follows that CV aR(1,U)0.05 (X) = (20−100)+·0.4+(20−20)+·0.6

0.05 − 20 = −20.Similarly, we obtain that CV aR

(1,D)0.05 (X) = 40 and

CV aR(0)0.05(−CV aR

(1)0.05(X)) = 40 �= 22 = CV aR

(0)0.05(X), (12.20)

i.e. the dynamic version of CV aR is not time-consistent.3. We compute now ρ0(X) = supQ∈{P,Q1,Q2,Q3} EQ[−X]. It is easy to checkwhat follows:

ωi P Q1 Q2 Q3

ω1 0.32 0.4 0.4 0.32ω2 0.48 0.4 0.4 0.48ω3 0.12 0.08 0.12 0.08ω4 0.06 0.08 0.06 0.08ω5 0.02 0.04 0.02 0.04


Hence, we obtain

EP [−X] = −100 · 0.32− 20 · 0.48 + 0 + 10 · 0.06 + 40 · 0.02 = −39.2EQ1 [−X] = −100 · 0.4− 20 · 0.4 + 0 + 10 · 0.08 + 40 · 0.04 = −45.6EQ2 [−X] = −100 · 0.4− 20 · 0.4 + 0 + 10 · 0.06 + 40 · 0.02 = −46.6EQ1 [−X] = −100 · 0.32− 20 · 0.48 + 0 + 10 · 0.08 + 40 · 0.04 = −39.2,

soρ0(X) = sup

Q∈{P,Q1,Q2,Q3}EQ[−X] = −39.2. (12.21)

In order to find ρ1(X), we proceed as in items 1.-2. so to obtain that ρ1(X)takes the following values at U and D, respectively,

ρU1 (X) = sup{ −100 · 0.4− 20 · 0.6;−100 · 0.5− 20 · 0.5;

−100 · 0.5− 20 · 0.5;−100 · 0.4− 20 · 0.6}= −52;

ρD1 (X) = sup{ 0 + 10 · 0.3 + 40 · 0.1; 0 + 10 · 0.4 + 40 · 0.2;

0 + 10 · 0.3 + 40 · 0.1; 0 + 10 · 0.4 + 40 · 0.2}= 12.

Proceeding backwards once more, we deduce that

ρ0(−ρ1(X)) = −39.2. (12.22)

By the arguments above and by (12.21), it follows that

ρ0(−ρ1(X)) = ρ0(X) = −39.2.

Note that one can prove that the dynamic risk measure (ρt)t=0,1,2 definedin (12.17) is time-consistent. In such a case, indeed, the set Q contains anyprobability measure obtained by “pasting” other probability measures of Q.Note that stability under pasting is not always satisfied by the set of general-ized scenarios in the representation of CV aR.


Exercise 12.11. Consider two stocks (A and B) whose daily returns arejointly normal. Assume that the correlation between their returns is −0.8,that stock A has current price 2 euros, negligible daily drift and daily volatil-ity of 1.2%, and that stock B has current price 1 euro, negligible daily driftand daily volatility of 1.6%.Rank the following financial positions, based on the riskiness evaluated

with respect to V aR at 2% and on a period of one day:

• 40 shares of stock A;


• 10 shares of stock B;• a portfolio composed by 40 shares of A and by 10 shares of B.

Is diversification of risk encouraged in the present setting?


X =

⎧⎪⎪⎨⎪⎪⎩

−4000; on ω1

−48; on ω2

0; on ω3

640; on ω4

where P (ω1) = 0.02, P (ω2) = 0.08, P (ω3) = 0.8 and P (ω4) = 0.1.

1. Compute the Value at Risk at 5% and the Tail Conditional Expectation at5% of X.

2. In X, replace the value −4000 with x. Find the largest possible loss x forwhich the margin to be deposited according to TCE at 5% does not exceedtwice the corresponding V aR.

Exercise 12.13. Consider two portfolios whose Profit & Loss are representedby the following random variables:

X =

⎧⎪⎪⎨⎪⎪⎩

−160; on ω1

−20; on ω2

0; on ω3

120; on ω4

Y =

⎧⎪⎪⎨⎪⎪⎩

−240; on ω1

0; on ω2

160; on ω3

−280; on ω4

where P (ω1) = 0.01, P (ω2) = 0.02, P (ω3) = 0.57 and P (ω4) = 0.40.

1. Compute the Value at Risk at 2% ofX, of Y and of X+Y . Is diversificationof risk encouraged or not?

2. Compute the Conditional Value at Risk at 2% of X, of Y and of X + Y .Is diversification of risk encouraged or not?


X =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−800; on ω1

−200; on ω2

0; on ω3

120; on ω4

400; on ω5

1200; on ω6

with P (ω1) = 0.02, P (ω2) = 0.08, P (ω3) = 0.64, P (ω4) = 0.16, P (ω5) = 0.06and P (ω6) = 0.04.


Let ρQ be the coherent risk measure generated by the set Q = {P, Q1, Q2}of generalized scenarios, where

Q1(ω1) = Q1(ω2) = Q1(ω3) = Q1(ω4) = Q1(ω5) = Q1(ω6) = 1/6Q2(ω1) = Q2(ω2) = 0.2; Q2(ω3) = Q2(ω4) = Q2(ω5) = Q2(ω6) = 0.15

Compute ρQ(X) and establish which between ρQ and the expected value oflosses is the stronger (or more conservative) risk measure.


X =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

−480; with prob. 0.02−20; with prob. 0.04

0; with prob. 0.6440; with prob. 0.2

800; with prob. 0.1

1. Compute the Value at Risk at 10% and the Conditional Value at Risk at10% of X.

2. What is the sign of CV aR and V aR of a position having as P & L the oneabove increased by an amount of 80 euros? Is it possible to obtain CV aRand V aR of such a new position with one direct computation only?

Exercise 12.16. Consider the filtrated probability space (Ω,F , (Ft)t=0,1,2, P )given by

Ω = {ω1, ω2, ω3, ω4}U = {ω1, ω2}D = {ω3, ω4}F0 = {∅, Ω}F1 = {∅, U, D, Ω}F2 = P(Ω)

and a financial investment whose Profit and Loss in 2 years is represented bythe random variable

X =

⎧⎪⎪⎨⎪⎪⎩

20, on ω1

10, on ω2

0, on ω3

−30, on ω4

,

defined on the space above. Consider now the dynamic risk measures(ρRt )t=0,1,2 and (ρSt )t=0,1,2 defined as

ρRt (X) � ess supQ∈R EQ[−X| Ft] (12.23)

ρSt (X) � ess supQ∈S EQ[−X| Ft], (12.24)


where R = {P, Q1, Q2}, S = {P, Q1, Q2, Q3} and the probability measuresare as follows:

event P Q1 Q2 Q3

ω1|U 0.5 0.5 0.5 0.5ω2|U 0.5 0.5 0.5 0.5ω3|D 0.5 0.8 0.8 0.5ω4|D 0.5 0.2 0.2 0.5

U 0.5 0.4 0.5 0.4D 0.5 0.6 0.5 0.6

Verify if ρR0 (X) = ρR0 (−ρR1 (X)) and if ρS0 (X) = ρS0 (−ρS1 (X)). Explain anddiscuss the results obtained.

References

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2. Artzner, P., Delbaen, F., Eber, J.M., Heath, D., Ku, H.: Coherent multiperiodrisk adjusted values and Bellman’s principle. Annals of Operations Research152, 5–22 (2007)

3. Barucci, E. : Financial Markets Theory. Springer-Verlag, London (2003)4. Barucci, E., Marsala, C., Nencini, M., Sgarra, C.: Ingegneria finanziaria.Un’introduzione quantitativa. Egea, Milano (2009)

5. Bates, D.: Jumps and stochastic volatility: the exchange rate processes implicitin Deutschemark options. Review of Financial Studies 9, 69–107 (1996)

6. Björk, T. : Arbitrage Theory in Continuous Time. Oxford University Press,Oxford, 3rd edition (2009)

7. Bollerslev, T.: General Autoregressive Conditional Heteroskedasticity. Journalof Econometrics 31, 307–327 (1986)

8. Brigo, D., Mercurio, F.: Interest Rate Models. Theory and Practice. Springer-Verlag, Berlin Heidelberg, 2nd Edition (2006)

9. Capiński, M., Zastawniak, T.: Mathematics for Finance. Springer-Verlag, Lon-don, 2nd Edition (2011)

10. Cheridito, P., Stadje, M.: Time-inconsistency of VaR and time-consistent alter-natives. Finance Research Letters 6, 40–46 (2009)

11. D’Ecclesia, R.L., Gardini, L.: Appunti di Matematica Finanziaria II. Giap-pichelli Editore, Torino, Italy (1998)

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Index

affine-term structure, 202aggregational Gaussianity, 233arbitrage, 32– free of, 32, 63– opportunity, 59, 62ARCH model, 235asset or nothing, 173attainable, 33, 62

beta, 20binomial, 14– distribution, 2– model, 31– process, 2Black formula, 206Black-Scholes– equation, 102– formula, 122, 124– model, 123– partial differential equation, 102bond, 31Brownian– local time, 87Brownian motion, 3– geometric, 4– standard, 3

Capital Asset Pricing Model, 20caplet, 206Central Limit Theorem, 3characteristic function, 237completeness, 33Compound Poisson process, 234

Conditional Value at Risk, 249contingent claim, 32coupon bond, 201– option on, 207

Delta, 126– hedging, 32, 126– neutral, 127derivative– price, 32, 67diffusion equation, 103– fundamental solution, 115diffusion parameter, 3Dirac Delta distribution, 88distribution– Bernoulli, 1– binomial, 2– Dirac Delta, 88– exponential, 2– normal, 3– Pareto, 2– Poisson, 2dividend, 124– continuous, 125– discrete, 126– rate, 125drift, 3, 123

early exercise, 153– premium, 168efficient– portfolio, 18efficient frontier, 19

EmanuelaRosazzaGianin, Carlo Sgarra:Mathematical Finance:Theory Review and Exercises.UNITEXT – La Matematica per il 3+2 70DOI 10.1007/978-3-319-01357-2, © Springer International Publishing Switzerland 2013

276 Index

entropy, 74– criterion, 80

fat tails, 233Feynman-Kac Representation Theorem,103

filtration, 1– generated by, 1formula– of Black, 206– of Black-Scholes, 122, 124– of Itô, 86– of Tanaka, 87forward measure, 206Fourier transform, 234free boundary problem, 153free lunch, 62Fundamental Theorem of Asset Pricing,34, 62

Gamma, 127– hedging, 127– neutral, 127GARCH model, 235

hedging strategy, 126Ho-Lee model, 203Hull-White model, 203

instantaneous forward rate, 201interest rate, 201– instantaneous forward rate, 201– short rate, 201Itô– formula, 86– lemma, 86– stochastic integral, 85

jump, 234– -size distribution, 235Jump-Diffusion model, 234

leverage effect, 233LIBOR, 206– forward rate, 206– spot, 232– spot rate, 206local time, 87

market– complete, 33, 62– free of arbitrage, 32– incomplete, 62– portfolio, 20martingale, 4– measure, 32, 62model– binomial, 31– Black-Scholes, 123– Ho-Lee, 203– Hull-White, 203– jump-diffusion, 234– stochastic volatility, 234– Vasiček, 203– with jumps, 234Mutual Funds Theorem, 20

no-arbitrage interval, 63numéraire, 205– change of, 205

occupation time, 87optimal stopping, 153option, 34– American, 34, 153– Asian, 174– average rate, 174– average strike, 174– Barrier, 174– binary, 173– cash or nothing, 173– chooser, 174– European, 34, 124– exotic, 173– Lookback, 176– on a coupon bond, 207– on a zero-coupon bond, 204– path-dependent, 173– perpetual, 167– price, 32, 63ordinary differential equation, 101

partial differential equation, 101– backward parabolic, 102– elliptic, 102– forward parabolic, 102– hyperbolic, 102– parabolic, 102

Index 277

– semilinear, 101– solution, 101portfolio, 17– efficient, 19– of minimum variance, 19– optimal, 24– optimization, 18price– of a derivative, 32, 63process– adapted, 1– binomial, 2– Compound Poisson, 234– continuous-time, 1– discrete-time, 1– log-normal, 4– martingale, 4– Poisson, 2– stochastic, 1Put-Call Parity, 34, 125

random variable, 1replicating strategy, 32, 63Rho, 127– neutral, 127risk measures, 247– acceptance set, 250– coherent, 249– Conditional Value at Risk, 249– dynamic, 264– Tail Conditional Expectation, 249– time-consistent, 266– Value at Risk, 247risk-neutral measure, 32, 62

self-financing, 64short rate, 201– dynamics, 202– Ho-Lee model, 203– Hull-White model, 203– Vasiček model, 203similarity– method, 103

– solution, 107spread, 138– bear, 139– bull, 151– butterfly, 139stationary, 244stochastic– differential equation, 86– integral, 85– process, 85– volatility model, 234stock, 31stop-loss strategy, 150strategy– hedging, 165– replicating, 32, 63– self-financing, 64– stop-loss, 150– trading, 61

Tail Conditional Expectation, 249Tanaka formula, 87Theta, 127– neutral, 127time series, 244

utility, 18– maximization, 18

Value at Risk, 247– Delta approximation, 248Vasiček model, 203Vega, 127– neutral, 127volatility, 123– clustering, 233– implied, 233– model, 234– smile, 233

zero-coupon bond, 201– option on, 204

Collana Unitext – La Matematica per il 3+2

Series Editors:A. Quarteroni (Editor-in-Chief)L. AmbrosioP. BiscariC. CilibertoG. van der GeerG. RinaldiW.J. Runggaldier

Editor at Springer:F. [email protected]

As of 2004, the books published in the series have been given a volume num-ber. Titles in grey indicate editions out of print.As of 2011, the series also publishes books in English.

A. Bernasconi, B. CodenottiIntroduzione alla complessità computazionale1998, X+260 pp, ISBN 88-470-0020-3

A. Bernasconi, B. Codenotti, G. RestaMetodi matematici in complessità computazionale1999, X+364 pp, ISBN 88-470-0060-2

E. Salinelli, F. TomarelliModelli dinamici discreti2002, XII+354 pp, ISBN 88-470-0187-0

S. BoschAlgebra2003, VIII+380 pp, ISBN 88-470-0221-4

S. Graffi, M. Degli EspostiFisica matematica discreta2003, X+248 pp, ISBN 88-470-0212-5

S. Margarita, E. SalinelliMultiMath – Matematica Multimediale per l’Università2004, XX+270 pp, ISBN 88-470-0228-1

A. Quarteroni, R. Sacco, F.SaleriMatematica numerica (2a Ed.)2000, XIV+448 pp, ISBN 88-470-0077-72002, 2004 ristampa riveduta e corretta(1a edizione 1998, ISBN 88-470-0010-6)

13. A. Quarteroni, F. SaleriIntroduzione al Calcolo Scientifico (2a Ed.)2004, X+262 pp, ISBN 88-470-0256-7(1a edizione 2002, ISBN 88-470-0149-8)

14. S. SalsaEquazioni a derivate parziali - Metodi, modelli e applicazioni2004, XII+426 pp, ISBN 88-470-0259-1

15. G. RiccardiCalcolo differenziale ed integrale2004, XII+314 pp, ISBN 88-470-0285-0

16. M. ImpedovoMatematica generale con il calcolatore2005, X+526 pp, ISBN 88-470-0258-3

17. L. Formaggia, F. Saleri, A. VenezianiApplicazioni ed esercizi di modellistica numericaper problemi differenziali2005, VIII+396 pp, ISBN 88-470-0257-5

18. S. Salsa, G. VerziniEquazioni a derivate parziali – Complementi ed esercizi2005, VIII+406 pp, ISBN 88-470-0260-52007, ristampa con modifiche

19. C. Canuto, A. TabaccoAnalisi Matematica I (2a Ed.)2005, XII+448 pp, ISBN 88-470-0337-7(1a edizione, 2003, XII+376 pp, ISBN 88-470-0220-6)

20. F. Biagini, M. CampaninoElementi di Probabilità e Statistica2006, XII+236 pp, ISBN 88-470-0330-X

21. S. Leonesi, C. ToffaloriNumeri e Crittografia2006, VIII+178 pp, ISBN 88-470-0331-8

22. A. Quarteroni, F. SaleriIntroduzione al Calcolo Scientifico (3a Ed.)2006, X+306 pp, ISBN 88-470-0480-2

23. S. Leonesi, C. ToffaloriUn invito all’Algebra2006, XVII+432 pp, ISBN 88-470-0313-X

24. W.M. Baldoni, C. Ciliberto, G.M. Piacentini CattaneoAritmetica, Crittografia e Codici2006, XVI+518 pp, ISBN 88-470-0455-1

25. A. QuarteroniModellistica numerica per problemi differenziali (3a Ed.)2006, XIV+452 pp, ISBN 88-470-0493-4(1a edizione 2000, ISBN 88-470-0108-0)(2a edizione 2003, ISBN 88-470-0203-6)

26. M. Abate, F. TovenaCurve e superfici2006, XIV+394 pp, ISBN 88-470-0535-3

27. L. GiuzziCodici correttori2006, XVI+402 pp, ISBN 88-470-0539-6

28. L. RobbianoAlgebra lineare2007, XVI+210 pp, ISBN 88-470-0446-2

29. E. Rosazza Gianin, C. SgarraEsercizi di finanza matematica2007, X+184 pp,ISBN 978-88-470-0610-2

30. A. MachìGruppi – Una introduzione a idee e metodi della Teoria dei Gruppi2007, XII+350 pp, ISBN 978-88-470-0622-52010, ristampa con modifiche

31 Y. Biollay, A. Chaabouni, J. StubbeMatematica si parte!A cura di A. Quarteroni2007, XII+196 pp, ISBN 978-88-470-0675-1

32. M. ManettiTopologia2008, XII+298 pp, ISBN 978-88-470-0756-7

33. A. PascucciCalcolo stocastico per la finanza2008, XVI+518 pp, ISBN 978-88-470-0600-3

34. A. Quarteroni, R. Sacco, F. SaleriMatematica numerica (3a Ed.)2008, XVI+510 pp, ISBN 978-88-470-0782-6

35. P. Cannarsa, T. D’AprileIntroduzione alla teoria della misura e all’analisi funzionale2008, XII+268 pp, ISBN 978-88-470-0701-7

36. A. Quarteroni, F. SaleriCalcolo scientifico (4a Ed.)2008, XIV+358 pp, ISBN 978-88-470-0837-3

37. C. Canuto, A. TabaccoAnalisi Matematica I (3a Ed.)2008, XIV+452 pp, ISBN 978-88-470-0871-3

38. S. GabelliTeoria delle Equazioni e Teoria di Galois2008, XVI+410 pp, ISBN 978-88-470-0618-8

39. A. QuarteroniModellistica numerica per problemi differenziali (4a Ed.)2008, XVI+560 pp, ISBN 978-88-470-0841-0

40. C. Canuto, A. TabaccoAnalisi Matematica II2008, XVI+536 pp, ISBN 978-88-470-0873-12010, ristampa con modifiche

41. E. Salinelli, F. TomarelliModelli Dinamici Discreti (2a Ed.)2009, XIV+382 pp, ISBN 978-88-470-1075-8

42. S. Salsa, F.M.G. Vegni, A. Zaretti, P. ZuninoInvito alle equazioni a derivate parziali2009, XIV+440 pp, ISBN 978-88-470-1179-3

43. S. Dulli, S. Furini, E. PeronData mining2009, XIV+178 pp, ISBN 978-88-470-1162-5

44. A. Pascucci, W.J. RunggaldierFinanza Matematica2009, X+264 pp, ISBN 978-88-470-1441-1

45. S. SalsaEquazioni a derivate parziali – Metodi, modelli e applicazioni (2a Ed.)2010, XVI+614 pp, ISBN 978-88-470-1645-3

46. C. D’Angelo, A. QuarteroniMatematica Numerica – Esercizi, Laboratori e Progetti2010, VIII+374 pp, ISBN 978-88-470-1639-2

47. V. MorettiTeoria Spettrale e Meccanica Quantistica – Operatori in spazi di Hilbert2010, XVI+704 pp, ISBN 978-88-470-1610-1

48. C. Parenti, A. ParmeggianiAlgebra lineare ed equazioni differenziali ordinarie2010, VIII+208 pp, ISBN 978-88-470-1787-0

49. B. Korte, J. VygenOttimizzazione Combinatoria. Teoria e Algoritmi2010, XVI+662 pp, ISBN 978-88-470-1522-7

50. D. MundiciLogica: Metodo Breve2011, XII+126 pp, ISBN 978-88-470-1883-9

51. E. Fortuna, R. Frigerio, R. PardiniGeometria proiettiva. Problemi risolti e richiami di teoria2011, VIII+274 pp, ISBN 978-88-470-1746-7

52. C. PresillaElementi di Analisi Complessa. Funzioni di una variabile2011, XII+324 pp, ISBN 978-88-470-1829-7

53. L. Grippo, M. SciandroneMetodi di ottimizzazione non vincolata2011, XIV+614 pp, ISBN 978-88-470-1793-1

54. M. Abate, F. TovenaGeometria Differenziale2011, XIV+466 pp, ISBN 978-88-470-1919-5

55. M. Abate, F. TovenaCurves and Surfaces2011, XIV+390 pp, ISBN 978-88-470-1940-9

56. A. AmbrosettiAppunti sulle equazioni differenziali ordinarie2011, X+114 pp, ISBN 978-88-470-2393-2

57. L. Formaggia, F. Saleri, A. VenezianiSolving Numerical PDEs: Problems, Applications, Exercises2011, X+434 pp, ISBN 978-88-470-2411-3

58. A. MachìGroups. An Introduction to Ideas and Methods of the Theory of Groups2011, XIV+372 pp, ISBN 978-88-470-2420-5

59. A. Pascucci, W.J. RunggaldierFinancial Mathematics. Theory and Problems for Multi-period Models2011, X+288 pp, ISBN 978-88-470-2537-0

60. D. MundiciLogic: a Brief Course2012, XII+124 pp, ISBN 978-88-470-2360-4

61. A. MachìAlgebra for Symbolic Computation2012, VIII+174 pp, ISBN 978-88-470-2396-3

62. A. Quarteroni, F. Saleri, P. GervasioCalcolo Scientifico (5a ed.)2012, XVIII+450 pp, ISBN 978-88-470-2744-2

63. A. QuarteroniModellistica Numerica per Problemi Differenziali (5a ed.)2012, XVIII+628 pp, ISBN 978-88-470-2747-3

64. V. MorettiSpectral Theory and Quantum MechanicsWith an Introduction to the Algebraic Formulation2013, XVI+728 pp, ISBN 978-88-470-2834-0

65. S. Salsa, F.M.G. Vegni, A. Zaretti, P. ZuninoA Primer on PDEs. Models, Methods, Simulations2013, XIV+482 pp, ISBN 978-88-470-2861-6

66. V.I. ArnoldReal Algebraic Geometry2013, X+110 pp, ISBN 978-3-642–36242-2

67. F. Caravenna, P. Dai PraProbabilità. Un’introduzione attraverso modelli e applicazioni2013, X+396 pp, ISBN 978-88-470-2594-3

68. A. de Luca, F. D’AlessandroTeoria degli Automi Finiti2013, XII+316 pp, ISBN 978-88-470-5473-8

69. P. Biscari, T. Ruggeri, G. Saccomandi, M. VianelloMeccanica Razionale2013, XII+352 pp, ISBN 978-88-470-5696-3

70. E. Rosazza Gianin, C. SgarraMathematical Finance: Theory Review and Exercises. From BinomialModel to Risk Measures2013, X+278pp, ISBN 978-3-319-01356-5

The online version of the books published in this series is available atSpringerLink.For further information, please visit the following link:http://www.springer.com/series/5418

from binomial model to risk measures unitext · 2020. 1. 17. · emanuelarosazzagianin dipartimento...

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