qmf lectures 2013-14
DESCRIPTION
QMF lectures 2013-14.TRANSCRIPT
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QUANTITATIVE METHODS FOR FINANCE
Room 221, 2nd oor, Via Necchi 9
Phone: +39-02-72342345
O�ce hours: 14:30-16:30 on Tuesday
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TEXTBOOK
I Lecture notes.
I A. BATTAUZ - F. ORTU, Arbitrage Theory in Discrete and Continuous
Time (Cod. 8444), Edizioni EGEA, last edition (available at Libreria EGEA, Via
Bocconi, 8).
I K. SYDSAETER - P. J. HAMMOND, Mathematics for economic analysis,
Prentice-Hall, 1995.
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THE COURSE MATERIAL IS ON BLACKBOARD (BB)
I Enroll in a BB course by entering your I-Catt URL (www.i-catt.it).
I At http://www.unicatt.it/faq/faq.asp?id=146 you'll �nd detailed inform-
ation on how to enroll in BB courses.
I You enter BB via http://blackboard.unicatt.it.
I Your BB username is: surname.IDnumber. Your BB password is: utente.
I For any problem, contact [email protected].
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COURSE OUTLINE
I No-arbitrage pricing of assets replicated by existing traded securities.
I No-arbitrage pricing of unreplicable assets.
I Constrained optimization with applications to �nance/economics.
I Equilibrium valuation of assets.
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LINEAR ALGEBRA (REVIEW)
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I IS THE NEUTRAL ELEMENT OF THE MULTIPLICATION
I The n�n identity matrix I has 1s down the main diagonal and 0s elsewhere.The 2� 2 identity matrix, for example, is
I =
"1 00 1
#.
"1 00 1
# "3 4 05 2 2
#=
"3 4 05 2 2
#.
264 5 18 31 0
375 "1 00 1
#=
264 5 18 31 0
375 .
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TRANSPOSITION AND MULTIPLICATION
I If you transpose a matrix product you multiply the transposed matrices inreverse order:
0BBBBBBBB@
Az }| {"1 2 01 1 1
#bz }| {264 581
375
1CCCCCCCCA
T
=
"2114
#T
m
bTz }| {h5 8 1
iATz }| {264 1 12 10 1
375 =h21 14
i
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INVERSE AND INVERTIBLE MATRICES
I If it exists, the inverse of an n�n matrix C is an n�n matrix C�1 satisfying
CC�1 = I and C�1C = I :
I For example,
Cz }| {"1 21 3
#C�1z }| {"3 �2�1 1
#=
"1 00 1
#
and
C�1z }| {"3 �2�1 1
#Cz }| {"1 21 3
#=
"1 00 1
#:
I An n� n matrix C is invertible if its inverse matrix C�1 exists.
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DETERMINANTS OF SQUARE MATRICES 1
I C is invertible if and only if (i�) it is non-singular, that is, its determinant
is non-zero.
det
"1 21 3
#!= 1 � (�1)1+1 � det (3)| {z }
the 1-1 cofactor
+
1 � (�1)2+1 � det (2)| {z }the 2-1 cofactor
= 1 � 3 + 1 � (�2) = 1 .
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DETERMINANTS OF SQUARE MATRICES 2
det
0B@264 1 2 32 3 23 3 2
3751CA = 1 � (�1)1+1 � det
"3 23 2
#!| {z }
the 1-1 cofactor
+
2 � (�1)2+1 � det "
2 33 2
#!| {z }
the 2-1 cofactor
+
3 � (�1)3+1 � det "
2 33 2
#!| {z }
the 3-1 cofactor
= 1 � (0) + 2 � 5 + 3 � (�5) = �5
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CONSTRUCTING THE INVERSE MATRIX 1
"1 21 3
#�1=
1
det
"1 21 3
#!| {z }
=1
matrix of cofactors of
"1 21 3
# !T
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CONSTRUCTING THE INVERSE MATRIX 2
matrix of cofactors of
"1 21 3
#=
266666666664
(�1)1+1 � det (3)| {z }the 1-1 cofactor
(�1)1+2 � det (1)| {z }the 1-2 cofactor
(�1)2+1 � det (2)| {z }the 2-1 cofactor
(�1)2+2 � det (1)| {z }the 2-2 cofactor
377777777775
=
264 3 �1
�2 1
375
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CONSTRUCTING THE INVERSE MATRIX 3
"1 21 3
#�1=
1
1
"3 �1�2 1
#T
=
"3 �2�1 1
#
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A FORMULA FOR INVERTING 2� 2 MATRICES
I Given ad� cb 6= 0, consider the square matrix
"a bc d
#.
I Its inverse is
1
ad� cb
"d �b�c a
#.
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INVERTING A 3� 3 MATRIX
I Show that
266666641 2 3
2 3 2
3 3 2
37777775
�1
=1
�5
266666640 5 �5
2 �7 4
�3 3 �1
37777775 .
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SYSTEMS OF LINEAR EQUATIONS (REVIEW)
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A FIRST EXAMPLE 1
I Consider the following linear system in the unknowns x1, x2, and x3 :
8>>>>>><>>>>>>:
x1 + 2x2 + 3x3
2x1 + 3x2 + 2x3
3x1 + 3x2 + 2x3
=
=
=
6
�1
0
I We can rewrite it as follows:
264 1 2 32 3 23 3 2
375264 x1x2x3
375 =
264 6�10
375
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A FIRST EXAMPLE 2
I We can also rewrite the system as a `constrained' linear combination:
x1
264 123
375+ x2264 233
375+ x3264 322
375 =
264 6�10
375
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A FIRST EXAMPLE 3
I The system admits a solution i� the vector
264 6�10
375can be seen as a linear combination of the vectors
264 123
375 ,
264 233
375 , and
264 322
375
with proper weights x�1, x�2, and x
�3.
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A FIRST EXAMPLE 4
I This can happen i� (Rouch�e-Capelli Theorem)
the number of independent columns of
264 1 2 32 3 23 3 2
375
equals
the number of independent columns of
264 1 2 3 62 3 2 �13 3 2 0
375| {z }complete matrix
.
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A FIRST EXAMPLE 5
I The rank of a matrix measures how many indipendent columns the matrix
has.
I The rank of a matrix is the highest order of non-singular square submatrices.
264 1 2 32 3 23 3 2
375 is non-singular so that
264 1 2 3 62 3 2 �13 3 2 0
375| {z }complete matrix
has maximum rank (3) .
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A FIRST EXAMPLE 6
I We can work out the solution x�1, x�2, and x
�3 by means of the Cramer's
Theorem:
x�1 = det
0B@264 6 2 3�1 3 20 3 2
3751CA = (�5) =
�5�5
= 1 ;
x�2 = det
0B@264 1 6 32 �1 23 0 2
3751CA = (�5) =
19
�5= �3:8 ;
x�3 = det
0B@264 1 2 62 3 �13 3 0
3751CA = (�5) =
�21�5
= 4:2 :
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A FIRST EXAMPLE 7
I We can work out x�1, x�2, and x
�3 also by means of inversion (given non-
singularity):
264 1 2 32 3 23 3 2
375�1
�
264 1 2 32 3 23 3 2
375 �264 x�1x�2x�3
375 =
264 1 2 32 3 23 3 2
375�1
�
264 6�10
375
m
264 1 0 00 1 00 0 1
375 �264 x�1x�2x�3
375 =
264 1 2 32 3 23 3 2
375�1
�
264 6�10
375 .
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A FIRST EXAMPLE 8
26666664x�1
x�2
x�3
37777775 =1
�5
266666640 5 �5
2 �7 4
�3 3 �1
37777775 �266666646
�1
0
37777775 =
266666641
�195215
37777775
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A SECOND EXAMPLE 1
I Consider the following linear system in the unknowns x1, x2, x3 and x4 :
8>>>>>><>>>>>>:
x1 � x2 + 2x3 � x4
2x1 � x2 � x3 + 2x4
�x1 + 2x2 + 2x3 + x4
=
=
=
1
0
1
I Let's express it as follows:
x1
264 12�1
375+ x2264 �1�12
375+ x3264 2�12
375+ x4264 �121
375 =
264 101
375 .
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A SECOND EXAMPLE 2
I The �rst three vectors are independent:
det
0B@264 1 �1 22 �1 �1�1 2 2
3751CA = 9
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A SECOND EXAMPLE 3
I If the fourth vector is brought on the right-hand side,
x1
264 12�1
375+ x2264 �1�12
375+ x3264 2�12
375 =
264 101
375� x4264 �121
375 ,
the solution (with one degree of freedom) can be written as
26666664x�1
x�2
x�3
37777775 =266666641 �1 2
2 �1 �1
�1 2 2
37777775
�10BBBBBB@
266666641
0
1
37777775� x426666664�1
2
1
37777775
1CCCCCCA =
26666664
13 �
53x4
29 �
169 x4
49x4 +
49
37777775 .
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A THIRD EXAMPLE 1
I Consider the following linear system in the unknowns x1, x2, and x3 :
8>>>>>>>>>><>>>>>>>>>>:
2x1 + 7x2 + x3
5x1 + 6x2 + 8x3
9x2
7x1 + 6x2 + 7x3
=
=
=
=
15
3
27
�3
I Let's state it as follows:
x1
266642507
37775+ x2266647696
37775+ x3266641807
37775 =
2666415327�3
37775 .
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A THIRD EXAMPLE 2
I The �rst three vectors are independent because
det
0B@264 2 7 15 6 80 9 0
3751CA = �99 .
I The last vector linearly depends on the �rst three vectors as
det
0BBB@266642 7 1 155 6 8 30 9 0 277 6 7 �3
377751CCCA = 0 .
I Hence, the system admits solution.
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A THIRD EXAMPLE 3
I We can work out x�1, x�2, and x
�3 by focusing on the �rst three equations
(given non-singularity):
x�1
264 250
375+ x�2264 769
375+ x�3264 180
375 =
264 15327
375m
264 x�1x�2x�3
375 =
264 2 7 15 6 80 9 0
375�1 264 153
27
375 =
264 �330
375 .
I The last equation is met by construction:
x�1 � 7 + x�2 � 6 + x�3 � 7 = �21 + 18 + 0 = � 3 .
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ONE-PERIOD FINANCIAL MARKETS
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TIMING
I Investors face two trading dates only, namely t = 0 and t = 1.
I At time t = 0, investors choose their investment strategy.
I At time t = 1 they receive the liquidation value of their strategy.
I At time 0, investors can choose among N + 1 securities, for which we use
the index j, with j = 0; : : : ; N .
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THE RISKLESS SECURITY
I The security with j = 0 represents a riskless security.
I B(0) � 1 denotes the time-0 price of the riskless security.
I B(1) � 1 + r denotes its time-1 price.
I The quantity r is the risk-free rate, with r > 0.
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THE RISKY SECURITIES
I For the N securities with j > 0, Sj(0) denotes their time-0 price.
I eSj(1) is a random variable that denotes their time-1 price, with j = 1; : : : ; N .
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UNCERTAINTY
I By time 1, the market uncertainty will resolve in one of K possible states
of the world.
I !k indicates the generic k-th state of the world at time 1.
I The !k's are relevant economic/�nancial scenarios.
I indicates the set of all states of the world, i.e. = f!1; : : : ; !Kg.
I Sj(1)(!k) indicates the time-1 price of the j-th security in scenario !k.
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THE PAYOFF MATRIX M
I The payo� matrix M has K + 1 rows and N + 1 columns.
I Each column j of M represents the out ows/in ows from buying 1 unit of
the j-th security (j = 0; 1; :::; N):
M �
26666666666666664
�1 �S1(0) �S2(0) � � � �SN(0)
1 + r S1(1)(!1) S2(1)(!1) � � � SN(1)(!1)
1 + r S1(1)(!2) S2(1)(!2) � � � SN(1)(!2)
... ... ... ...
1 + r S1(1)(!K) S2(1)(!K) � � � SN(1)(!K)
37777777777777775
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EXEMPLIFYING A MARKET WITH 3 SECURITIES AND 3 STATES 1
I Consider a one-period market with N = 2 and K = 3.
I Assume that B(1) = 1:05, so that r = 5%.
I The time-0 prices of the risky securities are S1(0) = 0:5 and S2(0) = 2:5.
I The time-1 prices of the risky securities are
S1(1)(!1) = 0
S1(1)(!2) = 2
S1(1)(!3) = 0
S2(1)(!1) = 0
S2(1)(!2) = 0
S2(1)(!3) = 5
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EXEMPLIFYING A MARKET WITH 3 SECURITIES AND 3 STATES 2
I The payo� matrix is then the following 4� 3 matrix:
M =
266666666664
�1 �0:5 �2:5
1:05 0 0
1:05 2 0
1:05 0 5
377777777775
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INVESTMENT STRATEGIES 1
I #0 is the position in the risk-free security.
I #1; #2 are the positions in the two risky securities.
I The positions represent the units of each security bought, or sold short, at
time 0.
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INVESTMENT STRATEGIES 2
#1 = 3 (buying 3 units of security 1)
initial out ow: 3 (�0:5) = �1:5
�nal in ows: 3
266666640
2
0
37777775 =
266666640
6
0
37777775
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INVESTMENT STRATEGIES 3 (LENDING)
#0 = 10 (buying 10 units of the riskless security)
initial out ow: 10 (�1) = �10
�nal in ows: 10
266666641:05
1:05
1:05
37777775 =
2666666410:5
10:5
10:5
37777775
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INVESTMENT STRATEGIES 4
#2 = �2 (selling short 2 units of security 2)
initial in ow: �2 (�2:5) = 5
�nal out ows: �2
266666640
0
5
37777775 =
266666640
0
�10
37777775
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INVESTMENT STRATEGIES 5 (BORROWING)
#0 = �7 (selling short 7 units of the riskless security)
initial in ow: �7 (�1) = 7
�nal out ows: �7
266666641:05
1:05
1:05
37777775 =
26666664�7: 35
�7: 35
�7: 35
37777775
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INVESTMENT STRATEGIES 6
I The generic investment strategy is a (column) vector # 2 R3:
# =
0B@ #0#1#2
1CA :
I Let's look at the following strategy #�:
#� =
0BBBBBB@4
5
2
1CCCCCCA =0BBBBBB@buying 4 units of the riskless security
buying 5 units of security 1
buying 2 units of security 2
1CCCCCCA :
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INVESTMENT STRATEGIES 7
I The market is competitive: no one has price impact.
I The market is frictionless:
B no indivisibilities (#is are not only integers);
B no short-selling constraint (#is can be negative);
B no trading limits (#is are unbounded);
B no margin requirements;
B no bid-ask spreads;
B no taxation on capital gains.
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THE INITIAL COST OF AN INVESTMENT STRATEGY
I At time 0, the quantity V# (0) is the cost an investor must face to set up
the investment strategy #.
I Let's assess the initial cost V#� (0) for the strategy #�:
#� =
0BBBBBB@4
5
2
1CCCCCCA :
I The corresponding initial out ow is �V#� (0) :
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THE FINAL PROCEEDS OF AN INVESTMENT STRATEGY
I The �nal proceed V# (1) (!k) is the time-1 cash ow in the state !k from
liquidating the investment strategy #.
I Let's assess the �nal proceeds of the strategy #�.
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THE STRATEGY FOR GIVEN FINAL PROCEEDS
I Let's �nd the strategy # whose �nal proceeds are:
26666664V# (1) (!1)
V# (1) (!2)
V# (1) (!3)
37777775 =
266666640:5
0:5
0:5
37777775 :
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BACK-ENGINEERING STRATEGIES
I Find the strategy # whose initial cost and �nal proceeds are:
V# (0) = 0;
V# (1) (!2) = 4;
V# (1) (!3) = 2.
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`FREE LUNCHES'?
I They are possible if there are:
I violations of the law of one price;
I arbitrage opportunities of the �rst type;
I arbitrage opportunities of the second type.
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VIOLATIONS OF THE LAW OF ONE PRICE
Two strategies have di�erent costs at time 0
despite
providing the same �nal proceeds in any state at time 1.
I In our market M there are no such violations.
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ARBITRAGE OPPORTUNITIES OF THE FIRST TYPE
There is a strategy with non-positive initial costthat
never yields negative �nal proceeds
while
granting strictly-positive �nal proceedsin
at least one of the states at time 1:
I In our market M there are no such opportunities.
I For example, let's work out the initial cost of a strategy that pays o� nothing
but 1 cent in the state !3:
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ARBITRAGE OPPORTUNITIES OF THE SECOND TYPE
There is a strategy with strictly-negative initial cost
that
never yields negative �nal proceeds.
I In our market M there are no such opportunities.
I For example, let's calculate the �nal proceed prevailing in the state !3 of a
strategy that costs �10 cents and pays o� nothing in the states !1 and !2.
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`FREE LUNCHES' ARE IMPLAUSIBLE
I Any non-satiated investor (in the sense that she always prefers more wealth
to less) would exploit these situations.
I There would be no equilibrium.
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THE NO-ARBITRAGE ASSUMPTION
I A market satis�es the no-arbitrage condition if it does not permit
I violations of the law of one price;
I arbitrage opportunities of the �rst type;
I arbitrage opportunities of the second type.
I Our market
M =
26664�1 �0:5 �2:51:05 0 01:05 2 01:05 0 5
37775is arbitrage-free.
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THE RISK-NEUTRAL PROBABILITY MEASURE Q 1
I A risk-neutral probability measure Q is made ofK probability masses Q (!k)
such that:
Q (!k) > 0 for any k = 1; ::::;K .
1 The masses are strictly-positive .
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THE RISK-NEUTRAL PROBABILITY MEASURE Q 2A
Q (!1) + :::+Q (!K) = 1 .
2 The masses sum to 1 .
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THE RISK-NEUTRAL PROBABILITY MEASURE Q 2B
B (0) =1
1 + rEQ [B (1)]
=1
1 + r[Q (!1)B (1) + :::+Q (!K)B (1)]
= 1 .
2 The masses sum to 1 .
I For the riskless security, the price is the discounted Q-expectation of the
payo�.
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THE RISK-NEUTRAL PROBABILITY MEASURE Q 3
Sj (0) =1
1 + rEQ
h eSj (1)i
=1
1 + r
hQ (!1)Sj (1) (!1) + :::+Q (!K)Sj (1) (!K)
i, j = 1; :::; N .
3 Prices are discounted Q-expectations of payo�s.
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WHY RISK-NEUTRAL? 1
I The statement
Sj (0) =1
1 + rEQ
h eSj (1)i , j = 1; :::; N ,
is equivalent to the statement
EQ" eSj (1)� Sj (0)
Sj (0)
#= r, j = 1; :::; N .
I Q-expected returns equal the risk-free rate.
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WHY RISK-NEUTRAL? 2
I UnderQ, the initial cost of a strategy # equals the discountedQ-expectationof its �nal proceed:
V# (0) =1
1 + rEQ
h eV# (1)i .
I The Q-expected return from # equals the risk-free rate.
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THE FIRST FUNDAMENTAL THEOREM OF ASSET PRICING
I The following statements are equivalent:
I the no-arbitrage condition holds;
I a risk-neutral probability measure Q exists.
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OUR ARBITRAGE-FREE MARKET ADMITS A Q
I The risk-neutral probability masses are:
26666664Q (!1)
Q (!2)
Q (!3)
37777775 =
266666640:212 5
0:262 5
0:525
37777775 .
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A LONG FORWARD CONTRACT ON SECURITY 2
I A long forward contract is an agreement to buy the risky security at date 1
at a pre-speci�ed price (the delivery price E = 1).
I The payo� of the contract is
26666664f (1) (!1)
f (1) (!2)
f (1) (!3)
37777775 =
26666664S2 (1) (!1)� E
S2 (1) (!2)� E
S2 (1) (!3)� E
37777775 =
26666664�1
�1
4
37777775 .
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NO-ARBITRAGE PRICING OF THE LONG FORWARD 1
I The unique strategy that replicates the long forward payo� ef (1) is
#f =
264 �0:952 380 9501
375 .
I The no-arbitrage price of the long forward equals the initial cost of the
replicating strategy #f :
V#f (0) = 1:547 619 1 .
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NO-ARBITRAGE PRICING OF THE LONG FORWARD 2
I The no-arbitrage price of the long forward is
f (0) =1
1 + rEQ
h ef (1)i
=1
1 + rEQ
h eS2 (1)� Ei
= 1:547 619 1 .
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NO-ARBITRAGE PRICING OF A EUROPEAN CALL
I A European call option on the risky security 2 with strike price E = 1 is the
right to buy the risky security at date 1 at the pre-speci�ed strike price.
I Find its no-arbitrage price c (0).
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NO-ARBITRAGE PRICING OF A EUROPEAN PUT
I A European put option on the risky security 2 with strike price E = 1 is the
right to sell the risky security at date 1 at the pre-speci�ed strike price.
I Find its no-arbitrage price p (0).
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THE PUT-CALL PARITY
c (0) � p (0) = f (0)
= S2 (0)�E
1 + r.
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MARKET COMPLETENESS
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COMPLETE MARKETS 1
I A payo� fX (1) is
replicable
if there exists a strategy # such that
V# (1) (!k) = X (1) (!k) for all k = 1; : : : ;K:
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COMPLETE MARKETS 2
I Our market M is
complete
if every possible payo� is replicable, that is, if there is a # such that
266666641:05 0 0
1:05 2 0
1:05 0 5
37777775
26666664#0
#1
#2
37777775 =
26666664X (1) (!1)
X (1) (!2)
X (1) (!3)
37777775
for any X (1) (!1), X (1) (!2), and X (1) (!3):
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COMPLETE MARKETS 3
I M is complete:
26666664#0
#1
#2
37777775 =
266666641:05 0 0
1:05 2 0
1:05 0 5
37777775
�1 26666664X (1) (!1)
X (1) (!2)
X (1) (!3)
37777775 .
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THE SECOND FUNDAMENTAL THEOREM OF ASSET PRICING
I The following statements are equivalent:
I Both no-arbitrage and market completeness hold;
I There exists one, and only one, risk-neutral probability Q.
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A UNIQUE Q FOR OUR COMPLETE M 1
I There is a unique triplet of strictly-positive masses Q (!1), Q (!2), andQ (!3) that meets the following constraints:
discounted Q�expected �nal pricesz }| {
11+0:05
266666641:05 0 0
1:05 2 0
1:05 0 5
37777775
T 26666664Q (!1)
Q (!2)
Q (!3)
37777775=
initial pricesz }| {266666641
0:5
2: 5
37777775.
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A UNIQUE Q FOR OUR COMPLETE M 2
I The unique triplet is
26666664Q (!1)
Q (!2)
Q (!3)
37777775 = 1:05
0BBBBBBB@
266666641:05 0 0
1:05 2 0
1:05 0 5
37777775
T1CCCCCCCA
�1 266666641
0:5
2: 5
37777775
=
266666640:212 5
0:262 5
0:525
37777775 .
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AN INCOMPLETE MARKET 1
I The market
M 0 =
266666666664
�1 �0:5
1:02 0
1:02 2
1:02 0:10
377777777775, N + 1 = 2, K = 3,
is incomplete.
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AN INCOMPLETE MARKET 2
I Indeed, the payo�
26666664X (1) (!1)
X (1) (!2)
X (1) (!3)
37777775 =
266666641
1:5
2
37777775
cannot be replicated:
det
0BBBBBB@
266666641:02 0 1
1:02 2 1:5
1:02 0:10 2
37777775
1CCCCCCA = 1: 989 .
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OUR INCOMPLETE M 0 IS ARBITRAGE-FREE 1
I M 0 supports many risk-neutral probability measures:
discounted Q-expected �nal pricesz }| {
11+0:02
264 1: 02 1: 02 1: 02
0 2 0:1
37526666664Q (!1)
Q (!2)
Q (!3)
37777775=
intial pricesz }| {264 1
0:5
375
m
264 1: 020
375Q (!1) +264 1: 02
2
375Q (!2) +264 1: 020:10
375Q (!3) = 1:02
264 1
0:5
375
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OUR INCOMPLETE M 0 IS ARBITRAGE-FREE 2
264 1: 020
375Q (!1) +264 1: 02
2
375Q (!2) = 1:02
264 1
0:5
375 �
264 1: 020:10
375 qz }| {Q (!3)
m
264 1: 02 1: 02
0 2
375264 Q (!1)Q (!2)
375 =
264 1: 02� 1:02q0:51� 0:10q
375m
264 Q (!1)Q (!2)
375 =
264 1: 02 1: 02
0 2
375�1 264 1: 02� 1:02q
0:51� 0:10q
375 .
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OUR INCOMPLETE M 0 IS ARBITRAGE-FREE 3
I We have
264 Q (!1)Q (!2)
375 =
266411:02 �12
0 12
3775264 1: 02� 1:02q0:51� 0:10q
375 =
264 0:745 � 0:95q0:255 � 0:05q
375
with the strict-positivity constraints
8>>>>>><>>>>>>:
Q (!1) = 0:745 � 0:95q > 0
Q (!2) = 0:255 � 0:05q > 0
Q (!3) = q > 0
() q 2 (0; 0:784 210 53) .
.
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NO-ARBITRAGE PRICING OF AN UNREPLICABLE PAYOFF 1
I The no-arbitrage prices of the unreplicable payo�
26666664X (1) (!1)
X (1) (!2)
X (1) (!3)
37777775 =
266666641
1:5
2
37777775are
1
1 + rEQ
hfX (1)i=
1
1:02
266666641 � (0:745 � 0:95q) +
1:5 � (0:255 � 0:05q) +
2 � q
37777775= 0:95588235q + 1:1053922 .
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NO-ARBITRAGE PRICING OF AN UNREPLICABLE PAYOFF 2
I The no-arbitrage prices range from
infQ
1
1 + rEQ
hfX (1)i= inf
q2(0;0:784 210 53)[0:955 882 35q + 1:1053922] = 1:1053922
to
supQ
1
1 + rEQ
hfX (1)i= sup
q2(0;0:784 210 53)[0:955 882 35q + 1:1053922] = 1:8550052 .
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WHERE DOES THE UPPER BOUND COME FROM?
1: 855 005 2 = supQ
1
1 + rEQ
hfX (1)i
= V#u (0)
where
#u = argmin#
V# (0) subject to V# (1) (!k) � X (1) (!k) in each state k:
I #u denotes a minimum-cost super-replicating strategy.
I V#u (0) is the minimum cost of super-replicating fX (1) .
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MINIMUM-COST SUPER-REPLICATION OF fX (1) 1
I The candidate super-replicating strategies # = [#0; #1]T solve the system
8>>>>>><>>>>>>:
1:02 � #0 + 0 � #1 � 1
1:02 � #0 + 2 � #1 � 1:5
1:02 � #0 + 0:10 � #1 � 2
;
the solution of which can be found graphically.
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MINIMUM-COST SUPER-REPLICATION OF fX (1) 2
8>>>>>><>>>>>>:
r1 : 1:02 � #0 + 0 � #1 = 1
r2 : 1:02 � #0 + 2 � #1 = 1:5
r3 : 1:02 � #0 + 0:10 � #1 = 2
;
5 4 3 2 1 1 2 3 4 5 6 7 8 9 10
4
2
2
4
6
8
10
12
14
theta_0
theta_1
r_1 r_3
r_2
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MINIMUM-COST SUPER-REPLICATION OF fX (1) 3
5 4 3 2 1 1 2 3 4 5 6 7 8 9 10
4
2
2
4
6
8
10
12
14
theta_0
theta_1
r_1 r_3
(1.5 , 8)
r_2sup
errep
licati
ng X
(1)
I The strategy [#0; #1]T = [1:5; 8]T super-replicates fX (1) .
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MINIMUM-COST SUPER-REPLICATION OF fX (1) 4
I The initial-cost function of any strategy # is
V#(0) = #0 + #1 � 0:5 .
I Its level curve c is the straight line #1 =c0:5 �
#00:5 :
5 4 3 2 1 1 2 3 4 5 6
5
4
3
2
1
1
2
3
4
5
6
7
8
theta_0
theta_1
r_1 r_3
r_2
super
replic
ating
X(1)
c = 1.855
c = 0
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MINIMUM-COST SUPER-REPLICATION OF fX (1) 5
I The minimum-cost super-replicating strategy [#u0 ; #u1 ]T is the point of in-
tersection between r2 and r3:
I Hence, we must solve the system
8><>:1:02 � #u0 + 2 � #u1 = 1:5
1:02 � #u0 + 0:10 � #u1 = 2
to obtain
264 #u0#u1
375 =
264 1: 986 58
�0:263 158
375 .
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WHERE DOES THE LOWER BOUND COME FROM?
1: 105 392 2 = infQ
1
1 + rEQ
hfX (1)i
= � V#l(0)
where
#l = argmax#
� V# (0) subject to V# (1) (!k) � �X (1) (!k) in each state k:
I #l denotes a maximum-in ow strategy that generates a liability lighter than�fX (1).
I �V#l(0) is the maximum in ow from super-replicating �fX (1) .
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MAXIMUM-INFLOW SUPER-REPLICATION OF �fX (1) 18>>>>>><>>>>>>:
h1 : 1:02 � #0 + 0 � #1 = �1
h2 : 1:02 � #0 + 2 � #1 = �1:5
h3 : 1:02 � #0 + 0:10 � #1 = �2
.
3.0 2.5 2.0 1.5 1.0 0.5 0.5 1.0 1.5 2.0 2.5 3.0
5
4
3
2
1
1
2
3
4
5
theta_0
theta_1
h_1
h_2
h_3
super
replic
ating
X(1)
I The strategy [#0; #1]T = [�0:25;�0:5]T super-replicates �fX (1) .
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MAXIMUM-INFLOW SUPER-REPLICATION OF �fX (1) 2
I The initial-in ow function of any strategy # is
f#(0) = �#0 � #1 � 0:5 ( = � V#(0) ) .
I Its level curve f is the straight line #1 = � f0:5 �
#00:5 :
3.0 2.5 2.0 1.5 1.0 0.5 0.5 1.0 1.5 2.0 2.5 3.0
5
4
3
2
1
1
2
3
4
5
theta_0
theta_1
h_1
h_2
h_3
super
replic
ating
X(1)
f = 1.10539
f = 0
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MAXIMUM-INFLOW SUPER-REPLICATION OF �fX (1) 3
I The liability-type strategy [#l0; #l1]T is the point of intersection between h1
and h2:
I Hence, we must solve the system
8><>:1:02 � #l0 + 0 � #l1 = �1
1:02 � #l0 + 2 � #l1 = �1:5
to obtain
264 #l0#l1
375 =
264 �0:980 392�0:25
375 .
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REPLICABLE PAYOFFS ADMIT A UNIQUE NO-ARBITRAGE PRICE
I Consider the following replicable payo�:26666664X (1) (!1)
X (1) (!2)
X (1) (!3)
37777775 =
266666642: 04
4: 04
2: 14
37777775 .
I Its no-arbitrage price does not depend on q :
1
1 + rEQ
hfX (1)i
=1
1:02
266666640:745 � 0:95q
0:255 � 0:05q
q
37777775
T 266666642: 04
4: 04
2: 14
37777775 = 2: 5 .
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OPTIMIZATION: AN INTRODUCTION
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WEALTH ALLOCATION AND FINAL WEALTH
I An investor allocates her initial wealth of 100 between a riskless security
( its net return is r ) and a risky security ( its net return is er ).
I The proportion of initial wealth allocated to the risky security is w .
I The �nal wealth achieved with the allocation w is
fW = 100 ( 1� w ) (1 + r) + 100 w (1 + er)
= 100 ( (1 + r) + w (er � r) ) .
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PORTFOLIO RETURNS AND BORROWING
I The portfolio based on the allocation w yields a net return of
fW � 100100
= r + w (er � r) .
I If the investor invests more than her initial wealth in the risky security
(w > 1), she needs to borrow:
1� w| {z }proportion allocated to the riskless security
< 0 .
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THE RETURN ON THE RISKLESS SECURITY
I Assume for now that
r = 0% .
I Hence, a purely riskless allocation (w = 0) implies
fW = 100 with probability 1 .
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THE RETURN ON THE RISKY SECURITY
er =
8><>:g = 30% with probability mass p
b = �10% with probability mass 1� p
I If p = 0:25 , the expectation is
E [ er ] = 0% .
I Hence, a purely risky allocation (w = 1) implies
Eh fW i
= 100 .
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THE INVESTOR IS RISK AVERSE 1
I If the investor were to pursue a purely riskless allocation (w = 0), she would
get the following utility out of the �nal wealth fW = 100 (with probability 1)
U ( 100 ) = log ( 100 ) .
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THE INVESTOR IS RISK AVERSE 2
I If the investor were to pursue a purely risky allocation (w = 1), she would
get the following expected utility out of the �nal wealth fW = 100 ( 1 + er ):
E [ log ( 100 (1 + er) ) ] = 0:25 � log ( 130 ) + 0:75 � log ( 90 )
< log ( 0:25 � 130 + 0:75 � 90 )
= log ( 100 ) .
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THE INVESTOR IS RISK AVERSE 3
70 75 80 85 90 95 100 105 110 115 120 125 130 135 140 145 1504.50
4.52
4.54
4.56
4.58
4.60
4.62
4.64
4.66
4.68
4.70
4.72
4.74
4.76
4.78
4.80
4.82
4.84
4.86
wealth W
(expected) utility
log ( W
)E [ l
og ( W
) ]
log ( 100 )
p = 0
p = 1
p = 0.25
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THE RETURN ON THE RISKLESS SECURITY
I Assume
r = 2% .
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A POSITIVE RISK PREMIUM ON THE RISKY SECURITY
I Consider
p = 0:5
so that the risk premium is
E [ er ] � r = 10% � 2% = 8% .
I The standard deviation of the risky return is
std [ er ] = 20% .
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OPTIMAL ALLOCATION WITHOUT PORTFOLIO CONSTRAINTS 1
I Recall that the �nal wealth is
fW = 100 ( ( 1 + r ) + w ( er � r ) ) .
I The optimal allocation is w� = argmaxw
Ehlog
� fW � i.
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OPTIMAL ALLOCATION WITHOUT PORTFOLIO CONSTRAINTS 2
w� = 242:86% ()
8>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>:
ddw E
hlog
� fW � i> 0 for w < w�
ddw E
hlog
� fW � i���w = w�
= 0 (stationarity)
ddw E
hlog
� fW � i< 0 for w > w�
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OPTIMAL ALLOCATION WITHOUT PORTFOLIO CONSTRAINTS 3
I Recall that log x is a real number only if x > 0 .
I w must be such that the �nal wealth fW is always strictly positive:
8>>><>>>:100 ( (1 + r) + w (g � r) ) = 102 + w28 > 0
100 ( (1 + r) + w (b� r) ) = 102� w12 > 0
() w 2�� 5114;17
2
�.
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OPTIMAL ALLOCATION WITHOUT PORTFOLIO CONSTRAINTS 4
d
dwEhlog
� fW � i� 0
m
1
2� 1
102 + w28� 28 +
1
2� 1
102� w12� ( � 12 ) � 0
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OPTIMAL ALLOCATION WITHOUT PORTFOLIO CONSTRAINTS 5
1
2� 1
102 + w28� 28 +
1
2� 1
102� w12� ( � 12 ) � 0
m
14 ( 102 � w 12 ) � 6 ( 102 + w 28)
( 102 + w 28) ( 102 � w 12 )� 0
m
816 � 336 w
( 102 + w 28) ( 102 � w 12 )� 0
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OPTIMAL ALLOCATION WITHOUT PORTFOLIO CONSTRAINTS 6
816 � 336w
( 102 + w28 ) ( 102� w12 )� 0
I The denominator is striclty positive for w 2�� 5114;
172
�.
I The numerator is non-negative for w 2 ( �1; 177 ] with
17
7= 242:86% :
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OPTIMAL ALLOCATION WITHOUT PORTFOLIO CONSTRAINTS 7
5 4 3 2 1 1 2 3 4 5 6 7 8 9 10
1.0
0.5
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
allocation w
expected utility
w *
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w� WITH A PORTFOLIO CONSTRAINT ( w � 1 ) 1
I The investor cannot borrow to allocate more than 100 into the risky security:
w� = 1
5 4 3 2 1 1 2 3 4 5 6 7 8 9 10
1
1
2
3
4
5
allocation w
expected utility
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w� WITH A PORTFOLIO CONSTRAINT ( w � 1 ) 2
I The portfolio constraint is binding at the optimum:
8>>>>>>><>>>>>>>:
ddw E
hlog
� fW � i���w = w�
> 0
w� � 1 = 0
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w� WITH A PORTFOLIO CONSTRAINT ( w � 1 ) 3
I Let's introduce the Lagrangian function
L (w; l) = Ehlog
� fW � i� l ( w � 1 ) .
I l is the Lagrange multiplier with
l � 0 .
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w� WITH A PORTFOLIO CONSTRAINT ( w � 1 ) 4
I Let's use the Lagrangian function to re-express the optimality conditions:
8>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>:
@@w L
���w = w�; l = l�
= 0
l� � 0
@@l L
���w = w�
� 0
l� � @@l L
���w = w�
= 0
=)
8>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>:
ddw E
hlog
�fW�i���w = w�
� l� = 0
l� = 0:041026 > 0
� ( w� � 1 ) = 0
l� � ( � ( w� � 1 ) ) = 0
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w� WITH A PORTFOLIO CONSTRAINT ( w � 1 ) 5
I If the constraint is relaxed ( say to w � 1:01 ), the optimized expected
utility goes up by
d
dwEhlog
� fW � i����w = w�
� 0:01 = l� � 0:01 .
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w� WITH A PORTFOLIO CONSTRAINT ( w � 1 ) 6
0.960 0.965 0.970 0.975 0.980 0.985 0.990 0.995 1.000 1.005 1.010 1.015 1.020 1.025 1.030 1.035 1.0404.6820
4.6825
4.6830
4.6835
4.6840
4.6845
4.6850
allocation w
expected utility
about 0.00041
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w� WITH A PORTFOLIO CONSTRAINT ( w � 1 ) 7
I The Lagrange multiplier l� is a shadow price :
l� = 0:041026 .
( optimized-expected-utility gain per unit of constraint relaxation )
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UNCONSTRAINED OPTIMIZATION
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A PROFIT FUNCTION
I A �rm produces two outputs x and y (they can be sold at the �xed prices
20 and 30, respectively).
I The production costs are quadratic:
C (x; y) = x2 + 2y2 � 2xy + 20 .
I The pro�t function remains quadratic:
P (x; y) = 20x+ 30y ��x2 + 2y2 � 2xy + 20
�.
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P 'S LEVEL CURVES 1
I The level curve p = 300 is the set of output pairs (x; y) such that the
pro�t P (x; y) equals 300.
0 10 20 30 40 50 60 700
10
20
30
40
50
60
70
x
y
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P 'S LEVEL CURVES 2
I The level curve p = 600 is in black
0 10 20 30 40 50 60 700
10
20
30
40
50
60
70
x
y
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P 'S LEVEL CURVES 3
I The level curve p = 705 is the singleton (x� = 35; y� = 25), which isthe maximum-pro�t production in the absence of constrains.
0 10 20 30 40 50 60 700
10
20
30
40
50
60
70
x
y
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FIRST PARTIAL DERIVATIVE WITH RESPECT TO x 1
Px (35; 25) = limh!0
P (35 + h; 25)� P (35; 25)h
= 0
I It is the slope of P 's graph in (x = 35; y = 25) along the direction (x; y = 25)
423040
3836
output x26
3424
output y
32
28
302228
20
550
600
700
650
profit level z
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FIRST PARTIAL DERIVATIVE WITH RESPECT TO x 2
Px (35; 25) =@
@x
�20x+ 30y � x2 � 2y2 + 2xy � 20
�����( x=35; y=25 )
= (20� 2x+ 2y)j ( x=35; y=25 )
= 0
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FIRST PARTIAL DERIVATIVE WITH RESPECT TO y 1
Py (35; 25) = limk!0
P (35; 25 + k)� P (35; 25)k
= 0
I It is the slope of P 's graph in (x = 35; y = 25) along the direction (x = 35; y)
423040
26 363828
output x
3424
output y
323022
2820
550
600
profit level z
650
700
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FIRST PARTIAL DERIVATIVE WITH RESPECT TO y 2
Py (35; 25) =@
@y
�20x+ 30y � x2 � 2y2 + 2xy � 20
������( x=35; y=25 )
= (30� 4y + 2x)j ( x=35; y=25 )
= 0
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THE (NECESSARY) FIRST-ORDER CONDITIONS FOR OPTIMALITY
I P : X � R2 �! R admits �rst partial derivatives at the interior point
(x�; y�) 2 X.
I If P has a local/global maximum at the interior point (x�; y�), then (x�; y�)is a stationary point for P :
Px ( x�; y� ) = Py ( x
�; y� ) = 0 .
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THE HESSIAN MATRIX
I If P : X � R2 �! R admits second partial derivatives at the interior
point (x0; y0) 2 X, then P 's Hessian matrix at (x0; y0) is
264 Pxx (x0; y0) Pxy (x0; y0)
Pyx (x0; y0) Pyy (x0; y0)
375 .
I Schwartz's Theorem: Pxy (x0; y0) = Pyx (x0; y0) .
I In our case, the Hessian is264 �2 2
2 �4
375 .
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TAYLOR'S FORMULA STOPPED AT THE SECOND ORDER
I Given any point (x0; y0), our quadratic pro�t function can be re-expressed
as follows:
P (x; y) = P (x0; y0) +
Px (x0; y0) (x� x0) + Py (x0; y0) (y � y0) +
1
2
264 x� x0y � y0
375T 264 Pxx (x0; y0) Pxy (x0; y0)
Pyx (x0; y0) Pyy (x0; y0)
375264 x� x0y � y0
375 .
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P 'S HESSIAN AT THE MAX. POINT (x� = 35; y� = 25) 1
P (x; y) = 705 +
0 (x� 35) + 0 (y � 25) + ( stationarity )
1
2
264 x� 35y � 25
375T 264 �2 2
2 �4
375264 x� 35y � 25
375| {z }
< 0 for any x 6= 35 or y 6= 25 (negative de�nite)
.
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P 'S HESSIAN AT THE MAX. POINT (x� = 35; y� = 25) 2
Pxx (x�; y�) = � 2 < 0
det
0B@264 Pxx (x�; y�) Pxy (x�; y�)
Pyx (x�; y�) Pyy (x�; y�)
3751CA = det
0B@264 �2 2
2 �4
3751CA = 4 > 0
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THE (SUFFICIENT) SECOND-ORDER CONDITIONS FOR A LOCAL
MAXIMUM
I P : X � R2 �! R admits continuous �rst and second partial derivatives
in an open ball of the interior point (x�; y�) 2 X.
I If (x�; y�) is a stationary point for P and if
I P 's Hessian matrix at (x�; y�) is negative de�nite,
I equivalently, Pxx (x�; y�) < 0 and
det
0B@264 Pxx (x�; y�) Pxy (x�; y�)
Pyx (x�; y�) Pyy (x�; y�)
3751CA > 0
then (x�; y�) is a local maximum point for P .
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CONVEX SETS
I X � R2 is convex when the segment that connects two arbitrary points ofthe set does not lie outside the set.
I R2 is strictly convex (the connecting segment lies strictly inside R2).
In( x; y ) 2 R 2 : x � 0 ^ y � 0
ois convex:
2 1 1 2 3 4
2
1
1
2
3
4
x
y
here
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P 'S STRICT CONCAVITY 1
I P : X � R2 �! R is strictly concave in its strictly convex domain X when
the segment that connects two arbitrary points of P 's graph lies strictly below the
graph.
profit level z
600
400
200
00
10
output y
20
0
20
output x40
60
806050
4030
I 20x+ 30y ��x2 + 2y2 � 2xy + 20
�is strictly concave.
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P 'S STRICT CONCAVITY 2
I P : X � R2 �! R is strictly concave in its strictly convex domain X
if and only if the non-empty sets f (x; y) 2 X : P (x; y) � a g are strictly
convex .
0 10 20 30 40 50 60 700
10
20
30
40
50
x
ylevel 300
level 600
level 700
I 20x+ 30y ��x2 + 2y2 � 2xy + 20
�is strictly concave.
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P 'S STRICT CONCAVITY 3
I Assume that P : X � R2 �! R is twice derivable in all the points of its
open and strictly convex domain X.
I P is strictly concave in X if and only if the Hessian matrix is negative
de�nite in any point of X.
I 20x+ 30y ��x2 + 2y2 � 2xy + 20
�is strictly concave:
its Hessian
264 �2 2
2 �4
375 is always negative de�nite.
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STRICT CONCAVITY AND THE UNIQUE GLOBAL MAXIMUM
I P : X � R2 �! R is strictly concave in its strictly convex domain X.
I If a local maximum point exists for P , it is also of global maximum andunique.
profit level z
600
400
200
00
10
output y
20
0
20
output x40
60
806050
4030
I (35; 25) is the unique global maximum point.
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C'S STRICT CONVEXITY 1
I C : X � R2 �! R is strictly convex in its strictly convex domain X whenthe segment that connects two arbitrary points of C's graph lies strictly above thegraph.
0
2
0
54
output x
6
output y
10
158
20
010
40cost level z60
80
100
I The cost function C (x; y) = x2+2y2�2xy+20 is strictly convex.
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C'S STRICT CONVEXITY 1
I C : X � R2 �! R is strictly convex in its strictly convex domain X if
and only if �C is strictly concave there.
106
25y x
4
0 0
810
100
80
60z40
20
15
0
I ��x2 + 2y2 � 2xy + 20
�is strictly concave.
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C'S STRICT CONVEXITY 3
I C : X � R2 �! R is strictly convex in its strictly convex domain X if and
only if the non-empty sets f (x; y) 2 X : C (x; y) � a g are strictly convex .
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150
1
2
3
4
5
6
7
8
9
10
x
y
level 40
level 80
level 120
I x2 + 2y2 � 2xy + 20 is strictly convex.
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MAXIMIZATION WITH EQUALITY CONSTRAINTS
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RECALLING THE COST FUNCTION
I A �rm produces two outputs x and y (they can be sold at the �xed prices
20 and 30, respectively) and its cost function is:
C (x; y) = x2 + 2y2 � 2xy + 20 .
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A COST CONSTRAINT THAT DEFINES AN IMPLICIT FUNCTION 1
I The output pair (x = 10; y = 10) commands a cost of 120. In a neighbor-hood of x = 10, the cost constraint
C (x; y) = 120
implies the existence of a unique function
y = � (x)
such that
10 = � (10) and ddx� (x) = �
C x ( x; �(x) )
C y ( x; �(x) ).
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A COST CONSTRAINT THAT DEFINES AN IMPLICIT FUNCTION 2
5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
4
2
2
4
6
8
10
12
14
output x
output y
( level curve ) cost = 120
( 10 , 10 )
10 = � (10) and ddx� (x)
���x = 10
= �C x ( 10; 10 )
C y ( 10; 10 )= 0
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PROFIT MAXIMIZATION WITH A COST CONSTRAINT 1
I The �rm's pro�t is
P (x; y) = 20x + 30y � C (x; y)
I The optimization problem is
maxx; y
P (x; y) subject to C (x; y) = a
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PROFIT MAXIMIZATION WITH A COST CONSTRAINT 2
I From the constraint equation
C (x; y) = a
we can work out y from x via the implicit function
y = � (x) :
I By plugging y = � (x) into P , the problem becomes:
maxx
P ( x; � (x) ) .
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PROFIT MAXIMIZATION WITH A COST CONSTRAINT 3
I If P ( x; � (x) ) exhibits an interior extremum point, then such a point
must be stationary:
dP
dx= Px + Py �
d
dx� = 0
m
Px
Py= � d
dx�
m
Px
Py= �
�CxCy
!.
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PROFIT MAXIMIZATION WITH A COST CONSTRAINT 4
I The necessary condition for optimality
Px = Py = Cx = Cy
holds if and only if there exists a real number l (the Lagrange multiplier )
such that
Px = l Cx ;
Py = l Cy :
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PROFIT MAXIMIZATION WITH A COST CONSTRAINT 5
I By introducing the Lagrangian function
L (x; y; l; a) = P (x; y) � l (C (x; y)� a) ;
the necessary condition for optimality and the constraint can be expressed as
follows (stationarity for L):
8>>>>>><>>>>>>:
Lx = Px � lCx = 0
Ly = Py � lCy = 0
Ll = � (C � a) = 0 (constraint)
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PROFIT MAXIMIZATION WITH A COST CONSTRAINT 6
8>>><>>>:Lx = 0 , 20� 2x+ 2y � l (2x� 2y) = 0
Ly = 0 , 30� 4y + 2x� l (4y � 2x) = 0
,
8>>>><>>>>:x� = 35
l+1
y� = 25l+1
��x�2 + 2y�2 � 2x�y� + 20 � a
�= 0| {z }
Ll = 0
,
8>>>>>><>>>>>>:
l = +�725a�20
�12 � 1 (select)
l = ��725a�20
�12 � 1 (discard)
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PROFIT MAXIMIZATION WITH A COST CONSTRAINT 7
l� =�725
a� 20
�12 � 1
x� =35
l� + 1= 35
�a� 20725
�12
y� =25
l� + 1= 25
�a� 20725
�12
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PROFIT MAXIMIZATION WITH A COST CONSTRAINT 8
L (x�; y�; l�; a) = P (x�; y�) � l� (C (x�; y�)� a)| {z }= 0
= P (x�; y�)
= 2 (725)12 (a� 20)
12 � a .
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THE LAGRANGE MULTIPLIER l� IS A SHADOW PRICE 1
dda P ( x
�; y� ) =d
da
�2 (725)
12 (a� 20)
12 � a
�
=�725
a� 20
�12 � 1 = l�
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THE LAGRANGE MULTIPLIER l� IS A SHADOW PRICE 2
I If we �x the cost constraint at a = 120 , then the shadow price is
l� = 1:6926 .
I The �rm is willing to surrender up to 1:6926 cents of pro�ts to obtain a
1-cent relaxation of the cost constraint ( da = 0:01 ):
d
daP ( x�; y� ) � 0:01 = l� � 0:01 ' 1:6926 cents .
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GRAPHICAL ANALYSIS WITH a = 120 1
I The optimal production is x� = 12:9987 and y� = 9:2848. The constrained-maximum pro�t is P (x�; y�) = 418: 5165:
5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
4
2
2
4
6
8
10
12
14
output x
output y
( constraint ) cost = 120
( level curve ) profit = 418.52
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GRAPHICAL ANALYSIS WITH a = 120 2
I At the point (x�; y�) there is tangency between the constrained-cost
level curve and the maximum-pro�t level curve:
� P x (x�; y�) = P y (x�; y�) = � C x (x�; y�) = C y (x�; y�)
5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
4
2
2
4
6
8
10
12
14
output x
output y
( constraint ) cost = 120
( level curve ) profit = 418.52
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OPTIMIZATION WITH INEQUALITY CONSTRAINTS
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THE PORTFOLIO RETURN
I The portfolio return is
erp = w1 er1 + w2 er2 + (1� w1 � w2) r .
wj = fraction of initial wealth in the risky asset j ( j = 1; 2 ) ,
erj = net return on the the risky asset j ( j = 1; 2 ) ,
r = net return on the riskless asset .
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THE EXPECTED RETURN ON A PORTFOLIO
I The expectation E [ erp ] is
T (w1; w2) = r + w1 ( r1 � r ) + w2 ( r2 � r ) ,
Eh erj i = rj ( j = 1; 2 ) .
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T 'S LEVEL CURVES 1
I The allocation pairs (w1; w2) such that T (w1; w2) = 8% are in black.
1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 1.2 1.4
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
1.2
1.4
w_1
w_2
r1 = 12%; r2 = 8%; r = 2%:
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T 'S LEVEL CURVES 2
I The allocation pairs (w1; w2) such that T (w1; w2) = 10% are in black.
1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 1.2 1.4
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
1.2
1.4
w_1
w_2
r1 = 12%; r2 = 8%; r = 2%:
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VARIANCES AND CORRELATION
V arh erj i = �2j ( j = 1; 2 )
Cov [ er1 , er2 ] = ( �1 �2 ) = �
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THE VARIANCE OF A PORTFOLIO RETURN
I The variance V ar (erp) is
V (w1; w2) =
264 w1w2
375T 264 �21 �1�2�
�1�2� �22
375264 w1w2
375
= �21w21 + 2��1�2w1w2 + �22w
22
� 0 for any portfolio choice (w1; w2) .
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V 'S LEVEL CURVES 1
I The allocation pairs (w1; w2) such that V (w1; w2) = (20%)2 are inblack.
1 1
1
1
w_1
w_2
�1 = 30%, �2 = 20%,
� = �0:5
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V 'S LEVEL CURVES 2
I The allocation pairs (w1; w2) such that V (w1; w2) = (10%)2 are inblack.
1 1
1
1
w_1
w_2
�1 = 30%, �2 = 20%,
� = �0:5
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V 'S LEVEL CURVES 3
I The allocation pair (w1; w2) such that V (w1; w2) = 0 is the point�w�1 = 0; w
�2 = 0
�.
1 1
1
1
w_1
w_2
�1 = 30%, �2 = 20%,
� = �0:5
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MAXIMIZING THE RISK-ADJUSTED EXPECTED RETURN 1
I The investor's risk aversion is measured by the parameter A > 0 .
I The risk-adjusted expected return on the portfolio is
J (w1; w2) = T (w1; w2) � A
2V (w1; w2) .
I The problem is
maxw1; w2
J (w1; w2) sub
( total risk exposure must not exceed q )z }| {w1 + w2 � q � 0 .
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MAXIMIZING THE RISK-ADJUSTED EXPECTED RETURN 2
I If q = 3 , the constraint is painless and not binding with w�1 = 107:4%and w�2 = 155:6% (the shadow price of the constraint is l� = 0).
1.0 0.5 0.5 1.0 1.5 2.0 2.5 3.0
1
1
2
3
w_1
w_2
h e r ew_1 + w_2 = 3
J's level curves
A = 2 ; �1 = 30%; �2 = 20%; � = �0:5; r1 = 12%; r2 = 8%; r = 2%:
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MAXIMIZING THE RISK-ADJUSTED EXPECTED RETURN 3
I Given L (w1; w2; l) = J (w1; w2) � l (w1 + w2 � q) , the �rst-order
conditions for constrained optimality can be written �a la Kuhn-Tucker :8>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>:
L w1
�w�1; w
�2; l
��= 0
L w2
�w�1; w
�2; l
��= 0
l� � 0
L l
�w�1; w
�2; l
��� 0
l� � L l
�w�1; w
�2; l
��= 0
q = 3
=)
8>>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>>:
J w1
�w�1; w
�2;�= l�
J w2
�w�1; w
�2;�= l�
l� = 0w�1 + w
�2 � 3 < 0
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MAXIMIZING THE RISK-ADJUSTED EXPECTED RETURN 4
I The feasibility conditions are:
8>><>>:l� � 0
L l
�w�1; w
�2; l
��� 0 ( w�1 + w
�2 � q � 0 )
I The complementary slackness condition is:
l� � L l
�w�1; w
�2; l
��
= 0
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MAXIMIZING THE RISK-ADJUSTED EXPECTED RETURN 5
I If q = 1 , the constraint is painful and binding with w�1 = 47:4% and
w�2 = 52:6% (the shadow price of the constraint is l� = 4:63%).
1.0 0.5 0.5 1.0 1.5 2.0 2.5 3.0
1
1
2
3
w_1
w_2
[ 1.074 , 1.556 ]
h e r e
w_1 + w_2 = 1
J's level curves
A = 2 ; �1 = 30%; �2 = 20%; � = �0:5; r1 = 12%; r2 = 8%; r = 2%:
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MAXIMIZING THE RISK-ADJUSTED EXPECTED RETURN 6
I Given L (w1; w2; l) = J (w1; w2) � l (w1 + w2 � q) , the �rst-order
conditions for constrained optimality can be written �a la Kuhn-Tucker :8>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>:
L w1
�w�1; w
�2; l
��= 0
L w2
�w�1; w
�2; l
��= 0
l� � 0
L l
�w�1; w
�2; l
��� 0
l� � L l
�w�1; w
�2; l
��= 0
q = 1
=)
8>>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>>:
J w1
�w�1; w
�2
�= l�
J w2
�w�1; w
�2
�= l�
l� > 0
w�1 + w�2 � 1 = 0
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STOCK PRICING
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THE DIVIDEND PROCESS
I The current date is t
I The stock pays out the dividend Xdt every `second' (a period of length dt)
I The dynamics of the dividend process is
dX = a (X; t) � dt| {z }expected change Et [dX]
+ b (X; t) � dz| {z }unexpected change
,
where dz is Gaussian with Et [dz] = 0 and Ethdz2
i= dt
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A FUNCTION OF THE DIVIDEND X AND ITS DYNAMICS
I S (X) is a function of X
I Given the notation
SX =d
dXS and SXX =
d2
dX2S ,
we have
dS = SXdX +1
2SXXb
2dt
=�SXa+
1
2SXXb
2�� dt| {z }
expected change Et [dS]
+ SXb � dz| {z }unexpected change
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EQUILIBRIUM STOCK PRICING 1
I The stock price is S (X)
I The per-annum expected total gain on the stock is
1
dtEt [dS] + X = expected capital gain + income gain
I In equilibrium, the following restriction holds true:
1dtEt [dS] + X = Sr + SX b � � � �
I Sr is the cost of borrowing expressed in Euro (r is the riskfree rate)
I SX b � � � � is the risk premium expressed in Euro
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EQUILIBRIUM STOCK PRICING 2
I SX b is the equity risk expressed in Euro
I � is the premium per unit of systematic risk
I � is the fraction of equity risk that is systematic.
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TRENDING DIVIDENDS
I X's dynamics is
dX = X� � dt+X� � dz
I X is expected to grow at the rate �
I X = 0 is an absorbing boundary
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STOCK PRICING WITH TRENDING DIVIDENDS 1
I The equilibrium restriction is
SXX�+1
2SXXX
2�2| {z }= 1
dtEt [dS]
+X = Sr + SXX���
I If X = 0, the stock is worthless (it is unable to generate dividends):
S (0) = 0 (boundary condition)
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STOCK PRICING WITH TRENDING DIVIDENDS 2
I The solution is
S (X) =X
r + ���� �(non-negative and meaningful i� r + ���� � > 0)
I Given the parabola
y ( ) =1
2�2 2 +
��� ���� 1
2�2� � r
we impose
y (1) = � (r + ���� �) < 0
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STOCK PRICING WITH TRENDING DIVIDENDS 3
I The elasticity of S (X) with respect to X is
SXSX = 1
0@ =dSSdXX
1A
I A 1% change in the dividend implies a 1% change in the stock price
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STOCK PRICING WITH TRENDING DIVIDENDS 4
I The total return on the stock is
dS + Xdt
S=
�r +
SXX
S���
�dt
| {z }expected total return
+SXX
S� dz| {z }
unexpected total return
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STOCK PRICING WITH TRENDING DIVIDENDS 5
I The per-annum expected total return on the stock is
1
dt
1
SEt [dS] +
X
S|{z}per-annum dividend yield
= r +SXSX���
= r + ���
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THE EXPECTED TOTAL RETURN ON A PORTFOLIO
I Our initial capital is H = 100 Euro. We invest 60 Euro in the stock and
40 Euro in the riskfree asset
I The per-annum expected total gain is
1
dtEt [dH] =
60
S
�1
dtEt [dS] +X
�+ 40r
= 60 (r + ���) + 40r
I The per-annum expected total return is
1
dt
1
HEt [dH] = r +
60
100���
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RATIONAL BUBBLES 1
I Another solution to
SXX�+1
2SXXX
2�2 +X = Sr + SXX��� with S (0) = 0
is given by
S (X) =X
r + ���� �| {z }fundamental value
+ c �X 1| {z }bubble
c = positive constant
1 = ���� ����2
� 12
�+
"��� ����2
� 12
�2+ 2
r
�2
#12
> 1
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RATIONAL BUBBLES 2
I The elasticity of S (X) with respect to X is `excessive':
SXSX =
S + ( 1 � 1) cX 1S
> 1
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THE COEFFICIENT 1
I Given the parabola y ( ) = 12�2 2 +
��� ���� 1
2�2� � r we have
y (1) = � (r + ���� �) < 0 and y ( 1) = 0
gamma
y
gamm
a_1
( r + sigma*lambda*rho mu ) < 0
1
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`POWER' STOCK PRICING 1
I The stock that pays out X2dt every `second' has the following equilibrium price:
P (X) =X2
r + 2�����2�+ �2
�
I Given the parabola
y ( ) =1
2�2 2 +
��� ���� 1
2�2� � r
we impose
y (2) = ��r + 2����
�2�+ �2
��< 0
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`POWER' STOCK PRICING 2
I The total return on the stock is
dP + X2dt
P=
�r +
PXX
P���
�dt
| {z }expected total return
+PXX
P� dz| {z }
unexpected total return
= ( r + 2��� ) dt + 2� dz
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MEAN-REVERTING DIVIDENDS
I X's dynamics is
dX = �k (X �m) � dt+X� � dz
I The long-run mean is E [X] = m > 0
I k > 0 is the speed of mean reversion
I If k �! +1, X remains �xed at the level m
I X = 0 is a re ecting boundary
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STOCK PRICING WITH MEAN-REVERTING DIVIDENDS 1
I The per-annum expected capital gain is
1
dtEt [dS] = SX (�k (X �m)) + 1
2SXXX
2�2
I The equilibrium stock price solves
1
dtEt [dS] + X = Sr + SXX���
and is
S (X) =m
r� k
r + ���+ k+
X
r + ���+ k
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STOCK PRICING WITH MEAN-REVERTING DIVIDENDS 2
I The absence of mean reversion brings us back to the trending-dividends case
(without growth):
limk!0
S (X) =X
r + ���
I A massive mean reversion brings us back to the riskless case (without growth):
limk!+1
S (X) =m
r
I Systematic risk and risk aversion (��� > 0) do have a long-run impact:
E [ S (X) ] =m
r� k + r
r + ���+ k<
m
r
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APPENDIX
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THE CAPITAL ASSET PRICING MODEL 1
I The stock market portfolio is worth M with dynamics
dM +XMdt = M ( r + �M � ) dt + M �M dzM
I The premium for systematic risk (expressed in Euro) is M�M � �
I The CAPM states
1
dtEt [dM ] +XM = Mr +
covt [dM; dM ]
vart [dM ]| {z }`beta' equal to 1
� M �M � �
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THE CAPITAL ASSET PRICING MODEL 2
I For the single stock, the CAPM states
1
dtEt [dS] +X = Sr +
covt [dS; dM ]
vart [dM ]| {z }`beta' equal to SX b � �
M �M
� M �M � �
= Sr + SX b � � � �
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