a stochastic process is a variable that evolves over …minsley/ec659f13/ec659_secii.pdfexample of a...
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Section II: Stochastic processes and Ito’s lemma
1 Introduction to stochastic processes
• A stochastic process is a variable that evolvesover time in a way that is at least in partrandom.
• eg. Temperature in Waterloo: a stochasticprocess with a deterministic component.
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• continuous time process versus a discrete time process
• continuous variable versus a discrete variable
A stochastic process is defined by a probability law for theevolution of a variable, say x, over time. We can calculate theprobability that x1, x2, x3, ... lie in some specified range attimes t1, t2, t3, ...:
prob(a1 < x1 ≤ b1, a2 < x2 ≤ b2, ...)
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• Example of a discrete-time, discrete state randomvariable: a random walk.
• Let xt denote a random variable that begins at x0 and attimes t = 1, 2, 3 takes a jump up of size ∆h withprobability p and a jump down of size ∆h withprobability q(= 1− p).
• x0 → x0 + ∆h with prob p
• x0 → x0 −∆h with prob q
• x follows a Markov process; Jumps are independent ofeach other.
• Sketch a lattice to show paths for x over time for threetime periods.
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• Let ∆x be the change in x over the interval t→ t+ ∆t.x can take only one jump over this interval.
• Determine E(∆x), E((∆x)2), and Var(∆x)
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• Consider the distribution of x after n moves. Note thatt = n∆t
• Consider the probability of j upward jumps. There are(nj
)ways in which exactly j upward jumps could occur.
• The probability of each of these is pjqn−j
• Pr(n, j) denotes the probability of j up moves and n− jdown moves.
• A binomial distribution.
pr(n, j) =
(n
j
)pjqn−j =
n!
j!(n− j)!pjqn−j (1)
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• Let xn be the value of x after n steps:
E(xn − x0) = nE(∆x) (2)
V ar(xn − x0) = nV ar(∆x) (3)
• Sub in for E(∆x) and V ar(∆x) and note that n = t/∆t
E(xn − x0) =t
∆t(p− q)∆h =
t
∆t(2p− 1)∆h (4)
(5)
V ar(xn − x0) =t
∆t4pq∆h2 =
t
∆t4p(1− p)∆h2 (6)
• Note that the variance of xn − x0 increases with t. Anon-stationary process.
• What about the mean?
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• Let the size of the jump be a continuous randomvariable, eg. ∆h ∼ N(0, σ2). This would be a discretetime, continuous state process.
• A first order AR(1) process is another example of adiscrete time, continuous state random variable
xt = δ + ρxt−1 + εt (7)
−1 < ρ < 1, ε ∼ N(0, σ)
• Random walks and AR(1) processes are Markovprocesses which means that only the current value isrelevant for predicting the future. Past history isirrelevant.
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2 The Wiener Process (BrownianMotion)
• One of the most useful stochastic processes used in manyfields - social sciences, science, engineering - namedafter English botanist Robert Brown who discovered it in1827
• The motion exhibited by a small particle totallyimmersed in liquid or gas
• A concise definition was given by Norbert Wiener in aseries of papers beginning in 1918
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Some useful definitions. Consider a random variable, X(t).
• X(t) has independent increments if for allt1 < t2 < ... < tn
X(tn)−X(tn−1), X(tn−1)−X(tn−2), ..., X(t1)−X(t0) (8)
are independent.
• X(t) has stationary increments if the distribution ofX(t+ s)−X(t) does not depend on t.
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• A Markov process is one which satisfies the Markovproperty. X(t) satisfies the Markov property if for alls, t ≥ 0, and non-negative integers i,j,x(u), 0 ≤ u < s
P{X(t+ s) = j| X(s) = i,X(u) = x(u), 0 ≤ u < s}= P{X(ts) = j|X(s) = i}
• In other words the conditional distribution of the futureX(t+ s) given the present X(s) and the past X(u),0 ≤ u < s, depends only on the present and isindependent of the past.
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• The Weiner process is a continuous time stochasticprocess
• A Markov process with mean zero and variance rate of 1
• If z(t) is a Wiener process then any ∆z satisfies∆z = ε
√∆t where ε ∼ N(0, 1).
• It follows that E(∆z) = 0 and
V ar(∆z) = V ar(ε√
∆t) (9)
= V ar(ε)(√
∆t2) = ∆t (10)
• The ∆z for any two short intervals, ∆t, are independent.
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• Consider the change in z during a relatively long periodof time, t. Denote this change as z(t)− z(0).
• Can be regarded as the sum of changes in z in n smallintervals of length ∆t. n = t/∆t.
• z(t)− z(0) =∑n
i=1 εi√
∆t
• εi (i = 1, 2, ..., n) are N(0, 1) and are independent
• Then z(t)− z(0) is normally distributed with mean of 0and variance of t
• z is a Wiener process. Its variance grows linearly withthe time horizon.
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• Let ∆t→ 0. Then
dz = εt√dt (11)
E(dz) = 0 (12)
V ar(dz) = E[(dz)2] = dt (13)
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• A Wiener process has no time derivative. As ∆t→ 0
derivative of z with respect to t goes to infinity.
• The expected length of the path followed by z in anytime interval is infinite.
• The expected number of times z equals any particularvalue at any time interval is infinite.
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Suppose z1 and z2 are Wiener processes. What is E(dz1dz2)?
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• Use the Wiener process as a building block for ageneralized Wiener process that allows for adeterministic drift term.
• The drift rate is the mean change per unit of time. Thevariance rate is the variance per unit of time.
• The basic Wiener process has a drift of zero and variancerate of 1.
• Generalized Wiener Process with constants a and b
dx = adt+ bdz (14)
• Solve for the value of x if b = 0
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• In a small interval ∆t, ∆x = a∆t+ bε√
∆t
• What are the mean and variance of ∆x? What are themean and variance of a change in x over a longer intervalt?
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3 Linking Brownian motion &random walks
• The generalized Wiener process
dx = adt+ bdz (15)
dz = εt√dt (16)
E(dx) = adt (17)
V ar(dx) = b2dt (18)
• The discrete random walk. x = x0 at t = 0. At t = ∆t,x0 → x0 + ∆h with prob p, and x0 → x0 −∆h withprob q.
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• Distribution of x after n moves (t = n∆t)
E(xn − x0) =t
∆t(p− q)∆h (19)
V ar(xn − x0) =t
∆t4pq∆h2 (20)
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• We want to take the limit as ∆t→ 0 in such a a way thatthe mean and variance of x after a finite time t isindependent of ∆t, and we would like to recoverEquations (15), (17), and (18).
• From Equation (20), we should choose∆h = constant
√∆t. Why is this so?
• Show that for other choices of ∆h the variance ofxn − x0 is either zero or infinite in finite time.
• Choose ∆h = b√
∆t for some constant b and sub intoEquations (19) and (20) to get the mean and variance.
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• From Equation (19) we see that for the mean of therandom walk to be independent of ∆t as ∆t→ 0 wemust have (p− q) = constant
√∆t.
• Choose p− q = ab
√∆t. Solve this for p and q using
p+ q = 1.
• We get
p =1
2
(1 +
a
b
√∆t)
(21)
q =1
2
(1− a
b
√∆t)
(22)
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Sub into Equations (19) and (20) for p and q and let ∆t→ 0
E(xn − x0) =tb√∆t
a
b
√∆t = at (23)
V ar(xn − x0) = t4pqb2 = t(1− a2
b2∆t)b2 (24)
= tb2 as ∆t→ 0. (25)
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• Note as ∆t→ 0, the number of steps approaches infinityand the binomial distribution converges to a normaldistribution.
• Now suppose xn − x0 becomes very small so thatxn − x0 ≈ dx and t = tn − t0 ≈ dt.
• Then E(dx) = adt and V ar(dx) = b2dt. We haveaccomplished our goal. We started with a random walkand have recovered a generalized Wiener process(Equations (15),(16) (17), (18))
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4 Ito Processes
• Use the Wiener process as a building block.
dx = a(x, t)dt+ b(x, t)dz (26)
• a(x, t) and b(x, t) are known functions. a(x, t) is theinstantaneous drift rates. b2(x, t) is the instantaneousvariance rate.
• E(dx) = a(x, t)dt and V ar(dx) = b2(x, t)dt
• Notice both the expected drift rate and variance rate maychange over time.
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• In the small time interval between t and t+ ∆t, xchanges to x+ ∆x where
∆x = a(x, t)∆t+ b(x, t)ε√
∆t (27)
• Assumes drift and variance rate of x remain constant andequal to a(x, t) and b(x, t)2 between t and t+ ∆t.
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4.1 Geometric Brownian Motion
• Let a(x, t) = αx and b(x, t) = σx, α and σ areconstants. Then,
dx = αxdt+ σxdz (28)
• Note that dxx∼ N or d[lnx] ∼ N
• GBM is often used to model stock prices.
• Suppose x represents a stock’s price and that volatility iszero. In this case we can solve for xt by directintegration. The result is that xt = x0e
αt
• It can also be shown that for a given x0, E[xt] = x0eαt.
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• Note that for GBM the variability of the proportionatereturn, dx/x, in any ∆t is the same regardless of thelevel of x.
• Example: A stock pays no dividends, has a volatility of30% per annum, and provides an expected return of 15%per year with continuous compounding.
• The stochastic differential equation describing this stockis dx
x= 0.15dt+ 0.3dz.
• A discrete time approximation would be:∆xx
= 0.15∆t+ 0.3ε√
∆t, ε ∼ N(0, 1) where ∆x is theincrease in stock price over the next interval.
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Consider a time interval of 1 week and the initial x0 = 100.What are the mean and standard deviation of ∆x?
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• It can be shown that the variance of x is
var(x(t)) = x20e
2αt(eσ2t − 1) (29)
We will derive this later.
• Determine a formula for the expected present discountedvalue of x(t)?
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4.2 Mean reverting processes
• More appropriate for commodities than GBM. Why?
• Ornstein-Uhlenbeck process
dx = η(x− x)dt+ σdz (30)
η is the speed of mean reversion
• Equation (30) is the limit of an AR(1) process as∆t→ 0.
xt − xt−1 = x(1− e−η) + (e−η − 1)xt−1 + εt (31)
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• It can be shown that
E(xt) = x+ (x0 − x)e−ηt (32)
var(xt − x) =σ2
2η(1− e−2ηt) (33)
• A Markov process, but does not have independentincrements.
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• Alternative mean reverting processes
dx = η(x− x)dt+ σxdz (34)
dx = ηx(x− x)dt+ σxdz (35)
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5 Ito’s lemma
5.1 Derivation
• Normal rules of calculous do not apply to stochasticdifferential equations. We need a result from stochasticcalculous.
• A general Ito process
dx = a(x, t)dt+ b(x, t)dz (36)
• Consider a function G = G(x, t). After dt,G→ G+ dG.
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• Ito’s lemma
dG =(a(x, t)∂G
∂x+ b2(x,t)
2∂2G∂x2
+ ∂G∂t
)dt
+b(x, t)∂G∂xdz (37)
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• It will be useful to find E(dz2).
• dz2 = ε2dt.
• We know that E(ε) = 0 and V ar(ε) = 1
• But V ar(ε) = E(ε2)− (E(ε))2 = 1. This implies thatE(ε2) = 1
• It follows that
E(dz2) = E(ε2dt) = E(ε2)E(dt) = dt (38)
• It can be shown that as dt→ 0, ε2dt becomesnon-stochastic, so that with probability 1, dz2 → dt asdt→ 0
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• Use a Taylor series to approximate the function G(x, t)
dG =∂G
∂xdx+
∂G
∂tdt+
1
2
∂2G
∂x2dx2+
1
6
∂3G
∂x3dx3+... (39)
• Denoting a(x, t) ≡ a and b(x, t) ≡ b,
(dx)2 = (adt+ bdz)2 = a2dt2 + 2(adt)(bdz) + b2dz2
= a2dt2 + 2(adt)bε√dt+ b2ε2dt
= a2dt2 + 2abε(dt3/2) + b2ε2dt
(40)
• The last term is the largest for small dt.
• dt3/2 and dt2 go to zero faster than dt as it becomesinfinitesimally small. These terms can be ignored.
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• We are left with
dx2 = b2dz2 = b2dt, as dt→ 0 (41)
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• In Equation (39) every term associated with theexpansion of dx3 will include dt raised to a powergreater than 1 and will go to zero faster than dt in thelimit. These terms can be ignored.
• Sub in for dx and dx2 in Equation (39)
dG =∂G
∂x(a(x, t)dt+b(x, t)dz)+
∂G
∂tdt+
1
2
∂2G
∂x2b2(x, t)dt
(42)
• Rearranging give Ito’s lemma.
dG =
[aGx +Gt +Gxx
b2
2
]dt+Gxbdz (43)
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• Ito’s lemma may be extended to two or more processes.Consider two stochastic variables, x and y.
dx = ax(x, t)dt+ bx(x, t)dzx (44)
dy = ay(y, t)dt+ by(y, t)dzy (45)
• For the function G(x, y) Ito’s lemma is as follows:
dG = [Gt + axGx + ayGy +Gxxb2x2
+Gyyb2y2
+ρxybxbyGxy]dt+Gxbxdzx +Gybydzy (46)
where ρxy = E[dzxdzy] is the correlation coefficientbetween x and y.
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Note that:
E[dzxdzy] = E[εx√dtεy√dt] (47)
= E[εx√dt]E[εy
√dt] + cov(εxεy)dt
= cov(εxεy)dt
= ρxydt
The last two lines follow because εi and εj have standardnormal distributions.
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Examples
1. x is described by dx = αxdt+ σxdz. SupposeG(x) = log(x). What stochastic process is followed byG(x)?
2. Consider correlated processes, x and y with correlationcoefficient ρ:
dx = αxxdt+ σxxdzx (48)
dy = αyydt+ σyydzy (49)
Suppose F (x, y) = xy. Use Ito’s lemma to find anexpression for dF .
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5.2 An Application of Ito’s lemma
• Previously we saw that a stochastic process such asdS = µdt+ σdz where µ and σ are constants can beintegrated exactly from t = 0 to t = tn to give
S(tn)− S(0) = µtn + σ[z(tn)− z(0)]
z(tn)− z(0) ∼ N(0, tn)
• It is only when µ and σ are constants that we can exactlyintegrate the stochastic differential equation.
• In other cases we can use Ito’s lemma to derive theproperties of a stochastic variable.
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• Suppose dS = µSdt+ σSdz, with µ and σ as constants.
• In this case the mean of S can be shown to be:
E[dS] = dS = E[µS]dt
S = S0eµt (50)
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• Now let G(S) = S2. Then GS = 2S and GSS = 2. FromIto’s lemma:
dG = E[2µS2 + σ2S2]dt+ E[2S2σdz]
= E[2µS2 + σ2S2]dt = [2µ+ σ2]Gdt (51)
• We can integrate directly to find G:
G = G0e(2µ+σ2)t
E(S2) = S20e
(2µ+σ2)t (52)
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• From Equations (50) and (52) we have:
V ar(S) = E(S2)− [E(S)]2
= S20e
(2µ+σ2)t − S20e
2µt
= S0e2µt[eσ
2t − 1]
= S2(eσ2t − 1) (53)
• So we have used Ito’s lemma to find the mean andvariance of a stochastic variable S that follow geometricBrownian motion.
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6 Jump Processes
• Some economic variables of interest make suddendiscrete jumps.
• Price of oil - mixed Brownian motion-jump process
• Poisson process - a process subject to jumps of fixed orrandom size. Each jump is called an event.
• Let φ denote the mean arrival rate of an event during dt.It represents the probability of an event over theinfinitesimal interval dt
• Denote the size of the jump as u. This can be a randomvariable.
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An aside on the Poisson distribution
• Let Z represent the number of events in a fixed timeinterval ∆t. If Z has a Poisson distribution theprobability density function is
f(Z,∆t) =(φ∆t)Ze−φ∆t
Z!, Z = 0, 1, 2, ...
(54)
• The number of events in non-overlapping intervals areindependent.
• Show that the expected time until the first event is 1/φ.
• The prob of Z = 1 in ∆t is φ∆t. (This can be foundusing a Taylor series approximation, ignoring terms of
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∆t2 and higher.)
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• q is a Poisson process
dq = 0 with probability 1− φdt= 1 with probability φdt (55)
• Let x be described by the following stochasticdifferential equation
dx = f(x, t)dt+ g(x, t)dq (56)
where f(x, t) and g(x, t) are known non-randomfunctions.
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• Suppose when x jumps it jumps to Jx where J is theproportional size of the jump. Restrict J to beingnon-negative.
dxjump = Jx− x = (J − 1)x (57)
Therefore g(x, t) ≡ (J − 1)x and
dx = f(x, t)dt+ (J − 1)xdq (58)
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• Assume the jump size has some known pdf, g(J). Thisimplies ∫ ∞
−∞g(J)dJ =
∫ ∞0
g(J)dJ = 1 (59)
• Suppose we have an investment whose value depends onx and t. Denote its value as V (x, t).
• The change in V due to jumps and the deterministic driftare, respectively:
dVjumps = [V (Jx, t)− V (x, t)]dq
dVdeter = [Vtdt+ Vxf(x, t)]dt
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• Adding these two gives
dV = [Vt + Vxf(x, t)]dt+ [V (Jx, t)− V (x, t)]dq (60)
• Take the expected value of dV , assuming the probabilityof a jump and the jump size are independent:
E[dV ] = [Vt+Vxf(x, t)]dt+EJ [V (Jx, t)−V (x, t)]E[dq]
(61)
• We can use Equation (61) to value projects which dependon a variable that exhibit jumps. This will be shown laterin the course.
• Look at the examples from Dixit and Pindyck, pages86-87.
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