Math 635: An Introduction to Brownian Motion and Stochastic Calculus

Math 635: An Introduction to Brownian Motion and Stochastic Calculus 1. Introduction and review

11. Stochastic differential equations

2. Notions of convergence and results from measure theory

12. Markov property

3. Review of Markov chains

13. SDEs and partial differential equations 14. Change of measure and asset pricing

4. Change of measure 5. Information and conditional expectations 6. Martingales

15. Martingale representation completeness

and

16. Applications and examples

7. Brownian motion 8. Stochastic integrals

17. Stationary distributions and forward equations

9. Black-Scholes and other models

18. Processes with jumps

10. The multidimensional stochastic calculus

19. Assignments 20. Problems

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February 22 review • Independence • Conditional expectations: Basic properties • Jensen’s inequality • Functions of known and unknown random variables • Filtrations and martingales • Optional sampling theorem • Doob’s inequalities


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1.

Introduction and review • The basic concepts of probability: Models of experiments • Sample space and events • Probability measures • Random variables • The distribution of a random variable • Definition of the expectation • Properties of expectations • Jensen’s inequality


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Experiments Probability models experiments in which repeated trials typically result in different outcomes. As a means of understanding the “real world,” probability identifies surprising regularities in highly irregular phenomena. If we roll a die 100 times we anticipate that about a sixth of the time the roll is 5. If that doesn’t happen, we suspect that something is wrong with the die or the way it was rolled.


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Probabilities of events Events are statements about the outcome of the experiment: {the roll is 6}, {the rat died}, {the television set is defective} The anticipated regularity is that P (A) ≈

#times A occurs #of trials

This presumption is called the relative frequency interpretation of probability.


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“Definition” of probability The probability of an event A should be #times A occurs in first n trials n→∞ n

P (A) = lim

The mathematical problem: Make sense out of this. The real world relationship: Probabilities are predictions about the future.


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Random variables In performing an experiment numerical measurements or observations are made. Call these random variables since they vary randomly. Give the quantity a name: X {X = a} and {a < X < b} are statements about the outcome of the experiment, that is, are events


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The distribution of a random variable If Xk is the value of X observed on the kth trial, then we should have #{k ≤ n : Xk = a} n→∞ n

P {X = a} = lim

If X has only finitely many possible values, then X P {X = a} = 1. a∈R(X)

This collection of probabilities determine the distribution of X.


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Distribution function More generally, n

1X 1(−∞,x] (Xk ) P {X ≤ x} = lim n→∞ n k=1

FX (x) ≡ P {X ≤ x} is the distribution function for X.


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The law of averages If R(X) = {a1 , . . . , am } is finite, then m

m

l=1

l=1

X #{k ≤ n : Xk = al } X X1 + · · · + Xn = lim al = al P {X = al } lim n→∞ n→∞ n n


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More generally, if R(X) ⊂ [c, d], −∞ < c < d < ∞, then X

xl P {xl < X ≤ xl+1 } =

l

lim

m X

n→∞

l=1

xl

#{k ≤ n : xl < Xk ≤ xl+1 } n

X1 + · · · + Xn ≤ lim n→∞ n m X #{k ≤ n : xl < Xk ≤ xl+1 } ≤ lim xl+1 n→∞ n X l=1 = xl+1 P {xl < X ≤ xl+1 } =

l X

xl+1 (FX (xl+1 ) − FX (xl ))

l

Z →

d

xdFX (x) c


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The expectation as a Stieltjes integral If R(X) ⊂ [c, d], define Z E[X] =

d

xdFX (x). c

If the relative frequency interpretation is valid, then X1 + · · · + Xn = E[X]. n→∞ n lim


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A random variable without an expectation Example 1.1 Suppose P {X ≤ x} =

x , 1+x

x ≥ 0.

Then m X X1 + · · · + Xn ≥ lim lP {l < X ≤ l + 1} n→∞ n

= =

l=0 m X

l=0 m X l=0

l(

l l+1 − ) l+2 l+1

l → ∞ as m → ∞ (l + 2)(l + 1)

One could say E[X] = ∞, and we will. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

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Review of basic calculus Definition 1.2 A sequence {xn } ⊂ R converges to x ∈ R (limn→∞ xn = x) if and only if for each > 0 there exists an n > 0 such that n > n implies |xn − x| ≤ . Pn P Let {ak } ⊂ R. The series ∞ a converges if lim k n→∞ k=1 ak ∈ R k=1 P∞ exists.PThe series converges absolutely if k=1 |ak | converges. (Or we write ∞ k=1 |ak | < ∞.) Examples of things you should know: lim (axn + byn ) = a lim xn + b lim yn

n→∞

n→

n→∞

if the two limits on the right exist. If limn,m→∞ |xn − xm | = 0 ({xn } is a Cauchy sequence), then limn→∞ xn exists. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

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Examples If α > 1, then

∞ X 1 < ∞. kα k=1

For α = 1,

∞ X 1 k=1

however ∞ X (−1)k k=1

k

= lim

n→∞

n X (−1)k k=1

k

k

= lim

m→∞

= ∞;

m X l=1

∞

X 1 1 1 (− + )=− 2l − 1 2l 2l(2l − 1) l=1


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The sample space The possible outcomes of the experiment form a set Ω called the sample space. Each event (statement about the outcome) can be identified with the subset of the sample space for which the statement is true.


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The collection of events If A = {ω ∈ Ω : statement I is true for ω} B = {ω ∈ Ω : statement II is true for ω} Then A ∩ B = {ω ∈ Ω : statement I and statement II are true for ω} A ∪ B = {ω ∈ Ω : statement I or statement II is true for ω} Ac = {ω ∈ Ω : statement I is not true for ω} Let F be the collection of events. Then A, B ∈ F should imply that A ∩ B, A ∪ B, and Ac are all in F. F is an algebra of subsets of Ω. In fact, we assume that F is a σ-algebra (closed under countable unions and complements). •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

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The probability measure Each event A ∈ F is assigned a probability P (A) ≥ 0. From the relative frequency interpretation, we must have P (A ∪ B) = P (A) + P (B) for disjoint events A and B and by induction, if A1 , . . . , Am are disjoint m X m P (∪k=1 Ak ) = P (Ak ) finite additivity k=1

In fact, we assume countable additivity: If A1 , A2 , . . . are disjoint events, then ∞ X ∞ P (∪k=1 Ak ) = P (Ak ). k=1

P (Ω) = 1. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

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A probability space is a measure space A measure space (M, M, µ) consists of a set M , a σ-algebra of subsets M, and a nonnegative function µ defined on M that satisfies µ(∅) = 0 and countable additivity. A probability space is a measure space (Ω, F, P ) satisfying P (Ω) = 1.


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Random variables If X is a random variable, then we must know the value of X if we know that outcome ω ∈ Ω of the experiment. Consequently, X is a function defined on Ω. The statement {X ≤ c} must be an event, so {X ≤ c} = {ω : X(ω) ≤ c} ∈ F. In other words, X is a measurable function on (Ω, F, P ). R(X) will denote the range of X Ω}

R(X) = {x ∈ R : x = X(ω), ω ∈


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Distributions Definition 1.3 The Borel subsets B(R) is the smallest σ-algebra of subsets of R containing (−∞, c] for all c ∈ R.

Definition 1.4 The distribution of a R-valued random variable X is the Borel measure defined by µX (B) = P {X ∈ B}, B ∈ B(R).

µX is called the measure induced by the function X.


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Discrete distributions Definition 1.5 A random variable is discrete or has a discrete distribution if and only if R(X) is countable. If X is discrete, the distribution of X is determined by the probability mass function pX (x) = P {X = x}, x ∈ R(X). Note that

X

P {X = x} = 1.

x∈R(X)


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Examples Binomial distribution n k P {X = k} = p (1 − p)n−k , k

k = 0, 1, . . . , n

for some postive integer n and some 0 ≤ p ≤ 1 Poisson distribution P {X = k} = e−λ

λk , k!

k = 0, 1, . . .

for some λ > 0.


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Absolutely continuous distributions Definition 1.6 The distribution of X is absolutely continuous if and only if there exists a nonnegative function fX such that Z b P {a < X ≤ b} = fX (x)dx, a < b ∈ R. a

Then fX is the probability density function for X.


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Examples Normal distribution 2 1 − (x−µ) 2 2σ e fX (x) = √ 2πσ

Exponential distribution fX (x) =

λe−λx 0

x≥0 x 0 lim P {|Xn − X| ≥ } = 0.

n→∞

Convergence in distribution: The sequence {Xn } converges to X is distribution if and only if for each x such that FX is continuous at x lim FXn (x) = FX (x).

n→∞


34

Examples Theorem 2.2 (The strong law of large numbers) Suppose ξ1 , ξ2 , . . . are independent and identically distributed with E[|ξi |] < ∞. Then ξ1 + · · · + ξn = E[ξ] a.s. n→∞ n lim

Theorem 2.3 (The weak law of large numbers) Suppose ξ1 , ξ2 , . . . are iid with E[ξi2 ] < ∞. Then for > 0, P {|

ξ1 + · · · + ξn V ar(ξ) → 0. − E[ξ]| ≥ } ≤ n n2


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Theorem 2.4 (Central limit theorem) Let ξ1 , ξ2 , . . . be iid with E[ξi ] = µ and V ar(ξi ) = σ 2 < ∞. Then Pn Z x y2 ξ − nµ 1 i √ e− 2 dy lim P { i=1√ ≤ x} = n→∞ nσ 2π −∞


36

Relationship among notions of convergence Lemma 2.5 Almost sure convergence implies convergence in probability. Convergence in probability implies convergence in distribution. Proof. Suppose Xn → X a.s. Then for > 0, An ≡ {sup |Xk − X| ≥ } ⊃ {|Xn − X| ≥ } k≥n

A1 ⊃ A2 ⊃ · · ·, so limn→∞ P (An ) = P (∩∞ k=1 Ak ). ∩∞ k=1 Ak = {ω : lim supn→∞ |Xn −X| ≥ } and hence limn→∞ P (An ) = 0.


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Suppose Xn → X in probability. Then FXn (x) ≤ FX (x+)+P {Xn ≤ x, X > x+} ≤ FX (x+)+P {|Xn −X| ≥ }. Therefore, lim sup FXn (x) ≤ lim FX (x + ) = FX (x), n→∞

→0

and similarly lim inf FXn (x) ≥ lim FX (x − ) = FX (x−). n→∞

→0


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Limit theorems for expectations and integrals Theorem 2.6 (Bounded convergence theorem) Let {Xn } be a sequence of random variables such that Xn → X in distribution. Suppose that there exists C > 0 such that P {|Xn | > C} = 0 for all n. Then limn→∞ E[Xn ] = E[X]. Proof. Let a0 ≤ −C < a1 < · · · < C ≤ am be points of continuity for FX . Then m−1 m−1 X X al P {al < Xn ≤ al+1 } ≤ E[Xn ] ≤ al+1 P {al < Xn ≤ al+1 }. l=0

l=0

Then the left and right sides converge to the left and right sides of m−1 X l=0

al P {al < X ≤ al+1 } ≤ E[X] ≤

m−1 X

al+1 P {al < X ≤ al+1 }.

l=0

and lim supn→∞ |E[Xn ] − E[X]| ≤ maxl (al+1 − al ).


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Lemma 2.7 If X ≥ 0 a.s., then limK→∞ E[X ∧ K] = E[X]. Proof. If K >

m n,

then

E[X] ≥ E[X ∧ K] ≥

m−1 X k=1

k k + 1 m→∞ bnXc k P{ ≤ X < } → E[ ]. n n n n

More generally, Theorem 2.8 (Monotone convergence theorem) If 0 ≤ X1 ≤ X2 ≤ · · · a.s. and X = limn→∞ Xn a.s., then E[X] = lim E[Xn ] n→∞

allowing ∞ = ∞.


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Lemma 2.9 (Fatou’s lemma) If P {Xn ≥ 0} = 1 and Xn converges in distribution to X, then lim inf E[Xn ] ≥ E[X] n→∞

Proof. lim inf E[Xn ] ≥ lim E[Xn ∧ K] = E[X ∧ K] → E[X] n→∞

n→∞

Let P {Xn = n} = 1−P {Xn = 0} = n−1 . Then Xn → 0 in distribution, and E[Xn ] = 1 for all n.


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Theorem 2.10 (Dominated convergence theorem) Suppose there exist Y such that |Xn | ≤ Y a.s. with E[Y ] < ∞ and limn→∞ Xn = X a.s. Then limn→∞ E[Xn ] = E[X]. Proof. E[Y ] − lim sup E[Xn ] = lim inf (E[Y ] − E[Xn ]) n→∞

n→∞

= lim inf E[Y − Xn ] n→∞

≥ E[Y − X] = E[Y ] − E[X] E[Y ] + lim inf E[Xn ] = lim inf (E[Y ] + E[Xn ]) n→∞

n→∞

= lim inf E[Y + Xn ] n→∞

≥ E[Y + X] = E[Y ] + E[X] •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

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Computation of expectations Recall µX (B) = P {X ∈ B}. g is Borel measurable if {x : g(x) ≤ c} ∈ B(R) for all c ∈ R. Note that (R, B(R), µX ) is a probability space and a Borel measurable function g is a “random variable” on this probability space. Theorem 2.11 If X is a random variable on (Ω, F, P ) and g is Borel measurable, then g(X) is a random variable and if E[|g(X)|] < ∞, Z g(x)µX (dx) E[g(X)] = R


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Lebesgue measure Lebesgue measure is the unique measure L on B(R) such that for a < b, L((a, b)) = b − a. Integration with respect to L is denoted Z Z g(x)dx ≡ g(x)L(dx) R

If f (x) =

Pm

i=1 ai 1Ai ,

R

Ai ∈ B(R) with L(Ai ) < ∞, ai ∈ R, then Z m X f (x)dx ≡ ai L(Ai ). R

i=1

Such an f is called a simple function.


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General definition of Lebesgue integral For g ≥ 0, Z

Z g(x)dx ≡ sup{

R

If 0,

R

R |g(x)|dx

f (x)dx : f simple, 0 ≤ f ≤ g}. R

< ∞, then setting g + (x) = g(x)∨0 and g − (x) = (−g(x))∨ Z Z Z g(x)dx ≡ g + (x)dx − g − (x)dx. R

R

R


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Riemann integrals Definition 2.12 Let −∞ < a < b < ∞. g defined on [a, b] is Riemann integrable if there exists I ∈ R so that for each > 0 there exists a δ > 0 such that a = t0 < t1 < · · · < tm = b, sk ∈ [tk , tk+1 ] and max |tk+1 − tk | ≤ δ implies m−1 X | g(sk )(tk+1 − tk ) − I| ≤ . k=0

Then I is denoted by Z

b

g(x)dx a

Pm−1

g(sk )(tk+1 − tk ) is called a Riemann sum. The Riemann integral is the limit of Riemann sums. k=0


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Some properties of Lebesgue integrals Theorem 2.13 If g is Riemann integrable, then g is Lebesgue integrable and the integrals agree. Fatou’s lemma, the monotone convergence theorem, and the dominated convergence theorem all hold for the Lebesgue integral.


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Distributions with Lebesgue densities Suppose Z µX (B) =

1B (x)fX (x)dx,

B ∈ B(R).

R

Then fX is a probability density function for X. Theorem 2.14 Let X be a random variable on (Ω, F, P ) with a probability density function fX . If g is Borel measurable and E[|g(X)|] < ∞, then Z g(x)fX (x)dx E[g(X)] = R


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3.

Review of Markov chains • Elementary definition of conditional probability • Independence • Markov property • Transition matrix • Simulation of a Markov chain


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Elementary definition of conditional probability For two events A, B ∈ F, the definition of the conditional probability of A given B (P (A|B)) is intended to capture how we would reassess the probability of A if we knew that B occurred. The relative frequency interpretation suggests that we only consider trials of the experiment on which B occurs. Consequently, # times that A and B occur # times B occurs # times that A and B occur/# trials = # times B occurs/# trials P (A ∩ B) ≈ P (B)

P (A|B) ≈

leading to the definition P (A|B) = P P(A∩B) (B) Note that for each fixed B ∈ F, P (·|B) is a probability measure on F. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

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Independence If knowing that B occurs doesn’t change our assessment of the probability that A occurs (P (A|B) = P (A)), then we say A is independent of B. By the defintion of P (A|B), independence is equivalent to P (A ∩ B) = P (A)P (B). {Ak , k ∈ I} are mutually independent if P (Ak1 ∩ · · · ∩ Akm ) =

m Y

P (Aki )

i=1

for all choices of {k1 , . . . , km } ⊂ I.


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Independence of random variables Random variables {Xk , k ∈ I} are mutually independent if P {Xk1 ∈ B1 , . . . , Xkm ∈ Bm } =

m Y

P {Xki ∈ Bi }

i=1

for all choices of {k1 , . . . , km } ⊂ I and B1 , . . . , Bm ∈ B(R).


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Joint distributions The collection of Borel subsets of Rm (B(Rm )) is the smallest σ-algebra containing (−∞, c1 ] × · · · × (−∞, cm ]. The joint distribution of (X1 , . . . , Xm ) is the measure on B(Rm ) defined by µX1 ,...,Xm (B) = P {(X1 , . . . , Xm ) ∈ B},

B ∈ B(Rm ).

Lemma 3.1 The joint distribution of (X1 , . . . , Xm ) is uniquely determined by the joint cdf FX1 ,...,Xm (x1 , . . . , xm ) = P {X1 ≤ x1 , . . . , Xm ≤ xm }.


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The Markov property Let X0 , X1 , . . . be positive, integer-valued random variables. Then P {X0 = i0 , . . . , Xm = im } = P {Xm = im |X0 = i0 , . . . , Xm−1 = im−1 } ×P {Xm−1 = im−1 |X0 = i0 , . . . , Xm−2 = im−2 } × · · · × P {X1 = i1 |X0 = i0 }P {X0 = i0 } X satisfies the Markov property if for each m and all choices of {ik }, P {Xm = im |X0 = i0 , . . . , Xm−1 = im−1 } = P {Xm = im |Xm−1 = im−1 }


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The transition matrix Suppose P {Xm = j|Xm−1 = i} = pij . (The transition probabilities do not depend on m.) The matrix P = ((pij )) satisfies X pij = 1. j

P {Xm+2 = j|Xm = i} X = P {Xm+2 = j, Xm+1 = k|Xm = i} = =

k X k X

P {Xm+2 = j|Xm+1 = k, Xm = i}P {Xm+1 = k|Xm = i} (2)

pik pkj ≡ pij

k

Then

(2)

((pij )) = P 2 . •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

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The joint distribution of a Markov chain The joint distribution of X0 , X1 , . . . is determined by the transition matrix and the initial distribution, νi = P {X0 = i}. P {X0 = i0 , . . . , Xm = im } = νi0 pi0 i1 pi1 i2 · · · pim−1 im . In general, (n)

P {Xm+n = j|Xm = i} = pij , where

(n)

((pij )) = P n .


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Simulation of a Markov chain Let E be the state space (the set of values the chain can assume) of the Markov chain. For simplicity, assume E = {1, . . . , N } of {1, 2, . . .}. Define H : E × [0, 1] → E by j−1 X

H(i, u) = j,

pik ≤ u
Ye } (Y − Ye )] = 0 = E[1{Y 0, E[X(t)] P {sup X(s) ≥ c} ≤ c s≤t Proof. Let τc = inf{t : X(t) ≥ x}. Then E[X(t)] ≥ E[X(t ∧ τ )] ≥ cP {τ ≤ t}. (Actually, for right continuous processes, τc may not be a stopping time unless one first modifies the filtration {Ft }, but the modification can be done without affecting the statement of the theorem.)


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Doob’s inequalities Theorem 6.10 Let X be a nonnegative submartingale. Then for p > 1, p p p E[sup X(s) ] ≤ E[X(t)p ] p−1 s≤t Corollary 6.11 If M is a square integrable martingale, then E[sup M (s)2 ] ≤ 4E[M (t)2 ]. s≤t


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7.

Brownian motion • Moment generating functions • Gaussian distributions • Characterization by mean and covariance • Conditions for independence • Conditional expectations • Central limit theorem • Construction of Brownian motion • Functional central limit theorem • The binomial model and geometric Brownian motion • Martingale properties of Brownian motion • Lévy’s characterization • Relationship to some partial differential equations


103

• First passage time • Quadratic variation • Nowhere differentiability • Law of the iterated logarithim • Modulus of continuity • Markov property • Transition density and operator • Strong Markov property • Reflection principle


104

Moment generating functions Definition 7.1 Let X be a Rd -valued random variable. Then the moment generating function for X (it it exists) is given by ϕX (λ) = E[eλ·X ],

λ ∈ Rd .

Lemma 7.2 If ϕX (λ) < ∞ for all λ ∈ Rd , then ϕX uniquely determines the distribution of X.


105

Gaussian distributions Definition 7.3 A R-valued random variable X is Gaussian (normal) with mean µ and variance σ 2 if fX (x) = √

2 1 − (x−µ) 2 2σ e 2πσ

The moment generating function for X is λX

E[e

σ 2 λ2 ] = exp{ + µλ} 2


106

Multidimensional Gaussian distributions Definition 7.4 A Rd -valued random variable X is Gaussian if λ · X is Gaussian for every λ ∈ Rd . Note that if E[X] = µ and cov(X) = E[(X − µ)(X − µ)T ] = Σ = ((σij )), then E[λ · X] = λ · µ,

var(λ · X) = λT Σλ.

Note that is X = (X1 , . . . , Xd ) is Gaussian in Rd , then for any choice of i1 , . . . , im ∈ {1, . . . , d}, (Xi1 , . . . Xim ) is Gaussian in Rm . Lemma 7.5 If X is Gaussian in Rd and C is a m × d matrix, then Y = CX

(7.1)

is Gaussian in Rm . •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

107

Characterization by mean and covariance If X is Gaussian in Rd , then λT Σλ ϕX (λ) = exp{ + λ · µ} 2 Since the distribution of X is determined by its moment generating function, the distribution of a Gausian random vector is determined by its mean and covariance matrix. Note that for Y given by (7.1), E[Y ] = Cµ and cov(Y ) = CΣC T .


108

Conditions for independence Lemma 7.6 If X = (X1 , X2 ) ∈ R2 is Gaussian, then X1 and X2 are independent if and only if cov(X1 , X2 ) = 0. In general, if X is Gaussian in Rd , then the components of X are independent if and only if Σ is diagonal. Proof. If R-valued random variables U and V are independent, then cov(U, V ) = 0, so if the components of X are independent then Σ is diagonal. If Σ is diagonal, then d Y

λ2i σi2 ϕX (λ) = exp{ + λi µ}, 2 i=1 and since the moment generating function of X determines the joint distribution, the components of X must be independent. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

109

Conditional expectations Lemma 7.7 Let X = (X1 , . . . , Xd ) be Gaussian, then E[Xd |X1 , . . . , Xd−1 ] = a +

d−1 X

bi Xi ,

i=1

where σdk =

d−1 X

bi σik ,

k = 1, . . . , d − 1

i=1

and µd = a +

d−1 X

b i µi .

i=1

Proof. The bi are selected so that Xd − (a + of X1 , . . . , Xd−1 .

Pd−1 i=1

bi Xi ) is independent


110

Example Suppose X1 , X2 , and X3 are independent, Gaussian, mean zero and variance one. Then X1 + X2 . E[X1 |X1 + X2 ] = 2 Let e2 = X1 + X2 − √1 X3 . e1 = X1 + X2 + √1 X3 , X X (7.2) 2 2 2 2 e1 and X e2 are independent, Gaussian, mean zero, and variance Then X one.


111

Central limit theorem If {ξi } are iid with E[ξi ] = 0 and V ar(ξi ) = 1, then Zn = satisfies Z b 1 2 1 √ e− 2 x dx. P {a < Zn ≤ b} → 2π a If we define [nt] 1 X Zn (t) = √ ξi n i=1

√1 n

Pn

i=1 ξi

then the distribution of (Zn (t1 ), . . . , Zn (tm )) is approximately Gaussian with cov(Zn (ti ), Zn (tj )) → ti ∧ tj , Note that for t0 < t1 < · · · < tm , Zn (tk ) − Zn (tk−1 ), k = 1, . . . , m are independent.


112

The CLT as a process


113

Interpolation lemma Lemma 7.8 Let Y and Z be independent, R-valued Gaussian random variables with E[Y ] = E[Z] = 0, V ar(Z) = a, and V ar(Y ) = 1. Then √ √ a a Z Z U= + Y, V = − Y 2 2 2 2 are independent Gaussian random variables satisfying Z = U + V . Proof. Clearly, Z = U +V . To complete the proof, check that cov(U, V ) = 0.


114

Construction of Brownian motion We can construct standard Brownian motion by repeated application of the Interpolation Lemma 7.8. Let ξk,n be iid, Gaussian, E[ξk,n ] = 0 and V ar(ξk,n ) = 1. Define W (1) = ξ1,0 , and define W ( 2kn ) inductively by k−1 W ( k+1 k−1 1 k 2n ) − W ( 2n ) + (n+1)/2 ξk,n W( n) = W( n ) + 2 2 2 2 1 1 1 k−1 k+1 = W ( n ) + W ( n ) + (n+1)/2 ξk,n 2 2 2 2 2

for k odd and 0 < k < 2n .


115

By the Interpolation Lemma, W(

k−1 1 k+1 1 k−1 1 k ) − W ( ) = W ( ) − W ( ) + ξk,n 2n 2n 2 2n 2 2n 2(n+1)/2

and k+1 k 1 k+1 1 k−1 1 ) − W ( ) = W ( ) − W ( ) − ξk,n 2n 2n 2 2n 2 2n 2(n+1)/2 are independent, Gaussian random variables. W(


116

Convergence to a continuous process Theorem 7.9 Let Γ be the set of diadic rationals in [0, 1]. Define Wn (t) = W (

k k n k+1 k ) + (t − )2 (W ( ) − W ( )), 2n 2n 2n 2n

k k+1 ≤ t ≤ . 2n 2n

Then sup |W (t) − Wn (t)| → 0 a.s., t∈Γ

and hence W extends to a continuous process on [0, 1].


117

Continuous functions on C[0, 1] Definition 7.10 Let C[0, 1] denote the continuous, real-valued functions on [0, 1]. g : C[0, 1] → R is continuous if xn , x ∈ C[0, 1] satisfying sup0≤t≤1 |xn (t) − x(t)| → 0 implies limn→∞ g(xn ) = g(x). For example, g(x) = sup x(t) 0≤t≤1

Z g(x) =

1

x(s)ds 0


118

Functional central limit theorem Theorem 7.11 (Donsker invariance principle) Let {ξi } be iid with E[ξi ] = 0 and V ar(ξi ) = σ 2 . Define [nt]

k √ 1 X ξi + (t − ) nξk+1 , Zn (t) = √ n n i=1

k+1 k ≤t≤ . n n

Then for each continuous g : C[0, 1] → R, g(Zn ) converges in distribution to g(σW ).


119

The binomial model and geometric Brownian motion In the binomial model for a stock price, at each time step the price either goes up a fixed percentage or down a fixed percentage. Assume the up-down probabilities are 12 − 21 and the time step is n1 . σ k+1 σ 1 k+1 k k ) = (1+ √ )Xn ( )} = P {Xn ( ) = (1− √ )Xn ( )} = . n n n n 2 n n P[nt] Then log Xn (t) = i=1 log(1+ √σn ξi ) where P {ξi = 1} = P {ξi = −1} = 1 2 . By Taylor’s formula P {Xn (

[nt]

[nt]

σ X 1 X 2 1 Xn (t) ≈ √ ξi − σ ≈ σW (t) − σ 2 t 2n i=1 2 n i=1 and

1 log Xn (t) ≈ exp{σW (t) − σ 2 t} 2 •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

120

Martingales Recall that an {Ft }-adapted process M is a {Ft }-martingale if and only if E[M (t + r)|Ft ] = M (t) for all t, r ≥ 0. Note that this requirement is equivalent to E[M (t + r) − M (t)|Ft ] = 0.


121

Convergence of conditional expectations Lemma 7.12 Suppose Xn , X are R-valued random variables and E[|Xn − X|] → 0. Then for a sub-σ-algebra D, E[|E[Xn |D] − E[X|D]|] → 0 Proof. By Jensen’s inequality E[|E[Xn |D] − E[X|D]|] = E[|E[Xn − X|D]|] ≤ E[E[|Xn − X||D]] = E[|Xn − X|]


122

Martingale properties of Brownian motion Let Ft = FtW = σ(W (s) : s ≤ t). Since W has independent increments, we have E[W (t + r)|Ft ] = E[W (t + r) − W (t) + W (t)|Ft ] = E[W (t + r) − W (t)|Ft ] + W (t) = W (t), and W is a martingale. Similarly, let M (t) = W (t)2 − t. Then E[M (t + r) − M (t)|Ft ] = E[(W (t + r) − W (t))2 + 2(W (t + r) − W (t))W (t) − r|Ft ] = E[(W (t + r) − W (t))2 |Ft ] + 2W (t)E[W (t + r) − W (t)|Ft ] − r =0 and hence M is a martingale.


123

An exponential martingale Let M (t) = exp{W (t) − 12 t}, then 1 E[M (t + r)|Ft ] = E[exp{W (t + r) − W (t) − r}M (t)|Ft ] 2 − 21 r = M (t)e E[exp{W (t + r) − W (t)}|Ft ] = M (t)


124

A general family of martingales Let f ∈ Cb3 (R) (the bounded continuous functions with three bounded continuous derivatives), and t = t0 < · · · < tm = t + r. Then E[f (W (t + r)) − f (W (t))|Ft ] X = E[ f (W (ti+1 )) − f (W (ti ))|Ft ] X = E[ f (W (ti+1 )) − f (W (ti )) − f 0 (W (ti ))(W (ti+1 ) − W (ti )) 1 − f 00 (W (ti ))(W (ti+1 ) − W (ti ))2 |Ft ] 2 X1 +E[ f 00 (W (ti ))(ti+1 − ti )|Ft ] 2 = Z1 + Z2


125

Note that since Fti ⊃ Ft , E[f 0 (W (ti ))(W (ti+1 ) − W (ti ))|Ft ] = E[E[f 0 (W (ti ))(W (ti+1 ) − W (ti ))|Fti ]|Ft ] = E[f 0 (W (ti ))E[(W (ti+1 ) − W (ti ))|Fti ]|Ft ] =0 E[f 00 (W (ti ))(W (ti+1 ) − W (ti ))2 |Ft ] = E[E[f 00 (W (ti ))(W (ti+1 ) − W (ti ))2 |Fti ]|Ft ] = E[f 00 (W (ti ))E[(W (ti+1 ) − W (ti ))2 |Fti ]|Ft ] = E[f 00 (W (ti ))(ti+1 − ti )|Ft ] Z t+r X1 1 00 E[| f 00 (W (ti ))(ti+1 − ti ) − f (W (s))ds|] → 0 2 2 t X X b E[|Z1 |] ≤ CE[ |W (ti+1 ) − W (ti )|3 ] ≤ C (ti+1 − ti )3/2 → 0 Rt so f (W (t)) − f (0) − 0 21 f 00 (W (s))ds is a {Ft }-martingale. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

126

Lévy’s characterization Theorem 7.13 Suppose M and Z given by Z(t) = M (t)2 − t are continuous, {Ft }-martingales, and M (0) = 0. Then M is a standard Brownian motion. Lemma 7.14 Under the assumptions of the theorem, E[(M (t+r)−M (t))2 |Ft ] = E[M (t+r)2 −2M (t)M (t+r)+M (t)2 |Ft ] = r, if τ is a {Ft }-stopping time, then E[(M ((t + r) ∧ τ ) − M (t ∧ τ ))2 |Ft ] = E[(t + r) ∧ τ − t ∧ τ |Ft ] ≤ r, and for f ∈ Cb3 (R), Z Mf (t) = f (M (t)) − f (M (0)) − 0

t

1 00 f (M (s))ds 2

(7.3)

is a {Ft }-martingale. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

127

Proof. Taking f (x) = eiθx , iθM (t)

e

iθM (0)

−e

t

Z + 0

1 2 iθM (s) θ e ds 2

is a martingale, and hence iθM (t+r)

E[e

iθM (t)

|Ft ] = e

Z − t

t+r

1 2 θ E[eiθM (s) |Ft ]ds 2

and iθ(M (t+r)−M (t))

ϕt,θ (r) ≡ E[e

Z

r

|Ft ] = 1 − 0

Therefore

1 2 θ ϕt,θ (s)ds 2

1 2

E[eiθ(M (t+r)−M (t)) |Ft ] = e− 2 θ r . It follows that (M (t + r) − M (t)) is Gaussian and independent of Ft .


128

Proof.[of lemma] To prove that (7.3) is a martingale, the problem is to show X E[ f (M (ti+1 )) − f (M (ti )) − f 0 (M (ti ))(M (ti+1 ) − M (ti )) 1 − f 00 (M (ti ))(M (ti+1 ) − M (ti ))2 |Ft ] 2 converges to zero. Let , δ > 0, and assume that ti+1 − ti ≤ δ. Define τ,δ = inf{t : sup |M (t) − M (s)| ≥ } t−δ≤s≤t

Then X X 3 |M (ti+1 ∧ τ,δ ) − M (ti ∧ τ,δ )| ≤ (M (ti+1 ∧ τ,δ ) − M (ti ∧ τ,δ ))2 , and the expectation of the term on the left is bounded by t. Select n → 0 and δn → 0 so that τn ,δn → ∞ a.s.


129

Applications of the optional sampling theorem Let −a < 0 < b, and define γa,b = inf{t : W (t) ∈ / (−a, b). Then E[W (γa,b ∧ t)] = 0 and E[γa,b ∧ t] = E[W (γa,b ∧ t)2 ] ≤ a2 ∨ b2 . Let t → ∞. E[γa,b ] < ∞, so 0 = E[W (γa,b )] = −aP {W (γa,b ) = −a} + bP {W (γa,b ) = b} = −a + (a + b)P {W (γa,b ) = b} Consequently, P {W (γa,b ) = b} =

a a+b

and

E[γa,b ] = E[W (γa,b )2 ] = a2

b a + b2 a+b a+b


130

Relationship to some partial differential equations Let W = (W1 , W2 ) where W1 and W2 are independent, R-valued standard Brownian motions. For f ∈ Cb3 (R2 ), Z t 1 ∆f (W (s))ds Mf (t) = f (W (t)) − f (0) − 0 2 is a martingale. Let Xx (t) = x + W (t) (standard Brownian motion starting at x). Then Z t 1 Mfx (t) = f (Xx (t)) − f (x) − ∆f (Xx (s))ds 2 0 is a martingale. Let D ⊂ R2 be bounded, open, and have a smooth boundary. Suppose ∆h(y) = 0, t ∈ D, and h(y) = ϕ(y), y ∈ ∂D (the / D}. Then boundary of D). Define τDx = inf{t : Xx (t) ∈ h(x) = E[ϕ(Xx (τDx ))] •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

131

Finiteness of the exit time Lemma 7.15 For ρ > 0, let γρ = inf{t : |W (t)| ≥ ρ}. Then E[γρ ] =

ρ2 2.

Proof. Let f (x) = |x|2 for |x| ≤ ρ. then ∆f (x) = 4 for |x| ≤ ρ, and the optional sampling theorem implies E[|W (γρ ∧ t)|2 ] = E[2(γρ ∧ t)]. Letting t → ∞ gives the result.


132

Laplace transforms for nonnegative random variables For a nonnegative random variable X, define the Laplace transform of its distribution by Z e−αx µX (dx), α ≥ 0. FX (α) = E[e−αX ] = R

Lemma 7.16 The distribution of a nonnegative random variable is uniquely determined by its Laplace transform.


133

First passage time For c > 0, let τc = inf{t : W (t) ≥ c}. Since 1 Ma (t) = exp{aW (t) − a2 t} 2 The optional sampling theorem implies 1 1 1 = E[exp{aW (t ∧ τc ) − a2 t ∧ τc }] → E[exp{ac − a2 τc }] 2 2 since τc < ∞ a.s. Consequently, 1 2

E[e− 2 a τc ] = e−ac and

√

Fτc (α) = E[e−ατc ] = e−

2αc

.


134

Quadratic variation For s < t, E[(W (t) − W (s))2 ] = t − s and 2 2 E[ (W (t) − W (s))2 − (t − s) ] = (t − s)2 E[ W (1)2 − 1 ]. Consequently, for 0 = t0 < · · · < tm = t, !2 m m X 2 X 2 2 (ti − ti−1 )2 (W (ti ) − W (ti−1 )) − t ] = E[ W (1) − 1 ] E[ i=1

i=1

which converges to zero as max(ti − ti−1 ) → 0. We say that [W ]t = t.


135

Covariation For independent standard Brownian motions W1 , W2 , a similar calculation implies [W1 , W2 ]t = lim

m X

(W1 (ti ) − W1 (ti−1 ))(W2 (ti ) − W2 (ti−1 )) = 0

i=1

where the convergence is in probability.


136

Nowhere differentiability Rt Rt Suppose that f (t) = 0 g(s)ds, where 0 |g(s)|ds < ∞ for all t > 0. Then Z t m X 2 (f (ti ) − f (ti−1 )) ≤ max |f (ti ) − f (ti−1 )| |g(s)|ds → 0 i=1

i

0

as max(ti − ti−1 ) → 0. Definition 7.17 A function f : [0, ∞) → R is nowhere differentiable if its derivative does not exist at any point t ∈ [0, ∞). Let ΓN D denote the collection of nowhere differentiable funtions. Theorem 7.18 P {W ∈ ΓN D } = 1.


137

Law of the iterated logarithim lim sup √ t→∞

W (t) = 1 a.s. 2t log log t

c (t) = tW (1/t) is Brownian motion. V ar(W c (t)) = t2 1 = t Therefore W t W (1/t)

lim sup p t→0

2t−1 log log 1/t

c (t) W

= lim sup p t→0

2t log log 1/t

= 1 a.s.

Consequently, W (t + h) − W (t) lim sup p = 1 a.s. 2h log log 1/h h→0


138

Modulus of continuity Definition 7.19 A function f : [a, b] → R is Hölder continuous with exponent 0 ≤ α ≤ 1, if |f (t) − f (s)| < ∞. |t − s|α a≤s τ on {τ < ∞}, and [2n t] [2n t] {τn ≤ t} = {τn ≤ n } = {τ < n } ∈ Ft . 2 2 •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

142

Conditioning up to discrete stopping times Lemma 7.24 Let E[|Z|] < ∞, and let τ be a discrete {Ft }-stopping time (R(τ ) = {t1 , t2 , . . .} ∪ {∞}). Then X E[Z|Fτ ] = E[Z|Ftk ]1{τ =tk } + Z1{τ =∞} . k


143

Strong Markov property Theorem 7.25 Let τ be a {FtX }-stopping time with P {τ < ∞} = 1. Then E[f (X(τ + t))|Fτ ] = T (t)f (X(τ )).

(7.5)

More generally, define Wτ by Wτ (t) = W (τ + t) − W (τ ). Then Wτ is a standard Brownian motion and Wτ is independent of Fτ .


144

Proof. Prove first for discrete stopping times and take limits. Let τn be as above. Then X E[f (X(τn + t))|Fτn ] = E[f (X(τn + t))|Fk2−n ]1{τn =k2−n } = =

k X k X

E[f (X(k2−n + t))|Fk2−n ]1{τn =k2−n } T (t)f (X(k2−n ))1{τn =k2−n }

k

= T (t)f (X(τn )). Assume that f is continuous so that T (t)f is continuous. Then E[f (X(τn + t))|Fτ ] = E[T (t)f (X(τn ))|Fτ ] and passing to the limit gives (7.5).


145

Lemma 7.26 If γ ≥ 0 is Fτ -measurable, then E[f (X(τ + γ))|Fτ ] = T (γ)f (X(τ )). Proof. First, assume that γ is discrete. Then X E[f (X(τ + γ))|Fτ ] = E[f (X(τ + γ))|Fτ ]1{γ=r} r∈R(γ)

=

X

E[f (X(τ + r))|Fτ ]1{γ=r}

r∈R(γ)

=

X

T (r)f (X(τ ))1{γ=r}

r∈R(γ)

= T (γ)f (X(τ )). Assuming that f is continuous, general γ can be approximated by discrete γ.


146

Reflection principle Lemma 7.27 P {sup W (s) ≥ c} = 2P {W (t) > c} s≤t

Proof. Let τ = t ∧ inf{s : W (s) ≥ c}, and γ = (t − τ ). Then setting f = 1(c,∞) , 1 E[f (W (t))|Fτ ] = E[f (W (τ + γ))|Fτ ] = T (γ)f (W (τ )) = 1{τ c}. Since {τ < t} ∪ {W (t) = c} = {sups≤t W (s) ≥ c} and P {W (t) = c} = 0, P {sups≤t W (s) ≥ c} = P {τ < t}. Corollary 7.28 The cdf of τc = inf{t : W (t) ≥ c} is Z ∞ x2 2 e− 2t dx. Fτc (t) = 2P {W (t) > c} = √ 2πt c •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

147

Robert Brown’s original paper http://sciweb.nybg.org/science2/pdfs/dws/Brownian.pdf


148

8.

Stochastic integrals • Definitions • Martingale properties • Continuous integrators and cadlag integrands • Stochastic integrals as integrators • Interated integrals • Quadratic variation of a stochastic integral ˆ formula • Ito’s • Solution of a stochastic differential equation


149

Notation for limits through partitions Let H({ti }) denote a functional of a partition 0 = t0 < t1 < · · · satisfying limm→∞ tm = ∞. For example X f (ti )(g(T ∧ ti+1 ) − g(T ∧ ti )) H({ti }) = i

The assertion that lim H({ti }) = H0

{ti }→∗

(8.1)

will mean that for each > 0 there exists a δ > 0 such that maxi (ti+1 − ti ) < δ implies |H({ti }) − H0 | < . If H({ti }) and H0 are random variables, then the limit in (8.1) exists in probability if and only if for each > 0, there exists a δ > 0 such that maxi (ti+1 − ti ) < δ implies P {|H({ti }) − H0 | > } ≤ .


150

Definitions Definition 8.1 A Brownian motion W is a {Ft }-Brownian motion if W is {Ft }-adapted and W (t + s) − W (t) is independent of Ft for all s, t ≥ 0. Note that FtW ⊂ Ft and that any {Ft }-Brownian motion is a {FtW }Brownian motion. In what follows, we will assume that there is a fixed filtration {Ft } and that W is a {Ft }-Brownian motion. All adapted processes are adapted to this filtration.


151

Simple integrands Let 0 = r0 < r1 < · · · < rm and rm+1 = ∞. X(t) =

m X

ξi 1[ri ,ri+1 ) .

(8.2)

i=0

Then X is a simple process. Lemma 8.2 The simple process X given by (8.2) is {Ft }-adapted if and only if for each i, ξi is Fri -measurable. Definition 8.3 Let X be an adapted, simple process. Then Z t m X I(t) = X(s)dW (s) ≡ ξi (W (t ∧ ri+1 ) − W (t ∧ ri )). 0

i=0


152

Martingale properties Lemma 8.4 Let X and I be as above. Suppose that for each i, E[|ξi |] < ∞. Then I is a martingale. Proof. Suppose s > t. Then, if ri ≥ t, E[ξi (W (s ∧ ri+1 ) − W (s ∧ ri ))|Ft ] = 0 = ξi (W (t ∧ ri+1 ) − W (t ∧ ri )), and if ri < t, E[ξi (W (s ∧ ri+1 ) − W (s ∧ ri ))|Ft ] = ξi E[(W (s ∧ ri+1 ) − W (s ∧ ri ))|Ft ] = ξi (W (t ∧ ri+1 ) − W (t ∧ ri )).


153

Itoˆ isometry Theorem 8.5 Let X and I be as above. Suppose that for each i, E[ξi2 ] < ∞. Then Z t

2

X 2 (s)ds]

E[I (t)] = E[ 0

Proof. The martingale property of W implies that for i 6= j, E[ξi (W (t ∧ ri+1 ) − W (t ∧ ri ))ξj (W (t ∧ rj+1 ) − W (t ∧ rj ))] = 0. Consequently, 2

E[I (t)] = =

m X i=0 m X i=0

E[ξi2 (W (ri+1 ∧ t) − W (ri ∧ t))2 ] Z t E[ξi2 (ri+1 ∧ t − ri ∧ t))] = E[ X 2 (s)ds] 0


154

An approximation lemma Rt Lemma 8.6 Let X be adapted and E[ 0 X 2 (s)ds] < ∞ for each t > 0. Then there exists a sequence {Xn } of adapted, simple processes such that Z t lim E[ (X(s) − Xn (s))2 ds] = 0. (8.3) n→∞

0

Note that (8.3) implies Z t lim E[ (Xm (s) − Xn (s))2 ds] ≤ n,m→∞

0

Z t lim (2E[ (X(s) − Xn (s))2 ds] n,m→0 Z0 t +2E[ (X(s) − Xm (s))2 ds]) = 0. 0


155

Extension of the definition of the integral By Doob’s inequality and the martingale properties of the stochastic integral Z t Z t Z t 2 E[sup | Xn dW − Xm dW | ] = E[sup | (Xn − Xm )dW |2 ] t≤T

0

0

t≤T

Z

0 T

≤ 4E[| (Xn − Xm )dW |2 ] Z 0T = 4E[ (Xm (s) − Xn (s))2 ds] 0

Consequently, there exists stochastic process I such that R t a continuous 2 limn→∞ E[supt≤T |I(t) − 0 Xn dW | ] = 0. (Note Rthat the proof of this fact requires additional machinery.) We define XdW = I.


156

Properties of the stochastic integral Theorem R t 2 8.7 Suppose that X1R, X2 are adapted and that for t > 0, E[ 0 Xi (s)ds] < ∞. Let Ii = Xi dW . a) Ii is adapted. Rt b) For each t ≥ 0, E[Ii2 (t)] = E[ 0 Xi2 (s)ds]. c) Ii is a martingale. R d) For a, b ∈ R, (aX1 + bX2 )dW = aI1 + bI2 .


157

L2 convergence of stochastic integrals Rt Lemma 8.8 Let {Xn }, X be adapted processes such that E[ 0 Xn2 (s)ds] < Rt ∞ for each n and t > 0 and for each t > 0 limn→∞ E[ 0 |Xn (s)−X(s)|2 ds] = 0. Then for each T > 0, Z t Z t lim E[sup | Xn dW − XdW |2 ] = 0. n→∞

t≤T

0

0

Proof. By Doob’s inequality and the isometry, Z t Z t Z T E[sup | Xn dW − XdW |2 ] ≤ 4E[ |Xn (s) − X(s)|2 ds] t≤T

0

0

0


158

Continuous integrators and cadlag integrands Let Y be a continuous, adapted process and X be a cadlag, adapted process. We define Z t X X(s−)dY (s) = lim X(ti ∧ t)(Y (ti+1 ∧ t) − Y (ti ∧ t)), (8.4) 0

{ti }→∗

if the limit on the right exists in probability for each t. Lemma 8.9 Suppose Y = W . If X is cadlag, adapted, and bounded by a constant, the two definitions of the stochastic integral are consistent.


159

R Proof. Note that the sum on the right of (8.4) is X {ti } dW for X {ti } X (t) = X(ti )1(ti ,ti+1 ] (t) and that

Z

t

|X {ti } (s) − X(s−)|2 ds = 0.

lim

{ti }→∗

0

By the bounded convergence theorem, Z t lim E[ |X {ti } (s) − X(s−)|2 ds] = 0, {ti }→∗

0

and the lemma follows by Lemma 8.8.


160

Elimination of the moment condition Theorem R t 2 8.10 Suppose that X is cadlag and adapted and that for each t > 0, 0 X (s)ds < ∞ a.s. Then Z t X X(s−)dW (s) ≡ lim X(ti ∧ t)(W (ti+1 ∧ t) − W (ti ∧ t)), (8.5) {ti }→∗

0

in probability. Proof. Let τc = inf{t : |X(t)|∨|X(t−)| ≥ c}. Then X τc (t) = X(t)1[0,τc ) (t) Rt is cadlag and adapted and satisfies E[ 0 |X τc (s)|2 ds] ≤ c2 t. Consequently, by Lemma 8.9, X lim X(ti ∧ t)1[0,τc ) (ti ∧ t)(W (ti+1 ∧ t) − W (ti ∧ t)) {ti }→∗

exists in probability. Since limc→∞ τc = ∞ a.s., (8.5) holds.


161

Convergence of stochastic integrals Theorem 8.11 Suppose that Xn , X are cadlag and adapted and that for each t > 0, Z t lim |Xn (s) − X(s)|2 ds = 0 n→∞

0

in probability. Then Z lim sup |

n→∞ t≤T

t

Z Xn (s)dW (s) −

0

t

X(s)dW (s)| = 0 0

in probability.


162

Rt Rt Proof. Let γcn = inf{t : 0 Xn2 (s)ds ≥ c} and γc = inf{t : 0 X 2 (s)ds ≥ c}. Then for each t > 0, Z t lim E[ |1[0,γcn ) (s)Xn (s) − 1[0,γc ) X(s)|2 ] = 0 n→∞

0

and hence Z lim E[sup |

n→∞

t≤T

0

t

Z 1[0,γcn ) (s)Xn (s)dW (s) −

t

1[0,γc ) X(s)dW (s)|2 ] = 0

0

and Z lim E[1{γcn ∧γc >T } sup |

n→∞

t≤T

t

Z Xn (s)dW (s) −

0

t

X(s)dW (s)|2 ] = 0.

0

For > 0 and c sufficiently large, lim supn→∞ P {γcn ∧ γc ≤ T } ≤ and the lemma follows.


163

Stochastic integrals as integrators Theorem 8.12 Suppose that X and Z are cadlag and adapted and Y (t) = Rt 0 Z(s−)dW (s). Then Z t X X(s−)dY (s) ≡ lim X(ti )(Y (ti+1 ∧ t) − Y (ti ∧ t)) {ti }→∗ 0 Z t Z t = X(s−)Z(s−)dW (s) = X(s)Z(s)dW (s). 0

0

Rt More generally, if U is cadlag and adapted and Y (t) = 0 Z(s−)dW (s) + Rt 0 U (s)ds, then Z t Z t Z t X(s)U (s)ds. X(s−)dY (s) = X(s)Z(s)dW (s) + 0

0

0


164

Proof. First check the result P for X simple. Then, for X cadlag and {ti } adapted, define X (t) = X(ti )1(ti ,ti+1 ] (t), so Z t X X(ti )(Y (ti+1 ∧ t) − Y (ti ∧ t)) = X {ti } (s−)dY (s) Z0 t = X {ti } (s−)Z(s−)dW (s). 0

Then, by Theorem 8.11, X lim X(ti )(Y (ti+1 ∧ t) − Y (ti ∧ t)) {ti }→∗ Z t X {ti } (s−)Z(s−)dW (s) = lim {ti }→∗ 0 Z t X(s−)Z(s−)dW (s). = 0


165

Interated integrals Corollary and adapted, and Rt R t 8.13 Suppose XR1 ,t X2 , U , and Z are cadlag Y (t) = 0 Z(s−)dW (s) + 0 U (s)ds and V (t) = 0 X1 (s−)dY (s). Then Z t X2 (s−)dV (s) 0 Z t = X2 (s−)X1 (s−)dY (s) Z0 t Z t = X2 (s−)X1 (s−)Z(s−)dW (s) + X2 (s−)X1 (s−)U (s)ds 0 0 Z t Z t X2 (s)X1 (s)Z(s)dW (s) + X2 (s)X1 (s)U (s)ds = 0

0


166

Quadratic variation as a stochastic integral Rt

X(s)dX(s) exists, and Z t 2 2 [X]t = X (t) − X (0) − 2 X(s)dX(s).

Lemma 8.14 [X]t exists if and only if

0

0

Proof. Noting that (b − a)2 = b2 − a2 − 2a(b − a), X (X(ti+1 ∧ t) − X(ti ∧ t))2 X = (X 2 (ti+1 ∧ t) − X 2 (ti ∧ t)) X −2 X(ti ∧ t)(X(ti+1 ∧ t) − X(ti ∧ t)) X 2 2 = X (t) − X (0) − 2 X(ti ∧ t)(X(ti+1 ∧ t) − X(ti ∧ t)) •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

167

Quadratic variation of a stochastic integral Rt Rt Lemma 8.15 If X and Z are adapted, 0 X 2 (s)ds < ∞ and 0 |Z(s)|ds < ∞ for each t > 0, and Z t Z t Y (t) = X(s)dW (s) + Z(s)ds, 0

then

0

Z [Y ]t =

t

X 2 (s)ds.

0

Proof. Check first for simple X and then extend by approximation.


168

Ito’s ˆ formula Theorem 8.16 Let Z

t

Z X(s)dW (s) +

Y (t) = Y (0) + 0

t

Z(s)ds 0

and f ∈ C 2 (R). Then Z

t

Z

t

1 00 f (Y (s))d[Y ]s 0 0 2 Z t Z t 0 = f (Y (0)) + f (Y (s))X(s)dW (s) + f 0 (Y (s))Z(s)ds 0 Z0 t 1 00 + f (Y (s))X 2 (s)ds 0 2

f (Y (t)) = f (Y (0)) +

0

f (Y (s))dY (s) +


169

Proof. Assume that f ∈ Cb3 (R). Then X f (Y (t)) = f (Y (0)) + (f (Y (ti+1 )) − f (Y (ti ))) X = f (Y (0)) + f 0 (Y (ti ))(Y (ti+1 ) − Y (ti )) X1 + f 00 (Y (ti ))(Y (ti+1 ) − Y (ti ))2 2 X +O( |Y (ti+1 ) − Y (ti )|3 ) P P Note that if lim (Y (ti+1 ) − Y (ti )2 exists, lim |Y (ti+1 ) − Y (ti )|3 = 0. The second term on the right converges by Theorem 8.12 and the third by Problem 10.


170

Ito’s ˆ formula take 2 Theorem 8.17 Let Z Y (t) = Y (0) +

t

Z X(s)dW (s) +

0

t

Z(s)ds 0

and f ∈ C 2,1 (R × R). Then Z

t

Z

t

f (Y (t), t) = f (Y (0), 0) + fx (Y (s), s)dY (s) + fs (Y (s), s)ds 0 0 Z t 1 + fxx (Y (s), s)d[Y ]s 0 2 Z t = f (Y (0), 0) + fx (Y (s), s)X(s)dW (s) Z t0 + (fx (Y (s), s)Z(s) + fs (Y (s), s))ds Z0 t 1 + fxx (Y (s), s)X 2 (s)ds 0 2 •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

171

Solution of a stochastic differential equation A solution of the stochastic differential equation dX = σXdW + µXdt is an adapted stochastic process satisfying Z t Z t X(t) = X(0) + σX(s)dW (s) + µX(s)ds. 0

(8.6)

0

ˆ formula suggests applying a substitution method similar to that Ito’s used to solve linear ordinary differential equaions, that is, look for a solution of the form X(t) = X(0) exp{aW (t) + bt}

(8.7)


172

Identification of solution ˆ formula, Setting Y (t) = aW (t) + bt so that [Y ]t = a2 t, applying Ito’s we have Z t Z t 1 X(s)dY (s) + X(s)d[Y ]s X(t) = X(0) + 0 0 2 Z t Z t 1 = X(0) + aX(s)dW (s) + (b + a2 )X(s)ds 2 0 0 Consequently, setting a = σ and b = µ − 21 σ 2 , (8.7) satisfies (8.6), that is 1 X(t) = X(0) exp{σW (t) + (µ − σ 2 )t} 2


173

Vasicek interest rate model=Ornstein-Uhlenbeck process dR = (α − βR)dt + σdW or

Z

t

(α − βR(s))ds + σW (t)

R(t) = R(0) + 0

Employ an integrating factor setting Q(t) = eβt R(t). Then Z t Z t Q(t) = R(0) + αeβs ds + σ eβs dW (s) 0

0

Hence −βt

R(t) = e

α R(0) + (1 − e−βt ) + σe−βt β

Z

t

eβs dW (s)

0


174

Covariation Theorem 8.18 Let Z

t

Y1 (t) = Y1 (0) +

Z

t

X1 (s)dW (s) + 0

Z Y2 (t) = Y2 (0) +

U1 (s)ds 0

t

Z X2 (s)dW (s) +

0

t

U2 (s)ds. 0

Y1 and Y2 are called Itoˆ processes. Then X [Y1 , Y2 ]t ≡ lim (Y1 (ti+1 ∧ t) − Y1 (ti ∧ t))(Y2 (ti+1 ∧ t) − Y2 (ti ∧ t)) {ti }→∗ Z t Z t Y1 (s)dY2 (s) − Y2 (s)dY1 (s) = Y1 (t)Y2 (t) − Y1 (0)Y2 (0) − 0 0 Z t = X1 (s)X2 (s)ds 0


175

Proof. X lim (Y1 (ti+1 ∧ t) − Y1 (ti ∧ t))(Y2 (ti+1 ∧ t) − Y2 (ti ∧ t)) {ti }→∗ X = (Y1 (ti+1 ∧ t)Y2 (ti+1 ∧ t) − Y1 (ti ∧ t)Y2 (ti ∧ t)) X − lim Y1 (ti ∧ t)(Y2 (ti+1 ∧ t) − Y2 (ti ∧ t)) {ti }→∗ X − lim Y2 (ti ∧ t)(Y1 (ti+1 ∧ t) − Y1 (ti ∧ t)) {ti }→∗ Z t Z t = Y1 (t)Y2 (t) − Y1 (0)Y2 (0) − Y1 (s)dY2 (s) − Y2 (s)dY1 (s)

[Y1 , Y2 ]t ≡

0

0

To verify the second equality, first assume X1 and X2 are simple. Then approximate.


176

Integration by parts Corollary 8.19 Let Y1 and Y2 be Itô processes. Then Z t Z t Y1 (s)dY2 (s) = Y1 (t)Y2 (t) − Y1 (0)Y2 (0) − Y2 (s)dY1 (s) − [Y1 , Y2 ]t . 0

0


177

9.

Black-Scholes and other models • Derivation of the Black-Scholes model • A population model • Value of a portfolio • Hedging a contract • Pricing contracts using the Black-Scholes model


178

Derivation of the Black-Scholes model First consider a discrete time model of the value of a stock. At each time step, the value increases or decreases by a certain percentage, so the value becomes Yk+1 = Yk (1 + ξk+1 ) If the time steps are small, the percentage change is small suggesting 1 1 n Yk+1 = Ykn (1 + √ ξk+1 + α) n n where we assume the {ξk } are independent with E[ξk ] = 0

V ar(ξk ) = σ 2


179

n . Then Define Sn (t) = Y[nt] [nt]−1

log Sn (t) = log Sn (0) +

X k=0 [nt]−1

1 1 log(1 + √ ξk+1 + α) n n

[nt]−1 X 1 [nt] 1 X 1 1 √ ξk+1 + α− ( √ ξk+1 + α)2 ≈ log Sn (0) + n 2 n n n k=0

k=0

1 ≈ log S(0) + σW (t) + αt − σ 2 t 2 and

1 S(t) = S(0) exp{σW (t) + (α − σ 2 )t} 2 S is called geometric Brownian motion.

(9.1)


180

An alternative derivation n Since Yk+1 = Ykn (1 +

√1 ξk+1 n

+ n1 α) = Ykn + Ykn ( √1n ξk+1 + n1 α), we have

[nt]−1 n Y[nt]

=

Y0n

+

X

n (Yk+1

[nt]−1

−

Ykn )

=

Y0n

+

k=0

X

1 Ykn √

n

k=0

[nt]−1

ξk+1 +

X

αYkn

k=0

1 n

P[nt]

An (t) = [nt] n , Z t Rt αSn (s−)dAn (s) Sn (t) = Sn (0) + 0 σSn (s−)dWn (s) +

n Setting Sn (t) = Y[nt] , Wn (t) =

1 √

σ n

k=1 ξk ,

0

suggesting a limit satisfying Z t Z t S(t) = S(0) + σS(s)dW (s) + αS(s)ds 0

(9.2)

0

We know that (9.1) satisfies (9.2), so this derivation looks plausible. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

181

A population model Consider the following discrete-time branching process. {ξik , i ≥ 1, k ≥ 0} iid nonnegative, integer-valued with E[ξik ] = 1 and V ar(ξik ) = σ 2 . ξik gives the number of offspring of the ith individual in the kth generation, so Zk+1 =

Zk X

ξik

= Zk +

i=1

Zk X

(ξik − 1).

i=1

Consider a sequence of branching processes with Z0n = n and define Xn (t) =

1 n Z . n [nt]


182

Derivation of limiting equation n

[nt]−1 Zk 1 X n 1 X X n Xn (t) = 1 + (Zk+1 − Zk ) = 1 + (ξik − 1) n n i=1 [nt]−1

k=0

k=0

Define Wn (t) = Claim: Wn and Wn (t)2 −

[nt] X

Zn

k X 1 (ξik − 1) √ p n σ n Zk i=1

k=1 [nt] n are

Z

martingales, and t

σ

Xn (t) = 1 +

p

Xn (s−)dWn (s)

0

suggesting that Xn converges to a solution of Z t p X(t) = 1 + σ X(s)dW (s) 0


183

Value of a portfolio Suppose that a market consists of a stock with price given by Z t Z t S(t) = S(0) + σS(s)dW (s) + αS(s)ds 0

0

and a money market account paying continuously compounded interest at rate r. ∆(t) denotes the number of shares of stock held by an investor at time t with the remainder of the investors wealth, X(t) − ∆(t)S(t) in the money market. The portfolio value satisfies Z t Z t X(t) = X(0) + ∆(s)dS(s) + r(X(s) − ∆(s)S(s))ds 0 0 Z t Z t (rX(s) + (α − r)∆(s)S(s))ds = X(0) + σ∆(s)S(s)dW (s) + 0

0


184

Hedging a contract Suppose a bank sells a contract that pays, at time T , an amount based on the the stock price at time T . That is, the bank pays h(S(T )). The contract can be hedged if there exists a portfolio whose value X at time T equals h(S(T )). We want an initial value X(0) and an investment strategy ∆ such that X(T ) = h(S(T )).


185

Derivation of the hedging strategy Suppose X(t) = q(t, S(t)). Then q(t, S(t)) Z

t

= q(0, S(0)) + qx (s, S(s))σS(s)dW (s) 0 Z t σ2 + (qx (s, S(s))αS(s) + qxx (s, S(s))S 2 (s) + qs (s, S(s)))ds 2 0 Comparing this identity to Z t Z t X(t) = X(0) + ∆(s)dS(s) + r(X(s) − ∆(s)S(s))ds 0 0 Z t Z t σ∆(s)S(s)dW (s) + (rX(s) + (α − r)∆(s)S(s))ds = X(0) + 0

0


186

A PDE for q Matching coefficients gives ∆(s) = qx (s, S(s)) σ2 qx (s, S(s))αS(s) + qxx (s, S(s))S 2 (s) + qs (s, S(s)) 2 = rq(s, S(s)) + (α − r)qx (s, S(s))S(s). and hence qx (s, S(s))rS(s) +

σ2 qxx (s, S(s))S 2 (s) + qs (s, S(s)) 2 = rq(s, S(s)).


187

Pricing contracts using the Black-Scholes model We want q to satisfy the partial differential equation 1 qs (s, x) + rxqx (s, x) + σ 2 x2 qxx (s, x) = rq(s, x) 2 with terminal condition q(T, x) = h(x). Then the price of the contract should be X(0) = q(0, S(0)). Holding ∆(s) = qx (s, S(s)) shares of stock at each time s and the remainder in the money market, the value of the portfolio at time T is q(T, S(T )) = h(S(T )).


188

10.

Multidimensional stochastic calculus

• Multidimensional Brownian motion • Itoˆ processes • Covariation • Covariation for Itoˆ processes • Multidimensional Itoˆ formula


189

Multidimensional Brownian motion Definition 10.1 W = (W1 , . . . , Wd ) is a d-dimensional standard {Ft }Brownian motion if each W1 , . . . , Wd are independent (1-dimensional) standard Brownian motions adapted to {Ft } and for each t ≥ 0, W (t+·)−W (t) is independent of Ft . We will view W as a column vector.


190

Stochastic integration with respect to W Let Mm×d denote the collection m × d matrices. Let X be Mm×d valued, cadlag and adapted. Then Z t X X(s)dW (s) = lim X(ti )(W (ti+1 ∧ t) − W (ti ∧ t)). {ti }→∗

0

Note that Z(t) =

Rt 0

X(s)dW (s) ∈ Rm and that Zi (t) =

d Z X j=1

t

Xij (s)dWj (s).

0


191

Itoˆ processes Definition 10.2 If X is Mm×d -valued, cadlag and adapted and U is Rm valued, cadlag and adapted, then Z t Z t Z(t) = X(s)dW (s) + U (s)ds 0

0

is an Itoˆ process.


192

Covariation Definition 10.3 The covariation of Y1 and Y2 is X [Y1 , Y2 ]t = lim (Y1 (ti+1 ∧ t) − Y1 (ti ∧ t))(Y2 (ti+1 ∧ t) − Y2 (ti ∧ t)) {ti }→∗

Lemma 10.4 Z [Y1 , Y2 ]t = Y1 (t)Y2 (t) − Y1 (0)Y2 (0) −

t

Z Y1 (s)dY2 (s) −

0

t

Y2 (s)dY1 (s) 0

if the two stochastic integrals exist. Lemma 10.5 The covariation is bilinear in the sense that [aY1 + bY2 , cZ1 + dZ2 ]t = ac[Y1 , Z1 ]t + ad[Y1 , Z2 ]t + bc[Y2 , Z1 ]t + bc[Y2 , Z2 ]t


193

Covariation for Itoˆ processes To compute the covariations for an Itoˆ process, it is enough to know the following: Lemma 10.6 If Z

t

Y1 (t) =

Z

t

X1 (s)dW1 (s) +

U1 (s)ds

0

Z

0 t

Y2 (t) =

Z

t

X2 (s)dW2 (s) + Z0 t

Y3 (t) =

U2 (s)ds Z0 t

X2 (s)dW1 (s) +

U2 (s)ds

0

0

then

Z [Y1 , Y2 ]t = 0

[Y1 , Y3 ]t =

t

X1 (s)X2 (s)ds 0


194

Covariation for Itoˆ processes Lemma 10.7 If Y1 and Y2 are R-valued Itô processes with [Y1 , Y2 ]t = X1 and X2 are cadlag and adapted, and Z t Z t Z1 (t) = X1 (s)dY1 (s) Z2 (t) = X2 (s)dY2 (s), 0

Rt 0

R(s)ds,

0

then Z [Z1 , Z2 ]t =

t

Z X1 (s)X2 (s)d[Y1 , Y2 ]s =

0

t

X1 (s)X2 (s)R(s)ds 0


195

Matrix notation for covariation Let Z(t) =

Rt

0 X(s)dW (s) +

Rt

[Zi , Zj ]t =

0

U (s)ds. Then

d Z X k=1

t

Xik (s)Xjk (s)ds,

0

or thinking of the covariation as defining a matrix [Z]t = (([Zi , Zj ]t )), Z t [Z]t = X(s)X(s)T ds 0

where M T denotes the transpose of the matrix M .


196

Multidimensional Itoˆ formula Theorem 10.8 Let Z be an m-dimensional Itô process Z t Z t Z(t) = Z(0) + X(s)dW (s) + U (s)ds 0

0

and let f ∈ C 2 (Rm ). Define the Hessian matrix Hf (x) = ((∂i ∂j f (x) )). Then Z t Z 1X t T f (Z(t)) = f (Z(0)) + ∇f (Z(s)) dZ(s) + ∂i ∂j f (Z(s))d[Zi , Zj ]s 2 i,j 0 0 Z t Z t ∇f (Z(s))T X(s)dW (s) + ∇f (Z(s))T U (s)ds = f (Z(0)) + 0 0 Z t 1 + trace(X(s)T Hf (Z(s))X(s))ds 2 0


197

Examples t

Z

1 ∇f (W (s)) dW (s) + 2 T

f (W (t)) = f (0) + 0

Z

t

∆f (W (s))ds 0

where ∆ is the Laplacian ∆f (x) =

d X

∂k2 f (x)

k=1 k X

X(t) = X(0) exp{

k=1

d

1X 2 σk )t} σk Wk (t) + (µ − 2 k=1

satisfies the stochastic differential equation Z t Z t T µX(s)ds X(t) = X(0) + X(s)σ dW (s) + 0

0


198

Diffusion equations If X satisfies Z

t

Z σ(X(s))dW (s) +

X(t) = X(0) + 0

t

b(X(s))ds, 0

then Z

T

Z

t

∇f (X(s)) σ(X(s))dW (s) +

f (X(t)) = f (X(0)) + 0

where

t

Lf (X(s))ds 0

1X aij (x)∂i ∂j f (x) + b(x) · ∇f (x) Lf (x) = 2 i,j

and a(x) = σ(x)σ(x)T .


199

11.

Stochastic differential equations

• Brownian bridge • Convergence of empirical distribution functions • Kolmogorov-Smirnov test • Conditioned Brownian motion • An SDE for Brownian bridge • Stochastic differential equations • Moment estimates • Gronwall inequality • Schwarz inequality • Lipschitz conditions for existence and uniqueness •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

200

Brownian bridge Recall, that if (Y1 , . . . , Yl , X1 , . . . , Xm ) is jointly Gaussian and Cov(Yi , Xj ) = 0 of i = 1, . . . , l and j = 1, . . . , m, then Y = (Y1 , . . . , Yl ) and X = (X1 , . . . , Xm ) are independent. Let W be a scalar, standard Brownian motion. By Problem 6, W (1) is independent of B defined by B(t) = W (t) − tW (1), 0 ≤ t ≤ 1. Note that Cov(B(t), B(s)) = s ∧ t − ts. Lemma 11.1 σ(B(t) : 0 ≤ t ≤ 1) is independent of σ(W (1)). Proof. Since E[B(t)W (1)] = 0, (B(t1 ), . . . , B(tm )) is independent of W (1) for all choices of t1 , . . . , tm ∈ [0, 1]. Independence of the σalgebras follows by a standard Math 831 argument.


201

Convergence of empirical distribution functions Let ξ1 , ξ2 , . . . be iid uniform [0, 1] random variables, and define the empirical distribution function n

1X Fn (t) = 1(−∞,t] (ξi ). n i=1 E[Fn (t)] = Fξ (t) and Cov(Fn (t), Fn (s)) =

1 (t ∧ s − ts) n

√ Define Bn (t) = n(Fn (t) − Fξ (t)), 0 ≤ t ≤ 1, then (Bn (t1 ), . . . , Bn (tm )) converges in distribution to (B(t1 ), . . . , B(tm )), t1 , . . . , tm ∈ [0, 1].


202

A class of continuous functions Definition 11.2 Let D[0, 1] denote the cadlag, real-valued functions on u [0, 1]. Let CD be the collection of functions g : D[0, 1] → R such that if xn ∈ D[0, 1] and x ∈ C[0, 1] satisfy sup0≤t≤1 |xn (t) − x(t)| → 0, then limn→∞ g(xn ) = g(x). For example, g(x) = sup x(t) 0≤t≤1

Z g(x) =

1

x(s)ds 0


203

Functional limit theorem u Theorem 11.3 For each g ∈ CD , g(Bn ) converges in distribution to g(B).

In particular, lim P { sup

n→∞

0≤t≤1

√

n(Fn (t) − Fξ (t)) ≤ x} = P { sup B(t) ≤ x} 0≤t≤1

and √ lim P { sup | n(Fn (t) − Fξ (t))| ≤ x} = P { sup |B(t)| ≤ x}

n→∞

0≤t≤1

0≤t≤1


204

More general empirical distributions Let X1 , X2 , . . . be iid with continuous distribution function FX . DeP n 1 fine FnX (x) = n i=1 1(−∞,x] (Xi ) and √ Dn (x) = n(FnX (x) − FX (x)). Since

n

FnX (x)

1X = 1(−∞,FX (x)] (FX (Xi )) a.s. n i=1

and FX (Xi ) is uniform [0, 1], Dn has the same distribution as Bn ◦ FX . In particular, supx Dn (x) (supx |Dn (x)|) has the same distribution as sup0≤t≤1 Bn (t) (sup0≤t≤1 |Bn (t)|).


205

Kolmogorov-Smirnov test Corollary 11.4 lim P {sup

n→∞

z

√

n(FnX (z) − FX (z)) ≤ x} = P { sup B(t) ≤ x} 0≤t≤1

and √ lim P {sup | n(FnX (z) − FX (z))| ≤ x} = P { sup |B(t)| ≤ x}

n→∞

z

0≤t≤1

√ Consequently, supz | n(FnX (z) − FX (z))| provides a test statistic to test the hypothesis that FX is the cdf for {Xi }.


206

Conditioned Brownian motion Note that E[g( sup W (t))|W (1)] = E[g( sup (B(t) + tW (1)))|W (1)], 0≤t≤1

0≤t≤1

and since W (1) is independent of B, if we define h(y) = E[g( sup (B(t) + ty))], 0≤t≤1

we have E[g( sup W (t))|W (1)] = h(W (1)) 0≤t≤1

The same analysis works for any function of W restricted to [0, 1], and we say that the conditional distribution of W given that W (1) = b is the distribution of Bb defined by Bb (t) = B(t) + tb.


207

Linear transformations of W Lemma 11.5 Let K : {(t, u) : 0 ≤ u ≤ t} → ∞ satisfy ∞, for all t ≥ 0, and define Z t Z(t) = K(t, u)dW (u).

Rt 0

K(t, u)2 du
0 such that |σ(x)| + |b(x)| ≤ K(1 + |x|) and that E[|X(0)|m ] < ∞. ˆ formula, Proof. Applying Ito’s Z t |X(t)|m = |X(0)|m + m|X(s)|m−2 X(s)T σ(X(s))dW (s) Z t 0 1 + m|X(s)|m−2 (X(s) · b(X(s)) + tracea(X(s)))ds 2 0 Z t 1 + m(m − 2)|X(s)|m−4 X(s)T a(X(s))X(s)ds 2 0


213

b > 0 such that Let τc = inf{t : |X(t)| ≥ c}. Then there exists K Z t∧τc m m b + |X(s)|m )ds] E[|X(t ∧ τc )| ] ≤ E[|X(0)| ] + E[ K(1 0 Z t b b + E[ K|X(s ∧ τc )|m ds] ≤ E[|X(0)|m ] + Kt 0

which by Gronwall’s inequality and Fatou’s lemma implies b Kt . E[|X(t)|m ] ≤ lim inf E[|X(t ∧ τc )|m ] ≤ (E[|X(0)|m ] + Kt)e b

c→∞


214

Gronwall’s inequality Lemma 11.8 Suppose that A be continuous and nondecreasing, X is cadlag, and that Z t (11.2) 0 ≤ X(t) ≤ + X(s)dA(s) . 0

Then X(t) ≤ eA(t) .


215

Proof. Iterating (11.2), we have Z t X(t) ≤ + X(s)dA(s) 0 Z tZ s ≤ + A(t) + X(u)dA(u)dA(s) 0 0 Z t Z tZ sZ ≤ + A(t) + A(s)dA(s) + 0

0

Checking that Z t 1 A(s)dA(s) = A(t)2 , 2 0 and in general Z t Z t1

Z

0

0

0

s

X(v)dA(v)dA(u)dA(s).

0

1 A(u)dA(u)dA(s) = A(t)3 6

tn−2

A(tn−1 )dA(tn−1 ) · · · dA(t1 ) =

... 0

Z tZ

0

u

0

we see that X(t) ≤ eA(t) .

1 A(t)n , n!


216

Schwarz inequality Lemma 11.9 Suppose E[X 2 ], E[Y 2 ] < ∞. Then p E[XY ] ≤ E[X 2 ]E[Y 2 ]. Proof. Note that 2ab ≤ z −2 a2 + z 2 b2 , so 2E[XY ] ≤ z −2 E[X 2 ] + z 2 E[Y 2 ]. Minimizing the right side with respect to z gives the desired inequality. The same argument gives E[X · Y ] ≤

p

E[|X|2 ]E[|Y |2 ]

for random vectors. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

217

Lipschitz conditions for existence and uniqueness Theorem 11.10 Suppose that |σ(x) − σ(y)| + |b(x) − b(y)| ≤ K|x − y|. Then there exists a unique solution of Z t Z t (11.3) X(t) = X(0) + σ(X(s))dW (s) + b(X(s))ds. 0

0


218

e are solutions. Then by Ito’s ˆ Proof.[of uniqueness] Suppose X and X formula Z t T e 2 = e e |X(t) − X(t)| 2(X(s) − X(s)) (σ(X(s)) − σ(X(s)))dW (s) 0 Z t e e + 2(X(s) − X(s)) · (b(X(s)) − b(X(s)))ds Z0 t T e e + trace(σ(X(s)) − σ(X(s)))(σ(X(s)) − σ(X(s))) d 0

e As in the moment estimate, let τc = inf{t : |X(t) − X(t)| ≥ c}. Then b such that there exists K Z t 2 e ∧ τc )| ] ≤ b e ∧ τc )|2 ]ds, E[|X(t ∧ τc ) − X(t KE[|X(s ∧ τc ) − X(s 0

e ∧ τc )|2 ] = 0. which implies E[|X(t ∧ τc ) − X(t


219

Proof.[of existence] Assume, in addition, that σ and b are bounded. For h > 0, define ηh (t) = h[h−1 t], and let Xh satisfy Z ηh (t) Z ηh (t) Xh (t) = X(0) + σ(Xh (s))dW (s) + b(Xh (s))ds. 0

0

Note that Xh is constant on the interval [kh, (k + 1)h), and that Xh ((k+1)h) = Xh (kh)+σ(Xh (kh))W ((k+1)h)−W (kh))+b(Xh (kh))h. (known as the Euler-Maruyama scheme in the SDE literature). Define Z

t

Dh (t) =

Z

t

σ(Xh (s))dW (s) + ηh (t)

b(Xh (s))ds, ηh (t)

so Z Xh (t) + Dh (t) = X(0) +

t

Z σ(Xh (s))dW (s) +

0

t

b(Xh (s))ds. 0


220

b such that Then there exists K Z

2

b E[|Xh (t) + Dh (t) − Xh0 (t) − Dh0 (t)| ] ≤ K

t

E[|Xh (s) − Xh0 (s)|2 ]ds,

0

so E[|Xh (t) − Xh0 (t)|2 ] ≤ −2E[(Xh (t) − Xh0 (t)) · (Dh (t) − Dh0 (t))] Z t b +K E[|Xh (s) − Xh0 (s)|2 ]ds 0 p p ≤ 2 E[|Xh (t) − Xh0 (t)|2 ] E[|Dh (t) − Dh0 (t)|2 ] Z t b +K E[|Xh (s) − Xh0 (s)|2 ]ds 0

and hence p p b E[|Xh (t)−Xh0 (t)|2 ] ≤ 2 E[|Xh (t) − Xh0 (t)|2 ] E[|Dh (t) − Dh0 (t)|2 ]eKt . Rt But E[|Dh (t)|2 ] ≤ 2(E[ ηh (t) tracea(Xh (s))ds] + h2 supx |b(x)|2 ]) → 0 and {Xh (t), h > 0} is Cauchy in L2 as h → 0. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

221

Uniform moment bounds Suppose 2xb(x) + σ 2 (x) ≤ K1 − ε|x|2 . (For example, consider the Rt equation X(t) = X(0) − 0 αX(s)ds + W (t).) Then εt

2

e |X(t)|

Z

t

eεs 2X(s)σ(X(s))dW (s) ≤ |X(0)| + Z t 0 Z t εs 2 + e [2X(s)b(X(s)) + σ (X(s))]ds + εeεs |X(s)|2 ds 0 0 Z t Z t ≤ |X(0)|2 + eεs 2X(s)σ(X(s))dW (s) + eεs K1 ds 0 Z0 t K 1 εt ≤ |X(0)|2 + eεs 2X(s)σ(X(s))dW (s) + (e − 1), 2 0 2

and hence eεt E[|X(t)|2 ] ≤ E[|X(0)|2 ] +

K1 εt [e − 1]. ε

Therefore, we have the uniform bound E[|X(t)|2 ]] ≤ e−εt E[|X(0)|2 ] +

K1 (1 − e−εt ). ε


222

Vector case In the vector case Z X(t) = X(0) +

t

Z σ(X(s))dW (s) +

0

2

|X(t)|

Z

2

t

t

b(X(s))ds, 0

T

= |X(0)| + 2X(s) σ(X(s))dW (s) Z t 0 T + (2X(s) · b(X(s)) + trace(σ(X(s))σ(X(s)) ))ds . 0

As in the univariate case, if we assume, T

2x · b(x) + trace(σ(x)σ(x) ) ≤ K1 − ε|x|2 , then E[|X(s)|2 ] is uniformly bounded.


223

12.

Markov property

• Definition • Solution of SDE as a function of the initial position • Flow property • Markov property • Strong Markov property • Differentiability with respect to initial position


224

Definition Recall the definition of the Markov property: Definition 12.1 If X is a stochastic process adapted to a filtration {Ft }, then X is {Ft }-Markov (or Markov with respect to {Ft }) if and only if for s, t ≥ 0, E[f (X(t + s))|Ft ] = E[f (X(t + s))|X(t)].


225

Solution of SDE as a function of the initial position For t ≥ r, let Z

r+t

Z σ(Xr (s − r, x))dW (s) +

Xr (t, x) = x + r

r+t

b(Xr (s − r.x))ds. r

Define Wr (t) = W (r + t) − W (r). Then Z t Z t Xr (t, x) = x + σ(Xr (s, x))dWr (s) + b(Xr (s.x))ds. 0

(12.1)

(12.2)

0

Assuming that uniqueness holds, since the distribution of Wr does not depend on r, the distribution of Xr does not depend on r. In particular, for f ∈ B(Rm ), the bounded, Borel-measurable functions on Rm , T (t)f (x) ≡ E[f (Xr (t, x))] does not depend or r. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

226

Flow property Write X(t, x) = X0 (t, x). Theorem 12.2 Let σ and b be such that solutions of the SDE are unique for all choices of the initial condition (for example, if σ and b are Lipschitz). Then X(r + t, x) = Xr (t, X(r, x)). Proof. By subtraction Z

r+t

X(r + t, x) = X(r, x) +

Z

r+t

b(X(s, x))ds(12.3)

σ(X(s, x))dW (s) + Zr t

r

Z σ(X(r + s, x))dWr (s) +

= X(r, x) + 0

t

b(X(r + s, x))ds. 0

Comparing this equation to (12.2), the proof becomes “clear”, although there are a number of technicalities. For example, we need to know that Xr (t, x) is at least a measurable function of x. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

227

Corollary 12.3 Under the assumptions of the theorem, if X(0) is independent of W and X satisfies (11.1), then X(r + t) = X(r + t, X(0)) = Xr (t, X(r)).


228

Markov property Theorem 12.4 Let σ and b be such that solutions of the SDE are unique for all choices of the initial condition (for example, if σ and b are Lipschitz). If X is a solution of (11.1), then X is Markov. Proof. By uniqueness, X(r) is independent of Wr and hence of Xr (t, x). Consequently, for f ∈ B(Rm ), E[f (X(r + t))|Fr ] = E[f (Xr (t, X(r))|Fr ] = T (t)f (X(r)).


229

Strong Markov property Theorem 12.5 Let σ and b be such that solutions of the SDE are unique for all choices of the initial condition (for example, if σ and b are Lipschitz). If X is a solution of (11.1), then for each finite {Ft }-stopping time τ and each f ∈ B(Rm ), E[f (X(τ + t))|Fτ ] = T (t)f (X(τ )). Proof. By the strong Markov property for Brownian motion, Wτ defined by Wτ (t) = W (τ + t) − W (τ ) is a Brownian motion independent of X(τ ), and Xτ satisfying Z t Z t Xτ (t, x) = x + σ(Xτ (s, x))dWτ (s) + b(Xτ (s.x))ds. (12.4) 0

0

has the same distribution as Xr (t, x). As with the Markov property, E[f (X(τ + t))|Fτ ] = E[f (Xτ (t, X(τ ))|Fr ] = T (t)f (X(τ )). •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

230

Continuity with respect to the initial condition Theorem 12.6 Let k ≥ 2. Under Lipschitz conditions, there exists a constant Mk such that E[|X(t, x) − X(t, y)|k ] ≤ |x − y|k eMk t . Proof. Let Zx,y (t) = X(t, x) − X(t, y). As in the proof of uniqueness k

|Zx,y (t)|

Z

k

t

= |x − y| + k|Zx,y (s)|k−2 (Zx,y (s))T (σ(X(s, x)) − X(s, y))dW (s) Z t 0 + x|Zx,y (s)|k−2 (Zx,y (s) · (b(X(s, x)) − b(X(s, y)) 0

1 + trace(σ(X(s, x)) − σ(X(s, y))(σ(X(s, x) − X(s, y))T )ds 2 1 + 2

Z

t

k(k − 2)|Zx,y (s)|k−4 |σ(X(s, x)) − σ(X(s, y)))T Zx,y (s)|2 ds

0


231

Differentiability with respect to initial position Assume d = m = 1. X(t, x) − X(t, y) x−y Z t σ(X(s, x)) − σ(X(s, y)) = 1+ ∆x,y (s)dW (s) X(s, x) − X(s, y) 0 Z t b(X(s, x)) − b(X(s, y)) + ∆x,y (s)ds X(s, x) − X(s, y) 0

∆x,y (t) ≡

and Z ∂x X(t, x) = 1 +

t

σ 0 (X(s, x))∂x X(s, x)dW (s) 0 Z t + b0 (X(s, x))∂x X(s, x)ds 0


232

13.

SDEs and partial differential equations

• Generator for a diffusion process • One dimensional exit distributions • Exit times • Dirichlet problems • Parabolic equations • Feynman-Kac formula • Pricing derivatives


233

Differential operators and diffusion processes Consider Z

t

Z σ(X(s))dW (s) +

X(t) = X(0) + 0

t

b(X(s))ds, 0

where X is Rd -valued, W is an m-dimensional standard Brownian motion, σ is a d × m matrix-valued function and b is an Rd -valued function. For a C 2 function f , d Z t X f (X(t)) = f (X(0)) + ∂i f (X(s))dX(s) i=1

0

Z t 1 X ∂i ∂j f (X(s))d[Xi , Xj ]s . + 2 0 1≤i,j≤d


234

Computation of covariation The covariation satisfies Z tX Z t σi,k (X(s))σj,k (X(s))ds = ai,j (X(s))ds, [Xi , Xj ]t = 0

0

k

where a = ((ai,j )) = σ · σ T , that is ai,j (x) =

P

k

σik (x)σkj (x).


235

Definition of the generator Let Lf (x) =

d X

bi (x)∂i f (x) +

i=1

1X ai,j (x)∂i ∂j f (x), 2 i,j

then Z

t T

Z

∇f (X(s))σ(X(s))dW (s) +

f (X(t)) = f (X(0)) + 0

t

Lf (X(s))ds . 0

Since a = σ · σT , X

ξi ξj ai,j = ξ T σσ T ξ = |σ T ξ|2 ≥ 0,

a is nonnegative definite, and L is an elliptic differential operator. L is called the generator for the corresponding diffusion process.


236

The generator for Brownian motion If X(t) = X(0) + W (t), then ((ai,j (x))) = I, and Lf (x) = 21 ∆f (x).


237

Exit distributions in one dimension If d = m = 1 and a(x) = σ 2 (x), then 1 Lf (x) = a(x)f 00 (x) + b(x)f 0 (x) 2 Suppose Lf (x) = 0 Then Z t f (X(t)) = f (X(0)) + f 0 (X(s))σ(X(s))dW (s). 0

Fix α < β, and define τ = inf{t : X(t) ∈ / (α, β)}. If supα0} ds) Xn (t) = Xn (0) + √ Ya (nλn t) − √ Yd (nµn n n 0 Z t √ √ + n(λn − µn )t + nµn 1{Xn (s)=0} ds 0

If λn → λ and µn → µ, then √ 1 Wan (t) ≡ √ Yea (nλn t) ⇒ λW1 (t) n 1 √ Wdn (t) ≡ √ Yed (nµn t) ⇒ µW2 (t)), n where W1 and W2 are standard Brownian motions.


322

Scaled equation Defining √ n(λn − µn ) Z t Bn (t) ≡ 1{Xn (s)>0} ds cn ≡

0

Λn (t) ≡

√

nµn (t − Bn (t)) =

√

Z nµn

t

1{Xn (s)=0} ds, 0

we can rewrite Xn (t) as Xn (t) = Xn (0) + Wan (t) − Wdn (Bn (t)) + cn t + Λn (t). Λn is nondecreasing and increases only when Xn is zero.


323

Skorohod problem Lemma 15.11 For w ∈ DR [0, ∞) with w(0) ≥ 0, there exists a unique pair (x, λ) satisfying x(t) = w(t) + λ(t) (15.1) such that λ(0) = 0, x(t) ≥ 0∀t, and λ is nondecreasing and increases only when x = 0. The solution is given by setting λ(t) = 0 ∨ sups≤t (−w(s)) and defining x by (15.1).


324

Proof.[of uniqueness] For t < τ0 = inf{s : w(s) ≤ 0}, λ(t) = 0 and hence x(t) = w(t). For t ≥ τ0 , the nonnegativity of x implies λ(t) ≥ −w(t), and λ(t) nondecreasing implies λ(t) ≥ sup(−w(s)). s≤t

If t is a point of increase of λ, then x(t) = 0, so we must have λ(t) = −w(t) ≤ sup(−w(s)).

(15.2)

s≤t

Since the right side of (15.2) is nondecreasing, we must have λ(t) ≤ sups≤t (−w(s)) for all t > τ0 , and the result follows.


325

Convergence to reflecting Brownian motion Λn (t) = 0 ∨ (− inf (Xn (0) + Wan (s) − Wdn (Bn (s)) + cn s)) s≤t

Consequently, if Xn (0) + Wan (t) − Wdn (Bn (t)) + cn t converges, so does Λn and Xn along with it. Assuming that cn → c, the limit will satisfy √ √ X(t) = X(0) + λW1 (t) − λW2 (t) + ct + Λ(t) √ Λ(t) = 0 ∨ sup(−(X(0) + λ(W1 (s) − W2 (s)) + ct)). s≤t

Defining

√1 (W1 2

− W2 ), √

X(t) = X(0) + X(t) ≥ 0 ∀t,

2λW (t) + ct + Λ(t)

where Λ is nondecreasing and Λ increases only when X(t) = 0. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

326

16.

Stationary distributions and forward equations

• Forward equations • Stationary distributions • Reflecting diffusions


327

Equations for probability distributions If

Z

t

Z

t

b(X(s))ds,

(16.1)

a = σσ T , f ∈ Cc2 (Rd ), and 1X aij (x)∂i ∂j f (s) + b(x) · ∇f (x), Lf (x) = 2 Z t f (X(t)) − Lf (X(s)) ds

(16.2)

X(t) = X(0) +

σ(X(s))dW (s) + 0

0

0

is a martingale.


328

Forward equation Since Z

t

E [f (X(t))] = E [f (X(0)] + E Lf (X(s))ds 0 Z t = E [f (X(0))] + E [Lf (X(s))] ds, 0

defining νt (Γ) = P {X(t) ∈ Γ}, for f ∈ Cc2 (Rd ), Z Z Z tZ f dνt = f dν0 + Lf dνs ds,

(16.3)

0

which is a weak form of the equation d νt = L∗ νt . dt


329

Uniqueness for the forward equation Theorem 16.1 Let Lf be given by (16.2) with a and b continuous, and let {νt } be probability measures on Rd satisfying (16.3) for all f ∈ Cc2 (Rd ). If (16.1) has a unique solution for each initial condition, then P {X(0) ∈ ·} = ν0 implies P {X(t) ∈ ·} = νt .


330

The adjoint operator In nice situations, νt (dx) = pt (x)dx. Then L∗ should be a differential operator satisfying Z Z pLf dx = f L∗ pdx. Rd

Rd


331

Example 16.2 Let d=1. Integrating by parts, we have Z ∞ 1 p(x) a(x)f 00 (x) + b(x)f 0 (x) dx 2 −∞ ∞ Z ∞ 1 1 d = p(x)a(x)f 0 (x) − f 0 (x) (a(x)p(x)) − b(x)p(x) dx. 2 2 dx −∞ −∞ The first term is zero, and integrating by parts again we have Z ∞ d 1 d (a(x)p(x)) − b(x)p(x) dx f (x) dx 2 dx −∞ so

d L p= dx ∗

1 d (a(x)p(x)) − b(x)p(x) . 2 dx

Example 16.3 Let Lf = 12 f 00 (Brownian motion). Then L∗ p = 12 p00 , that is, L is self adjoint.


332

Stationary distributions Suppose

R

Lf dπ = 0 for all f in Cc2 (Rd ). Then Z Z Z tZ f dπ = f dπ + Lf dπds, 0

and hence νt ≡ π gives a solution of (16.3). Under the R t conditions of Theorem 16.1, if P {X(0) ∈ ·} = π and f (X(t)) − 0 Lf (X(s))ds is a martingale for all f ∈ Cc2 (Rd ), then P {X(t)} ∈ ·} = π, i.e. π is a stationary distribution for X.


333

Computation of a stationary distribution if d = 1 Assuming π(dx) = π(x)dx d 1 d (a(x)π(x)) − b(x)π(x) = 0, dx 2 dx {z } | this is a constant: let the constant be 0 so 1 d (a(x)π(x)) = b(x)π(x). 2 dx Rx Applying the integrating factor exp(− 0 2b(z)/a(z)dz) Rx 1 − R0x 2b(z) dz d − 0 2b(z) a(z) a(z) dz π(x) = 0 e (a(x)π(x)) − b(x)e 2 dx −

a(x)e

Rx 0

π(x) =

2b(z) a(z) dz

π(x) = C

C R0x 2b(z) e a(z) dz . a(x)


334

Existence of a stationary distribution Lemma 16.4 Assume a(x) > 0 for all x. If Z ∞ 1 R0x 2b(z) e a(z) dz dx < ∞, C= −∞ a(x) then

R x 2b(z) 1 π(x) = e 0 a(z) dz Ca(x) is a stationary density for (16.1).


335

Diffusion with a boundary Suppose Z X(t) = X(0) +

t

Z σ(X(s))dW (s) +

0

t

b(X(s))ds + Λ(t) 0

with X(t) ≥ 0, and that Λ is nondecreasing and increasing only when X(t) = 0. Then Z t Z t T f (X(t)) = f (X(0)) + ∇f (X(s))σ(X(s))dW (s) + Lf (X(s))ds 0 0 Z t + f 0 (X(s))dΛ(s), 0

so

Z f (X(t)) −

t

Lf (X(s))ds 0

is a martingale, if f ∈ Cc2 and f 0 (0) = 0. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

336

Adjoint operator for reflecting process Z 0

∞

∞ Z ∞ 1 1 d 0 0 p(x)Lf (x)dx = f p(x)a(x)f (x) − (a(x)p(x)) − b(x)p(x) dx 2 2 dx 0 0 {z } | =0 ∞ Z ∞ 1 d (a(x)p(x)) − b(x)p(x) + f (x)L∗ p(x)dx = −f (x) 2 dx 0 0

and hence d L∗ p(x) = dx

1 d (a(x)p(x)) − b(x)p(x) 2 dx

for p satisfying

1 0 1 a (0) − b(0) p(0) + a(0)p0 (0) = 0 . 2 2

(16.4)


337

Forward equation The density for the distribution of the process should satisfy d p t = L∗ p t dt and the boundary condition (16.4). The stationary density satisfies d 1 d (a(x)π(x)) − b(x)π(x) = 0 dx 2 dx and the boundary condition implies 1 d (a(x)π(x)) − b(x)π(x) = 0 2 dx giving π(x) =

c R0x 2b(z) e a(z) dz , x ≥ 0. a(x)


338

Stationary distribution for reflecting Brownian motion Example 16.5 Let X(t) = X(0) + σW (t) − bt + Λ(t), where a = σ 2 and b > 0 are constant. Then 2b 2b π(x) = 2 e− σ2 x , σ so the stationary distribution is exponential.


339

17.

Processes with jumps

• Poisson process • Martingale properties of the Poisson process • Strong Markov property for the Poisson process • Marked and compound Poisson processes • Stochastic equations with jumps • General counting processes • Intensities


340

Poisson process A Poisson process is a model for a series of random observations occurring in time. For example, the process could model the arrivals of customers in a bank, the arrivals of telephone calls at a switch, or the counts registered by radiation detection equipment. x x

x

x

x

x

x

x

t Let N (t) denote the number of observations by time t. In the figure above, N (t) = 6. Note that for t < s, N (s) − N (t) is the number of observations in the time interval (t, s].


341

Basic assumptions 1) Observations occur one at a time. 2) Numbers of observations in disjoint time intervals are independent random variables, i.e., if t0 < t1 < · · · < tm , then N (tk ) − N (tk−1 ), k = 1, . . . , m are independent random variables. 3) The distribution of N (t + a) − N (t) does not depend on t.


342

Characterization of a Poisson process Theorem 17.1 Under assumptions 1), 2), and 3), there is a constant λ > 0 such that, for t < s, N (s) − N (t) is Poisson distributed with parameter λ(s − t), that is, (λ(s − t))k −λ(s−t) e . P {N (s) − N (t) = k} = k! Proof. Let Nn (t) be the number of time intervals ( nk , k+1 n ], k = 0, . . . , [nt] that contain at least one observation. Then Nn (t) is binomially distributed with parameters n and pn = P {N ( n1 ) > 0}. Then (1 − pn )n = P {Nn (1) = 0} = P {N (1) = 0} and npn → λ ≡ − log P {N (1) = 0}, and the rest follows by standard Poisson approximation of the binomial.


343

Interarrival times Let Sk be the time of the kth observation. Then P {Sk ≤ t} = P {N (t) ≥ k} = 1 −

k−1 X (λt)i i=0

i!

e−λt ,

t ≥ 0.

Differentiating to obtain the probability density function gives 1 k−1 −λt e t≥0 (k−1)! λ(λt) fSk (t) = 0 t 1, Tk = Sk −Sk−1 . Then T1 , T2 , . . . are independent and exponentially distributed with parameter λ.


344

Martingale properties of the Poisson process Theorem 17.3 (Watanabe) If N is a Poisson process with parameter λ, then N (t) − λt is a martingale. Conversely, if N is a counting process and N (t) − λt is a martingale, then N is a Poisson process.


345

Proof. E[eiθ(N (t+r)−N (t)) |Ft ] n−1 X =1+ E[(eiθ(N (sk+1 )−N (sk ) − 1 − (eiθ − 1)(N (sk+1 ) − N (sk ))eiθ(N (sk )−N (t)) |Ft ] k=0 n−1 X

+

λ(sk+1 − sk )(eiθ − 1)E[eiθ(N (sk )−N (t)) |Ft ]

k=0

The first term converges to zero by the dominated convergence theorem, so we have Z r iθ(N (t+r)−N (t)) iθ E[e |Ft ] = 1 + λ(e − 1) E[eiθ(N (t+s)−N (t)) |Ft ]ds 0 iθ

and E[eiθ(N (t+r)−N (t)) |Ft ] = eλ(e

−1)t

.


346

Strong Markov property A Poisson process N is compatible with a filtration {Ft }, if N is {Ft }adapted and N (t + ·) − N (t) is independent of Ft for every t ≥ 0. Lemma 17.4 Let N be a Poisson process with parameter λ > 0 that is compatible with {Ft }, and let τ be a {Ft }-stopping time such that τ < ∞ a.s. Define Nτ (t) = N (τ + t) − N (τ ). Then Nτ is a Poisson process that is independent of Fτ and compatible with {Fτ +t }.


347

Proof. Let M (t) = N (t) − λt. By the optional sampling theorem, E[M ((τ + t + r) ∧ T )|Fτ +t ] = M ((τ + t) ∧ T ), so E[N ((τ +t+r)∧T )−N ((τ +t)∧T )|Fτ +t ] = λ((τ +t+r)∧T −(τ +t)∧T ). By the monotone convergence theorem E[N (τ + t + r) − N (τ + t)|Fτ +t ] = λr which gives the lemma.


348

Marked Poisson processes Let N be a Poisson process with parameter λ and associate with each jump time Sk a mark ξk in a space E, called the mark space. Assume that the {ξk } are iid and independent of N .


349

An independence lemma Lemma 17.5 Let N be a Poisson process and {ξk } be independent and identically distributed and independent of N . Then Ft = σ(N (s), ξN (s) : s ≤ t) is independent of G t = σ(N (s) − N (t), ξN (t)+k : s ≥ t, k = 1, 2, . . .). Proof. The proof is left as an exercise. The lemma may look obvious, but it in general, it is false if the assumption that the ξk are identically distributed is dropped.


350

Compound Poisson processes PN (t)

Suppose E = R and let Q(t) = Poisson process.

k=1

ξk . Then Q is a compound

Lemma 17.6 Suppose E[ξk ] = β. Then Q has independent increments, E[Q(t)] = βλt, and M (t) = Q(t) − βλt is a martingale. P Proof. Since E[ nk=1 ξk ] = nβ, E[Q(t)|N (t)] = N (t)β and E[Q(t)] = E[E[Q(t)|N (t)]] = λtβ. N (t)+N (t+s)−N (t)

E[Q(t + s) − Q(t)|Ft ] = E[

X

ξk |Ft ] = λsβ

k=N (t)+1


351

A stochastic equation for a jump process Theorem 17.7 Let X(0) be independent of the marked Poisson process, let H : Rd × E → Rd , and let X satisfy Z N (t) X X(t) = X(0)+ H(X(s−), ξk ) = X(0) +

H(X(s−), z)N (ds × dz)

[0,t]×E

k=1

Then X is a Markov process and Z t Z f (X(t)) − λ (f (X(s) + H(X(s), z)) − f (X(s)))µξ (dz)ds 0

E

is a martingale for each f ∈ B(Rd ).


352

Proof. Since N (t+s)

X(t + s) = X(t) +

X

H(X(s−), ξk )

k=N (t)+1

is a function of X(t), N (t + s) − N (t), and (ξN (t)+1 , ξN (t)+2 , . . .), the Markov property follows by Lemma 17.5. Noting that E[f (X(t + h)) − f (X(t))|Ft ] = E[1{N (t+h)−N (t)=1} (f (X(t) + H(X(t), ξN (t)+1 )) − f (X(t))|Ft ] + o(h) Z −λh = λhe (f (X(t) + H(X(t), z)) − f (X(t))µξ (dz) + o(h) E

X E[f (X(t + s) − f (X(t))|Ft ] = E[f (X(ti+1 )) − f (X(ti ))|Ft ] Z t+s Z ≈ E[ λ (f (X(r) + H(X(r), z)) − f (X(r)))µξ (dz)dr t

E


353

Martingales associated with a marked Poisson process Lemma 17.8 If (N, {ξk }) is a marked Poisson process, then N (t+s)

E[

X

Z G(k − 1, ξk )|Ft ] = E[

t+s

Z G(N (s), z)µξ (dz)|Ft ],

λ E

t

k=N (t)+1

and hence MG (t) =

N (t) X k=1

Z G(k − 1, ξk ) −

t

Z G(N (s), z)µξ (dz)

λ 0

E

is a martingale.


354

Lemma 17.9 Let X be cadlag and adapted and G be a bounded, continuous function. Define Z Z(t) =

t

G(X(s−), ξN (s) )dN (s) = 0

N (t) X

G(X(Sk −), ξk ).

k=1

Then Z E[Z(t + s) − Z(t)|Ft ] = E[ t

and hence

Z Z(t) −

t

Z G(X(s), z)µξ (dz)ds|Ft ]

λ E

Z G(X(s), z)µξ (dz)ds

λ 0

t+s

E

is a martingale.


355

Proof. Note that Z(t) = lim

h→0

[t/h] X

G(X(kh), ξN (kh)+1 )(N ((k + 1)h) − N (kh))

k=0

and Z E[G(X(kh), ξN (kh)+1 )(N ((k+1)h)−N (kh))|Fkh ] =

G(X(kh), z)µξ (dz)λh E


356

Change of measure

Theorem 17.10 Let G be bounded and continuous, G ≥ −1, X cadlag and adapted, and Z satisfy Z t Z t Z Z(t) = 1+ Z(s−)G(X(s−), ξN (s) )dN (s)− Z(s)λ G(X(s), z)µξ (dz)ds. 0

E

0

Then Z

t

Z(t) = exp{ 0

Z t Z log(1+G(X(s−), ξN (s) ))dN (s)− λ G(X(s), z)µξ (dz)ds} 0

E

is a martingale.


357

Transformation of martingales under a change of measure Lemma 17.11 Let Z be as in Theorem 17.10, and suppose dPe|Ft = Z(t)dP|Ft . Rt If M is a {Ft }-local martingale under P and [Z, M ]t = 0 V (s−)dN (s). Then Z t V (s−) dN (s) Y (t) = M (t) − Z(s) 0 is a {Ft }-local martingale under Pe.


358

Proof. Z

t

Z(t)Y (t) = Z(0)Y (0) +

Z

t

Y (s−)dZ(s) + 0

Z

Z(s−)dY (s) + [Z, Y ]t 0

t

Z

t

= Z(0)Y (0) + Y (s−)dZ(s) + Z(s−)dY (s) + [Z, M ]t 0 0 Z t V (s−)G(X(s−), ξN (s) ) dN (s) − 1 + G(X(s−), ξN (s) ) 0 Z t Z t = Z(0)Y (0) + Y (s−)dZ(s) + Z(s−)dM (s) 0

0


359

General counting processes N is a counting process if N (0) = 0, N is right continuous, and N is constant except for jumps of +1. N is determined by its jump times 0 < σ1 < σ2 < · · ·. If N is adapted to Ft , then the σk are Ft -stopping times.


360

Intensity for a counting process If N is a Poisson process with parameter λ and N is compatible with {Ft }, then P {N (t + ∆t) > N (t)|Ft } = 1 − e−λ∆t ≈ λ∆t. For a general counting process N , at least intuitively, a nonnegative, {Ft }-adapted stochastic process λ(·) is an {Ft }-intensity for N if Z t+∆t P {N (t + ∆t) > N (t)|Ft } ≈ E[ λ(s)ds|Ft ] ≈ λ(t)∆t. t

Definition 17.12 λ is an {Ft }-intensity for N if and only if for each n = 1, 2, . . .. Z t∧σn

N (t ∧ σn ) −

λ(s)ds 0

is a {Ft }-martingale. •First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit

361

Modeling with intensities Let X be a stochastic process (cadlag, E-valued for simplicity) that models “external noise.” Let Dc [0, ∞) denote the space of counting paths (zero at time zero and constant except for jumps of +1). Condition 17.13 λ : [0, ∞) × DE [0, ∞) × Dc [0, ∞) → [0, ∞) and assume that if X is cadlag and adapted, then s → λ(s, Z, N ) is cadlag and adapted.


362

Change of measure Theorem 17.14 On (Ω, F, P ), let N be a unit Poisson process, X be cadlag and adapted, λ satisfy Condition 17.13, and λ be bounded by a constant. Let Z satisfy Z t Z t Z(t) = 1+ (λ(s−, X, N )−1)Z(s−)dN (s)− Z(s)(λ(s, X, N )−1)ds. 0

0

Then Z is a martingale, and if dPe|Ft = Z(t)dP|Ft , then under Pe, Z t N (t) − λ(s, X, N )ds 0

is a martingale, that is, N is counting process with intensity λ(s, X, N ).


363

18.

Assignments

1. Due February 1: Exercises 1.5 and 1.8 2. Due February 8: Exercises 1.14 and 1.15 3. Due February 15: Exercise 2.9 and Problem 2 4. Due February 22: Problems 6, 7, 8 5. Due March 13: Problems 9, 10 6. Due April 19: Problems 11, 12 7. Due April 26: Exercises 5.8 and 5.10 8. Due May 3: Exercises 5.11 and 6.1 9. Due May 10: Exercise 5.14


364

19.

Problems

1. Let X, Y , and Z be random variables. X and Y are conditionally independent given Z if and only if E[f (X)g(Y )|Z] = E[f (X)|Z]E[g(Y )|Z] for all bounded measurable f and g. (a) Show that X and Y are conditionally independent given Z if and only if E[f (X)|Y, Z] = E[f (X)|Z]

(19.1)

for all bounded measurable f . (b) Let X, Y , and Z be independent, R-valued random variables, and let ψ : R2 → R and ϕ : R2 → R be Borel measurable functions. Define U = ψ(X, Z) and V = ϕ(Y, Z). Show that U and V are conditionally independent given Z. 2. Let X be uniformly distributed on [0, 1], and let D be the σ-algebra generated by {{X ≤ 21 }, { 12 < X ≤ 3 }, { 34 < X ≤ 1}}. Give an explicit representation of E[X|D] in terms of X. What is the probability 4 distribution of E[X|D]? 3. Let Q be a probability measure on (Ω, F ) that is absolutely continuous with respect to P and let L = dQ dP be the corresponding Radon-Nikodym derivative. Let PD and QD be the restrictions of P and Q to the sub-σ-algebra D. Show that QD is absolutely continuous with respect to PD and that dQD = E[L|D] . dPD


365

4. Suppose X and Y are independent random variables, and let µX and µY denote the distributions of X and Y . Let h be a measurable function satisfying Z ∞ Z ∞ |h(x, y)|µX (dx)µY (dy) < ∞. −∞

−∞

Define Z = h(X, Y ) and

Z

∞

g(y) =

h(x, y)µX (dx). −∞

Show that E[Z|Y ] = g(Y ). 5. Let X1 , X2 , . . . be independent random variables with E[Xk ] = 1, k = 1, 2, . . .. Let F0 = {∅, Ω} and for n = 1, 2, . . ., let Fn = σ(X1 , . . . , Xn ), that is, Fn is the smallest σ-algebra with respect to which Q X1 , . . . , Xn are measurable. Let M0 = 1 and for n = 1, 2, . . ., let Mn = n k=1 Xk . Show that E[Mn+1 |Fn ] = Mn . 6. Let W be a standard Brownian motion and define B(t) = W (t) − tW (1),

0 ≤ t ≤ 1.

Show that B is independent of W (1). (It is enough to show that B(t1 ), . . . , B(tm ) is independent of W (1) for each choice of 0 ≤ t1 < · · · < tm ≤ 1. 7. Use the result in Problem 6 to compute E[eλW (t) |W (1)].


366

8. Let

f (t) = tW ( 1 ). W t

f is a standard Brownian motion. Show that W 9. Let X1 and X2 be adapted, simple processes. Show that for a, b ∈ R, Z = aX1 + bX2 is an adapted, simple process and that Z Z Z ZdW = a X1 dW + b X2 dW. (Note that if U and V are D-measurable, then aU + bV is D-measurable.) n n 10. For n = 1, 2, . . ., let 0 = tn 0 < t1 < · · ·, and let {ai } ⊂ [0, ∞). Suppose that for each t > 0, X n lim ai = t. n→∞

tn i ≤t

Show that for f ∈ C[0, ∞), lim

n→∞

X tn i ≤t

n f (tn i )ai =

t

Z

f (s)ds. 0

(In fact, the result holds for all cadlag f .)


367

11. Consider

t

Z X(t) = 1 +

σ

p X(s)dW (s) + bt

0

and let τ = inf{t : X(t) = 0}. Give necessary and sufficient conditions on σ and b > 0 for P {τ < ∞} > 0. 12. Consider

t

Z X(t) = 1 +

t

Z σ(X(s))dW (s) +

0

b(X(s))ds 0

where d = m = 1, σ and b are continuous, and inf x∈R σ(x) > 0. Give necessary and sufficient conditions on σ and b for P {limt→∞ X(t) = ∞} = 1.


368

List of topics 1. Review of basic concepts of probability

16. Multivariable Itoˆ formula

2. From probability to measure theory

17. Brownian bridge

3. Review of Markov chains 4. Information and sigma algebras

18. Convergence of empirical distribution functions

5. Conditional expectations

19. Stochastic differential equations

6. Martingales

20. Examples and the Markov property

7. The central limit theorem and Brownian motion

21. SDEs and partial differential equations

8. Martingale properties of Brownian motion

23. Multidimensional SDEs

9. Sample path properties of Brownian motion

24. Examples

22. Examples

10. The Markov property

25. Change of probability measure

11. Definition of the Ito integral

26. Girsanov formula

12. Properties and examples

27. Pricing and risk neutral measures

ˆ formula 13. Ito’s

28. Martingale representation

14. Examples and applications

29. Fundamental theorems of asset pricing

15. Black-Scholes formula

30. Applications


369

Math 635: An Introduction to Brownian Motion and Stochastic Calculus

Math 635: An Introduction to Brownian Motion and Stochastic Calculus

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