Math 635: An Introduction to Brownian Motion and Stochastic Calculus

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1 First Prev Next Go To Go Back Full Screen Close Quit 1 Math 635: An Introduction to Brownian Motion and Stochastic Calculus 1. Introduction and review 2. Notions of convergence and results from measure theory 3. Review of Markov chains 4. Change of measure 5. Information and conditional expectations 6. Martingales 7. Brownian motion 8. Stochastic integrals 9. Black-Scholes and other models 1. The multidimensional stochastic calculus 11. Stochastic differential equations 12. Markov property 13. SDEs and partial differential equations 14. Change of measure and asset pricing 15. Martingale representation and completeness 16. Applications and examples 17. Stationary distributions and forward equations 18. Processes with jumps 19. Assignments 2. Problems

2 First Prev Next Go To Go Back Full Screen Close Quit 2 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

3 First Prev Next Go To Go Back Full Screen Close Quit 3 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

4 First Prev Next Go To Go Back Full Screen Close Quit 4 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 1 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.

5 First Prev Next Go To Go Back Full Screen Close Quit 5 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 probability. interpretation of

6 First Prev Next Go To Go Back Full Screen Close Quit 6 Definition of probability The probability of an event A should be P (A) = lim n #times A occurs in first n trials n The mathematical problem: Make sense out of this. The real world relationship: Probabilities are predictions about the future.

7 First Prev Next Go To Go Back Full Screen Close Quit 7 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

8 First Prev Next Go To Go Back Full Screen Close Quit 8 The distribution of a random variable If X k is the value of X observed on the kth trial, then we should have P {X = a} = lim n #{k n : X k = a} n If X has only finitely many possible values, then P {X = a} = 1. a R(X) This collection of probabilities determine the distribution of X.

9 First Prev Next Go To Go Back Full Screen Close Quit 9 Distribution function More generally, 1 P {X x} = lim n n n 1 (,x] (X k ) k=1 F X (x) P {X x} is the distribution function for X.

10 The law of averages If R(X) = {a 1,..., a m } is finite, then X X n m #{k n : X k = a l } lim = lim a l n n n n l=1 = m a l P {X = a l } l=1 First Prev Next Go To Go Back Full Screen Close Quit 1

11 First Prev Next Go To Go Back Full Screen Close Quit 11 More generally, if R(X) [c, d], < c < d <, then l x l P {x l < X x l+1 } = lim n = l m l=1 x l #{k n : x l < X k x l+1 } n X X n lim n n m lim n l=1 x l+1 #{k n : x l < X k x l+1 } n x l+1 P {x l < X x l+1 } = l x l+1 (F X (x l+1 ) F X (x l )) d c xdf X (x)

12 First Prev Next Go To Go Back Full Screen Close Quit 12 The expectation as a Stieltjes integral If R(X) [c, d], define E[X] = d c xdf X (x). If the relative frequency interpretation is valid, then X X n lim n n = E[X].

13 First Prev Next Go To Go Back Full Screen Close Quit 13 A random variable without an expectation Example 1.1 Suppose Then X X n lim n n P {X x} = = = x 1 + x, x. m lp {l < X l + 1} l= m l= m l= l( l + 1 l + 2 One could say E[X] =, and we will. l l + 1 ) l (l + 2)(l + 1) as m

14 First Prev Next Go To Go Back Full Screen Close Quit 14 Review of basic calculus Definition 1.2 A sequence {x n } R converges to x R (lim n x n = x) if and only if for each ɛ > there exists an n ɛ > such that n > n ɛ implies x n x ɛ. Let {a k } R. The series k=1 a k converges if lim n n k=1 a k R exists. The series converges absolutely if k=1 a k converges. (Or we write k=1 a k <.) Examples of things you should know: lim (ax n + by n ) = a lim x n + b lim y n n n n if the two limits on the right exist. If lim n,m x n x m = ({x n } is a Cauchy sequence), then lim n x n exists.

15 First Prev Next Go To Go Back Full Screen Close Quit 15 Examples If α > 1, then For α = 1, however ( 1) k k=1 k = lim n n ( 1) k k=1 k k=1 k=1 = lim m 1 k α <. 1 k = ; m l=1 ( 1 2l l ) = 1 2l(2l 1) l=1

16 First Prev Next Go To Go Back Full Screen Close Quit 16 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.

17 First Prev Next Go To Go Back Full Screen Close Quit 17 The collection of events If Then A = {ω Ω : statement I is true for ω} B = {ω Ω : statement II is true for ω} 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} A c = {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 A c 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).

18 First Prev Next Go To Go Back Full Screen Close Quit 18 The probability measure Each event A F is assigned a probability P (A). 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 A 1,..., A m are disjoint m P ( m k=1a k ) = P (A k ) finite additivity k=1 In fact, we assume countable additivity: If A 1, A 2,... are disjoint events, then P ( k=1a k ) = P (A k ). P (Ω) = 1. k=1

19 First Prev Next Go To Go Back Full Screen Close Quit 19 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( ) = and countable additivity. A probability space is a measure space (Ω, F, P ) satisfying P (Ω) = 1.

20 First Prev Next Go To Go Back Full Screen Close Quit 2 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

21 First Prev Next Go To Go Back Full Screen Close Quit 21 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 µ X (B) = P {X B}, B B(R). µ X is called the measure induced by the function X.

22 First Prev Next Go To Go Back Full Screen Close Quit 22 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 p X (x) = P {X = x}, x R(X). Note that x R(X) P {X = x} = 1.

23 First Prev Next Go To Go Back Full Screen Close Quit 23 Examples Binomial distribution P {X = k} = ( ) n p k (1 p) n k, k for some postive integer n and some p 1 Poisson distribution k =, 1,..., n for some λ >. P {X = k} = e λλk, k =, 1,... k!

24 First Prev Next Go To Go Back Full Screen Close Quit 24 Absolutely continuous distributions Definition 1.6 The distribution of X is absolutely continuous if and only if there exists a nonnegative function f X such that P {a < X b} = b a f X (x)dx, a < b R. Then f X is the probability density function for X.

25 First Prev Next Go To Go Back Full Screen Close Quit 25 Examples Normal distribution f X (x) = 1 e (x µ)2 2σ 2 2πσ Exponential distribution f X (x) = { λe λx x x <

26 First Prev Next Go To Go Back Full Screen Close Quit 26 Expectations If X is discrete, then letting R(X) = {a 1, a 2,...}, X = i a i 1 Ai where A i = {X = a i }. If i a i P (A i ) <, then E[X] = i a i P {X = a i } = i a i P (A i )

27 First Prev Next Go To Go Back Full Screen Close Quit 27 For general X, let Y n = nx n, Z n = nx n. Then Y n X Z n, so we must have E[Y n ] E[X] E[Z n ]. Specifically, if k k P {k < X k + 1} <, which is true if and only if E[ Y n ] < and E[ Z n ] < for all n (we will say that X is integrable), then define Notation E[X] lim n E[Y n ] = lim n E[Z n ]. (1.1) E[X] = Ω XdP = Ω X(ω)P (dω).

28 First Prev Next Go To Go Back Full Screen Close Quit 28 Properties Lemma 1.7 (Monotonicity) If P {X Y } = 1 and X and Y are integrable, then E[X] E[Y ]. Lemma 1.8 (Positivity) If P {X } = 1, and X is integrable, then E[X]. Lemma 1.9 (Linearity) If X and Y are integrable and a, b R, then ax+ by is integrable and E[aX + by ] = ae[x] + be[y ].

29 First Prev Next Go To Go Back Full Screen Close Quit 29 Jensen s inequality Lemma 1.1 Let X be a random variable and ϕ : R R be convex. If E[ X ] < and E[ ϕ(x) ] <, then ϕ(e[x]) E[ϕ(X)]. Proof. If ϕ is convex, then for each x, ϕ + ϕ(y) ϕ(x) (x) = lim y x+ y x and ϕ(y) ϕ(x) + ϕ + (x)(y x). Setting µ = E[X], exists E[ϕ(X)] E[ϕ(µ) + ϕ + (µ)(x µ)] = ϕ(µ) + ϕ + (µ)e[x µ] = ϕ(µ).

30 First Prev Next Go To Go Back Full Screen Close Quit 3 Consequences of countable additivity P (A c ) = 1 P (A) If A B, then P (B) P (A). If A 1 A 2, then P ( k=1 A k) = lim n P (A n ). P ( k=1a k ) = P ( k=1(a k A c k 1)) = = lim n n k=1 P (A k A c k 1) k=1 P (A k A c k 1) = lim n P (A n ) If A 1 A 2, then P ( k=1 A k) = lim n P (A n ). A n = k=1a k ( k=na k A c k+1)

31 First Prev Next Go To Go Back Full Screen Close Quit 31 Properties of cumulative distribution functions If X is a R-valued random variable (P { < X < } = 1) lim x F X (x) = lim n P {X n} = P {X < } = 1. lim x F X (x) = lim n P {X n} = P ( n {X n}) =. F X (x) F X (x ) = lim (F X (x) F X (x n 1 )) n = lim P {x n 1 < X x} = P {X = x}. n Since F X (x) F X (x ) = P {X = x}, there can be only finitely many discontinuities with F X (x) F X (x ) n 1, n = 1, 2,.... Consequently, a cdf has at most countably many discontinuities.

32 First Prev Next Go To Go Back Full Screen Close Quit 32 Expectations of nonnegative functions If P {X } = 1 and l= lp {l < X l + 1} =, we will define E[X] =. Note, however, whenever I write E[X] I mean that E[X] is finite unless I explicitly allow E[X] =.

33 First Prev Next Go To Go Back Full Screen Close Quit Notions of convergence and results from measure theory Three kinds of convergence of random variables Limit theorems for expectations and integrals Computation of expectations

34 First Prev Next Go To Go Back Full Screen Close Quit 34 Three kinds of convergence of random variables Let {X n } be a sequence of random variables and X another random variable. Let F Xn (x) = P {X n x} denote the cdf for X n. Definition 2.1 Almost sure convergence: lim n X n = X almost surely (a.s.) if and only if P {ω : lim n X n (ω) = X(ω)} = 1. Convergence in probability: The sequence {X n } converges to X in probability if and only if for each ɛ > lim P { X n X ɛ} =. n Convergence in distribution: The sequence {X n } converges to X is distribution if and only if for each x such that F X is continuous at x lim F X n n (x) = F X (x).

35 First Prev Next Go To Go Back Full Screen Close Quit 35 Examples Theorem 2.2 (The strong law of large numbers) Suppose ξ 1, ξ 2,... are independent and identically distributed with E[ ξ i ] <. Then ξ ξ n lim n n = E[ξ] a.s. Theorem 2.3 (The weak law of large numbers) Suppose ξ 1, ξ 2,... are iid with E[ξi 2 ] <. Then for ɛ >, P { ξ ξ n n E[ξ] ɛ} V ar(ξ) nɛ 2.

36 First Prev Next Go To Go Back Full Screen Close Quit 36 Theorem 2.4 (Central limit theorem) Let ξ 1, ξ 2,... be iid with E[ξ i ] = µ and V ar(ξ i ) = σ 2 <. Then n i=1 ξ i nµ x x} = nσ lim P { n 1 2π e y2 2 dy

37 First Prev Next Go To Go Back Full Screen Close Quit 37 Relationship among notions of convergence Lemma 2.5 Almost sure convergence implies convergence in probability. Convergence in probability implies convergence in distribution. Proof. Suppose X n X a.s. Then for ɛ >, A n {sup X k X ɛ} { X n X ɛ} k n A 1 A 2, so lim n P (A n ) = P ( k=1 A k). k=1 A k = {ω : lim sup n X n X ɛ} and hence lim n P (A n ) =.

38 First Prev Next Go To Go Back Full Screen Close Quit 38 Suppose X n X in probability. Then F Xn (x) F X (x+ɛ)+p {X n x, X > x+ɛ} F X (x+ɛ)+p { X n X ɛ}. Therefore, and similarly lim sup F Xn (x) lim F X (x + ɛ) = F X (x), n ɛ lim inf n F X n (x) lim ɛ F X (x ɛ) = F X (x ).

39 First Prev Next Go To Go Back Full Screen Close Quit 39 Limit theorems for expectations and integrals Theorem 2.6 (Bounded convergence theorem) Let {X n } be a sequence of random variables such that X n X in distribution. Suppose that there exists C > such that P { X n > C} = for all n. Then lim n E[X n ] = E[X]. Proof. Let a C < a 1 < < C a m be points of continuity for F X. Then m 1 l= a l P {a l < X n a l+1 } E[X n ] m 1 l= a l+1 P {a l < X n a l+1 }. Then the left and right sides converge to the left and right sides of m 1 l= a l P {a l < X a l+1 } E[X] m 1 l= and lim sup n E[X n ] E[X] max l (a l+1 a l ). a l+1 P {a l < X a l+1 }.

40 First Prev Next Go To Go Back Full Screen Close Quit 4 Lemma 2.7 If X a.s., then lim K E[X K] = E[X]. Proof. If K > m n, then E[X] E[X K] m 1 k=1 k n P {k n X < k + 1 n } m E[ nx n ]. More generally, Theorem 2.8 (Monotone convergence theorem) If X 1 X 2 a.s. and X = lim n X n a.s., then allowing =. E[X] = lim n E[X n ]

41 First Prev Next Go To Go Back Full Screen Close Quit 41 Lemma 2.9 (Fatou s lemma) If P {X n } = 1 and X n converges in distribution to X, then Proof. lim inf n E[X n] E[X] lim inf n E[X n] lim n E[X n K] = E[X K] E[X] Let P {X n = n} = 1 P {X n = } = n 1. Then X n in distribution, and E[X n ] = 1 for all n.

42 First Prev Next Go To Go Back Full Screen Close Quit 42 Theorem 2.1 (Dominated convergence theorem) Suppose there exist Y such that X n Y a.s. with E[Y ] < and lim n X n = X a.s. Then lim n E[X n ] = E[X]. Proof. E[Y ] lim sup E[X n ] = lim inf (E[Y ] E[X n]) n n E[Y ] + lim inf n = lim inf n E[Y X n] E[Y X] = E[Y ] E[X] E[X n] = lim inf n (E[Y ] + E[X n]) = lim inf n E[Y + X n] E[Y + X] = E[Y ] + E[X]

43 First Prev Next Go To Go Back Full Screen Close Quit 43 Computation of expectations Recall µ X (B) = P {X B}. g is Borel measurable if {x : g(x) c} B(R) for all c R. Note that (R, B(R), µ X ) 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) ] <, E[g(X)] = g(x)µ X (dx) R

44 First Prev Next Go To Go Back Full Screen Close Quit 44 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 g(x)dx g(x)l(dx) R If f(x) = m i=1 a i1 Ai, A i B(R) with L(A i ) <, a i R, then m f(x)dx a i L(A i ). R R i=1 Such an f is called a simple function.

45 First Prev Next Go To Go Back Full Screen Close Quit 45 General definition of Lebesgue integral For g, R g(x)dx sup{ f(x)dx : f simple, f g}. R If R g(x) dx <, then setting g+ (x) = g(x) and g (x) = ( g(x)), g(x)dx g + (x)dx g (x)dx. R R R

46 First Prev Next Go To Go Back Full Screen Close Quit 46 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 ɛ > there exists a δ > such that a = t < t 1 < < t m = b, s k [t k, t k+1 ] and max t k+1 t k δ implies Then I is denoted by g(s k )(t k+1 t k ) I ɛ. m 1 k= b a g(x)dx m 1 k= g(s k)(t k+1 t k ) is called a Riemann sum. The Riemann integral is the limit of Riemann sums.

47 First Prev Next Go To Go Back Full Screen Close Quit 47 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.

48 First Prev Next Go To Go Back Full Screen Close Quit 48 Distributions with Lebesgue densities Suppose µ X (B) = R 1 B (x)f X (x)dx, B B(R). Then f X 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 f X. If g is Borel measurable and E[ g(x) ] <, then E[g(X)] = g(x)f X (x)dx R

49 First Prev Next Go To Go Back Full Screen Close Quit Review of Markov chains Elementary definition of conditional probability Independence Markov property Transition matrix Simulation of a Markov chain

50 First Prev Next Go To Go Back Full Screen Close Quit 5 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, P (A B) = # 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) leading to the definition P (A B) = P (A B) P (B) Note that for each fixed B F, P ( B) is a probability measure on F.

51 First Prev Next Go To Go Back Full Screen Close Quit 51 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). {A k, k I} are mutually independent if P (A k1 A km ) = for all choices of {k 1,..., k m } I. m P (A ki ) i=1

52 First Prev Next Go To Go Back Full Screen Close Quit 52 Independence of random variables Random variables {X k, k I} are mutually independent if P {X k1 B 1,..., X km B m } = m P {X ki B i } i=1 for all choices of {k 1,..., k m } I and B 1,..., B m B(R).

53 First Prev Next Go To Go Back Full Screen Close Quit 53 Joint distributions The collection of Borel subsets of R m (B(R m )) is the smallest σ-algebra containing (, c 1 ] (, c m ]. The joint distribution of (X 1,..., X m ) is the measure on B(R m ) defined by µ X1,...,X m (B) = P {(X 1,..., X m ) B}, B B(R m ). Lemma 3.1 The joint distribution of (X 1,..., X m ) is uniquely determined by the joint cdf F X1,...,X m (x 1,..., x m ) = P {X 1 x 1,..., X m x m }.

54 First Prev Next Go To Go Back Full Screen Close Quit 54 The Markov property Let X, X 1,... be positive, integer-valued random variables. Then P {X = i,..., X m = i m } = P {X m = i m X = i,..., X m 1 = i m 1 } P {X m 1 = i m 1 X = i,..., X m 2 = i m 2 } P {X 1 = i 1 X = i }P {X = i } X satisfies the Markov property if for each m and all choices of {i k }, P {X m = i m X = i,..., X m 1 = i m 1 } = P {X m = i m X m 1 = i m 1 }

55 First Prev Next Go To Go Back Full Screen Close Quit 55 The transition matrix Suppose P {X m = j X m 1 = i} = p ij. (The transition probabilities do not depend on m.) The matrix P = ((p ij )) satisfies p ij = 1. j P {X m+2 = j X m = i} = k P {X m+2 = j, X m+1 = k X m = i} = k = k P {X m+2 = j X m+1 = k, X m = i}p {X m+1 = k X m = i} p ik p kj p (2) ij Then ((p (2) ij )) = P 2.

56 First Prev Next Go To Go Back Full Screen Close Quit 56 The joint distribution of a Markov chain The joint distribution of X, X 1,... is determined by the transition matrix and the initial distribution, ν i = P {X = i}. P {X = i,..., X m = i m } = ν i p i i 1 p i1 i 2 p im 1 i m. In general, where P {X m+n = j X m = i} = p (n) ij, ((p (n) ij )) = P n.

57 First Prev Next Go To Go Back Full Screen Close Quit 57 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 [, 1] E by j 1 j H(i, u) = j, p ik u < and V : [, 1] E by V (u) = j, k=1 j 1 ν k u < k=1 k=1 p ik j ν k. k=1 If ξ is uniform [, 1], then j 1 P {V (ξ) = j} = P { k=1 ν k ξ < j ν k } = ν j. k=1

58 First Prev Next Go To Go Back Full Screen Close Quit 58 Theorem 3.2 Let ξ, ξ 1,... be iid uniform [, 1] random variables, and define X = V (ξ ), X n+1 = H(X n, ξ n+1 ). Then {X n } is a Markov chain with initial distribution {ν k } and transition matrix ((p ij )). Proof. As noted above, X has distribution {ν k }. Note that X k is a function of {ξ,..., ξ k } and hence is independent of ξ k+1, ξ k+2,... P {X m = i m X = i,..., X m 1 = i m 1 } = P {H(X m 1, ξ m ) = i m X = i,..., X m 1 = i m 1 } = P {H(i m 1, ξ m ) = i m X = i,..., X m 1 = i m 1 } = P {H(i m 1, ξ m ) = i m } = p im 1 i m.

59 First Prev Next Go To Go Back Full Screen Close Quit 59 Stationary distributions Definition 3.3 A probability distribution π = {π k } is a stationary distribution for a Markov chain with transition matrix P = ((p ij )) if k π kp kj = π j Lemma 3.4 If π is a stationary distribution for the Markov chain {X n } and X has distribution π, then for each n, X n has distribution π. Proof. Noting that P {X 1 = j} = k P {X 1 = j, X = k} = k π k p kj = π j, the lemma follows by induction. Note that π T P = π T, so that π T eigenvalue 1. is a left eigenvector of P for the

60 First Prev Next Go To Go Back Full Screen Close Quit 6 The ergodic theorem Definition 3.5 A Markov chain is irreducible if for each i, j E, there exists n such that p (n) ij >. Theorem 3.6 If {X n } is irreducible and and has stationary distribution π, then for each bounded function f on E, f(x ) + + f(x n ) lim n n + 1 = k π k f(k).

61 First Prev Next Go To Go Back Full Screen Close Quit Change of measure Defining a change of measure Expectations under the new measure Absolute continuity and the Radon-Nikodym theorem Equivalent measures

62 First Prev Next Go To Go Back Full Screen Close Quit 62 Defining a change of measure Lemma 4.1 Let Z be a nonnegative random variable on (Ω, F, P ) such that E[Z] E P [Z] = 1. Define Then P is a probability measure on F. P (A) = E P [1 A Z], A F. (4.1)

63 First Prev Next Go To Go Back Full Screen Close Quit 63 Proof. P (Ω) = E P [1 Ω Z] = E P [Z] = 1. Suppose A 1, A 2,... F are disjoint. Then 1 i=1 A i Z = 1 Ai Z and i=1 P ( i=1a i ) = E P [1 i=1 A i Z] = E P [ = lim n E P [ = i=1 P (A i ) n i=1 1 Ai Z] i=1 1 Ai Z] = lim n n E P [1 Ai Z] The third equality follows by the monotone convergence theorem. i=1

64 First Prev Next Go To Go Back Full Screen Close Quit 64 Expectations under the new measure Lemma 4.2 Let P be given by (4.1). Suppose E P [ X Z] <. Then E P [X] = E P [XZ]. Proof. Since E P [1 Ai ] = P (A i ) = E P [1 Ai Z], if X = m i=1 a i1 Ai, then m m E P [X] = a i E P [1 Ai ] = a i E P [1 Ai Z] = E P [XZ]. i=1 i=1 The result follows for general X by approximation.

65 First Prev Next Go To Go Back Full Screen Close Quit 65 Absolute continuity and the Radon-Nikodym theorem Definition 4.3 Let (Ω, F, P ) be a probability space and let P be another probability measure defined on F. Then P is absolutely continuous with respect to P if and only if P (A) = implies P (A) =. Theorem 4.4 (Radon-Nikodym theorem) P is absolutely continuous with respect to P if and only if there exists a nonnegative random variable Z such that P (A) = E P [1 A Z], A F.

66 First Prev Next Go To Go Back Full Screen Close Quit 66 Equivalent measures Definition 4.5 Probability measures P and P on F are equivalent if and only if P is absolutely continuous with respect to P and P is absolutely continuous with respect to P. Lemma 4.6 P and P are equivalent if and only if there exists a nonnegative random variable Z such that P {Z > } = 1 and P (A) = E P [1 A Z], A F. In addition, P (A) = E P [1 A Z 1 ].

67 First Prev Next Go To Go Back Full Screen Close Quit 67 Example Let X be a standard normal random variable on (Ω, F, P ), and for θ R, define Z θ = exp{θx 1 2 θ2 }. Then E P [Z θ ] = 1. Define P θ (A) = E P [1 A Z θ ]. Then P θ {X x} = E P [1 (,x] (X)Z θ ] = E P [1 (,x] (X) exp{θx 1 2 θ2 }] = = x 1 (,x] (z) exp{θz 1 2 θ2 } 1 2π e 1 2 z2 dz 1 2π exp{ 1 2 (z θ)2 }dz so that under P θ, X is normally distributed with E P θ [X] = θ and V ar P θ (X) = 1.

68 First Prev Next Go To Go Back Full Screen Close Quit Information and conditional expectations Modeling information Information obtained by observing a random variable Information evolving in time Information obtained by observing a stochastic process Discrete observations Approximating random variables using available information Conditional expectations Independence Properties of conditional expectations Definition of a Markov process

69 First Prev Next Go To Go Back Full Screen Close Quit 69 Modeling information Recall, events correspond to statements, the event being the set of outcomes for which the statement is true. Available information corresponds to the collection of statements whose truth can be checked with that information. We model information by the collection D of events corresponding to those statements. Clearly, D is closed under finite intersections, finite unions, and complements, that is, D is an algebra. We assume that it is a σ-algebra.

70 First Prev Next Go To Go Back Full Screen Close Quit 7 Information obtained by observing a random variable If X is a random variable and the available information is the information obtained by observing the value of X, then that information corresponds to σ(x), the smallest σ-algebra with respect to which X is measurable. Lemma 5.1 σ(x) = {{X B} : B B(R)}. Proof. Check that the right side is a σ-algebra.

71 First Prev Next Go To Go Back Full Screen Close Quit 71 Information evolving in time We will assume that time is continuous and identified with [, ). As time evolves, more information is obtained and assuming that nothing is forgotten, the information available at time t, F t, includes the information available at time s for all s < t, that is, for s < t, F s F t. Definition 5.2 A filtration is an increasing family {F t } of σ-algebras indexed by [, ). < s < t implies F s F t.

72 First Prev Next Go To Go Back Full Screen Close Quit 72 Information obtained by observing a stochastic process A (continuous time) stochastic process is a family of random variables {X(t), t [, )} indexed by [, ). Definition 5.3 The natural filtration corresponding to a stochastic process X is the filtration given by F X t = σ(x(s), s t), t. σ(x(s), s t) is the smallest σ-algebra with respect to which X(s) is measurable for each s t. Definition 5.4 A stochastic process is adapted to a filtration {F t } if and only if for each t, X(t) is F t -measurable. In particular, F X t F t.

73 First Prev Next Go To Go Back Full Screen Close Quit 73 Discrete observations Let {D k, k = 1, 2,...} be a partition of Ω, that is, a countable collection of disjoint sets whose union is Ω. σ({d k }) = { k K D k : K {1, 2,...}} (Check that the right side is a σ-algebra.)

74 First Prev Next Go To Go Back Full Screen Close Quit 74 Approximating random variables using available information Let X be a random variable and D a σ-algebra modeling the available information. We want to approximate X using the available information. If Y is the approximation, what must be true about Y? For example, {Y c} must be in D, that is, Y is D-measurable. If D = σ({d k }), the for each k, we know whether or not ω D k. For each k, there must be a constant d k such if ω D k, then Y (ω) = d k, that is Y = d k 1 Dk. k

75 First Prev Next Go To Go Back Full Screen Close Quit 75 Conditional expectations We want Y to be a good approximation, that is, we want the error X Y to be small, at least on average. Assuming E[X 2 ] <, select Y to minimize Then for D-measurable Z, E[(X Y ) 2 ]. E[(X (Y + ɛz)) 2 ] = E[(X Y ) 2 ] 2ɛE[Z(X Y )] + ɛ 2 E[Z] 2 has its minimum at ɛ =. Differentiating, we must have E[Z(X Y )] =.

76 First Prev Next Go To Go Back Full Screen Close Quit 76 Definition of conditional expectation Definition 5.5 Y = E[X D] (the conditional expectation of X give D) if a) Y is D-measurable. b) For each D D, E[1 D X] = E[1 D Y ]. Note that the definition makes sense for all X with E[ X ] <.

77 First Prev Next Go To Go Back Full Screen Close Quit 77 Existence and uniqueness of the conditional expectation Existence follows by the Radon-Nikodym theorem, since for nonnegative X, Q(D) = E[1 D X] defines a measure on D that is absolutely continuous with respect to P on D. Uniqueness follows by the observation that if Y and conditions of the definition, then Ỹ satisfy the E[1 {Y > Ỹ } (Y Ỹ )] = = E[1 {Y <Ỹ }(Y Ỹ )]. It follows that P {Y = Ỹ } = 1.

78 First Prev Next Go To Go Back Full Screen Close Quit 78 Example Suppose that {D k } is a partition and D = σ({d k }). Then E[X D] = k E[X1 Dk ] P (D k ) 1 D k.

79 First Prev Next Go To Go Back Full Screen Close Quit 79 Independence As in the elementary definition of independence, we could say that a random variable X is independent of the (information) σ-algebra D if E[g(X) D] = E[g(X)] for every bounded, measure g (every g B(R)). This identity holds for all bounded measurable g if and only if To be precise: E[g(X)1 D ] = E[g(X)]P (D), g B(R), D D. Definition 5.6 σ-algebras D 1 and D 2 are independent if and only if P (D 1 D 2 ) = P (D 1 )P (D 2 ). A random variable X is independent of a σ-algebra D if and only if P ({X B} D) = P {X B}P (D), B B(R), D D, that is, the σ-algebras D and σ(x) are independent.

80 First Prev Next Go To Go Back Full Screen Close Quit 8 Consequences of independence Lemma 5.7 If X and Y are independent and g, h : R R are Borel measurable, then g(x) and h(y ) are independent. If in addition, g(x) and h(y ) are integrable, then E[g(X)h(Y )] = E[g(X)]E[h(Y )]. (5.1) Proof. The first part of the lemma follows from the fact that {g(x) B} = {X g 1 (B)}, {h(y ) C} = {Y h 1 (C)} where g 1 (B) = {x : g(x) B} is Borel for Borel B by the definition of Borel measurable. To prove (5.1), check first for indicator functions, then for simple functions, and complete the proof by approximation. (In other words, apply what Shreve calls the standard machine. )

81 First Prev Next Go To Go Back Full Screen Close Quit 81 Joint distributions of independent random variables Lemma 5.8 If (X 1,..., X m ) is independent of (Y 1,..., Y n ) and C R m, D R n, then µ X1,...,X m,y 1,...,Y n (C D) = µ X1,...,X m (C)µ Y1,...,Y n (D)

82 First Prev Next Go To Go Back Full Screen Close Quit 82 The Tonelli and Fubini theorems Theorem 5.9 Suppose X and Y are independent and ψ : R 2 [, ). Define ϕ X (y) = E[ψ(X, y)] and ϕ Y (x) = E[ψ(x, Y )]. Then E[ψ(X, Y )] = E[ϕ X (Y )] = E[ϕ Y (X)] (5.2) = ψ(x, y)µ X (dx)µ Y (dy) = ψ(x, y)µ Y (dy)µ X (dx) R R If ψ : R 2 R is Borel measurable and E[ ψ(x, Y ) ] <, then (5.2) holds. Note that the theorem will also hold for independent random vectors. R R

83 First Prev Next Go To Go Back Full Screen Close Quit 83 Conditioning on random variables If X 1,..., X m are random variables, then we will write E[Y σ(x 1,..., X m )] = E[Y X 1,..., X m ] Lemma 5.1 Suppose that Z is σ(x 1,..., X m ). Then there exists B(R m )- measurable h such that Z = h(x 1,..., X m ). Consequently, if Y is integrable, there exists a B(R m )-measurable h Y such that E[Y X 1,..., X m ] = h Y (X 1,..., X m )

84 First Prev Next Go To Go Back Full Screen Close Quit 84 Properties of conditional expectations Theorem 5.11 Let D, G F be σ-algebras, X, Y integrable random variables, and a, b R. Then a) [Linearity] E[aX + by D] = ae[x D] + be[y D]. b) [Positivity] If P {X } = 1, then P {E[X D] } = 1. c) [Monotonicity] If P {X Y } = 1, then P {E[X D] E[Y D]} = 1. d) [Factoring out known quantity] If Y is D-measurable and X and XY are integrable, E[XY D] = Y E[X D]. If Y is D-measurable, then E[Y D] = Y.

85 First Prev Next Go To Go Back Full Screen Close Quit 85 e) [Iterated conditioning] If D G, E[E[X G] D] = E[X D]. f) [Independence] If X is independent of D, then E[X D] = E[X].

86 First Prev Next Go To Go Back Full Screen Close Quit 86 Proof. In each case, check the measurability in Part (a) of the definition of conditional expectation and they verify the integral identity in Part (b). a) E[(aE[X D] + be[y D])1 D ] = ae[e[x D]1 D ] + be[e[y D]1 D ] = ae[x1 D ] + be[y 1 D ] = E[(aX + by )1 D ] d) We want E[Y E[X D]1 D ] = E[Y X1 D ] First, let Y be an indicator, then a simple D-measurable random variable, and then complete the argument by approximation.

87 First Prev Next Go To Go Back Full Screen Close Quit 87 e) For D D G, E[E[X D]1 D ] = E[X1 D ] = E[E[X G]1 D ], where the last equality follows from the fact that D G. f) For D D E[E[X]1 D ] = E[X]E[1 D ] = E[X1 D ]

88 First Prev Next Go To Go Back Full Screen Close Quit 88 Jensen s inequality Theorem 5.12 If ϕ is convex and X and ϕ(x) are integrable, E[ϕ(X) D] ϕ(e[x D]). Proof. As before, if ϕ is convex, then for each x, ϕ + (x) = lim y x+ ϕ(y) ϕ(x) y x exists and ϕ(y) ϕ(x) + ϕ + (x)(y x). Setting Y = E[X D], E[ϕ(X) D] E[ϕ(Y ) + ϕ + (Y )(X Y ) D] = ϕ(y ) + ϕ + (Y )E[X Y D] = ϕ(y ).

89 First Prev Next Go To Go Back Full Screen Close Quit 89 Functions of known and unknown random variable Lemma 5.13 Suppose that X is independent of D and Y is D-measurable. Suppose that ϕ : R 2 R and E[ ϕ(x, Y ) ] <. Define ψ(y) = E[ϕ(X, y)]. Then E[ϕ(X, Y ) D] = ψ(y ). Proof. Since Y is D-measurable, ψ(y ) is D-measurable. For D D, X is independent of (Y, 1 D ), Consequently, by the Fubini theorem, E[ϕ(X, Y )1 D ] = ϕ(x, y)zµ X (dx)µ Y,1D (dy dz) = E[ψ(Y )1 D ]. R 2 R

90 First Prev Next Go To Go Back Full Screen Close Quit 9 Bayes formula Theorem 5.14 Let P be absolutely continuous with respect to P and d P = LdP. Let X be a bounded random variable. Then E P [X D] = EP [XL D] E P [L D] Proof. Let D D. Then [ E P E P ] [XL D] E P [L D] 1 D [ E = E P P ] [XL D] E P [L D] 1 DL [ E = E P P [XL D] E P [L D] 1 DE[L D] = E P [E P [XL D]1 D ] = E P [XL1 D ] = E P [X1 D ] ]

91 First Prev Next Go To Go Back Full Screen Close Quit 91 Recursive generation of Markov chains Theorem 5.15 Let X, ξ 1, ξ 2,... be independent R-valued random variables. For n = 1, 2,..., let H n : R 2 R, and define X n = H n (X n 1, ξ n ). Then {X n } is Markov with respect to F n = σ(x, ξ 1,..., ξ n ). Proof. Let f be bounded and measurable, and define T n f(x) = f(h n (x, z))µ ξn (dz) By Lemma 5.13, R E[f(X n ) F n 1 ] = E[f(H n (X n 1, ξ n )) F n 1 ] = T n f(x n 1 ).

92 First Prev Next Go To Go Back Full Screen Close Quit 92 Definition of a Markov process Definition 5.16 If X is a stochastic process adapted to a filtration {F t }, then X is {F t }-Markov (or Markov with respect to {F t }) if and only if for s, t, E[f(X(t + s)) F t ] = E[f(X(t + s)) X(t)].

93 First Prev Next Go To Go Back Full Screen Close Quit Martingales Definitions of martingale and sub/supermartingale Stopping times Optional sampling theorem Doob s inequalities

94 First Prev Next Go To Go Back Full Screen Close Quit 94 Definitions of martingale and sub/supermartingale Definition 6.1 Let {F t } be a filtration and X a stochastic process adapted to {F t } such that X(t) is integrable for each t. a) X is a {F t }-martingale if and only if E[X(s) F t ] = X(t), t s b) X is a {F t }-submartingale if and only if E[X(s) F t ] X(t), t s c) X is a {F t }-supermartingale if and only if E[X(s) F t ] X(t), t s

95 First Prev Next Go To Go Back Full Screen Close Quit 95 Examples Let ξ 1, ξ 2,... be iid, and define S n = n k=1 ξ i and F n = σ(ξ 1,..., ξ n ). If E[ξ k ] =, then S n is a {F n }-martingale. If ρ = E[e ξ i ] <, then X is a {F n }-martingale. Also, see Problem 5. n = es n ρ n

96 First Prev Next Go To Go Back Full Screen Close Quit 96 Applications of Jensen s inequality Lemma 6.2 Let X be a martingale and ϕ be a convex function. If Y (t) = ϕ(x(t)) is integrable for each t, then Y is a submartingale. Let X be a submartingale and ϕ be a convex, nondecreasing function. If Z(t) = ϕ(x(t)) is integrable for each t, then Z is a submartingale. Proof. Suppose X is a martingale. By Jensen s inequality, for s t, E[ϕ(X(s)) F t ] ϕ(e[x(s) F t ]) = ϕ(x(t)). Suppose X is a submartingale and ϕ is nondecreasing. Then E[ϕ(X(s)) F t ] ϕ(e[x(s) F t ]) ϕ(x(t)).

97 First Prev Next Go To Go Back Full Screen Close Quit 97 Stopping times Definition 6.3 Let {F t } be a filtration. Then a nonnegative (or more generally [, ]-valued) random variable τ is a {F t }-stopping time if and only if {τ t} F t for each t. If τ is the time that an alarm clock goes off and {F t } is the information available to an observer, then the observer hears the alarm clock go off.

98 First Prev Next Go To Go Back Full Screen Close Quit 98 Examples Lemma 6.4 Let X be a continuous, R-valued process adapted to {F t }, and define τ c = inf{t : X(t) c}. Then τ c is a {F t }-stopping time. Proof. For t, {τ c t} = n s Q,s t {X(s) c n 1 } F t. Lemma 6.5 Let X be a continuous, R d -valued process adapted to {F t }, and let K R d be closed. Define τ K = inf{t : X(t) K}. Then τ K is a {F t }-stopping time. Proof. Let O n = {x : inf y K x y < n 1. For t, {τ K t} = n s Q,s t {X(s) O n } F t.

99 First Prev Next Go To Go Back Full Screen Close Quit 99 Information up to a stopping time Definition 6.6 Let τ be a {F t }-stopping time. Then Lemma 6.7 F τ = {A F : A {τ t} F t, t } a) If τ is a {F t }-stopping time, then F τ is a σ-algebra. b) If τ 1 and τ 2 are {F t }-stopping times, then F τ1 F τ2. c) If X is right continuous and {F t }-adapted and τ is a {F t }-stopping time, then X(t τ) is F t -measurable and F τ -measurable.

100 First Prev Next Go To Go Back Full Screen Close Quit 1 Optional sampling theorem Theorem 6.8 Let X be a continuous {F t }-submartingale and τ 1 and τ 2 be {F t }-stopping times. Then E[X(t τ 2 ) F τ1 ] X(t τ 1 τ 2 ). Note that if X is a supermartingale, the inequality is reversed and if X is a martingale, the inequality can be replaced by equality.

101 First Prev Next Go To Go Back Full Screen Close Quit 11 Doob s inequalities Theorem 6.9 Let X be a right continuous, nonnegative submartingale. Then for c >, P {sup X(s) c} E[X(t)] s t c Proof. Let τ c = inf{t : X(t) x}. Then E[X(t)] E[X(t τ)] cp {τ t}. (Actually, for right continuous processes, τ c may not be a stopping time unless one first modifies the filtration {F t }, but the modification can be done without affecting the statement of the theorem.)

102 First Prev Next Go To Go Back Full Screen Close Quit 12 Doob s inequalities Theorem 6.1 Let X be a nonnegative submartingale. Then for p > 1, ( ) p p E[sup X(s) p ] E[X(t) p ] s t p 1 Corollary 6.11 If M is a square integrable martingale, then E[sup M(s) 2 ] 4E[M(t) 2 ]. s t

103 First Prev Next Go To Go Back Full Screen Close Quit 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

104 First Prev Next Go To Go Back Full Screen Close Quit 14 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

105 First Prev Next Go To Go Back Full Screen Close Quit 15 Moment generating functions Definition 7.1 Let X be a R d -valued random variable. Then the moment generating function for X (it it exists) is given by ϕ X (λ) = E[e λ X ], λ R d. Lemma 7.2 If ϕ X (λ) < for all λ R d, then ϕ X uniquely determines the distribution of X.

106 First Prev Next Go To Go Back Full Screen Close Quit 16 Gaussian distributions Definition 7.3 A R-valued random variable X is Gaussian (normal) with mean µ and variance σ 2 if f X (x) = 1 e (x µ)2 2σ 2 2πσ The moment generating function for X is E[e λx ] = exp{ σ2 λ 2 + µλ} 2

107 First Prev Next Go To Go Back Full Screen Close Quit 17 Multidimensional Gaussian distributions Definition 7.4 A R d -valued random variable X is Gaussian if λ X is Gaussian for every λ R d. Note that if E[X] = µ and then cov(x) = E[(X µ)(x µ) T ] = Σ = ((σ ij )), E[λ X] = λ µ, var(λ X) = λ T Σλ. Note that is X = (X 1,..., X d ) is Gaussian in R d, then for any choice of i 1,..., i m {1,..., d}, (X i1,... X im ) is Gaussian in R m. Lemma 7.5 If X is Gaussian in R d and C is a m d matrix, then is Gaussian in R m. Y = CX (7.1)

108 First Prev Next Go To Go Back Full Screen Close Quit 18 Characterization by mean and covariance If X is Gaussian in R d, then ϕ X (λ) = exp{ λt Σλ 2 + λ µ} 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.

109 First Prev Next Go To Go Back Full Screen Close Quit 19 Conditions for independence Lemma 7.6 If X = (X 1, X 2 ) R 2 is Gaussian, then X 1 and X 2 are independent if and only if cov(x 1, X 2 ) =. In general, if X is Gaussian in R d, 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 ) =, so if the components of X are independent then Σ is diagonal. If Σ is diagonal, then ϕ X (λ) = d i=1 exp{ λ2 i σ2 i 2 + λ i µ}, and since the moment generating function of X determines the joint distribution, the components of X must be independent.

110 First Prev Next Go To Go Back Full Screen Close Quit 11 Conditional expectations Lemma 7.7 Let X = (X 1,..., X d ) be Gaussian, then d 1 E[X d X 1,..., X d 1 ] = a + b i X i, i=1 where d 1 σ dk = b i σ ik, k = 1,..., d 1 i=1 and d 1 µ d = a + b i µ i. Proof. The b i are selected so that X d (a + d 1 i=1 b ix i ) is independent of X 1,..., X d 1. i=1

111 First Prev Next Go To Go Back Full Screen Close Quit 111 Example Suppose X 1, X 2, and X 3 are independent, Gaussian, mean zero and variance one. Then Let X 1 = X 1 + X 2 2 E[X 1 X 1 + X 2 ] = X 1 + X X 3, X2 = X 1 + X X 3. (7.2) Then X 1 and X 2 are independent, Gaussian, mean zero, and variance one.

112 First Prev Next Go To Go Back Full Screen Close Quit 112 Central limit theorem If {ξ i } are iid with E[ξ i ] = and V ar(ξ i ) = 1, then Z n = 1 n n i=1 ξ i satisfies b 1 P {a < Z n b} e 1 2 x2 dx. 2π If we define a Z n (t) = 1 [nt] n then the distribution of (Z n (t 1 ),..., Z n (t m )) is approximately Gaussian with cov(z n (t i ), Z n (t j )) t i t j, Note that for t < t 1 < < t m, Z n (t k ) Z n (t k 1 ), k = 1,..., m are independent. i=1 ξ i

113 The CLT as a process First Prev Next Go To Go Back Full Screen Close Quit 113

114 First Prev Next Go To Go Back Full Screen Close Quit 114 Interpolation lemma Lemma 7.8 Let Y and Z be independent, R-valued Gaussian random variables with E[Y ] = E[Z] =, V ar(z) = a, and V ar(y ) = 1. Then U = Z a Y, V = Z a 2 2 Y are independent Gaussian random variables satisfying Z = U + V. Proof. Clearly, Z = U+V. To complete the proof, check that cov(u, V ) =.

115 First Prev Next Go To Go Back Full Screen Close Quit 115 Construction of Brownian motion We can construct standard Brownian motion by repeated application of the Interpolation Lemma 7.8. Let ξ k,n be iid, Gaussian, E[ξ k,n ] = and V ar(ξ k,n ) = 1. Define and define W ( k 2 n ) inductively by W (1) = ξ 1,, W ( k 1 2n) = W (k 2 ) + W (k+1 2 ) W ( k 1 n 2 ) 1 n + n 2 2 (n+1)/2ξ k,n = 1 2 W (k 1 2 ) + 1 n 2 W (k ) + 1 n 2 (n+1)/2ξ k,n for k odd and < k < 2 n.

116 First Prev Next Go To Go Back Full Screen Close Quit 116 By the Interpolation Lemma, and W ( k 1 2n) W (k 2 ) = 1 n 2 W (k ) 1 n 2 W (k 1 2 ) + 1 n 2 (n+1)/2ξ k,n W ( k n ) W ( k 2 n) = 1 2 W (k n ) 1 2 W (k 1 2 n ) 1 2 (n+1)/2ξ k,n are independent, Gaussian random variables.

117 First Prev Next Go To Go Back Full Screen Close Quit 117 Convergence to a continuous process Theorem 7.9 Let Γ be the set of diadic rationals in [, 1]. Define W n (t) = W ( k 2 n) + (t k 2 n)2n (W ( k n ) W ( k 2 n)), k 2 n t k n. Then sup W (t) W n (t) a.s., t Γ and hence W extends to a continuous process on [, 1].

118 First Prev Next Go To Go Back Full Screen Close Quit 118 Continuous functions on C[, 1] Definition 7.1 Let C[, 1] denote the continuous, real-valued functions on [, 1]. g : C[, 1] R is continuous if x n, x C[, 1] satisfying sup t 1 x n (t) x(t) implies lim n g(x n ) = g(x). For example, g(x) = sup x(t) t 1 g(x) = 1 x(s)ds

119 First Prev Next Go To Go Back Full Screen Close Quit 119 Functional central limit theorem Theorem 7.11 (Donsker invariance principle) Let {ξ i } be iid with E[ξ i ] = and V ar(ξ i ) = σ 2. Define Z n (t) = 1 [nt] n i=1 ξ i + (t k n ) nξ k+1, k n t k + 1 n. Then for each continuous g : C[, 1] R, g(z n ) converges in distribution to g(σw ).

120 First Prev Next Go To Go Back Full Screen Close Quit 12 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 and the time step is 1 n. P {X n ( k + 1 n ) = (1+ σ n )X n ( k n )} = P {X n( k + 1 n ) = (1 σ n )X n ( k n )} = 1 2. Then log X n (t) = [nt] i=1 log(1+ σ n ξ i ) where P {ξ i = 1} = P {ξ i = 1} =. By Taylor s formula 1 2 and X n (t) σ [nt] n i=1 ξ i 1 2n [nt] σ 2 σw (t) 1 2 σ2 t i=1 log X n (t) exp{σw (t) 1 2 σ2 t}

121 First Prev Next Go To Go Back Full Screen Close Quit 121 Martingales Recall that an {F t }-adapted process M is a {F t }-martingale if and only if E[M(t + r) F t ] = M(t) for all t, r. Note that this requirement is equivalent to E[M(t + r) M(t) F t ] =.

122 First Prev Next Go To Go Back Full Screen Close Quit 122 Convergence of conditional expectations Lemma 7.12 Suppose X n, X are R-valued random variables and E[ X n X ]. Then for a sub-σ-algebra D, E[ E[X n D] E[X D] ] Proof. By Jensen s inequality E[ E[X n D] E[X D] ] = E[ E[X n X D] ] E[E[ X n X D]] = E[ X n X ]

123 First Prev Next Go To Go Back Full Screen Close Quit 123 Martingale properties of Brownian motion Let F t = Ft W = σ(w (s) : s t). Since W has independent increments, we have E[W (t + r) F t ] = E[W (t + r) W (t) + W (t) F t ] = E[W (t + r) W (t) F t ] + W (t) = W (t), and W is a martingale. Similarly, let M(t) = W (t) 2 t. Then E[M(t + r) M(t) F t ] = E[(W (t + r) W (t)) 2 + 2(W (t + r) W (t))w (t) r F t ] = E[(W (t + r) W (t)) 2 F t ] + 2W (t)e[w (t + r) W (t) F t ] r = and hence M is a martingale.

124 First Prev Next Go To Go Back Full Screen Close Quit 124 An exponential martingale Let M(t) = exp{w (t) 1 2t}, then E[M(t + r) F t ] = E[exp{W (t + r) W (t) 1 2 r}m(t) F t] = M(t)e 1 2 r E[exp{W (t + r) W (t)} F t ] = M(t)

125 First Prev Next Go To Go Back Full Screen Close Quit 125 A general family of martingales Let f Cb 3 (R) (the bounded continuous functions with three bounded continuous derivatives), and t = t < < t m = t + r. Then E[f(W (t + r)) f(w (t)) F t ] = E[ f(w (t i+1 )) f(w (t i )) F t ] = E[ f(w (t i+1 )) f(w (t i )) f (W (t i ))(W (t i+1 ) W (t i )) +E[ 1 2 f (W (t i ))(t i+1 t i ) F t ] = Z 1 + Z f (W (t i ))(W (t i+1 ) W (t i )) 2 F t ]

126 First Prev Next Go To Go Back Full Screen Close Quit 126 Note that since F ti F t, E[f (W (t i ))(W (t i+1 ) W (t i )) F t ] = E[E[f (W (t i ))(W (t i+1 ) W (t i )) F ti ] F t ] = E[f (W (t i ))E[(W (t i+1 ) W (t i )) F ti ] F t ] = E[f (W (t i ))(W (t i+1 ) W (t i )) 2 F t ] = E[E[f (W (t i ))(W (t i+1 ) W (t i )) 2 F ti ] F t ] = E[f (W (t i ))E[(W (t i+1 ) W (t i )) 2 F ti ] F t ] = E[f (W (t i ))(t i+1 t i ) F t ] E[ 1 2 f (W (t i ))(t i+1 t i ) +r t 1 2 f (W (s))ds ] E[ Z 1 ] CE[ W (t i+1 ) W (t i ) 3 ] Ĉ (t i+1 t i ) 3/2 so f(w (t)) f() 1 2 f (W (s))ds is a {F t }-martingale.

127 First Prev Next Go To Go Back Full Screen Close Quit 127 Lévy s characterization Theorem 7.13 Suppose M and Z given by Z(t) = M(t) 2 t are continuous, {F t }-martingales, and M() =. Then M is a standard Brownian motion. Lemma 7.14 Under the assumptions of the theorem, E[(M(t+r) M(t)) 2 F t ] = E[M(t+r) 2 2M(t)M(t+r)+M(t) 2 F t ] = r, if τ is a {F t }-stopping time, then E[(M((t + r) τ) M(t τ)) 2 F t ] = E[(t + r) τ t τ F t ] r, and for f C 3 b (R), is a {F t }-martingale. M f (t) = f(m(t)) f(m()) 1 2 f (M(s))ds (7.3)

128 First Prev Next Go To Go Back Full Screen Close Quit 128 Proof. Taking f(x) = e iθx, is a martingale, and hence and Therefore e iθm(t) e iθm() + E[e iθm(t+r) F t ] = e iθm(t) +r ϕ t,θ (r) E[e iθ(m(t+r) M(t)) F t ] = 1 t 1 2 θ2 e iθm(s) ds 1 2 θ2 E[e iθm(s) F t ]ds r E[e iθ(m(t+r) M(t)) F t ] = e 1 2 θ2r. 1 2 θ2 ϕ t,θ (s)ds It follows that (M(t + r) M(t)) is Gaussian and independent of F t.

129 First Prev Next Go To Go Back Full Screen Close Quit 129 Proof.[of lemma] To prove that (7.3) is a martingale, the problem is to show E[ f(m(t i+1 )) f(m(t i )) f (M(t i ))(M(t i+1 ) M(t i )) 1 2 f (M(t i ))(M(t i+1 ) M(t i )) 2 F t ] converges to zero. Let ɛ, δ >, and assume that t i+1 t i δ. Define τ ɛ,δ = inf{t : sup M(t) M(s) ɛ} t δ s t Then M(ti+1 τ ɛ,δ ) M(t i τ ɛ,δ ) 3 ɛ (M(t i+1 τ ɛ,δ ) M(t i τ ɛ,δ )) 2, and the expectation of the term on the left is bounded by ɛt. Select ɛ n and δ n so that τ ɛn,δ n a.s.

130 First Prev Next Go To Go Back Full Screen Close Quit 13 Applications of the optional sampling theorem Let a < < b, and define γ a,b = inf{t : W (t) / ( a, b). Then and Let t. E[γ a,b ] <, so E[W (γ a,b t)] = E[γ a,b t] = E[W (γ a,b t) 2 ] a 2 b 2. = E[W (γ a,b )] = ap {W (γ a,b ) = a} + bp {W (γ a,b ) = b} Consequently, P {W (γ a,b ) = b} = = a + (a + b)p {W (γ a,b ) = b} a a+b and E[γ a,b ] = E[W (γ a,b ) 2 ] = a 2 b a + b + a b2 a + b