Stochastic Analysis. Prof. Dr. Nina Gantert. Lecture at TUM in WS 2011/2012. June 13, Produced by Leopold von Bonhorst and Nina Gantert

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1 Stochastic Analysis Prof. Dr. Nina Gantert Lecture at TUM in WS 211/212 June 13, 212 Produced by Leopold von Bonhorst and Nina Gantert

2 Contents 1 Definition and Construction of Brownian Motion 3 2 Some properties of Brownian Motion 7 3 The Cameron-Martin Theorem and the Paley-Wiener stochastic integral 13 4 Brownian Motion as a continuous martingale 17 5 Stochastic integrals with respect to Brownian Motion 19 6 Itô s formula and examples 24 7 Pathwise stochastic integration with respect to continuous semimartingales 27 8 Cross-variation and Itô s product rule 32 9 Stochastic Differential Equations 35 1 Girsanov transforms 38 2

3 1 Definition and Construction of Brownian Motion 1.1 Historic origin Brown (1827): Movement of a pollen in a liquid Bachelier (19): Model for stock market fluctuations Einstein (195): Motion of a particle 1.2 Heuristic description with symmetric random walks Y 1,Y 2,... iid with P[Y i = 1] = 1 2 = P[Y i = 1], S k = k i=1 Y i, k = 1,2,..., S =. Fix N N Rescale the process to the time interval [,1]: X k N = 1 N S k, k =,1,2...,N Then, (i) X =. (ii) For t < t 1 < t 2 <... < t m 1,t i = k i N,k i {,1,...,N},X ti X ti 1 are independent and E[X ti X ti 1 ] = and Var(X ti X ti 1 ) = Var( 1 N k i j=k i 1 +1 Y j ) = 1 N (k i k i 1 ). Due to the CLT, the laws of (X ti X ti 1 ) converge to N(,t i t i 1 ) for N (and k i N N t i [,1]). This motivates the following definition: 1.3 Basic Definitions Definition 1.1 Brownian Motion (BM) is a stochastic process B t (ω),( t 1) on a probability space (Ω,A,P) such that (i) B = P-a.s. (ii) For t < t 1 < t 2 <... < t n 1, the increments B ti B ti 1, 1 i n, are independent with law N(,t i t i 1 ) (iii) t B t (ω) is continuous for P-a.a.ω. Definition 1.2 Let (B t ) t 1 be a BM on the probability space (Ω,A,P). Then, the image measure of P under the map Ω C[,1] 3

4 ω (B t (ω)) t 1 is the Wiener measure. The Wiener measure is a prob. measure on (C[,1],F), with F = σ({g t : t 1}), where g t : C[,1] R,g t (x) = x(t) (x C[,1]). Interpretation: Def. 1.1: t B t (ω) is a stochastic evolution in time. Def. 1.2: (B t ) t 1 is a random variable with values in C[,1]. Theorem 1.3 Brownian motion exists. There are different proofs of Theorem 1.3. We give here a proof, which constructs linear interpolations on the sets D n = { k 2n : k 2 n }. We follow the proof of Theorem 1.3. in [4]. Proof: Let D = n D n and let (Ω,A,P) be a probability space such that {Z t,t D} are iid random variables on (Ω,A,P), with law N(,1). Let B := and B 1 := Z 1. For each n N, we define the random variables B s,s D n such that: (1) For r < s < t,r,s,t D n,b t B s is independent of B s B r, and B t B s has the law N(,t s) (2) The vectors (B s,s D n ) and (Z t,t D \D n ) are independent. For D = {,1}, we are done. Proceeding inductively, assume that we followed the construction for some n 1. We then define B s for s D n \D n 1 by B s = 1 2 (B s 1 2 n +B s+ 1 2 n)+ 1 2 n+1 2 The first term is the linear interpolation of B at the neighboring points of s in D n 1. Z s. Therefore, B s is independent of (Z t,t D \D n ) and (2) is satisfied. Moreover, since 1 2 (B s+ 1 2 n B s 1 2 n) 4

5 1 depends only on (Z t,t D n 1 ), it is independent of Z 2 n+1 s. By induction assumption, both terms have law N(, ). Hence, their sums B n+1 s B s 1 and their difference 2n B s+ 1 B 2 n s are iid with law N(, 1 ). 2 n Exercise: X and Y iid random variables with law N(,σ 2 ) X +Y,X Y are iid random variables with law N(,2σ 2 ). To see that all increments B s B s 1 2 n,s D n \{} are independent, it suffices to show that they are pairwise independent, since the vector of increments is Gaussian. We saw that B s B s 1 2 n, B s+ 1 2 n B s (with s D n \D n 1 ) are independent. The other possibility is that the increments are over intervals separated by some s D n 1. Choose s D j with this property and j minimal, so that the two intervals are contained in [s 1 2 j,s] and [s,s+ 1 2 j ]. By induction hypothesis, the increments over these two intervals of length 1 2 j are independent, and the increments over the intervals of lengths 1 are constructed from the 2 n independent increments B s B s 1 and B s+ 2 j 1 B s, respectively, using disjoint sets of 2 random variables (Z j t,t D n ). Hence they are independent (1) is satisfied. This completes the induction. Now, we interpolate between the dyadic points. More precisely, let Z 1 t = 1 f (t) = t = linear in between. and for each n 1, 2 n+1 2 Z t t D n \D n 1 f n (t) = t D n 1 linear between consecutive points in D n f,f 1,f 2,...are continuous functions and, n and s D n n B s := f j (s) = f j (s). (1.1) j= j= 5

6 We prove (1.1) by induction. (1.1) holds for n =. Suppose it holds for n 1. Let s D n \D n 1. Since for j n 1, the function f j is linear on [s 1 2 n,s+ 1 2 n ] we get n 1 f j (s) = j= n 1 j=1 f j (s 1 )+f 2 n j (s+ 1 ) 2 n = (B s 1 +B 2 n s+ 1 n). 2 Since f n (s) = 1 Z 2 n+1 s, this gives (1.1). 2 Since Z d s = N(,1), we have for c > 1, and n large enough, P[ Z 1 c n] e c2 n 2 (since x e u 2 2 du 1 x e x2 2, Proof: exercise). the series P[ s D n with Z s c n] n= n= s D n P[ Z s c n] n= (2 n +1)e c2 n 2 converges if c > 2log2. Fix c > 2log2. Apply the Borel-Cantelli lemma : N (ω) < s.t. for n N (ω), and s D n we have Z s < c n For n N (ω), f n < c n 1 2 n 2 For P-a.a. ω, the sequence B m t = m n= f n(t) converges uniformly in t [,1] for m B t := lim m Bm t is continuous in t. We check that the increments of B have the right finite-dimensional distributions: Assume t 1 < t 2 <... < t n 1. Then, we find t 1,k t 2,k... t n,k 1 with t i,k D and lim k t i,k = t i and since t B t is continuous for P-a.a. ω, B ti+1 B ti = lim k (B t i+1,k B ti,k ) P-a.s. Since lim E[B t i+1,k B ti,k ] = and k lim Cov(B t i+1,k B ti,k,b tj+1,k B tj,k ) = lim I {i=j} (t i+1,k t i,k ) = I {i=j} (t i+1 t i ), k k the increments B ti+1 B ti,i = 1,2,...,n, are independent Gaussian random variables with mean and the variance t i+1 t i, using Lemma 1.4 (s. below) Lemma 1.4 (X n ) n N sequence of Gaussian random vectors and lim X n = X P-a.s.. If b := lim E[X n ] and C := lim Cov(X n ) exist, then X is Gaussian with mean b and Covariance Matrix C. Proof: See [4], Prop

7 Definition 1.5 Astochastic process (B t ) t on some prob. space (Ω,A,P) is a Brownian Motion if: (i) B = P-a.s. (ii) For t < t 1 <... < t n, the increments B ti B ti 1 are independent with law N(,t i t i 1 ) (iii) t B t (ω) is continuous for P-a.a. ω. We obtain (B t ) t from a sequence of iid BMs (B t ) t 1 as follows: B t = B t t t + t 1 i.e. by glueing the paths (B i t) t 1 together. B1 i (t ) i= Then (B t ) t is a BM. Proof: exercise Definition 1.6 A stochastic process (V t ) t is a Gaussian process if for all t 1 < t 2 <... < t n, the vector (V t1,...,v tn ) is a Gaussian random vector. (B t ) t is a Gaussian process. Proof: See exercises. 2 Some properties of Brownian Motion The paths of BM are random fractals in the follows sense: Lemma 2.1 (Scaling invariance) Let (B t ) t be a BM and let a >. Then the process (X t ) t given by X t = 1B a a 2 t (t ) is also a BM. 7

8 Proof: Independence and stationarity of the increments and continuity of the paths persist under the scaling. It remains to show that X t X s = 1 a (B a 2 t B a 2 s) has the law N(,t s). But X t X s is a Gaussian RV with the expectation and variance 1 E[(B a 2 a 2 t B a 2 s) 2 ] = 1 a 2 (t s) = t s. a 2 Example 2.2 (1) Let a < < b and consider T a,b with T a,b := inf{t : B t {a,b}} the first exit time of BM from the intervall (a,b). Then, with X t = 1 a B a 2 t, E[T a,b ] = a 2 E[inf{t : X t { 1, In particular E[T b,b ] = b 2 E[T 1,1 ] = const b 2. b a }}] = a2 E[T 1, b ] a (2) Ruin probabilities: P[(B t ) t exits (a,b) at a] = P[(X t ) t exits ( 1, depends only on the ratio b a. b ) at 1] a Theorem 2.3 (Time inversion) Let (B t ) t be a BM. Then the process (X t ) t given by t = X t = tb1 t > t is again a BM. Proof: Recall that (B t1,...,b tn ), t 1 < t 2 <... t n are Gaussian random vectors and are therefore characterized by their expectations and their Covariances Cov(B ti,b tj ) = t i t j (2.1) Proof of (2.1): Let t i < t j. Then E[B ti B tj ] = E[B ti (B tj B ti )]+E[B 2 t i ] = +t i = t i 8

9 (X t ) t is also a Gaussian process (check!) and the Gaussian random vectors (X t1,...,x tn ) have expectations E[X ti ] =,1 i n. For t >,h, the Covariance of X t and X t+h is given by Cov(X t,x t+h ) = Cov(tB1,(t+h)B 1 ) = t(t+h)cov(b1,b 1 ) = t t t+h t t+h Hence, the laws of all the finite vectors (X t1,...,x tn ), t 1 < t 2 <... t n, are the same as for BM. The paths t X t are clearly continuous for all t > (for P-a.a. ω). For t =, we use the following two facts: (1) Since Q is countable, (X t,t,t Q) has the same law as (B t,t,t Q) lim tց,t Q X t = P-a.s. (2) Q (, ) is dense in (, ) and (X t ) t is continuous on (, ) (for P-a.a. ω) so that = lim tց,t Q X t = lim t X t = P-a.s. Example 2.4 (Ornstein-Uhlenbeck-process) Let (B t ) t be a BM, and set X t = e t B e 2t,(t R). Then X d t = N(,1) t Proof: X t is a Gaussian RV with E[X t ] =,Var(X t ) = e 2t e 2t = 1 Further (X t ) t R has the same law as (X t ) t R. Proof: Set X t = X t,(t R) and tb1 t > t B t = t =. Then X t = e t B e 2t = e t Be 2te 2t = e t Be 2t. Since ( B t ) t is a BM, ( X t ) t R d = (X t ) t R. (X t ) t R is a Gaussian process with E[X t ] =, t and Cov(X s,x t ) = E[X s,x t ] = e (s+t) E[B e 2sB e 2t] (2.1) = e (s+t) e 2(s t) = e t s. Later we will see that ( 1 2 X t ) t is a (weak) solution of the stochastic differential equation dx t = db t X t dt. Corollary 2.5 (Law of large numbers) (B t ) t BM. Then, 1 lim t t B t = P-a.s. 9

10 Remark: 1 lim n B n = P-a.s. since B n = n i=1 (B i B i 1 ) = n i=1 Y i, (Y i ) i 1 iid with law N(,1). Proof: Let tb1 t > t X t = t =. 1 Then lim B t t t = lim X1 t t = X = P-a.s. Remark 2.6 The law of the iterated logarithm says that lim sup t B t 2tloglogt = 1 P-a.s. (2.2) In particular, lim sup t lim inf t B t 2tloglogt = 1 P-a.s. (2.3) B t t = +,liminf t B t t = P-a.s. Theorem 2.7 (Paley, Wiener, Zygmund) Let (B t ) t be a BM. Then P[{ω : t B t (ω)is nowhere differentiable}] = 1 Proof: Assume that there is t [,1] such that t B t (ω) is diff. in t. Then, there is a constant M < such that sup s [,1] B t +s B t s M (2.4) If t [ ] k 1, k 2 n 2 for some n > 2,k 2 n, then we have for 1 j n, n B k+j B k+j 1 2 n 2 n M(2j +1) 1 (2.5) 2 n Proof of (2.5): B k+j 1 2 n 2 n B k+j B 2 n t + B t B k+j 1 2 n B k+j M j +1 2 n +M j 2 n M(2j +1) 1 2 n. 1

11 Let A n,k C[,1] be the collection of functions satisfying (2.5) for j = 1,2,3. Claim: Proof of (2.6): and B k+j 2 n B k+j 1 Hence, P[A n,k ] P[ B 1 7M 2 n ]3 (2.6) B k+j 2 n B k+j 1 2 n d = N(, 1 2 n) d = 1 2 n B 1,j = 1,2,3 are independent. Further, 2 n [ P B 1 7M 3 2 n] P [ 2 n k=1 A n,k ] P n=2 Therefore, using the Borel-Cantelli lemma, ( 7M 2 n) 3. ( ) 3 7M 2 n 2 = (7M)3 n 2 n [ 2 n k=1 P[(2.4) holds for some t [,1]] P = P =. A n,k ] <. [ 2 n k=1 [ A n,k 2 n m=2 n=mk=1 happens for infinitely n A n,k ] ] Corollary 2.8 For P-a.a. ω, the function t B t (ω) is not of bounded variation on any interval. Recall that for g which is right-continuous on [a,b], we set { m } V g [a,b] = sup g(t k ) g(t k 1 ) : a t < t 1 <... < t m b,m N. k=1 V g [a,b] is the variation of g on [a,b]. We say that g is of bounded variation (BV) on [a,b] if V g [a,b] <. Example: If t g(t) is increasing on [a,b], g is BV on [a,b], and V g [a,b] = g(b) g(a). Lemma 2.9 Assume g is BV on [a,b] and right-continuous g 1,g 2 : [a,b] R,g 1,g 2 increasing and right-continuous such that g = g 1 g 2 (2.7) 11

12 Proof: See literature. Proof of Corollary 2.8:AtheorembyLebesguesaysthatanincreasingfunctiong : [a,b] R is differentiable for λ-a.a. s [a,b]. Remark 2.1 Let g be right-continuous and increasing on [a,b]. Then, g defines a measure ν g on [a,b] by ν g ((a 1,b 1 ]) = g(b 1 ) g(a 1 ), a a 1 < b 1 b. If g is BV on [a,b], and f C[a,b] we can define b a f(s)dg(s) := b a f(s)ν g1 (ds) b a f(s)ν g2 (ds) (withg 1,g 2 from(2.7)). ForaBM(B t ) t, thisprocedurecannotbeapplied-nevertheless, we will be able to define b a f(s)db s. Remark 2.11 A continuous function g which is BV on [,1] has quadratic variation, i.e. for E n [,1],E n = {,t 1,...,t n,1},( t 1 <... < t n 1) we have (g(t i+1 ) g(t i )) 2 max g(t i+1 ) g(t i ) g(t i+1 ) g(t i ) t i E n t i E n t i E n if s(e n ), where s(e n ) := sup ti E n t i+1 t i. is the mesh of E n. Theorem 2.12 (B t ) t BM. Let (E n ) be a sequence of partitions with s(e n ),E n E n+1 E n+2... Then, t >, Proof: (1) Convergence in L 2 : V n t := t i E n, (B ti+1 B ti ) 2 t P-a.s. and in L 2. E[V n t ] = t i E n, Using the independence of the increments, Var(V n t ) = t i E n, Var(Vt n), see Remark 2.11 Proof of (*): Y = d N(,σ 2 ) (t i+1 t i ) t. Var((B ti+1 B ti ) 2 ) = 2(t i+1 t i ) 2 ( ) t i E n, Var(Y 2 ) = E[Y 4 ] E[Y 2 ] 2 = 3σ 4 σ 4 = 2σ 4. 12

13 (2) We first show P-a.s. convergence along dyadic partitions E n = {, 1 2 n, 2 2 n, 3 2 n,...,1}. For these partitions, Var(V n t ) = 2 2n ( 1 2 n ) 2 = 2 2 n n=1 Var(V n t ) <. Using Lemma 2.13 below we conclude that V n t t P-a.s.. (3) For general sequence of partitions, use (2) and an approximation argument. Lemma 2.13 Y 1,Y 2,... RVs with n=1 Var(Y n) <. Then, (Y n E[Y n ]) P-a.s.. Proof: See exercises. 3 The Cameron-Martin Theorem and the Paley-Wiener stochastic integral We know several facts about paths of BM, for instance: P[ t [,1] s.t. t B t differentiable in t ] = Do these properties remain true for (B t +ct) t 1 or, more generally, for (B t +h(t)) t 1 where h C[,1]? We denote by { H = h C[,1] : there is f L 2 [,1] s.t. h(t) = the Cameron-Martin space (or Dirichlet space). } f(s)ds, t 1 Given h H, f is uniquely determined as an element of L 2 [,1] and we write Example: h = f h = f is true λ-a.s. (λ = Lebesgue measure on [,1]). 13

14 Recall that for two measures µ,ν on (Ω,A) we write µ ν and say µ and ν are singular if there is A A with µ(a) =,ν(a c ) =. We write ν µ and say ν is absolutely continuous w.r.t. µ if, A A, µ(a) = ν(a) =. We write ν µ and say ν and µ are equivalent if ν µ and µ ν. Let µ be Wiener measure on (C[,1],F) and µ h be the law of (B t +h(t)) t 1. Theorem 3.1 Assume h C[,1] and h() =. (i) If h / H then µ h µ. (ii) If h H then µ h µ. For the proof, we will need the following quantity: ( ( ) ( )) 2n 2 j j 1 Q n (h) := 2 n h h 2 n 2 n (n = 1,2,...) (3.1) j=1 Lemma 3.2 Q n (h),n = 1,2,... is an increasing sequence and Moreover, if h H, then h H supq n (h) <. n Q n (h) 1 h (s) 2 ds = 1 Proof: The general inequality (a+b) 2 2a 2 +2b 2 gives [ ( ) ( )] 2 ( ) ( j j 1 2j 1 j 1 h h 2[ h h 2 n 2 n 2 n+1 2 n f(s) 2 ds. )] 2 +2[ h ( ) j h 2 n ( )] 2 2j 1 Summing this inequality over j {1,2,...,n} gives Q n (h) Q n+1 (h) Q n (h) is increasing in n. For h H with h = f, we have, using Jensen s inequality 2 n+1 ( 2n j Q n (h) = 2 n 2 n j=1 f(s)ds j 1 2 n ) 2 2 n j=1 j 2 n j 1 2 n f(s) 2 ds = 1 f(s) 2 ds Hence, h H sup n Q n (h) <. Proof of : Let t U([,1]), then there exists a sequence of intervals I n (t) = [a n,b n ] = [ kn 1, kn 2 n ] 2 s.t. t n In (t), n. Given I 1 (t),...i n (t), the interval I n+1 (t) is, with probability 1 the left or the right half of I 2 n(t). Let M n = M n (t) = 2 n (h(b n )) h(a n )), then (M n ) n 1 is a martingale w.r.t. σ(i n (t) (on ([,1],B [,1],λ)). Furthermore: 2n ( ( ) ( )) 2 k k 1 E[Mn 2 1 ] = 22n h h 2 n 2 n 2 = Q n(h) n k=1 Ifsup n Q n (h) <,(M n ) n 1 isamartingalewhich isboundedin L 2, i.e. sup n E[M 2 n] <. We prove later: 14

15 Lemma 3.3 Assume (M n ) n 1 is a martingale on (Ω,A,P) and (M n ) n 1 is bounded in L 2. Then there exists a RV X s.t. M n X P-a.s. and in L 2 By Lemma 3.3, M n X a.s. and in L 2. Let For j,m fixed, we have n: h ( ) ( j 2 = g j m X(s) λ-a.s. h H and 2 m ) g(s) := s X(t)dt ( ) j j 2 h = m M 2 m n (t)dt j 2 m X(t)dt j,m and by continuity, g(s) = h(s) s [,1] and h (s) = Q n (h) = E[M 2 n ] E[X2 ] = 1 (h (t)) 2 dt. Proof of Lemma 3.3: M n is bounded in L 1 and by the martingale convergence theorem (Theorem in Probability Theory lecture notes) M n X P-a.s. and X L 1. We have for m > n: E[(M m M n ) 2 ] = E[M 2 m ] E[M2 n ] (3.2) since E[M m M n ] = E[E[M m M n A n ]] = E[M n E[M m A n ]] = E[M 2 n ] Fatou s Lemma implies from (3.2) [ ] E[(X M n ) 2 ] E lim inf (M m M n ) 2 liminf m m E[M2 m ] E[M2 n ] The last expression tends to a.s. for n, since E[Mn 2 ] is increasing in n: E[Mn 2 ] (3.2) = n E[(M k M k 1 ) 2 ]+E[M1 2 ]. k=2 Lemma 3.4 (The Paley-Wiener stochastic integral) Let (B t ) t be a BM and h H. Then ξ n := 2 n ( ) ( )) j j 1 2 (h n h (B 2 n 2 n j ) 2 n Bj 1 2 n j=1 converges a.s. and in L 2. The limit is denoted by 1 h db. 15

16 Proof: Recall from the construction of BM that B2j 1 = 1 ( ) 2 n B2j 2 +B 2j +σ 2 2n n 2 n Z2j 1 2 n where σ n = 2 (n+1)/2 and Z t are iid standard normal. Therefore: 2n 1 ξ n ξ n 1 = 2 n σ n j=1 ( 2h ( ) ( ) ( )) 2j 1 2j 2 2j h h 2 n which implies that (ξ n ) n 1 is a martingale. 2 n 2 n Z2j 1 2 n (E[ξ n A n 1 ] = E[ξ n ξ n 1 A n 1 ]+E[ξ n 1 A n 1 ] = E[ξ n 1 A n 1 ]) 2n ( ( ) ( )) 2 [ E[ξn] 2 2n j j 1 ( ) ] 2 = 2 h h E B 2 n 2 n j = Q 2 n Bj 1 n (h). 2 n j=1 Hence, for h H, the convergence follows from Lemma 3.3. Proof of the Cameron-Martin Theorem: Let µ n and µ h,n denote the finite-dimensional distributions on the set D n. Then the Radon-Nikodym derivative dµ h,n dµ n is the ratio of the two Lebesgue densities. For x C[,1] and j x = (n) j x = x ( ) ( j 2 x j 1 ) n 2, n dµ h,n dµ n (x) = 2 n j=1 exp ( ( ) ( ) jx j h) 2 ( j x) 2 exp = exp( H n (x)) 2 1 n with H n (x) = 2 n 1 n j=1 (( jh) 2 2 j x j h). By Theorem 14.5 in the Probability Theory lecture notes, exp( H n (x)) is a martingale 2 1 n under µ, since it is non-negative it converges a.s. to a finite RV X. We show later: µ h (A) = Xdµ+µ h (A {X = }) for A F. This implies: A µ(x = ) = 1 µ µ h µ(x > ) = 1 µ µ h WehaveE µ [H n ] = H n (x)µ(dx) = 1 Q 2 n(h)andvar µ (H n ) = 2 2n n j=1 ( jh) 2 Var µ ( j x) = Q n (h). By Chebyshev s inequality, we get P µ (H n 1 ) ( 1 4 Q n(h) = P µ 2 Q n(h) H n 1 ) 4 Q n(h) Q n(h) ( 1 Q 4 n(h) ) 2 = 16 Q n (h). Now, if h / H, then by Lemma 3.2, H n and x = µ-a.s.. For the converse, suppose h H. By Lemma 3.4 (and the second part of Lemma 3.2), 1 H(x) 2 h h db. Therefore x > µ-a.s. and µ µ h. Finally note that µ h µ µ µ h. 16

17 4 Brownian Motion as a continuous martingale Definition 4.1 Consider a probability space (Ω,F,P). (i) A filtration is a sequence of σ-fields (F t ) t with F s F t F, s < t. (ii) Astochasticprocess(X t ) t on(ω,f,p)isadaptedto(f t ) t ifx t isf t -measurable, t. Suppose (X t ) t is a stochastic process on (Ω,F,P). Then we can define a filtration (F t ) t by taking F t := σ({x s, s t}), i.e. F t is the σ-field generated by {X s, s t}. Then, (X t ) t is adapted to (F t ) t. Definition 4.2 A real-valued stochastic process (X t ) t is a martingale with respect to a filtration (F t ) t if it is adapted to (F t ) t and E[ X t ] < t (4.1) and, for s t E[X t F s ] = X s P-a.s. (4.2) Theprocess (X t ) t isasubmartingale withrespect toafiltration(f t ) t ifit isadapted to (F t ) t, (4.1) holds and for s t E[X t F s ] X s P-a.s. (4.3) and it is a supermartingale with respect to a filtration(f t ) t if it is adapted to (F t ) t, (4.1) holds and, for s t E[X t F s ] X s P-a.s. (4.4) Remark 4.3 If (X t ) t is a martingale with respect to (F t ) t, X t is in general not a martingale but a submartingale. More generally, if (X t ) t is a martingale with respect to (F t ) t and f : R R is a convex function such that E[ f(x t ) ] <, t, then f(x t ) t is a submartingale with respect to (F t ) t. Proof: E[f(X t ) F s ] Jensen f(e[x t F s ]) = f(x s ) P-a.s. hence (4.2) holds. Remark 4.4 Let (B t ) t be a BM and F t = σ({x s, s t}). Then (B t ) t is a martingale with respect to (F t ) t. Proof: E[B t F s ] = E[B t B s F s ]+E[B s F s ] = +B s P-a.s. (See exercise 3.3 (ii): B t B s is independent of F s, hence E[B t B s F s ] =.) 17

18 Definition 4.5 Let (Ω,F,P) be a probability space with a filtration (F t ) t. A RV T with values in [, ] is a stopping time with respect to (F t ) t if {T t} F t, t. Example 4.6 Let (B t ) t be a BM and F t = σ({x s, s t}), t. (i) Let y and T = inf{t : B t = y}. Then, T is a stopping time with respect to (F t ) t. Proof: {T t} = {B s U(y, 1 n )} F t, where n=1 s:s Q (,t) U(y, 1 n ) := {z R : z y < 1 n } (ii) Let I = (a,b), < a < b and T = inf{t : B t I}. Then, T is not a stopping time because {T t} / F t. Proof: See [4]. Definition 4.7 Assume T is a stopping time with respect to the filtration(f t ) t. Define F T := {A F : A {T t} F t, t }. F T is called the σ-field of events observable until time T. A martingale (X t ) t is a continuous martingale if P[t X t (ω) is continuous] = 1. Theorem 4.8 (Optional stopping) Suppose (X t ) t is a continuous martingale and S T stopping times. If the process (X t T ) t is dominated by an integrable RV X, i.e. X t T X, t a.s. and E[ X ] <, then E[X T F S ] = X S P-a.s. Proof: This can be derived from the result for martingales in discrete time (see Theorem 14.9 in the Probability Theory lecture notes for the case S = ). See [4] for details. Theorem 4.9 (Wald s Lemma for BM) Let (B t ) t be a BM and T a stopping time such that either (a) E[T] < or (b) (B t T ) t is dominated by an integrable RV. Then, we have E[B T ] =. Remark 4.1 One does need a condition on T, as the following example shows: Let T = inf{t : B t = 1}. (Then T < P-a.s., see exercise 5.2). Clearly, E[B T ] = 1. We conclude from Theorem 4.9 that E[T] = and that (B t T ) t is not dominated by an integrable RV. 18

19 Proof of Theorem 4.9: We show that (a) implies (b). Suppose E[T] <, and define M k = max t 1 B t+k B k and M = T k=1 M k. Then T E[M] = E[ M k ] = = k=1 E[I {T>k 1} M k ] k=1 P[T > k 1]E[M k ] k=1 E[M ]E[T +1] But E[M ] = E[max t 1 B t ] < (Proof: exercise). If (b) is satisfied, we can apply the optional stopping theorem (Theorem 4.8) with S =, giving E[X T F ] = X, P-a.s. which yields that E[B T ] =. 5 Stochastic integrals with respect to Brownian Motion Let (B t ) t be a BM on some probability space (Ω,F,P) and F t the completion of σ({b s,s t}) (see Theorem 1.32 in the Probability Theory lecture notes). Then, (B t ) t is adapted to (F t ) t. Definition 5.1 A process {X t (ω) : t,ω Ω} is progressively measurable if for each t the mapping X : [,t] Ω R is measurable with respect to the σ-field B [,t] F t. Lemma 5.2 Any process (X t ) t which is adapted and either right-continuous or leftcontinuous is also progressively measurable. Proof: Assume (X t ) t is right-continuous. Fix t >. For n N, s t define X (n) (ω) = X (ω) and X (n) kt (k +1)t s (ω) = X(k+1)t(ω) for < s k =,1,2,...,2 n 1. 2n 2n 2 n The mapping (s,ω) X s (n) (ω) is B [,t] F t -measurable. By right-continuity, we have lim X s (n) (ω) = X s (ω) for all s [,t] and ω Ω, hence the limit mapping (s,ω) X s (ω) is also B [,t] F t -measurable. The left-continuous case is analogous. 19

20 We construct the integral by starting with simple progressively measurable processes H t (ω) and then proceeding to more complicated ones. Consider first step processes {H t (ω) : t,ω Ω} of the form H t (ω) = k A j (ω)i (tj,t j+1 ](t) for t 1 <... < t k+1 j=1 where A j is F tj -measurable, 1 j k. We define the integral as H s db s := k A j (B tj+1 B tj ) j=1 NowletH beaprogressively measurableprocesssatisfyinge [ H 2 s ds] <.SupposeH canbeapproximatedbyasequence ofprogressively measurablestepprocesses H (n),n 1, then we define H s db s := lim More precisely, let H 2 2 := E [ H 2 sds ]. We will show that H (n) s db s. (5.1) (1) Every progressively measurable H satisfying E [ H 2 s ds] < can be approximated in the 2 - norm by progressively measurable step processes. (2) For each approximating sequence, the limit in (5.1) exists in the L 2 -sense. (3) This limit does not depend on the approximating sequence of step processes. We start with (1): Lemma 5.3 For every progressively measurable process {H s (ω) : s,ω Ω} satisfying E [ H 2 s ds] < there exists a sequence (H (n) ) n N of progressively measurable step processes such that lim H (n) H 2 =. Proof: We approximate the progressively measurable process successively by a bounded progressively measurable process a bounded, almost surely continuous progressively measurable process a progressively measurable step process. Let H = {H s (ω),s,ω Ω} be a progressively measurable process with H 2 <. First define H (n) s (ω) = H s (ω) s n otherwise. 2

21 Clearly, lim H (n) H 2 =. Second, approximate any progressively measurable process H on a finite interval by truncating its values, i.e. define H (n) by H (n) s = (H s (ω) n) ( n). Clearly, H (n) is progressively measurable and H (n) H 2. Now we approximate any uniformly bounded progressively measurable H by bounded, almost surely continuous progressively measurable processes. Let h = 1 and n H s (n) (ω) = 1 s H t (ω)dt h s h (where we set H s (ω) = H (ω) for s < ). H (n) is again progressively measurable (since we only take averages over the past). H (n) is almost surely continuous. Further, ω Ω and almost every (with respect to Lebesgue measure) s [,t], 1 lim h h s s h H t (ω)dt = H s (ω). Since H is uniformly bounded, we obtain that lim H (n) H 2 =. Finally, a bounded, amost surely continuous, progressively measurable process H can be approximated by a sequence of progressively measurable step processes H (n) by taking ( ) j H s (n) = H n,ω for j n s j +1 n. The process H (n) are again progressively measurable and one easily sees that This completes the proof of Lemma 5.3. lim H(n) H 2 =. Lemma 5.4 Let H be a progressively measurable step process and E [ H 2 sds ] <. Then [ ( ) ] 2 [ ] E H s db s = E Hsds 2 Proof: Let H = k i=1 A ii (ai,a i+1 ] be a progressively measurable step process. Then [ ( ) ] [ 2 k ] E H s db s = E A i A j (B ai+1 B ai )(B aj+1 B aj ) = k E[A 2 i(b ai+1 B ai ) 2 ]+2 i=1 = k i,j=1 k i=1 j=i+1 E[A i A j (B ai+1 B ai ) E[(B aj+1 B aj ) F aj ]] k [ E[A 2 i]e[(b ai+1 B ai ) 2 ] = E i=1 ] Hsds 2 21

22 Corollary 5.5 Suppose (H (n) ) n N is a sequence of progressively measurable step processes such that Then E[ E[( (H (n) s H (m) s ) 2 ds] n,m. (H (n) s H (m) s )db s ) 2 ] n,m. Proof: Because the difference of two progressively measurable step processes is again a progressively measurable step process, Lemma 5.4 can be applied to H (n) H (m) and yields the claim. We showed (1). The following theorem addresses (2) and (3). Theorem 5.6 Suppose (H (n) ) n N is a sequence of progressively measurable step processes and H a progressively measurable process such that Then lim E[ (H s (n) H s ) 2 ds] = lim H (n) s db s =: H s db s exists as a limit in the L 2 -sense and does not depend on the choice of (H (n) ) n N. Moreover, we have E[ (H (n) s H (m) s ) 2 db s ] n,m (5.2) Proof: ( By the triangle inequality (H (n) ) n N satisfies the assumptions of Corollary 5.5 and ) hence H (n) s db s is a Cauchy sequence in L 2. Since L 2 is complete, the limit exists n N and does not depend on the choice of the approximating sequence. Finally, (5.2) follows from Lemma 5.4, applied to H (n), taking the limit as n. This completes the construction of the stochastic integral H 2 sdb s for progressively measurable processes H with E [ H 2 s ds] <. Remark 5.7 If the sequence of step processes in Theorem 5.6 is chosen such that then by (5.2) we get n=1 and therefore, almost surely, [ ] E (H s (n) H s ) 2 ds n=1 <, [ ( ) ] 2 E (H s (n) H s )db s < ( n=1 ) 2 H s (n) B s H s db s <. 22

23 This implies that, almost surely, lim H (n) s db s = H s db s. We now want to describe the stochastic integral as a process in time. We will see that it will be a continuous martingale. Definition 5.8 Suppose H = {H s (ω) : s,ω Ω} is progressively measurable with E [ H 2 s ds] <. Define the progressively measurable process {Hs t (ω),s,ω Ω} H t s (ω) := H s(ω)i {s t}. Then, the stochastic integral of H up to time t is defined H s db s := H t s db s. Remark 5.9 We have seen already in Lemma 3.4 that for g L 2 [,1] we can define 1 g(s)db s. Provided that both integrals exist, the Paley-Wiener integral from Lemma 3.4 agrees with the stochastic integral just defined. Proof: See exercises. Definition 5.1 A stochastic process (X t ) t is a modification of a stochastic process (Y t ) t if, for every t, we have P[X t = Y t ] = 1. [ ] t Theorem5.11 Assume that(h s (ω)) s is progressivelymeasurableand E H s(ω) 2 ds < (, t. Then there exists a modification (M t ) t of H sdb s such that )t P[t M t (ω) is continuous] = 1. Further, (M t ) t is a martingale and hence [ ] E H s db s = t. Proof: Fix t N and let H (n) be a sequence of progressively measurable step processes such that E [ ( H (n) H t 2 ) ] 2 (H s (n) H t s )db s. For s t the random variable s H(n) u db u is F s -measurable and E[ s H(n) u ( db u F s ] = (proof: exercise!) which implies that the process H(n) u db u is a martingale, n. ) t t By Doob s maximal inequality, see below, for p = 2, [ ( ) 2 ] [ ( E H s (n) db s H s (m) db s 4E sup t t 23 (H (n) s H (m) s )db s ) 2 ].

24 This implies that M (n) t := H(n) s db s, t t, n =,1,2,... defines a Cauchy sequence in the space of continuous functions on [,t ] (equipped with the supremum norm). We denote the limit of this Cauchy sequence by (M t ) t t. Hence, the process (M t ) t t is almost surely a uniform limit of continuous processes and therefore almost surely continuous. Due to Theorem 5.6, [ P M t = H s db s ] = 1. For fixed t [,t ], the random variable H sdb s is the limit (in L 2 ) of H(n) s db s, hence it is F t -measurable, and H t s db s has conditional expectation E[ H t s db s F t ] =. Therefore, H sdb s is a conditional expectation of M t, given F t, i.e. [ ] M t = E H s db s F t. Therefore, (M t ) t t is a martingale, as a process of successive predictions, (see (14.4) in Probability Theory lecture notes). Doob s maximal inequality Suppose (X t ) t is a continuous martingale and p > 1. Then, for any t ( ) p p E[ sup X s p ] E[ X t p ]. s t p 1 Proof: See literature. 6 Itô s formula and examples Let f C 1 (R) (f continuously differentiable) and x : [, ) R x continuous and BV on [,t]. Then f(x(t)) f(x()) = f (x(s))dx(s) Example: x(s) = s, s. Then f(t) f() = f (s)ds. Itô s formula gives an analogue for the case when x is replaced by a BM B t. The crucial difference is that the second derivative of f is needed. Theorem 6.1 (Itô s formula I) Let f : R R be twice continuously differentiable such that E[ (f (B s )) 2 ds] < for some t >, where (B t ) t is a BM. Then, almost surely for all s [,t] f(b s ) f(b ) = s f (B u )db u s f (B u )du. 24

25 Example 6.2 Let f(x) = x 2. We have E[ s B2 u du] = s udu <, s >. Hence s s Bs 2 = 2 B u db u +s Bs 2 s = 2 B u db u. We conclude from Theorem 5.11 that (B 2 s s) s is a martingale. Example 6.3 Let f(x) = x 3. We have E[ s B4 u du] = s E[B4 u ]du <, s >. Hence Let B 3 s = 3 s B 2 udb u +3 s B u du. s M s = Bs 3 3 B u du, s. We conclude from Theorem 5.11 that (M s ) s is a martingale. To prove Theorem. 6.1, we will need the following. Theorem 6.4 Let f : R R is continuous, t >, and = t (n) 1 <... < t (n) n partitions such that their mesh max 1 i n 1 t(n) i+1 t(n) i. = t are Then n 1 j=1 f(b (n) t )(B (n) j t j+1 B (n) t ) 2 f(b s )ds j Proof: For f 1, see Theorem For general f, see Theorem 7.12 in Mörter/Peres. Proof of Theorem 6.1: We write w(δ,m) for the modulus of continuity of f on [ M,M]: w(δ,m) = sup f (s) f (t) s,t [ M,M], s t <δ Using Taylor s formula, for any x,y [ M,M],x < y with x y < δ, f(y) f(x) f (x)(y x)+ 1 2 f (x)(y x) 2 w(δ,m)(y x) 2. Take a sequence = t (n) 1 <... < t (n) n = t. We write = t 1 <... < t n = t for simplicity. With and we get δ B := max 1 i n 1 B t i+1 B ti M B := max s t B s, n 1 (f(b ti+1 ) f(b ti )) i=1 n f (B ti )(B ti+1 B ti ) i=1 n 1 i=1 1 2 f (B ti )(B ti +1 B ti ) 2 25

26 n 1 w(δ B,M B ) (B ti+1 B ti ) 2 i=1 Now, n 1 i=1 f(b t i+1 ) f(b ti ) = f(b t ) f(b ), and there is a sequence of partitions with mesh going to s.t. n 1 f (B ti )(B ti+1 B ti ) i=1 n 1 f (B ti )(B ti+1 B ti ) 2 i=1 f (B s )db s f (B s )ds n 1 (B ti+1 B ti ) 2 t P a.s. i=1 P a.s. P a.s. By continuity of the Brownian path, w(δ B,M B ) converges almost surely to. This proves Itô s formula for fixed t, or indeed almost surely for all s Q [,t]. Since all the terms in Itô s formula are continuous almost surely, we get the result simultaneously for all s [,t]. We next state Itô s formula for functions f which can depend also on time. Theorem 6.5 (Itô s formula II) Let f : R R R, (x,t) f(x,t) be twice continuously differentiable in the x-coordinate and once continuously differentiable in the t-coordinate. Assume that E[ ( xf(b s,s)) 2 ds] < for some t >. Then, almost surely for all s [,t]: f(b s,s) f(b,) = Proof: See Mörter/Peres. Example 6.6 Fix α >. s Then f(b s,s) 1 = x f(b u,u)db u + s d t f(b u,u)du+ 1 2 f(b t,t) = e αbt 1 2 α2t = M t (M = 1) s αm u db u + s M s M = s s ( 1 2 α2 )M u du+ 1 2 αm u db u. s xx f(b u,u)du α 2 M u du We conclude from Theorem 5.11 that (M t ) t is a martingale (details: exercise). M solves the stochastic differential equation dm s = αm s db s, M = 1 which can be written in integral form M s = 1+ s M u db u, s. 26

27 Definition 6.7 (B t ) t = (B (1) t,...,b (d) t ) t is a d-dim. BM if (B (1) t ) t,...,(b (d) t ) t are iid one-dimensional BMs. For H s = (H s (1),...,H s (d) ) we write H s db s = d i=1 H (i) s db(i) s. Theorem 6.8 (Multidimensional Itô formula) Let (B t ) t be a d-dim. BM and f : R d+1 R be such that the partial derivatives i f and jk f exist for all 1 i d+1,1 j,k d and are continuous. If for some t > [ E ] x f(b s,s) 2 ds < where x f = ( 1 f,..., d f) then, almost surely, for all s t f(b s,s) f(b,) = where x f = d j=1 jjf. s x f(b u,u)db u + s d+1 f(b u,u)du+ 1 2 s x f(b u,u)du 7 Pathwise stochastic integration with respect to continuous semimartingales We saw that for a continuous process H with E[ H2 sds] <, H sdb s can be defined as an almost sure limit, see Remark 5.7. Can we replace (B t ) t with another continuous martingale? Definition 7.1 Let E n = { = t (n) 1 <... < t (n) n } be a sequence of partitions with s(e n ) = sup t (n) i+1 t(n) i. Then, the function X t has continuous quadratic variation along the sequence E n if exists P-a.s. X t = lim (X tj+1 X tj ) 2 (7.1) t i E n Remark 7.2 X t is increasing and continuous, hence it defines a measure ν on (R,B) given by ν((a,b]) = X b X a. In particular, if f : R R is continuous, f(s)d X s is well-defined. In analogy to Theorem 6.4, we have 27

28 Theorem 7.3 Assume that X t has continuous quadratic variation X t and f : R R is continuous. Then, for t >, f(x ti )(X tj+1 X tj ) 2 t i E n Proof: Clear for f 1 due to (7.1). Rest see literature. f(x s )d X s Note that the assumption X t has continuous quadratic variation can only be satisfied for continuous processes. Theorem 7.4 (Itô s formula for (deterministic) functions with continuous quadratic variation) Assume that the function X t has continuous quadratic variation X t and let f : R R be twice continuous differentiable. Then, where f(x t ) f(x ) = f (X u )dx u f (X u )dx u = lim t i E n Sketch of proof: As in the proof of Theorem 6.1, we have f (X u )d X u (7.2) f (X ti )(X ti+1 X ti ). f(x ti+1 ) f(x ti ) f (X ti )(X ti+1 X ti ) t i E n t i E n t i E n w(δ X,M X ) (X ti+1 X ti ) f (X ti )(X ti+1 X ti ) 2 where and and Now, and t i E n δ X = max X ti+1 X ti t i E n w(δ,m) = sup M X = max s t X s s,t [ M,M] s t <δ f (s) f (t). (f(x ti+1 ) f(x ti )) f(x t ) f(x ) t i E n f (X ti )(X ti+1 X ti ) 2 t i E n f (X s )d X s 28

29 and (X ti+1 X ti ) 2 X t. t i E n Moreover, w(δ X,M X ) δ. Hence, f (X ti )(X ti+1 X ti ) t i E n has to converge as well and (7.2) holds. Remark 7.5 (1) Theorem 7.4 gives a pathwise version of Itô s formula, without probability. (2) If X is BV on [,t], X t = and (7.2) becomes Short notation: f(x t ) f(x ) = f (X u )dx u. df(x) = f (X)dX classical differential If X has continuous quadratic variation X and X t, then df(x) = f (X)dX f (X)d X Itô differential We have defined f (X u )dx u for all continuous X t with continuous quadratic variation X t. Which stochastic processes X t have a.s. continuous quadratic variation X t? Lemma 7.6 (i) Assume X t has continuous quadratic variation X t. Then, if f is continuously differentable, f(x t ) has continuous quadratic variation f(x) t = f (X s ) 2 d X s (ii) Let X t = M t +A t,t, where M t has quadratic variation M t and A t has quadratic variation A t =, then X t has quadratic variation X t and X t = M t. (iii) Let f C 1 (R) and assume X t has continuous quadratic variation X t. Then, M t = f(x s)dx s has continuous quadratic variation M t = f(x s ) 2 d X s. (iv) Let f C 1 (R 2 ) and assume X t has continuous quadratic variation X t. Then, g(t) = f(x t,t) has continuous quadratic variation ( ) 2 f g t = x (X s,s) d X s. 29

30 Examples: 1. (B t ) t BM, α >,Z t = e αbt,t. Then, Lemma 7.4 (i) implies that Z t = α2 e 2αBs ds = α2 Zs 2ds P-a.s. Note that Z t is random (whereas B t = t, t P-a.s.). 2. (B t ) t BM,α >,M t = e αbt 1 2 α2t,t. Then, weknowthatm t = 1+ αm sdb s,t. Hence, Lemma 7.4 (iv) implies that M t = α2 Ms 2ds P-a.s. Note that M t is random (whereas B t = t, t P-a.s.). Proof of Lemma 7.4: (i) i X := X ti+1 X ti. Then and Hence, f(x ti+1 ) f(x ti ) = f(x ti ) i X +R i R i sup f (X s ) f (X u ) i X s,u [t i,t i+1 ] (f(x ti+1 ) f(x ti )) 2 = f (X ti ) 2 ( i X) 2 + Ri 2 +2 f (X ti ) i X R i t i E n But due to Theorem 7.3. Rest: Exercise. t i E n f (X ti ) 2 ( i X) 2 t i E n (ii) i M = M ti+1 M ti, i A = A ti+1 A ti. Then Hence i:t i E n ( i X) 2 = t i E n f (X s ) 2 d X s ( i X) 2 = ( i M) 2 +( i A) 2 +2 i M i A. i:t i E n ( i M) 2 + i:t i E n ( i A) 2 +2 i:t i E n t i E n i M i A M t, since ( i A) 2 and (details: exercise). i:t i E n i:t i E n i M i A 3

31 (iii) Due to (7.2), taking g such that g = f, M t = g(x t ) g(x ) 1 2 g (X s )d X s. But g (X s )d X s is continuous and BV as a function of t. Using (ii) and (i), M t = g(x) t = (iv) See [5], Remark g (X s ) 2 d X s = f(x s ) 2 d X s Theorem 7.7 If (X t ) t is a continuous martingale with X = and E[X 2 t] <, t there exists a unique process X t with X = which is continuous, adapted and increasing, such X 2 t X t is a martingale. Moreover, X t has quadratic variation X t. Proof: If (B t ) t is a BM and (H t ) t progressively measurable and X t = H sdb s, then X t = H2 sds P-a.s. (see Lemma 5.4 and Lemma 7.4(i).) General case: See literature. X t is called quadratic variation or bracket or compensator of (X t ). The first part of Theorem 7.6 has a discrete time analogue. A stochastic process (Y n ) is previsible w.r.t. a filtration (A n ) if Y n is A n 1 -measurable n. Theorem 7.8 (Doob decomposition) Suppose (X n ) n=,1,2,... is a martingale w.r.t. the filtration (A n ) and E[X 2 n] <, n. Then, there is a unique previsible increasing process (A n ) with A = such that (X 2 n A n ) n=,1,2,... is a martingale w.r.t. (A n ). Proof: Let A = and define, for n 1,A n = A n 1 +E[X 2 n A n 1] X 2 n 1. Clearly, A n is A n 1 -measurable n. Since (X 2 n) n=,1,2,... is a submartingale w.r.t. A n, A n is increasing. Further, E[X 2 n A n A n 1 ] = E[X 2 n 1 A n 1 A n 1 ] = X 2 n 1 A n 1 (X 2 n A n) n=,1,2,... is a martingale w.r.t. (A n ). To prove the uniqueness, assume that (A n ) and (B n ) both fulfill the requirements. Then A n B n is a previsible martingale starting at, and we infer that A n B n = E[A n B n A n 1 ], hence, by induction, A n = B n n. Definition 7.9 A continuous semimartingale w.r.t. a filtration (F t ) t is a process (X t ) t which is adapted to (F t ) t and which has a decomposition X t = X + M t + A t, t P-a.s. where (M t ) t is a continuous martingale and (A t ) t is a continuous adapted process which is BV (on each interval [,t]). If E[M 2 t] <, t, we define for f C 1 (R) f(s)dx s := f(s)dm s + 31 f(s)da s

32 8 Cross-variation and Itô s product rule Definition 8.1 The cross-variation X, Y is given by (provided that the limit exists). Clearly, X,X = X. X,Y t = lim t i E n Lemma 8.2 The following statements are equivalent (i) X,Y exists and t X,Y t is continuous. (ii) X +Y exists and t X +Y t is continuous. If (i) and (ii) hold, then (X ti+ X ti )(Y ti+ Y ti ) X,Y = 1 ( X +Y X Y ) (8.1) 2 In particular, in this case g(s)d X,Y s is well-defined for g C[, ). Proof: (X ti+1 X ti )(Y ti+1 Y ti ) = 1 ( ) 2 ((Xti+1 +Y ) (X +Y )) 2 X ti+1 ti ti (Xti+1 ti )2 (Y Y ti+1 ti )2. Example 8.3 Let (X t ) t and (Y t ) t be independent BMs. Claim: For P-a.a. ω Proof: X(ω),Y(ω) t = t X,Y = 1 ( X +Y X Y ) 2 and we have X t = t, t, and Y t = t, t P-a.s. Suffices to show X+Y = 2t t P-a.s. Z t := 1 2 (X t +Y t ), t is again a BM (Proof: exercise). Therefore, Z t = t t P-a.s. implying that X +Y = 2t, t P-a.s. Lemma 8.4 X, X continuous as before, f,g C 1 (R), Then Y t = f(x s )dx s, Z t = Y,Z t = g(x s )dx s f(x s )g(x s )d X s (8.2) 32

33 Proof: Define F and G by F = f, G = g, F() = G() =. Then, Y t = F(X t )+A t where and Z t = G(X t ) B t where A t = F(X ) 1 2 F (X s )d X s B t = G(X ) 1 2 G (X s )d X s. A t,b t are continuous with quadratic variation. Hence, Y,Z t = 1 2 ( Y +Z t Y t Z t ) = 1 ( (f(x s )+g(x s )) 2 d X s 2 = f(x s )g(x s )d X s f(x s ) 2 d X s g(x s ) 2 d X s ) Theorem 8.5 (Itô s formula in d dimensions) X = (X (1),...,X (d) ) : [, ) R d where X (i), 1 i d are continuous with continuous quadratic variation X (i) t and continuous cross-variations X (i),x (k) t, 1 i, k d. Let f C 2 (R d ). Then, f(x t ) f(x ) = ( f,dx s )+ 1 2 d i,k=1 2 f x i x k (X s )d X (i),x (k) s where ( f,dx s ) = d k=1 f x i (X s )dx (i) s. Proof: Analogous to the proof of Theorem 6.1: Taylor formula forf(x ti+1 ) f(x ti ). See literature. Note that Theorem 6.8 follows, using Example 8.3. Corollary 8.6 (Itô s product rule) Assume that X,Y, X, Y and X,Y are continuous. Then, t > Short notation: X t Y t = X Y + Y s dx s + X s dy s + X,Y t (8.3) d(x Y) = YdX +XdY +d X,Y (8.4) Proof: Apply Theorem 8.5 with d = 2,f(x,y) = x y. 33

34 Example 8.7 (Ornstein-Uhlenbeck process) Let α > and (B t ) t a BM, x R. Then, we say that X t = e αt x +e αt e αs db s, t (8.5) is an Ornstein-Uhlenbeck process with parameter α and starting point x. Claim: (X t ) t solves the stochastic differential equation dx t = db t αx t dt X t = B t i.e. αx sds X = x X = x Proof: X t = e αt x +e αt e αs db s dx t = αx e αt dt α(e αt e αs db s )dt+e αt e αt db t = αx t dt+db t, where we applied Itô s product rule to e αt eαs db s. Claim: If X d = N(, 1 2α ), X independent of (B t ) t, then X t = e αt X +e αt e αs db s is a Gaussian process with E[X t ] =, t and Cov(X s,x t ) = 1 2α e α t s. Hence, for any t, E[Xt] 2 = 1 X d 2α t = N(, 1 ). 2α Inparticular, fortheprocessz t = e t B e 2t,t, inexample2.4, ( 1 2 Z t ) t isanornstein- Uhlenbeck process with α = 1 and X = N(, 1 2 ). Proof: Clearly, E[X t ] =, t. Assume s t. Then, E[X s X t ] = E[X 2 ]e αt e αs +++e α(s+t) E[ s e αu db u e αv db v ] For any martingale (M t ) t, E[M t M s ] = E[Ms 2 ] for s t.(proof: exercise). Hence, applying this to the martingalem t = eαu db u, E[X s X t ] = 1 2α e α(t+s) +e α(t+s) E[( = 1 2α e α(t+s) +e α(t+s) s s e αv db v ) 2 ] e 2αv dv = 1 2α e α(t+s) +e α(t+s) 1 2α (e2αs 1) = 1 2α e α(t s). 34

35 9 Stochastic Differential Equations In this chapter we want to study stochastic differential equations (SDE) of the form: X = ξ dx t = σ(t,x t ) db t +b(t,x t )dt (9.1) (X t ) t = (X (1) t,x (2) t,...,x (n) t ) t is an unknown R-valued process, (B t ) t = (B (1) t,b (2) t,...,b (m) t ) t is an m-dimensional Brownian motion and b(t,x) and σ(t,x) are measurable functions of (t,x) R + R n. The drift vector b(t,x) is R n -valued and the dispersion matrix σ(t,x) is an n m-matrix valued. Further ξ is a random variable with values in R n which is independent of (B t ) t. Definition 9.1 We say the SDE (9.1) has the strong solution (X t ) t if the following conditions hold: (i) (X t ) t is adapted to the filtration (F ξ t) t where F ξ t is the completion of σ(ξ,{b s : s t}) for t. (ii) (X t ) t satisfies the following integral equation: for t,i = 1,2,...n. X (i) t = ξ (i) + m j=1 σ ij db (j) s + b i (s,x s )ds (9.2) Remark 9.2 (1) Processes which fulfill the integral equations in (9.2) and are defined on a possibly enlarged probability space, but do not have to be adapted to the filtration (F ξ t) t, are called weak solutions. (2) The second property of a solution of a SDE includes the requirement that the integrals in (9.2) are well defined. Example 9.3 For α,β R consider the SDE X = 1 dx t = αx t db t +βx t dt for a one-dimensional BM (B t ) t. This SDE has the (unique) strong solution X t = exp(αb t +(β α2 2 )t). Proof: exercise Remark The SDE from Example 9.3 is used for the Black-Scholes model. 35

36 Remark 9.4 In the following Theorem we use x 2 = n (x i ) 2 for x = (x 1,...,x n ) R n, i=1 σ 2 = n m (σ ij ) 2 for σ = (σ ij ) ij R n m. i=1 j=1 Theorem 9.5 (Existence and uniqueness) Assume that b : R + R n and σ : R + R n R n m are measurable and satisfy the Lipschitz-condition as well as the growth condition σ(t,x) σ(t,y) + b(t,x) b(t,y) K x y (9.3) σ(t,x) 2 + b(t,x) 2 K 2 (1+ x 2 ) (9.4) for a constant K >, for all t and x,y R n. Let ξ be a random variable with values in R n which is independent of the m-dimensional BM (B t ) t such that E[ ξ 2 ] <. Then the SDE (9.1) has a unique, continuous, strong solution (X t ) t such that [ T ] E X t 2 dt < T > (9.5) Remark 9.6 Uniqueness intheorem9.5meansthatfortwocontinuoussolutions(x t ) t, (X t) t of the SDE (9.1) which fulfill the properties of Theorem 9.5 we have P(X t = X t t ) = 1. Example 9.7 To illustrate that we need some conditions like (9.3) and (9.4) let us look at the following examples from ODEs: dx t dt = (X t ) 2, X = 1 corresponding to b(x) = x 2 (which does not satisfy (9.4)) has the unique solution Another example: X t = 1 1 t dx t dt for t < 1. = 3(X t ) 2/3, X = where b(x) = 3x 2/3 does not satisfy (9.3) at x = and, for t a, X t = (t a) 3, for t > a. are solutions for all a >. 36

37 Sketch of the proof of Theorem 9.5: Uniqueness: Uses the following lemma Lemma 9.8 Gronwall inequality Let f : [,T] R be integrable and A R,C > such that Then Proof: see literature. f(t) A+C f(t) A+e Ct f(s)ds t [,T], t [,T]. Consider two continuous solutions (X t ) t,(x t ) t of the SDE (9.1) which fulfill condition (9.5) X t X t = (σ(s,x s ) σ(s,x s )) db s + Therefore (using (a+b) 2 2a 2 +2b 2 ), X t X t 2 (σ(s,x s ) σ(s,x s)) db s 2 +2 (b(s,x s ) b(s,x s ))ds (b(s,x s ) b(s,x s))ds 2 A short calculation for the first term (see Theorem 5.6 for n = m = 1) and an application of the Cauchy-Schwarz inequality shows E[ X t X t 2 ] 2 E[ (σ(s,x s ) σ(s,x s)) 2 ]db s +2t E[ (b(s,x s ) b(s,x s)) 2 ]ds Therefore we have for f(t) := E[ X t X t 2 ] and C := 2(T + 1)K 2 (for T > ) due to condition (9.3) f(t) C f(s)ds for t T and the Gronwall inequality implies f. We can conclude due to the continuity of (X t ) t,(x t ) t that we have P(X t = X t t ) = 1. (using the fact that Q + is countable and dense in R + ). Existence: We define for n N X (n+1) t := ξ + σ(s,x (n) s ) db s + b(s,x (n) s )ds for t inductively where X t := ξ. Using the growth condition we can show inductively that for T > T E[ X (n) s 2 ]ds <, i.e. the stochastic integral is well defined in every step of the recursion. One can show that (X (n) ) n N converges a.s. uniformly on [,T] for every T > and the limit solves the SDE (9.1) and has the properties of Theorem 9.5 (more details can be found in the literature). 37

38 1 Girsanov transforms Goal: Construction of a stochastic process (X t ) t with dx t = b(x t,t)dt+db t where (B t ) t is a BM, i.e. X t = x + Interpretation: X = x b(x s,s)ds+b t (1.1) Deterministic process X t with d X dt t = b(x t,t) with additional noise (B t ) t. Possible strategies: 1) Construction of a strong solution: For a given BM (B t ) t on (Ω,A,P), solve (1.1). Example 1.1 (Ornstein-Uhlenbeck process) dx t = db t αx t dt X = x has the strong solution see Example 8.8. X t = e αt (x + e αs db s ), (1.2) Drawback: a strong solution does not always exist. 2) Construction of a weak solution: Find a BM (B t ) t and a process (X t ) t on some probability space (Ω,A,P) such that (1.1) holds, i.e. find (X t ) t such that is a BM. B t = X t x b(x s,s)ds General Girsanov transformation (Ω,A,P) probability space, (F t ) t filtration, P probability measure on (Ω,A). Assume that for all t, P Ft P Ft. Then, there are Radon-Nikodym derivatives Z t := d P dp F t = d P Ft dp Ft (1.3) 38

39 and (Z t ) t is a martingale with respect to (F t ) t and P (see Probability Theory lecture notes, ) We always assume that (Z t ) t has continuous paths and that inf{t : Z t (ω) = } = P-a.s. Definition 1.2 (M t ) t is a local martingale (up to ) if there is a sequence of stopping times T 1 T 2... such that i) supt n = n P-a.s. ii) (M t Tn ) t is a martingale, n. Each martingale is a local martingale but there are local martingales which are not martingales. We will use the following important fact: If M is a continuous local martingale and (Y t ) t a continuous adapted process, then Y sdm s, t is again a continuous local martingale. See the literature, for instance [5], Proposition Lemma 1.3 In the above setup, with Z t := d P dp F t, the following holds. (i) For s t and a function g t which is F t -measurable and bounded, we have Ẽ[g t F s ] = 1 Z s E[g t Z t F s ] P-a.s. where Ẽ denotes expectation with respect to P. Proof: (ii) For M = ( M t ) t continuous and adapted, the following two statements are equivalent: a) ( M t ) t is a local martingale with respect to P. b) ( M t Z t ) t is a local martingale with respect to P. (i) Assume g s is F s -measurable and bounded. Ẽ[g s g t ] = E[g s g t Z t ] = E[g s E[g t Z t F s ]] ] = E [g s Z s E[g t Z t F s ] 1Zs [ ] = Ẽ 1 g s E[g t Z t F s ] Z s Ẽ[g t F s ] = 1 Z s E[g t Z t F s ] 39

40 (ii) Assume that ( M t Z t ) t is a martingale with respect to P. Then Ẽ[ M t F s ] = (i) 1 Z s E[ M t Z t F s ] = 1 Z s Ms Z s = M s, P-a.s. ( M t ) t is a martingale with respect to P, hence b) a) for martingales. Rest of the proof: Exercise. Theorem 1.4 Assume that (M t ) is a continuous local martingale with respect to P and Z t is defined as in (1.3). Then, M t = M t is a continuous local martingale with respect to P. Short notation: 1 Z s d M,Z s dm = d M + 1 Z d M,Z Proof: Due to Lemma 1.1, it suffices to show that ( M t Z t ) t is a continuous local martingale with respect to P. Let A t = 1 Z s d M,Z s. Then, M t Z t = Z t (M t A t ) The definition of A implies that Itô s product rule = Z M + (M s A s )dz s Z s da s = Z,M t. Z s da s + Z,M t (1.4) Due to (1.4), Mt Z t M Z is a stochastic integral, hence again a continuous local martingale with respect to P. Remark 1.5 logz t = logz + Z continuous local martingale Y t = Y = 1 Z 2 sd Z s, hence 1 t dz s Z s 1 Z 2 s d Z s 1 Z s dz s is a continuous local martingale with logz t = logz +Y t 1 2 Y t Z t = Z e Yt 1 2 Y t and Z solves dz = ZdY. Using M,Z t = Z sd M,Y s, see Lemma 1.6 below we get from Theorem 1.3 that dm = d M +d M,Y (1.5) 4