Brownian Motion and Stochastic Calculus

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1 ETHZ, Spring 17 D-MATH Prof Dr Martin Larsson Coordinator A Sepúlveda Brownian Motion and Stochastic Calculus Exercise sheet 6 Please hand in your solutions during exercise class or in your assistant s box in HG E65 no latter than April 7th Exercise 61 Let (B t t be a Brownian motion and define the process (M t t by M t sup s t B s Show that for any fixed t Law M t B t B t Law M t (1 That is, show that the random variables have the same density functions Solution 61 For any g : R R bounded Borel measurable function, we know that E g( B t 1 g( x e x /(t dx g(x πt πt e x /(t dx ( Thus, we see that the probability density function of B t on R is given by the function x 1 x πt e x /(t From Corollary 55 of the script, we know that the probability density function of the joint law of (B t, M t where M t : sup s t B s is given by the function (x, y (y x πt 3 e (y x /(t 1 {y, x y} (3 Take any g : R R bounded Borel measurable function We deduce from (3 that E g(m t B t (y x g(y x e (y x /(t dx dy πt 3 y, y x By a change of variable u : y x v : y we get that E g(m t B t g(u (u + v e (u+v /(t du dv (4 πt3 u, v By another change of variable n : u and m : u + v and as x e cx / dx e cx / c, we get that E g(m t B t g(n πt 3 m e m /(t dm dn n g(n πt e n /(t dn (5 Law Comparing ( with (5 yields that M t B t B t Now, from (3, we deduce for any g : R R bounded Borel measurable function that E g(m t y, y x g(y (y x πt 3 e (y x /(t dx dy Updated: March 8, 17 1 / 7

2 Brownian Motion and Stochastic Calculus, Spring 17 By a change of variable u : y and v : y x we get that E g(m t g(u (u + v e (u+v /(t du dv (6 πt3 u, v Comparing (4 with (6 yields M t B t Law M t Updated: March 8, 17 / 7

3 Brownian Motion and Stochastic Calculus, Spring 17 Exercise 6 Let (B t t be a Brownian motion and denote by G t : σ(b u, u t, t Define R f(x f(x and R t f(x 1 f(y exp ( 1t (y πt x + exp ( 1t (y + x dy, t > Let us consider the process (X t t by X t : B t Show that E f(x t+h G t Rh f(x t P -as for f bb(r and t, h Solution 6 Fix any t, h and f bb(r The case where h is trivial, therefore, let h > From the lecture (cf Example 3 in Section 3 in the lecture notes, we know that Brownian motion is a Markov process with transition semigroup given by R f(x f(x and 1 R h f(x πh Therefore, we get for f(x : f( x bb(r that R E f(x t+h Gt Rh f(bt 1 πh (y x f(y exp ( dy when h >, f bb(r h 1 πh R f(y exp ( (y B t dy h f(y exp ( (y B t dy h f( y exp ( (y B t dy (7 h, + 1 πh (, By a change of variables and by observing that {} is a null set, we deduce from (7 that E f(x t+h 1 Gt f(y exp ( (y B t dy πh, h + 1 f(y exp ( (y + B t dy (8 πh h R h f(b t, By symmetry of the expression in (8, we see that E f(x t+h Gt Rh f( B t and thus E f(x t+h Gt Rh f(x t Updated: March 8, 17 3 / 7

4 Brownian Motion and Stochastic Calculus, Spring 17 Exercise 63 Let (B t t be a Brownian motion For any a > consider the stopping times T a : inf { t > Bt a }, Show that the Laplace transform of T a has value: E exp( µt a exp ( a µ, µ > and show that P T a < 1 Hint: Consider the martingale M λ t ( exp λb t λ t Solution 63 We know that for any λ R, the process M λ : (M λ t t defined by M λ t exp (λb t λ t is a continuous F t -martingale, where we denote by (F t t the filtration generated by B Moreover, for any n N, T a n is a bounded stopping time Thus, applying the stopping theorem (ie Serie 6 Exercise 1 we get E F M λ 1 P -as By taking expectations, we get that M λ T a n Now, on the event {T a < } we have exp (λb Ta n λ (T a n E MT λ a n 1 ( n exp (λb Ta λ T a e λa exp λ T a On the event {T a } we have B t a for any t and thus we get for any λ > that exp (λb Ta n λ (T a n exp (λb n λ n n We conclude that for any λ > exp (λb Ta n λ (T a n Observe that for any n N we have exp ( n e λa exp λ T a 1 {Ta< } P -as (9 (λb Ta n λ (T a n e λa Thus, we deduce from (9, by applying dominated convergence theorem, that for any λ > 1 E ( MT λ n a n e λa E exp λ T a 1 {Ta< } and so, for any λ > ( e λa E exp λ T a 1 {Ta< } 1 (1 Take any positive, decreasing sequence (λ n n N converging to We deduce from (1 and the monotone convergence theorem that P T a < ( lim n eλna E exp λ n T a 1 {Ta< } 1 Updated: March 8, 17 4 / 7

5 Brownian Motion and Stochastic Calculus, Spring 17 which proves the first part Thus, as we now know that P T a < 1 we get from (1 that for any λ > ( e λa E exp λ T a 1 (11 Fix any µ > For λ : µ, (11 yields the desired result Updated: March 8, 17 5 / 7

6 Brownian Motion and Stochastic Calculus, Spring 17 Exercise 64 Let (F n n N be a decreasing sequence of sub-σ-fields of F (ie F n+1 F n F, n N and let (X n n N be a backward submartingale, ie E X n <, X n is F n -measurable and EX n F n+1 X n+1 P -as for every n N (a Show that for any n m, N, M >, E X n 1 { Xn M} E Xm E N M E Xn (b Show that lim n EX n > implies that the sequence (X n n N is uniformly integrable Hint: use a to conclude that (X n n N is uniformly integrable Solution 64 (a For any n m, N, M >, by the backward submartingale property, using that the set { X n > M} F n, we obtain that E X n 1 { Xn M} E Xn E Xn 1 { Xn<M} E X n E Xm 1 { Xn<M} E X n E Xm + E Xm 1 { Xn M} E X m X n + E Xm 1 { Xn M, X m N} + E X m 1 { Xn M,< X m<n} + E Xm 1 { Xn M, X m } E X m X n + E Xm 1 { Xn M, X m N} + E Xm 1 { Xn M,< X m<n} E X m E N M E X n 1 { Xn M} E X m E N M E X n 1 { Xn } E X m E N M E Xn (b First, as (X n n N is a backward submartingale, we obtain that for any n X + n X n + X n EX F n + X n Thus, as (EX F n n N is uniformly integrable, if we can show that (X n n N is uniformly integrable, we can conclude that also (X + n n N is uniformly integrable which then implies that (X n n N is uniformly integrable Now, (X n n N being uniformly integrable is equivalent having that for any ε > we find M such that sup E X n 1 { Xn M} ε n N Fix any ε > Due to the assumption that lim n EX n >, we can find m N such that for any n m EX m EX n ε/4 Since X n is integrable for every n N we can find M m such that for any M M m max E X n 1 { Xn M} ε/4 n<m As sup E X n 1 { Xn M} max E X n 1 { Xn M} + sup E X n 1 { Xn M} n N n<m n m Updated: March 8, 17 6 / 7

7 Brownian Motion and Stochastic Calculus, Spring 17 it suffices to find M M m such that sup E X n 1 { Xn M} 3ε/4 n m We first observe that (X n + n N is also a backward submartingale Indeed, due to Jensen s inequality, we obtain that E X n + F n+1 EXn F n+1 + X n+1 + Now, we claim that sup n N EX n < Assume by contradiction that sup n N EX n Then, we can find a subsequence (n k k N such that lim k EX n k sup EXn Using that (X + n n N is also a backward submartingale, we obtain for any k that n N E X n k E X + nk E Xnk E X + E Xnk E X E X nk But as X is integrable and as lim k E X nk > by assumption, we obtain that lim k EX n k E X lim k E X nk < which gives us a contradiction Thus we have proved that sup u N EXu < For any n m, N, M >, by a, by the choice of m and as sup k N EX k <, we obtain that E X n 1 { Xn M} E Xm E N ε/4 + E N X m 1 { Xm N} + M E Xn ε/4 + E N X m 1 { Xm N} + M sup u N E Xu M E Xn Since X u is integrable for any u N we can find N big enough such that second term above is smaller than ε/4 After having chosen N we can find M M m such that that the last term above is smaller than ε/4 Thus, we get for that chosen M that hence we get the result sup E X n 1 { Xn M} 3ε/4 n m Updated: March 8, 17 7 / 7