Homework..7. (a). The relation T n T n 1 = T(B n ) shows that T n T n 1 is an identically sequence with common law as T. Notice that for any n 1, by Theorem.16 the Brownian motion B n (t) is independent of F + (T n 1 ). Consequently, T n T n 1 = T(B n ) is independent of T 1,,T n 1. Hence, T n T n 1 is an i.i.d. sequence. By The law of large numbers, So T n B(T n ) B(T n ) = ET and n T n T n B(T n ) = T n = 0 a.s. = 0 a.s. (b). Write By independence and stationarity of the Brownian increments, B(T n ) B(T n 1 is an i.i.d. sequence with common distribution as B(T). By the conclusion of part (a), By Borel-Cantelli lemma, B(T n ) B(T n 1 ) n=1 = 0 a.s. P B(T n ) B(T n 1 ) n < Or, n=1 P B(T) n < This leads to E B(T) <. By the representation B(T n ) = n k=1 B(T k ) B(T k 1 ) (Under the convention B(T 0 ) = 0) By the law of large numbers and part (b) B(T n ) Comparing this to Part (a), EB(T) = 0. = EB(T) a.s..10. Define the filtration A(t);t 0 as: A(t) = F + (e t ). Notice that for any s,t > 0, X(s+t) = e B(e (s+t) (s+t) ) B(e s ) +e t X(s) 1
Conditioning on A(s), X(s) is a constant and B(e (s+t) ) B(e s ) is a random variable with the same distribution as B(e (s+t) e s ) with B(t) starting at 0. Our job is reduce to find the distribution of e (s+t) B(e (s+t) e s )+e t x d = e (s+t) e (s+t) e s U +e t x where U N(0,1). It is easy (by checking the expectation and the variance) to see that the distribution is N(e t x, 1 e t ). Therefore, the transition density is p(t,x,y) = 1 π(1 e t ) exp (y e t x) (1 e t ) Prove do the second part, consider the Brownian motion B(t) = 0 t = 0 tb(1/t) t > 0 x,y R We have X( t) = e t B(e t ) = e t B(e t ) t 0 By what have been proved (with B(t) being replaced by B(t)), X( t); t 0 has the same distribution as X(t); t 0..15. (a) Write X(t) = exp σb(t) σ t t 0 Clearly, X(t) is adaptive with respect to the filtration F + (t). Notice that X(t) 0 and For any 0 s < t, EX(t) = exp E [ X(t) F + (s) ] = exp = exp = exp So X(t) is a Martingale. σ t E exp σb(t) = 1 < σ t E σ t exp σb(s) σ t exp σb(s) exp [ ] F E exp σb(t) + (s) [ ] F exp σb(t) B(s) + (s) σ (t s) = X(s)
(b). Due to similarity, we only show that B(t) t is a martingale. For s < t, by part (a), [ E exp σb(t) σ t F (s)] + = exp σb(s) σ s Taking second derivatives with respective to σ on the both sides we have, respectively, [ (B(t) σt ) E t exp σb(t) σ t F (s)] + (B(t) σt ) = t exp σb(s) σ s To justify it, a usual way of dominated convergence is needed (as the differentiation is defined by it). Taking σ = 0 on the both sides, E [ B(t) t F + (s) ] = B(s) s This, together with the obvious integrability, shows that B(t) t is a martingale. (c). We consider the martingale X(t) = B(t) 4 6tB(t) +3t. Given N > 0, write T N = T N. By the fact that for any t > 0, B(t T N ) max a,b, we have that X(t T N ) max a 4,b 4 +6max a,b +3N By Proposition.4. (by the condition a < 0 < b, I believe that the authors consider the case B(0) = 0), EX(T N ) = E 0 X(0) = 0, or EB(T N ) 4 6ET N B(T N ) +3ET N = 0 We now let N. By the fact that B(T N ) max a,b and dominated convergence, EB(T N) 4 = EB(T) 4 = a 4 PB(T) = a+b 4 PB(T) = b = a 4 b a a +b +b4 a +b = a b( a 3 +b 3 ) a +b = a b( a a b+b ) where the third step follows from the calculation in the proof of Theorem.49. By the bound0 T N B(T N ) T max a,b,thefactthatet < (why?) andthedominated convergence, ET NB(T N ) = ETB(T) = a ET1 B(T)=a +b ET1 B(T)=b = a ET +(b a )ET1 B(T)=b = a 3 b+(b a )ET1 B(T)=b 3
where the last step follows from Theorem.49. By monotonic convergence, ET N = ET Therefore, we conclude that ) a b( a a b+b ) 6 ( a 3 b+(b a )ET1 B(T)=b +3ET = 0 ( ) With the same argument to the martingale B(t) 3 3tB(t), we have EB(T) 3 3ETB(T) = 0. Notice that EB(T) 3 = a 3 PB(T) = a+b 3 PB(T) = b = a3 b+b 3 a a +b = a b(b a ) ETB(T) = aet1 B(T)=a +bet1 B(T)=b Combine our computation, = aet +(b a)et1 B(T)=b = a b+(b+ a )ET1 B(T)=b ET1 B(T)=b = a b(b+ a ) 3(b+ a ) Bring this back to (*), ET = a b3 + a 3 b+3 a b 3.16. Define the stopping time T = infs > 0; B(s) a+bt When Brownian motion starts at 0, there is positive chance that T =. By continuity of the Brownian curve, P 0 B(t) = a+bt for some t > 0 = P 0 T < Let N > 0 and write T N = T N and consider the process X(t) = expbb(t) b t t 0 4
Taking σ = b in Problem.15, X(t) is a martingale. WriteT N = N T forn > 0. Foranyt 0, B(t T N ) a+b(t T N ). Consequently, 0 X(t T N ) exp b ( a+b(t T N ) ) b (t T N ) expab By Proposition.4, therefore, E 0 expbb(t N ) b T N = E 0 X(T) = E 0 X(0) = 1 ( ) Write E 0 expbb(t N ) b T N = E 0 expbb(t N ) b T N 1 T< +E 0 expbb(t N ) b T N 1 T= On the event T <, bb(tn ) b T N = bb(t) b T = ab a.s. By the bound expbb(t N ) b T N expab and dominated convergence, E 0expbB(T N ) b T N 1 T< = expabp 0 T < In addition, E 0 expbb(t N ) b T N 1 T= = E 0 expbb(n) b N1 T= According to the law of large numbers, Consequently, B(N) N = 0 a.s. expbb(n) b N = 0 a.s. On the event T = we have the bound Hence, by dominated convergence, expbb(n) b N expab E 0expbB(N) b N1 T= = 0 ( ) 5
Summarizing our argument since (**), we have expabp 0 T < = 1 Warning. The argument for (***) collapses without the indicator of T =, as E 0 expbb(n) b N = 1 N > 0.19. (a). Under the interpretation of this exercise, B(t) = B 1 (t) +ib (t), where B 1 (t) and B (t) are two independent linear Brownianmotions with B 1 (0) = 0 and B (0) = 1. Notice that the function f(x,y) = e λy cosλx and g(x,y) = e λy sinλx satisfies f(x,y) = 0and g(x,y) = 0. ByCorollary.53theprocessese λb (t) cosλb 1 (t) and e λb (t) sinλb 1 (t) are martingales. Hence, the requested conclusion follows from the relation e iλb(t) = e λb (t) cosλb 1 (t)+ie λb (t) sinλb 1 (t) (b). Extra assumption: We have to assume that λ 0. Indeed, B(T) = B 1 (T) is a real random variable. What we try to prove is the characteristic function of B 1 (T) is equal to e λ which would be greater than 1 when λ < 0. BydefinitionT = infs > 0; B (s) = 0. Ifweareallowedtouse Optional stopping theorem, Ee iλb(t) = Ee iλb(0) = e λ where the last step follows from the fact that B(0) = i. We now justify the use of Proposition.4 (Optional stopping rule). First notice that for any t > 0, e iλb(t t) e λb (T t) 1 where the last step follows from the assumption that λ 0 and the fact that B (T t) 0. Hence, Proposition.4 applies. Remark. First,B(T) = B 1 (T). Second, B 1 (0) = 0impliesthatB 1 (t)issymmetric. Third, by definition T is independent of B 1 (t). Therefore, B 1 (T) is symmetric. So the characteristic function of B 1 (T) is real and even function. Thus, for any λ R, Ee iλb(t) = Ee iλb 1(T) = e λ This result shows that the real random variable B 1 (T) obeys Cauchy distribution, as pointed out in Theorem.37. 6