March 15th, 2005, 6:10 am
QuoteOriginally posted by: OMDCovariance of stopping time and value at the stopped time:This is really cool. I am really enjoying this stuff right now I must say!Let B_t be Brownian motion. Let A and B be greater than zero.Let T be the stopping time given by T = inf{t : B_t = A or B_t = -B}Through elementary techniques (though creative) or through a really nice first use of martingales and the canonical theorems that people prove about them in conjunction with stopping times, it can be shown that:P(B_T = A) = B/(A + B)E(T) = ABThis alone constitutes a good exercise if you have not done this, but what really made my jaw drop in fascination was a beautiful technique shown in Steele's stochastic calculus book for calculating Cov(B_T,T).Obviously this comes down to calculating E(T * B_T) since E(B_T) = 0.It was a really nice application of Ito's lemma, a small PDE result that follows immediately from Ito's lemma, the fact that Ito integrals are local martingales, the fact that a bounded local martingale is a martingale, along with Doob's stopping time theorem for martingales. Without looking this up, can you calculuate E(T * B_T) and show work?Sorry! I just had to share this because I thought it was really neat!I guess I will just throw up the solution:Consider f(t,x) where we replace x with B_t when we use Ito's formula.We would like f(t,x) = tx + g(t,x) where f(t,B_t) is a martingale.Note that Ito integrals are local martingales.By Ito, we have:edited - to say that forgot to put the (1/2) in the last integral above! D'OH!To make f(t,B_t) a local martingale, we just need to kill off the integrals with respect to ds.Therefore we need to find f(t,x) = tx + g(t,x) such that f_t + (1/2)f_xx = 0.This can easily be done by taking f(t,x) = tx - (x^3)/3Therefore we have M_t = tB_t - ((B_t)^3)/3 is a local martingale.When we introduce the stopping time T discussed in the problem, we know by a variation of Doob's stopping time theorem that if we let u = min(T,t), then M_u is also a local martingale. Furthermore, M_u is both bounded and continuous, so it is indeed an honest martingale.Therefore, E(M_u) = M_0 = 0 for all u.Note that M_u is dominated by the integrable function max(A,B).Therefore, by the dominated convergence theorem, we get lim_{u---> \infty} E(M_u) = E(M_T) = 0 by the martingale property.So, E(M_T) = E(T*B_T) - E((B_T)^3)/3 = 0So, E(T*B_T) = E((B_T)^3)/3 = P(B_T = A)*(A^3)/3 + P(B_T = -B)*(-B^3)/3 and since P(A) = B/(A+B) and P(B) = 1-P(A), we getE(T*B_T) = (B(A^3)/3 - A(B^3)/3)/(A+B) = AB(A-B)/3.QED - very nice I think!For some reason, I could only use the equation feature for posting once - it would not work the subsequent times Itried to use them.
Last edited by
OMD on March 14th, 2005, 11:00 pm, edited 1 time in total.