Stochastic processes-Brownian Motion

Question

I hope someone can help me with the following exercise... Show that $ \int_0^t s \, dB_s =tB_t-\int_0^t B_s \, ds $, for each $t>0$, where $B$ is a Brownian motion. Thanks in advance!!!

Since you now accepted a remarkably succinct answer, I hope you checked for yourself that Itô's formula did not introduce supplementary terms in the differentials involved. To wit, let me remind you that the "integration by parts" formula $X_sdB_s=d(X_sB_s)-B_sdX_s$ is not true for every process $(X_s)$... — Did, May 02 '13 at 18:25

score 0 · Accepted Answer · answered Apr 29 '13 at 22:29

0

I think you want to start with integration by parts:

$sdB_s=d(sB_s)-B_sds$

answered Apr 29 '13 at 22:29

bob.sacamento

4,028

No prob. Glad to be of help. – bob.sacamento May 02 '13 at 17:02

Stochastic processes-Brownian Motion

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