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Let $\{B_{t}\}_{t\geq0}$ be Brownian motion. What is the variance of $B_{t}B_{s}$?

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up vote 7 down vote accepted

Assuming that $t\geq s$, write $B_t B_s=(B_t-B_s)B_s+B_s^2$. Taking the expectation, we find that $\mathbb{E}(B_t B_s)=s$. On the other hand $$(B_t B_s)^2=(B_t-B_s)^2B_s^2+2(B_t-B_s)B_s^3+B_s^4,$$ so taking expectation this time gives $$\mathbb{E}((B_t B_s)^2)=(t-s)s+0+3s^2.$$

Finally, taking the difference of these we get $$\mbox{Var}(B_t B_s)=\mathbb{E}((B_t B_s)^2)-\mathbb{E}(B_t B_s)^2=(t+s)s.$$

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Thanks, just a confusion, is it not true that $B_{t}$ is Normal with mean 0 and variance $t$? I think your derivation is based on variance $t^2$ – V.A. Apr 4 '12 at 16:00
Yes, I made a mistake that I will correct now. – Byron Schmuland Apr 4 '12 at 16:06
Thanks for spotting my error! – Byron Schmuland Apr 4 '12 at 16:09
The same error also appeared on the first term for the second moment . I have edited your answer. Can you double check that? – V.A. Apr 4 '12 at 18:47
Yes, you are right. Thanks for the correction. – Byron Schmuland Apr 4 '12 at 18:55

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