Covariance and Brownian motion

Question

B(t) is a Brownian motion and $0<s<t<u<v$. Calculate $Cov[(B(t)-B(s))^2,(B(v)-B(u))^2]$

I know the formula for the expected value of a product of 2 B-terms. But here I need the expected value of a product of 4 B-terms. Has anybody an idea how to proceed here?

score 2 · Accepted Answer · answered Sep 13 '19 at 14:16

2

By definition of the Brownian Motion, since $0<s<t<u<v,$ $(B_s,B_t-B_s,B_u-B_t,B_v-B_u)$ is an independent vector. Hence, $(B_t-B_s)^2$ is independent of $(B_v-B_u)^2,$ so the covariance is $0$.

answered Sep 13 '19 at 14:16

WoolierThanThou

13,480

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In a more general setting, simply expand and look up the moments of the relevant Gaussian distributions. – WoolierThanThou Sep 13 '19 at 14:17

Covariance and Brownian motion

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