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?
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 $0$.