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Let $B(t)$ and $B(s)$ are brownian-motion I want to show $$E(B(t)-B(s))^4=3(t-s)^2$$
thanks for help.
If $B(t)$, $t\ge 0$ is a Brownian motion, then $B(t)-B(s)$ has $N(0,t-s)$ distribution. From there, you just need to figure out how to compute the 4th moment of a Gaussian with given variance, which I trust you can do.
Required, but never shown