I want to show $E(B(t)-B(s))^4=3(t-s)^2$

Question

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.

Answer 1

Hint: assuming $t\geq s$,

$$B(t)-B(s) \sim B(t-s) \sim \sqrt{t-s}N(0,1). $$

$N(0,1)$ is normal with mean zero and variance $1$. The expected values of $N(0,1)^n$ to various positive integers (such as $4$) are well-known (they are zero if $n$ is odd). Try Googling "normal distribution". One of the first hits should be a Wikipedia article citing many properties of the standard normal distribution $Z\equiv N(0,1)$, including $E(Z^2)$ and $E(Z^4)$.

I haven't checked your equation to check if it's correct. But this is a way to find the expected value of $(B(t)-B(s))^4$.