Assume that $\{B_t \}_{t \geq 0}$ is Wiener Process. Show that $E|B_t - B_s|^4 = 3|t-s|^2$.
I managed to show this without absolute value using Moment-generating function of normal distribution, so I have just $E(B_t - B_s)^4 = 3(t-s)^2$ and I don't know how to get this result from it.
It is not true in general that $|B_t - B_s|^4 = (B_t - B_s)^4$.
If you have a one-dimensional Wiener process, this will indeed be the case and you are done. But, then you might be wondering? Why do we write absolute values? That happens if you are dealing with $n$-dimensional Wiener processes. It is not too hard to derive the result for an $n$-dimensional Wiener process, but it doesn't hurt if you see it once.
So, we have $$|B_t-B_s|^4 = \left(\sum_{i=1}^{n}(B_{ti} - B_{si})^2\right)^2.$$
You want to take the expectation, so you should note that per definition the $i$th component and the $j$th component are independent if $i\neq j$. Hence,
$$ E|B_t-B_s|^4 = \sum_{i=1}^{n}\sum_{j=1}^{n} E((B_{ti} - B_{si})^2(B_{tj} - B_{sj})^2) = \sum_{i=1}^{n} E((B_{ti} - B_{si})^4 + \sum_{i=1}^{n}\sum_{j=1,j\neq i}^{n} E(B_{ti} - B_{si})^2 E(B_{tj} - B_{sj})^2 = \\ n\cdot 3 (t-s)^2 + n(n-1)\cdot (t-s)\cdot(t-s) = \left(n^2 + 2n\right)\cdot(t-s)^2 $$