Prove $E[(\Delta B_j)^4]=3(\Delta t_j)^2$ where the Delta stands for the change of something i.e $B_j-B_{j-1}=\Delta B_j$ and the $B_j$ stand for the standard Brownian motion
I won't show my step here since i know it is a few step computation, i think my mistake on conceptual got me wrong from start
Here is the second degree example with answer: https://math.stackexchange.com/questions/1308705/show-a-lemma-of-brownian-motion-is-valid
The difficulty is that i can't just use the second degree example to convert it into four degree solution...
This is merely the fact that if $X\sim N(0,\sigma^2)$ then $\operatorname{E}(X^4)=3\sigma^4$. If $Z=X/\sigma\sim N(0,1)$ then clearly $\operatorname{E}(Z^4)=\operatorname{E}(X^4)/\sigma^4$, so it is enough to prove $\operatorname{E}(Z^4)=3$. $$ \operatorname{E}(Z^4)=\frac 1 {\sqrt{2\pi}}\int_{-\infty}^\infty z^4 e^{-z^2/2}\,dz. $$ Let $u=z^2/2$ so that $du = z\ dz$. Then we have $$ 2\cdot\frac 1{\sqrt{2\pi}}\int_0^\infty (2u)^{3/2} e^{-u}\,du=2\cdot\frac 1 {\sqrt{2\pi}} \cdot2\sqrt2\cdot \Gamma(5/2) = 2\cdot\frac 1 {\sqrt{2\pi}} \cdot\frac32\cdot\frac12\cdot\Gamma(1/2) $$ $$ =2\cdot\frac 1 {\sqrt{2\pi}}\cdot 2\sqrt2 \cdot\frac32\cdot\frac22\cdot\sqrt\pi = 3. $$