I am trying to calculate the K-th moment for Standard Brownian Motion: $$Z(t) \sim N(0,t)$$
I'm given that the second moment is $t$, but I'm having trouble seeing how that was arrived at.
I thought to use $M_X(t) = E[e^{tX}] = e^{\mu t +\frac{1}{2} \sigma^2t^2}$, but had no luck.
Ultimately, I have to repeat the process for Arithmetic and Geometric Brownian Motion as well. So some pointers there would be helpful too.
For geometric Brownian motion, just set $t=k$ in the MGF.
For arithmetic, you integrate against the standard normal to get moments of $N(0,1).$ Scale by $t^{1/2}$ to get moments of $W_t.$
So we have $$ \frac{1}{\sqrt{2\pi}} \int x^{k} e^{-x^2/2} dx = \frac{1}{\sqrt{2\pi}} \int x^{k-1} \; xe^{-x^2/2} \; dx $$ and integrate by parts.