I am given 3 things: $Z$ follows a normal distribution $N(0,1)$ $Y=e^{X}$ $X=3-2Z$ What is the moment generation function of $X$ and the $r^{th}$ moment of $Y$ ($E[Y^{r}]$)? My attempt: I know that $M_{X}(t)=E[e^{tX}]=E[e^{t(\mu+\sigma Z)}]=e^{\mu t + (\sigma ^2 t^2)/2}$. So by $X=3-2Z$, $3$ is $\mu$ and $-2$ is $\sigma$. Therefore, $M_X(t)=e^{3t+2t^2}$. And since $E[Y^{r}]=E[e^{rX}]=M_X(r)$, $E[Y^{r}]= e^{3r+2r^2}$? Login To add answer/comment