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I am given 3 things:

  1. $Z$ follows a normal distribution $N(0,1)$
  2. $Y=e^{X}$
  3. $X=3-2Z$

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 $E[Y^{r}]=E[e^{rX}]=M_X(r)$, $E[Y^{r}]= e^{3r+2r^2}$?

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Since $Z \sim N(0,1)$, it's M.G.F. is given by

$M_Z(t)=e^\frac{t^2}{2}$

Make the necessary transformations and use the result as given by L.V.Rao, you'll surely get it.

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    $\begingroup$ The answer you refer to was deleted long ago. $\endgroup$ – Glen_b Sep 9 '17 at 8:44

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