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I've got a question about the distribution of the random variable $Y = X_{1}X_{2}$, where $X_i \sim \operatorname{N}(0,1)$, and $X_1$, $X_2$ independent.

Specifically, I need to find the MGF of $Y$.

Here's what I've tried:

The $r^\text{th}$ moment of $Y$ is given by:

$$\begin{align} \operatorname{E} (Y^r) & = \operatorname{E} (X_1^r X_2^r) \\ & = \operatorname{E} (X_1^r) \operatorname{E} (X_2^r) & \text{because } X_1 \text{ and } X_2 \text{ are independent} \\ \end{align}$$

Since $X_i \sim \operatorname{N}(0,1)$, I know that the $r^\text{th}$ central moment is given by:

$$ \operatorname{E}(X_i^r) = \begin{cases} 0 & r \text{ odd} \\ (2r-1)! & r \text{ even} \\ \end{cases}$$

Because $\operatorname{E}(X_i^r)$ does not depend on $i$, $\operatorname{E} (X_1^r) = \operatorname{E} (X_2^r)$ and so

$$\operatorname{E} (X_1^r) \operatorname{E} (X_2^r) = ((2r-1)!)^2 \quad \text{for } r \text{ even}$$

But I can't really derive a formula for the $r^{\text{th}}$ integral, especially if I don't know how the formula for $\operatorname{E}(X_i^r)$ looks before plugging in $t=0$.

Are there other ways to derive the MGF of $Y$?

EDIT:

I also need to show that $Y$ can be expressed as a difference of two independent, Gamma-distributed variables.

Alex Lostado
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    Note that, for every $t$ and every $s<1$, $$E(e^{tX_2})=e^{t^2/2}\qquad E(e^{sX_1^2/2})=\frac1{\sqrt{1-s}}$$ hence, for every $|t|<1$, $$E(e^{tX_1X_2})=E(E(e^{tX_1X_2}\mid X_1))=E(e^{t^2X_1^2/2})=\frac1{\sqrt{1-t^2}}$$ – Did Apr 18 '18 at 13:38
  • @Did How did you get $E(e^{sX_1^2/2}) = \frac{1}{\sqrt(1-s)}$? – Alex Lostado Apr 18 '18 at 13:43
  • Using the PDF of $X_1$ and a trivial change of variable. – Did Apr 18 '18 at 13:43
  • Are you sure about your formula of $\mathbb E [X^r]$ ? – Kroki Apr 18 '18 at 18:57
  • Plenty of similar posts like https://math.stackexchange.com/questions/2026175/finding-the-m-g-f-of-product-of-two-random-variables/ or https://math.stackexchange.com/questions/581229/finding-the-moment-generating-function-of-the-product-of-two-standard-normal-dis. – StubbornAtom Apr 18 '18 at 20:00
  • @StubbornAtom Thanks, will look into them! – Alex Lostado Apr 18 '18 at 20:16

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