Characteristic function of $Y=X_1X_2$, where $X_1$ and $X_2$ are standard normal Let $X_1$ and $X_2$ be real-valued independent random variables with a standard normal distribution. Let $ Y = X_1X_2$. Find the characteristic function of Y.
Attempt:
$\phi_{Y}(t)$ = $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} exp(it(x_1x_2))exp(-1/2(x_1x_2)^2)dx_1dx_2$
I know that the standard normal characteristic function is:
$\phi_{X_1}(t)$ = $exp(\frac{-t^2}{2})$ and likewise for $X_2$
However, I'm not sure what the resulting product is. I realize that I can use a Jacobian, but I'd like to keep it simple and see if there is a way to derive the characteristic function with something easier. 
 A: Let $X$ and $Y$ two independent standardnormal random variables.
The characteristic Standardnormal function is $\mathrm E(\exp(\mathrm itX))=\exp(-\frac12t^2)$, hence:
$$\phi_{XY}(t)=\mathrm E(\exp(\mathrm itXY))=E(\mathrm E(\exp(\mathrm itXY)\mid Y))=E(\exp(-\frac12t^2Y^2))$$
Now,
$$
\mathrm E(\exp(-{\textstyle\frac12}t^2Y^2))=\frac1{\sqrt{2\pi}}\int_{\mathbb R}\mathrm e^{-\frac12t^2y^2}\mathrm e^{-\frac12y^2}\mathrm dy\\
=\frac{\sqrt{t^2+1}}{\sqrt{t^2+1}}\int \frac{1}{\sqrt{2\pi}}e^{-{1\over2}y^2(t^2+1)}dy\\
=\frac{1}{\sqrt{t^2+1}}\int\frac{1}{\sqrt{2\pi}\sqrt{\frac{1}{t^2+1}}}\exp\left({-\frac{1}{2}\frac{y^2}{\left(\frac{1}{\sqrt{t^2+1}}\right)^2}}\right)dy\\=\frac{1}{\sqrt{t^2+1}}\int \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\frac{y^2}{\sigma^2}}dy
$$
One can see that the latter integral being a Normal density over $\mathbb{R}$ with $\left(\mu=0,\sigma=\sqrt{\frac{1}{t^2+1}}\right)$ hence integrates to 1, so finally:
$$
\phi_{XY}(t)=\frac1{\sqrt{1+t^2}}
$$
The Normal Product distribution can be found here.
A: Here's the correctly set up integral to get you on the right track: $$\begin{align} 
\mathbb{E}\exp\{itY\}&=\mathbb{E}\exp\{itX_1X_2\}\\
&=\int\int \exp\{itx_1x_2\}f_{X_1,X_2}dx_1dx_2\\
&=\int\int \exp\{itx_1x_2\}f_{X_1}f_{X_2}dx_1dx_2\\
&= \frac{1}{2\pi}\int\int \exp\{itx_1x_2\}\exp\{-x^2_1/2\}\exp\{-x^2_2/2\}dx_1dx_2
\end{align}$$
Here, $f_{X_1,X_2}$ is the density of $(X_1,X_2)$ and $f_{X_i}$ is the density of $X_i$.
