# Characteristic function of $Y=X_1X_2$, where $X_1$ and $X_2$ are standard normal [duplicate]

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.

• There are some minor mistakes in your formulas to start with. Note that the density of $X_i$ is proportional to $\exp\{-x^2/2\}$ so the density of $X_1X_2$ should be...? There is also a minor error (typo?) in $\phi_{X_1}$
– KOE
Jun 25, 2014 at 8:54
• Density is then $exp{-x^4/2}$? I know that the characteristic function of the sum of two real-valued random variables is the product of the characteristic funcitons. Jun 25, 2014 at 8:58
• I know that the characteristic function of the sum of two real-valued random variables is the product of their characteristic functions. However, I don't know what the characteristic function is of the product of two real-valued random variables. Is the product just $exp{{\frac{-(x_1x_2)^2}{2}}}$? Jun 25, 2014 at 9:01
• Ok, let me give you a hint in an answer below to clear some things up. By the way, this looks like a self-study type of question? Maybe add that tag.
– KOE
Jun 25, 2014 at 9:02
• I wrote out the integral, so the answer should be complete now. Jun 25, 2014 at 12:05

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.

• I'm not quite sure how you arrived at $\sqrt{\sigma^2}$, or rather $\frac{1}{\sqrt{1+t^2}}$ Jun 25, 2014 at 11:22
• So by dividing the $y^2$ by $\sigma^2$ you normalize it, making it basically a standard normal once again. But I don't see how you obtain the final result. Jun 25, 2014 at 11:33
• Ok I will write it out in a sec. Jun 25, 2014 at 11:34
• If you don't want to compute the integral, you can "cheat" and note that $Y^2$ is $\chi^2(1)$ and use its moment generating function to obtain the last result.
– KOE
Jun 25, 2014 at 11:35

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 $X_i$.

• So the probability density function of the product of two random variables equals the product of their probability density functions? Whatever happened to the fact that $f_Y = f_{X_1X_2}$ has only one argument while $f_{X_1}f_{X_2}$ has two arguments? Jun 25, 2014 at 10:51
• Thanks @DilipSarwate, missed a comma (and carried the error on). Better now I hope.
– KOE
Jun 25, 2014 at 11:28