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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.

emcor
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statsguyz
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  • 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 $\phi_{X_1}$ – ekvall Jun 25 '14 at 08:54
  • 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. – statsguyz Jun 25 '14 at 08:58
  • 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 $exp{{\frac{-(x_1x_2)^2}{2}}}$? – statsguyz Jun 25 '14 at 09:01
  • 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. – ekvall Jun 25 '14 at 09:02
  • I wrote out the integral, so the answer should be complete now. – emcor Jun 25 '14 at 12:05
  • Because the CF is given by the MGF evaluated at a purely imaginary argument (and there are no technical issues of convergence, etc, in dealing with these rapidly-decreasing Gaussians), I take it that finding the MGF and CF are essentially equivalent problems. The present question then reduces to the simplest case of the duplicate, which deals with the sum of products of Normal variables. – whuber Jun 25 '14 at 15:06

2 Answers2

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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.

emcor
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  • I'm not quite sure how you arrived at $\sqrt{\sigma^2}$, or rather $\frac{1}{\sqrt{1+t^2}}$ – statsguyz Jun 25 '14 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. – statsguyz Jun 25 '14 at 11:33
  • Ok I will write it out in a sec. – emcor Jun 25 '14 at 11:34
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    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. – ekvall Jun 25 '14 at 11:35
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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 $X_i$.

ekvall
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    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? – Dilip Sarwate Jun 25 '14 at 10:51
  • Thanks @DilipSarwate, missed a comma (and carried the error on). Better now I hope. – ekvall Jun 25 '14 at 11:28