Gaussian convolution

Question

$X$ and $Y$ are standard normal independent random variables, let $Z=XY$. Is the distribution of $Z$ equal to $N(0,\frac{1}{2})$?

I tried using convolution to get the distribution of $Z$. Is it correct?

Answer 1

There is a similar formula for the distribution of the product $Z=XY$ of two nicely behaved random variables: $$ f_Z(z)=\int_{-\infty}^\infty f_X(t)f_Y(z/t)\frac{1}{|t|}dt.$$ So in this case with two standard normals, that's $$ f_Z(z)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-\frac{1}{2}(t^2+\frac{z^2}{t^2})}\frac{dt}{|t|}.$$ There's no way to integrate this in terms of elementary functions, but we can get it into a well-known integral representation of a well-known special function:

If we let $u= t/\sqrt{|z|},$ this gives $$ f_Z(z)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-|z|\frac{1}{2}(u^2+1/u^2)}\frac{du}{|u|}$$ and letting $w=2\ln(u)$ gives $$ f_Z(z) = \frac{1}{\pi}\int_{0}^\infty e^{-|z|\cosh(w)}dw = \frac{1}{\pi}K_0(|z|)$$ where $K_0$ is the modified Bessel function. (The relevant integral formula is both on the wikipedia page and here.)

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You mentioned in the comments going the route of characteristic equations, and that works too. We have $$ \phi_Z(t)=E(e^{itZ}) = E(e^{itXY}) =\frac{1}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty e^{itxy-x^2/2-y^2/2}dxdy.$$ You can complete the square to $$ \phi_Z(t)=\frac{1}{2\pi}\int_{-\infty}^\infty dy e^{-y^2(1+t^2)/2}\int_{-\infty}^\infty e^{-\frac{1}{2}(x-ity)^2}dx.$$ The $x$ integral is just a shifted standard Gaussian (shifted in an imaginary direction, but that doesn't wind up mattering), so is just $\sqrt{2\pi},$ so we get $$ \phi_Z(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{1}{2}(1+t^2) dy} = \frac{1}{\sqrt{1+t^2}}\int_{-\infty}^\infty e^{-\frac{1}{2}u^2} \frac{du}{\sqrt{2\pi}} = \frac{1}{\sqrt{1+t^2}}.$$

But then, if you want the PDF, there's the work of doing the inverse Fourier transform to compute it from the characteristic function. Using this answer, we obtain $$ F_Z(z) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-izt}\frac{1}{\sqrt{1+t^2}}dt = \frac{1}{2\pi} 2K_0(|z|) = \frac{1}{\pi}K_0(|z|),$$ exactly as above.

Answer 2

The sum Z=X+Y is Gaussian distributed, with the variance being the sum of variances of X and Y. It can "easily" be shown using the characteristic function (Fourier of the probability density function).

I don't think the product is Gaussian.