# Finding the M.G.F of product of two random variables.

We are given two independent standard normal random variables $X$ and $Y$. We need to find out the M.G.F of $XY$.

I tried as follows : \begin{align} M_{XY}(t)&=E\left(e^{(XY)t}\right)\\&=\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(XY)t}f_X(x)f_y(y)dxdy \\ &=\dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(xy)t}e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &= \dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-\dfrac{(x-ty)^2}{2}}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dy \\ &= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{-y^{2}(\frac{1}{2}+t^2)}dy \\ &= \dfrac{1}{\sqrt{1+2t^2}} \end{align} Is this correct ?

• Small typos in second equality: Exponent should have $(xy)t$ (not uppercase) whereas both subscripts of the PDFs should be capitalized. – Semiclassical Nov 22 '16 at 19:25
• @Semiclassical I'll keep that in mind for next time. Thanks :) – User9523 Nov 22 '16 at 19:29

In the $4^{th}$ line as we complete the square: \begin{align} & \dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-\dfrac{(x-ty)^2}{2}}e^{\dfrac{t^2y^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{\dfrac{t^2y^2}{2}}e^{\frac{-y^2}{2}}dy \\ &= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{-\frac{y^{2}(1-t^2)}2}dy \\ &= \dfrac{1}{\sqrt{1-t^2}} \end{align}
• There is additional mistake in the calculation. Final result should be $\frac{1}{\sqrt{1 - t^2}}$. Convergence radius is $t^2 < 1$. Easier way to arrive the same conclusion is here: stats.stackexchange.com/questions/51699/… – tmrlvi Jun 7 '17 at 8:50
• Funny enough, the solution is $f(t)=1/\sqrt{(1-t^2)}=\arcsin'(t)$. I wonder whether this can be related to angles somehow. – ndrizza Apr 4 at 15:06