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 ?
$\endgroup$ 31 Answer
$\begingroup$Careless mistake at second last line: \begin{align} \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^{\dfrac{-(t^2+1)y^2}{2}}dy\\ &=\frac{1}{\sqrt{t^2+1}} \end{align}
Edit:
There is actually a mistake earlier. Thanks, tmrlvi for pointing out.
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}
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