X and Y are i.i.d. Normal Random Variables: $ \mathcal N(0,1) $.
Find Moment Generating Function of 'XY'.
I could come till:
$$ \iint e^{txy} * \frac{e^{-\frac{x^{2}}{2}}}{\sqrt{2\pi}} * \frac{e^{-\frac{y^{2}}{2}}}{\sqrt{2\pi}} dx dy $$
$ x = r*cos(\theta), y = r*sin(\theta) $
$$\iint e^{tr^{2}sin(\theta)cos(\theta)} * \frac{e^{-\frac{r^{2}}{2}}}{2\pi} * r dr d\theta
$$
While integrating this first with r, I can't put directly $r=\infty $ for upper limit to take that term as zero.
This is because for $r=\infty $, there is corresponding $\theta = 0.5*arcsin(\frac{1}{t})$.
Now, how to proceed in this integration?
Best Answer
The $\theta$-dependent exponent is missing a factor of $t$. Since$$\int_0^\infty re^{-r^2(1/2-t\sin\theta\cos\theta)}dr=\frac{1}{1-t\sin2\theta},$$you need to evaluate $\frac{1}{2\pi}\int_0^{2\pi}\frac{d\theta}{1-t\sin2\theta}$, which as a sanity check is $1$ if $t=0$. Convergence requires $|t|<1$, which makes sense as the PDF of $XY$ is asymptotic to $\sqrt{\frac{\pi}{2|z|}}e^{-|z|}$ for large $|z|$.