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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?

Shreyas Sarda
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  • You can also do it without polar coordinates by linearly transforming $x,,y$ to write $x^2-2txy+y^2$ as a sum of two squares. – J.G. Apr 01 '21 at 11:41
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    https://math.stackexchange.com/q/2026175/321264, https://math.stackexchange.com/q/581229/321264 – StubbornAtom Apr 01 '21 at 11:42

1 Answers1

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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 $|z|$.

J.G.
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  • Could you please explain how z being asymptotic imply |t| < 1? – Shreyas Sarda Apr 01 '21 at 12:07
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    @ShreyasSarda Let $f$ denote the PDF. The $\int_0^\infty e^{tz}f(z)dz$ part of the MGF diverges for $t\ge1$, because the integrand is at least $O(1/z)$. A similar argument with the $\int_{-\infty}^0$ part addresses $t\le-1$. – J.G. Apr 01 '21 at 12:53