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Let $X, Y$ be two i.i.d.$\sim\mathcal{N}(0,1)$ random variables, I need to compute $E(\exp(sXY))$ for $s \geq 0$.

Since they are independent the joint pdf $f_{XY}= \frac{1}{2\pi}\exp(-\frac{x^2+y^2}{2})$, but how do I compute $$ \frac{1}{2\pi}\int_{\mathbb{R}}\int_{\mathbb{R}} \exp(sxy)\exp(-\frac{x^2+y^2}{2})dxdy$$ This is my current approach, but I am uncertain about a few steps: $ \begin{align} \frac{1}{2\pi} \int_{\mathbb{R}}\exp(-\frac{y^2}{2})\int_{\mathbb{R}}\exp{\frac{-(x+sy)^2}{2}}\exp{\frac{s^2y^2}{2}}dxdy &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\exp{(\frac{-y^2}{2})}\exp{\frac{s^2y^2}{2}}dy \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\exp{(-y^2\cdot(\frac{1-s^2}{2}))}\\ &=\frac{1}{\sqrt{1-s^2}} \end{align}$

Is it correct? Can I just consider sy fixed in the first integration and pretend it is the expectation of the inner normal r.v.?

user1868607
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  • Maybe you could apply the moment generating function to XY – New_to_this Jun 26 '18 at 15:59
  • Isn't that what I am computing? – user1868607 Jun 26 '18 at 16:08
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    I'd suggest trying a change of variables, namely from Cartesian to polar coordinates. – Niki Di Giano Jun 26 '18 at 16:15
  • Yes I was a bit too fast - sorry. Just thought of expectation of XY. Have you tried applying the law of total expectation. – New_to_this Jun 26 '18 at 16:15
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    First integrate with respect to x, keeping y fixed just as with a normal distribution, then you will se that you can integrate y again with a normal distribution function if $|s| < 1$ – Stefan Jun 26 '18 at 16:30
  • Plenty of duplicates actually: https://math.stackexchange.com/q/2026175/321264, https://math.stackexchange.com/q/581229/321264, https://math.stackexchange.com/q/74013/321264 among others. – StubbornAtom Jun 26 '18 at 19:51

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