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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 ?

Siong Thye Goh
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User9523
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1 Answers1

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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}

Siong Thye Goh
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    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: https://stats.stackexchange.com/questions/51699/moment-generating-function-of-the-inner-product-of-two-gaussian-random-vectors – tmrlvi Jun 07 '17 at 08:50
  • @tmrlvi Thanks for pointing out the mistake. The alternative method is cool. – Siong Thye Goh Jun 07 '17 at 15:40
  • 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 04 '19 at 15:06