$X$ and $Y$ are two dependent Gaussian random variables with finite means $\mu_x,\mu_y$, variances $\sigma_x^2,\sigma_y^2$, and covariance $\rho$. Prove $\mathbb{E}[X^2Y^2]$ is finite.

My effort: It is equivalent to show that $\text{Var}[XY]$ is finite. There is a bound on the variance of $XY$ which does not work here since $X$ and $Y$ are not bounded. Another way might be to use this pdf of the product and show that the integral is finite. Although it works for zero means, we can say even if the mean is not zero we are OK. I thought there might be a better solution. Any idea?

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    $\begingroup$ Cauchy-Schwartz would allow you to bound this using the fourth moments, so hopefully they are finite (which indeed you would need anyway since if X=Y, the expression reduces to the fourth moment of X) $\endgroup$ – E-A Aug 29 '18 at 20:49
  • $\begingroup$ @E-A Thanks for your comment. Can you please expand on that a little? Fourth moments of what? If you are referring to en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality , that gives the reverse. $\endgroup$ – Susan_Math123 Aug 29 '18 at 20:51
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    $\begingroup$ @E-A If you mean fourth moments of $X$ and $Y$, of course, they are finite since they are Gaussians with finite mean and variance. $\endgroup$ – Susan_Math123 Aug 29 '18 at 20:54
  • $\begingroup$ With the information given, the joint density function $f(x,y)$ can be expressed and the integral of $x^2y^2f(x,y)$ can be calculated. $\endgroup$ – herb steinberg Aug 30 '18 at 0:01
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    $\begingroup$ As E-A pointed out you just use the Cauchy-Schwartz inequality, which gives you $\mathbb{E}[X^2Y^2] \leq \sqrt{\mathbb{E}[X^4]\mathbb{E}[Y^4]}$, and the RHS is finite as mentioned. This is most general - only use the fact that they are marginally normal, and it does not require them to have a joint normality. $\endgroup$ – BGM Aug 30 '18 at 9:41

I should probably write it as a full answer so that the question is answered: the result immediately follows from Cauchy-Schwartz inequality (as BGM wrote out in the comments) $\mathbb{E}X^2Y^2 \leq \sqrt{\mathbb{E}X^4\mathbb{E}Y^4}$, and for Gaussians the fourth moment is finite.


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