# Calculating $E[X^2Y^2]$ given $E[X^2]$, $E[Y^2]$, $E[X]$, $E[Y]$, and that $X$, $Y$ are Gaussian. [closed]

Suppose $E[X]=E[Y]=0$, and $E[X^2]=E[Y^2]=1$. Can you show that $E[X^2Y^2] = 1 + 2\operatorname{cov}(X,Y)^2$? I am not even sure if this expression is correct, I found it in a geostatistics paper, which used this result to show something else. (Note that under these conditions $\operatorname{cov}(X,Y)$ is simply equal to $E[XY]$).

edit: $X$ and $Y$ are Gaussian random variables. Also, it might possibly be useful to consider $E[E[X^2Y^2|X]]$.

-

## closed as too localized by Andrés E. Caicedo, Bill Thurston, Martin Brandenburg, Douglas Zare, Qiaochu YuanJan 2 '11 at 14:09

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Hmm... the variables $X$ and $Y$ are Gaussian, although I can't see this helping... – Azure Jan 1 '11 at 20:28
Doesn't look like it. Suppose $$X = \begin{cases} 1/\sqrt{2} & \text{with probability } 1/2, \\ -1/\sqrt{2} & \text{with probability } 1/2. \end{cases}$$ And $Y$ has the same distribution and is independent of $X$. Then $E(X^2 Y^2) = 1/4$ and $1 + 2\operatorname{cov}(X,Y)^2 = 1$. – Michael Hardy Jan 1 '11 at 20:28
You should really edit the information that the variables are gaussian into the question. Gaussians are far more constrained than general random variables. That said, I suspect this will wind up closed as too elementary; you'd probably do better on math.stackexchange.com – David Speyer Jan 1 '11 at 20:32
I consider this to be an exercise. The point is that it can all be worked out with explicit formulas. – Deane Yang Jan 1 '11 at 21:04
From an ongiong conversation with the questioner on another question, I have the strong impression that this was done with good intentions (a misconception of the etiquette here). – user9072 Jul 4 '11 at 21:04

The result holds if, additionnally to the conditions of the post, one assumes that the vector $(X,Y)$ is Gaussian. Then, $Y=aX+\sqrt{1-a^2}Z$ with $a=\mathrm{cov}(X,Y)$ and $Z$ standard Gaussian such that $X$ and $Z$ are independent. Using $E(X^4)=3$, $E(Z)=0$, $E(X^2)=E(Z^2)=1$ and the independence of $X$ and $Z$, one gets indeed that $E(X^2Y^2)=3a^2+0+1-a^2=1+2a^2$.
Otherwise, that is, under the conditions of the post only, the result is false. For example, if $Y=UX$ with $U=\pm1$ centered and such that $U$ and $X$ are independent. all the hypotheses of the post are met but $\mathrm{cov}(X,Y)=0$ and $E(X^2Y^2)=3$.