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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 $E[E[X^2Y^2|X]]$.

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    $\begingroup$ Hmm... the variables $X$ and $Y$ are Gaussian, although I can't see this helping... $\endgroup$
    – Azure
    Commented Jan 1, 2011 at 20:28
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    $\begingroup$ 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 $1 + 2\operatorname{cov}(X,Y)^2 = 1$. $\endgroup$
    – Michael Hardy
    Commented Jan 1, 2011 at 20:28
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    $\begingroup$ 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 $\endgroup$
    – David E Speyer
    Commented Jan 1, 2011 at 20:32
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    $\begingroup$ I consider this to be an exercise. The point is that it can all be worked out with explicit formulas. $\endgroup$
    – Deane Yang
    Commented Jan 1, 2011 at 21:04
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    $\begingroup$ 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). $\endgroup$
    – user9072
    Commented Jul 4, 2011 at 21:04

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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 $E(X^2Y^2)=3$.

A well known motto in such situations is: "Joint laws, or else..."

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  • $\begingroup$ Excellent! Thank you. In the paper I am reading, it does hold that vector (X,Y) is Gaussian, so that was a hidden assumption in my question. $\endgroup$
    – Azure
    Commented Jan 2, 2011 at 0:17

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