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Let $(X_1 ,Y_1) , ...(X_n ,Y_n)$ be IID and have a normal distribution with $ E(X_1)=E(Y_1)=0$ and $ var(X_1)=var(Y_1)=1$ and $Cov(X_1 ,Y_1)=\theta$.

I need to find $E(X_1^2Y_1^2)$.
Can i generalize , $Var(X) =E(X^2) - (E(X))^2$ to this ?.

I am not sure whether it is correct or not.

So in that case,

$Cov(X^2_1 ,Y^2_1)$ = $E(X_1^2Y_1^2)$ -$E(X^2_1)E(Y^2_1)$.

Then my next problem is how to find $Cov(X^2_1 ,Y^2_1)$ .

Can anyone help me figure out that ?

Thank you.

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    $\begingroup$ $$E(X_1^2Y_1^2)=E\,( E(X_1^2Y_1^2\mid Y_1))=E(Y_1^2 E(X_1^2\mid Y_1))$$ And the conditional distribution $X_1\mid Y_1$ is known. $\endgroup$
    – StubbornAtom
    Commented Nov 9, 2018 at 19:08

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Define $Z_1:=aX_1+bY_1$ so $Z_1$ has zero mean and $\text{var}Z_1=a^2+b^2+2ab\theta$, while $\text{cov}(X_1,\,Z_1)=a+b\theta$. Simultaneously solving $a^2+b^2+2ab\theta=1,\,a+b\theta=0$, viz. $a=-\frac{\theta}{\sqrt{1-\theta^2}},\,b=\frac{1}{\sqrt{1-\theta^2}}$, ensures $X_1,\,Z_1$ are uncorrelated mean-$0$ variance-$1$ variables. Since $(X_1,\,Y_1)$ is multivariate normal, so is $(X_1,\,Z_1)$; and (as can easily be proven from the pdf), these variables being uncorrelated implies they're independent. Then $$E(X_1^2 Y_1^2)=E\Bigg(X_1^2\bigg(\frac{Z_1-aX_1}{b}\bigg)^2\Bigg)=\frac{a^2E(X_1^4)-2abE(X_1^3Z_1)+E(X_1^2Z_1^2)}{b^2}=\frac{3a^2+1}{b^2},$$which you can write in terms of $\theta$.

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  • $\begingroup$ Thank you for your answer. I want to know how you derive $a^2 + b^2 + 2ab\theta=1$ and $a + b\theta=0 $. ? $\endgroup$
    – student_R123
    Commented Nov 9, 2018 at 18:37
  • $\begingroup$ @student_R123 As I said, I'm choosing $a,\,b$ so that $Z_1$ has mean $0$, variance $1$. $\endgroup$
    – J.G.
    Commented Nov 9, 2018 at 18:56

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