# Expectation of product of squared random variables

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.

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

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