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Consider jointly normally distributed random variables $X,Y\sim N(0,1)$ that have $\text{Corr}(X,Y)=\rho$. Show that $\text{Corr}(X^2,Y^2)=\rho^2$.

(Hint: Consider $X,U\sim N(0,1)$ where they are independent then we have $Y=\rho X+\sqrt{1-\rho^2}U$).

I sense that a transform is required for both $X$ and $Y$ but not sure where to go next.

edit: I am more interested in knowing how the hint is used.

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    $\begingroup$ For fun you might try to find $X,Y\sim N(0,1)$ uncorrelated but having $X^2= Y^2$, i.e. corr$(X,Y) = 0$ and corr$(X^2,Y^2) = 1$ $\endgroup$ – P.Windridge Jun 19 '18 at 12:26
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    $\begingroup$ Have you tried using the definition of correlation? You have to find a couple of moments, other than that this is straightforward. And this is without using the hint. $\endgroup$ – StubbornAtom Jun 19 '18 at 13:06
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    $\begingroup$ The hint only helps you to find $E(X^2Y^2)$ as far as I can tell. You need to know $E(X^4)$ apart from that , and the rest of the moments are given. Nothing vastly different from my hint though. $\endgroup$ – StubbornAtom Jun 20 '18 at 10:33
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Hint:

Let $(X,Y)$ be jointly normal variables with zero means and unit variances and $\text{Corr}(X,Y)=\rho$.

By definition, \begin{align}\text{Corr}(X^2,Y^2)=\frac{\text{Cov}(X^2,Y^2)}{\sqrt{\text{Var}(X^2)\text{Var}(Y^2)}}\end{align}

where $\text{Cov}(X^2,Y^2)=\mathbb E(X^2Y^2)-\mathbb E(X^2)\mathbb E(Y^2)$, and

$\text{Var}(X^2)=\mathbb E(X^4)-(\mathbb E(X^2))^2=\text{Var}(Y^2)$.

For finding $\mathbb E(X^2Y^2)$ quickly, note that $\mathbb E(X^2Y^2)=\mathbb E(\mathbb E(X^2Y^2\mid X))=\mathbb E(X^2\,\mathbb E(Y^2\mid X))$.

And we know that $Y\mid X\sim\mathcal{N}(\rho X,1-\rho^2)$.

So, $\mathbb E(Y^2\mid X)=\text{Var}(Y\mid X)+(\mathbb E(Y\mid X))^2=\cdots$.

I think you can find the moments now.

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You ask how to use the hint. One way is to focus on computing the covariances that go into the correlation formula. There are two. I will do the algebra for you to reduce the problem to simpler calculations requiring a little statistical thought.

The easy covariance calculation (because it involves only one variable at a time and both variables have standard Normal distributions and we know their first four moments are $0,1,0,3$) is $$\operatorname{Var}(X^2) = \operatorname{Var}(Y^2) = E[(Y^2)^2]-E[Y^2]^2=3-1=2.\tag{*}$$

This implies you would like to prove

$$\operatorname{Cov}(X^2,Y^2) = \sqrt{\operatorname{Var}(X^2)}\sqrt{\operatorname{Var}(Y^2)} \operatorname{Cor}(X^2,Y^2) = 2\rho^2.\tag{**}$$

To compute this, let's just blindly apply the hint by making the substitution for $Y,$ expanding $Y^2$ algebraically, and exploiting the linearity of $\operatorname{Cov}$ in its second argument to break the resulting expression into three simpler ones:

$$\eqalign{ \operatorname{Cov}(X^2,Y^2) &= \operatorname{Cov}(X^2, (\rho X + \rho^\prime U)^2) \\ &= \operatorname{Cov}(X^2, \rho^2 X^2 + 2\rho\rho^\prime XU + (\rho^\prime)^2 U^2)\\ &= \rho^2 \operatorname{Cov}(X^2, X^2) + 2\rho\rho^\prime \operatorname{Cov}(X^2, XU) + (\rho^\prime)^2 \operatorname{Cov}(X^2, U^2). }$$

(To make the patterns clearer to see, I have written $\sqrt{1-\rho^2}=\rho^\prime.$)


That should be good enough, but I'll take you a little closer to the end so you can see where this is all going. We have already found (at $*$) that

$$\operatorname{Cov}(X^2, X^2) = \operatorname{Var}(X^2) = 2.$$

Plugging this value into the preceding expression and comparing it to $(**)$ shows we would like to demonstrate

$$2\rho^2 = \operatorname{Cov}(X^2, Y^2) = 2\rho^2 + 2\rho\rho^\prime \operatorname{Cov}(X^2, XU) + (\rho^\prime)^2 \operatorname{Cov}(X^2, U^2).$$

Evidently, no matter what value $\rho$ might have, the sum of the last two terms needs to be zero. This insight reminds us that $X$ and $U$ are independent. Exploit that fact to show that each of the remaining covariances is zero. Explicitly, prove

$$\operatorname{Cov}(X^2, XU) = 0 = \operatorname{Cov}(X^2, U^2).$$

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