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.

  • 1
    $\begingroup$ 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$(X^2,Y^2) = 1$ $\endgroup$ Jun 19, 2018 at 12:26
  • 2
    $\begingroup$ 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. $\endgroup$ Jun 19, 2018 at 13:06

2 Answers 2



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.


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.