# Prove that \text{Corr}(X^2,Y^2)=\rho^2\text{Corr}(X^2,Y^2)=\rho^2 where X,YX,Y are jointly N(0,1)N(0,1) variables with correlation \rho\rho

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

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

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

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

Hint:

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

By definition,

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.