# Correlation between square of normal random variables

Suppose I have $$X,Y$$ bivariate normal with correlation coefficient $$\rho \in (0,1)$$ . Then , what is the correlation between $$X^2$$ and $$Y^2$$ ?

I am aware of the fact that the square of the normal follows a chi square distribution . So , I can find out $$Var(X^2)$$ and $$Var(Y^2)$$ . However , I am unable to calculate the covariance between $$X^2$$ and $$Y^2$$ .

• Try writing $X,Y$ as appropriate linear combinations of two iid standard normals $Z,W$. Then the covariance is the expectation of some polynomial in $Z,W$ which should be easy to find, since the expectation of each monomial $Z^i W^j$ factors as $E[Z^i] E[W^j]$, and the moments of the standard normal are well known. – Nate Eldredge Apr 28 '19 at 4:22

You can do what Nate Eldredge suggested. Otherwise, you can use the moment generating function $$M$$ of the bivariate normal distribution $$N(\mu_1,\mu_2,\sigma_1^2,\sigma_2^2,\rho)$$ of $$(X,Y)$$ given by $$M(t_1,t_2)=\exp(\boldsymbol\mu'\boldsymbol t+\tfrac12\,\boldsymbol t'\Sigma\boldsymbol t)$$ for $$\boldsymbol t=[t_1,t_2]'$$, $$\boldsymbol\mu=[\mu_1,\mu_2]'$$, $$\Sigma=\begin{bmatrix}\sigma_1^2&\rho\sigma_1\sigma_2\\\rho\sigma_1\sigma_2&\sigma_2^2 \end{bmatrix}$$, where $$'$$ denotes the transpose of a matrix. Then $$EX^2Y^2=\frac{\partial^4}{\partial^2t_1\,\partial^2t_2}\,M(t_1,t_2)\Big|_{t_1=t_2=0}= 4 \mu _1 \mu _2 \rho \sigma _1 \sigma _2+\left(\mu _1^2+\sigma _1^2\right) \left(\mu _2^2+\sigma _2^2\right)+2 \rho ^2 \sigma _1^2 \sigma _2^2,$$ whence $$Cov(X^2,Y^2)=EX^2Y^2-EX^2\,EY^2=4 \mu _1 \mu _2 \rho \sigma _1 \sigma _2+2 \rho ^2 \sigma _1^2 \sigma _2^2.$$ Special cases of this formula are $$Var(X^2)=Cov(X^2,X^2)=4 \mu _1^2 \sigma _1^2+2 \sigma _1^4$$ and $$Var(Y^2)=Cov(Y^2,Y^2)=4 \mu _2^2 \sigma _2^2+2 \sigma _2^4.$$ So, the correlation between $$X^2$$ and $$Y^2$$ is $$\frac{Cov(X^2,Y^2)}{\sqrt{Var(X^2)Var(Y^2)}} =\frac{4 \mu _1 \mu _2 \rho \sigma _1 \sigma _2+2 \rho ^2 \sigma _1^2 \sigma _2^2}{\sqrt{\left(4 \mu _1^2 \sigma _1^2+2 \sigma _1^4\right) \left(4 \mu _2^2 \sigma _2^2+2 \sigma _2^4\right)}} =\frac{2 \mu _1 \mu _2 \rho \sigma _1 \sigma _2+ \rho ^2 \sigma _1^2 \sigma _2^2}{\sqrt{\left(2 \mu _1^2 \sigma _1^2+ \sigma _1^4\right) \left(2 \mu _2^2 \sigma _2^2+ \sigma _2^4\right)}}.$$ In particular, the correlation between $$X^2$$ and $$Y^2$$ is $$\rho^2$$ when $$\mu_1=\mu_2=0$$.