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 $Y^2$ .

  • 2
    $\begingroup$ 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. $\endgroup$ Apr 28, 2019 at 4:22

1 Answer 1


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 $\mu_1=\mu_2=0$.


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