Let $X\sim\mathcal N(μ_1,σ_1^2)$ and $Y\sim\mathcal N(μ_2,σ_2^2)$ and $\mathsf{Cov}(X,Y)=c$ How can we compute $\mathsf {Cov}(X^2,Y^2)$?

I think the answer should be something like $c^2$ and I think the joint PDF is really not necessary here. I think the approach is using some independent standard normal random variables by variable changing. But I can't make it out.



That's the way to goal.

Let $X= \mu_1+\sigma_1~U~,~ Y= \mu_2+\sigma_2~(\rho~U+\sqrt{1-\rho^2}~V)$ where $U,V\mathop{\sim}\limits^\textrm{iid}\mathcal N(0,1)$ and $\rho=c/\sigma_1\sigma_2$

$$\begin{align}\mathsf {Cov}(X^2, Y^2) ~=~& \mathsf {Cov}\Big(\big(\mu_1+\sigma_1U\big)^2, \big(\mu_2+\sigma_2(\rho~U+\sqrt{1-\rho^2}~V)\big)^2\Big) \\[1ex] ~=~& \mathsf{Cov}\Big(\mu_1^2+2\mu_1\sigma_1U+U^2~,~ \mu_2^2+2\mu_2\sigma_2(\rho~U+\sqrt{1-\rho^2}~V)+\sigma_2^2\big(\rho^2U^2+2\rho\sqrt{1-\rho^2}~UV+(1-\rho^2)V^2\big)\Big) \end{align}$$

Simplify with the bilinearity of covariance, making use of: $\mathsf {Cov}(U,U) = 1, \mathsf {Cov}(U,V)=0, \mathsf{Cov}(U, U^2)=0,\mathsf {Cov}(U,UV)=0$ etc. (...why?)

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  • 2
    $\begingroup$ An additional point. This is the Cholesky factorisation of $\Sigma$, where $\Sigma$ is positive definite. One can obtain any bivariate normal by linearly tranforming two independent standard normal variables as $Y=\mu+AZ$ where $A$ is a lower triangular matrix. Then $\Sigma=AA^T$. $\endgroup$ – theoGR Aug 17 '16 at 12:22

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