Covariance of X^2 Y^2 when Cov(X,Y) = 0?

Question

If you have that X and Y are independent random variables and their covariance is equal to zero, then is the $Cov(X^2, Y^2)$ zero as well? I feel like this is too good to be true, but I cannot come up with a conclusion.

If $X$ and $Y$ are independent, then $X^2$ and $Y^2$ are independent. See e.g. https://math.stackexchange.com/questions/8742/are-functions-of-independent-variables-also-independent. — Minus One-Twelfth, Oct 03 '19 at 23:25

score 1 · Answer 1 · answered Oct 03 '19 at 23:41

Yes that is true.

If $X$ and $Y$ are independent it implies their correlation, and therefore covariance are zero (but the converse is not necessarily true).

Assuming this knowledge, then by squaring the RVs separately we are not inducing any new correlational information, and thus they still remain uncorrelated and independent.

In this way we conclude $Cov(X,Y) = Cov(X^2,Y^2) = 0$.

score 1 · Answer 2 · answered Oct 03 '19 at 23:44

That the covariance of $X$ and $Y$ is zero is a consequence of $X$ and $Y$ being independent. So to say "$X$ and $Y$ are independent random variables and their covariance is equal to zero" is slightly redundant. As Minus One-Twelfth said, if $X$ and $Y$ are independent then $X^2$ and $Y^2$ are independent, and therefore the covariance of $X^2$ and $Y^2$ would be zero.

Independence is necessary. If $X$ has a continuous uniform on the interval $[-1,1]$, and we define $Y = X^2$, then $X$ and $Y$ are of course dependent on each other, but $\text{Cov}[X,Y]=0$ while $\text{Cov}[X^2,Y^2] \neq 0$.

Covariance of X^2 Y^2 when Cov(X,Y) = 0?

2 Answers2