If we have that $Cov(Y, X-Y) = 0$, does this imply $Y$ is independent of $X-Y$?

Question

If we have that $Cov(Y, X-Y) = 0$, does this imply $Y$ is independent of $X-Y$ for random variables $X$ and $Y$? Intuition tells me that normally this is not true, but we have the variable $Y$ appear twice so I am not so sure. Could anyone lend me a hand here? thanks!

Answer 1

You are correct that uncorrelated does not imply independence. Two notable exceptions:

1. They jointly follows a multivariate normal.
2. They both follow a Bernoulli distribution.

Answer 2

Counterexample:

Let it be that $X=ZY$ for some non-degenerate random variable $Z$ s.t. $Y$ and $Z$ are independent.

Then $Y$ and $X-Y=(Z-1)Y$ are not independent.

Also let it be that $\mathbb EZ=1$ and $\mathbb EY=0$, and consequently $\mathbb EX=\mathbb EY\mathbb EZ=0$.

Then $\mathbb EX=\mathbb EY\mathbb EZ=0$ and $\mathbb E(X-Y)=0$ so that:

$\text{Cov}(Y,X-Y)=\mathbb EY(X-Y)=\mathbb EY^2(Z-1)=\mathbb EY^2\mathbb E(Z-1)=0$

The last line under condition that $\mathbb EY^2<\infty$.