Let $X$, $Y,$ and $Z$ be random variables. (There are no restrictions on these variables, but you may assume that these are continuous random variables if you want.) Suppose that $X$ and $Z$ are independent, and also suppose that $Y$ and $Z$ are independent. Does it follow that $\mathrm{cov}(XY,Z)=0$? (I understand that $XY$ and $Z$ may not be independent, but this does not rule out the zero covariance.)
Under the assumption that $X$ and $Z$ are independent and that $Y$ and $Z$ are independent, I am able to show that $$\mathrm{cov}(X,YZ) = \mathrm{cov}(Y,XZ) = \mathrm{cov}(XY,Z) + \mathrm{E}[Z]\cdot \mathrm{cov}(X,Y).$$ Showing this is fairly straightforward: $\mathrm{cov}(XY,Z)=\mathrm{E}[XYZ]-\mathrm{E}[XY]\cdot\mathrm{E}[Z]$; in addition, $\mathrm{cov}(X,YZ)=\mathrm{E}[XYZ]-\mathrm{E}[X]\cdot\mathrm{E}[YZ]=\mathrm{E}[XYZ]-\mathrm{E}[X]\cdot\mathrm{E}[Y]\cdot\mathrm{E}[Z]$, implying that $\mathrm{cov}(X,YZ)=\mathrm{cov}(XY,Z)+\mathrm{E}[XY]\cdot\mathrm{E}[Z]-\mathrm{E}[X]\cdot\mathrm{E}[Y]\cdot\mathrm{E}[Z]$, which is obviously equal to $\mathrm{cov}(XY,Z) + \mathrm{E}[Z]\cdot\mathrm{cov}(X,Y).$ And, of course, $\mathrm{cov}(X,YZ) = \mathrm{E}[XYZ]-\mathrm{E}[X]\cdot\mathrm{E}[Y]\cdot\mathrm{E}[Z]= \mathrm{cov}(Y,XZ)$, proving the above result.
However, to proceed further, my intuition tells me that $\mathrm{cov}(XY,Z) = 0$ and thus that $\mathrm{cov}(X,YZ) = \mathrm{cov}(Y,XZ) = \mathrm{E}[Z]\cdot\mathrm{cov}(X,Y)$. Am I wrong in thinking that $\mathrm{cov}(XY,Z) = 0$? But if it is true that $\mathrm{cov}(XY,Z) = 0$, is there a simple proof that does not possibly involve measure theory? Thanks.
Best Answer
Edit: The original question asked about independence, and was answered by the example below. This example also settles the modified question about covariance, since $\text{Cov}(XY,Z)\ne 0$.
Toss a fair coin twice. Let $X=1$ if we have head on the first toss, and $0$ otherwise. Let $Y=1$ if we have head on the second, and $0$ otherwise. Let $Z=1$ if the two tosses give different results, and $0$ otherwise. Then the random variables $X$, $Y$, and $Z$ are pairwise independent.
The random variables $XY$ and $Z$ are not independent.