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.

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    $\begingroup$ The revised suggestion is wrong as well, try $(X,Y)$ uniform on $\{-1,1\}^2$ and $Z=XY$. Then any two random variables from $X$, $Y$ and $Z$ are independent and $\mathrm{cov}(XY,Z)=1$. $\endgroup$ – Did Feb 7 '16 at 14:54
  • $\begingroup$ @Did:I don't see why this is the case. $P(X\in [-1/2,1/2]|Z>0.5)=0\neq P(X\in [-1/2,1/2])$. So $X$ and $Z$ are not independent in this case. $\endgroup$ – adrido Mar 5 '18 at 0:29
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    $\begingroup$ @adrido Did you get that $X$, $Y$ and $Z$ are all two-valued random variables, always equal to $+1$ or $-1$? Then yes, $X$ and $Z$ are independent. (By the way, this counterexample is, under different but equivalent presentations, ultra classical.) $\endgroup$ – Did Mar 5 '18 at 6:31
  • $\begingroup$ ops! Sorry I didn't. nice example! $\endgroup$ – adrido Mar 16 '18 at 0:56

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.

  • $\begingroup$ I have edited my question to ask about the zero covariance instead of independence between XY and Z. If X and Z are independent and Y and Z are independent random variables, is it true that cov(XY, Z) = 0? If true, is there a simple proof of this? Thanks. $\endgroup$ – g123 Feb 7 '16 at 9:26
  • $\begingroup$ Sorry this is off topic but can you tell me what will be the formula for $Cov(X,Y|Z)$? $\endgroup$ – Ronald Feb 3 '18 at 9:57

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