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I know that $X$ and $Y$ are independent as well as $Z$ and $W$. I am trying to find a formula for $\text{Cov}(XY,ZW)$. I started by writing it down as $$\begin{align}\text{Cov}(XY,ZW)&=E(XYZW)-E(XY)E(ZW)\\[2ex]&=E(XYZW)-E(X)E(Y)E(Z)E(W)\end{align}$$ and I am stuck. I am not sure if I can write $E(XYZW)$ as $E(XZ)E(YW)$, and even if I can, I am not sure if that would be helpful.

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  • $\begingroup$ You can't and it wouldn't be. $\endgroup$ – Jake Westfall Mar 16 '17 at 14:21
  • $\begingroup$ @Matthew Cant he? $X $ and $Y $ are independent. $\endgroup$ – Łukasz Grad Mar 16 '17 at 14:37
  • $\begingroup$ @ŁukaszGrad My mistake: I read it too fast. if $X$ and $Y$ are independent then obviously $\operatorname{Cov}(X,Y) = 0$ and $\operatorname{E}[XY]=\operatorname{E}[X]\operatorname{E}[Y]$. That's fine. Continuing forward though, $\operatorname{E}[XYZW] = \operatorname{Cov}(XZ,YW) + \operatorname{E}[XZ]\operatorname{E}[YW]$ won't be particularly helpful. $\endgroup$ – Matthew Gunn Mar 16 '17 at 15:00
  • $\begingroup$ @ŁukaszGrad The substitution requires not that $X$ be independent of $Y$, which is assumed, but that $XY$ be independent of $ZW$, which does not follow from the assumptions (at least not as I have read and understood them -- I understand them to be just $X \perp Y$ and $Z \perp W$.) Maybe the OP can clarify. $\endgroup$ – Jake Westfall Mar 16 '17 at 20:36
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I'm afraid that knowing only $X \perp Y$ and $Z \perp W$ will not get you very far. The best you can do is probably:

$$\text{Cov}(XY,ZW) = \text{E}(X)\text{E}(Z)\text{Cov}(Y,W) + \text{E}(X)\text{E}(W)\text{Cov}(Y,Z) + \text{E}(Y)\text{E}(Z)\text{Cov}(X,W) + \text{E}(Y)\text{E}(W)\text{Cov}(X,Z) + \text{E}(X)\text{E}[\Delta_Y\Delta_Z\Delta_W] + \text{E}(Y)\text{E}[\Delta_X\Delta_Z\Delta_W] + \text{E}(Z)\text{E}[\Delta_X\Delta_Y\Delta_W] + \text{E}(W)\text{E}[\Delta_X\Delta_Y\Delta_Z] + \text{E}[\Delta_X\Delta_Y\Delta_Z\Delta_W]$$

where $\Delta_A = A - \text{E}[A]$.

To improve this, it's obviously useful to know more joint independences or if some of the random variables have mean zero.

For details, see: Bohrnstedt, G. W., & Goldberger, A. S. (1969). On the Exact Covariance of Products of Random Variables. Journal of the American Statistical Association, 64(328), 1439-1442.

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