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I have the following covariance term:

$Cov(x,yz)$

with $x$, $y$, and $z$ being random variables with a mean and variance.

I found a paper by Bohrnstedt and Goldberger from 1969, On The Exact Covariance Of Products Of Random Variables, who show that

$Cov(x,yz) = E[y]Cov(x,z) + E[z]Cov(x,y) + E[(x-E[x])(y-E[y])(z-E[z])]$

Now, I know that $y$ and $z$ are independent ($x$ instead is dependent on $y$ and $z$). $z$ is i.i.d. standard normal. Is there a way to simplify the last equation in this case?

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    $\begingroup$ Where is the conditioning? $\endgroup$ – Dilip Sarwate Jun 28 '18 at 16:25
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    $\begingroup$ Please be more specific about which "normal covariances" you are willing to use. At the most basic level any formula will require information about moments of $y_{t+1}$ and $z_{t+1}$ through the fourth order, which can scarcely be called "covariances" any more, but if you're willing to think of $y_{t+1}^2$ as if it were a single random variable, then arguably such a formula could be found. In that case, though, why would you complicate the statement of the problem by writing $y_{t+1}^2$ when you could more simply refer to it as a positive random variable (say $U$)? $\endgroup$ – whuber Jun 28 '18 at 16:25
  • $\begingroup$ If you mean that z is independent of both x and y, then you can use your formula (and whuber's suggestion) to simplify right down to a single pairwise covariance term. $\endgroup$ – Glen_b Jun 29 '18 at 4:46
  • $\begingroup$ Thank you all so far! With normal covariances, I mean covariances as $Cov(x_{t+1},y^2_{t+1})$, so the squared term is fine in my case. I just would like to get rid of the product $y^2_{t+1}z^2_{t+1}$. The only thing I know is that z is i.i.d. with mean zero and variance 1. If I would assume a distribution for $y^2$ and know that $z$ is i.i.d., do I need to assume anything for $z^2$ to solve this? Or would the first and third term drop out anyways? $\endgroup$ – tucker.crowe Jun 29 '18 at 9:28
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    $\begingroup$ $z$ being "iid" would be a statement about the joint distribution of $z_1,z_2,\ldots$ but it's unclear whether this implies anything about the joint distribution of $z_{t+1}$ and $x_{t+1},y_{t+1}$. Additionally, the relevance of the $t$-indices is unclear - could the notation be simplified to ask about $Cov(x, y^2\,z^2)$? (If not, then I think some information is missing) $\endgroup$ – Juho Kokkala Jun 29 '18 at 16:46
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Sometimes you can work these sorts of things out--and gain some insight--by considering the simplest possible non-trivial cases. That would mean that $X,Y,$ and $Z$ would have only two possible values. Let's make the calculations easy by supposing that their means are all zero and they have unit variances. This easily implies all three variables have equal chances of being $\pm 1$: they are Rademacher variables.

There aren't too many such situations, as we can count by observing

  1. The trivariate distribution of $(X,Y,Z)$ is determined by eight probabilities associated with the eight possible non-negative values $(\pm1,\pm1,\pm1).$
  2. Those eight values sum to unity (a linear constraint).
  3. Setting three means to zero adds three more linear constraints.

That still leaves $8-3-1=4$ parameters. For simplicity, let's suppose $Y$ and $Z$ are independent (one additional restriction) and that $X$ has the same correlation coefficient with each of $Y$ and $Z.$ There's still a one-parameter family of mathematical solutions. There's no assurance any of them give probability distributions, because the solutions might always have negative components.

But let's push ahead anyway. If we let $8\rho$ denote the covariance of $X$ with $YZ$ and $4\tau$ the common covariances of $X$ with $Y$ and $Z,$ the mathematical solutions for the eight probabilities are given in this table:

$$\begin{array}{cccc} X & Y & Z & \text{Probability} \\ -1 & -1 & -1 & 1/8 - \rho + \tau \\ -1 & -1 & 1 & 1/8+ \rho \\ -1 & 1 & -1 & 1/8+ \rho \\ -1 & 1 & 1 & 1/8-\rho-\tau \\ 1 & -1 & -1 & 1/8 + \rho - \tau \\ 1 & -1 & 1 & 1/8- \rho \\ 1 & 1 & -1 & 1/8- \rho \\ 1 & 1 & 1 & 1/8+\rho+\tau \\ \end{array} $$

You can easily compute that all the linear constraints hold, no matter what the numbers $\rho$ and $\tau$ might be. It's also obvious that when $\pm\rho \pm \tau \le 1/8,$ all these probabilities are non-negative. This gives a range of possible values of $8\rho=\operatorname{Cov}(X, YZ)$ from $-1 + |4\tau|$ to $1 - |4\tau|.$

Just as a check, note that with $\tau=0,$ the two extreme examples $8\rho=\pm 1$ give these two distributions:

  • $8\rho=-1:$ $\Pr(-1,-1,-1)=\Pr(-1,1,1)=\Pr(1,-1,1)=\Pr(1,1,-1)=1/4.$

  • $8\rho=1:$ $\Pr(-1,-1,1)=\Pr(-1,1,-1)=\Pr(1,-1,-1)=\Pr(1,1,1)=1/4.$

In both cases you can easily check that when you select any two of the three variables, all four pairs of their values $(\pm1,\pm1)$ are equally likely, whence in both cases these variables are pairwise independent. Nevertheless, in the first case $YZ$ always has the opposite sign of $X$ and in the second case it always has the same sign.

This example shows that unless you add some more conditions, you might not be able to say anything at all about $\operatorname{Cov}(X,YZ)$ in terms of the expectations and covariances of the individual variables!

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