I have standard normally distributed random variables $X, Y, V, Z$ and want to find $$Var(XY + VZ)$$ but have the following conditions: $$Cov(Y,Z)>0$$ $$Cov(X,Y)=Cov(X,V)=Cov(X,Z)=Cov(Y,V)=Cov(V,Z)=0$$

That is, the multiplicands $Y$ and $Z$ are dependent. I assume this complicates the proposed variance, but am unsure how to proceed.

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    $\begingroup$ Please explain what you mean by "proposed sum" and tell us what information you have about these variables. If the two conditions you have supplied are the only ones, then there's extremely little one can say. In particular, this variance depends on the multivariate fourth moments, but you haven't specified anything about those. $\endgroup$ – whuber Jul 3 '17 at 18:53
  • $\begingroup$ New edit that reflects both comments. @whuber, I don't know anything beyond the first two moments. $\endgroup$ – Jayden Nord Jul 3 '17 at 19:45
  • $\begingroup$ Now that you have stated these are "standard normally distributed" RVs, you know a lot about the moments: "standard" tells us the means are zero and the variances are unity. "Normal" tells us the third moments are zero and implies the fourth moments are $3$. Moreover, if you also assume $(X,Y,Z,V)$ is multivariate Normal, then the three zero conditions on the covariances imply pairwise independence in each case. Where, then, lies the problem in computing the variance of $XY+VZ$? $\endgroup$ – whuber Jul 3 '17 at 19:52
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    $\begingroup$ Given how much the question has changed, I deleted my answer. $\endgroup$ – Mark L. Stone Jul 3 '17 at 19:59
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    $\begingroup$ @ wolfies When you write up your answer, you might want to explain why $XY$ is uncorrelated with $VZ$, despite $Y$ being correlated with $Z$, and therefore that the value of $Cov(Y,Z)$ is irrelevant to the answer. $\endgroup$ – Mark L. Stone Jul 3 '17 at 21:26

we have \begin{align} E(XY+VZ)&=E(XY)+E(VZ)\\ &=EXEY+EVEZ=0. \end{align} and hence \begin{align} E(XY+VZ)^2&=E(X^2Y^2)+E(V^2Z^2)+2EXYVZ\\ &=EX^2EY^2+EV^2EZ^2+2EXEVEXY\\ &=2. \end{align}

  • $\begingroup$ +1 Small typo in your 2nd last line: 2EXEVEXY should be 2 EX EV EYZ. The proof also assumes independence (not stated by the OP, but I am sure intended) ... in the case of Normality, if I recall, independence and zero covariance are IFF anyway ... but that might best still be mentioned to take the E operator inside. $\endgroup$ – wolfies Jul 7 '17 at 15:57

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