1

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.

Jayden Nord
  • 41
  • 2
  • 1
    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,](https://stats.stackexchange.com/questions/155007) but you haven't specified anything about those. – whuber Jul 03 '17 at 18:53
  • New edit that reflects both comments. @whuber, I don't know anything beyond the first two moments. – Jayden Nord Jul 03 '17 at 19:45
  • 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 $XY+VZ$? – whuber Jul 03 '17 at 19:52
  • @whuber, my concern is with the covariance between $Y$ and $Z$. Shouldn't that covariance be factored into the calculation of the variance of $XY+VZ$? – Jayden Nord Jul 03 '17 at 19:59
  • 2
    Given how much the question has changed, I deleted my answer. – Mark L. Stone Jul 03 '17 at 19:59
  • Yes, Jayden, that is correct. Unless you can provide more information about the other two covariances (for $(X,Z)$ and $(Y,V)$), all one can do is provide some bounds for $\operatorname{Var}(XY+VZ)$--and I'm afraid they will be fairly broad. – whuber Jul 03 '17 at 20:20
  • @whuber, I senselessly forgot the additional covariances - I somehow thought there were only 4 possible combinations. Edited question should have all possible covariances listed. – Jayden Nord Jul 03 '17 at 20:25
  • I suspected the question might change :-). Life is now easy, because your assumptions imply the independence of $X, V,$ and $(Y,Z)$. Just write down the standard formula for the covariance: you have all the information you need (and the independence facts will make most of the terms equal to zero). – whuber Jul 03 '17 at 20:29
  • I don't have time to write up an answer at the moment, but in the case of standard joint Normals, I am getting: $Var(X Y + VZ) = \mu _{0,0,2,2}+2 \mu _{1,1,1,1}+\mu _{2,2,0,0}$. And under the above assumptions, $\mu _{1,1,1,1} =0$ and $\mu _{2,2,0,0} = 1$ and $\mu _{0,0,2,2} = 1$, so unless I have made a booboo, the answer would be 2. Rhymes too. – wolfies Jul 03 '17 at 20:40
  • 1
    @ 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. – Mark L. Stone Jul 03 '17 at 21:26

1 Answers1

2

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}

Math-fun
  • 223
  • 1
  • 7
  • +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. – wolfies Jul 07 '17 at 15:57