# From $\mathbb{Cov}(X^2,Y)$ to $\mathbb{Cov}(X,Y)$

Say I have the expression for $\mathrm{Cov}(X^2,Y)>0$, where both $X$ and $Y$ are known non-negative random variables. What if anything can we say about $\mathrm{Cov}(X,Y)$ without having to derive it from scratch? I think we can say that $\mathrm{Cov}(X,Y)>0$, but can we say anything more, given the expression for $\mathrm{Cov}(X^2,Y)$?

## 1 Answer

I think it can say very poor . For example if you take $Y=X^2$, then $Cov(X^2,Y)=1$. But if you take positive values for X then $Cov(X,Y)>0$, and for negative values for X, $Cov(X,Y)<0$. Posibly if you take any type of values, $Cov(X,Y)\cong 0$.

• Thanks, but note that I wrote: both X and Y are known non-negative random variables. – Omri May 31 '12 at 14:48
• I was trying to find a counterexample by picking Y=c/X + X$^2$ +D. I am thinking that since we have so much freedom picking a distribution for X Let D be a large enough constant so that Y is always > 0 Then we can allow C to be negative. Part of Y is negatively correlated with X and X$^2$ and the other part is positively correlated. So I figured that I could find a distribution for X and constants C and D such that Y>0 covarince of Y with X$^2$ is positive but Cov(X, Y)<0. I was not able to do this with D=0 but think that it may be possible with large D. – Michael Chernick May 31 '12 at 17:24