# Expanding cov(cov(x,y),y) to compare to cov(x,y)

I am trying to compare the following two terms and determine which one is larger than the other.

Term 1: $${\operatorname{cov}(X,Y)} \over {\operatorname{var}(Y)}$$

Term 2: $${\operatorname{cov}(\operatorname{cov}(X,Y), Y)} \over {(\operatorname{var}(Y))^2}$$

If x and y are uncorrelated, then $$\operatorname{cov}(X,Y) = 0$$, so both of these terms would be $$0$$. What would happen if they are positively correlated? I think I'm missing some information on the properties of covariances that is preventing me from manipulating the second term.

• With respect to Term 2: the covariance of $x$ and $y$ is a number, e.g., 4.25, and as such, has no covariance with anything else. Feb 3, 2020 at 20:08
• @jbowman Would that make it equal to 0? Feb 3, 2020 at 20:12
• Yes, so Term 2 is equal to 0 also. Feb 3, 2020 at 20:14
• Could you please explain the second numerator? Assuming $X$ and $Y$ refer to random variables, then $\operatorname{Cov}(X,Y)$ is either a fixed number or matrix, whence its covariance with $Y$ cannot be anything other than zero. That strongly suggests you mean something else by your symbols, but what is it?
– whuber
Feb 3, 2020 at 20:49
• @whuber I meant exactly what I wrote. This was the result of a proof I'm working on. I didn't realize it would be 0. Feb 3, 2020 at 21:30

Statistics like $$\operatorname{cov}(X,Y)$$, $$\operatorname{var}(X), E[X], E[Y]$$ are all constants. Only when we use random variables on givens side, they become functions (in general) of RVs: e.g. $$E[X|Y], \operatorname{var}(X|Y)$$.
So, since $$\operatorname{cov}(X,Y)=a$$ is a constant, $$\operatorname{cov}(\operatorname{cov}(X,Y), Y)=\operatorname{cov}(a,Y)=0$$, which would make your second term $$0$$. Based on your positive/negative correlation, Term 1 is either greater or smaller than Term 2.