$X$ and $Y$ are two arbitrary random variables with fixed, identical mean $\mu$ and identical variance $\sigma^2$. Note that they are not necessarily independent. So what is the minimum/maximum value that cov$(X,Y)$ can attain, and under what equivalent conditions?
I can make two examples (just for thoughts):
- $\mu=0$, $Y=-X$, then cov$(X,Y)=\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)=-\mathbb{E}(X^2)-0=-\sigma^2$.
- $Y=X$, then cov$(X,Y)=\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)=\mathbb{E}(X^2)-\mu^2=\sigma^2$.
But apparently this is not the answer, because
- violates the condition that $\mu$ is fixed: we cannot assume $\mu=0$. The result should be expressed as function of $\mu$ and $\sigma$.
- can't prove that they are indeed min/max.
So could anyone please give me some insights on how to determine the min/max and prove them? Thanks :)
The maximum is $\sigma^{2}$ . This is an easy consequence of Cauchy-Schwarz inequality: $|E(X-EX)(Y-EY)|^{2} \leq var (X) var (Y)$.
To show that this value is attained take $Z \sim N(0,1)$ and $X=Y=\mu+\sigma Z$. This gives the upper bound.