# What is $\text{cov}(Y,Y)$ given $\text{cov}(X,Y)$ and $\text{cov}(X,X)$

Just some definitions:

$X,Y$ are random variables

$\mu_X = E[X]$

$\text{cov}(X,Y) = E[(X-\mu_X)(Y-\mu_Y)]$

Given $\text{cov}(X,X)$ and $\text{cov}(X,Y)$, what can be said about $\text{cov}(Y,Y)$?

For example, if $X$ has low variance, and $\text{cov}(X,Y)$ is high, then I would expect $Y$ to have low variance too.

• Hint: Think $X$ a normal with media 0 and low variance, $Y = 1000000 X$ – leonbloy Jul 5 '11 at 22:07

By the Cauchy-Schwarz Inequality, $$(\text{Cov}(X,Y))^2 \le \text{Var}(X)\text{Var}(Y).$$
I have used the more standard name $\text{Var}(U)$ where you use $\text{Cov}(U,U)$.
• And you can't do any better than Cauchy-Schwarz. That is, for any real numbers $s, t, u$ with $s \ge 0$, $t \ge 0$ and $u^2 \le s t$, there are random variables $X$, $Y$ with $\hbox{Var}(X) = s$, $\hbox{Var}(Y) = t$ and $\hbox{Cov}(X,Y) = u$. For example, if $U$ and $V$ are independent random variables of variance 1 and $u = \sqrt{s t} \cos (\theta)$, take $X =\sqrt{s} U$, $Y = \sqrt{t} (U \cos(\theta) + V \sin(\theta))$. – Robert Israel Jul 6 '11 at 1:27