# 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

## 1 Answer

By the Cauchy-Schwarz Inequality, $$(\text{Cov}(X,Y))^2 \le \text{Var}(X)\text{Var}(Y).$$

This result appears not to quite match your expectations. It more or less says that we cannot have high covariance when the variances are both low. But in some ways this is an artefact, a consequence of how variance and covariance scale when the random variables are scaled.

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