Square of Covariance is Less Than or Equal to Product of Variances

Theorem
Let $X$ and $Y$ be random variables.

Let the variances of $X$ and $Y$ exist and be finite.

Then:


 * $\paren {\map {\operatorname {Cov} } {X, Y} }^2 \le \var X \var Y$

where $\map {\operatorname {Cov} } {X, Y}$ denotes the covariance of $X$ and $Y$.

Proof
We have, by the definition of variance, that both:


 * $\expect {\paren {X - \expect X}^2}$

and:


 * $\expect {\paren {Y - \expect Y}^2}$

exist and are finite.

Therefore: