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:

$\paren {\cov {X, Y} }^2 \le \var X \, \var Y$

where $\cov {X, Y}$ denotes the covariance of $X$ and $Y$.

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:

$\ds \paren {\cov {X, Y} }^2$

$=$

$\ds \paren {\expect {\paren {X - \expect X} \paren {Y - \expect Y} } }^2$

Definition of Covariance

$\ds $

$\le$

$\ds \expect {\paren {X - \expect X}^2} \expect {\paren {Y - \expect Y}^2}$

$\ds $

$=$

$\ds \var X \, \var Y$

Definition of Variance

$\blacksquare$