# Square of Expectation of Product is Less Than or Equal to Product of Expectation of Squares

## Theorem

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

Let the expectation of $X Y$, $\expect {X Y}$, exist and be finite.

Then:

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

## Proof

Note that:

- $\map \Pr {Y^2 \ge 0} = 1$

so we have by Expectation of Non-Negative Random Variable is Non-Negative:

- $\expect {Y^2} \ge 0$

First, take $\expect {Y^2} > 0$.

Let $Z$ be a random variable with:

- $Z = X - Y \dfrac {\expect {X Y} } {\expect {Y^2} }$

Note that we have:

- $\map \Pr {Z^2 \ge 0} = 1$

so again applying Expectation of Non-Negative Random Variable is Non-Negative, we have:

- $\expect {Z^2} \ge 0$

That is:

\(\ds 0\) | \(\le\) | \(\ds \expect {Z^2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \expect {\paren {X - Y \frac {\expect {X Y} } {\expect {Y^2} } }^2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \expect {X^2 - 2 X Y \frac {\expect {X Y} } {\expect {Y^2} } + Y^2 \frac {\paren {\expect {X Y} }^2} {\paren {\expect {Y^2} }^2} }\) | Square of Sum | |||||||||||

\(\ds \) | \(=\) | \(\ds \expect {X^2} - 2 \expect {X Y} \frac {\expect {X Y} } {\expect {Y^2} } + \expect {Y^2} \frac {\paren {\expect {X Y} }^2} {\paren {\expect {Y^2} }^2}\) | Expectation is Linear | |||||||||||

\(\ds \) | \(=\) | \(\ds \expect {X^2} - \frac {\paren {\expect {X Y} }^2 } {\expect {Y^2} }\) |

We therefore have:

- $\dfrac {\paren {\expect {X Y} }^2} {\expect {Y^2} } \le \expect {X^2}$

giving:

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

as required.

It remains to address the case $\expect {Y^2} = 0$.

Note that since $\map \Pr {Y^2 \ge 0} = 1$, from Condition for Expectation of Non-Negative Random Variable to be Zero we necessarily have:

- $\map \Pr {Y^2 = 0} = 1$

That is:

- $\map \Pr {Y = 0} = 1$

This implies that the random variable $X Y$ has:

- $\map \Pr {X Y = 0} = 1$

From which, we have that:

- $\map \Pr {X Y \ge 0} = 1$

So, applying Expectation of Non-Negative Random Variable is Non-Negative again we have:

- $\expect {X Y} = 0$

With that, we have:

- $\paren {\expect {X Y} }^2 = 0$

and:

- $\expect {X^2} \expect {Y^2} = \expect {X^2} \times 0 = 0$

So the inequality:

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

also holds in the case $\expect {Y^2} = 0$, completing the proof.

$\blacksquare$

## Also see

- Hölder's Inequality, of which this is an instantiation

## Sources

- 2011: Morris H. DeGroot and Mark J. Schervish:
*Probability and Statistics*(4th ed.): $4.6$: Covariance and Correlation: Theorem $4.6.2$