Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Expected value ▷ Squared expectation of a product

Theorem: Let $X$ and $Y$ be two random variables with expected values $\mathrm{E}(X)$ and $\mathrm{E}(Y)$ and let $\mathrm{E}(XY)$ exist and be finite. Then, the square of the expectation of the product of $X$ and $Y$ is less than or equal to the product of the expectation of the squares of $X$ and $Y$:

$\label{eq:EXY2-EX2-EY2} \left[ \mathrm{E}(XY) \right]^2 \leq \mathrm{E}\left( X^2 \right) \mathrm{E}\left( Y^2 \right) \; .$

Proof: Note that $Y^2$ is a non-negative random variable whose expected value is also non-negative:

$\label{eq:mean-Y2-nonneg} \mathrm{E}\left( Y^2 \right) \geq 0 \; .$

1) First, consider the case that $\mathrm{E}\left( Y^2 \right) > 0$. Define a new random variable Z as

$\label{eq:Z} Z = X - Y \, \frac{\mathrm{E}(XY)}{\mathrm{E}\left( Y^2 \right)} \; .$

Once again, because $Z^2$ is always non-negative, we have the expected value:

$\label{eq:mean-Z2-nonneg} \mathrm{E}\left( Z^2 \right) \geq 0 \; .$

Thus, using the linearity of the expected value, we have

$\label{eq:mean-prodsqr-v1} \begin{split} 0 &\leq \mathrm{E}\left( Z^2 \right) \\ &\leq \mathrm{E}\left( \left( X - Y \, \frac{\mathrm{E}(XY)}{\mathrm{E}\left( Y^2 \right)} \right)^2 \right) \\ &\leq \mathrm{E}\left( X^2 - 2 X Y \, \frac{\mathrm{E}(XY)}{\mathrm{E}\left( Y^2 \right)} + Y^2 \, \frac{\left[ \mathrm{E}(XY) \right]^2}{\left[ \mathrm{E}\left( Y^2 \right) \right]^2} \right) \\ &\leq \mathrm{E}\left( X^2 \right) - 2 \, \mathrm{E}(XY) \, \frac{\mathrm{E}(XY)}{\mathrm{E}\left( Y^2 \right)} + \mathrm{E}\left( Y^2 \right) \, \frac{\left[ \mathrm{E}(XY) \right]^2}{\left[ \mathrm{E}\left( Y^2 \right) \right]^2} \\ &\leq \mathrm{E}\left( X^2 \right) - 2 \, \frac{\left[ \mathrm{E}(XY) \right]^2}{\mathrm{E}\left( Y^2 \right)} + \frac{\left[ \mathrm{E}(XY) \right]^2}{\mathrm{E}\left( Y^2 \right)} \\ &\leq \mathrm{E}\left( X^2 \right) - \frac{\left[ \mathrm{E}(XY) \right]^2}{\mathrm{E}\left( Y^2 \right)} \; , \\ \end{split}$

giving

$\label{eq:EXY2-EX2-EY2-qed-v1} \left[ \mathrm{E}(XY) \right]^2 \leq \mathrm{E}\left( X^2 \right) \mathrm{E}\left( Y^2 \right)$

as required.

2) Next, consider the case that $\mathrm{E}\left( Y^2 \right) = 0$. In this case, $Y$ must be a constant with mean $\mathrm{E}(Y) = 0$ and variance $\mathrm{Var}(Y) = 0$, thus we have

$\label{eq:Pr-Y-0} \mathrm{Pr}(Y = 0) = 1 \; .$

This implies

$\label{eq:Pr-XY-0} \mathrm{Pr}(XY = 0) = 1 \; ,$

such that

$\label{eq:EXY} \mathrm{E}(XY) = 0 \; .$

Thus, we can write

$\label{eq:mean-prodsqr-v2} 0 = \left[ \mathrm{E}(XY) \right]^2 = \mathrm{E}\left( X^2 \right) \mathrm{E}\left( Y^2 \right) = 0 \; ,$

giving

$\label{eq:EXY2-EX2-EY2-qed-v2} \left[ \mathrm{E}(XY) \right]^2 \leq \mathrm{E}\left( X^2 \right) \mathrm{E}\left( Y^2 \right)$

as required.

Sources:

Metadata: ID: P359 | shortcut: mean-prodsqr | author: JoramSoch | date: 2022-10-11, 01:39.