Index: The Book of Statistical ProofsGeneral TheoremsProbability theoryExpected value ▷ Squared expectation of a product

Theorem: Let 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.