Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Expected value ▷ (Non-)Multiplicativity

Theorem:

1) If two random variables $X$ and $Y$ are independent, the expected value is multiplicative, i.e.

$\label{eq:mean-mult} \mathrm{E}(X\,Y) = \mathrm{E}(X) \, \mathrm{E}(Y) \; .$

2) If two random variables $X$ and $Y$ are dependent, the expected value is not necessarily multiplicative, i.e. there exist $X$ and $Y$ such that

$\label{eq:mean-nonmult} \mathrm{E}(X\,Y) \neq \mathrm{E}(X) \, \mathrm{E}(Y) \; .$

Proof:

1) If $X$ and $Y$ are independent, it holds that

$\label{eq:ind} p(x,y) = p(x) \, p(y) \quad \text{for all} \quad x \in \mathcal{X}, y \in \mathcal{Y} \; .$

Applying this to the expected value for discrete random variables, we have

$\label{eq:mean-mult-disc} \begin{split} \mathrm{E}(X\,Y) &= \sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} (x \cdot y) \cdot f_{X,Y}(x,y) \\ &\overset{\eqref{eq:ind}}{=} \sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} (x \cdot y) \cdot \left( f_X(x) \cdot f_Y(y) \right) \\ &= \sum_{x \in \mathcal{X}} x \cdot f_X(x) \, \sum_{y \in \mathcal{Y}} y \cdot f_Y(y) \\ &= \sum_{x \in \mathcal{X}} x \cdot f_X(x) \cdot \mathrm{E}(Y) \\ &= \mathrm{E}(X) \, \mathrm{E}(Y) \; . \\ \end{split}$

And applying it to the expected value for continuous random variables, we have

$\label{eq:mean-mult-cont} \begin{split} \mathrm{E}(X\,Y) &= \int_{\mathcal{X}} \int_{\mathcal{Y}} (x \cdot y) \cdot f_{X,Y}(x,y) \, \mathrm{d}y \, \mathrm{d}x \\ &\overset{\eqref{eq:ind}}{=} \int_{\mathcal{X}} \int_{\mathcal{Y}} (x \cdot y) \cdot \left( f_X(x) \cdot f_Y(y) \right) \, \mathrm{d}y \, \mathrm{d}x \\ &= \int_{\mathcal{X}} x \cdot f_X(x) \, \int_{\mathcal{Y}} y \cdot f_Y(y) \, \mathrm{d}y \, \mathrm{d}x \\ &= \int_{\mathcal{X}} x \cdot f_X(x) \cdot \mathrm{E}(Y) \, \mathrm{d}x \\ &= \mathrm{E}(X) \, \mathrm{E}(Y) \; . \\ \end{split}$

2) Let $X$ and $Y$ be Bernoulli random variables with the following joint probability mass function

$\label{eq:joint} \begin{split} p(X=0, Y=0) &= 1/2 \\ p(X=0, Y=1) &= 0 \\ p(X=1, Y=0) &= 0 \\ p(X=1, Y=1) &= 1/2 \end{split}$

and thus, the following marginal probabilities:

$\label{eq:marg} \begin{split} p(X=0) = p(X=1) &= 1/2 \\ p(Y=0) = p(Y=1) &= 1/2 \; . \end{split}$

Then, $X$ and $Y$ are dependent, because

$\label{eq:dep} p(X=0, Y=1) \overset{\eqref{eq:joint}}{=} 0 \neq \frac{1}{2} \cdot \frac{1}{2} \overset{\eqref{eq:marg}}{=} p(X=0) \, p(Y=1) \; ,$

and the expected value of their product is

$\label{eq:mean-prod} \begin{split} \mathrm{E}(X\,Y) &= \sum_{x \in \left\lbrace 0,1 \right\rbrace} \sum_{y \in \left\lbrace 0,1 \right\rbrace} (x \cdot y) \cdot p(x,y) \\ &= (1 \cdot 1) \cdot p(X=1, Y=1) \\ &\overset{\eqref{eq:joint}}{=} \frac{1}{2} \end{split}$

while the product of their expected values is

$\label{eq:prod-mean} \begin{split} \mathrm{E}(X) \, \mathrm{E}(Y) &= \left( \sum_{x \in \left\lbrace 0,1 \right\rbrace} x \cdot p(x) \right) \cdot \left( \sum_{y \in \left\lbrace 0,1 \right\rbrace} y \cdot p(y) \right) \\ &= \left( 1 \cdot p(X=1) \right) \cdot \left( 1 \cdot p(Y=1) \right) \\ &\overset{\eqref{eq:marg}}{=} \frac{1}{4} \end{split}$

and thus,

$\label{eq:mean-nonmult-qed} \mathrm{E}(X\,Y) \neq \mathrm{E}(X) \, \mathrm{E}(Y) \; .$
Sources:

Metadata: ID: P55 | shortcut: mean-mult | author: JoramSoch | date: 2020-02-17, 21:51.