Condition for Independence from Product of Expectations

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Theorem

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.

Let $\expect X$ denote the expectation of $X$.


Then $X$ and $Y$ are independent if and only if:

$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$

for all functions $g, h: \R \to \R$ for which the latter two expectations exist.


Corollary

Let $X$ and $Y$ be independent discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.

Then:

$\expect {X Y} = \expect X \expect Y$

assuming the latter expectations exist.


Proof

Sufficient Condition

Suppose that $X$ and $Y$ were not independent.

That is:

$\map \Pr {X = a, Y = b} \ne \map \Pr {X = a} \map \Pr {Y = b}$

for some $a, b \in \R$.


Now, suppose that

$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$

for all functions $g, h: \R \to \R$ for which the latter two expectations exist.


Let us define $g$ and $h$ as examples of such functions as follows:

$(1): \quad \map g x := \begin{cases} 1 & : x = a \\ 0 & : x \ne a \end{cases}$
$(2): \quad \map h y := \begin{cases} 1 & : y = b \\ 0 & : y \ne b \end{cases}$

where $a \in \R$ and $b \in \R$ are arbitrary real numbers.

Then:

\(\ds \expect {\map g x \map h y}\) \(=\) \(\ds \map \Pr {X = a, Y = b}\)
\(\ds \expect {\map g X} \expect {\map h Y}\) \(=\) \(\ds \map \Pr {X = a} \map \Pr {Y = b}\)


So by hypothesis:

$\map \Pr {X = a, Y = b} = \map \Pr {X = a} \map \Pr {Y = b}$

But also by hypothesis:

$\map \Pr {X = a, Y = b} \ne \map \Pr {X = a} \map \Pr {Y = b}$

This contradicts our supposition that:

$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$

for all functions $g, h: \R \to \R$ for which the latter two expectations exist.

So, if the above supposition holds, then $X$ and $Y$ have to be independent.

$\Box$


Necessary Condition

Suppose $X$ and $Y$ are independent.

Let $g, h: \R \to \R$ be any real functions such that $\expect {\map g x}$ and $\expect {\map h y}$ exist.


Then:

\(\ds \expect {\map g x \map h y}\) \(=\) \(\ds \sum_{x, y} \map g x \map h y \map \Pr {X = x, Y = y}\)
\(\ds \) \(=\) \(\ds \sum_{x, y} \map g x \map h y \map \Pr {X = x} \map \Pr {Y = y}\) by definition of independent random variables
\(\ds \) \(=\) \(\ds \sum_x \map g x \map \Pr {X = x} \sum_y \map h y \map \Pr {Y = y}\)
\(\ds \) \(=\) \(\ds \expect {\map g x} \expect {\map h y}\)

thus proving that

$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$

whatever $g$ and $h$ are.

$\blacksquare$


Sources