Theorem: Expectation of Product of Independent Random Variables is Product of Expectations

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ and $Y$ be non-negative real-valued random variables that are independent.
Then:
:$\expect {X Y} = \expect X \expect Y$

Proof:
We first prove the claim in the case that $X = \chi_A$ for $A \in \Sigma$ and $Y = \chi_B$ for $B \in \Sigma$.
In particular, we have $A \in \map \sigma X$ and $B \in \map \sigma Y$ where $\map \sigma X$ and $\map \sigma Y$ are the $\sigma$-algebras generated by $A$ and $B$ respectively.
Then, we have: 
{{begin-eqn}}
{{eqn	| l = \expect {X Y}
		| r = \expect {\chi_A \cdot \chi_B}
}}
{{eqn	| r = \expect {\chi_{A \cap B} }
		| c = Characteristic Function of Intersection
}}
{{eqn	| r = \map \Pr {A \cap B}
		| c = Integral of Characteristic Function
}}
{{eqn	| r = \map \Pr A \map \Pr B
		| c = {{Defof|Independent Sigma-Algebras}}
}}
{{eqn	| r = \expect {\chi_A} \expect {\chi_B}
		| c = Integral of Characteristic Function
}}
{{eqn	| r = \expect X \expect Y
}}
{{end-eqn}}
Now let $X$ and $Y$ be general non-negative real-valued random variables.
Let $X'$ and $Y'$ be positive simple real-valued random variables that are $\map \sigma X$ and $\map \sigma Y$ measurable respectively.
From Simple Function has Standard Representation, there exists:
:a finite sequence $a_1, \ldots, a_n$ of real numbers
:a partition $A_1, A_2, \ldots, A_n$ of $\map \sigma X$-measurable sets
with:
:$\ds X' = \ds \sum_{i \mathop = 1}^n a_i \chi_{A_i}$
and:
:a finite sequence $b_1, \ldots, b_m$ of real numbers
:a partition $B_1, B_2, \ldots, B_m$ of $\map \sigma Y$-measurable sets
with:
:$\ds Y' = \ds \sum_{j \mathop = 1}^m b_j \chi_{B_j}$
Then we have:
{{begin-eqn}}
{{eqn	| l = \expect {X' Y'}
		| r = \expect {\paren {\sum_{i \mathop = 1}^n a_i \chi_{A_i} } \paren {\sum_{j \mathop = 1}^m b_j \chi_{B_j} } }
}}
{{eqn	| r = \expect {\sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^m a_i b_j \chi_{A_i} \chi_{B_j} }
}}
{{eqn	| r = \expect {\sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^m a_i b_j \chi_{A_i \cap B_j} }
		| c = Characteristic Function of Intersection
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^m a_i b_j \map \Pr {A_i \cap B_j}
		| c = Integral of Characteristic Function
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^m a_i b_j \map \Pr {A_i} \map \Pr {B_j}
		| c = {{Defof|Independent Sigma-Algebras}}
}}
{{eqn	| r = \paren {\sum_{i \mathop = 1}^n a_i \map \Pr{A_i} } \paren {\sum_{j \mathop = 1}^m b_j \map \Pr {B_j} }
}}
{{eqn	| r = \expect {\sum_{i \mathop = 1}^n a_i \chi_{A_i} } \expect {\sum_{j \mathop = 1}^m b_j \chi_{B_j} }
		| c = Integral of Characteristic Function, Expectation is Linear
}}
{{eqn	| r = \expect {X'} \expect {Y'}
}}
{{end-eqn}}
Now, from Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequences of positive simple random variables $\sequence {X_n}_{n \mathop \in \N}$, $\sequence {Y_n}_{n \mathop \in \N}$ with: 
:$\ds \map X \omega = \lim_{n \mathop \to \infty} \map {X_n} \omega$
and:
:$\ds \map Y \omega = \lim_{n \mathop \to \infty} \map {Y_n} \omega$
for all $\omega \in \Omega$.
Furthermore, we can show that $X_n \in \map \sigma X$ and $Y_n \in \map \sigma Y$ for each $n \in \N$.
Note that from Measurable Function is Pointwise Limit of Simple Functions, we have:
:$\map {X_n} \omega = \ds \sum_{k \mathop = 0}^{n 2^n} k 2^{-n} \map {\chi_{ {A_k}^n} } \omega$
where:
:<nowiki>${A_k}^n := \begin{cases}
 \set {k 2^{-n} \le X < \paren {k + 1} 2^{-n} } & : k \ne n 2^n \\
 \set {X \ge n} & : k = n 2^n
\end{cases}$</nowiki>
Each ${A_k}^n$ is $\map \sigma X$-mesurable, since by Characterization of Sigma-Algebra Generated by Collection of Mappings $\map \sigma X$ is generated by the preimages of $X$ under Borel sets.
So from Characteristic Function Measurable iff Set Measurable and Pointwise Sum of Measurable Functions is Measurable, we have that $X_n$ is $\map \sigma X$-mesurable for each $n \in \N$.
Similarly, $Y_n$ is $\map \sigma Y$-mesurable for each $n \in \N$.
So, for each $n \in \N$ we have from our previous computation on positive simple random variables:
:$\expect {X_n Y_n} = \expect {X_n} \expect {Y_n}$ 
We have that $\sequence {X_n Y_n}_{n \mathop \in \N}$ is increasing, so we have:
:$\expect {X_n Y_n} \to \expect {X Y}$
by the Monotone Convergence Theorem (Measure Theory).
Also applying the Monotone Convergence Theorem (Measure Theory) we have:
:$\expect {X_n} \to \expect X$
and:
:$\expect {Y_n} \to \expect Y$
So we have:
:$\expect {X_n} \expect {Y_n} \to \expect X \expect Y$ 
and so:
:$\expect {X Y} = \expect X \expect Y$ 
as desired.
{{qed}}
Category:Expectation of Product of Independent Random Variables is Product of Expectations
Category:Expectation
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