Condition for Independence from Product of Expectations/Corollary/General Result

Corollary to Condition for Independence from Product of Expectations
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. Let $X_1, X_2, \ldots, X_n$ be independent discrete random variables.

Then:
 * $\displaystyle \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$

assuming the latter expectations exist.

Proof
Proof by induction:

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
 * $\displaystyle \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$

Basis for the Induction
$\map P 1$ is the case:
 * $\displaystyle \expect {\prod_{k \mathop = 1}^1 {X_k} } = \expect {X_1} = \prod_{k \mathop = 1}^1 \expect {X_k}$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:
 * $\displaystyle \expect {\prod_{k \mathop = 1}^r {X_k} } = \prod_{k \mathop = 1}^r \expect {X_k}$

from which it is to be shown that:
 * $\displaystyle \expect {\prod_{k \mathop = 1}^{r + 1} {X_k} } = \prod_{k \mathop = 1}^{r + 1} \expect {X_k}$

Induction Step
This is the induction step:

We have:

So $\map P r \implies \map P {r + 1}$ and thus it follows by the Principle of Mathematical Induction that:


 * $\displaystyle \forall n \in \Z_{> 0}: \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$