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

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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:

\(\ds \expect {\prod_{k \mathop = 1}^{r + 1} {X_k} }\) \(=\) \(\ds \expect {X_{r + 1} \prod_{k \mathop = 1}^r {X_k} }\)
\(\ds \) \(=\) \(\ds \expect {X_{r + 1} } \expect {\prod_{k \mathop = 1}^r {X_k} }\) Corollary to Condition for Independence from Product of Expectations
\(\ds \) \(=\) \(\ds \expect {X_{r + 1} } \prod_{k \mathop = 1}^r \expect {X_k}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 1}^{r + 1} \expect {X_k}\)

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

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

$\blacksquare$