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:
- $\ds \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$
assuming the latter expectations exist.
Proof
Proof by induction:
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
- $\ds \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$
Basis for the Induction
$\map P 1$ is the case:
- $\ds \expect {\prod_{k \mathop = 1}^1 {X_k} } = \expect {X_1} = \prod_{k \mathop = 1}^1 \expect {X_k}$
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:
- $\ds \expect {\prod_{k \mathop = 1}^r {X_k} } = \prod_{k \mathop = 1}^r \expect {X_k}$
from which it is to be shown that:
- $\ds \expect {\prod_{k \mathop = 1}^{r + 1} {X_k} } = \prod_{k \mathop = 1}^{r + 1} \expect {X_k}$
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:
- $\ds \forall n \in \Z_{> 0}: \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$
$\blacksquare$