Sum of Expectations of Independent Trials/Proof 2

Proof
The proof proceeds by induction on the number of terms $n$ in the sum.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\ds \expect {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \expect {X_j}$

Basis for the Induction
$\map P 1$ is the case:
 * $\ds \expect {\sum_{j \mathop = 1}^1 X_j} = \sum_{j \mathop = 1}^1 \expect {X_j}$

That is:
 * $\expect {X_1} = \expect {X_1}$

which is tautologically true.

This is our basis for the induction.

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

So this is the induction hypothesis: $\ds \expect {\sum_{j \mathop = 1}^k X_j} = \sum_{j \mathop = 1}^k \expect {X_j}$

from which it is to be shown that: $\ds \expect {\sum_{j \mathop = 1}^{k + 1} X_j} = \sum_{j \mathop = 1}^{k + 1} \expect {X_j}$

Induction Step
This is the induction step:

Denote the random variable $Y = \ds \sum_{j \mathop = 1}^k X_j$.

Then we compute:

The result follows by the Principle of Mathematical Induction.