Sum of Expectations of Independent Trials

Theorem
Let $\mathcal E_1, \mathcal E_2, \ldots, \mathcal E_n$ be a sequence of experiments whose outcomes are independent of each other.

Let $X_1, X_2, \ldots, X_n$ be discrete random variables on $\mathcal E_1, \mathcal E_2, \ldots, \mathcal E_n$ respectively.

Let $E \left({X_j}\right)$ be the expectation of $X_j$ for $j \in \left\{{1, 2, \ldots, n}\right\}$.

Then we have, whenever both sides are defined:
 * $\displaystyle E \left({\sum_{j \mathop = 1}^n X_j}\right) = \sum_{j \mathop = 1}^n E \left({X_j}\right)$

That is, the sum of the expectations equals the expectation of the sum.

Proof 1
We will use induction on $n$, that is, on the number of terms in the sum.

Basis for the Induction
The case $n = 1$ is tautologically true.

This is our basis for the induction.

Induction Hypothesis
It follows that our induction hypothesis is:


 * $\displaystyle E \left({\sum_{j \mathop = 1}^{n-1} X_j}\right) = \sum_{j \mathop = 1}^{n-1} E \left({X_j}\right)$

Induction Step
Finally, this is our induction step:

Denote $Y = \displaystyle \sum_{j \mathop = 1}^{n-1} X_j$. Then we compute:

The result follows by the Principle of Mathematical Induction.

Proof 2
We will use induction on $n$, that is, on the number of terms in the sum.

Basis for the Induction
The case $n = 1$ is tautologically true.

This is our basis for the induction.

Induction Hypothesis
It follows that our induction hypothesis is:


 * $\displaystyle E \left({\sum_{j \mathop = 1}^{n-1} X_j}\right) = \sum_{j \mathop = 1}^{n-1} E \left({X_j}\right)$

Induction Step
Finally, this is our induction step:

Denote the random variable $Y = \displaystyle \sum_{j \mathop = 1}^{n-1} X_j$. Then we compute:

The result follows by the Principle of Mathematical Induction.