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:

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

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

Proof

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:

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

Induction Step

Finally, this is our induction step:

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle E \left({\sum_{j=1}^n X_j}\right)\)	\(=\)	\(\displaystyle E \left({Y + X_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{y + x_n \in \R} \left({y + x_n}\right) \Pr \left({Y + X_n = y + x_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of expectation
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{y \in \R} \sum_{x_n \in \R} \left({y + x_n}\right) \Pr \left({Y = y, X_n = x_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of joint probability
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{y \in \R} \sum_{x_n \in \R} \left({y + x_n}\right) \Pr \left({Y = y}\right) \Pr \left({X_n = x_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Independence of $Y$ and $X_n$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{y \in \R} \sum_{x_n \in \R} y \Pr \left({Y = y}\right) \Pr \left({X_n = x_n}\right) + \sum_{y \in \R} \sum_{x_n \in \R} x_n \Pr \left({Y = y}\right) \Pr \left({X_n = x_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Splitting the summation
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{y \in \R} y \Pr \left({Y = y}\right) + \sum_{x_n \in \R} x_n \Pr \left({X_n = x_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Because $\displaystyle \sum_{x_n \in \R} \Pr \left({X_n = x_n}\right) = 1 = \sum_{y \in \R} \Pr \left({Y = y}\right)$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle E \left({Y}\right) + E \left({X_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of expectation
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{j=1}^n E \left({X_j}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the Induction Hypothesis

The result follows by the Principle of Mathematical Induction.

$\blacksquare$