Sum of Expectations of Independent Trials/Proof 1

Theorem

Let $\EE_1, \EE_2, \ldots, \EE_n$ 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 $\EE_1, \EE_2, \ldots, \EE_n$ respectively.

Let $\expect {X_j}$ denote the expectation of $X_j$ for $j \in \set {1, 2, \ldots, n}$.

Then we have, whenever both sides are defined:

$\ds \expect {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \expect {X_j}$

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

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 our induction step:

Denote $Y = \ds \sum_{j \mathop = 1}^k X_j$

Then we compute:

\(\ds \expect {\sum_{j \mathop = 1}^{k + 1} X_j}\)

\(=\)

\(\ds \expect {Y + X_{k + 1} }\)

\(\ds \)

\(=\)

\(\ds \sum_{y \mathop + x_{k + 1} \mathop \in \R} \paren {y + x_{k + 1} } \map \Pr {Y + X_{k + 1} = y + x_{k + 1} }\)

Definition of Expectation

\(\ds \)

\(=\)

\(\ds \sum_{y \mathop \in \R} \sum_{x_{k + 1} \mathop \in \R} \paren {y + x_{k + 1} } \map \Pr {Y = y, X_{k + 1} = x_{k + 1} }\)

Definition of Joint Probability Mass Function

\(\ds \)

\(=\)

\(\ds \sum_{y \mathop \in \R} \sum_{x_{k + 1} \mathop \in \R} \paren {y + x_{k + 1} } \map \Pr {Y = y} \, \map \Pr {X_{k + 1} = x_{k + 1} }\)

Independence of $Y$ and $X_{k + 1}$

\(\ds \)

\(=\)

\(\ds \sum_{y \mathop \in \R} \sum_{x_{k + 1} \mathop \in \R} y \, \map \Pr {Y = y} \, \map \Pr {X_{k + 1} = x_{k + 1} } + \sum_{y \mathop \in \R} \sum_{x_{k + 1} \mathop \in \R} x_{k + 1} \, \map \Pr {Y = y} \, \map \Pr {X_{k + 1} = x_{k + 1} }\)

splitting the summation

\(\ds \)

\(=\)

\(\ds \sum_{y \mathop \in \R} y \, \map \Pr {Y = y} + \sum_{x_{k + 1} \mathop \in \R} x_{k + 1} \, \map \Pr {X_{k + 1} = x_{k + 1} }\)

because $\ds \sum_{x_{k + 1} \mathop \in \R} \map \Pr {X_{k + 1} = x_{k + 1} } = 1 = \sum_{y \mathop \in \R} \map \Pr {Y = y}$

\(\ds \)

\(=\)

\(\ds \expect Y + \expect {X_{k + 1} }\)

Definition of Expectation

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 1}^{k + 1} \expect {X_j}\)

by the Induction Hypothesis

The result follows by the Principle of Mathematical Induction.

$\blacksquare$

This article, or a section of it, needs explaining. In particular: When splitting the summation, I use that both of the expressions exist (which they do by the assumption I added - when they exist). This should be mentioned Moved this discussion into talk page. -- PM You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.