Linearity of Expectation¶

Proposition¶

For integrable random variables $X_1,\ldots,X_n$ and real constants $a_1,\ldots,a_n$ , linearity of expectation states

\mathbb{E}\!\left[\sum_{i=1}^n a_iX_i\right]=\sum_{i=1}^n a_i\mathbb{E}X_i.

Independence is not required. In particular, $\mathbb{E}[aX+bY]=a\mathbb{E}X+b\mathbb{E}Y$ $b E Y$ .

Interpretation¶

The average of a finite linear combination is the same linear combination of averages. Dependence does not affect this rule.

Three number lines for X, Y and X + Y with their possible values as dots and their means 2, 4 and 6 marked by small triangles.

The independent variables shown have means $2$ and $4$ , so their sum has mean $6$ . Here the sum takes the values $3$ , $5$ , $7$ and $9$ , each with probability $1/4$ ; the identity for the means itself does not require independence.

Addition and sum forms the finite weighted sum of the variables. Expectation then averages it: distributing the probability weights across its summands explains why taking this average commutes with the finite sum.

Key Results¶

For integrable $X_1,\ldots,X_n$ , $\displaystyle \mathbb{E}[\sum_ia_iX_i]=\sum_ia_i\mathbb{E}X_i$ $i \sum a_{i} E X_{i}$ .
$\mathbb{E}[a+bX]=a+b\mathbb{E}X$ $b E X$ .

Examples¶

If $\mathbb{E}X=2$ and $\mathbb{E}Y=5$ $5$ , then $\mathbb{E}[3X-Y]=1$ regardless of dependence.
If $Y=X$ and $\mathbb{E}X=4$ $4$ , then $\mathbb{E}[X+Y]=8$ despite perfect dependence.

Prerequisites¶

Sources¶

Probability Theory for EOR · Week 4
Nicky van Foreest · Probability Distributions · Refresher

Aliases: linearity of expected value, additivity of expectation Keywords: expectation is linear Symbols: sum

Last updated: 17 October 2026 21:53