---
tag: 0V0778DCW9
kind: statement
role: theorem
name: Variance of a sum of independent random variables
aliases: [bienayme-formula]
msc: [60A05]
uses: [0D86GJEZ39, 077GN9HDR6, 0XCE070V8B]
sources:
  - url: https://en.wikipedia.org/wiki/Variance
    locator: Sum of uncorrelated variables
---

## Statement

Let $X_1, \dots, X_n$ be [[independent|0XCE070V8B]] real random variables with $E[X_i^2] < \infty$ for each $i$. Then
$$\operatorname{Var}(X_1 + \dots + X_n) = \operatorname{Var}(X_1) + \dots + \operatorname{Var}(X_n).$$

## Remarks

The proof uses the [[variance of a sum|0D86GJEZ39]]: independent variables are [[uncorrelated|077GN9HDR6]], so each covariance term is $0$.
So the formula holds more generally for pairwise uncorrelated $X_i$.
For i.i.d. $X_i$ with variance $\sigma^2$, the sample mean $\bar{X}_n = (X_1 + \dots + X_n)/n$ has variance $\sigma^2 / n$.
For $X \sim \operatorname{Bin}(n, p)$, the formula gives $\operatorname{Var}(X) = np(1 - p)$, as $X$ is a sum of $n$ independent Bernoulli variables.

## History

The formula is also called the Bienaymé formula, after Irénée-Jules Bienaymé.

## Explanations

### L1 human

For independent variables, variances simply add.
This is why an average of many independent measurements is more precise than one measurement: its variance gets smaller like $1/n$.