Variance of a sum of independent random variables

Statement

Let $X_1, \dots, X_n$ be independent 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: independent variables are uncorrelated, 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 $1/n$.