---
tag: TCV030YM49
kind: proof
name: Proof of the formula for the variance of a sum of independent random variables
msc: [60A05]
proves: 0V0778DCW9
uses: [0XCE070V8B, 0E9502NT92, 0D86GJEZ39, 077GN9HDR6]
sources:
  - url: https://en.wikipedia.org/wiki/Variance
    locator: Sum of uncorrelated variables
---

## Proof

We use the general formula for the variance of a sum and show that all covariances vanish.

1. Let $X_1, \dots, X_n$ be [[independent|0XCE070V8B]] real random variables with $E[X_i^2] < \infty$ for each $i$.
2. Fix $i < j$. Apply the definition of independence to the set $J = \set{i, j}$. For all Borel sets $A$ and $B$, it gives $P(X_i \in A, X_j \in B) = P(X_i \in A) \, P(X_j \in B)$.
3. So $X_i$ and $X_j$ are [[independent as a pair|0E9502NT92]].
4. Both have finite second moments. Since [[independent random variables are uncorrelated|077GN9HDR6]], $\operatorname{Cov}(X_i, X_j) = 0$.
5. By the [[formula for the variance of a sum|0D86GJEZ39]],
$$\operatorname{Var}(X_1 + \dots + X_n) = \sum_{i=1}^n \operatorname{Var}(X_i) + 2 \sum_{1 \le i < j \le n} \operatorname{Cov}(X_i, X_j).$$
6. By step 4, every covariance term is $0$. So $\operatorname{Var}(X_1 + \dots + X_n) = \operatorname{Var}(X_1) + \dots + \operatorname{Var}(X_n)$.

## Idea

The variance of a sum is the sum of the variances plus twice the sum of all covariances. Independence of each pair kills every covariance, so only the variances remain.