CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Variance adds for every finite pairwise-independent family

Statement

If $(X_{i})_{i \in I}$ is a finite pairwise-independent family, then $Var ⁣ (\sum_{i \in I} X_{i}) = \sum_{i \in I} Var (X_{i}) .$ Pairwise independence, rather than mutual independence, is sufficient.

Facts & Assumptions

Given: A finite pairwise-independent family $(X_{i})_{i \in I}$ .

[L1]

For two independent random variables, expectation of their product is the product of their expectations (Expectation factors over a finite product of mutually independent random variables).

[L2]

Variance of a finite sum is the sum of variances and twice all pairwise covariances (Variance of a finite sum as the sum of all variances and covariances).

Proof

technique · direct

1.1

For distinct $i, j$ , pairwise independence and [L1] give $E [X_{i} X_{j}] = E [X_{i}] E [X_{j}]$ , so $Cov (X_{i}, X_{j}) = 0$ .

L1algebra

2.1

Substitute step 1.1 into [L2]; every off-diagonal term vanishes, leaving the displayed formula. The empty and singleton cases are included.

step 1.1L2algebra∎

Depends on

Used by

Dependency tree · two levels

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Sources

C. M. Grinstead and J. L. Snell, Introduction to Probability, 2nd ed., Theorem 6.4 (standard reference, not scraped)
J. Matousek and J. Vondrak, The Probabilistic Method, Section 6.1 (standard reference, not scraped)