CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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

A Bernoulli(p) variable has mean p and variance p(1-p); a binomial(n,p) variable has mean np and variance np(1-p) Lemma

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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)