Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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A Bernoulli(p) variable has mean p and variance p(1p); a binomial(n,p) variable has mean np and variance np(1p)

Statement

If X is Bernoulli(p), then E[X]=p and Var(X)=p(1p). If S is binomial(n,p), then E[S]=np,Var(S)=np(1p). These formulas include p=0, p=1, and n=0.

Facts & Assumptions

Given: A Bernoulli(p) variable X and a binomial(n,p) variable S=i<nXi.

[L1]

Bernoulli and binomial variables have the distributions and independent-sum form in the Definition (Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials).

[L4]

Variance adds for finite pairwise-independent families (Variance adds for every finite pairwise-independent family).

Proof

technique · direct
1.1

Since X is 1 with probability p and 0 otherwise, E[X]=p and E[X2]=p.

L1
2.1

Hence Var(X)=pp2=p(1p).

step 1.1L3algebra
3.1

Linearity gives E[S]=i<np=np, and pairwise-independent variance additivity gives Var(S)=i<np(1p)=np(1p).

step 1.1step 2.1L1L2L4algebra
4.1

At p=0 or p=1 the variance formula is zero, and at n=0 both finite sums are empty and equal zero.

step 3.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 39 results over 18 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