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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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

Statement

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

Facts & Assumptions

Given: A Bernoulli $(p)$ variable $X$ and a binomial $(n, p)$ variable $S = \sum_{i < n} X_{i}$ .

[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).

[L2]

Expectation is linear without independence (Expectation is linear for every finite family of random variables, without any independence hypothesis).

[L3]

$Var (X) = E [X^{2}] - E [X]^{2}$ ( $Var (X) = E [X^{2}] - E [X]^{2}$ and $Cov (X, Y) = E [X Y] - E [X] E [Y]$ ).

[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 $E [X^{2}] = p$ .

2.1

Hence $Var (X) = p - p^{2} = p (1 - p)$ .

step 1.1L3algebra

3.1

Linearity gives $E [S] = \sum_{i < n} p = n p$ , and pairwise-independent variance additivity gives $Var (S) = \sum_{i < n} p (1 - p) = n p (1 - p)$ .

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 · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

C. M. Grinstead and J. L. Snell, Introduction to Probability, 2nd ed., Section 6.2 (standard reference, not scraped)
H. Pishro-Nik, Introduction to Probability, Statistics, and Random Processes, Sections 3.1.5 and 3.2.4 (standard reference, not scraped)

A Bernoulli(p)(p) variable has mean pp and variance p(1−p)p(1−p); a binomial(n,p)(n,p) variable has mean npnp and variance np(1−p)np(1−p)