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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

Work on a common probability space; the independent Bernoulli variables used to represent a binomial sum are independent in the sense of Independent random elements. The original finite-space atom-independence convention is included.

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 $S = \sum_{i < n} X_{i}$ on one common probability space, with mutually independent Bernoulli summands. All these variables and their finite sums are bounded almost surely, hence square-integrable and integrable.

[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 (Linearity, monotonicity, and the modulus bound for expectation).

[L3]

$Var (X) = E [X^{2}] - E [X]^{2}$ and $Cov (X, Y) = E [X Y] - E [X] E [Y]$ (Variance and covariance identities for random variables).

[L4]

For square-integrable variables on one common probability space, variance of a finite sum is the sum of its individual variances plus twice the sum of the covariances over distinct unordered pairs. (Variance of a finite sum as the sum of all variances and covariances)

[L5]

General independence factors inverse-image event probabilities. On an original finite probability space, attained-value independence agrees with this general notion. (Independent random elements, The general rectangle criterion agrees with the published finite random-variable definition)

[L6]

The expectation of the indicator of a measurable event is its probability, and expectations of integrable variables agree under almost-sure equality. (The expectation of an indicator is the probability of the event, Expectation depends only on the almost-everywhere class)

Proof

technique · direct

1.1L1L6

Since $X$ is $1$ with probability $p$ and $0$ otherwise, both $X$ and $X^{2}$ agree almost surely with the indicator of the measurable event ${X = 1}$ . By [L6], $E [X] = E [X^{2}] = P (X = 1) = p$ .

2.1step 1.1L3algebra

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

3.1step 1.1step 2.1L1L2L3L4L5L6algebra

For distinct $i, j$ , the independent Bernoulli variables in [L1] and [L5] have $X_{i} X_{j} = 1_{{X_{i} = X_{j} = 1}}$ almost surely, so [L6] and independence give $E [X_{i} X_{j}] = P (X_{i} = X_{j} = 1) = p^{2}$ . Their means are $p$ by step 1.1, so [L3] gives $Cov (X_{i}, X_{j}) = p^{2} - p^{2} = 0$ . Linearity gives $E [S] = \sum_{i < n} p = n p$ , and [L4] with these zero covariances gives $Var (S) = \sum_{i < n} p (1 - p) = n p (1 - p)$ .

4.1step 3.1∎

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

Depends on

Used by

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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)