Expectation with respect to the binomial distribution

Statement

Let $n \ge 1$ be an integer, let $p \in [0, 1]$ , and let $X$ have the binomial distribution $\operatorname{Bin}(n, p)$ . Then $E[X] = \sum_{k=0}^{n} k \binom{n}{k} p^k (1 - p)^{n - k} = np.$

Remarks

A short proof: $X$ has the same law as $Y_1 + \dots + Y_n$ with independent Bernoulli $Y_i$ with $E[Y_i] = p$ . Then linearity of expectation gives $np$ . A direct proof uses $k \binom{n}{k} = n \binom{n-1}{k-1}$ and the binomial theorem. The variance of $X$ is $np(1 - p)$ .

Tag	`0PVCTCKD3C`
URI	`m2://0PVCTCKD3C`
Aliases	`mean-of-binomial-distribution`
Kind	statement, proposition
Writing	W0 stub
MSC	60E05
Status	This copy of the site has no build data. Run `m2 status 0PVCTCKD3C`.
Uses	Binomial distribution, Expected value, Linearity of expectation, Bernoulli distribution

Tag

0PVCTCKD3C

URI

m2://0PVCTCKD3C

Aliases

mean-of-binomial-distribution

Kind

statement, proposition

Writing

W0 stub

MSC

60E05

Status

This copy of the site has no build data. Run m2 status 0PVCTCKD3C.

Uses

Binomial distribution, Expected value, Linearity of expectation, Bernoulli distribution

Expectation with respect to the binomial distribution

Statement

Remarks

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