Mean of the geometric distribution

Statement

Let $p \in (0, 1]$ and let $X$ have the geometric distribution with parameter $p$ , so $P(X = k) = (1 - p)^k p$ for $k = 0, 1, 2, \dots$ . Then $E[X] = \frac{1 - p}{p}.$ For the number $T = X + 1$ of trials up to and including the first success, $E[T] = 1/p$ .

Remarks

A proof uses the tail sum formula: $P(X \ge n) = (1 - p)^n$ , so $E[X] = \sum_{n \ge 1} (1 - p)^n = (1 - p)/p$ . The variance of $X$ and of $T$ is $(1 - p)/p^2$ .

Tag	`0DB0R4N5XX`
URI	`m2://0DB0R4N5XX`
Kind	statement, proposition
Writing	W0 stub
MSC	60E05
Status	This copy of the site has no build data. Run `m2 status 0DB0R4N5XX`.
Uses	Geometric distribution, Expected value, Tail sum formula for expectation

Tag

0DB0R4N5XX

URI

m2://0DB0R4N5XX

Kind

statement, proposition

Writing

W0 stub

MSC

60E05

Status

This copy of the site has no build data. Run m2 status 0DB0R4N5XX.

Uses

Geometric distribution, Expected value, Tail sum formula for expectation

Mean of the geometric distribution

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