Mean of the Poisson distribution

Statement

Let $\lambda > 0$ and let $X$ have the Poisson distribution with parameter $\lambda$ . Then $E[X] = \sum_{k=0}^{\infty} k \, e^{-\lambda} \frac{\lambda^k}{k!} = \lambda.$

Remarks

Proof: the term with $k = 0$ is zero. For $k \ge 1$ , $k \lambda^k / k! = \lambda \cdot \lambda^{k-1}/(k-1)!$ . So the sum is $\lambda e^{-\lambda} \sum_{j \ge 0} \lambda^j / j! = \lambda e^{-\lambda} e^{\lambda} = \lambda$ . The variance of $X$ is also $\lambda$ . So the parameter $\lambda$ is the mean number of events.

Tag	`086HEVXR9Z`
URI	`m2://086HEVXR9Z`
Kind	statement, proposition
Writing	W0 stub
MSC	60E05
Status	This copy of the site has no build data. Run `m2 status 086HEVXR9Z`.
Uses	Poisson distribution, Expected value

Tag

086HEVXR9Z

URI

m2://086HEVXR9Z

Kind

statement, proposition

Writing

W0 stub

MSC

60E05

Status

This copy of the site has no build data. Run m2 status 086HEVXR9Z.

Uses

Poisson distribution, Expected value

Mean of the Poisson distribution

Statement

Remarks

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