Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate discrete distributions ▷ Poisson distribution ▷ Mean

Theorem: Let $X$ be a random variable following a Poisson distribution:

\[\label{eq:poiss} X \sim \mathrm{Poiss}(\lambda) \; .\]

Then, the mean or expected value of $X$ is

\[\label{eq:poiss-mean} \mathrm{E}(X) = \lambda \; .\]

Proof: The expected value of a discrete random variable is defined as

\[\label{eq:mean} \mathrm{E}(X) = \sum_{x \in \mathcal{X}} x \cdot f_X(x) \; ,\]

such that, with the probability mass function of the Poisson distribution, we have:

\[\label{eq:poiss-mean-s1} \begin{split} \mathrm{E}(X) &= \sum_{x=0}^\infty x \cdot \frac{\lambda^x \, e^{-\lambda}}{x!} \\ &= \sum_{x=1}^\infty x \cdot \frac{\lambda^x \, e^{-\lambda}}{x!} \\ &= e^{-\lambda} \cdot \sum_{x=1}^\infty \frac{x}{x!} \lambda^x \\ &= \lambda e^{-\lambda} \cdot \sum_{x=1}^\infty \frac{\lambda^{x-1}}{(x-1)!} \; . \end{split}\]

Substituting $z = x-1$, such that $x = z+1$, we get:

\[\label{eq:poiss-mean-s2} \mathrm{E}(X) = \lambda e^{-\lambda} \cdot \sum_{z=0}^\infty \frac{\lambda^z}{z!} \; .\]

Using the power series expansion of the exponential function

\[\label{eq:exp-ps} e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \; ,\]

the expected value of $X$ finally becomes

\[\label{eq:poiss-mean-s3} \begin{split} \mathrm{E}(X) &= \lambda e^{-\lambda} \cdot e^{\lambda} \\ &= \lambda \; . \end{split}\]
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Metadata: ID: P151 | shortcut: poiss-mean | author: JoramSoch | date: 2020-08-19, 06:09.