Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate discrete distributions ▷ Poisson distribution ▷ Variance

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

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

Then, the variance of $X$ is

$\label{eq:poiss-var} \mathrm{Var}(X) = \lambda \; .$

Proof: The variance can be expressed in terms of expected values as

$\label{eq:var-mean} \mathrm{Var}(X) = \mathrm{E}(X^2) - \mathrm{E}(X)^2 \; .$ $\label{eq:poiss-mean} \mathrm{E}(X) = \lambda \; .$

Let us now consider the expectation of $X \, (X-1)$ which is defined as

$\label{eq:mean} \mathrm{E}[X \, (X-1)] = \sum_{x \in \mathcal{X}} x \, (x-1) \cdot f_X(x) \; ,$

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

$\label{eq:poiss-x2x-mean-s1} \begin{split} \mathrm{E}[X \, (X-1)] &= \sum_{x=0}^\infty x \, (x-1) \cdot \frac{\lambda^x \, e^{-\lambda}}{x!} \\ &= \sum_{x=2}^\infty x \, (x-1) \cdot \frac{\lambda^x \, e^{-\lambda}}{x!} \\ &= e^{-\lambda} \cdot \sum_{x=2}^\infty x \, (x-1) \cdot \frac{\lambda^x}{x \cdot (x-1) \cdot (x-2)!} \\ &= \lambda^2 \cdot e^{-\lambda} \cdot \sum_{x=2}^\infty \frac{\lambda^{x-2}}{(x-2)!} \; . \end{split}$

Substituting $z = x-2$, such that $x = z+2$, we get:

$\label{eq:poiss-x2x-mean-s2} \mathrm{E}[X \, (X-1)] = \lambda^2 \cdot e^{-\lambda} \cdot \sum_{z=0}^\infty \frac{\lambda^z}{z!} \; .$

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 $X \, (X-1)$ finally becomes

$\label{eq:poiss-x2x-mean-s3} \mathrm{E}[X \, (X-1)] = \lambda^2 \cdot e^{-\lambda} \cdot e^{\lambda} = \lambda^2 \; .$

Note that this expectation can be written as

$\label{eq:poiss-x2-mean-s1} \mathrm{E}[X \, (X-1)] = \mathrm{E}(X^2 - X) = \mathrm{E}(X^2) - \mathrm{E}(X) \; ,$

such that, with \eqref{eq:poiss-x2x-mean-s3} and \eqref{eq:poiss-mean}, we have:

$\label{eq:poiss-x2-mean-s2} \mathrm{E}(X^2) - \mathrm{E}(X) = \lambda^2 \quad \Rightarrow \quad \mathrm{E}(X^2) = \lambda^2 + \lambda \; .$

Plugging \eqref{eq:poiss-x2-mean-s2} and \eqref{eq:poiss-mean} into \eqref{eq:var-mean}, the variance of a Poisson random variable finally becomes

$\label{eq:poiss-var-qed} \mathrm{Var}(X) = \lambda^2 + \lambda - \lambda^2 = \lambda \; .$
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Metadata: ID: P230 | shortcut: poiss-var | author: JoramSoch | date: 2021-04-29, 09:59.