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Derivation of the Mean and Variance of the Poisson Distribution

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The Formal Theorem

Let X X be a discrete random variable following a Poisson distribution with parameter λ>0 \lambda > 0 , denoted as X∼Poisson(λ) X \sim \text{Poisson}(\lambda) . Its probability mass function (PMF) is defined as:
P(X=k)=e−λλkk!,k=0,1,2,… P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad k = 0, 1, 2, \dots
The mean E[X] E[X] and variance Var(X) \text{Var}(X) are both equal to the rate parameter λ \lambda :
E[X]=λandVar(X)=λ E[X] = \lambda \quad \text{and} \quad \text{Var}(X) = \lambda

Analytical Intuition.

Imagine the Poisson distribution as the mathematical heartbeat of a continuous stream of independent events—raindrops striking a windowpane or photons hitting a sensor. The parameter λ \lambda acts as the 'intensity' of this phenomenon, representing the average frequency in a fixed interval. When we derive the mean, we are performing a weighted average over an infinite horizon of possible counts. This process relies on the Taylor series expansion of eλ e^\lambda , which serves as the invisible machinery bridging discrete counts to continuous rates. As we move to the variance, we explore the 'jitter' or volatility of this stream. By calculating the second factorial moment E[X(X−1)] E[X(X-1)] , we uncover a profound and rare symmetry: the variance perfectly mirrors the mean. In the cinematic lens of probability, this implies that in a state of pure randomness (where events occur at a constant average rate), the spread of our uncertainty grows in perfect lockstep with the expected signal. The higher the rate, the higher the fluctuation, yet both are governed by the singular, elegant force of λ \lambda .
CAUTION

Institutional Warning.

Students often stumble when shifting the summation index during the derivation. In calculating E[X] E[X] , the term for k=0 k=0 is zero, so the sum must begin at k=1 k=1 . Similarly, for the second moment, the sum must begin at k=2 k=2 to prevent invalid factorials.

Academic Inquiries.

01

Why do we use the factorial moment E[X(X−1)] E[X(X-1)] instead of calculating E[X2] E[X^2] directly?

The factorial moment E[X(X−1)] E[X(X-1)] is algebraically superior because the k(k−1) k(k-1) term cancels perfectly with the first two factors of k! k! in the denominator, simplifying the infinite series into a recognizable exponential form.

02

Is the equality of mean and variance unique to the Poisson distribution?

Yes, in the context of common discrete distributions, this 'equidispersion' property (E[X]=Var(X) E[X] = \text{Var}(X) ) is a defining characteristic of the Poisson distribution and is often used as a test for Poisson data.

03

How does the identity eλ=∑λkk! e^\lambda = \sum \frac{\lambda^k}{k!} fit into the proof?

This identity is the engine of the derivation. By factoring out λ \lambda or λ2 \lambda^2 from the summation, we transform the remaining sum into the power series for eλ e^\lambda , which then cancels with the e−λ e^{-\lambda} term in the PMF.

Standardized References.

  • Definitive Institutional SourceCasella, G., & Berger, R. L. (2002). Statistical Inference. Duxbury Press.

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Institutional Citation

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NICEFA Visual Mathematics. (2026). Derivation of the Mean and Variance of the Poisson Distribution: Visual Proof & Intuition. Retrieved from https://nicefa.org/library/applied-statistics/derivation-of-the-mean-and-variance-of-the-poisson-distribution

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