Intermediate26 min

The Poisson Distribution

Rare events, derivation as a binomial limit, the Poisson process, and additivity.

Assumes you know

Bernoulli and Binomial Distributions

The Poisson Distribution

Intuition first

The Poisson counts how many times a rare event happens in a fixed window — emails per hour, defects per metre of cable, earthquakes per decade, server errors per minute.

It arises from the binomial in a specific limit. Imagine chopping an hour into millions of tiny instants, each with a minuscule chance of an email arriving. You have a binomial with enormous $n$ and tiny $p$ , and only the product $\lambda = np$ matters — the individual $n$ and $p$ become unidentifiable. The Poisson is what that limit converges to.

Its signature is that mean and variance are equal, both $\lambda$ . That gives you an immediate diagnostic: compute the sample mean and sample variance of your counts, and if the variance is much larger, the events are clustering and a Poisson model will understate your uncertainty.

Definition

X \sim \text{Poisson}(\lambda), \qquad p(k) = \frac{e^{-\lambda}\lambda^k}{k!}, \qquad k = 0, 1, 2, \dots

\E[X] = \lambda, \qquad \Var(X) = \lambda, \qquad M_X(t) = \exp\!\big(\lambda(e^t - 1)\big)

Notation used in this lesson
Symbol	Meaning	Read aloud
λ	Mean number of events in the window — the rate times the window length	lambda
k	Observed count, any non-negative integer	k
e	Euler's number, ≈ 2.71828	e

Derivation as a binomial limit

From binomial to PoissonAdvanced

Take $X \sim \text{Binomial}(n, p)$ and let $n \to \infty$ with $p = \lambda/n$ held so the mean $np = \lambda$ stays fixed.

\Prob(X = k) = \binom{n}{k}\left(\frac{\lambda}{n}\right)^k\left(1 - \frac{\lambda}{n}\right)^{n-k}

Expand the binomial coefficient and separate the factors:

= \frac{n(n-1)\cdots(n-k+1)}{k!}\cdot\frac{\lambda^k}{n^k}\cdot\left(1-\frac{\lambda}{n}\right)^{n}\left(1-\frac{\lambda}{n}\right)^{-k}

= \frac{\lambda^k}{k!} \cdot \underbrace{\frac{n(n-1)\cdots(n-k+1)}{n^k}}_{\to\, 1} \cdot \underbrace{\left(1-\frac{\lambda}{n}\right)^{n}}_{\to\, e^{-\lambda}} \cdot \underbrace{\left(1-\frac{\lambda}{n}\right)^{-k}}_{\to\, 1}

Each limit in turn:

The first fraction is a product of $k$ terms each tending to 1, since $k$ is fixed while $n \to \infty$ .
The second is the standard limit $\lim_{n\to\infty}(1 - x/n)^n = e^{-x}$ .
The third has a fixed exponent $-k$ and a base tending to 1.

Therefore

\Prob(X = k) \;\longrightarrow\; \frac{e^{-\lambda}\lambda^k}{k!}

The practical rule: use Poisson to approximate a binomial when $n \geq 20$ and $p \leq 0.05$ , or whenever $n \geq 100$ and $np \leq 10$ .

Mean and variance from the MGFAdvanced

M_X(t) = \exp\!\big(\lambda(e^t - 1)\big)

First derivative, by the chain rule:

M_X'(t) = \lambda e^t \exp\!\big(\lambda(e^t-1)\big) \quad\Longrightarrow\quad \E[X] = M_X'(0) = \lambda e^0 e^0 = \lambda

Second derivative, product rule on $\lambda e^t$ and the exponential:

M_X''(t) = \lambda e^t\exp\!\big(\lambda(e^t-1)\big) + \big(\lambda e^t\big)^2\exp\!\big(\lambda(e^t-1)\big)

\E[X^2] = M_X''(0) = \lambda + \lambda^2

Hence

\Var(X) = \E[X^2] - (\E[X])^2 = \lambda + \lambda^2 - \lambda^2 = \lambda

The MGF also gives the additivity property immediately. For independent $X \sim \text{Poisson}(\lambda_1)$ and $Y \sim \text{Poisson}(\lambda_2)$ :

M_{X+Y}(t) = \exp\!\big(\lambda_1(e^t-1)\big)\exp\!\big(\lambda_2(e^t-1)\big) = \exp\!\big((\lambda_1+\lambda_2)(e^t-1)\big)

which is the Poisson MGF with parameter $\lambda_1 + \lambda_2$ . So $X + Y \sim \text{Poisson}(\lambda_1 + \lambda_2)$ .

Scaling the window

Rates are per unit time or space, so the parameter scales with the window:

\lambda_{\text{window}} = \text{rate} \times \text{window length}

Twelve calls per hour means $\lambda = 12$ for an hour, $\lambda = 2$ for ten minutes, and $\lambda = 288$ for a day. Getting this scaling wrong is the most common error in Poisson problems.

Solved problem 1 · Calls at a help desk

A help desk receives calls at an average rate of 3 per 10 minutes. Find the probability of (a) exactly 2 calls in 10 minutes, (b) no calls in 5 minutes, (c) at least 3 calls in 20 minutes.

Step 1 — part (a): the window matches the stated rate

$\lambda = 3$ for a 10-minute window.

\Prob(X = 2) = \frac{e^{-3}\,3^2}{2!} = \frac{e^{-3} \times 9}{2}

e^{-3} = 0.049787

\Prob(X = 2) = \frac{0.049787 \times 9}{2} = \frac{0.448083}{2} = 0.224042

Step 2 — part (b): rescale to 5 minutes

Half the window, so half the rate:

\lambda = 3 \times \frac{5}{10} = 1.5

\Prob(X = 0) = \frac{e^{-1.5}(1.5)^0}{0!} = e^{-1.5} = 0.223130

Step 3 — part (c): rescale to 20 minutes

Double the window:

\lambda = 3 \times \frac{20}{10} = 6

Step 4 — compute the complement

\Prob(X \geq 3) = 1 - \Prob(X \leq 2) = 1 - \big[p(0) + p(1) + p(2)\big]

e^{-6} = 0.00247875

p(0) = e^{-6} = 0.00247875

p(1) = e^{-6} \times 6 = 0.01487251

p(2) = \frac{e^{-6} \times 36}{2} = e^{-6} \times 18 = 0.04461753

\Prob(X \leq 2) = 0.00247875 + 0.01487251 + 0.04461753 = 0.06196879

\Prob(X \geq 3) = 1 - 0.06196879 = 0.93803

Step 5 — sanity check

With $\lambda = 6$ expected calls in 20 minutes, observing at least 3 should be very likely, and $0.938$ is consistent ✓.

Answer

(a) $\approx 0.2240$ ; (b) $\approx 0.2231$ ; (c) $\approx 0.9380$ .

Solved problem 2 · Detecting overdispersion

A website records daily error counts over 30 days with sample mean $\bar{x} = 4.2$ and sample variance $s^2 = 18.6$ . Is a Poisson model appropriate?

Step 1 — the Poisson requirement

Poisson demands $\E[X] = \Var(X)$ , so the sample mean and variance should be close.

Step 2 — compute the dispersion index

\text{dispersion} = \frac{s^2}{\bar{x}} = \frac{18.6}{4.2} \approx 4.43

Under a true Poisson this ratio should be near 1.

Step 3 — test it formally

The dispersion statistic

D = \frac{(n-1)s^2}{\bar{x}} = \frac{29 \times 18.6}{4.2} = \frac{539.4}{4.2} \approx 128.4

is approximately $\chi^2$ with $n - 1 = 29$ degrees of freedom under the Poisson hypothesis. The critical value at the 1% level is about $49.6$ , and $128.4$ vastly exceeds it.

Step 4 — conclude and diagnose

Strongly overdispersed. The data varies about 4.4 times more than Poisson allows.

Plausible causes: errors arrive in bursts, because one failure triggers cascading retries; the underlying rate varies by day of week or with traffic; or the days are not independent because an unresolved fault persists.

Step 5 — what to use instead

A negative binomial model, which is Poisson with a gamma-distributed rate and therefore has variance exceeding its mean by a fitted amount. Or a Poisson model with covariates capturing the systematic variation — day of week, traffic volume — which may reduce the residual dispersion to near 1.

Answer

No. Dispersion index $\approx 4.4$ against a required 1, and a dispersion test rejecting overwhelmingly. Fitting Poisson anyway would produce confidence intervals roughly $\sqrt{4.4} \approx 2.1$ times too narrow.

The Poisson process

If events occur at constant rate $\lambda$ independently, then:

Counts in a window of length $t$ are $\text{Poisson}(\lambda t)$ .
Waiting times between consecutive events are $\text{Exponential}(\lambda)$ .
Counts in disjoint windows are independent.

That second point links this lesson to the next: the Poisson counts events, the exponential times the gaps, and they describe the same process from two angles.

python

import numpy as np
from scipy import stats

X = stats.poisson(mu=3)
print(f"(a) P(X=2), λ=3      {X.pmf(2):.6f}")
print(f"(b) P(X=0), λ=1.5    {stats.poisson(1.5).pmf(0):.6f}")
print(f"(c) P(X>=3), λ=6     {stats.poisson(6).sf(2):.6f}   <- sf(2), not sf(3)")

# Poisson as a binomial limit.
lam = 3
print("\nbinomial → Poisson as n grows with np = 3:")
for n in (10, 100, 1000, 100_000):
    b = stats.binom(n=n, p=lam/n).pmf(2)
    print(f"  n={n:<7} P(X=2) = {b:.6f}   Poisson {stats.poisson(lam).pmf(2):.6f}")

# Dispersion check on simulated data.
rng = np.random.default_rng(0)
pure = rng.poisson(4.2, 30)
bursty = rng.negative_binomial(1.2, 1.2/(1.2+4.2), 30)   # same mean, more variance
for name, d in (("true Poisson", pure), ("overdispersed", bursty)):
    print(f"\n{name}: mean {d.mean():.2f}  var {d.var(ddof=1):.2f}  "
          f"dispersion {d.var(ddof=1)/d.mean():.2f}")

Exercise 1

A book has on average 0.5 typos per page. What is the probability a 10-page chapter contains no typos?

Show solution

Scale the rate to the window:

\lambda = 0.5 \times 10 = 5

\Prob(X = 0) = \frac{e^{-5} \cdot 5^0}{0!} = e^{-5} \approx 0.006738

About 0.67% — very unlikely. A common error is using $\lambda = 0.5$ and reporting $e^{-0.5} \approx 0.607$ , which answers a different question: the probability a single page is clean.

Worth noting the consistency: the probability all 10 pages are independently clean is $(e^{-0.5})^{10} = e^{-5}$ , the same answer, because the Poisson's additivity over disjoint windows is exactly this multiplication.

Exercise 2

A factory has 2,000 components, each failing independently with probability 0.0015 per day. Approximate the probability that more than 4 fail on a given day, and justify the approximation.

Show solution

Exactly, $X \sim \text{Binomial}(2000, 0.0015)$ . Check the Poisson approximation conditions: $n = 2000 \geq 100$ and $np = 3 \leq 10$ ✓ — large $n$ , small $p$ , moderate product.

So $X \approx \text{Poisson}(\lambda = 3)$ .

\Prob(X > 4) = 1 - \Prob(X \leq 4) = 1 - \sum_{k=0}^{4}\frac{e^{-3}3^k}{k!}

With $e^{-3} = 0.049787$ :

p(0) = 0.049787, \quad p(1) = 0.149361, \quad p(2) = 0.224042

p(3) = \frac{0.049787 \times 27}{6} = 0.224042, \quad p(4) = \frac{0.049787 \times 81}{24} = 0.168031

\Prob(X \leq 4) = 0.049787 + 0.149361 + 0.224042 + 0.224042 + 0.168031 = 0.815263

\Prob(X > 4) \approx 1 - 0.815263 = 0.1847

About 18.5%. The exact binomial value is $0.1846$ , so the approximation is accurate to four decimal places — and far easier, since the exact calculation involves terms like $\binom{2000}{4}$ .

Note $p(2) = p(3)$ here, which happens whenever $\lambda$ is an integer: the Poisson is bimodal at $\lambda - 1$ and $\lambda$ .

Next: Uniform and Exponential Distributions, the first continuous families and the waiting-time partner of the Poisson.