5.2.3 The Poisson distribution

Counting rare events

Many counts have no fixed number of trials. Emails arriving in an hour, flaws in a metre of cable, goals in a football match: each can be 0, 1, 2 or more.

The Poisson distribution fits such counts when three things hold. Events come one at a time, independently of each other, at a steady average rate. The rate $\lambda$ (lambda) is the average number of events per interval, a count per hour or per metre.

The Poisson pmf

Rule: Poisson distribution

\begin{gathered} P(X = k) = e^{-\lambda}\,\frac{\lambda^k}{k!} \\[4pt] k = 0, 1, 2, \ldots \end{gathered}

The number of events in an interval where they occur at average rate $\lambda > 0$ . We write $X \sim \operatorname{Poisson}(\lambda)$ . Recall $0! = 1$ (lesson 5.1.2), so $P(X = 0) = e^{-\lambda}$ .

For $\lambda = 2$ , to four decimals:

$k$	0	1	2	3	4	5	6	7
$P(X = k)$	0.1353	0.2707	0.2707	0.1804	0.0902	0.0361	0.0120	0.0034

The distribution function of lesson 5.2.1 adds these up: $F(3) = 0.8571$ , so $P(X \ge 4) = 1 - F(3) = 0.1429$ . The values never reach 0, but they shrink fast: $k = 8$ has probability 0.0009.

Why it adds up to 1

The events $\{X = 0\}, \{X = 1\}, \{X = 2\}, \ldots$ never overlap, and the list never ends. Together they cover every outcome. So by the third axiom for an endless list of disjoint events (lesson 5.1.1), their probabilities add up to $P(\Omega) = 1$ .

By lesson 2.7.2, $1 + \lambda + \frac{\lambda^2}{2!} + \frac{\lambda^3}{3!} + \cdots = e^\lambda$ for every $\lambda$ . Multiply by $e^{-\lambda}$ , and they do:

\sum_{k=0}^{\infty} e^{-\lambda}\frac{\lambda^k}{k!} = e^{-\lambda}e^{\lambda} = 1

For $\lambda = 2$ , the partial sums of $\sum \frac{2^k}{k!}$ are 5, 7.2667 and 7.389 after the terms up to $k = 2$ , 5 and 10. They settle on $e^2 = 7.3891$ .

F(3)

import math from scipy.stats import poisson lam = 2 print([round(math.exp(-lam) * lam**k / math.factorial(k), 4) for k in range(9)]) print([round(float(poisson.pmf(k, lam)), 4) for k in range(9)]) print(round(poisson.cdf(3, lam), 4), round(poisson.sf(3, lam), 4)) print(round(sum(math.exp(-lam) * lam**k / math.factorial(k) for k in range(30)), 10)) print([round(sum(lam**k / math.factorial(k) for k in range(n + 1)), 4) for n in [2, 5, 10]], round(math.exp(lam), 4))

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The shape

Each probability is the one before times $\frac{\lambda}{k}$ :

\frac{P(X = k)}{P(X = k - 1)} = \frac{\lambda}{k}

While $k < \lambda$ the ratio is above 1 and the bars rise; once $k > \lambda$ they fall. So the most likely count is the whole-number part of $\lambda$ . When $\lambda$ is a whole number, $\lambda - 1$ and $\lambda$ tie.

For $\lambda = 2$ the ratios are 2, 1, $\frac{2}{3}$ , …, so $P(X = 1) = P(X = 2) = 0.2707$ . For $\lambda = 2.5$ the peak is at 2, with 0.2565. For $\lambda = 0.5$ the first ratio is already $\frac{1}{2}$ , so the bars fall from the start: $P(X = 0) = 0.6065$ .

\frac{2}{k}

import numpy as np from scipy.stats import poisson print([round(2 / k, 4) for k in range(1, 6)]) for lam in [0.5, 2, 2.5]: pmf = poisson.pmf(np.arange(12), lam) print(lam, [round(float(v), 4) for v in pmf[:5]], int(pmf.argmax()), round(float(pmf.max()), 4))

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$\lambda$ is the long-run average count, and lesson 5.3.1 makes "average" precise; below, the dashed line sits at $\lambda$ .

Set $\lambda = 2$ and the interval from 0 to 3, and read $P(X \le 3) = 0.8571$ . Then move $\lambda$ through 2.5, 3 and 3.5 and watch the tallest bar.

Explore

Each bar is the probability of one value. Change the parameters, then set an interval: the shaded bars add up to its probability.

The dashed line is the mean.

λ, rate = 3

Interval from a = 0

to b = 3

Probability: P(0 ≤ X ≤ 3) = 0.6472
Mean: 3
Variance: 3, so the standard deviation is 1.7321

Changing the interval

The rate belongs to an interval length. Events at rate $\lambda$ per hour come at rate $\lambda t$ per $t$ hours, so the count over $t$ hours is Poisson with parameter $\lambda t$ .

Rule: Counts over a time t

\begin{gathered} X_t \sim \operatorname{Poisson}(\lambda t) \\[4pt] P(\text{no event by } t) = e^{-\lambda t} \end{gathered}

$X_t$ is the number of events in a stretch of length $t$ , with $\lambda$ per unit of length. The rate and the length must use the same unit. The second line is $P(X_t = 0)$ .

Emails arrive at 3 an hour on average. Over 20 minutes, $t = \frac{1}{3}$ hour and $\lambda t = 1$ , so no email at all has probability $e^{-1} = 0.3679$ . Over a whole hour that falls to $e^{-3} = 0.0498$ ; over two hours, $\lambda t = 6$ and exactly six emails has probability 0.1606.

$e^{-\lambda t}$ is also the chance that the first event comes after $t$ . Lesson 5.4.2 turns it into the exponential distribution of waiting times, and case study 5.2.5 counts defaults with the Poisson.

Try it One hour is the unit, so 20 minutes is t = 1/3 . The cell prints the three probabilities above, then five or more emails in one hour: 0.1847.

import math from scipy.stats import poisson rate = 3 print(round(math.exp(-rate * (1 / 3)), 4), round(math.exp(-rate * 1), 4)) print(round(poisson.pmf(6, rate * 2), 4)) print(round(poisson.sf(4, rate * 1), 4))

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The Poisson as a limit of the binomial

A Poisson count is like a binomial count with very many trials, each with a tiny chance. Split an hour into $n$ short slots, each holding an email with probability $p = \frac{\lambda}{n}$ . The number of emails is $\operatorname{Bin}\left(n, \frac{\lambda}{n}\right)$ , and the average count $np = \lambda$ stays fixed as $n$ grows.

Here is $P(X = 2)$ for $\lambda = 2$ as the slots get shorter.

$n$	10	100	1,000	10,000	Poisson
$p = \frac{2}{n}$	0.2	0.02	0.002	0.0002
$P(X = 2)$	0.3020	0.2734	0.2709	0.2707	0.2707

Algebra shows where $e^{-\lambda}$ comes from. Write the binomial pmf with $p = \frac{\lambda}{n}$ and split it into four factors:

\begin{gathered} \binom{n}{k}\left(\frac{\lambda}{n}\right)^k\left(1 - \frac{\lambda}{n}\right)^{n - k} \\[4pt] = \frac{n(n - 1)\cdots(n - k + 1)}{n^k} \cdot \frac{\lambda^k}{k!} \\[4pt] \cdot \left(1 - \frac{\lambda}{n}\right)^n \cdot \left(1 - \frac{\lambda}{n}\right)^{-k} \end{gathered}

The first factor has $k$ terms on top, each close to $n$ when $n$ is large, so it tends to 1. The third tends to $e^{-\lambda}$ , the limit of case study 2.1.5 with the negative rate $-\lambda$ . The last tends to 1, since $k$ stays fixed.

$n$	first factor	$\left(1 - \frac{2}{n}\right)^n$	$\left(1 - \frac{2}{n}\right)^{-2}$	$P(X = 2)$
10	0.9	0.1074	1.5625	0.3020
100	0.99	0.1326	1.0412	0.2734
1,000	0.999	0.1351	1.004	0.2709

With $\lambda = 2$ and $k = 2$ , the middle factor is $\frac{2^2}{2!} = 2$ . The third column approaches $e^{-2} = 0.1353$ , and the last column approaches $2 \times 0.1353 = 0.2707$ .

n

import math from scipy.stats import binom for n in [10, 100, 1000]: first, third, last = n * (n - 1) / n**2, (1 - 2 / n)**n, (1 - 2 / n)**-2 print(n, round(first, 4), round(third, 4), round(last, 4), round(first * 2 * third * last, 4), round(binom.pmf(2, n, 2 / n), 4)) print(round(math.exp(-2), 4))

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Rule: Poisson approximation

\operatorname{Bin}(n, p) \approx \operatorname{Poisson}(np)

For many trials with a small chance each. It replaces two numbers, $n$ and $p$ , by one, $\lambda = np$ .

Binomial counts with $np = 2$ against the Poisson with $\lambda = 2$ . With $p = 0.2$ the two differ by up to 0.0313 at one value; with $p = 0.02$ , by at most 0.0027.

When the approximation is good

What matters is that $p$ is small, not only that $n$ is large. With $\lambda = 2$ fixed, the largest gap between the two pmfs is 0.0313 for $p = 0.2$ , 0.0027 for $p = 0.02$ and 0.0003 for $p = 0.002$ . Each tenfold drop in $p$ cuts it about ten times.

With $n = 20$ and $p = 0.5$ the approximation fails: the largest gap is 0.0511. The binomial gives $P(X = 10) = 0.1762$ , the Poisson with $\lambda = 10$ only 0.1251. The Poisson also puts probability 0.0016 on more than 20 events, which 20 trials cannot produce.

P(X = 2)

import numpy as np from scipy.stats import binom, poisson for n in [10, 100, 1000, 10000]: print(n, round(binom.pmf(2, n, 2 / n), 4)) print(round(poisson.pmf(2, 2), 4)) for n, p in [(10, 0.2), (100, 0.02), (1000, 0.002), (20, 0.5)]: k = np.arange(n + 1) print(n, p, round(np.abs(binom.pmf(k, n, p) - poisson.pmf(k, n * p)).max(), 4)) print(round(binom.pmf(10, 20, 0.5), 4), round(poisson.pmf(10, 10), 4), round(poisson.sf(20, 10), 4)) rng = np.random.default_rng(56) X = rng.binomial(1000, 0.002, size=200_000) print([round(float((X == k).mean()), 4) for k in range(5)]) print([round(float(poisson.pmf(k, 2)), 4) for k in range(5)])

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Common mistakes

Match $\lambda$ to the interval. Three an hour is one per 20 minutes: use $\lambda t$ , with $t$ in the rate's unit.
Keep both $e^{-\lambda}$ and $k!$ . Without them the values do not add up to 1.
The Poisson has no upper limit; the binomial stops at $n$ . That is one reason the approximation fails when $p$ is not small.
Check the conditions. Events that come in bursts, such as calls to a help desk during an outage, are not independent. The Poisson then understates large counts.
A whole-number $\lambda$ has two most likely counts. For $\lambda = 2$ , both 1 and 2.
poisson.sf(k, lam) is $P(X > k)$ . For $P(X \ge k)$ use sf(k - 1, lam).

The Poisson distribution

0 of 12 correct

1.
$X \sim \operatorname{Poisson}(2)$ . Find $P(X = 0)$ (type e for $e$ )
2.
$X \sim \operatorname{Poisson}(2)$ . Find $P(X = 3)$ exactly
3.
$X \sim \operatorname{Poisson}(1)$ . Find $P(X \le 1)$ exactly
4.
$X \sim \operatorname{Poisson}(3)$ . Find $P(X \ge 1)$ exactly
5.
What is $\sum_{k=0}^{\infty} \frac{3^k}{k!}$ ?
6.
$X \sim \operatorname{Poisson}(2)$ . Find $\frac{P(X = 5)}{P(X = 4)}$
7.
What is the most likely value of a Poisson count with $\lambda = 4.6$ ?
8.
Calls come at 6 a day on average. What is $\lambda$ for a 4-hour shift?
9.
Emails arrive at 3 an hour. Find the probability of none in 20 minutes (type e for $e$ )
10.
Approximate $\operatorname{Bin}(200, 0.01)$ by a Poisson. What is $\lambda$ ?
11.
Use the Poisson approximation for $P(X = 0)$ when $X \sim \operatorname{Bin}(500, 0.004)$ (type e for $e$ )
12.
poisson.sf(2, 4) gives $P(X \ge k)$ for which $k$ ?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

P(X = k) = e^{-\lambda}\frac{\lambda^k}{k!}

The Poisson distribution

Counting rare events

The Poisson pmf

Rule: Poisson distribution

Why it adds up to 1

The shape

Changing the interval

Rule: Counts over a time t

The Poisson as a limit of the binomial

Rule: Poisson approximation

When the approximation is good

Common mistakes

The Poisson distribution

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