Poisson Distribution

The Poisson distribution counts how many times something happens in a fixed stretch of time or space: emails arriving in an hour, flaws in a metre of cable, goals in a match, bacteria in a drop of water. It is the first topic on Paper 6 and the one every later topic leans on, because Poisson probabilities turn up again in sums of random variables, normal approximations and hypothesis tests. Almost every S2 paper has at least one full question built on it.

Counting events, not successes

In S1 the binomial distribution counted successes in a fixed number of trials. There was a definite $n$: ten throws, twenty patients. Many real counts have no such $n$. Nobody can say how many "trials" there are for an email to arrive in an hour. There is only a rate: on average, $4.5$ emails per hour.

Imagine slicing the hour into $3600$ seconds. In each second an email either arrives or it does not, with a tiny probability $p = 4.5/3600$. The count is then roughly $B(3600, 0.00125)$. Slice more finely, into milliseconds, and the count is roughly $B(3\,600\,000, 0.00000125)$. The mean $np = 4.5$ never changes. As the slices get thinner, these binomial distributions settle down to a single limiting distribution that depends only on the mean. That limit is the Poisson distribution.

This picture explains everything about the Poisson:

• it needs only one parameter, the mean number of events in the interval, written $\lambda$ (lambda);
• the count can be $0, 1, 2, 3, \dots$ with no upper limit, because there is no fixed $n$;
• it applies when events happen singly, independently and at random at a constant average rate, the same conditions that make the binomial slices valid.

The probability formula

The formula is in the formula booklet (MF19), so you do not need to memorise it, but you must be fluent with it. The first few terms are worth knowing by sight:

$r$	$0$	$1$	$2$	$3$	$4$
$P(X = r)$	$e^{-\lambda}$	$e^{-\lambda}\lambda$	$e^{-\lambda}\dfrac{\lambda^2}{2}$	$e^{-\lambda}\dfrac{\lambda^3}{6}$	$e^{-\lambda}\dfrac{\lambda^4}{24}$

The probabilities add up to $1$ because $e^{\lambda} = 1 + \lambda + \dfrac{\lambda^2}{2!} + \dfrac{\lambda^3}{3!} + \cdots$, so

$$\sum_{r=0}^{\infty} e^{-\lambda}\frac{\lambda^r}{r!} = e^{-\lambda}e^{\lambda} = 1.$$

What the distribution looks like

For small $\lambda$ the distribution is piled up near zero and strongly skewed to the right. As $\lambda$ grows, the peak moves right, spreads out and becomes more symmetrical. The two graphs below draw each probability $P(X = r)$ as a bar of width $1$ centred on $r$.

$X \sim \text{Po}(1.5)$:

$X \sim \text{Po}(6)$:

The second shape is already close to a bell curve. That observation is the basis of the normal approximation to the Poisson, used when $\lambda$ is large.

Mean and variance

The binomial picture makes this believable: $B(n, p)$ has variance $np(1 - p)$, and when $p$ is tiny, $1 - p \approx 1$, so the variance is almost exactly $np$, the mean.

The equality of mean and variance is the fingerprint of a Poisson distribution. If a set of data has a mean of $3.1$ and a variance of $3.0$, a Poisson model is plausible. If the variance is $9$, it is not. This is examined directly; see modelling with the Poisson distribution.

Calculating probabilities

The formula booklet for 9709 contains no cumulative Poisson tables. Every Poisson probability on Paper 6 is calculated from the formula, so questions are designed so that you only need a handful of terms. Your job is to choose the shortest route.

Changing the interval

The parameter $\lambda$ is the mean for the interval in the question, not necessarily the interval in which the rate was given. Because events occur at a constant average rate, the mean scales in proportion to the length of the interval.

This is the most common source of lost marks in Poisson questions. Always ask, before writing a single probability: what interval is this question about, and what is the mean for that interval?

Working backwards to find $\lambda$

Some questions give a probability and ask for the mean. Two patterns cover nearly all of them.

From $P(X = 0)$. Since $P(X = 0) = e^{-\lambda}$, taking logarithms gives $\lambda = -\ln P(X = 0)$. This is the only Poisson probability you can invert exactly, which is why examiners use it.

From two neighbouring probabilities. If $P(X = r) = P(X = r + 1)$ then, by the recurrence, $\dfrac{\lambda}{r + 1} = 1$, so $\lambda = r + 1$. More generally, write both probabilities with the formula and cancel $e^{-\lambda}$ and common powers.

Combining the Poisson with other ideas

Exam-hard Poisson questions rarely stop at a single probability. Three combinations come up again and again.

Repeated intervals. If $p$ is the probability of some event in one hour, and hours are independent, then the number of hours (out of $n$) in which it happens is $B(n, p)$. You find $p$ with the Poisson, then switch to the binomial.

Conditional probability. "Given that at least one car passes, find the probability that exactly two pass" is $P(X = 2 \mid X \ge 1) = \dfrac{P(X = 2)}{P(X \ge 1)}$, because $X = 2$ is already inside $X \ge 1$.

Unknown interval length. "Find the shortest time for which the probability of at least one event exceeds $0.99$" leads to $1 - e^{-mt} > 0.99$, which you solve with logarithms.

Practice questions

Answers

1. $X \sim \text{Po}(3.2)$. (a) $P(X = 4) = e^{-3.2}\dfrac{3.2^4}{24} = 0.178$. (b) $P(X < 3) = P(X \le 2) = e^{-3.2}\left(1 + 3.2 + \dfrac{3.2^2}{2}\right) = e^{-3.2}(9.32) = 0.380$. (c) $P(X \ge 2) = 1 - e^{-3.2}(1 + 3.2) = 1 - 0.1712 = 0.829$.

2. (a) $X \sim \text{Po}(0.8)$: $P(X = 0) = e^{-0.8} = 0.449$. (b) For 5 pages the mean is $5 \times 0.8 = 4$, so $Y \sim \text{Po}(4)$. $P(Y > 2) = 1 - e^{-4}\left(1 + 4 + \dfrac{4^2}{2}\right) = 1 - 13e^{-4} = 1 - 0.2381 = 0.762$.

3. $\text{Var}(X) = E(X^2) - [E(X)]^2$, so $\lambda = 12 - \lambda^2$, giving $\lambda^2 + \lambda - 12 = 0$, $(\lambda + 4)(\lambda - 3) = 0$. Since $\lambda > 0$, $\lambda = 3$. $P(X = 3) = e^{-3}\dfrac{27}{6} = 0.224$.

4. $e^{-\lambda} = 0.15 \Rightarrow \lambda = -\ln 0.15 = 1.897$ (4 s.f.). $P(X > 2) = 1 - e^{-1.897}\left(1 + 1.897 + \dfrac{1.897^2}{2}\right) = 1 - 0.15(4.6966) = 1 - 0.7045 = 0.296$ (using $e^{-\lambda} = 0.15$ exactly).

5. Let $X \sim \text{Po}(2.6)$. $P(X \ge 2) = 1 - e^{-2.6}(1 + 2.6) = 1 - 0.26739 = 0.73262$. The periods are independent, so the probability for all three is $0.73262^3 = 0.393$.

6. $e^{-\mu}\dfrac{\mu^2}{2} = 3e^{-\mu}\mu$. Cancel $e^{-\mu}\mu$: $\dfrac{\mu}{2} = 3$, so $\mu = 6$. $P(Y = 3) = e^{-6}\dfrac{216}{6} = 36e^{-6} = 0.0892$.

7. Weekly demand $X \sim \text{Po}(7.5)$. With stock $n$ we need $P(X > n) < 0.05$, that is $P(X \le n) > 0.95$. Building the cumulative sum term by term: $P(X \le 11) = 0.9208$, so $P(X > 11) = 0.0792$ (too large); $P(X \le 12) = 0.9573$, so $P(X > 12) = 0.0427 < 0.05$. The smallest stock is $12$ cameras. (Showing both $P(X \le 11)$ and $P(X \le 12)$ is essential: the examiner needs to see that $12$ is the smallest value that works.)

8. (a) For area $A$, the number of flaws is $\text{Po}(0.4A)$. $e^{-0.4A} = 0.75 \Rightarrow A = \dfrac{-\ln 0.75}{0.4} = 0.719\ \text{m}^2$. (b) For $5\ \text{m}^2$, $X \sim \text{Po}(2)$. $P(X = 1 \mid X \ge 1) = \dfrac{P(X = 1)}{1 - P(X = 0)} = \dfrac{2e^{-2}}{1 - e^{-2}} = \dfrac{0.27067}{0.86466} = 0.313$.