The Poisson distribution

Random events and the Poisson model Paper 6

Many counts in real life are counts of random events in a fixed interval: the number of emails arriving in an hour, the number of flaws in a metre of cloth, the number of accidents at a junction in a month. The Poisson distribution models such counts when

• the events occur singly (two cannot happen at exactly the same instant or place),
• the events occur independently of each other,
• the events occur at random, at a constant average rate $\lambda$ for the interval used.

If $X$ is the number of events in the interval, we write $X\sim \mathrm{Po}(\lambda )$, and
$$\mathrm{P}(X=r)={\mathrm{e}}^{-\lambda }\frac{{\lambda }^r}{r!},r=0,1,2,\dots$$
This formula is printed in MF19, and its derivation is not required. Unlike the binomial distribution there is no upper limit on $r$: in principle any number of events could occur, although large values become very unlikely.

Each condition matters in context. Customers arriving at a shop in groups break "singly"; breakdowns that make further breakdowns more likely break "independently"; a rate that is higher at lunchtime than at midnight breaks "constant rate" unless the interval is chosen so that the rate is steady. In an exam, an assumption must be stated in the context of the question, for example "the emails arrive independently of one another", not just "independent".

Calculating Poisson probabilities Paper 6

Work from the formula, term by term. The terms are linked: $\mathrm{P}(X=r+1)=\mathrm{P}(X=r)\times \frac{\lambda }{r+1}$, which is a quick way to build a list.

For a range of values, add terms or use the complement:

• $\mathrm{P}(X⩽2)={\mathrm{e}}^{-\lambda }(1+\lambda +\frac{{\lambda }^2}{2})$;
• $\mathrm{P}(X⩾3)=1-\mathrm{P}(X⩽2)$, because $X⩾3$ has no upper limit;
• $\mathrm{P}(X>3)=1-\mathrm{P}(X⩽3)$ and $\mathrm{P}(X<3)=\mathrm{P}(X⩽2)$: read "more than" and "fewer than" carefully, because $X$ only takes whole numbers.

Changing the interval. The rate scales with the length of the interval. If calls arrive at an average of 3.2 per 5 minutes, then in 10 minutes the number of calls is $\mathrm{Po}(6.4)$, and in 1 minute it is $\mathrm{Po}(0.64)$. This works because events in separate stretches of time are independent, so the counts add, and the mean of a sum is the sum of the means.

Mean and variance, and judging the model Paper 6

If $X\sim \mathrm{Po}(\lambda )$, then
$$\mathrm{E}(X)=\lambda \text{and}\mathrm{Var}(X)=\lambda \mathrm{.}$$
So the standard deviation is $\sqrt{\lambda }$. The mean and variance being equal is a fingerprint of the Poisson distribution, and it gives a quick test of whether a Poisson model is reasonable for some data: work out the mean and variance of the data. If they are roughly equal, a Poisson model is plausible; if the variance is much larger than the mean (events cluster) or much smaller (events are too regular), it is not.

For example, if 60 days of breakdowns give a mean of 2.1 per day but a variance of 4.6, the breakdowns are more spread out than a Poisson distribution allows, perhaps because one breakdown tends to cause others. The independence condition is in doubt, and $\mathrm{Po}(2.1)$ would be a poor model.

When a model is accepted, its expected frequencies are found by multiplying probabilities by the number of trials: out of 80 days, the expected number of days with no events is $80\mathrm{P}(X=0)$.

The distribution $\mathrm{Po}(2.5)$. It is skewed to the right, with its peak just below the mean 2.5. As $\lambda$ grows, the peak moves right and the shape becomes more symmetrical.

The Poisson approximation to the binomial Paper 6

Suppose $X\sim \mathrm{B}(n,p)$ with $n$ large and $p$ small: 2% of bulbs are faulty and a box holds 150. Faulty bulbs are then rare events among many trials, which is exactly the Poisson situation, and
$$X\text{is approximately}\mathrm{Po}(np)\mathrm{.}$$
The mean is kept the same, $\lambda =np$. The binomial variance $np(1-p)$ is close to $np$ when $p$ is small, so the variances agree too.

Conditions. You must know them: $n$ is large and $p$ is small, approximately
$$n>50\text{and}np<5.$$
State the check with numbers: "$n=150>50$ and $np=3<5$". If $np$ is larger than 5 and $nq$ is also larger than 5, the normal approximation from Paper 5 is the one to use instead. No continuity correction is needed here: both distributions are discrete.

The normal approximation to the Poisson Paper 6

When $\lambda$ is large, $\mathrm{Po}(\lambda )$ becomes close to symmetrical and bell-shaped, and adding many terms by hand becomes slow. Then
$$X\sim \mathrm{Po}(\lambda )\text{is approximately}\mathrm{N}(\lambda ,\lambda ),$$
keeping both the mean and the variance $\lambda$. The condition to know is that $\lambda$ is large: $\lambda >15$, approximately.

A Poisson variable takes whole numbers only, so a continuity correction is needed, exactly as for the binomial: each whole number $r$ stands for the interval from $r-0.5$ to $r+0.5$. Write the event with $⩽$ or $⩾$ first, then widen it by 0.5:

• $\mathrm{P}(X⩽20)\approx \mathrm{P}(Y<20.5)$;
• $\mathrm{P}(X<20)=\mathrm{P}(X⩽19)\approx \mathrm{P}(Y<19.5)$;
• $\mathrm{P}(X⩾30)\approx \mathrm{P}(Y>29.5)$;
• $\mathrm{P}(X>30)=\mathrm{P}(X⩾31)\approx \mathrm{P}(Y>30.5)$.

Remember to standardise with the standard deviation $\sqrt{\lambda }$, not with $\lambda$.