Sums of Poisson Variables

A2 · S2 · 10 min

If calls arrive at random at one desk and, independently, at random at another, then the total number of calls is also a random count, and it is Poisson too, with the two means added. This one fact is the last of the linear-combination results on the syllabus. It appears in Paper 6 questions about two sources of events (two machines, two types of fault, cars and lorries), often combined with a rescaled interval, a normal approximation or a conditional probability.

The result

Picture two independent streams of random events, say emails from friends at $1.2$ per hour and work emails at $3.1$ per hour. Merge them into one inbox. The merged stream still has events arriving singly, independently and at random, now at the combined rate of $4.3$ per hour. So the merged count satisfies all the conditions of a Poisson model.

Key result

If $X \sim \text{Po}(\lambda)$ and $Y \sim \text{Po}(\mu)$ are independent, then

X + Y \sim \text{Po}(\lambda + \mu).

This extends to any number of independent Poisson variables: the total is Poisson with the sum of the means. Proofs are not required.

The parameters agree with the general rules: $E(X + Y) = \lambda + \mu$ and, by independence, $\text{Var}(X + Y) = \lambda + \mu$ . Mean equals variance, exactly as a Poisson distribution needs.

Where the result comes from

For the curious (not examinable): $P(X + Y = n) = \sum_{r=0}^{n} P(X = r)P(Y = n - r) = \sum_{r=0}^{n} e^{-\lambda}\dfrac{\lambda^r}{r!}\,e^{-\mu}\dfrac{\mu^{n-r}}{(n-r)!}$ . Taking out $\dfrac{e^{-(\lambda + \mu)}}{n!}$ leaves $\sum \binom{n}{r}\lambda^r\mu^{n-r} = (\lambda + \mu)^n$ by the binomial theorem, which gives exactly the $\text{Po}(\lambda + \mu)$ formula.

What does not stay Poisson

Only sums of independent Poisson variables are Poisson. Two other combinations look similar but are not.

Multiples. $2X$ can only take even values $0, 2, 4, \dots$ , so it cannot be Poisson. Its numbers confirm this: $E(2X) = 2\lambda$ but $\text{Var}(2X) = 4\lambda$ , and a Poisson variable must have mean equal to variance. Contrast this with $X_1 + X_2$ , two separate intervals, which is $\text{Po}(2\lambda)$ .

Differences. $X - Y$ can be negative, so it cannot be Poisson. Its mean is $\lambda - \mu$ and its variance is $\lambda + \mu$ ; you can still calculate these, but not Poisson probabilities.

Combination	Mean	Variance	Poisson?
$X + Y$ (independent)	$\lambda + \mu$	$\lambda + \mu$	Yes, $\text{Po}(\lambda + \mu)$
$X_1 + X_2$ (two separate intervals)	$2\lambda$	$2\lambda$	Yes, $\text{Po}(2\lambda)$
$2X$	$2\lambda$	$4\lambda$	No
$X - Y$	$\lambda - \mu$	$\lambda + \mu$	No

Totals of Poisson counts

Find the mean of each count for the interval in the question (rescale each rate first).
Check that the counts are independent; say so if asked to state an assumption.
Add the means: total $\sim \text{Po}(\lambda + \mu)$ .
Calculate the probability directly, or with a normal approximation (and continuity correction) if the total mean is greater than $15$ .

Two sources of calls

Calls to a company's sales line arrive at random at an average rate of $2.1$ per minute, and calls to its support line arrive independently at random at an average rate of $1.4$ per minute. Find the probability that the total number of calls in a given minute is at least $3$ .

Solution

Let $S \sim \text{Po}(2.1)$ and $T \sim \text{Po}(1.4)$ , independent. Then $S + T \sim \text{Po}(3.5)$ .

P(S + T \ge 3) = 1 - e^{-3.5}\left(1 + 3.5 + \frac{3.5^2}{2}\right) = 1 - e^{-3.5}(10.625) = 1 - 0.3208 = 0.679

Different rates, different lengths

Faults occur at random in cable of type A at an average rate of $0.3$ per $100$ m, and in cable of type B at an average rate of $0.5$ per $100$ m, independently. An engineer lays $400$ m of type A and $200$ m of type B. Find the probability that there are fewer than $3$ faults in total.

Solution

Rescale each rate to its own length first.

Type A: $400$ m gives mean $4 \times 0.3 = 1.2$ . Type B: $200$ m gives mean $2 \times 0.5 = 1.0$ .

The total $F \sim \text{Po}(1.2 + 1.0) = \text{Po}(2.2)$ .

P(F < 3) = P(F \le 2) = e^{-2.2}\left(1 + 2.2 + \frac{2.2^2}{2}\right) = e^{-2.2}(5.62) = 0.623

A common error is to add the rates first ( $0.8$ per $100$ m) and multiply by the total length ( $6$ ), giving $\text{Po}(4.8)$ . That would only be right if both cables had the same length.

Sums with a normal approximation

Customers arrive at the two entrances of a shop independently and at random, at average rates of $8$ per hour at the north entrance and $9$ per hour at the south entrance. Use a suitable approximation to find the probability that more than $20$ customers arrive in a given hour.

Solution

The total $T \sim \text{Po}(8 + 9) = \text{Po}(17)$ .

Since $17 > 15$ , approximate by $N(17, 17)$ , with a continuity correction:

P(T > 20) = P(T \ge 21) \approx P\left(Z > \frac{20.5 - 17}{\sqrt{17}}\right) = P(Z > 0.849) = 1 - 0.8021 = 0.198

Splitting a known total

Sometimes the total is given and the question asks how it splits between the two sources. Use the definition of conditional probability, and the independence of the two counts:

P(X = r \mid X + Y = n) = \frac{P(X = r)\,P(Y = n - r)}{P(X + Y = n)}.

The numerator uses independence; the denominator uses the sum result. A lot cancels. In fact the answer is always the binomial probability $\binom{n}{r}p^r(1 - p)^{n-r}$ with $p = \dfrac{\lambda}{\lambda + \mu}$ . This is a useful check, though the direct method is all you need.

An exam-style conditional question

Cars pass a point on a road at random at an average rate of $1.5$ per minute, and lorries pass independently at random at an average rate of $0.5$ per minute.

(a) Find the probability that exactly $4$ vehicles pass in a given minute.

(b) Given that exactly $4$ vehicles pass in a given minute, find the probability that exactly $3$ of them are cars.

Solution

Let $C \sim \text{Po}(1.5)$ and $L \sim \text{Po}(0.5)$ , independent, so $V = C + L \sim \text{Po}(2)$ .

(a)

P(V = 4) = e^{-2}\frac{2^4}{4!} = \frac{2}{3}e^{-2} = 0.0902

(b) Three cars and one lorry:

P(C = 3 \mid V = 4) = \frac{P(C = 3)P(L = 1)}{P(V = 4)} = \frac{e^{-1.5}\dfrac{1.5^3}{6} \times e^{-0.5}(0.5)}{e^{-2}\dfrac{16}{24}}

The $e^{-2}$ cancels:

= \frac{0.5625 \times 0.5}{0.6667} = 0.422

Check: $\binom{4}{3}(0.75)^3(0.25) = 0.421875$ , which agrees.

P(C \ge 1)P(L \ge 1) = (1 - e^{-1.5})(1 - e^{-0.5}) = 0.77687 \times 0.39347 = 0.306

Common mistakes

Treating $2X$ as $\text{Po}(2\lambda)$ . Two separate intervals give $\text{Po}(2\lambda)$ ; doubling one count does not.
Writing $X - Y \sim \text{Po}(\lambda - \mu)$ . A difference is not Poisson, and its variance is $\lambda + \mu$ .
Adding rates before rescaling. When the two sources are observed over different lengths or times, scale each mean separately, then add.
Forgetting independence. The result needs independent counts. If one source triggers the other (a fault in one machine causing faults in a second), it fails.
Multiplying instead of conditioning. $P(C = 3 \mid V = 4)$ is not $P(C = 3)$ ; you must divide by $P(V = 4)$ .

Exam tip

State the combined distribution explicitly: " $X + Y \sim \text{Po}(3.5)$ ". This is usually a method mark on its own.
"State an assumption" for adding Poisson variables: the two counts are independent, in context ("the number of cars passing is independent of the number of lorries").
Check whether the combined mean is above $15$ ; if it is, and the question says "use a suitable approximation", switch to the normal with a continuity correction.
In conditional questions, write the event in words first ("3 cars and 1 lorry"), then the probability. It prevents the most common slip, forgetting the second variable's probability in the numerator.

Summary

Independent $X \sim \text{Po}(\lambda)$ and $Y \sim \text{Po}(\mu)$ : $X + Y \sim \text{Po}(\lambda + \mu)$ .
Extends to any number of independent Poisson counts.
$2X$ and $X - Y$ are not Poisson; their variances are $4\lambda$ and $\lambda + \mu$ .
Rescale each source to its own interval, then add the means.
Large total mean ( $> 15$ ): use $N(\lambda + \mu, \lambda + \mu)$ with continuity correction.
Given the total $n$ : $P(X = r \mid X + Y = n) = \dfrac{P(X = r)P(Y = n - r)}{P(X + Y = n)}$ .

Practice questions

Question

$X \sim \text{Po}(3)$ and $Y \sim \text{Po}(2.5)$ are independent. Find $P(X + Y = 4)$ and $P(X + Y \le 2)$ .
A person receives personal emails at an average rate of $1.2$ per hour and work emails at an average rate of $3.1$ per hour, independently and at random. Find the probability of receiving at least $3$ emails in a 30-minute period.
$X \sim \text{Po}(4)$ and $Y \sim \text{Po}(3)$ are independent. Find $E(X - Y)$ and $\text{Var}(X - Y)$ , and give two reasons why $X - Y$ does not have a Poisson distribution.
A shop sells type A phones at an average rate of $2.5$ per day and type B phones at an average rate of $1.5$ per day, independently and at random. Use a suitable approximation to find the probability that more than $25$ phones in total are sold in a 5-day period.
$X \sim \text{Po}(2)$ and $Y \sim \text{Po}(1)$ are independent. Given that $X + Y = 3$ , find the probability that $X = 3$ .
$X \sim \text{Po}(\lambda)$ and $Y \sim \text{Po}(2\lambda)$ are independent, and $P(X + Y = 0) = 0.05$ . Find $\lambda$ .
Two radioactive sources emit particles independently and at random, at average rates of $0.3$ and $0.2$ per second. Find the probability that, in a 10-second period, (a) at least $2$ particles are emitted in total, (b) at least one particle is emitted by each source.
In football matches, the numbers of goals scored by the home team and the away team are modelled by independent distributions $\text{Po}(1.6)$ and $\text{Po}(1.1)$ . (a) Find the probability that exactly $3$ goals are scored in a match. (b) Given that exactly $3$ goals are scored, find the probability that the home team scores more goals than the away team.

Answers

$X + Y \sim \text{Po}(5.5)$ . $P(X + Y = 4) = e^{-5.5}\dfrac{5.5^4}{24} = 0.156$ . $P(X + Y \le 2) = e^{-5.5}\left(1 + 5.5 + 15.125\right) = 0.0884$ .
In 30 minutes: personal $\text{Po}(0.6)$ , work $\text{Po}(1.55)$ , total $\text{Po}(2.15)$ . $P(T \ge 3) = 1 - e^{-2.15}\left(1 + 2.15 + \dfrac{2.15^2}{2}\right) = 1 - e^{-2.15}(5.46125) = 1 - 0.6361 = 0.364$ .
$E(X - Y) = 4 - 3 = 1$ ; $\text{Var}(X - Y) = 4 + 3 = 7$ . Not Poisson because $X - Y$ can take negative values, and because its mean ( $1$ ) is not equal to its variance ( $7$ ).
In 5 days: $\text{Po}(12.5)$ and $\text{Po}(7.5)$ , total $T \sim \text{Po}(20)$ . Since $20 > 15$ , use $N(20, 20)$ . $P(T > 25) = P(T \ge 26) \approx P\left(Z > \dfrac{25.5 - 20}{\sqrt{20}}\right) = P(Z > 1.230) = 1 - 0.8907 = 0.109$ .
$P(X = 3 \mid X + Y = 3) = \dfrac{P(X = 3)P(Y = 0)}{P(X + Y = 3)} = \dfrac{e^{-2}\frac{8}{6} \cdot e^{-1}}{e^{-3}\frac{27}{6}} = \dfrac{8}{27} = 0.296$ .
$X + Y \sim \text{Po}(3\lambda)$ . $e^{-3\lambda} = 0.05 \Rightarrow 3\lambda = \ln 20 \Rightarrow \lambda = 0.999$ (3 s.f.).
In 10 seconds: $A \sim \text{Po}(3)$ , $B \sim \text{Po}(2)$ , total $\text{Po}(5)$ . (a) $1 - e^{-5}(1 + 5) = 1 - 6e^{-5} = 0.960$ . (b) $(1 - e^{-3})(1 - e^{-2}) = 0.95021 \times 0.86466 = 0.822$ .
(a) Total $\sim \text{Po}(2.7)$ : $P(T = 3) = e^{-2.7}\dfrac{2.7^3}{6} = 0.220$ . (b) Home team scores more when the split is $3$ – $0$ or $2$ – $1$ . $P(H = 3)P(A = 0) + P(H = 2)P(A = 1) = e^{-2.7}\left(\dfrac{1.6^3}{6} + \dfrac{1.6^2}{2}(1.1)\right) = e^{-2.7}(0.68267 + 1.408) = e^{-2.7}(2.09067)$ . Divide by $P(T = 3) = e^{-2.7}(3.2805)$ : $\dfrac{2.09067}{3.2805} = 0.637$ .

The result

What does not stay Poisson

Splitting a known total

Practice questions

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