SB4 - Combining Poisson variables

SB4 - Combining Poisson variables

Separate random event streams can sometimes be treated as one. You will combine independent Poisson counts, match parameters to observation intervals, and recognise why overlap or dependence can invalidate an apparently simple calculation.

Summing independent Poisson variables

Two independent detector counts are modelled as Poisson with means 2 and 3 per minute. Is their combined count Poisson with mean 5? Adding their means is easy; establishing the distribution needs one more condition.

For a total of two detections, one detector could record 0 and the other 2, or they could record 1 each, or 2 and 0. These are mutually exclusive possibilities. Independence lets us multiply each pair of probabilities, then add the results.

Independent Poisson sum

If X∼Po⁡(λ)X\sim\operatorname{Po}(\lambda) and Y∼Po⁡(μ)Y\sim\operatorname{Po}(\mu) are independent, then

X+Y∼Po⁡(λ+μ)X+Y\sim\operatorname{Po}(\lambda+\mu)

Here λ,μ>0\lambda,\mu>0, and both variables count contributions to the stated total. The rule extends to a finite collection of mutually independent Poisson variables. This sum rule is not separately printed in the formula booklet.

Here is why the rule gives the whole distribution, rather than just its mean. For any integer n≥0n\ge0, split according to the first count:

P⁡(X+Y=n)=∑k=0nP⁡(X=k)P⁡(Y=n−k)=e−(λ+μ)n!∑k=0n(nk)λkμn−k=e−(λ+μ)(λ+μ)nn!\begin{aligned} \operatorname{P}(X+Y=n) &=\sum_{k=0}^{n}\operatorname{P}(X=k)\operatorname{P}(Y=n-k)\\ &=\frac{\mathrm{e}^{-(\lambda+\mu)}}{n!} \sum_{k=0}^{n}\binom{n}{k}\lambda^k\mu^{n-k}\\ &=\frac{\mathrm{e}^{-(\lambda+\mu)}(\lambda+\mu)^n}{n!} \end{aligned}

The final step uses the binomial expansion. This is exactly a Poisson mass function with parameter λ+μ\lambda+\mu.

Worked example 1: a combined detector count

Question. Independent counts X∼Po⁡(2)X\sim\operatorname{Po}(2) and Y∼Po⁡(3)Y\sim\operatorname{Po}(3) refer to the same minute. Find the distribution of T=X+YT=X+Y, and P⁡(T=0)\operatorname{P}(T=0).

Plan. Check independence, add parameters, then use the zero-count formula.

T∼Po⁡(2+3)=Po⁡(5),P⁡(T=0)=e−5=0.00673795…T\sim\operatorname{Po}(2+3)=\operatorname{Po}(5),\qquad \operatorname{P}(T=0)=\mathrm{e}^{-5}=0.00673795\ldots

Answer. T∼Po⁡(5)T\sim\operatorname{Po}(5); the probability is 0.0067380.006738.

Check. Zero total requires both counts to be zero, giving e−2e−3=e−5\mathrm{e}^{-2}\mathrm{e}^{-3}=\mathrm{e}^{-5}.

Keep the total mean the same, but redistribute it between the two detectors. This tests whether the split matters.

Worked example 2: a different split of the same mean

Question. Independent counts A∼Po⁡(1.5)A\sim\operatorname{Po}(1.5) and B∼Po⁡(3.5)B\sim\operatorname{Po}(3.5) refer to one minute. Find the distribution of S=A+BS=A+B, and P⁡(S=0)\operatorname{P}(S=0).

Plan. Add the new parameters and compare the total with the previous example.

S∼Po⁡(1.5+3.5)=Po⁡(5),P⁡(S=0)=e−5=0.00673795…S\sim\operatorname{Po}(1.5+3.5)=\operatorname{Po}(5),\qquad \operatorname{P}(S=0)=\mathrm{e}^{-5}=0.00673795\ldots

Answer. The total has the same distribution and zero-count probability as before.

Check. Independently, Var⁡(S)=1.5+3.5=5\operatorname{Var}(S)=1.5+3.5=5, agreeing with its Poisson parameter. The surprising result is that the entire total distribution depends on the sum, not on how that mean is split.

Expected values always add when they exist, even without independence. That general expectation rule alone cannot identify a Poisson distribution.

Watch out

Do not write X+Y∼Po⁡(λ+μ)X+Y\sim\operatorname{Po}(\lambda+\mu) from the marginal distributions alone. Independence is part of the rule.

Now combine new streams.

Rates across intervals

A rate of 6 events per hour produces a mean of 3 in half an hour, provided the same constant-rate model remains valid. Always convert the interval before calculating a probability.

Rate and interval

For a Poisson process with constant mean rate r>0r>0 per unit time, the count NtN_t over a fixed interval of length t>0t>0 satisfies

Nt∼Po⁡(rt)N_t\sim\operatorname{Po}(rt)

Use matching units for rr and tt. Counts over disjoint intervals are independent under this process model. This interval rule is not separately printed in the formula booklet.

Joining disjoint intervals adds their counts and their mean counts. If the rate changes between intervals, separately model each interval and add the parameters only when the counts are independent. Do not replace two different rates by an unweighted average: interval lengths matter.

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