FS1 2.1 - Adding independent Poisson variables

FS1 2.1 - Adding independent Poisson variables

Independent Poisson counts combine by adding their parameters. You will use that property across time intervals and separate event streams, distinguish a new total from a scaled old count and calculate probabilities when a combined total is already known.

Adding counts rather than duplicating one count

Does observing arrivals for twice as long mean doubling the number you saw in the first minute? Suppose independent one-minute counts X1,X2X_1,X_2 each have distribution Po⁡(1)\operatorname{Po}(1). Two minutes produce X1+X2X_1+X_2; doubling the first minute's count produces 2X12X_1. Those are different random variables.

For a striking comparison,

P(X1+X2=0)=e−1e−1=e−2\mathrm{P}(X_1+X_2=0)=\mathrm{e}^{-1}\mathrm{e}^{-1}=\mathrm{e}^{-2}

whereas

P(2X1=0)=P(X1=0)=e−1\mathrm{P}(2X_1=0)=\mathrm{P}(X_1=0)=\mathrm{e}^{-1}

The probabilities are approximately 0.13530.1353 and 0.36790.3679. The second variable can only be even; the two-minute count can be odd. Doubling a time window and duplicating an observed count cannot be interchangeable.

You already know

For X∼Po⁡(λ)X\sim\operatorname{Po}(\lambda), λ\lambda describes the expected count in the stated interval, and P(X=0)=e−λ\mathrm{P}(X=0)=\mathrm{e}^{-\lambda}. Independent events have joint probability equal to the product of their probabilities.

The additive property describes genuinely separate, independent counts. They may count arrivals in disjoint time intervals, or arrivals from different independent sources over the same interval.

Adding independent Poisson variables

If

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

and X,YX,Y are independent, then

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

More generally, the sum of mutually independent Poisson variables has parameter equal to the sum of their parameters. No proof is needed.

Not in the formula booklet: learn it.

Independence is a condition, not a consequence of giving two marginal Poisson distributions. A common alarm may create linked message counts at two servers. Overlapping time windows also share events, so their counts must not be treated as independent without justification.

Key idea

Add the parameters of independent counts. Multiplying a single count by a constant does not create independent copies and does not generally give another Poisson distribution.

The following check targets the difference between two copies and one scaled variable.

Matching the parameter to the interval

Under a constant-rate Poisson event model, a longer interval combines more independent opportunities for events; multiply the rate by the interval length in matching units.

If one minute has parameter λ\lambda, five independent one-minute counts add to a five-minute count with parameter 5λ5\lambda. The same rate reasoning works for fractional intervals, lengths and areas when the relevant modelling assumptions hold.

Parameter from a constant rate

For a constant positive event rate ρ\rho per unit interval and an interval of size t>0t>0,

Nt∼Po⁡(ρt)N_t\sim\operatorname{Po}(\rho t)

Use matching units for ρ\rho and tt, with independent counts over disjoint intervals under the model. It is the parameter, not one realised count, that scales with tt.

Not in the formula booklet: learn it.

Write the parameter calculation before the probability. That line makes a unit conversion or a misplaced factor much easier to spot.

Worked example 1: a five-minute count

Question. Jobs reach a computer queue independently at a constant average rate of 0.80.8 per minute. Under a Poisson model, find the probability of at least three arrivals in five minutes, to four decimal places.

Plan. Find the five-minute parameter, then complement the counts 0,1,20,1,2.

λ5=0.8(5)=4rate times minutesN5∼Po⁡(4)state the new count modelP(N5≥3)=1−P(N5≤2)retain count 3=1−0.2381033…evaluate the cumulative probability=0.7618966…complement before rounding\begin{aligned} \lambda_5&=0.8(5)=4 &&\text{rate times minutes}\\ N_5&\sim\operatorname{Po}(4) &&\text{state the new count model}\\ \mathrm{P}(N_5\ge3)&=1-\mathrm{P}(N_5\le2) &&\text{retain count }3\\ &=1-0.2381033\ldots &&\text{evaluate the cumulative probability}\\ &=0.7618966\ldots &&\text{complement before rounding} \end{aligned}

Why this step? The given rate belongs to one minute. Using Po⁡(0.8)\operatorname{Po}(0.8) would answer a different question, however accurately its tail was calculated.

Answer. 0.76190.7619.

Check. A five-minute expected count of 44 makes three or more arrivals reasonably likely. The excluded cumulative event stops at 22, not 33.

Worked example 2 changes only the observation length, halving it from five minutes to two and a half. The tail threshold remains three, while the parameter halves.

Worked example 2: a fractional interval

Question. For the same arrival model, find the probability of at least three arrivals in 2.52.5 minutes.

Plan. Replace the interval length by 2.52.5, then use the same integer complement.

λ2.5=0.8(2.5)=2scale the parameterN2.5∼Po⁡(2)new observation windowP(N2.5≥3)=1−P(N2.5≤2)same count threshold=1−0.6766764…use parameter 2=0.3233235…evaluate\begin{aligned} \lambda_{2.5}&=0.8(2.5)=2 &&\text{scale the parameter}\\ N_{2.5}&\sim\operatorname{Po}(2) &&\text{new observation window}\\ \mathrm{P}(N_{2.5}\ge3)&=1-\mathrm{P}(N_{2.5}\le2) &&\text{same count threshold}\\ &=1-0.6766764\ldots &&\text{use parameter }2\\ &=0.3233235\ldots &&\text{evaluate} \end{aligned}

Why this step? Halving the time halves the parameter, not the tail probability. The probability is a non-linear function of the parameter.

Answer. 0.32330.3233 to four decimal places.

Check. The chance of at least three arrivals has fallen from 0.76190.7619, as a shorter observation window should make that event less likely. It is not half the old probability.

These time-specific distributions follow from the additive property and the constant-rate assumptions; their individual parameter calculations are not printed in the formula booklet. If the rate changes during the interval, split it into suitable independent portions and add their expected counts instead of using one unjustified constant rate.

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