Mathematics 9709 · Probability & Statistics 2

Combining Poisson distributions

Pool the rare events and you just pool the rates. Lions at 5/week and tigers at 3/week (independent) become big cats at 8/week: independent Poissons add their $\lambda$ 's into one Poisson. But only for adding, a difference or multiple of Poissons is not Poisson.

Add independent Poissons; nothing else stays Poisson

$X\sim\text{Po}(\lambda),\ Y\sim\text{Po}(\mu)\text{ indep}\ \Rightarrow\ X+Y\sim\text{Po}(\lambda+\mu)$

Independent Poissons ADD into one Poisson (rates add over the same interval). State the full Po(λ+μ), not just “Poisson”.

Worked example

Exact P(X + Y = k): pool, then one calculation

Independent $X\sim\text{Po}(7)$ , $Y\sim\text{Po}(5)$ . The slow way sums many products; the fast way pools first: $X+Y\sim\text{Po}(12)$ , then one plug-in.

$P(X+Y=2)=e^{-12}\dfrac{12^2}{2!}=0.000442$

Answer

\approx0.000442

Worked example

Add the rates, then a tail

Josh texts $\text{Po}(3.2)$ /week, Reuben $\text{Po}(2.5)$ /week, independent. Total $T\sim\text{Po}(5.7)$ . $P(T\ge5)=1-P(T\le4)$ :

$=1-e^{-5.7}\big(1+5.7+\tfrac{5.7^2}{2!}+\tfrac{5.7^3}{3!}+\tfrac{5.7^4}{4!}\big)=1-0.327$

Answer

P(T\ge5)\approx0.673

“At least 5” = 1 − P(T ≤ 4); “more than 5” = 1 − P(T ≤ 5). This off-by-one is a top mark-loser.

Worked example

Where the magic stops

$X\sim\text{Po}(2.4),\ Y\sim\text{Po}(3.6)$ , $T=2X-Y$ . You can still get the mean and variance, but check whether it is Poisson:

$E(T)=2(2.4)-3.6=1.2$ .
$\operatorname{Var}(T)=2^2(2.4)+(-1)^2(3.6)=9.6+3.6=13.2$ .
$E(T)\ne\operatorname{Var}(T)$ , and a Poisson must have equal mean and variance.

Answer

E=1.2,\ \operatorname{Var}=13.2\Rightarrow T\text{ is NOT Poisson}

A multiple or difference breaks the equal-mean-and-variance signature, so it cannot be Poisson.

Combination of independent Poissons	Still Poisson?
$X+Y$	Yes: Po(λ + μ)
$X+Y+Z$	Yes: Po(λ + μ + ν)
$X-Y$	No (mean ≠ variance)
$2X$	No (mean ≠ variance)

Your turn— tap to reveal the worked answer (combine)

Cars arrive $\text{Po}(2)$ /hour and lorries $\text{Po}(3)$ /hour, independent. Find the chance exactly four vehicles pass in an hour.

Total $\text{Po}(5)$ : $P=e^{-5}\tfrac{5^4}{4!}$ .

Answer

\approx0.175

Examiner note

Only a SUM of independent Poissons is Poisson; for a difference or multiple you can still give

E

and

\operatorname{Var}

(and use a normal if the means are large), but it is not

\text{Po}

. (9709/62/O/N/24 Q5)

Common mistake

Using the add rule on a difference or multiple (only sums of independent Poissons stay Poisson); adding rates over different intervals without rescaling first; and writing “Po” or “

\lambda=5.7

” alone instead of the full

\text{Po}(5.7)

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Combining Poisson distributions

Add independent Poissons; nothing else stays Poisson

$X\sim\text{Po}(\lambda),\ Y\sim\text{Po}(\mu)\text{ indep}\ \Rightarrow\ X+Y\sim\text{Po}(\lambda+\mu)$

Independent Poissons ADD into one Poisson (rates add over the same interval). State the full Po(λ+μ), not just “Poisson”.

Worked example

Exact P(X + Y = k): pool, then one calculation

Independent $X\sim\text{Po}(7)$ , $Y\sim\text{Po}(5)$ . The slow way sums many products; the fast way pools first: $X+Y\sim\text{Po}(12)$ , then one plug-in.

$P(X+Y=2)=e^{-12}\dfrac{12^2}{2!}=0.000442$

Answer

\approx0.000442

Worked example

Add the rates, then a tail

Josh texts $\text{Po}(3.2)$ /week, Reuben $\text{Po}(2.5)$ /week, independent. Total $T\sim\text{Po}(5.7)$ . $P(T\ge5)=1-P(T\le4)$ :

$=1-e^{-5.7}\big(1+5.7+\tfrac{5.7^2}{2!}+\tfrac{5.7^3}{3!}+\tfrac{5.7^4}{4!}\big)=1-0.327$

Answer

P(T\ge5)\approx0.673

“At least 5” = 1 − P(T ≤ 4); “more than 5” = 1 − P(T ≤ 5). This off-by-one is a top mark-loser.

Worked example

Where the magic stops

$X\sim\text{Po}(2.4),\ Y\sim\text{Po}(3.6)$ , $T=2X-Y$ . You can still get the mean and variance, but check whether it is Poisson:

$E(T)=2(2.4)-3.6=1.2$ .
$\operatorname{Var}(T)=2^2(2.4)+(-1)^2(3.6)=9.6+3.6=13.2$ .
$E(T)\ne\operatorname{Var}(T)$ , and a Poisson must have equal mean and variance.

Answer

E=1.2,\ \operatorname{Var}=13.2\Rightarrow T\text{ is NOT Poisson}

A multiple or difference breaks the equal-mean-and-variance signature, so it cannot be Poisson.

Combination of independent Poissons	Still Poisson?
$X+Y$	Yes: Po(λ + μ)
$X+Y+Z$	Yes: Po(λ + μ + ν)
$X-Y$	No (mean ≠ variance)
$2X$	No (mean ≠ variance)

Your turn— tap to reveal the worked answer (combine)

Cars arrive $\text{Po}(2)$ /hour and lorries $\text{Po}(3)$ /hour, independent. Find the chance exactly four vehicles pass in an hour.

Total $\text{Po}(5)$ : $P=e^{-5}\tfrac{5^4}{4!}$ .

Answer

\approx0.175

Examiner note

Only a SUM of independent Poissons is Poisson; for a difference or multiple you can still give

E

and

\operatorname{Var}

(and use a normal if the means are large), but it is not

\text{Po}

. (9709/62/O/N/24 Q5)

Common mistake

Using the add rule on a difference or multiple (only sums of independent Poissons stay Poisson); adding rates over different intervals without rescaling first; and writing “Po” or “

\lambda=5.7

” alone instead of the full

\text{Po}(5.7)