Mathematics 9709 · Probability & Statistics 2

The Poisson formula & scaling λ

Poisson is what the binomial becomes when events are rare but the chances are endless. Chop an hour into a million tiny slivers of time; in each one an event either happens or it doesn't, with a minuscule probability. You can't count the slivers ( $n\to\infty$ ) or the per-sliver chance ( $p\to0$ ) — only the average number per interval, $\lambda$ . Everything collapses onto that one number.

The formula, and picking the right λ

That single $\lambda$ is why a Poisson has only one parameter, and why its mean and variance are both $\lambda$ . So reading a question is really one job: pin down $\lambda$ for the interval you are asked about, then the formula does the rest.

$P(X=x)=e^{-\lambda}\dfrac{\lambda^x}{x!},\qquad E(X)=\operatorname{Var}(X)=\lambda$

λ is the whole story: the average count per interval, and both the mean and the variance. Pin down λ first; the formula does the rest.

Po(1.7): a rare-event distribution piles up at small x with a long thin right tail. The single knob λ is both the mean and the variance.

Worked example

Start simple: one probability

A consumer hotline takes calls at a mean rate of six per minute, $X\sim\text{Po}(6)$ . Find the chance of exactly nine calls in a minute — one clean plug-in:

$P(X=9)=e^{-6}\dfrac{6^9}{9!}=0.0688$

Answer

0.0688

Worked example

Build up: cumulative and 'at least'

Same $X\sim\text{Po}(6)$ . Now find $P(X\le3)$ , $P(X>1)$ and $P(X\ge1)$ .

“3 or fewer” includes 3 — sum and factor out $e^{-6}$ : $P(X\le3)=e^{-6}\!\left(1+6+\tfrac{6^2}{2!}+\tfrac{6^3}{3!}\right)=0.151$ .
“more than 1” excludes 1: $P(X>1)=1-e^{-6}(1+6)=0.983$ .
“at least 1” is the easy one: $P(X\ge1)=1-e^{-6}=0.998$ .

Answer

0.151,\quad 0.983,\quad 0.998

Factoring out $e^{-6}$ once keeps the arithmetic clean and is exactly how the mark scheme lays it out.

λ is a rate, so scale it to the interval

“Six per minute” is a rate: so many per so much. Double the interval, double $\lambda$ ; take a fifth of it, take a fifth of $\lambda$ . Always re-scale $\lambda$ to match the interval the question asks about before using the formula.

Worked example

Same rate, four different intervals

People arrive at a post office at two every 5 minutes. Find the probability of: (a) three in 10 min, (b) more than four in 30 min, (c) five in 4 min, (d) one in 1 min.

(a) 10 min is $2\times$ the base, so $\lambda=4$ : $P(X=3)=e^{-4}\tfrac{4^3}{3!}=0.195$ .
(b) 30 min is $6\times$ , so $\lambda=12$ : $P(X>4)=1-P(X\le4)=0.992$ .
(c) 4 min is $\tfrac{4}{5}\times$ , so $\lambda=1.6$ : $P(X=5)=e^{-1.6}\tfrac{1.6^5}{5!}=0.0176$ .
(d) 1 min is $\tfrac15\times$ , so $\lambda=0.4$ : $P(X=1)=e^{-0.4}(0.4)=0.268$ .

Answer

0.195,\quad 0.992,\quad 0.0176,\quad 0.268

Scale λ to the question's interval FIRST. λ need not be a whole number, and P(X=0) = e^(−λ).

Worked example

State the conditions, then count

A typesetter makes on average five errors per page. Model the error count and find how many of 200 pages have at most two errors.

Conditions: errors occur singly, independently, at random, at a constant average rate — so $X\sim\text{Po}(5)$ .
“at most two”: $P(X\le2)=e^{-5}\!\left(1+5+\tfrac{5^2}{2!}\right)=0.1247$ .
Turn a probability into a count with $N\times P$ : $200\times0.1247=24.9$ .

Answer

\approx25\text{ pages}

The four conditions — singly, independently, at random, constant rate — are the standard 1-mark “state the assumptions” answer.

Independent Poissons add

Two independent streams of rare events are just one bigger stream: if $X\sim\text{Po}(\lambda_1)$ and $Y\sim\text{Po}(\lambda_2)$ are independent, then $X+Y\sim\text{Po}(\lambda_1+\lambda_2)$ . Pool the rates, then use one Poisson.

Worked example

Pool two independent rates

Line A $\text{Po}(3)$ , line B $\text{Po}(2)$ , independent. Total $\text{Po}(5)$ : $P(X+Y=4)=e^{-5}\tfrac{5^4}{4!}=0.175$ . (9709/62/M/J/24 Q5)

Answer

\approx0.175

Is a Poisson even the right model?

Given raw data, you can check whether Poisson fits before trusting it. Its mean and variance are equal, so compute both — if they are close, use the mean as $\lambda$ and compare the expected frequencies $N\,P(X=r)$ with the observed.

Worked example

Does a Poisson fit the data?

A clinic logs patients arriving in 250 half-hour intervals (mean 1.70 per half-hour). Should arrivals be modelled by a Poisson distribution?

Mean $=1.70$ ; variance from the data $=1.61$ . They are close, so a Poisson is reasonable — take $\lambda=1.70$ .
Expected frequency at each $r$ is $250\times P(X=r)$ with $X\sim\text{Po}(1.7)$ (e.g. $250\times0.150=37$ at $r=3$ ).
Compare with the observed counts:

patients r	0	1	2	3	4
expected 250·P(X=r)	46	78	66	37	16
observed	45	81	58	40	22

Answer

\text{mean}\approx\text{variance},\ \ \text{expected}\approx\text{observed}

So a Poisson model, $\text{Po}(1.7)$ , fits the data.

To check a Poisson model: mean ≈ variance, then use the mean as λ and match expected frequencies N·P(X=r) to the observed.

Your turn— tap to reveal the worked answer (scaling λ)

Calls arrive at 3 per 2 minutes. Find P(exactly 4 in 6 minutes).

6 min is $3\times$ the base, so $\lambda=9$ : $P(X=4)=e^{-9}\tfrac{9^4}{4!}$ .

Answer

\approx0.0337

Two ways to scale give the same answer

Breakages run at $\text{Po}(0.8)$ per week. The chance of none in 6 weeks is either $[P(X=0)]^6=(e^{-0.8})^6=e^{-4.8}$ , or scale first to $\text{Po}(4.8)$ and read $P(X=0)=e^{-4.8}=0.00823$ . Identical — because raising $e^{-\lambda}$ to a power is scaling $\lambda$ .

Situation	Use
rare events at rate λ	$\text{Po}(\lambda)$
data with mean ≈ variance	$\text{Po}(\text{mean})$
B(n,p), n large, p small	$\text{Po}(np)$
Po(λ), λ>15, a tail	$N(\lambda,\lambda)+\text{c.c.}$
two independent Poissons	$\text{Po}(\lambda_1+\lambda_2)$

Examiner note

State

X\sim\text{Po}(\lambda)

with the scaled

\lambda

; for a sum, pool the rates BEFORE computing. Keep

e^{-\lambda}

symbolic until the end. (9709/62/M/J/23 Q4)

Common mistake

Forgetting to scale

\lambda

to the interval (the #1 error). Mind

P(X>k)

(excludes

k

) vs

P(X\ge k)

, and carry

e^{-\lambda}

symbolically — evaluate last.

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The Poisson formula & scaling λ

The formula, and picking the right λ

$P(X=x)=e^{-\lambda}\dfrac{\lambda^x}{x!},\qquad E(X)=\operatorname{Var}(X)=\lambda$

λ is the whole story: the average count per interval, and both the mean and the variance. Pin down λ first; the formula does the rest.

Po(1.7): a rare-event distribution piles up at small x with a long thin right tail. The single knob λ is both the mean and the variance.

Worked example

Start simple: one probability

A consumer hotline takes calls at a mean rate of six per minute, $X\sim\text{Po}(6)$ . Find the chance of exactly nine calls in a minute — one clean plug-in:

$P(X=9)=e^{-6}\dfrac{6^9}{9!}=0.0688$

Answer

0.0688

Worked example

Build up: cumulative and 'at least'

Same $X\sim\text{Po}(6)$ . Now find $P(X\le3)$ , $P(X>1)$ and $P(X\ge1)$ .

“3 or fewer” includes 3 — sum and factor out $e^{-6}$ : $P(X\le3)=e^{-6}\!\left(1+6+\tfrac{6^2}{2!}+\tfrac{6^3}{3!}\right)=0.151$ .
“more than 1” excludes 1: $P(X>1)=1-e^{-6}(1+6)=0.983$ .
“at least 1” is the easy one: $P(X\ge1)=1-e^{-6}=0.998$ .

Answer

0.151,\quad 0.983,\quad 0.998

Factoring out $e^{-6}$ once keeps the arithmetic clean and is exactly how the mark scheme lays it out.

λ is a rate, so scale it to the interval

Worked example

Same rate, four different intervals

People arrive at a post office at two every 5 minutes. Find the probability of: (a) three in 10 min, (b) more than four in 30 min, (c) five in 4 min, (d) one in 1 min.

(a) 10 min is $2\times$ the base, so $\lambda=4$ : $P(X=3)=e^{-4}\tfrac{4^3}{3!}=0.195$ .
(b) 30 min is $6\times$ , so $\lambda=12$ : $P(X>4)=1-P(X\le4)=0.992$ .
(c) 4 min is $\tfrac{4}{5}\times$ , so $\lambda=1.6$ : $P(X=5)=e^{-1.6}\tfrac{1.6^5}{5!}=0.0176$ .
(d) 1 min is $\tfrac15\times$ , so $\lambda=0.4$ : $P(X=1)=e^{-0.4}(0.4)=0.268$ .

Answer

0.195,\quad 0.992,\quad 0.0176,\quad 0.268

Scale λ to the question's interval FIRST. λ need not be a whole number, and P(X=0) = e^(−λ).

Worked example

State the conditions, then count

A typesetter makes on average five errors per page. Model the error count and find how many of 200 pages have at most two errors.

Conditions: errors occur singly, independently, at random, at a constant average rate — so $X\sim\text{Po}(5)$ .
“at most two”: $P(X\le2)=e^{-5}\!\left(1+5+\tfrac{5^2}{2!}\right)=0.1247$ .
Turn a probability into a count with $N\times P$ : $200\times0.1247=24.9$ .

Answer

\approx25\text{ pages}

The four conditions — singly, independently, at random, constant rate — are the standard 1-mark “state the assumptions” answer.

Independent Poissons add

Worked example

Pool two independent rates

Line A $\text{Po}(3)$ , line B $\text{Po}(2)$ , independent. Total $\text{Po}(5)$ : $P(X+Y=4)=e^{-5}\tfrac{5^4}{4!}=0.175$ . (9709/62/M/J/24 Q5)

Answer

\approx0.175

Is a Poisson even the right model?

Worked example

Does a Poisson fit the data?

A clinic logs patients arriving in 250 half-hour intervals (mean 1.70 per half-hour). Should arrivals be modelled by a Poisson distribution?

Mean $=1.70$ ; variance from the data $=1.61$ . They are close, so a Poisson is reasonable — take $\lambda=1.70$ .
Expected frequency at each $r$ is $250\times P(X=r)$ with $X\sim\text{Po}(1.7)$ (e.g. $250\times0.150=37$ at $r=3$ ).
Compare with the observed counts:

patients r	0	1	2	3	4
expected 250·P(X=r)	46	78	66	37	16
observed	45	81	58	40	22

Answer

\text{mean}\approx\text{variance},\ \ \text{expected}\approx\text{observed}

So a Poisson model, $\text{Po}(1.7)$ , fits the data.

To check a Poisson model: mean ≈ variance, then use the mean as λ and match expected frequencies N·P(X=r) to the observed.

Your turn— tap to reveal the worked answer (scaling λ)

Calls arrive at 3 per 2 minutes. Find P(exactly 4 in 6 minutes).

6 min is $3\times$ the base, so $\lambda=9$ : $P(X=4)=e^{-9}\tfrac{9^4}{4!}$ .

Answer

\approx0.0337

Two ways to scale give the same answer

Situation	Use
rare events at rate λ	$\text{Po}(\lambda)$
data with mean ≈ variance	$\text{Po}(\text{mean})$
B(n,p), n large, p small	$\text{Po}(np)$
Po(λ), λ>15, a tail	$N(\lambda,\lambda)+\text{c.c.}$
two independent Poissons	$\text{Po}(\lambda_1+\lambda_2)$

Examiner note

State

X\sim\text{Po}(\lambda)

with the scaled

\lambda

; for a sum, pool the rates BEFORE computing. Keep

e^{-\lambda}

symbolic until the end. (9709/62/M/J/23 Q4)

Common mistake

Forgetting to scale

\lambda

to the interval (the #1 error). Mind

P(X>k)

(excludes

k

) vs

P(X\ge k)

, and carry

e^{-\lambda}

symbolically — evaluate last.