Home page for accesible maths 5 Models for discrete random variables 5.5 Geometric random variables 5.6.1 Poisson approximation to the Binomial

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5.6 Poisson random variables

Unlike the other random variables discussed here, we define the Poisson random variable directly through its pmf:

The pmf of a Poisson random variable ${R}$ is $\displaystyle p_{R}(r)=\frac{\lambda^{r}\exp(-\lambda)}{r!},$ for $r=0,1,2,\dots$ , with $p_{R}(r)=0$ otherwise, where the parameter ${\lambda>0}$ . We say ${R\sim{\rm Pois}(\lambda)}$ .

The Poisson random variable arises physically in two ways:

1.

The number of events in a fixed time interval of a continuous time process where events occur at random at a given rate over time. (Covered in later courses.)
2.

The number of successes when the probability of success is very rare.

Examples of Poisson random variables are

•

the number of raindrops to land on your head in a given time interval,
•

the number of floods of a river in a year,
•

the number of deaths due to typhoid over a year, (assuming typhoid cases are independent),
•

the number of hits on a website in a given period of time.

Exercise 5.19.

Calculate the probabilities of ${0}$ , ${1}$ and ${2}$ deaths from typhoid in a year if the number ${R}$ has a Poisson random variable with ${\lambda=4.6}$ .

R hint: dpois(0:2,lambda=4.6)

Solution.

$\displaystyle p_{R}(r)$	$\displaystyle=$	$\displaystyle\frac{\lambda^{r}\exp(-\lambda)}{r!}$
$\displaystyle p_{R}(0)$	$\displaystyle=$	$\displaystyle\exp(-\lambda)$
	$\displaystyle=$	$\displaystyle\exp(-4.6)=0.01,$
$\displaystyle p_{R}(1)$	$\displaystyle=$	$\displaystyle\lambda\exp(-\lambda)$
	$\displaystyle=$	$\displaystyle 4.6\exp(-4.6)=0.046,$
$\displaystyle p_{R}(2)$	$\displaystyle=$	$\displaystyle\lambda^{2}\exp(-\lambda)/2$
	$\displaystyle=$	$\displaystyle 4.6^{2}\exp(-4.6)/2=0.106.$

Example 5.20.

Verify that ${p_{R}(r)=\frac{\lambda^{r}\exp(-\lambda)}{r!}}$ is a proper probability mass function. This proof makes use of the definition of the exponential function in terms of its series expansion.

Solution.

For each $r\geq 0$ ,

	$\displaystyle p_{R}(r)$	$\displaystyle=$	$\displaystyle\frac{\lambda^{r}\exp(-\lambda)}{r!}$
		$\displaystyle\geq$	$\displaystyle 0.$

$\displaystyle\sum_{r=0}^{\infty}p_{R}(r)$	$\displaystyle=$	$\displaystyle\sum_{r=0}^{\infty}\frac{\lambda^{r}\exp(-\lambda)}{r!}$
	$\displaystyle=$	$\displaystyle\exp(-\lambda)\sum_{r=0}^{\infty}\frac{\lambda^{r}}{r!}$
	$\displaystyle=$	$\displaystyle\exp(-\lambda)\exp(\lambda)$
	$\displaystyle=$	$\displaystyle 1.$

Exercise 5.21.

If ${R\sim{\rm Pois}(\lambda)}$ show that ${{\rm E}(R)=\lambda}$ . The technique is similar to that in the above example.

Solution.

$\displaystyle{\rm E}(R)$	$\displaystyle=$	$\displaystyle\sum_{r=0}^{\infty}r\,p(r)$
	$\displaystyle=$	$\displaystyle\sum_{r=0}^{\infty}r\,\frac{\lambda^{r}\exp(-\lambda)}{r!}$
	$\displaystyle=$	$\displaystyle 0+\sum_{r=1}^{\infty}r\,\frac{\lambda^{r}\exp(-\lambda)}{r!}$
	$\displaystyle=$	$\displaystyle\lambda\exp(-\lambda)\sum_{r=1}^{\infty}\frac{\lambda^{r-1}}{(r-1% )!}$
	$\displaystyle=$	$\displaystyle\lambda\exp(-\lambda)\exp(\lambda)$
	$\displaystyle=$	$\displaystyle\lambda.$

Example 5.22.

Find the variance of the Poisson rv by first computing ${{\rm E}[R(R-1)]}$ .

Solution.

$\displaystyle{\rm E}[R(R-1)]$	$\displaystyle=$	$\displaystyle\sum_{r=0}^{\infty}r(r-1)\,p_{R}(r)$
	$\displaystyle=$	$\displaystyle 0+0+\sum_{r=2}^{\infty}r(r-1)\,p_{R}(r)$
	$\displaystyle=$	$\displaystyle\sum_{r=2}^{\infty}r(r-1)\,\frac{\lambda^{r}\exp(-\lambda)}{r!}$
	$\displaystyle=$	$\displaystyle\lambda^{2}\exp(-\lambda)\sum_{r=2}^{\infty}\frac{\lambda^{r-2}}{% (r-2)!}$
	$\displaystyle=$	$\displaystyle\lambda^{2}\exp(-\lambda)\exp(\lambda)=\lambda^{2}.$

Hence

$\displaystyle{\rm Var}(R)$	$\displaystyle=$	$\displaystyle{\rm E}[R(R-1)]+{\rm E}[R]-({\rm E}[R])^{2}$
	$\displaystyle=$	$\displaystyle\lambda^{2}+\lambda-(\lambda)^{2}$
	$\displaystyle=$	$\displaystyle\lambda.$

Thus a key property of the Poisson pmf is that the expectation and variance are both equal to ${\lambda}$ .

Exercise 5.23.

Use R to give a barplot of the Poisson pmf on ${0,1,\dots,7}$ , when ${\lambda=0.5}$ and when ${\lambda=3}$ .

par(mfrow=c(1,2))
barplot( dpois(0:7,lambda=0.5),names.arg=c(0:7),ylim=c(0,1) )
barplot( dpois(0:7,lambda=3),names.arg=c(0:7),ylim=c(0,1)  )

5.6.1 Poisson approximation to the Binomial