Home page for accesible maths 3.1 Discrete Probability Models 3.1.4 Geometric Random Variables:

\operatorname{\mathsf{Geom}}(\theta)

{\operatorname{\mathsf{Unif}}}(a,b)

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

3.1.5 Poisson Random Variables: $\operatorname{\mathsf{Poisson}}(\lambda)$

In a Poisson process events occur independently at random times with a given average rate, $\phi>0$ per time unit. For example: the times of raindrops hitting your umbrella during a rain shower, the times of hits on a website or of arrivals at A&E. Here $\phi$ represents the intensity of the rain, hits or arrivals – the expected number of raindrops to hit your umbrella in a second, or hits on the website per second or arrivals at A&E per hour.

The Poisson process itself is studied rigorously and in detail in the third year module: Stochastic processes. In this module we will meet three distributions which arise from it.

The Poisson distribution is the distribution of the number of events from a Poisson process in an interval $[0,t]$ .

The sample space is $\{0,1,\ldots\}$ , and

\displaystyle p(r;\theta)=\frac{\lambda^{r}\exp(-\lambda)}{r!}

for $r=0,1,2,\dots$ , where $\theta=\lambda=\phi t$ and (see first year probability for proof)

$\operatorname{\mathsf{E}}\left[{R}\right]=\lambda$ ,
${\operatorname{\mathsf{Var}}}\left[{R}\right]=\lambda$ .

The Poisson distribution also has the interpretation as the limit distribution of a Binomial distribution with $n\to\infty$ and $n\theta\to\lambda$ (see first year probability for proof).

Example 3.1.1.

If $X$ is a Poisson random variable with expectation $\lambda$ , find the probability that $X$ is greater than 1 given that it is greater than 0.

Solution.

$\operatorname{\mathsf{P}}\left({X>1|X>0}\right)={\color[rgb]{0.76,0.01,0}% \operatorname{\mathsf{P}}\left({X>1\cap X>0}\right)/\operatorname{\mathsf{P}}% \left({X>0}\right)}$
$X>1\cap X>0={\color[rgb]{0.76,0.01,0}X>1}$
$\operatorname{\mathsf{P}}\left({X>0}\right)={\color[rgb]{0.76,0.01,0}1-% \operatorname{\mathsf{P}}\left({X=0}\right)=1-\exp(-\lambda)}$

	$\displaystyle\operatorname{\mathsf{P}}\left({X>1}\right)$	$\displaystyle={\color[rgb]{0.76,0.01,0}1-\operatorname{\mathsf{P}}\left({X=0% \cup X=1}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1-\exp(-\lambda)-\lambda\exp(-\lambda)}$

	$\displaystyle\operatorname{\mathsf{P}}\left({X>1\|X>0}\right)$	$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1-\exp(-\lambda)-\lambda\exp(-% \lambda)}{1-\exp(-\lambda)}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1-\frac{\lambda}{\exp(\lambda)-1}.}$

Example 3.1.2.

The probability that a randomly chosen electrical component is defective is 0.002. Assume that this probability is the same for all components that are manufactured, and they fail independently of one another.

(a)

What is the distribution of the number of defectives in a batch of size 1200? What is the probability of at most one defective?
(b)

Which other distribution can we use to approximate this? How well does it estimate the probability of at most one defective?
(c)

Your colleague tells you he tested the whole batch and found one defective. As far as you are concerned, what is the distribution of the number he had tested before he found this defective?
(d)

Imagine, now, an unending sequence of components, not just 1200. What is the distribution of the number of non-defectives before the first defective is found?

Solution.

(a)

$R\sim\operatorname{\mathsf{Bin}}(1200,0.002)$ . ${\color[rgb]{0.76,0.01,0}0.998^{1200}+1200\times 0.002\times 0.998^{1199}% \approx 0.3081.}$
(b)

Use the Poisson approximation to the Binomial, ${\color[rgb]{0.76,0.01,0}R\sim\operatorname{\mathsf{Poisson}}(1200\times 0.002% )=\operatorname{\mathsf{Poisson}}(2.4)}$ , appropriate for rare events. Estimate: $e^{-2.4}+2.4e^{-2.4}=0.3084$ – accurate to 3dp.
(c)

Discrete ${\operatorname{\mathsf{Unif}}}(0,1199)$ .
(d)

Geometric $\operatorname{\mathsf{Geom}}(0.002)$ .

Style control - access keys in brackets

3.1.5 Poisson Random Variables: 𝖯𝗈𝗂𝗌𝗌𝗈𝗇⁡(λ)

Example 3.1.1.

Example 3.1.2.

3.1.5 Poisson Random Variables: $\operatorname{\mathsf{Poisson}}(\lambda)$