Now we will learn Poisson distribution. In the previous article we learn Binomial distribution. So let’s start.

Poisson Distribution

Definition

Let X be a discrete random variable assuming the possible values 0, 1, …, n, …
\[\text{If}~P\left( X=k \right)=\frac{{{e}^{-\alpha }}{{\alpha }^{k}}}{k!},k=0,1,…,n,…\]
We say that X has a Poisson distribution with parameter α > 0.

Note: Let X be a binomially distributed variable with parameter p (based on n repetition of an experiment). That is
\[P\left( X=k \right)={}^{n}{{C}_{k}}{{p}^{k}}{{\left( 1-p \right)}^{n-k}}\]
Suppose that as n→∞, np=α (constant), or equivalently, as n→∞, p→0 such that np→α
Under these conditions we have, the Poisson distribution with parameter α
\[\underset{n\to \infty }{\mathop{\lim }}\,P\left( X=k \right)=\frac{{{e}^{-\alpha }}{{\alpha }^{k}}}{k!}\]

Mean

Let X is a random variable following Poisson distribution with parameter α then the mean of X is defined as
\[E\left( X \right)=\sum\limits_{k=0}^{\infty }{k.P\left( X=k \right)=}\sum\limits_{k=0}^{\infty }{k.\frac{{{e}^{-\alpha }}{{\alpha }^{k}}}{k!}}\]
\[=\alpha {{e}^{-\alpha }}\sum\limits_{k=1}^{\infty }{\frac{{{\alpha }^{k-1}}}{\left( k-1 \right)!}}=\alpha {{e}^{-\alpha }}.{{e}^{\alpha }}=\alpha \]

Variance

\[V\left( X \right)=E\left( {{X}^{2}} \right)-{{\left[ E\left( X \right) \right]}^{2}}\]
\[E\left( {{X}^{2}} \right)=E\left( X\left( X-1 \right) \right)+E\left( X \right)\]
\[=\sum\limits_{k=0}^{\infty }{k\left( k-1 \right)}\frac{{{e}^{-\alpha }}{{\alpha }^{k}}}{k!}+\alpha \]
\[={{\alpha }^{2}}{{e}^{-\alpha }}\sum\limits_{k=2}^{\infty }{\frac{{{\alpha }^{k-2}}}{\left( k-2 \right)!}+\alpha }\]
\[\therefore E\left( {{X}^{2}} \right)={{\alpha }^{2}}+\alpha \]
\[\therefore V\left( X \right)={{\alpha }^{2}}+\alpha -{{\alpha }^{2}}=\alpha \]

Mode

We have to find that value of the random variable X following Poisson distribution for which the probability of occurrence is maximum. Thus, if X=k is the modal value its definition provides
\[P\left( X=k \right)\ge P\left( X=k+1 \right)~\text{and}~P\left( X=k \right)\ge P\left( X=k-1 \right)\]
\[P\left( X=k \right)\ge P\left( X=k+1 \right)~\Rightarrow k+1\le \alpha \]
\[P\left( X=k \right)\ge P\left( X=k-1 \right)\Rightarrow k\ge \alpha \]
Combining both the equations
\[\alpha -1\le k\le \alpha \]
Case1. Let α be a positive integer then k can take two values α and α – 1 and the distribution will have two modes.
Case2. Let α be a fraction then k is the greatest integer less than α.

Moment generating function (mgf)

Let X is a Poisson random variable with the parameter α its moment generating function is defined as
\[{{M}_{X}}\left( t \right)=E\left( {{e}^{tX}} \right)=\sum\limits_{k=0}^{\infty }{{{e}^{tk}}.\frac{{{e}^{-\alpha }}{{\alpha }^{k}}}{k!}}\]
\[={{e}^{-\alpha }}\sum\limits_{k=0}^{\infty }{\frac{{{\left( \alpha t \right)}^{k}}}{k!}}={{e}^{-\alpha }}{{e}^{\alpha {{e}^{t}}}}\]
\[\therefore {{M}_{X}}\left( t \right)={{e}^{\alpha \left( {{e}^{t}}-1 \right)}}\]

Suppose that a container contains 10,000 particles. The probability that such a particle escapes from the container equals 0.0004. What is the probability that more than 5 such escapes occur?

Solution:

Let X = Number of particles escaping from the container
n = 10,000, p = 0.0004, α = np = 4
\[P\left( \text{more than 5 particles escapes} \right)=P\left( X>5 \right)\]
\[=1-P\left( X\le 5 \right)=1-\sum\limits_{k=0}^{5}{\frac{{{e}^{-4}}{{4}^{k}}}{k!}}\]
\[=1-\left[ \frac{{{e}^{-4}}{{4}^{0}}}{0!}+\frac{{{e}^{-4}}{{4}^{1}}}{1!}+\frac{{{e}^{-4}}{{4}^{2}}}{2!}+\frac{{{e}^{-4}}{{4}^{3}}}{3!}+\frac{{{e}^{-4}}{{4}^{4}}}{4!}+\frac{{{e}^{-4}}{{4}^{5}}}{5!} \right]\]

An insurance company has discovered that only about 0.1% of the population is involved in a certain type of accident each year. If its 10,000 policy holders were randomly selected from the population, what is the probability that not more than 5 of its clients are involved in such an accident next year?

Solution:

X = Number of people involved in an accident = 0, 1, 2, …, 10000
n = 10000, p = 0.001, α = np = 10
\[P\left( \text{not more than 5 people involved in an accident} \right)=P\left( X\le 5 \right)\]
\[=\sum\limits_{k=0}^{5}{\frac{{{e}^{-10}}{{10}^{k}}}{k!}}\]
\[=\frac{{{e}^{-10}}{{10}^{0}}}{0!}+\frac{{{e}^{-10}}{{10}^{1}}}{1!}+\frac{{{e}^{-10}}{{10}^{2}}}{2!}+\frac{{{e}^{-10}}{{10}^{3}}}{3!}+\frac{{{e}^{-10}}{{10}^{4}}}{4!}+\frac{{{e}^{-10}}{{10}^{5}}}{5!}\]

In a certain factory producing blades there is a small chance 1/500 for any blades to be defective. The blades are supplied in packets of 10. Calculate the approximate number of packets containing (i) No defective baldes (ii) one defective in a consignment of 20000 packets.

Solution:

Do it yourself.

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