# Poisson Distribution

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The Poisson distribution can describe an experiment in which an outcome can be observed a number of times (for example, accidental deaths). The random variable X denotes the outcomes and is discrete in nature. This random variable will be described by a probability density function referred to as a Poisson distribution with parameter ${\displaystyle \lambda }$:

${\displaystyle f_{PO}(x;\lambda )=\left\{{\begin{array}{ll}{\frac {\lambda ^{x}}{x!}}e^{-\lambda }\quad &{\text{for}}\ x=0,1,2,\dots ;\lambda >0\\&\\0\quad &{\text{otherwise}}\end{array}}\right.}$

The distribution function is:

${\displaystyle F_{PO}(x;\lambda )=\left\{{\begin{array}{ll}\sum \limits _{k=0}^{x}{\frac {\lambda ^{x}}{x!}}e^{-\lambda }\quad &{\text{for}}k\geq 0;\lambda >0\\&\\0\quad &{\text{for}}\ k\leq 0\end{array}}\right.}$

The expected value and variance of the Poisson distribution are: ${\displaystyle E(X)=\lambda \quad Var(X)=\lambda .}$ The table presents the Poisson distribution for different values of ${\displaystyle \lambda }$.

Properties of the poisson distribution:

- Reproductivity: Consider two independent variables ${\displaystyle X\sim PO(\lambda _{1})}$ a ${\displaystyle Y\sim PO(\lambda _{2})}$ , then the random variable Z = X+Y is a Poisson distributed with parameter ${\displaystyle {\lambda _{1}+\lambda _{2}}}$: ${\displaystyle Z\sim PO(\lambda _{1}+\lambda _{2})}$

-Poisson distribution for an arbitrary interval length: If the number of outcomes in a single interval is poisson distributed, then the number of outcomes in an interval of length t will also be Poisson distributed with parameter ${\displaystyle \lambda t}$: ${\displaystyle f_{PO}(x;\lambda \cdot t)={\frac {(\lambda t)^{x}}{x!}}e^{-\lambda t}}$

The Poisson distribution is dependent on the parameter ${\displaystyle \lambda ,}$ which influences its shape, position and variance. The following example allows you to explore the effect of the parameter ${\displaystyle \lambda }$ on shape of this distribution. In addition, you can also compute the probabilities associated with outcomes that have a Poisson distribution.

### Extended example of the Poisson distribution

Through experience, the customer service department of a major supermarket knows that it receives on the average, 1 customer per hour between 9 am and 2 pm, and 2 customers per hour between 2 pm and 7 pm. Since a request for service from any customer can be considered to be random, as well as independent of other customer requests, the random variable ${\displaystyle X_{1}}$ = number of customers per hour between 9 am and 2 pm will follow a Poisson distribution and have parameter ${\displaystyle \lambda _{1}}$ = 1. The random variable ${\displaystyle X_{2}}$ = number of customers between 2 pm and 7 pm will also follow a Poisson distribution but with parameter ${\displaystyle \lambda _{2}}$ = 2. For both time intervals, t = 5. Using this information, we can compute the probability of having a certain number of customers between 9 am and 2 pm. For example, if ${\displaystyle X_{1}}$ = 6; ${\displaystyle P(X_{1}=6)=f_{PO}(6;1\cdot 5)={\frac {(\lambda t)^{x}}{x!}}e^{-\lambda t}={\frac {(1\cdot 5)^{6}}{6!}}e^{-1\cdot 5}=0.1462}$ The probability of having more than 4 customers at the customer department will be ${\displaystyle P(X_{1}>4)=1-P(X_{1}\leq 4)=1-e^{-5}\left({\frac {5^{0}}{0!}}+{\frac {5^{1}}{1!}}+{\frac {5^{2}}{2!}}+{\frac {5^{3}}{3!}}+{\frac {5^{4}}{4!}}\right)=1-0.4405=0.5595}$ The Probability Density Function PO(5)

We can also obtain the probabilities for (${\displaystyle X_{2}}$ = 6, ${\displaystyle X_{2}>4}$) between 2 p.m and 7 p.m. ${\displaystyle P(X_{2}=6)=f_{PO}(6;2\cdot 5)={\frac {(\lambda t)^{x}}{x!}}e^{-\lambda t}={\frac {(2\cdot 5)^{6}}{6!}}e^{-2\cdot 5}=0.063}$ ${\displaystyle P(X_{2}>4)=1-P(X_{2}\leq 4)=1-e^{-10}\left({\frac {10^{0}}{0!}}+{\frac {10^{1}}{1!}}+{\frac {10^{2}}{2!}}+{\frac {10^{3}}{3!}}+{\frac {10^{4}}{4!}}\right)=1-0.0293=0.9707}$ Probability Density Function PO(10)

Using these results, we can determine whether the random variables for customer service requests are , i.e., ${\displaystyle X_{1}}$ and ${\displaystyle X_{2},}$ are independent. The probability of receiving more than 4 customers between 9 a.m and 2 p.m and 2 p.m and 7 p.m can be obtained as: ${\displaystyle P(X_{1}>4,X_{2}>4)=P(X_{1}>4)\cdot P(X_{2}>4)=0.5595\cdot 0.9707=0.5431.}$ To obtain the total number of customers between 9 a.m and 7 p.m, we create the random variable ${\displaystyle Y=X_{1}+X_{2}}$. Since ${\displaystyle X_{1}}$ and ${\displaystyle X_{2}}$ are independent, Y will also have a Poisson distribution with parameter ${\displaystyle \lambda _{1}+\lambda _{2}=1+2=3}$.

### An example of the Poisson distribution

A town has 20,000 inhabitants who need to be vaccinated. The probability that the vaccine provokes an adverse reaction in an inoculated person is 0.0001.

In fact, this is a Bernoulli experiment, where; 1. ${\displaystyle A}$ = Occurrence of adverse effect ${\displaystyle {\bar {A}}}$ = No adverse effects from vaccine 2. ${\displaystyle P(A)}$ = 0.0001 is constant. 3. Independence of trials, i.e. of vaccinations. To obtain the probabilities the number of adverse reactions, the binomial distribution could be used. However, the small probability associated with an outcome and the large number of trials suggest that the Poisson distribution could be used as an approximation, since ${\displaystyle n>30}$ and ${\displaystyle p\leq 0..05}$ . ${\displaystyle \lambda =np=20000\cdot 0.0001=2}$ This is the expected number of cases with adverse reactions. The probability density function PO(2) is plotted below:

The probability that no one suffers adverse effects is P(X = 0) = P(X ${\displaystyle \leq }$ 0) = F(0) = 0.1353 The probability that one person has a bad reaction to the vaccination is: P(X = 1) = P(X ${\displaystyle \leq }$ 1) - P(X ${\displaystyle \leq }$ 0) = F(1) - F(0) = 0.2707 The probability that more than 4 person have adverse effects is: P(X > 4) = 1 - F(4) The value of F(4) can be found in the tables for a Poisson distribution for ${\displaystyle \lambda }$ = 2 and X = 4: F(4) = 0.9473 P(X > 4) = 1 - 0.9473 = 0.0527

The following are some examples that would have a Poisson distribution:

• The number of printing defaults per page in books.
• The number of twining breaks a weaving machine in an interval of time.
• The number of incoming calls at a telephone center
• The number of vehicles that drive past an intersection per minute.
• The number of patients arriving at an emergency department per hour.
• The number of alpha-particles emitted by a radioactive substance in a specific time interval.
• The number of fish caught during a day
• The number of reported accidents to an insurance firm per year.
• The number of bank customers applying for credit in a month.

The following assumptions are needed.

• The possibility of occurrence is always based on an interval. The use of an appropriate scale will ascertain that the given size is made up of continuous interval units.
• The occurrence of an outcome is purely random in the sense that it cannot be predetermined.
• The independence of the outcomes, means that an occurrence (or non-occurrence) of an outcome can not influence the occurrence of the same outcome in another trial. Subsequently the number of outcomes in 2 disjoint intervals are independent.
• 2 outcomes can not occur at the same time, i.e. in any arbitrary interval, the possibility of obtaining more than one outcome should be 0.
• The ”intensity” of occurrence of an outcome must be constant with a parameter ${\displaystyle \lambda >0}$, i.e. the average number of outcomes in an interval must be independent of the interval chosen. Consequently, the probability of occurrence in a specific interval will only dependent on the size of the interval.

If these assumptions are true then the variable is described by a Poisson process. The Poisson distribution can also be derived using a binomial distribution using the following assumptions:

• The number of trials; n, is large.
• The probability of occurrence of an outcome A, P(A) = p, in a single trial is very small.
• E(X) = np = ${\displaystyle \lambda }$, then with increasing number of trials n; ((n ${\displaystyle \rightarrow \infty }$)), p will approach zero (p${\displaystyle {\rightarrow }}$0).

Consequently, the Poisson distribution PO(${\displaystyle \lambda }$ = np) can be used to approximate a binomial distribution. With large n and small p the Poisson distribution is often referred to as the distribution of rare occurrences. The rule of thumb for a Poisson approximation for a binomial distribution requires ${\displaystyle n>30}$ and p${\displaystyle \leq }$0,05. The following diagram presents a plots of Poisson probability density functions for ${\displaystyle \lambda }$ = 5 and ${\displaystyle \lambda }$ = 1. The smaller the value of ${\displaystyle \lambda }$, the more the Poisson distribution is skewed to the left. However, as ${\displaystyle \lambda }$ increases density function becomes more symmetric.