POISSON DISTRIBUTION
DEFINITION
The Poisson distribution is a discrete distribution under which the random variable counts the number of events occurring in a specified time interval. The Poisson distribution has the parameter $$\lambda$$ , which indicates the mean rate of occurrence of events.
The Poisson distribution can also be considered as the limiting distribution of a sequence of Binomial distributions as $$n\longrightarrow \infty$$ (n tends to infinity) and $$\theta \longrightarrow 0$$ such that $$n\theta$$ is held constant at $$\lambda$$.
Apart from being used to model the rate of occurrence of an event, the Poisson distribution is used to model the waiting time between events occurring as a Poisson process.

CALCULATOR

#### Enter the following details:

Mean rate of occurence:

Number of occurrences required:

Probability of occurences:

Probability of or less occurences:

Probability of more than occurences:

Mean of the specified poisson distribution:

Variance of the specified poisson distribution:

Probability generating function:

Moment generating function:

Cumulant generating function:

Probability distribution
 x p(X=x) P(X ≤ x)
FORMULA AND DERIVATION
Probability distribution:-

The Poisson distribution can be derived from the exponential function.
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\dots \dots \$
or
$e^{\lambda }=1+\lambda +\frac{{\lambda }^2}{2!}+\frac{{\lambda }^3}{3!}\dots \dots \dots$
Dividing both sides by $$e^{\lambda }$$, we get
$\frac{e^{\lambda }}{e^{\lambda }}=\ \frac{1}{e^{\lambda }}+\frac{\lambda }{e^{\lambda }}+\frac{{\lambda }^2}{e^{\lambda }2!}+\frac{{\lambda }^3}{e^{\lambda }3!}\dots \dots \dots \$ $1=\ \frac{{\lambda }^0e^{-\lambda }}{0!}+\frac{{\lambda }^1e^{-\lambda }}{1!}+\frac{{\lambda }^2e^{-\lambda }}{2!}+\frac{{\lambda }^3e^{-\lambda }}{3!}\dots \dots \dots$ $1=\ \sum^n_{x=0}{\frac{{\lambda }^xe^{-\lambda }}{x!}}$
Since the RHS sums up to 1, we can consider this as a probability function. Thus for each value of $$x=0,1,2,3\dots \dots$$ the probability function for the Poisson distribution can be written as follows:-
$P\left(X=x\right)=\ \frac{{\lambda }^xe^{-\lambda }}{x!}$
where, $$\lambda$$ is the mean rate of occurrence of an event.

Mean:-

The mean or expected value for the Poisson distribution can be derived from first principles as follows:-
$E\left(X\right)=\mu =\ \sum{xP\left(X=x\right)}$ $\mu =\ \sum{x.\frac{{\lambda }^xe^{-\lambda }}{x!}}$ $\ \ \ =\ \left(\frac{0\times \ {\lambda }^0e^{-\lambda }}{0!}\right)+\ \left(\frac{1\times {\lambda }^1e^{-\lambda }}{1!}\right)+\ \left(\frac{2\times \ {\lambda }^2e^{-\lambda }}{2!}\right)+\ \left(\frac{3\times {\lambda }^3e^{-1}}{3!}\right)\dots$ $\ \ \ =0+\lambda e^{-\lambda }+\frac{2{\lambda }^2e^{-\lambda }}{2\times 1!}+\ \frac{3{\lambda }^3e^{-1}}{3\times 2!}+\ \frac{4{\lambda }^4e^{-\lambda }}{4\times 3!}\dots \dots \dots$ $\ \ \ =0+\ {\lambda e}^{-\lambda }+\ {\lambda }^2e^{-\lambda }+\ \frac{{\lambda }^3e^{-\lambda }}{2!}+\ \frac{{\lambda }^4e^{-\lambda }}{3!}\dots \dots \dots \$
Taking $${\lambda e}^{-\lambda }$$ as a common term, we get
$\ \ \ =\ {\lambda e}^{-\lambda }\left[1+\lambda +\frac{{\lambda }^2}{2!}+\ \frac{{\lambda }^3}{3!}+\frac{{\lambda }^4}{4!}\dots \dots \dots \right]$
The terms in the bracket form the series expansion of the exponential function
$\ \ \ =\ {\lambda e}^{-\lambda }\left[e^{\lambda }\right]$ $\ \ \ =\lambda$
$$\lambda$$, the parameter of the distribution is itself the mean of the distribution. This is intuitive, because λ represents the mean rate of occurrence.

Variance:-

Variance of the Poisson distribution can be derived from first principles using the formula:
$$Var\left(X\right)=E\left[{\left(x-\mu \right)}^2\right]=\sum{{\left(x-\mu \right)}^2P\left(X=x\right)}$$
or, using a simpler formula:
$$Var\left(X\right)=E\left(X^2\right)-E^2\left(X\right)$$

$$E\left(X^2\right)$$ can be calculated as follows:-
$E\left(X^2\right)=\ \sum{x^2p\left(X=x\right)}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ =\ \left(\frac{0^2\times {\lambda }^0e^{-\lambda }}{0!}\right)+\ \left(\frac{1^2\times {\lambda }^1e^{-\lambda }}{1!}\right)+\ \left(\frac{2^2\times \ {\lambda }^2e^{-\lambda }}{2!}\right)\dots \dots \dots$ $\ \ \ \ \ \ \ \ \ \ \ \ \ =0+\ {\lambda e}^{-\lambda }+\ \frac{{4\lambda }^2e^{-\lambda }}{2!}+\ \frac{{9\lambda }^3e^{-\lambda }}{3!}+\ \frac{16{\lambda }^4e^{-\lambda }}{4!}\dots \dots \dots \$ $\ \ \ \ \ \ \ \ \ \ \ \ \ =\ {\lambda e}^{-\lambda }\left[1+\frac{4\lambda }{2!}+\ \frac{9{\lambda }^2}{3!}+\ \frac{16{\lambda }^3}{4!}\dots \dots \dots \right]$ $\ \ \ \ \ \ \ \ \ \ \ \ \ =\ {\lambda e}^{-\lambda }\left[1+2\lambda +\ \frac{3{\lambda }^2}{2!}+\ \frac{4{\lambda }^3}{4!}\dots \dots \dots \right]$ $\ \ \ \ \ \ \ \ \ \ \ \ \ =\ {\lambda e}^{-\lambda}\left[1+\lambda +\lambda +\frac{{\lambda }^2}{2!}+\frac{{2\lambda }^2}{2!}+\frac{{\lambda }^3}{3!}+\ \frac{3{\lambda }^3}{3!}\dots \dots \dots \right]$ $\ \ \ \ \ \ \ \ \ \ \ \ \ =\ {\lambda e}^{-\lambda}\left[1+\lambda +\frac{{\lambda }^2}{2!}+\ \frac{{\lambda }^3}{3!}\dots \dots +\ \lambda +\ \frac{{2\lambda }^2}{2!}+\frac{{3\lambda }^3}{3!}\dots \dots \right]$ $\ \ \ \ \ \ \ \ \ \ \ \ \ =\ {\lambda e}^{-\lambda }\left[e^{\lambda }+\ \lambda \left(1+\lambda +\frac{{\lambda }^2}{2!}\dots \dots \dots \right)\right]$ $\ \ \ \ \ \ \ \ \ \ \ \ \ =\lambda e^{-\lambda }\left[e^{\lambda }+{\lambda e}^{\lambda }\right]$ $\ \ \ \ \ \ \ \ \ \ \ \ \ =\ {\lambda e}^{-\lambda }e^{\lambda \ }+\ {\lambda e}^{-\lambda }{\lambda e}^{\lambda }$ $\ \ \ \ \ \ \ \ \ \ \ \ \ =\ \lambda +{\lambda }^2$
Now, substituting this in the formula for variance, we get
$Var\left(X\right)=\ {\sigma }^2=E\left(X^2\right)-E^2\left(X\right)$ ${\sigma }^2=\ \lambda +{\lambda }^2-{\lambda }^2$ $\ \ \ \ =\ \lambda$
The variance of the Poisson distribution is the same as its mean.

Probability Generating Function(PGF):-

The PGF for a Poisson distribution is
$G_X\left(t\right)=e^{\lambda \left(t-1\right)}$
Moment Generating Function(MGF):-

The MGF for a Poisson distribution is
$M_X\left(t\right)=\ e^{\lambda \left(e^t-1\right)}$
Cumulant Generating Function(CGF):-

The CGF for a Poisson distribution is
$C_X\left(t\right)=\ \lambda \left(e^t-1\right)$
EXAMPLES
Example 1

A person is counting the number of phone calls he gets every day from his friend, for three months. He calculates that on average, he gets 3 calls per day from his friend. On one particular day, he wishes to find out the probability that he would get 5 calls from his friend that day.

Here, the number of phone calls arriving can be modelled using the Poisson distribution, where the mean rate of occurrence is 3 calls per day.

So, $$\lambda =3$$

The probability of getting 5 calls would be:-
$P\left(X=x\right)=\ \frac{{\lambda }^xe^{-\lambda }}{x!}$ $P\left(X=5\right)=\ \frac{3^5e^{-3}}{5!}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.1008$
Example 2

An accident zone is being closely monitored for a few weeks and it has been found that accidents occur at an average of 2 per hour. People residing near this zone wish to find out the probability that there will be less than two accidents in any given hour.

Here, the mean rate of occurrence is 2 per hour.
So, $$\lambda =2$$

Less than two accidents could mean 1 accident or no accident.
Therefore, the required probability is the probability that there will be 0 or 1 accident.
$P\left(X<2\right)=P\left(X=0\right)+P\left(X=1\right)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{2^0e^{-2}}{0!}+\frac{2^1e^{-2}}{1!}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.406$
RELATED TOPICS
Binomial distribution