Poisson distribution is one of the important topics of statistics. In statistics, Poisson distribution is one of the discrete probability distribution. This distribution is used for calculating the possibilities for an event with the given average rate of value ie $\lambda$. A Poisson random variable, x, refers to the number of success in a Poisson experiment. Poisson distribution is a limiting process of binomial distribution. Poisson distribution occurs when there are events which do not occur as outcomes of a definite number of outcomes.

Poisson distribution is used under the following conditions:

- Number of trials 'n' tends to infinity
- Probability of success 'p' tends to zero and
- np = 1 is finite.

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A Poisson distribution is the probability distribution that results from a Poisson experiment. A Poisson experiment is a statistical experiment that can be classified as successes or failures.

f(x) = $\frac{e^{-\lambda}\lambda^x }{x!}$

where,

λ is an average rate of value.

x is a Poisson random variable.

e is the base of logarithm.

### Poisson Distribution Function

Probability distribution function for X is

f(x) = $\frac{e^{-\lambda}\lambda^x }{x!}$, where x = 0, 1, 2, 3, .................

Let X be discrete random variable that can take on the values 0, 1, 2,....... such that the probability function of X is given by

f(x) = P(X = x) = $\frac{\lambda^x e^{-1}}{x!}$, x = 0, 1, 2,..................

Where $\lambda$ is given positive constant.

### Continuous Poisson Distribution

If the observed random variable is continuous, the number of the possible values is infinitely large.The Poisson distribution for continuous event numbers X. This can be achieved by replacing the factorial by the gamma-function.

$P(X)$ = $\frac{\lambda^x e^{-1}}{x!}$ = $\frac{\lambda^x e^{-1}}{\gamma x}$

A random variable is said to have a Poisson distribution with parameter $\lambda$, where $\lambda$ is expected value of the Poisson distribution.

Expected value of the Poisson distribution

E(X) = $\mu$ = $\frac{d(e^{\lambda(t-1)})}{dt}$ at t = 1

= $e^{\lambda(t-1)}$ $\lambda$ at t = 1

= $\lambda$

The expected value (mean) and variance of the Poisson distribution is equal to $\lambda$.

P(x, $\lambda$) = $\frac{e^{- \lambda} \lambda^x}{x!}$

where x is the actual number of successes that result from the experiment.

$\lambda$ = the mean, the expected number of occurrence over the given span

e = is a constant, 2.71828

### Poisson Distribution Variance

In Poisson distribution, the mean and the variance are equal. The mean of the distribution is numerically equal to its variance.

### Solved Example

**Question: **Show that in Poisson distribution mean and variance are equal.

** Solution: **

**Step 1:**

**Mean**

skipped

### Solved Example

**Question: **If the variance of the Poisson distribution is 2, find the probabilities for r = 1, 2, 3, 4 and 5 from the recurrence relation of the Poisson distribution.

** Solution: **

### Solved Examples

**Question 1: **A random variable X has a Poisson distribution with parameter l such that P (X = 1) = (0.2) P (X = 2). Find P (X = 0).

** Solution: **

**Question 2: **

** Solution: **

=> P(N = 2) = $\frac{e^{-2}2^2}{2!}$

= 2$e^{-2}$

Step 2:

Let M be the number of minute, among the 5 min considered, during which exactly 2 calls will be received. M follows a binomial distribution with parameters n = 5 and p = 2$e{-2}$.

P(M = 5) = C(5, 5)$(2e^{-2})^5$$(1 - 2e^{-2})^{5 - 5}$

= 32 * $e^{-10}$

≈ 0.00145.

where,

λ is an average rate of value.

x is a Poisson random variable.

e is the base of logarithm.

Probability distribution function for X is

f(x) = $\frac{e^{-\lambda}\lambda^x }{x!}$, where x = 0, 1, 2, 3, .................

Let X be discrete random variable that can take on the values 0, 1, 2,....... such that the probability function of X is given by

f(x) = P(X = x) = $\frac{\lambda^x e^{-1}}{x!}$, x = 0, 1, 2,..................

Where $\lambda$ is given positive constant.

If the observed random variable is continuous, the number of the possible values is infinitely large.The Poisson distribution for continuous event numbers X. This can be achieved by replacing the factorial by the gamma-function.

$P(X)$ = $\frac{\lambda^x e^{-1}}{x!}$ = $\frac{\lambda^x e^{-1}}{\gamma x}$

We shall now deduce the Poisson distribution from the binomial distribution by assuming that n ® ¥ and p ® 0 such that the product np always remains finite, say l.

We shall now use a very important result of limits in Calculus. We state this result without proof:

where, e is a constant lying between the number 2 and 3 and e^{x} is defined by

with x a real number. From equation (1), we observe that each of (r - 1) factors,

that

Thus,

where, l is a finite number and is equal to np.

The sum of the probabilities P(X = r) or simply P(r) for r = 0, 1, 2, … is 1. This can be seen by putting r = 0, 1, 2, … in (4) and adding all the probabilities.

Also, each of the probabilities is a non-negative fraction. This leads to the distribution defined below:

A random variable X taking values 0, 1, 2, … is said to have a Poisson distribution with parameter l (finite), if its probability distribution is given by

There are many daily life situations where n is very large and p is very small. In such situations, the Poisson distribution can be more conveniently used as an approximation to binomial distribution which may prove cumbersome for large values of n. This is called Poisson approximation to binomial distribution.

The probability mass function of the Poisson distribution given by

**Note:**

= e^{-1} e^{1}^{}

= 1

A random variable is said to have a Poisson distribution with parameter $\lambda$, where $\lambda$ is expected value of the Poisson distribution.

Expected value of the Poisson distribution

E(X) = $\mu$ = $\frac{d(e^{\lambda(t-1)})}{dt}$ at t = 1

= $e^{\lambda(t-1)}$ $\lambda$ at t = 1

= $\lambda$

The expected value (mean) and variance of the Poisson distribution is equal to $\lambda$.

The compound Poisson distributions are object of study in probability, it has a number of useful properties. Formulae for the exponential and based on the aggregate claims simplify because mean and variance both are same and P(t) = exp($\lambda(e^t - 1))$. For large $\lambda$ the distribution of the compound Poisson can be approximated by a normal distribution with $\lambda$ E(x) and variance $\lambda E(x^2)$. And the quantile premium is given by:

=> P(a) = $\lambda$ E(x) + $\pi ^{-1}(1 - a)\sqrt{\lambda E(x^2)}$, a lies between (0, 1).

As with the binomial distribution there is a table that we can use under certain conditions that will make calculating probabilities a little easier when using the Poisson Distribution. Table is showing tabulated values of f(x) = P(X $\geq$ x), where X has a Poisson distribution with parameter $\lambda$. The table displays values of Poisson distribution.Cumulative Poisson Distribution is defined as the measure of the probability in which the average number of success estimated within a range of experiments. A cumulative Poisson distribution is used to calculate the probability of getting atleast n successes in a Poisson experiment.

A Poisson distribution is the probability distribution that results from a Poisson experiment. Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is $\lambda$. In Poisson distribution the mean of the distribution is represented by $\lambda$. Then, the Poisson probability is:Formula for cumulative Poisson distribution

F(x, $\lambda$) = $\sum_{n = 0}^x$$ \frac{e^{- \lambda} \lambda ^x}{n!}$.

P(x, $\lambda$) = $\frac{e^{- \lambda} \lambda^x}{x!}$

where x is the actual number of successes that result from the experiment.

$\lambda$ = the mean, the expected number of occurrence over the given span

e = is a constant, 2.71828

In Poisson distribution, the mean and the variance are equal. The mean of the distribution is numerically equal to its variance.

= e^{-1} e^{1}

= l**Step 2:Variance**

=> V(x) = $\lambda$

Confidence intervals for the probability that the number of occurrences in a specific period will be less than or equal to a specific value. Confidence interval for $\lambda$, the mean occurrence rate of the sampled population using the relationship between the chi-square and Poisson distributions.

=> $\frac{\chi^2(\frac{\alpha}{2}, 2k)}{2}$ $\leq$ $\lambda$ $\leq$ $\frac{\chi^2(1-\frac{\alpha}{2}, 2k + 2)}{2}$.

=> $\frac{\chi^2(\frac{\alpha}{2}, 2k)}{2}$ $\leq$ $\lambda$ $\leq$ $\frac{\chi^2(1-\frac{\alpha}{2}, 2k + 2)}{2}$.

Where, k = number of event occurrences in a given interval

$\chi^2$ = chi square deviate with lower area and degree of freedom.

A Poisson process is a particular random process in time is associate with two distributions. First is Poisson distribution and second is exponential distribution.

The Poisson distribution P_{n}(t) is the probability that after the elapse of time t for n events.

$P_n(t)$ = $\frac{e^{- \lambda t}(\lambda t)^n}{n!}$

In particular,

$P_0(t)$ = $e^{-\lambda t}$

If interval [0 t] is subdivided in small segments of size dt, the event E should not happen. The chance for no event in a single interval is 1 - $\lambda$ t. Therefore the chance it does not happen in any interval is

A random variable is said to have a Uniform distribution if the probability distribution function f (x) = 1/a - b , where x = a, a + h, a + 2h,…, a + (k - 1)h where a and h being fixed real numbers and k a fixed positive integer. This distribution will occur in practice if under the given experimental conditions, the different values of the random variable happen to be equally likely.$P_0(t) = (1 - \lambda dt)^N$, N = number of intervals with N = $\frac{t}{dt}$

When dt -> 0

$P_0(t) = \lim_{dt -> 0}(1 - \lambda dt)^{\frac{t}{dt}}$

= $e^{- \lambda t}$

Now, P(t)dt is the probability the event will happen in the small interval of the time [t, t + dt].

P(t) = $\lambda e^{\lambda t}$, where P(t) is the exponential distribution.

Let us derive a formula for P_{n}(t + dt) is the chance for n events to happen at time t + dt.

$P_n (t + dt) = P_n(t)(1 - \lambda dt) + P_{n - 1}(t) \lambda dt$

= $P_n(t)$ + $\frac{dP_n(t)}{dt}$ dt

$\frac{dP_n(t)}{dt}$ dt = $\lambda(-P_n(t) + P_{n - 1}(t)) dt$

$\frac{dP_n(t)}{dt}$ = $\lambda(-P_n(t) + P_{n - 1}(t))$

$\frac{dP_n(t)}{dt}$ = $\lambda( P_{n - 1}(t) - P_n(t))$

$\frac{dP_n(t)}{dt}$ = $\lambda( P_{n - 1}(t) - P_n(t))$

This is a recursive differential equation for P_{n}(t).

skipped

A gamma distribution is a type of statistical distribution arises naturally in processes for which the waiting times between Poisson distributed events are relevant. Gamma Poisson distribution is useful when the data is a combination of several Poisson distributions, each with a different $\lambda$.

### Truncated Poisson Distribution

Poisson distribution is truncated because there is no observations are available. For r $\geq$ 0 truncated Poisson distribution is defined as

Poisson distribution can be calculated by using Poisson distribution formula. Let us see with the help of example how to calculate Poisson distribution.Poisson distribution is truncated because there is no observations are available. For r $\geq$ 0 truncated Poisson distribution is defined as

P_{r} = $\frac{\lambda^r e^{- \lambda}}{r!}$, r = 1, 2, 3, ....................

### Zero Inflated Poisson Distribution

The zero inflected Poisson distribution is a Poisson related distribution with probability

density function P(x, $\theta, \lambda$) is given by

For x = 0

P(x, $\theta, \lambda$) = $\theta + (1 - \theta) e^{- \lambda}$

and for x = 1, 2, 3,................

P(x, $\theta, \lambda$) = $(1 - \theta) e^{- \lambda}$$ \frac{\lambda^x}{x!}$

The variance of the Poisson distribution = l = 2

Recurrence relation is given by

Given below are examples on Poisson distribution for your better understanding:

For the Poisson distribution, the probability function is given by

Given P (x = 1) = (0.2) P (X = 2)

Telephone calls arrive at an exchange according to Poisson process with rate $\lambda$ = 2/min. Calculate the probability that exactly 2 calls will be received during each of the first 5 min of the hour.

**Step 1:**

=> P(N = 2) = $\frac{e^{-2}2^2}{2!}$

= 2$e^{-2}$

Step 2:

Let M be the number of minute, among the 5 min considered, during which exactly 2 calls will be received. M follows a binomial distribution with parameters n = 5 and p = 2$e{-2}$.

P(M = 5) = C(5, 5)$(2e^{-2})^5$$(1 - 2e^{-2})^{5 - 5}$

= 32 * $e^{-10}$

≈ 0.00145.

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