Poisson Distribution

Poisson Distribution

Poisson distribution helps to describe the probability of occurrence of a number of events in some given time interval or in a specified region. The time interval may be of any length, such as a minutes, a day, a week etc.

Definition of Poisson Distribution

A discrete random variable $X$ is said to have Poisson distribution with parameter $\lambda$ if its probability mass function is

 \begin{align*} P(X=x)&= \begin{cases} \frac{e^{-\lambda}\lambda^x}{x!} , & x=0,1,2,\cdots; \lambda>0; \\ 0, & Otherwise. \end{cases} \end{align*}

The variate $X$ is called Poisson variate and $\lambda$ is called the parameter of Poisson distribution.

In notation, it can be written as $X\sim P(\lambda)$.

Key features of Poisson Distribution

• Let X denote the number of times an event occurs in a given interval. The interval can be time, area, volume or distance.
• The probability of occurrence of an event is same for each interval.
• The occurrence of an event in one interval is independent of the occurrence in other interval.

Poisson distribution as a limiting form of binomial distribution

In binomial distribution if $n\to \infty$, $p\to 0$ such that $np=\lambda$ (finite) then binomial distribution tends to Poisson distribution

Proof

Let $X\sim B(n,p)$ distribution. Then the probability mass function of $X$ is

 $$\begin{equation*} P(X=x)= \begin{cases} \binom{n}{x} p^x q^{n-x} & x=0,1,2,\cdots, n\\ & 0 < p < 1, q=1-p\\ 0 & \text{ otherwise }. \end{cases} \end{equation*}$$

Taking the $\lim$ as $n\to \infty$ and $p\to 0$, we have

 $$\begin{eqnarray*} P(x) &=& \lim_{n\to\infty \atop{p\to 0}} \binom{n}{x} p^x q^{n-x} \\ &=& \lim_{n\to\infty \atop{p\to 0}} \frac{n!}{(n-x)!x!} p^x q^{n-x} \\ &=& \lim_{n\to\infty \atop{p\to 0}} \frac{n(n-1)(n-2)\cdots (n-x+1)}{x!} p^x q^{n-x} \\ &=& \lim_{n\to\infty \atop{p\to 0}} \frac{n^x(-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{x-1}{n})}{x!} p^x (1-p)^{n-x} \\ &=& \lim_{n\to\infty \atop{p\to 0}} \frac{(np)^x(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{x-1}{n})}{x!}(1-p)^{n-x} \\ &=& \lim_{n\to\infty \atop{p\to 0}} \frac{\lambda^x}{x!}\big(1-\frac{1}{n}\big)\big(1-\frac{2}{n}\big)\cdots \big(1-\frac{x-1}{n}\big)\big(1-\frac{\lambda}{n}\big)^{n-x} \\ &=& \frac{\lambda^x}{x!}\lim_{n\to\infty }\bigg[\big(1-\frac{\lambda}{n}\big)^{-n/\lambda}\bigg]^{-\lambda}\big(1-\frac{\lambda}{n}\big)^{-x} \\ &=& \frac{\lambda^x}{x!}e^{-\lambda} (1)^{-x}\qquad (\because \lim_{n\to \infty}\big(1-\frac{\lambda}{n}\big)^{-n/\lambda}=e)\\ &=& \frac{e^{-\lambda}\lambda^x}{x!},\;\; x=0,1,2,\cdots; \lambda>0. \end{eqnarray*}$$

Clearly, $P(x)\geq 0$ for all $x\geq 0$, and

 $$\begin{eqnarray*} \sum_{x=0}^\infty P(x) &=& \sum_{x=0}^\infty \frac{e^{-\lambda}\lambda^x}{x!} \\ &=& e^{-\lambda}\sum_{x=0}^\infty \frac{\lambda^x}{x!} \\ &=& e^{-\lambda}\bigg(1+\frac{\lambda}{1!}+\frac{\lambda^2}{2!}+\cdots + \bigg)\\ &=& e^{-\lambda}e^{\lambda}=1. \end{eqnarray*}$$

Hence, $P(x)$ is a legitimate probability mass function.

Graph of Poisson Distribution

Following graph shows the probability mass function of Poisson distribution with parameter $\lambda = 5$.

Mean of Poisson Distribution

The expected value of Poisson random variable is $E(X)=\lambda$.

Proof

The expected value of Poisson random variable is

 $$\begin{eqnarray*} E(X) &=& \sum_{x=0}^\infty x\cdot P(X=x)\\ &=& \sum_{x=0}^\infty x\cdot \frac{e^{-\lambda}\lambda^x}{x!}\\ &=& 0 +\lambda e^{-\lambda}\sum_{x=1}^\infty \frac{\lambda^{x-1}}{(x-1)!}\\ &=& \lambda e^{-\lambda}\bigg(1+\frac{\lambda}{1!}+\frac{\lambda^2}{2!}+\cdots + \bigg)\\ &=& \lambda e^{-\lambda}e^{\lambda} \\ &=& \lambda. \end{eqnarray*}$$

Variance of Poisson Distribution

The variance of Poisson random variable is $V(X) =\lambda$.

Proof

The variance of random variable $X$ is given by

 $$V(X) = E(X^2) - [E(X)]^2$$

Let us find the expected value $X^2$.

 $$\begin{eqnarray*} E(X^2) & = & E[X(X-1)]+ E(X)\\ &=& \sum_{x=0}^\infty x(x-1)\cdot P(X=x)+\lambda\\ &=& \sum_{x=0}^\infty x(x-1)\cdot \frac{e^{-\lambda}\lambda^x}{x!} +\lambda\\ &=& 0 +0+\lambda^2 e^{-\lambda}\sum_{x=2}^\infty \frac{\lambda^{x-2}}{(x-2)!}+\lambda\\ &=& \lambda^2 e^{-\lambda}\bigg(1+\frac{\lambda}{1!}+\frac{\lambda^2}{2!}+\cdots + \bigg)+\lambda\\ &=& \lambda^2 e^{-\lambda}e^{\lambda} = \lambda^2+\lambda. \end{eqnarray*}$$

Thus, variance of Poisson random variable is

 $$\begin{eqnarray*} V(X) &=& E(X^2) - [E(X)]^2\\ &=& \lambda^2+\lambda-\lambda^2\\ &=& \lambda. \end{eqnarray*}$$

For Poisson distribution, Mean = Variance = $\lambda$.

Moment Generating Funtion of Poisson Distribution

The moment generating function of Poisson distribution is $M_X(t)=e^{\lambda(e^t-1)}, t \in R$.

Proof

The moment generating function of Poisson random variable $X$ is

 $$\begin{eqnarray*} M_X(t) &=& E(e^{tx}) \\ &=& \sum_{x=0}^\infty e^{tx} \frac{e^{-\lambda}\lambda^x}{x!}\\ &=& e^{-\lambda}\sum_{x=0}^\infty \frac{(\lambda e^{t})^x}{x!}\\ &=& e^{-\lambda}\cdot e^{\lambda e^t}\\ &=& e^{\lambda(e^t-1)}, \; t\in R. \end{eqnarray*}$$

Mean and variance of Poisson distribution from MGF

The moments of Poisson distribution can also be obtained from moment generating function.

The $r^{th}$ moment of Poisson random variable is given by

 $$\begin{equation*} \mu_r^\prime=\bigg[\frac{d^r M_X(t)}{dt^r}\bigg]_{t=0}. \end{equation*}$$

The mean and variance of Poisson distribution are respectively $\mu_1^\prime=\lambda$ and $\mu_2=\lambda$.

Proof

The moment generating function of Poisson distribution is $M_X(t) =e^{\lambda(e^t-1)}$.

Differentiating $M_X(t)$ w.r.t. $t$

 $$$$\label{p11} \frac{d M_X(t)}{dt}= e^{\lambda(e^t-1)}(\lambda e^{t}).$$$$

Putting $t=0$, we get

 $$\begin{eqnarray*} \mu_1^\prime &=& \bigg[\frac{d M_X(t)}{dt}\bigg]_{t=0} \\ &=& \bigg[e^{\lambda(e^t-1)}(\lambda e^{t})\bigg]_{t=0}\\ &=& \lambda = \text{ mean }. \end{eqnarray*}$$

Again differentiating \eqref{p11} w.r.t. $t$, we get

 $$\begin{equation*} \frac{d^2 M_X(t)}{dt^2}= e^{\lambda(e^t-1)}(\lambda e^{t})+(\lambda e^t)e^{\lambda(e^t-1)}(\lambda e^{t}). \end{equation*}$$

Putting $t=0$, we get

 $$\begin{eqnarray*} \mu_2^\prime &=& \bigg[\frac{d^2 M_X(t)}{dt^2}\bigg]_{t=0} \\ &=& \lambda+\lambda^2. \end{eqnarray*}$$

Hence,

 $$\begin{eqnarray*} \text{Variance }=\mu_2 &=& \mu_2^\prime - (\mu_1^\prime)^2\\ &=&\lambda+\lambda^2-\lambda^2\\ &=&\lambda. \end{eqnarray*}$$

Characteristics Funtion of Poisson Distribution

The characteristics function of Poisson distribution is $\phi_X(t)=e^{\lambda(e^{it}-1)}, t \in R$.

Proof

The characteristics function of Poisson random variable $X$ is

 $$\begin{eqnarray*} \phi_X(t) &=& E(e^{itx}) \\ &=& \sum_{x=0}^\infty e^{itx} \frac{e^{-\lambda}\lambda^x}{x!}\\ &=& e^{-\lambda}\sum_{x=0}^\infty \frac{(\lambda e^{it})^x}{x!}\\ &=& e^{-\lambda}\cdot e^{\lambda e^{it}}\\ &=& e^{\lambda(e^{it}-1)}, \; t\in R. \end{eqnarray*}$$

Probability generating function of Poisson Distribution

The probability generating function of Poisson distribution is $P_X(t)=e^{\lambda(t-1)}$.

Proof

Let $X\sim P(\lambda)$ distribution. Then the m.g.f. of $X$ is

 $$\begin{eqnarray*} P_X(t) &=& E(t^x) \\ &=& \sum_{x=0}^\infty t^x \frac{e^{-\lambda}\lambda^x}{x!}\\ &=& e^{-\lambda}\sum_{x=0}^\infty \frac{(\lambda t)^x}{x!}\\ &=& e^{-\lambda}\cdot e^{\lambda t}\\ &=& e^{\lambda(t-1)}. \end{eqnarray*}$$

The sum of two independent Poisson variates is also a Poisson variate.

That is, if $X_1$ and $X_2$ are two independent Poisson variate with parameters $\lambda_1$ and $\lambda_2$ respectively then $X_1+X_2 \sim P(\lambda_1+\lambda_2)$.

Proof

Let $X_1$ and $X_2$ be two independent Poisson variate with parameters $\lambda_1$ and $\lambda_2$ respectively.

Then the MGF of $X_1$ is $M_{X_1}(t) =e^{\lambda_1(e^t-1)}$ and the MGF of $X_2$ is $M_{X_2}(t) =e^{\lambda_2(e^t-1)}$.

Let $Y=X_1+X_2$. Then the MGF of $Y$ is

 $$\begin{eqnarray*} M_Y(t) &=& E(e^{tY}) \\ &=& E(e^{t(X_1+X_2)}) \\ &=& E(e^{tX_1} e^{tX_2}) \\ &=& E(e^{tX_1})\cdot E(e^{tX_2})\\ & & \quad \qquad (\because X_1, X_2 \text{ are independent })\\ &=& M_{X_1}(t)\cdot M_{X_2}(t)\\ &=& e^{\lambda_1(e^t-1)}\cdot e^{\lambda_2(e^t-1)}\\ &=& e^{(\lambda_1+\lambda_2)(e^t-1)}. \end{eqnarray*}$$

which is the m.g.f. of Poisson variate with parameter $\lambda_1+\lambda_2$.

Hence, by uniqueness theorem of MGF, $Y=X_1+X_2$ follows a Poisson distribution with parameter $\lambda_1+\lambda_2$.

Mode of Poisson distribution

The condition for mode of Poisson distribution is
$\lambda-1 \leq x\leq \lambda$.

Proof

The mode is that value of $x$ for which $P(x)$ is greater that or equal to $P(x-1)$ and $P(x+1)$, i.e., $P(x-1)\leq P(x) \geq P(x+1)$.

Now, $P(x-1)\leq P(x)$ gives

 $$\begin{eqnarray*} \frac{e^{-\lambda}\lambda^{(x-1)}}{(x-1)!} &\leq& \frac{e^{-\lambda}\lambda^{x}}{x!} \\ x &\leq & \lambda. \end{eqnarray*}$$

And, $P(x) \geq P(x+1)$ gives

 $$\begin{eqnarray*} \frac{e^{-\lambda}\lambda^{x}}{x!} &\leq& \frac{e^{-\lambda}\lambda^{(x+1)}}{(x+1)!} \\ \lambda-1 &\leq & x. \end{eqnarray*}$$

Hence, the condition for mode of Poisson distribution is

 $$\begin{equation*} \lambda-1 \leq x\leq \lambda. \end{equation*}$$

Recurrence relation for raw moments

The recurrence relation for raw moments of Poisson distribution is

 $$\begin{equation*} \mu_{r+1}^\prime = \lambda \bigg[ \frac{d\mu_r^\prime}{d\lambda} + \mu_r^\prime\bigg]. \end{equation*}$$

Recurrence relation for central moments

The recurrence relation for central moments of Poisson distribution is

 $$\begin{equation*} \mu_{r+1} = \lambda \bigg[ \frac{d\mu_r}{d\lambda} + r\mu_{r-1}\bigg]. \end{equation*}$$

Recurrence relation for probabilities

The recurrence relation for probabilities of Poisson distribution is

 $$\begin{equation*} P(X=x+1) = \frac{\lambda}{x+1}\cdot P(X=x), \; x=0,1,2\cdots. \end{equation*}$$

Conclusion

In this tutorial, you learned about theory of Poisson distribution like the probability mass function, mean, variance, moment generating function and other properties of Poisson distribution.

To read more about the step by step examples and calculator for Poisson distribution refer the link Poisson Distribution Calculator with Examples. This tutorial will help you to understand how to calculate mean, variance of Poisson distribution and you will learn how to calculate probabilities and cumulative probabilities for Poisson distribution with the help of step by step examples.