Particular Cases of Power Series Distribution

Particular Cases of PSD

In this tutorial, we will discuss some of the particular cases of Power Series Distributions. Some discrete distributions are particular cases of power series distribution, namely, Binomial distribution, Poisson distribution, Geometric distribution, Negative Binomial distribution and Logarithmic Series distribution.

Binomial Distribution

Binomial distribution is a particular case of Power series distribution.

Let $f(\theta)=(1+\theta)^n$, $0<\theta<1$ and $n>0$. Then

 $$\begin{eqnarray*} f(\theta) &=& (1+\theta)^n\\ &=& 1+\binom{n}{1}\theta + \binom{n}{2}\theta^2 + \cdots+ \binom{n}{n}\theta^n\\ &=&\sum_{x=0}^n \binom{n}{x}\theta^x \end{eqnarray*}$$

Let $a_x = \binom{n}{x}$. $\theta^x = \theta^x$ and $f(\theta) =(1+\theta)^n$. Then the power series distribution becomes,

 $$\begin{eqnarray*} P(X=x) & = & \frac{a_x\theta^x}{f(\theta)}\\ & = & \binom{n}{x}\frac{\theta^x}{(1+\theta)^n}\\ & = & \binom{n}{x}\bigg(\frac{\theta}{1+\theta}\bigg)^x\bigg(\frac{1}{1+\theta}\bigg)^{n-x} \end{eqnarray*}$$

Letting $p=\frac{\theta}{1+\theta}$, so $q=1-p = \frac{1}{1+\theta}$, we get

 $$\begin{eqnarray*} P(X=x) &=&\binom{n}{x}p^xq^{n-x}\\ & & \quad x=0,1,2,\cdots, n,\\ & & \quad 0 < p, q < 1, \; p+q=1. \end{eqnarray*}$$

which is the p.m.f. of Binomial distribution with parameter $n$ and $p$. Hence taking $f(\theta)=(1+\theta)^n$ and $p=\frac{\theta}{1+\theta}$, $q=1-p = \frac{1}{1+\theta}$, we get a Binomial distribution.

Mean and Variance of Binomial Distribution

The mean of P.S.D. is

 $$\begin{equation*} \mu_1^\prime = \frac{\theta f^\prime(\theta)}{f(\theta)}. \end{equation*}$$

Hence, mean of Binomial distribution is

 $$\begin{eqnarray*} \mu_1^\prime &=& \theta\cdot\frac{ n(1+\theta)^{n-1}}{(1+\theta)^n}\\ &=& n\frac{\theta}{1+\theta}\\ &=&np. \end{eqnarray*}$$

The variance of P.S.D. is

 $$\begin{eqnarray*} \mu_2 & = & \theta\cdot \frac{d\mu_1^\prime}{d\theta}\\ &=& n\theta\frac{d}{d\theta}\bigg(\frac{\theta}{1+\theta}\bigg)\\ &=& n\theta\frac{(1+\theta)\cdot 1- \theta\cdot 1}{(1+\theta)^2}\\ &=&\frac{n\theta}{(1+\theta)^2}\\ &=&npq. \end{eqnarray*}$$

Hence, variance of Binomial distribution is

 $$\begin{equation*} \mu_2 = \frac{n\theta}{(1+\theta)^2}=npq. \end{equation*}$$

M.G.F. of Binomial Distribution

The m.g.f. of P.S.D. is $M_X(t) = \dfrac{f(\theta e^t)}{f(\theta)}$.

The m.g.f. of Binomial distribution can be obtained by taking $f(\theta) = (1+\theta)^n$, we have

 $$\begin{eqnarray*} M_X(t) &=& \frac{(1+\theta e^t)^n}{(1+\theta)^n}\\ &=&\bigg(\frac{1+\theta e^t}{1+\theta}\bigg)^n \\ &=& \bigg(\frac{1}{1+\theta}+ e^t\frac{\theta}{1+\theta}\bigg)^n \\ & = &(q+pe^t)^n, \end{eqnarray*}$$

where $p=\dfrac{\theta}{1+\theta}$ and $q=\dfrac{1}{1+\theta}$.

Poisson Distribution

Poisson distribution is a particular case of Power series distribution.

Let $f(\theta)=e^\theta$, $\theta > 0$. Then

 $$\begin{eqnarray*} f(\theta) &=&e^\theta\\ &=& 1+\theta + \frac{\theta^2}{2!} + \frac{\theta^3}{3!} +\cdots\\ &=&\sum_{x=0}^\infty \frac{\theta^x}{x!} \end{eqnarray*}$$

Let $a_x = \frac{1}{x!}$. $\theta^x = \theta^x$ and $f(\theta)=e^\theta$. Then the power series distribution becomes,

 $$\begin{eqnarray*} P(X=x) & = & \frac{a_x \theta^x}{f(\theta)}\\ &=& \frac{\theta^x}{x! e^\theta}\\ &=& \frac{e^{-\theta}\theta^x}{x!}, \; x=0,1,2,\cdots \end{eqnarray*}$$

which is the p.m.f. of Poisson distribution. Hence taking $f(\theta)=e^\theta$ we get a Poisson distribution.

Mean and Variance of Poisson Distribution

The mean of P.S.D. is

 $$\begin{equation*} \mu_1^\prime = \frac{\theta f^\prime(\theta)}{f(\theta)}. \end{equation*}$$

Hence, mean of Poisson distribution is

 $$\begin{equation*} \mu_1^\prime = \frac{\theta e^\theta}{e^\theta} = \theta. \end{equation*}$$

The variance of P.S.D. is

 $$\begin{eqnarray*} \mu_2 & = & \theta\cdot \frac{d\mu_1^\prime}{d\theta}\\ &=&\theta\frac{d}{d\theta}\theta\\ &=&\theta. \end{eqnarray*}$$

Hence, variance of Poisson distribution is

 $$\begin{equation*} \mu_2 = \theta. \end{equation*}$$

M.G.F. of Poisson Distribution

The m.g.f. of P.S.D. is $M_X(t) = \dfrac{f(\theta e^t)}{f(\theta)}$.

The m.g.f. of Poisson distribution can be obtained by taking $f(\theta) = e^\theta$, we have

 $$\begin{eqnarray*} M_X(t) &=& \frac{e^{\theta e^t}}{e^\theta}\\ &=&e^{\theta(e^t-1)}. \end{eqnarray*}$$

Geometric Distribution

Geometric distribution is a particular case of Power series distribution.

Let $f(\theta)=(1-\theta)^{-1}$, $0 < \theta < 1$. Then

 $$\begin{eqnarray*} f(\theta) &=&(1-\theta)^{-1} = 1+\theta + \theta^2 + \cdots\\ &=&\sum_{x=0}^\infty \theta^x \end{eqnarray*}$$

Let $a_x = 1$. $\theta^x = \theta^x$ and $f(\theta) =(1-\theta)^{-1}$. Then the power series distribution becomes,

 $$\begin{eqnarray*} P(X=x) & = & \frac{a_x \theta^x}{f(\theta)}\\ &=& \theta^x\frac{1}{(1-\theta)^{-1}}\\ &=& \theta^x(1-\theta),\quad x=0,1,2,\cdots \end{eqnarray*}$$

which is the p.m.f. of Geometric distribution with probability of success $p=1-\theta$. Hence taking $f(\theta)=(1-\theta)^{-1}$ we get a Geometric distribution.

Mean and Variance of Geometric Distribution

The mean of P.S.D. is

 $$\begin{equation*} \mu_1^\prime = \frac{\theta f^\prime(\theta)}{f(\theta)}. \end{equation*}$$

Hence, mean of Geometric distribution is

 $$\begin{eqnarray*} \mu_1^\prime &=& \theta \cdot\frac{-(1-\theta)^{-2}(-1)}{(1-\theta)^{-1}}\\ &=& \frac{\theta}{1-\theta}\\ &=&\frac{q}{p}. \end{eqnarray*}$$

The variance of P.S.D. is

 $$\begin{eqnarray*} \mu_2 & = & \theta\cdot \frac{d\mu_1^\prime}{d\theta} \\ &=& \theta\frac{d}{d\theta}\bigg(\frac{\theta}{1-\theta}\bigg)\\ & = & \theta\cdot \frac{(1-\theta)\cdot 1 - \theta \cdot (-1)}{(1-\theta)^{2}}\\ & = & \frac{\theta}{(1-\theta)^2}\\ &=&=\frac{q}{p^2}. \end{eqnarray*}$$

Hence, variance of Geometric distribution is

 $$\begin{equation*} \mu_2 = \frac{\theta}{(1-\theta)^2}=\frac{q}{p^2}. \end{equation*}$$

M.G.F. of Geometric Distribution

The m.g.f. of P.S.D. is $M_X(t) = \dfrac{f(\theta e^t)}{f(\theta)}$.
The m.g.f. of Geometric distribution can be obtained by taking $f(\theta) = (1-\theta)^{-1}$, we have

 $$\begin{eqnarray*} M_X(t) &=& \frac{(1-\theta e^t)^{-1}}{(1-\theta)^{-1}}\\ &=& (1-\theta)(1-\theta e^t)^{-1}\\ &=& p(1-qe^t)^{-1}, \end{eqnarray*}$$

where $p=1-\theta$ and $q=\theta$.

Negative Binomial Distribution

Negative Binomial distribution is a particular case of Power series distribution.

Let $f(\theta)=(1-\theta)^{-r}$ $0 < \theta < 1$ and $r > 0$. Then

 $$\begin{eqnarray*} f(\theta) &=& (1-\theta)^{-r}\\ & = & 1+\binom{r}{1}\theta + \binom{r+1}{2}\theta^2 + \cdots+\binom{r+x-1}{x}\theta^x + \cdots\\ & = & \sum_{x=0}^\infty \binom{r+x-1}{x}\theta^x. \end{eqnarray*}$$

Let $a_x = \binom{r+x-1}{x}$. $\theta^x = \theta^x$ and $f(\theta)=(1-\theta)^{-r}$. Then the power series distribution becomes,

 $$\begin{eqnarray*} P(X=x) & = & \frac{a_x \theta^x}{f(\theta)}\\ &=& \binom{r+x-1}{x}\theta^x\frac{1}{(1-\theta)^{-r}}\\ &=& \binom{r+x-1}{x}\theta^x(1-\theta)^r\\ &=& \binom{x+r-1}{r-1}\theta^x(1-\theta)^r,\quad x=0,1,2,\cdots\\ &=& \binom{x+r-1}{r-1}q^xp^r,\quad x=0,1,2,\cdots \end{eqnarray*}$$

which is the p.m.f. of Negative Binomial distribution with probability of success $p=1-\theta$. Hence taking $f(\theta)=(1-\theta)^{-r}$ we get a Negative Binomial distribution.

Mean and Variance of Negative Binomial Distribution

The mean of P.S.D. is

 $$\begin{equation*} \mu_1^\prime = \frac{\theta f^\prime(\theta)}{f(\theta)}. \end{equation*}$$

Hence, mean of Negative Binomial distribution is

 $$\begin{eqnarray*} \mu_1^\prime &=& \frac{\theta}{(1-\theta)^{-r}} \cdot\big[-r(1-\theta)^{-r-1}(-1)\big]\\ &=&\frac{r\theta}{1-\theta}\\ &=&\frac{rq}{p}. \end{eqnarray*}$$

The variance of P.S.D. is

 $$\begin{eqnarray*} \mu_2 & = & \theta\cdot \frac{d\mu_1^\prime}{d\theta}\\ &=& \theta\frac{d}{d\theta}\bigg(\frac{r\theta}{1-\theta}\bigg)\\ & = & r\theta\cdot \frac{(1-\theta)\cdot 1 - \theta \cdot (-1)}{(1-\theta)^{2}}\\ & = & \frac{r\theta}{(1-\theta)^2}\\ &=& \frac{rq}{p^2}. \end{eqnarray*}$$

Hence, variance of Negative Binomial distribution is

 $$\begin{equation*} \mu_2 = \frac{r\theta}{(1-\theta)^2}=\frac{rq}{p^2}. \end{equation*}$$

M.G.F. of Negative Binomial Distribution

The m.g.f. of P.S.D. is $M_X(t) = \dfrac{f(\theta e^t)}{f(\theta)}$. The m.g.f. of Negative Binomial distribution can be obtained by taking $f(\theta) = (1-\theta)^{-r}$, we have

 $$\begin{eqnarray*} M_X(t) &=& \frac{(1-\theta e^t)^{-r}}{(1-\theta)^{-r}}\\ &=& (1-\theta)(1-\theta e^t)^{-r}\\ & = &p^r(1-qe^t)^{-r}, \end{eqnarray*}$$

where $p=1-\theta$ and $q=\theta$.

Logarithmic Series Distribution

Let $f(\theta)=-\log(1-\theta)$. Then

 $$\begin{eqnarray*} f(\theta) &=& -\log(1-\theta)\\ & = & \theta +\frac{\theta^2}{2} + \frac{\theta^3}{3} + \cdots,\quad 0 < \theta < 1\\ & = & \sum_{x=1}^\infty \frac{\theta^x}{x}. \end{eqnarray*}$$

Let $a_x = \frac{1}{x}$. $\theta^x = \theta^x$ and $f(\theta) =-\log(1-\theta)$. Then the power series distribution becomes,

 $$\begin{eqnarray*} P(X=x) & = & \frac{a_x \theta^x}{f(\theta)}\\ &=& \frac{\theta^x}{-x\log(1-\theta)},\quad x=1,2,\cdots \end{eqnarray*}$$

which is the p.m.f. of Logarithmic series distribution. Hence taking $f(\theta)=-\log(1-\theta)$ we get a Logarithmic series distribution.

Mean and Variance of Logarithmic series Distribution

The mean of P.S.D. is

 $$\begin{equation*} \mu_1^\prime = \frac{\theta f^\prime(\theta)}{f(\theta)}. \end{equation*}$$

Hence, mean of Logarithmic series distribution is

 $$\begin{equation*} \mu_1^\prime = \frac{-\theta}{\log(1-\theta)}\cdot\frac{1}{(1-\theta)}. \end{equation*}$$

The variance of P.S.D. is

 $$\begin{eqnarray*} \mu_2 & = & \theta\cdot \frac{d\mu_1^\prime}{d\theta}\\ &=& \theta\frac{d}{d\theta}\bigg(-\frac{\theta}{(1-\theta)\log(1-\theta)}\bigg)\\ & = & \theta\bigg(-\frac{\log(1-\theta)+\theta}{(1-\theta)^2[\log(1-\theta)]^2}\bigg)\\ & = & -\frac{\theta\log(1-\theta)+\theta^2}{(1-\theta)^2[\log(1-\theta)]^2}. \end{eqnarray*}$$

Hence, variance of Logarithmic Series distribution is

 $$\begin{equation*} \mu_2 = -\frac{\theta\log(1-\theta)+\theta^2}{(1-\theta)^2[\log(1-\theta)]^2}. \end{equation*}$$