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
$$ \begin{equation*} \mu_2 = -\frac{\theta\log(1-\theta)+\theta^2}{(1-\theta)^2[\log(1-\theta)]^2}. \end{equation*} $$
Hope you enjoyed this article.Do read step by step guide on Power Series Distribution.
If you have any doubt or queries on Particular Cases of Power Series Distribution feel free to post them in the comment section.