Poisson Xgamma distribution: A discrete model for count data analysis

Abstract

In this article, we attempt to introduce a count data model which is obtained by compounding Poisson distribution with Xgamma distribution. Important mathematical and statistical properties of the distribution have been derived and discussed. Parameter estimation is discussed using the maximum likelihood method of estimation followed by Monte Carlo simulation to investigate the behavior of the ML estimators. Finally, two real-life data sets are analyzed to investigate the suitability of the proposed distribution in modeling count data.

Keywords

Poisson distribution Xgamma distribution compounding count data

1. Introduction

Researchers obtain plethora of probability models for the sake of analyzing many types of data from various fields, such as medicine, transport, engineering, agriculture and so on. Lots of well known techniques are employed to serve the purpose of constructing new probability distributions. Some well known techniques like discretization, T-X family, and compounding technique provides a very powerful way to extend common parametric families of distributions to fit data sets not adequately fit by classical distributions. Regarding the compound of probability distributions, the work has been done in this particular area since 1920. It is well known that Greenwood and Yule (1920) established a relationship between Poisson distribution and a negative binomial distribution through compounding mechanism by treating the rate parameter in Poisson distribution as a gamma variate. Skellam (1948) derived a probability distribution from the binomial distribution by regarding the probability of success as a beta variable between sets of trials. Lindley (1958) suggested a one parameter distribution to illustrate the difference between fiducial distribution and posterior distribution. Dubey (1970) derived a compound gamma, beta and F distribution by compounding a gamma distribution with another gamma distribution and reduced it to the beta 1 ${}^{\rm st}$ and beta 2 ${}^{\rm nd}$ kind and to the F distribution by suitable transformations. Gerstenkorn (1993, 1996) proposed several compound distributions, he obtained compound of gamma distribution with exponential distribution by treating the parameter of gamma distribution as an exponential variate and also obtained compound of Pólya with beta distribution. Mahmoudi et al. (2010) generalized the Poisson-Lindley distribution of Sankaran (1970) and showed that their generalized distribution has more flexibility in analyzing count data. Zamani and Ismail (2010) constructed a new compound distribution by compounding negative binomial with one parameter Lindley distribution that provides good fit for count data where the probability at zero has a large value. A new generalized negative binomial distribution was proposed by Gupta and Ong (2004), this distribution arises from Poisson distribution if the rate parameter follows generalized gamma distribution; the resulting distribution so obtained was applied to various data sets and can be used as better alternative to negative binomial distribution. Adil et al. (2016) proposed a new competitive count data model by compounding negative binomial distribution with Kumaraswamy distribution that finds its application in biological sciences. Para and Jan (2018) introduced two count data models in medical sciences using compounding technique on discrete version of Weibull and inverse Weibull distribution by treating probability parameter as a random variable following beta distribution.

In this paper, we propose a new discrete count data model by compounding Poisson distribution with Xgamma distribution (Sen et al., 2016), as there is a need to find more flexible model for analyzing statistical data. The merit of the introduced distribution comes in its ability to describe all types of dispersions for data as shall be shown later. Also, the probability mass function of the distribution displays unimodal and right skewed shapes. Moreover, the capability of the distribution comparing to some of existing distribution is investigated, although existence of excess of zero, by fitting practical integer-valued data sets in application section.

2. Definition of proposed model (Poisson Xgamma distribution)

If $X|\lambda\sim P(\lambda)$ , where $\lambda$ is itself a random variable following Xgamma distribution (Sen et al., 2016) with parameter $\theta$ , then determining the distribution that results from marginalizing over $\lambda$ will be known as a compound of Poisson distribution with that of Xgamma distribution, which is denoted by $P$ -Xgamma $({X;\theta})$ .

Theorem 1. The probability mass function of a Poisson Xgamma distribution i.e., $P$ -Xgamma $({X;\theta})$ is given by

$\displaystyle P(X=x)=\frac{\theta^{2}}{2(1+\theta)^{x+4}}({2(1+\theta)^{2}+% \theta(x+1)(x+2)});\quad x=0,1,2,3,\ldots;\theta>0$

Proof The pmf of a Poisson Xgamma distribution, i.e., $P$ -Xgamma $({X;\theta})$ can be obtained by

$\displaystyle g(x|\lambda)=\frac{e^{-\lambda}\lambda^{x}}{x!};x=0,1,2,3,\ldots% ,\lambda>0,$

when its parameter $\lambda$ follows Xgamma distribution (XGD) with pdf

$\displaystyle h(\lambda;\theta)=\frac{\theta^{2}}{1+\theta}\left({1+\frac{% \theta}{2}\lambda^{2}}\right)e^{-\theta\lambda};\lambda>0,\theta>0.$

We have

$\displaystyle P(X=x)=\int_{0}^{\infty}{g(x|\lambda).h(\lambda;\theta)d\lambda}$ $\displaystyle P(X=x)=\frac{\theta^{2}}{x!(1+\theta)}\left({\int_{0}^{\infty}{e% ^{-(1+\theta)\lambda}\lambda^{x}d\lambda+\frac{\theta}{2}\int_{0}^{\infty}{e^{% -(1+\theta)\lambda}}\lambda^{x+2}d\lambda}}\right)$ $\displaystyle P(X=x)=\frac{\theta^{2}}{2(1+\theta)^{x+4}}({2(1+\theta)^{2}+% \theta(x+1)(x+2)});x=0,1,2,3,\ldots,\theta>0,$ (1)

which is the pmf of $P$ -Xgamma $({X;\theta})$ .

Figure 1 exhibits the pmf plot of the proposed model for different values of the parameter $\theta$ .

Figure 1.

Index of dispersion (IOD) plot of Poisson Xgamma distribution.

The corresponding cdf of Poisson Xgamma distribution is obtained as:

$\displaystyle F_{X}(x)=\sum\limits_{n=0}^{x}{\frac{\theta^{2}}{2(1+\theta)^{n+% 4}}({2(1+\theta)^{2}+\theta(n+1)(n+2)})}=1-\frac{\theta^{3}+\frac{(\theta x)^{% 2}}{2}+\frac{5\theta x^{2}}{2}+5\theta^{2}+x\theta+4\theta+1}{(1+\theta)^{x+4}% };\quad x=0,1,2,\ldots,\theta>0.$ (2)

Figure 2 exhibits the cdf plot of the proposed model for different values of the parameter $\theta$ .

Figure 2.

cdf plot of discrete Poisson Xgamma distribution.

2.1 Random data generation from Poisson Xgamma distribution

In order to simulate the data from Poisson Xgamma distribution, we employ the inverse cdf method. Simulating a sequence of random numbers $x_{1},x_{2},\ldots,x_{n}$ of the Poisson Xgamma random variable $X$ with pmf $p(X=x_{i})=p_{i},\sum_{i=0}^{k}{p_{i}=1}$ and a cdf $F(X;\theta)$ , where k may be finite or infinite can be described as in the following steps.

Step1: Generate a random number $u$ from uniform distribution U (0,1).

Step2: Generate random number $x_{i}$ based on

$\displaystyle\text{if }u\leqslant p_{0}=F(x_{0};\theta)\text{ then }X=x_{0}$ $\displaystyle\text{if }p_{0}<u\leqslant p_{0}+p_{1}=F(x_{1};\theta)\text{ then% }X=x_{1}\ldots$ $\displaystyle\text{if }\sum\limits_{j=0}^{k-1}{p_{j}}<u\leqslant\sum\limits_{j% =0}^{k}{p_{j}}=F(x_{k};\theta)\text{ then }X=x_{k}.$

In order to generate n random numbers $x_{1},x_{2},\ldots,x_{n}$ from $P$ -Xgamma $({X;\theta})$ distribution, repeat the step 1 to step $2∼{}n$ times. We have employed R software for running the simulation study of the proposed model.

3. Statistical properties

In this section, different structural properties of the Poisson Xgamma model have been evaluated. These include moments, index of dispersion, moment generating function and probability generating function.

3.1 Moments

3.1.1 Factorial moments

Using Eq. (1), the $r$ th factorial moment about origin of the $P$ -Xgamma $({X;\theta})$ can be obtained as

$\displaystyle\mu_{(r)}^{\prime}=E[E(X^{(r)}|\lambda)],\text{ where }X^{(r)}=X(% X-1)(X-2)\ldots(X-r+1)$ $\displaystyle\mu_{(r)}^{\prime}=\int_{0}^{\infty}\left[{\sum\limits_{x=0}^{% \infty}{x^{(r)}}\frac{e^{-\lambda}\lambda^{x}}{x!}}\right]\cdot\frac{\theta^{2% }}{1+\theta}\left({1+\frac{\theta}{2}\lambda^{2}}\right)e^{-\theta\lambda}d\lambda$ $\displaystyle\mu_{(r)}^{\prime}=\frac{\theta^{2}}{1+\theta}\int_{0}^{\infty}{% \left[{\lambda^{r}\left({\sum\limits_{x=r}^{\infty}{\frac{e^{-\lambda}\lambda^% {x-r}}{(x-r)!}}}\right)}\right]\left(1+\frac{\theta}{2}\lambda^{2}\right)e^{-% \theta\lambda}d\lambda}.$

Taking $u=x-r$ , we get

$\displaystyle\mu_{(r)}^{\prime}=\frac{\theta^{2}}{1+\theta}\int_{0}^{\infty}{% \left[{\lambda^{r}\left({\sum\limits_{u=0}^{\infty}{\frac{e^{-\lambda}\lambda^% {u}}{u!}}}\right)}\right]}\left(1+\frac{\theta}{2}\lambda^{2}\right)e^{-\theta% \lambda}d\lambda$ $\displaystyle\mu_{(r)}^{\prime}=\frac{\theta^{2}}{1+\theta}\left({\int_{0}^{% \infty}{\lambda^{r}e^{-\theta\lambda}d\lambda+\frac{\theta}{2}\int_{0}^{\infty% }{\lambda^{r+2}}e^{-\theta\lambda}d\lambda}}\right)$ $\displaystyle\mu_{(r)}^{\prime}=\frac{r!}{1+\theta}\left[{\frac{2\theta^{2}+(r% +1)(r+2)}{2\theta^{r}}}\right]$ (3)

Taking $r=1,2,3,4$ in Eq. (3), the first four factorial moments about origin of Poisson Xgamma distribution can be obtained as

$\displaystyle\mu_{(1)}^{\prime}=\frac{\theta^{2}+3}{\theta(\theta+1)}$ $\displaystyle\mu_{(2)}^{\prime}=\frac{2(\theta^{2}+6)}{\theta^{2}(1+\theta)}$ $\displaystyle\mu_{(3)}^{\prime}=\frac{6(\theta^{2}+10)}{\theta^{3}(1+\theta)}$ $\displaystyle\mu_{(4)}^{\prime}=\frac{24(\theta^{2}+15)}{\theta^{4}(1+\theta)}$

Figure 3.

Index of dispersion (IOD) plot of Poisson Xgamma distribution.

3.1.2 Moments about origin (raw moments)

The first four moments about origin, using the relationship between factorial moments about origin and the moments about origin, of $P$ -Xgamma $({X;\theta})$ can be obtained as

$\displaystyle\mu_{1}^{\prime}=\frac{\theta^{2}+3}{\theta(1+\theta)}$ $\displaystyle\mu_{2}^{\prime}=\frac{2(\theta^{2}+6)+\theta(\theta^{2}+3)}{% \theta^{2}(1+\theta)}$ $\displaystyle\mu_{3}^{\prime}=\frac{\theta^{4}+6\theta^{3}+45\theta+60}{\theta% ^{3}(1+\theta)}$ $\displaystyle\mu_{4}^{\prime}=\frac{\theta^{5}+14\theta^{4}-15\theta^{3}+162% \theta^{2}+360(\theta+1)}{\theta^{4}(1+\theta)}$

Mean, variance and index of dispersion of the proposed model are given by

$\displaystyle\text{Mean }=\mu_{1}^{\prime}=\frac{\theta^{2}+3}{\theta(1+\theta)}$ $\displaystyle\text{Variance }=\mu_{2}=\frac{3(1+5\theta+\theta^{3})-\theta^{2}% }{\theta^{2}(1+\theta)^{2}}$ $\displaystyle\text{Index of dispersion }=\frac{3(1+5\theta+\theta^{3})-\theta^% {2}}{\theta(1+\theta)(3+\theta^{2})}$

Figure 3 displays the index of dispersion (IOD) plot of Poisson Xgamma distribution, and shows that the distribution can describe all types of dispersions for data.

3.2 Probability generating function of the Poisson Xgamma distribution

We derive moment generating function and probability generating function of PXGD in this section.

Theorem 2. If $X$ has $P$ -Xgamma $({X;\theta})$ , then the probability generating function $P_{X}(t)$ has the following form

$\displaystyle P_{X}(t)=\frac{\theta^{2}}{2(1+\theta)^{3}}\left[{\frac{2[(1+% \theta)^{2}+2]}{(\theta+1-t)}+\frac{4t\theta^{2}-2t^{2}\theta+4t\theta}{(% \theta+1-t)^{3}}}\right]$

Proof We begin with the well known definition of the probability generating function given by

$\displaystyle P_{X}(t)=\sum\limits_{x=0}^{\infty}{t^{x}\left[{\frac{\theta^{2}% }{2(1+\theta)^{x+4}}({2(1+\theta)^{2}+\theta(x+1)(x+2)})}\right]}$ $\displaystyle P_{X}(t)=\frac{\theta^{2}}{2(1\!+\!\theta)^{4}}\left[{2(1\!+\!% \theta)^{2}\sum\limits_{x=0}^{\infty}{\left({\frac{t}{1\!+\!\theta}}\right)}^{% x}\!+\!\theta\sum\limits_{x=0}^{\infty}{x^{2}\left({\frac{t}{1+\theta}}\right)% ^{x}}\!+\!3\theta\sum\limits_{x=0}^{\infty}{x\left({\frac{t}{1\!+\!\theta}}% \right)^{x}}\!+\!2\theta\sum\limits_{x=0}^{\infty}{\left({\frac{t}{1\!+\!% \theta}}\right)^{x}}}\right]$ $\displaystyle P_{X}(t)=\frac{\theta^{2}}{2(1+\theta)^{3}}\left[{\frac{2(1+% \theta)^{2}}{(\theta+1-t)}+\frac{\theta(t(\theta+1-t)+2t^{2})}{(\theta+1-t)^{3% }}+\frac{3\theta t}{(\theta+1-t)^{2}}+\frac{2\theta}{(\theta+1-t)}}\right]$ $\displaystyle P_{X}(t)=\frac{\theta^{2}}{2(1+\theta)^{3}}\left[{\frac{2[(1+% \theta)^{2}+2]}{(\theta+1-t)}+\frac{(t\theta^{2}+\theta t^{2}+t\theta)}{(% \theta+1-t)^{3}}+\frac{3t\theta}{(\theta+1-t)^{2}}}\right]$ $\displaystyle P_{X}(t)=\frac{\theta^{2}}{2(1+\theta)^{3}}\left[{\frac{2[(1+% \theta)^{2}+2]}{(\theta+1-t)}+\frac{4t\theta^{2}-2t^{2}\theta+4t\theta}{(% \theta+1-t)^{3}}}\right]$

Theorem 3. If $X$ has $P$ -Xgamma $({X;\theta})$ , then the moment generating function $M_{X}(t)$ has the following form

$\displaystyle M_{X}(t)=\frac{\theta^{2}}{2(1+\theta)^{3}}\left[{\frac{2[(1+% \theta)^{2}+\theta]}{(\theta+1-e^{t})}+\frac{4e^{t}\theta^{2}-2e^{2t}\theta+4e% ^{t}\theta}{(\theta+1-e^{t})^{3}}}\right]$

Proof We begin with the well known definition of the probability generating function given by

$\displaystyle M_{X}(t)=\sum\limits_{x=0}^{\infty}{e^{tx}\left[{\frac{\theta^{2% }}{2(1+\theta)^{x+4}}\left({2(1+\theta)^{2}+\theta(x+1)(x+2)}\right)}\right]}$ $\displaystyle M_{X}(t)=\frac{\theta^{2}}{2(1+\theta)^{4}}\left[{2(1+\theta)^{2% }\sum\limits_{x=0}^{\infty}{\left({\frac{e^{t}}{1+\theta}}\right)}^{x}+\theta% \sum\limits_{x=0}^{\infty}{x^{2}\left({\frac{e^{t}}{1+\theta}}\right)^{x}}+3% \theta\sum\limits_{x=0}^{\infty}{x(\frac{e^{t}}{1+\theta})^{x}}+2\theta\sum% \limits_{x=0}^{\infty}{\left({\frac{e^{t}}{1+\theta}}\right)^{x}}}\right]$ $\displaystyle M_{X}(t)=\frac{\theta^{2}}{2(1+\theta)^{3}}\left[{\frac{2[(1+% \theta)^{2}+\theta]}{(\theta+1-e^{t})}+\frac{4e^{t}\theta^{2}-2e^{2t}\theta+4e% ^{t}\theta}{(\theta+1-e^{t})^{3}}}\right].$

4. Reliability analysis

In this section, we have been obtained the reliability function, hazard rate function and reverse hazard rate function of the proposed Poisson Xgamma model.

4.1 Reliability function R (x)

The reliability function is defined as the probability that a system survives beyond a specified time. It is also referred to as survival or survivor function of the distribution. It can be computed as complement of the cumulative distribution function of the model. The reliability function or the survival function of Poisson Xgamma distribution is calculated as:

$\displaystyle R(x,\theta)=P(X>x)=\frac{\theta^{3}+\frac{\theta^{2}(x-1)^{2}}{2% }+\frac{5\theta(x-1)^{2}}{2}+5\theta^{2}+\theta(x-1)+4\theta+1}{(1+\theta)^{x+% 3}}$

4.2 Hazard function

The hazard function is also known as hazard rate, instantaneous failure rate or force of mortality is given as:

$\displaystyle\text{H.R }=h(x,\theta)=\frac{P(x,\theta)}{R(x,\theta)}=\frac{% \theta^{2}\left({2(1+\theta)^{2}+\theta(x+1)(x+2)}\right)}{2\left({1+\theta}% \right)\left({\theta^{3}+\frac{\theta^{2}(x-1)^{2}}{2}+\frac{5\theta(x-1)^{2}}% {2}+5\theta^{2}+\theta(x-1)+4\theta+1}\right)}$

4.3 Reverse hazard rate function

The reverse hazard rate function of Poisson Xgamma distribution is given as:

$\displaystyle\text{R.H.R }=h_{r}(x,\theta)=\frac{P(x,\theta)}{F(x,\theta)}=% \frac{\theta^{2}\left({2(1+\theta)^{2}+\theta(x+1)(x+2)}\right)}{(1+\theta)^{x% +4}-(2\theta^{3}+\theta^{2}x^{2}+5\theta x^{2}+10\theta^{2}+2x\theta+8\theta+2)}$

5. Order statistics

Let $X_{(1)},X_{(2)},X_{(3)},\ldots,X_{(n)}$ be the ordered statistics of the random sample $X_{1},X_{2},X_{3},\ldots,X_{n}$ drawn from the discrete distribution with cumulative distribution function $F_{X}(x)$ and probability mass function $P_{X}(x)$ , then the probability mass function of rth order statistics $X_{(r)}$ is given by:

$\displaystyle f_{x(r)}({x,\theta})=\frac{n!}{({r-1})!(n-r)!}P(x)\left[{F(x)}% \right]^{r-1}\left[{1-F(x)}\right]^{n-r}.r=1,2,3,\ldots,n$

Using the Eqs (1) and (2), the probability density function of rth order statistics of Poisson Xgamma distribution is given by:

$\displaystyle f_{x(r)}({x,\theta})=\frac{n!}{(r-1)!(n-r)!}\frac{\theta^{2}}{2(% 1+\theta)^{x+4}}({2(1+\theta)^{2}+\theta(x+1)(x+2)})$ $\displaystyle\left[{1-\frac{\theta^{3}+\frac{(\theta x)^{2}}{2}+\frac{5\theta x% ^{2}}{2}+5\theta^{2}+x\theta+4\theta+1}{(1+\theta)^{x+4}}}\right]^{r-1}\left[{% \frac{\theta^{3}+\frac{(\theta x)^{2}}{2}+\frac{5\theta x^{2}}{2}+5\theta^{2}+% x\theta+4\theta+1}{(1+\theta)^{x+4}}}\right]^{n-r}$

Then, the pmf of first order $X_{(1)}$ Poisson Xgamma distribution is given by:

$\displaystyle f_{x(1)}({x,\theta})=\frac{n\theta^{2}}{2(1+\theta)^{x+4}}({2(1+% \theta)^{2}+\theta(x+1)(x+2)})\left[{\frac{\theta^{3}+\frac{(\theta x)^{2}}{2}% +\frac{5\theta x^{2}}{2}+5\theta^{2}+x\theta+4\theta+1}{(1+\theta)^{x+4}}}% \right]^{n-1}.$

and the pmf of nth order $X_{(n)}$ Poisson Xgamma model is given as:

$\displaystyle f_{x(n)}({x,\theta})=\frac{n\theta^{2}}{2(1+\theta)^{x+4}}({2(1+% \theta)^{2}+\theta(x+1)(x+2)})\left[{1-\frac{\theta^{3}+\frac{(\theta x)^{2}}{% 2}+\frac{5\theta x^{2}}{2}+5\theta^{2}+x\theta+4\theta+1}{(1+\theta)^{x+4}}}% \right]^{n-1}$

6. Estimation of parameters

In this section, we discuss the parameter estimation of the Poisson Xgamma distribution using method of maximum likelihood estimation and the frequentist approach such as the method of moments.

6.1 Method of Maximum Likelihood Estimation

This is one of the most useful methods for estimating the different parameters of the distribution. Let $X_{1},X_{2},X_{3},\ldots,X_{n}$ be the random sample of size $n$ , drawn from Poisson Xgamma distribution (PXGD), then the likelihood function of PXGD is given as

$\displaystyle L(x|\theta)=\frac{\theta^{2n}}{2(1+\theta)^{4n}}\prod\limits_{i=% 1}^{n}{\left({\frac{2(1+\theta)^{2}+\theta(x+1)(x+2)}{(1+\theta)^{x}}}\right)}$ $\displaystyle\log L=2n\log\theta-4n\log(2+2\theta)+\sum\limits_{i=1}^{n}{\log(% 2(1+\theta)^{2}+\theta(x+1)(x+2))-\sum\limits_{i=1}^{n}{x_{i}\log(1+\theta)}}$ $\displaystyle\frac{\partial}{\partial\theta}\log L=\frac{2n}{\theta}+\frac{4n}% {2(1+\theta)}+\sum\limits_{i=1}^{n}\frac{4(1+\theta)+(x+1)(x+2)}{2(1+\theta)^{% 2}+\theta(x+1)(x+2)-\frac{\sum_{i=1}^{n}{x_{i}}}{(1+\theta)}=0}$ (4)

The solution of the Eq. (4) is not possible in a closed form, so by using numerical computation, the solution of the log-likelihood Eq. (4) will provide the MLE of $\theta$ .

6.2 Regularity conditions of MLE in Poisson Xgamma distribution

Lemma 1. Suppose that the data $X_{1},X_{2},X_{3},\ldots,X_{n}$ is generated from a Poisson Xgamma distribution with unknown parameters $\Theta_{0}=\theta_{0}$ and $\hat{\Theta}=\hat{\theta}$ is the MLE. We have for any $\Theta=\theta$ , of Poisson Xgamma distribution, $L(\Theta)\leqslant L(\Theta_{0})$ . Moreover, the inequality is strict, $L(\Theta)<L(\Theta_{0})$ unless ${\rm P}_{\Theta_{0}}(f(X/\Theta)=f(X/\Theta_{0}))=1$ , which means that ${\rm P}_{\Theta}={\rm P}_{\Theta_{0}}$ .

Proof Let us consider the difference

$\displaystyle L(\Theta)-L(\Theta_{0})={\rm E}_{\Theta_{0}}(\log f(X/\Theta)-% \log f(X/\Theta_{0}))={\rm E}_{\Theta_{0}}\log\left({\frac{f(X/\Theta)}{f(X/% \Theta_{0})}}\right)$

Since $\log t\leqslant t-1$ , we can write

$\displaystyle{\rm E}_{\hat{\Theta}}\log\left({\frac{f(X/\Theta)}{f(X/\Theta_{0% })}}\right)\leqslant\frac{f(X/\Theta)}{f(X/\Theta_{0})}-1=\int{\left(\frac{f(x% /\Theta)}{f(x/\Theta_{0})}-1\right)}f(x/\Theta_{0})dx=\int{f(x/\Theta)}dx-\int% {f(x/\Theta_{0})}dx=0$

This proves that $L(\Theta)-L(\Theta_{0})\leqslant 0$ .

Theorem 4. Under some regularity conditions on the family of distributions, MLE $\hat{\Theta}=\hat{\theta}$ of Poisson Xgamma Distribution is consistent, i.e. $\hat{\Theta}\to\Theta_{0}$ as $n\to\infty$ .

Proof We have the following facts:

1.
$\hat{\Theta}$ is the maximizer of $L_{n}(\Theta)$ (by definition of ML estimation).
2.
$\Theta_{0}$ is the maximizer of $L(\Theta)$ (by Lemma 4.1).
3.
$\forall\Theta$ , we have $L_{n}(\Theta)\to L(\Theta)$ (by Law of Large numbers).

Therefore, since two functions $L_{n}(\Theta)$ and $L(\Theta)$ are getting closer, the points of maximum should also get closer which exactly means that $\hat{\Theta}\to\Theta_{0}$ .
6.2.1 Asymptotic normality of ML estimates of PXGD

We want to show the asymptotic normality of MLE in Poisson Xgamma Distribution, i.e. to show that $\sqrt{n}({\hat{\Theta}-\Theta_{0}})\to N_{2}(0,I_{x}^{-1}(\Theta_{0}))$ for some variance $I_{x}^{-1}(\Theta_{0})$ . This asymptotic variance in some sense measures the quality of MLE $.I_{x}(\Theta_{0})$ is the Fisher Information regarding $\hat{\Theta}=\hat{\theta}$ .

Proof Since MLE $\hat{\Theta}$ is maximizer of $L_{n}(\Theta)=\frac{1}{n}\sum_{i=1}^{n}\log f(X_{i}/\Theta)$ , we have $L^{\prime}_{n}(\Theta)=0$ .

Let us use the mean value theorem,

$\displaystyle\frac{f(a)-f(b)}{a-b}=f^{\prime}(c)\text{ or }f(a)=f(b)+f^{\prime% }(c)(a-b)\text{ for }c\in[a,b]$

$f(\Theta)=L^{\prime}_{n}(\Theta),a=\hat{\Theta},b=\Theta_{0},$ then we can write, $0=L^{\prime}_{n}(\hat{\Theta})=L^{\prime}_{n}(\Theta_{0})+L^{\prime\prime}_{n}% (\hat{\Theta}_{1})(\hat{\Theta}_{1}-\Theta_{0}),\text{ for }\hat{\Theta}_{1}% \in[\hat{\Theta},\Theta_{0}],$ from here we get that

$\displaystyle\hat{\Theta}_{1}-\Theta_{0}=-\frac{L^{\prime}_{n}(\Theta_{0})}{L^% {\prime\prime}_{n}(\hat{\Theta}_{1})}\text{ and }\sqrt{n}({\hat{\Theta}-\Theta% _{0}})=-\frac{\sqrt{n}L^{\prime}_{n}(\Theta_{0})}{L^{\prime\prime}_{n}(\hat{% \Theta}_{1})}$ (5)

Since by Lemma (1) in the previous section, we know that $\Theta_{0}$ is maximizer of $L(\Theta)$ , we $L^{\prime}(\Theta_{0})=E_{\Theta_{0}}l^{\prime}(X/\Theta_{0})=0$ .

Therefore, the numerator in

$\displaystyle\sqrt{n}L^{\prime}_{n}(\Theta_{0})=\sqrt{n}\left(\frac{1}{n}\sum% \limits_{i=1}^{n}{l^{\prime}(X_{i}/}\Theta_{0}-0)\right)=\sqrt{n}\left({\frac{% 1}{n}\sum\limits_{i=1}^{n}{l^{\prime}(X_{i}/}\Theta_{0})-E_{\Theta_{0}}l^{% \prime}(X_{1}/\Theta_{0})}\right)\to N(0,Var_{\Theta}(l^{\prime}(X_{1}/\Theta_% {0}))$ (6)

converges in distribution by central Limit theorem.

Next, let us consider the denominator in Eq. (5). First of all, we have that for all $\Theta$ , $L^{\prime\prime}_{n}(\Theta)=\frac{1}{n}\sum_{i=1}^{n}{l^{\prime\prime}(X_{i}/% }\Theta)$ $\to$ $E_{\Theta_{0}}l^{\prime\prime}(X_{1}/\Theta)$ by large law of numbers.

Also since, $\hat{\Theta}_{1}\in[\hat{\Theta},\Theta_{0}]$ and by consistency result in Theorem 4, $\hat{\Theta}\to\Theta_{0}$ , we have $\hat{\Theta}_{1}\to\Theta_{0}$ .

We have $L^{\prime\prime}_{n}(\hat{\Theta}_{1})=E_{\Theta_{0}}l^{\prime\prime}(X_{1}/% \Theta_{0})=-I(\Theta_{0})$ .

Combining this with Eq. (6) we get,

$\displaystyle-\frac{\sqrt{n}L^{\prime}_{n}(\Theta_{0})}{L^{\prime\prime}_{n}(% \hat{\Theta}_{1})}\to N\left({0,\frac{\text{Var}_{\Theta_{0}}l^{\prime}(X_{1}/% \Theta_{0})}{(I(\Theta_{0}))^{2}}}\right)$

Finally, the variance $\text{Var}_{\Theta_{0}}(l^{\prime}(X_{1}/\Theta_{0})=E_{\Theta_{0}}(l^{\prime}% (X_{1}/\Theta_{0})^{2}-(E_{\Theta_{0}}(l^{\prime}(X_{1}/\Theta_{0}))^{2}=I(% \Theta_{0})-0$ . Hence $\sqrt{n}(\hat{\Theta}-\Theta_{0})\to N_{2}(0,I_{x}^{-1}(\Theta_{0}))$ , where in the last equality, we used the definition of Fisher information.

6.3 Simulation study

In this section, we investigate the behavior of the ML estimators for a finite sample size n. Simulation study based on different samples from $P$ -Xgamma $({X;\theta})$ distribution is carried out. The random observations are generated by using the inverse cdf method presented in Section 2.1 from $P$ -Xgamma $({X;\theta})$ . A simulation study was carried out for six random parameter values as $\theta=(0.4,0.8,1.3,1.9,2.3,2.4)$ and the process was repeated 1000 times by taking different sample sizes $n=$ (10, 25, 75, 100, 300, 600). The simulated results are given in Table 1. We observe in Table 1 that the agreement between theory and practice improves as the sample size $n$ increases, as all the measures of comparison decrease by increasing the sample size. MSE and Variance of the estimators suggest that the estimators are consistent and the maximum likelihood method performs quite well in estimating the model parameters of the proposed distribution.

Table 1
Simulation study of ML estimators of Poisson Xgamma distribution

Sample size ( $n$ )	$\theta=0.4$				$\theta=0.8$
	Bias	Variance	MSE	Coverage probability (95%)	Bias	Variance	MSE	Coverage probability (95%)
10	0.01369	0.01107	0.01126	0.95200	0.03174	0.01486	0.01587	0.94900
25	0.01227	0.00413	0.00428	0.95800	0.02012	0.00314	0.00355	0.95800
75	0.00400	0.00108	0.00109	0.95800	0.01005	0.00166	0.00176	0.96700
100	0.00232	0.00072	0.00072	0.96700	0.00141	0.00064	0.00064	0.97100
300	0.00122	0.00031	0.00031	0.95300	0.00003	0.00021	0.00021	0.96900
600	0.00091	0.00015	0.00015	0.96200	0.00002	0.00006	0.00006	0.98100
Sample size ( $n$ )	$\theta=1.3$				$\theta=1.9$
	Bias	Variance	MSE	Coverage probability (95%)	Bias	Variance	MSE	Coverage probability (95%)
10	0.02381	0.05321	0.05378	0.94900	0.16088	0.01476	0.04065	0.94900
25	0.01676	0.04645	0.04673	0.95800	0.09607	0.00304	0.01227	0.95800
75	0.01103	0.03919	0.03931	0.96700	0.01179	0.00156	0.00170	0.96700
100	0.09031	0.02115	0.02931	0.97100	0.02803	0.00054	0.00132	0.97100
300	0.00759	0.00930	0.00936	0.96900	0.00931	0.00011	0.00019	0.96900
600	0.00128	0.00011	0.00011	0.98100	0.00293	0.00004	0.00005	0.98100
Sample size ( $n$ )	$\theta=2.2$				$\theta=3.4$
	Bias	Variance	MSE	Coverage probability (95%)	Bias	Variance	MSE	Coverage probability (95%)
10	0.23858	0.53261	0.58953	0.94900	1.07940	9.67985	10.84496	0.92900
25	0.16810	0.46495	0.49321	0.95800	0.59866	2.92271	3.28110	0.94800
75	0.11076	0.39242	0.40469	0.96700	0.05246	0.37736	0.38011	0.95100
100	0.09081	0.21201	0.22025	0.97100	0.06917	0.28800	0.29278	0.96800
300	0.00809	0.09350	0.09356	0.96900	0.06518	0.28009	0.28433	0.96600
600	$-$ 0.00278	0.00159	0.00159	0.98100	0.01618	0.09562	0.09588	0.97100

6.4 Moments method of estimation

In order to estimate the unknown parameter $\theta$ of discrete Poisson Xgamma model by the method of moments, we need to equate first sample moment with the corresponding population moment. $\tau_{1}=\gamma_{1},$ where $\gamma_{1}=\frac{1}{n}\sum_{i=1}^{n}{x_{i}}$ is the first sample moment and $\tau_{1}$ is the first population moment and the solution for $\hat{\theta}$ may be obtained by solving above equation through numerical methods. Equating the first population moment about origin with the first sample moment of $P$ -Xgamma $({X;\theta})$ , we have

$\displaystyle\frac{\theta^{2}+3}{\theta(1+\theta)}=\frac{1}{n}\sum\limits_{i=1% }^{n}{x_{i}}$ $\displaystyle\frac{\theta^{2}+3}{\theta(1+\theta)}=\gamma_{1}$ $\displaystyle(1-\gamma_{1})\theta^{2}-\gamma_{1}\theta+3=0$ $\displaystyle\hat{\theta}=\frac{\gamma_{1}\pm\sqrt{\gamma_{1}^{2}-12(1-\gamma_% {1})}}{2(1-\gamma_{1})},$

which is the moment estimate of the parameter $\theta$ . It should be noted that statistic for $\hat{\theta}$ has certain mathematical restrictions to be greater than zero. Hence, the method is not preferred for parameter estimation of $P$ -Xgamma $({X;\theta})$ .

7. Applications of Poisson Xgamma distribution

In this section, we fit our proposed distribution to two practical datasets, the first representing epileptic seizure counts (Chakraborty, 2010) to illustrate our claim that our proposed model fits well when compared to other competing models. The data set representing epileptic seizure counts has a long right tail and approaches to zero slowly. The data set is given in Table 2. The second data set will be shown in next paragraphs.

Table 2
Dataset representing epileptic seizure counts (Chakraborty, 2010)

Epileptic seizure (X)	0	1	2	3	4	5	6	7	8
Observed counts	126	80	59	42	24	8	5	4	3

In each of these distributions, the parameters are estimated by using the maximum likelihood method. We have analyzed the data using R software. Parameter estimates along with standard errors in brackets and model function of the fitted distributions are given in Table 3. Computationally, the maximum likelihood estimates for the parameters of interest were obtained by Newton Raphson method.

Table 3

Estimated Parameters by ML method for fitted distributions for dataset representing epileptic seizure counts

Distribution	Parameter estimates (standard error)	Model function
Poisson Xgamma distribution	$\hat{\theta}=1.23(0.06)$	$p(x)={\displaystyle\frac{\theta^{2}}{2(1+\theta)^{x+4}}}[{2(1+\theta)^{2}+% \theta(x+1)(x+2)}];\quad x=0,1,2,3,\ldots,\theta>0$
Poisson distribution	$\hat{\lambda}=1.54(0.06)$	$p(x)={\displaystyle\frac{e^{-\lambda}\lambda^{x}}{x!}}\lambda>0;\quad x=0,1,2,\ldots$
Zero inflated poisson	$\begin{array}[]{c}\hat{\alpha}=2.11,\hat{\lambda}=0.27\\ (0.10,0.031)\end{array}$	$p(x)=\begin{cases}\alpha+(1-\alpha){\displaystyle\frac{e^{-\lambda}\lambda^{x}% }{x!}},&\lambda>0;x=0\\ (1-\alpha){\displaystyle\frac{e^{-\lambda}\lambda^{x}}{x!}},&\lambda>0;x=0,1,2% ,\ldots\end{cases}\quad 0<\alpha<1;\lambda>0$
Geometric distribution	$\hat{p}=0.393(0.016)$	$p(x)=q^{x}p0<q<1;\quad q=1-p;\quad x=0,1,2,\ldots$
Negative binomial distribution	$\begin{array}[]{c}\hat{r}=1.55,\hat{p}=0.501\\ (0.27,0.047)\end{array}$	$p(x)=\begin{pmatrix}{x+r-1}\\ x\end{pmatrix}p^{r}q^{x},\quad x=0,1,2,\ldots,r>0\text{ and }0<p<1$
Discrete weibull	$\begin{array}[]{c}\hat{q}=0.66,\hat{\beta}=1.16\\ (0.024,0.059)\end{array}$	$p(x)=q^{x^{\beta}}-q^{(x+1)^{\beta}}\quad 0<q<1;x=0,1,2,\ldots$
Discrete lindley	$\begin{array}[]{c}\hat{p}=0.472,\hat{\theta}=0.615\\ (0.028,0.286)\end{array}$	$p(x)={\displaystyle\frac{p^{x}}{1+\theta}}({\theta(1-2p)+(1-p)(1-\theta x)})% \quad 0<p<1;\theta>0;x=0,1,2,\ldots$
Poisson quasi-lindley	$\begin{array}[]{c}\hat{\theta}=1.11,\hat{\alpha}=0.383\\ (0.332,0.129)\end{array}$	$p(x)={\displaystyle\frac{\theta}{\alpha+1}}\left({{\displaystyle\frac{\alpha(% \theta+1)+\theta(x+1)}{(\theta+1)^{x+2}}}}\right)\quad\theta>0;\alpha>-1;x=0,1% ,2,\ldots$

We compute the expected frequencies for fitting Poisson Xgamma, Poisson, Zero Inflated Poisson, Geometric, Negative Binomial, discrete Weibull (Nakagawa & Osaki, 1975), discrete Lindley (Bakouch et al., 2012) and Poisson Quasi Lindley (Altun, 2019) distributions with the help of R studio statistical software (R version 3.5.3, 2019) and Pearson’s chi-square test is applied to check the goodness of fit of the models. The expected counts and chi square $p$ -value for each fitted model are given in Table 4. Based on the chi-square $p$ -value, we observe that Poisson Xgamma distribution provides a satisfactorily better fit for the data set representing epileptic seizure counts compared to other distributions.

Table 4

Fitted proposed distribution and other competing models to a dataset representing epileptic seizure counts

Epileptic seizure (X)	Observed counts	Poisson Xgamma distribution	Poisson distribution	Zero inflated poisson	Geometric distribution	Negative binomial distribution	Discrete weibull	Discrete lindley	Discrete poisson quasi lindley
0	126	132.981	74.935	126.000	137.963	120.201	120.120	121.867	121.868
1	80	83.371	115.712	65.080	83.736	93.009	92.875	90.942	90.941
2	59	53.367	89.339	68.974	50.823	59.184	59.036	58.745	58.744
3	42	33.497	45.985	48.733	30.847	34.949	35.133	35.216	35.216
4	24	20.391	17.752	25.824	18.722	19.837	20.071	20.167	20.167
5	8	12.037	5.482	10.948	11.363	10.987	11.131	11.197	11.197
6	5	6.914	1.411	3.868	6.897	5.984	6.029	6.079	6.079
7	4	3.878	0.311	1.171	4.186	3.221	3.202	3.245	3.246
8	3	2.132	0.072	0.402	6.464	3.627	1.673	1.710	1.710
Degrees of freedom		6	4	4	6	5	5	5	5
$p$ -value		0.431	$<$ 0.001	0.012	0.144	0.259	0.261	0.331	0.332

Furthermore, in order to compare our proposed distribution and other competing models above, we consider the criteria like AIC (Akaike information criterion), AICC (corrected Akaike information criterion) and BIC (Bayesian information criterion). The better distribution corresponds to lesser AIC, AICC and BIC values.

$\displaystyle\text{AIC}=2k-2\log L,\text{AICC}=\text{AIC}+\frac{2k(k+1)}{n-k-1% }\text{ and BIC}=k\log n-2\log L$

where $k$ is the number of parameters in the statistical model, n is the sample size and $-2\log L$ is the maximized value of the log-likelihood function under the considered model. From Table 5, it has been observed that the Poisson Xgamma distribution has the lesser AIC and BIC values as compared to other competing models. Hence we conclude that the Poisson Xgamma distribution leads to a better fit than the other competing models for analyzing the data set given in Table 2.

Table 5

Model comparison criterion for fitted models to a dataset representing epileptic seizure counts

Criterion	Poisson Xgamma distribution	Poisson distribution	Zero inflated poisson	Geometric distribution	Negative binomial distribution	Discrete weibull	Discrete lindley	Discrete poisson quasi lindley
$-\log l$	595.343	636.045	599.637	598.396	594.942	594.749	594.482	594.482
AIC	1192.687	1274.091	1203.274	1198.791	1193.884	1193.499	1192.964	1192.964
BIC	1196.547	1277.952	1210.996	1202.652	1201.605	1201.220	1200.685	1200.685

In the second data set, analyze the data regarding the distribution of accidents to 647 women working on high explosive shells in 5 weeks, studied by Ghitany and Al-Mutairi (2009). The data set is given in Table 6.

Table 6

Accidents data of 647 women working on high explosive shells in 5 weeks

Number of accidents	0	1	2	3	4	5
Observed count	447	132	42	21	3	2

We also analyze the data in Table 6 using R software (R version 3.5.3, 2019). Parameter estimates along with standard errors in brackets and model function of the fitted distributions are given in Table 7. Computationally, the maximum likelihood estimates for the parameters of interest were obtained by Newton Raphson method.

Table 7

Estimated Parameters by ML method for fitted distributions for dataset in Table 6

Distribution	Parameter Estimates (standard error)	Model function
Poisson Xgamma distribution	$\hat{\theta}=3.17(0.178)$	$p(x)={\displaystyle\frac{\theta^{2}}{2(1+\theta)^{x+4}}}[{2(1+\theta)^{2}+% \theta(x+1)(x+2)}];\quad x=0,1,2,3,\ldots,\theta>0$
Poisson distribution	$\hat{\lambda}=0.47(0.026)$	$p(x)={\displaystyle\frac{e^{-\lambda}\lambda^{x}}{x!}}\lambda>0;\quad x=0,1,2,\ldots$
Zero inflated poisson	$\hat{\alpha}=0.88,\hat{\lambda}=0.47$ $(0.08,0.041)$	$p(x)=\begin{cases}\alpha+(1-\alpha){\displaystyle\frac{e^{-\lambda}\lambda^{x}% }{x!}},&\lambda>0;x=0\\ (1-\alpha){\displaystyle\frac{e^{-\lambda}\lambda^{x}}{x!}},&\lambda>0;x=0,1,2% ,\dots\end{cases}$ $0<\alpha<1;\lambda>0$
Geometric distribution	$\hat{p}=0.682(0.015)$	$p(x)=q^{x}p0<q<1;\quad q=1-p;x=0,1,2,\ldots$
Negative binomial distribution	$\hat{r}=0.86,\hat{p}=0.65$ $(0.18,0.051)$	$p(x)=\begin{pmatrix}{x+r-1}\\ x\end{pmatrix}p^{r}q^{x},\quad x=0,1,2,\ldots,$ $r>0\text{ and }0<p<1$
Discrete weibull	$\hat{q}=0.31,\hat{\beta}=0.96$ $(0.01,0.053)$	$p(x)=q^{x^{\beta}}-q^{(x+1)^{\beta}}$ $0<q<1;x=0,1,2,\ldots$
Discrete lindley	$\hat{p}=0.317,\hat{\theta}=0.001$ $(0.12,0.401)$	$p(x)={\displaystyle\frac{p^{x}}{1+\theta}}({\theta(1-2p)+(1-p)(1-\theta x)})$ $0<p<1;\theta>0;x=0,1,2,\ldots$
Poisson quasi-lindley	$\hat{\theta}=2.15,\hat{\alpha}=484.5$ $(0.161,63.57)$	$p(x)={\displaystyle\frac{\theta}{\alpha+1}}\left({{\displaystyle\frac{\alpha(% \theta+1)+\theta(x+1)}{(\theta+1)^{x+2}}}}\right)$ $\theta>0;\alpha>-1;x=0,1,2,\ldots$

We compute the expected frequencies for fitting Poisson Xgamma and the compared distributions mentioned before to the data set given in Table 6 with the help of R software (R version 3.5.3, 2019). Pearson’s chi-square test is applied to check the goodness of fit of the models discussed. The expected counts and chi square $p$ -value for each fitted model are given in Table 8. Based on the chi-square $p$ -value, we observe that Poisson Xgamma distribution provides a satisfactorily better fit for the data set regarding distribution of accidents to 647 women working on high explosive shells in 5 weeks, studied by Ghitany and Al-Mutairi (2008).

Table 8

Fitted probability models with expected frequencies and chi-square $p$ -value for dataset in Table 6

Number of accidents	Observed count	Poisson Xgamma	Poisson	Zero inflated poisson	Geometric	Negative binomial	Poisson lindley	Discrete weibull	Discrete lindley	Discrete poisson quasi lindley
0.00	447.00	442.56	406.31	447.00	441.57	445.89	439.45	445.54	441.57	441.56
1.00	132.00	138.52	189.03	124.60	140.20	134.90	142.76	135.36	140.20	140.21
2.00	42.00	44.85	43.97	54.95	44.52	43.99	44.97	44.00	44.52	44.52
3.00	21.00	14.47	6.82	16.15	14.13	14.69	13.85	14.62	14.13	14.14
4.00	3.00	4.58	0.79	3.56	4.49	4.96	4.19	4.92	4.49	4.49
5.00	2.00	1.42	0.07	0.63	1.42	1.69	1.25	1.68	1.42	1.43
Chi-square df		3	2	3	3	2	3	2	2	2
$p$ -value		0.302	$<$ 0.001	0.024	0.191	0.157	0.087	0.194	0.093	0.091

Using AIC and BIC criterion for model comparison, we observed that the Poisson Xgamma distribution has lesser AIC and BIC values as compared to other competing models (see Table 9). Hence we conclude that the Poisson Xgamma distribution leads to a better fit than the other competing models for analyzing the data set given in Table 6.

Table 9

Negative loglikelihood, AIC and BIC values for fitted models to data set in Table 6

Criterion	Poisson Xgamma	Poisson	Zero inflated poisson	Geometric	Negative binomial	Poisson lindley	Discrete weibull	Discrete lindley	Discrete poisson quasi lindley
$-\log l$	592.222	617.184	593.272	592.480	592.267	592.708	592.2961	592.4798	592.4797
AIC	1186.445	1236.369	1190.544	1186.960	1188.534	1187.416	1188.592	1188.96	1188.95
BIC	1190.917	1240.841	1199.489	1191.432	1197.479	1191.888	1197.537	1197.904	1197.902

8. Conclusion

A new discrete probability model is introduced using compounding technique. Some important probabilistic properties and the problem of estimation of its parameters are studied. In addition, the discrete Poisson Xgamma distribution is appropriate for modeling both over and under dispersed data, beside equi-dispersion, since, depending on the values of the parameters, its variance can be larger or smaller than the mean, which is not the case with some known standard classical discrete distributions. Applications in handling count data are shown to signify the suitability of the proposed discrete probability model.

Footnotes

Acknowledgments

The authors sincerely thank two anonymous referees and a member of the editorial board for their valuable comments and suggestions that led to improvement of this article.

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Poisson Xgamma distribution: A discrete model for count data analysis

Abstract

Keywords

1. Introduction

2. Definition of proposed model (Poisson Xgamma distribution)

3. Statistical properties

3.1 Moments

3.1.1 Factorial moments

3.2 Probability generating function of the Poisson Xgamma distribution

4. Reliability analysis

4.1 Reliability function R (x)

4.2 Hazard function

4.3 Reverse hazard rate function

5. Order statistics

6. Estimation of parameters

6.1 Method of Maximum Likelihood Estimation

Table 1 Simulation study of ML estimators of Poisson Xgamma distribution

7. Applications of Poisson Xgamma distribution

Table 2 Dataset representing epileptic seizure counts (Chakraborty, 2010)

Footnotes

Acknowledgments

References

Table 1
Simulation study of ML estimators of Poisson Xgamma distribution

Table 2
Dataset representing epileptic seizure counts (Chakraborty, 2010)