Comment on Aljohani et al. The Uniform Poisson–Ailamujia Distribution: Actuarial Measures and Applications in Biological Science. Symmetry 2021, 13, 1258

Abstract

In this note, we would like to point out that the uniform Poisson–Ailamujia introduced by Aljohani et al. is a reparametrized geometric distribution.

Recently, Aljohani and colleagues [1] proposed a new asymmetric distribution called the uniform Poisson–Ailamujia (UPA) distribution using a Poisson compounding scheme. In this scheme, the probability mass function (pmf) of a random variable $X$ can be obtained by taking

$$\mathrm{Pr}\left(X=x\right)={\displaystyle \sum _{n=x}^{\infty }\mathrm{Pr}\left(N=n\right)\mathrm{Pr}\left(X=x|N=n\right)},$$

(1)

where for UPA distribution, $X|N$ follows a discrete uniform distribution, $U\left(n\right)$ with $n>0$ and $N$ follows the Poisson–Ailamujia distribution, $PA\left(\alpha \right)$ with $\alpha >0$. The technique used by [1] is a powerful technique in obtaining new distribution. Therefore, the pmf of UPA distribution can be written as

$$\mathrm{Pr}\left(X=x|\alpha \right)=\frac{2\alpha }{{\left(1+2\alpha \right)}^{x+1}};x=0,1,2,\dots$$

(2)

However, we found that pmf in Equation (2) can be reparametrized by changing variable $\alpha$ via a function $\theta$, such that

$$\theta \left(\alpha \right)=\frac{2\alpha }{1+2\alpha }$$

(3)

yielding

$$\mathrm{Pr}\left(X=x|\theta \right)={\left(1-\theta \right)}^x\theta ;x=0,1,2,\dots ,0<\theta <1,$$

(4)

which is equivalent to a geometric distribution with parameter $\theta$. In geometric distribution, parameter $\theta$ is widely known as the probability of success, whereas parameter $\alpha$ is a scale parameter because it either stretches or shrinks the trend of the probability values. Being a scale parameter, $\alpha$ can take on any large positive values. Based on Equation (3), it is easy to show that $\underset{\alpha \to 0}{\lim }\theta \left(\alpha \right)=0$ and $\underset{\alpha \to \infty }{\lim }\theta \left(\alpha \right)=1$, which is equivalent to the range of $\theta$ in a geometric distribution.

The formulae of mean and maximum likelihood estimation (MLE) of $\alpha$ further confirm the equivalence of the UPA distribution to the geometric distribution. As stated by the authors in [1], the respective formulae of mean and MLE of $\alpha$ for the UPA distribution are given as

$$\mu =\frac{1}{2\alpha },$$

(5)

$$\widehat{\alpha }=\frac{1}{2\overline{x}},$$

(6)

respectively, where $\overline{x}$ is the sample mean. By rearranging (3), the following is obtained

$$2\alpha =\frac{\theta }{1-\theta }$$

(7)

By substituting (7) into (5), the standard formula of the mean for geometric distribution expressed as $\mu ={\theta }^{-1}\left(1-\theta \right)$ can be obtained. Similarly, by substituting (7) into (6), one can obtain the MLE of $\theta$ for the geometric distribution, which is expressed as $\widehat{\theta }={\left(1+\overline{x}\right)}^{-1}$.

We also fitted the dataset that corresponds to the number of chromatid aberrations [2], as adopted in [1] to a geometric distribution. The model fittings of the data using both the UPA and geometric distributions are identical, as shown in

Table 1. Model fittings of the number of chromatid aberrations to the UPA and geometric distributions.

Count	Observed	Distribution
		UPA	Geometric
0	268	264.03	264.03
1	87	89.75	89.75
2	26	30.51	30.51
3	9	10.37	10.37
4	10	3.53	5.34
Estimated parameter		$\widehat{\alpha }=0.9709$	$\widehat{\theta }=0.6601$
${\chi }^2$		5.33	5.06 *

* test statistics when ${\widehat{f}}_4^{*}=5.34$.

. Although $\mathrm{Pr}\left(X=4|\widehat{\alpha }\right)=\mathrm{Pr}(X=4|\widehat{\theta })=0.0088$ yields ${\widehat{f}}_4\approx 3.53$, ${\widehat{f}}_4$ does not bring the total of fitted data to $400$, as it does with the total observed data. Therefore, the frequency of data when $x=4, {\widehat{f}}_4^{*}$ can be calculated using the sum of the remaining fitted values, such that

$${\widehat{f}}_4^{*}=n-{\displaystyle \sum _{x=0}^3{\widehat{f}}_x}=n-\left({\widehat{f}}_0+{\widehat{f}}_1+{\widehat{f}}_2+{\widehat{f}}_3\right).$$

(8)

This will improve the ${\chi }^2$ statistics from 5.33 to 5.06. There is also an inconsistency with the ${\chi }^2$ value in Table 9 of [1] and the calculated ${\chi }^2$ value.

In their abstract, the authors proclaimed that the UPA distribution can be a good alternative to both Poisson and geometric distributions. However, the authors did not compare model fitting of the UPA distribution to the geometric distribution. It is suggested that the authors should compare both UPA and geometric distributions, especially regarding actuarial measures and different estimation techniques, and further solidify their findings.

In any case, the proposed work in [1] is built on the foundation of the geometric distribution. The extensive and comprehensive simulation study that was conducted to evaluate the performance of eight estimators significantly contributes to computational statistics. Some, if not all of, the properties and measures may have also contributed to statistical knowledge. We encourage the authors to respond to our comments, and we hope a fruitful discussion for the advancement of statistical knowledge can take place.