The geometric exponential Poisson distribution

Abstract

Many if not most lifetime distributions are motivated only by mathematical interest. Here, a new three-parameter distribution motivated mainly by lifetime issues is introduced. Some properties of the new distribution including estimation procedures, univariate generalizations and bivariate generalizations are derived. Two real data applications are described to show superior performance versus some known lifetime models.

Appendix

The calculations of the paper require the following lemmas.

Lemma 1

If a random variable $X$ has the GEP distribution then

$$\begin{aligned}&B (a, b, c)= E \left\{ \frac{\exp \left[ -a \lambda X - b \theta \exp (-\lambda X) \right] }{\left[ 1 - \exp (-\theta ) - \eta \left\{ 1 - \exp \left[ -\theta \exp (-\lambda X) \right] \right\} \right] ^c} \right\} \\&\quad = \frac{\theta (1 - \eta ) \left[ 1 - \exp (-\theta ) \right] }{\left[ 1 - \exp (-\theta ) - \eta \right] ^{2 + c}} \sum _{k = 0}^\infty {-2 - c \atopwithdelims ()k} \frac{\eta ^k \gamma \left( a + 2, (1 + b + k) \theta \right) }{\left[ 1 - \exp (-\theta ) - \eta \right] ^k (1 + b + k)^{a + 2} \theta ^{a + 2}}. \end{aligned}$$

Proof

Using the series expansion, (5), we can write

$$\begin{aligned} B (a, b, c)&= \theta \lambda (1 - \eta ) \left[ 1 - \exp (-\theta ) \right] \int \limits _0^\infty \frac{\exp \left[ -(a + 1) \lambda x - (b + 1) \theta \exp (-\lambda x) \right] }{\left[ 1 - \exp (-\theta ) - \eta \left\{ 1 - \exp \left[ -\theta \exp (-\lambda x) \right] \right\} \right] ^{2 + c}} dx \\&= \theta (1 - \eta ) \left[ 1 - \exp (-\theta ) \right] \int \limits _{0}^{1} \frac{y^{a + 1} \exp \left[ -(b + 1) \theta y \right] }{\left[ 1 - \exp (-\theta ) - \eta + \eta \exp (-\theta y) \right] ^{2 + c}} dy\\&= \frac{\theta (1 - \eta ) \left[ 1 - \exp (-\theta ) \right] }{\left[ 1 - \exp (-\theta ) - \eta \right] ^{2 + c}} \sum _{k = 0}^\infty {-2 - c \atopwithdelims ()k} \left[ \frac{\eta }{1 - \exp (-\theta ) - \eta } \right] ^{k} \\&\times \int \limits _{0}^{1} y^{a + 1} \exp \left[ -(b + k + 1) \theta y \right] dy. \end{aligned}$$

The result follows by the definition of the incomplete gamma function. $\square$

Lemma 2

If a random variable $X$ has the GEP distribution then

$$\begin{aligned} D (a, b, c, d)&= E \left\{ \frac{X^{d} \exp \left[ -a \lambda X - b \theta \exp (-\lambda X) \right] }{\left[ 1 - \exp (-\theta ) - \eta \left\{ 1 - \exp \left[ -\theta \exp (-\lambda X) \right] \right\} \right] ^{c}} \right\} \\&= \frac{\theta (1 - \eta ) \left[ 1 - \exp (-\theta ) \right] }{(-\lambda )^d \left[ 1 - \exp (-\theta ) - \eta \right] ^{2 + c}} \sum _{k = 0}^\infty {-2 - c \atopwithdelims ()k}\nonumber \\&\times \left. \frac{\partial ^d}{\partial s^d} \frac{\eta ^{k} \gamma \left( s + a + 2, (1 + b + k) \theta \right) }{\left[ 1 - \exp (-\theta ) - \eta \right] ^{k} (1 + b + k)^{a + 2} \theta ^{a + 2}} \right| _{s = 0}. \end{aligned}$$

Proof

Using the series expansion, (5), we can write

$$\begin{aligned} D (a, b, c, d)&= \theta \lambda (1 - \eta ) \left[ 1 \!-\! \exp (-\theta ) \right] \int \limits _{0}^{\infty } \frac{x^{d} \exp \left[ -(a \!+\! 1) \lambda x \!-\! (b\! + \!1) \theta \exp (-\lambda x) \right] }{\left[ 1 \!-\! \exp (-\theta ) \!-\! \eta \left\{ 1 \!-\! \exp \left[ -\theta \exp (-\lambda x) \right] \right\} \right] ^{2 \!+\! c}} dx \\&= \theta (-\lambda )^{-d} (1 - \eta ) \left[ 1 - \exp (-\theta ) \right] \int \limits _{0}^{1} \frac{(\log y)^d y^{a + 1} \exp \left[ -(b + 1) \theta y \right] }{\left[ 1 - \exp (-\theta ) - \eta + \eta \exp (-\theta y) \right] ^{2 + c}} dy \\&= \frac{\theta (1 - \eta ) \left[ 1 - \exp (-\theta ) \right] }{(-\lambda )^d \left[ 1 - \exp (-\theta ) - \eta \right] ^{2 + c}} \sum _{k = 0}^\infty {-2 - c \atopwithdelims ()k} \left[ \frac{\eta }{1 - \exp (-\theta ) - \eta } \right] ^{k} \\&\times \int \limits _0^1 (\log y)^d y^{a + 1} \exp \left[ -(b + k + 1) \theta y \right] dy \\&= \frac{\theta (1 - \eta ) \left[ 1 - \exp (-\theta ) \right] }{(-\lambda )^d \left[ 1 - \exp (-\theta ) - \eta \right] ^{2 + c}} \sum _{k = 0}^\infty {-2 - c \atopwithdelims ()k} \left[ \frac{\eta }{1 - \exp (-\theta ) - \eta } \right] ^{k} \\&\times \left. \frac{\partial ^{d}}{\partial s^{d}} \int \limits _{0}^{1} y^{s + a + 1} \exp \left[ -(b + k + 1) \theta y \right] dy \right| _{s = 0}. \end{aligned}$$

The result follows by the definition of the incomplete gamma function.