Characterizations of the geometric distribution via residual lifetime

Abstract

In this work, some characterizations of the geometric distribution based on two kinds of residual lifetime are presented. Firstly we characterize geometric distribution by using certain relationships of moments of the truncated life from below, that is $\max \{X-t,0\}$, where X is a positive integer-valued random variable. Secondly using certain relationships of moments of residual life $\gamma _{t}$ at t of a renewal process, we characterize the common distribution of the inter-arrival times $\{X_{i},i\ge 1\}$ to be geometrically distributed when $X_{1}$ is also a positive integer-valued random variable.

Appendix

For every $n\in \mathcal {N}$, let $\phi ^{\left( n\right) }(\theta )$ be the $n^{th}$ derivative of $\phi (\theta )$. However, we also denote $\phi ^{\left( 0\right) }(\theta )=\phi (\theta )$, $\phi ^{\left( 1\right) }(\theta )=\phi ^{\prime }(\theta ),$ $\phi ^{\left( 2\right) }(\theta )=\phi ^{\prime \prime }(\theta )$ and $\phi ^{\left( 3\right) }(\theta )=\phi ^{\prime \prime \prime }(\theta )$ for convenience. On the other hand, for q(t), $\forall t\in \mathcal {N}$, we denote L(q(t)) the Laplace–Stieltjes transform of q(t), that is

$$\begin{aligned} L(q(t))=\sum _{t=0}^{\infty }e^{-\theta t}q(t),\quad \theta \ge 0. \end{aligned}$$

Now, for $n\in \mathcal {N},$ define

$$\begin{aligned} q_{1}(t;n)= & {} \sum _{x\,=\,t+1}^{\infty }x^{n}P(X=x),\quad \forall t\in \mathcal {N}, \end{aligned}$$

(50)

$$\begin{aligned} q_{2}(t;n)= & {} t\sum _{x\,=\,t+1}^{\infty }x^{n}P(X=x),\quad \forall t\in \mathcal {N}, \end{aligned}$$

(51)

$$\begin{aligned} q_{3}(t;n)= & {} t^{2}\sum _{x\,=\,t+1}^{\infty }x^{n}P(X=x),\quad \forall t\in \mathcal { N}, \end{aligned}$$

(52)

and

$$\begin{aligned} q_{4}(t;n)=t^{3}\sum _{x=t+1}^{\infty }x^{n}P(X=x),\quad \forall t\in \mathcal {N}. \end{aligned}$$

(53)

In the following, we will calculate the Laplace–Stieltjes transforms of (50)–(53), respectively. These results are useful to characterize distribution in this work.

Lemma 2

$$\begin{aligned} L(q_{1}(t;n))=\frac{\mu _{n}-(-1)^{n}\phi ^{\left( n\right) }(\theta )}{ 1-e^{-\theta }},\quad \theta \ge 0. \end{aligned}$$

Proof

The proof is completed upon noting that

$$\begin{aligned} L(q_{1}(t;n))= & {} \sum _{t=0}^{\infty }e^{-\theta t}\left( \sum _{x=t+1}^{\infty }x^{n}P(X=x)\right) \\= & {} \sum _{x=1}^{\infty }\sum _{t=0}^{x-1}e^{-\theta t}x^{n}P(X=x) \\= & {} \sum _{x=1}^{\infty }\left( \frac{1-e^{-\theta x}}{1-e^{-\theta }}\right) x^{n}P(X=x) \\= & {} \frac{1}{1-e^{-\theta }}\left( \sum _{x=1}^{\infty }x^{n}P(X=x)-\sum _{x=1}^{\infty }e^{-\theta x}x^{n}P(X=x)\right) \end{aligned}$$

and

$$\begin{aligned} \phi ^{\left( n\right) }(\theta )=(-1)^{n}\sum _{x=1}^{\infty }e^{-\theta x}x^{n}P(X=x). \end{aligned}$$

(54)

$\square$

Lemma 3

$$\begin{aligned} L(q_{2}(t;n))=\frac{e^{-\theta }\mu _{n}-(-1)^{n}e^{-\theta }\phi ^{\left( n\right) }(\theta )+(-1)^{n}(1-e^{-\theta })\phi ^{\left( n+1\right) }(\theta )}{(1-e^{-\theta })^{2}},\quad \theta \ge 0. \end{aligned}$$

Proof

The proof is completed upon noting that

$$\begin{aligned} L(q_{2}(t;n))= & {} \sum _{t=0}^{\infty }e^{-\theta t}\left( t\sum _{x=t+1}^{\infty }x^{n}P(X=x)\right) \\= & {} \sum _{x=1}^{\infty }\left( \sum _{t=0}^{x-1}te^{-\theta t}\right) x^{n}P(X=x) \\= & {} \sum _{x=1}^{\infty }\left( \frac{e^{-\theta }-xe^{-\theta x}+(x-1)e^{-(x+1)\theta }}{(1-e^{-\theta })^{2}}\right) x^{n}P(X=x) \\= & {} \frac{1}{(1-e^{-\theta })^{2}}\left( \sum _{x=1}^{\infty }\left( e^{-\theta }{-}xe^{-\theta x}{+}xe^{-\theta x}e^{-\theta }{-}e^{-\theta x}e^{-\theta }\right) x^{n}P(X=x)\right) \\= & {} \frac{1}{(1-e^{-\theta })^{2}}\left( \sum _{x=1}^{\infty }\left( e^{-\theta }-(1-e^{-\theta })xe^{-\theta x}-e^{-\theta x}e^{-\theta }\right) x^{n}P(X=x)\right) \\= & {} \frac{1}{(1-e^{-\theta })^{2}}\left( e^{-\theta }\sum _{x=1}^{\infty }x^{n}P(X=x){-}(1-e^{-\theta })\sum _{x=1}^{\infty }e^{-\theta x}x^{n+1}P(X=x)\right. \\&\left. -e^{-\theta }\sum _{x=1}^{\infty }e^{-\theta x}x^{n}P(X=x)\right) \end{aligned}$$

and formula (54). $\square$

Lemma 4

$$\begin{aligned} L(q_{3}(t;n))= & {} \frac{1}{1-e^{-\theta }}\Big (\left. \mu _{n}-(-1)^{n}\phi ^{\left( n+2\right) }(\theta )-2(-1)^{n}\phi ^{\left( n+1\right) }(\theta )-(-1)^{n}\phi ^{\left( n\right) }(\theta )\right. \\&\left. +2L(q_{2}(t;n))-L(q_{1}(t;n))\right. \Big ),\quad \theta \ge 0. \end{aligned}$$

Proof

The proof is completed upon noting that

$$\begin{aligned}&L(q_{3}(t;n))=\sum _{t=0}^{\infty }e^{-\theta t}\left( t^{2}\sum _{x=t+1}^{\infty }x^{n}P(X=x)\right) \\&\quad =\sum _{x=1}^{\infty }\left( \sum _{t=0}^{x-1}t^{2}e^{-\theta t}\right) x^{n}P(X=x) \\&\quad =\sum _{x=1}^{\infty }\left( \frac{1-e^{-\theta x}(x-1)^{2}+\sum _{t=0}^{x-1}\left( 2t-1\right) e^{-\theta t}}{1-e^{-\theta }} \right) x^{n}P(X=x) \\&\quad =\frac{1}{1-e^{-\theta }}\left( \sum _{x=1}^{\infty }\left( 1-e^{-\theta x}(x-1)^{2}+2\sum _{t=0}^{x-1}te^{-\theta t}-\sum _{t=0}^{x-1}e^{-\theta t} \right) x^{n}P(X=x)\right) \\&\quad =\frac{1}{1-e^{-\theta }}\left( \sum _{x=1}^{\infty }x^{n}P(X=x)-\sum _{x=1}^{\infty }e^{-\theta x}(x-1)^{2}x^{n}P(X=x)\right. \\&\qquad +\left. 2\sum _{x=1}^{\infty }\sum _{t=0}^{x-1}te^{-\theta t}x^{n}P(X=x)-\sum _{x=1}^{\infty }\sum _{t=0}^{x-1}e^{-\theta t}x^{n}P(X=x)\right) \\&\quad =\frac{1}{1-e^{-\theta }}\left( \mu _{n}-\sum _{x=1}^{\infty }e^{-\theta x}(x-1)^{2}x^{n}P(X=x)+2L(q_{2}(t;n))-L(q_{1}(t;n))\right) \\&\quad =\frac{1}{1-e^{-\theta }}\left( \mu _{n}-\sum _{x=1}^{\infty }e^{-\theta x}x^{n+2}P(X=x)+2\sum _{x=1}^{\infty }e^{-\theta x}x^{n+1}P(X=x)\right. \\&\qquad -\left. \sum _{x=1}^{\infty }e^{-\theta x}x^{n}P(X=x)+2L(q_{2}(t;n))-L(q_{1}(t;n))\right) \\&\quad =\frac{1}{1-e^{-\theta }}\left( \mu _{n}-(-1)^{n+2}\phi ^{\left( n+2\right) }(\theta )+2(-1)^{n+1}\phi ^{\left( n+1\right) }(\theta )-(-1)^{n}\phi ^{\left( n\right) }(\theta )\right. \\&\qquad \left. +\,2L(q_{2}(t;n))-L(q_{1}(t;n))\right) \end{aligned}$$

and formula (54). $\square$

Remark 5

In particular, by Lemmas 2 and 3, we have that if $n=0,$ then

$$\begin{aligned} L(q_{3}(t;0))= & {} \frac{e^{-\theta }\left( 1+e^{-\theta }\right) }{(1-e^{-\theta })^{3}}-\frac{e^{-\theta }\left( 1+e^{-\theta }\right) }{(1-e^{-\theta })^{3} }\phi (\theta )\\&+\frac{2e^{-\theta }}{(1-e^{-\theta })^{2}}\phi ^{\prime }(\theta )-\frac{1}{1-e^{-\theta }}\phi ^{\prime \prime }(\theta ); \end{aligned}$$

if $n=1,$ then

$$\begin{aligned} L(q_{3}(t;1))= & {} \frac{\mu _{1}e^{-\theta }\left( 1+e^{-\theta }\right) }{ (1-e^{-\theta })^{3}}+\frac{e^{-\theta }\left( 1+e^{-\theta }\right) }{ (1-e^{-\theta })^{3}}\phi ^{\prime }(\theta )\\&-\frac{2e^{-\theta }}{ (1-e^{-\theta })^{2}}\phi ^{\prime \prime }(\theta )+\frac{1}{1-e^{-\theta }} \phi ^{\prime \prime \prime }(\theta ). \end{aligned}$$

Lemma 5

$$\begin{aligned} L(q_{4}(t;n))= & {} \frac{1}{1-e^{-\theta }}\left( -\mu _{n}+(-1)^{n}\phi ^{\left( n+3\right) }(\theta )\right. \\&+3(-1)^{n}\phi ^{\left( n+2\right) }(\theta )+3(-1)^{n}\phi ^{\left( n+1\right) }(\theta )\\&\left. +(-1)^{n}\phi ^{\left( n\right) }(\theta )+3L(q_{3}(t;n))-3L(q_{2}(t;n))+L(q_{1}(t;n))\right) ,\quad \theta \ge 0. \end{aligned}$$

Proof

The proof is completed upon noting that

$$\begin{aligned} L(q_{4}(t;n))= & {} \sum _{t=0}^{\infty }e^{-\theta t}\left( t^{3}\sum _{x=t+1}^{\infty }x^{n}P(X=x)\right) \\= & {} \sum _{x=1}^{\infty }\left( \sum _{t=0}^{x-1}t^{3}e^{-\theta t}\right) x^{n}P(X=x) \\= & {} \sum _{x=1}^{\infty }\left( \frac{-1-\left( x-1\right) ^{3}e^{-\theta x}+\sum _{t=0}^{x-1}\left( 3t^{2}-3t+1\right) e^{-\theta t}}{1-e^{-\theta }} \right) x^{n}P(X=x) \\= & {} \sum _{x=1}^{\infty }\left( \frac{-1-(x^{3}-3x^{2}+3x-1)e^{-\theta x}+\sum _{t=0}^{x-1}(3t^{2}-3t+1)e^{-\theta t}}{1-e^{-\theta }}\right) x^{n}P(X=x) \\= & {} \frac{1}{1-e^{-\theta }}\left( -\mu _{n}-(-1)^{n+3}\phi ^{\left( n+3\right) }(\theta )+3(-1)^{n+2}\phi ^{\left( n+2\right) }(\theta )-3(-1)^{n+1}\phi ^{\left( n+1\right) }(\theta )\right. \\&\left. +(-1)^{n}\phi ^{\left( n\right) }(\theta )+3L(q_{3}(t;n))-3L(q_{2}(t;n))+L(q_{1}(t;n))\right) \end{aligned}$$

and formula (54). $\square$

Remark 6

In particular, by Lemmas 2 and 4, we have that if $n=0,$ then

$$\begin{aligned} L(q_{4}(t;0))= & {} \frac{e^{-\theta }(1+4e^{-\theta }+e^{-2\theta })}{ (1-e^{-\theta })^{4}}-\frac{e^{-\theta }(1+4e^{-\theta }+e^{-2\theta })}{ (1-e^{-\theta })^{4}}\phi (\theta ) \\&+\frac{3e^{-\theta }(1+e^{-\theta })}{(1-e^{-\theta })^{3}}\phi ^{\prime }(\theta )-\frac{3e^{-\theta }}{(1-e^{-\theta })^{2}}\phi ^{\prime \prime }(\theta )+\frac{1}{(1-e^{-\theta })}\phi ^{\prime \prime \prime }(\theta ) \end{aligned}$$