On the Time-Dependent Delta-Shock Model Governed by the Generalized PóLya Process

Abstract

One of the widely discussed in the literature and relevant in practice shock models is the delta-shock model that is described by the constant time of a system’s recovery after a shock. However, in practice, as time progresses and due to deterioration of a system, this recovery time is gradually increasing. This important phenomenon was not discussed in the literature so far. Therefore, in this paper, we are considering a time-dependent delta-shock model, i.e., the recovery time becomes an increasing function of time. Moreover, we assume that shocks occur according to the generalized Pólya process that contains the homogeneous Poisson process, the non-homogeneous Poisson process and the Pólya process as particular cases. For the defined survival model, we derive the corresponding survival function and the mean lifetime and study the related optimal replacement policy along with some relevant stochastic properties.

Appendix

Proof of Theorem 3.1

Note that

$$ \bar F_{L}(t)=P(L > t ) = P(L > t , N(t) = 0) + \sum\limits_{m = 1}^{\infty} P(L > t , N(t) = m). $$

(A1)

Now, it is easy to verify that

$$ P(L > t , N(t) = 0) = \exp \{ - \beta {\Lambda}(t) \}. $$

(A2)

Further, note that the system survives n shocks in [0, t) provided $X_{1} > \delta _{0},X_{2} > \delta (T_{1}), \dots ,$ X_n > δ(T_n− 1), or equivalently, $T_{1} > \delta _{0} , T_{2} > T_{1} + \delta (T_{1}), \dots ,T_{n} > T_{n-1}+\delta (T_{n-1})$. As g(t) = t + δ(t), this condition can equivalently be written as $T_{1} > \delta _{0} , T_{2} > g(T_{1}),\dots ,T_{n} > g(T_{n-1})$. Further, this implies that t > T_n > g^n− 1(δ₀) as g(⋅) is a strictly increasing continuous function. This means that the probability of the event “the system survives n shocks till time t” is zero, for t ≤ g^n− 1(δ₀). Now, for 1 ≤ m ≤ M₀(t), we can write

$$ P(L > t , N(t) = m) = P(L > t | N(t) = m) P(N(t) = m) . $$

(A3)

Now, consider

$$ \begin{array}{@{}rcl@{}} && P(L > t | N(t) = m) \\ & &= P(X_{1} > \delta_{0}, X_{2} > \delta(T_{1}),\dots,X_{m} > \delta(T_{m-1}) | N(t) = m) \\ && = P(T_{1} > \delta_{0}, T_{2} > T_{1} + \delta(T_{1}),\dots,T_{m} > T_{m-1}+\delta(T_{m-1}) | N(t) = m) \\ && = P(T_{1} > \delta_{0}, T_{2} > g(T_{1}),\dots,T_{m} > g(T_{m-1}) | N(t) = m) \\ & & = {\int}_{g^{m-1}(\delta_{0})}^{t} {\int}_{g^{m-2}(\delta_{0})}^{g^{-1}(t_{m})} {\dots} {\int}_{g(\delta_{0})}^{g^{-1}(t_{3})} {\int}_{\delta_{0}}^{g^{-1}(t_{2})} f_{(T_{1},T_{2},\dots, T_{N(t)} | N(t))}(t_{1},t_{2},\dots,t_{m-1}t_{m} | m) \\&& \times dt_{1} dt_{2} {\dots} dt_{m-1}dt_{m} \\ && = \frac{m !}{ \left( \exp{ \{ \alpha {\Lambda}(t) \} } - 1\right)^{m}} {\int}_{g^{m-1}(\delta_{0})}^{t} {\int}_{g^{m-2}(\delta_{0})}^{g^{-1}(t_{m})} {\dots} {\int}_{g(\delta_{0})}^{g^{-1}(t_{3})} {\int}_{\delta_{0}}^{g^{-1}(t_{2})} \left (\prod\limits_{i = 1}^{m} \alpha \lambda(t_{i}) \exp \{ \alpha {\Lambda}(t_{i}) \} \right) \\&& \times dt_{1} dt_{2} {\dots} dt_{m-1} dt_{m}, \end{array} $$

where the last equality follows from Lemma 3.1(b). On using the above expression together with Lemma 3.1(a) in Eq. A3, we get

$$ \begin{array}{@{}rcl@{}} &&\!\!\!\!\!P(L > t,N(t) = m) \\ & &\!\!\!\!\!= \frac{\Gamma\left( \frac{\beta}{\alpha} + m\right)}{\Gamma\left( \frac{\beta}{\alpha}\right) m!} (1 - exp \{ - \alpha {\Lambda}(t) \} )^{m} (exp \{ - \alpha {\Lambda}(t)\})^{\frac{\beta}{\alpha}} \frac{m !}{ \left( \exp{ \{ \alpha {\Lambda}(t) \} } - 1\right)^{m}} \\ &&\!\times {\int}_{g^{m-1}(\delta_{0})}^{t} {\int}_{g^{m-2}(\delta_{0})}^{g^{-1}(t_{m})} {\dots} {\int}_{g(\delta_{0})}^{g^{-1}(t_{3})} {\int}_{\delta_{0}}^{g^{-1}(t_{2})} \left (\prod\limits_{i = 1}^{m} \alpha \lambda(t_{i}) \exp \{ \alpha {\Lambda}(t_{i}) \} \right) dt_{1} dt_{2} {\dots} dt_{m-1} dt_{m} \\ && \!\!\!\!\!= \exp \{ - \beta {\Lambda}(t)\} \frac{\Gamma\left( \frac{\beta}{\alpha} + m\right)}{\Gamma\left( \frac{\beta}{\alpha}\right)} \frac{1}{ (\exp{ \{\alpha {\Lambda}(t)\})^{m}} } \\ && \!\times {\int}_{g^{m-1}(\delta_{0})}^{t} {\int}_{g^{m-2}(\delta_{0})}^{g^{-1}(t_{m})} {\dots} {\int}_{g(\delta_{0})}^{g^{-1}(t_{3})} {\int}_{\delta_{0}}^{g^{-1}(t_{2})} \left( \prod\limits_{i = 1}^{m} \alpha \lambda(t_{i}) \exp \{ \alpha {\Lambda}(t_{i}) \} \right) dt_{1} dt_{2} {\dots} dt_{m-1}dt_{m}. \\ && \end{array} $$

(A4)

Finally, on using Eqs. A2 and A4 in A1, we get the result. □

Proof of Corollary 3.1

Since δ(t) = δ₀ + γt, we have g(t) = (1 + γ)t + δ₀ and $g^{-1}(t) = \frac {t - \delta _{0}}{(1 + \gamma )}$. Then

$$ \begin{array}{@{}rcl@{}} M_{0}(t) & =& max\{ n \geq 1 | g^{n-1}(\delta_{0}) < t \} \\ & =& max\{ n \geq 1 | \delta_{0} + (1 + \gamma) \delta_{0} + {\dots} + (1+\gamma)^{n-1} \delta_{0} < t \}\\ & =& max\{ n \geq 1 | \delta_{0}\left( \frac{ ((1 + \gamma )^{n} - 1) }{\gamma} \right) < t \} \\ & =& \left\lfloor \frac{\ln{ \left( \frac{ \delta_{0} + t \gamma }{\delta_{0} } \right) }}{ \ln{(1 + \gamma) }} \right\rfloor=N_{0}(t). \end{array} $$

(A5)

Now, consider the following integral.

$$ \begin{array}{@{}rcl@{}} B(t)&\underset{=}{\text{def}}& {\int}_{g^{m-1}(\delta_{0})}^{t} {\int}_{g^{m-2}(\delta_{0})}^{g^{-1}(t_{m})} \dots{\int}_{g(\delta_{0})}^{g^{-1}(t_{3})} {\int}_{\delta_{0}}^{g^{-1}(t_{2})}\left (\prod\limits_{i = 1}^{m} \alpha \lambda(t_{i}) \exp \{ \alpha {\Lambda}(t_{i}) \} \right) dt_{1} dt_{2} {\dots} dt_{m-1} dt_{m} \\ & = & {\int}_{\delta_{0} + (1 + \gamma) \delta_{0} + \cdot \cdot \cdot + (1+\gamma)^{m-1} \delta_{0}}^{t} {\int}_{\delta_{0} + (1 + \gamma) \delta_{0} + {\dots} + (1+\gamma)^{m-2} \delta_{0}}^{\frac{t_{m} - \delta_{0}}{1 + \gamma}} \cdot \cdot \cdot {\int}_{(2 + \gamma) \delta_{0}}^{\frac{t_{3} - \delta_{0}}{1 + \gamma}} {\int}_{\delta_{0}}^{\frac{t_{2} - \delta_{0}}{1 + \gamma}} \left( \frac{\alpha}{b}\right)^{m} \\ &&\times dt_{1} dt_{2} {\dots} dt_{m-1} dt_{m} \\ &=&{\int}_{\frac{ (1+\gamma)^{m} -1 }{\gamma} \delta_{0}}^{t}{\int}_{\frac{ (1+\gamma)^{m-1} -1 }{\gamma} \delta_{0}}^{\frac{t_{m} - \delta_{0}}{1+\gamma}} {\dots} {\int}_{(2 + \gamma) \delta_{0}}^{\frac{t_{3} - \delta_{0}}{1 + \gamma}} {\int}_{\delta_{0}}^{\frac{t_{2} - \delta_{0}}{1 + \gamma}} \left( \frac{\alpha}{b} \right)^{m} dt_{1} dt_{2} {\dots} dt_{m-1} dt_{m}. \end{array} $$

(A6)

We solve the above integration by iterative process. Note that

$$ \begin{array}{@{}rcl@{}} {\int}_{(2+\gamma)\delta_{0}}^{\frac{t_{3} - \delta_{0}}{1 + \gamma}} {\int}_{\delta_{0}}^{\frac{t_{2} - \delta_{0}}{1 + \gamma}} \left( \frac{\alpha}{b}\right)^{2} dt_{1} dt_{2} & = &\left( \frac{\alpha}{b}\right)^{2} {\int}_{(2+\gamma)\delta_{0}}^{\frac{t_{3} - \delta_{0}}{1 + \gamma}} \left( \frac{t_{2} - (2+\gamma)\delta_{0}}{1 +\gamma} \right) dt_{2} \\ & =& \frac{\alpha^{2}}{b^{2}(1+\gamma)^{1+2}} \frac{ \left[ t_{3} - (\delta_{0} + (1+\gamma) \delta_{0} + (1+\gamma)^{2} \delta_{0} ) \right]^{2} }{2} . \end{array} $$

By proceeding in a similar manner, we get

$$ \begin{array}{@{}rcl@{}} && {\int}_{\frac{ (1+\gamma)^{m-1} -1 }{\gamma} \delta_{0}}^{\frac{t_{m} - \delta_{0}}{1 + \gamma}} {\dots} {\int}_{(2 + \gamma) \delta_{0}}^{\frac{t_{3} - \delta_{0}}{1 + \gamma}} {\int}_{\delta_{0}}^{\frac{t_{2} - \delta_{0}}{1 + \gamma}} \left( \frac{\alpha}{b} \right)^{m} dt_{1} dt_{2} {\dots} dt_{m-1} \\ && = \frac{\alpha^{m}}{ b^{m}(1+\gamma)^{1+2+{\dots} +(m-1)} } \frac{ \left\{ t_{m} - (\delta_{0} + (1 + \gamma) \delta_{0} + {\dots} + (1+\gamma)^{m-1} \delta_{0}) \right\}^{(m-1)}}{(m-1) !} \\ && = \frac{\alpha^{m}}{ b^{m}(m-1) !(1+\gamma)^{\frac{m(m-1)}{2}} } \left[t_{m} - \left( \frac{ (1 + \gamma )^{m} - 1 }{\gamma} \right) \delta_{0} \right]^{(m-1)}. \end{array} $$

(A7)

On using Eqs. A7 in A6, we get

$$ \begin{array}{@{}rcl@{}} B(t) &= &\frac{\alpha^{m}}{ b^{m}(m-1) !(1+\gamma)^{\frac{m(m-1)}{2}} } {\int}_{\left( \frac{ (1 + \gamma )^{m} - 1 }{\gamma} \right) \delta_{0}}^{t} \left[t_{m} - \left( \frac{ (1 + \gamma )^{m} - 1 }{\gamma} \right) \delta_{0} \right]^{(m-1)} dt_{m} \\ & = & \frac{ \alpha^{m} }{ b^{m} m! (1+\gamma)^{\frac{m(m-1)}{2}}}\left[ t - \left( \frac{ (1 + \gamma )^{m} - 1 }{\gamma} \right) \delta_{0} \right]^{m}. \end{array} $$

(A8)

Finally, the result follows from Theorem 3.1 by using Eqs. A5 and A8. □

Proof of Theorem 3.2

Note that

$$ \begin{array}{@{}rcl@{}} E(L) & =& {\int}_{0}^{\infty} P(L>t) dt \\ & =& {\int}_{0}^{\delta_{0}} P(L>t) dt + {\int}_{\delta_{0}}^{g(\delta_{0})} P(L>t) dt + {\int}_{g(\delta_{0})}^{g^{2}(\delta_{0})} P(L>t) dt+\dots\\ & =& {\int}_{0}^{\delta_{0}} P(L >t, N(t) = 0) dt + {\int}_{\delta_{0}}^{g(\delta_{0})} \left[ P(L >t, N(t) = 0) + P(L >t, N(t) = 1) \right] dt \\ && + {\int}_{g(\delta_{0})}^{g^{2}(\delta_{0})} \left[ P(L >t, N(t) = 0) + P(L >t, N(t) = 1)+ P(L >t, N(t) = 2) \right] dt + {\dots} \\&=&\left( {\int}_{0}^{\delta_{0}} P(L >t, N(t) = 0) dt +{\int}_{\delta_{0}}^{g(\delta_{0})} P(L >t, N(t) = 0) dt+\dots\right) \\&&+\left( {\int}_{\delta_{0}}^{g(\delta_{0})} P(L >t, N(t) = 1) dt+{\int}_{g(\delta_{0})}^{g^{2}(\delta_{0})} P(L >t, N(t) = 1) dt+\dots\right) +\dots \\ & = &{\int}_{0}^{\infty} P(L >t, N(t) = 0) dt + {\int}_{\delta_{0}}^{\infty} P(L >t, N(t) = 1) dt + \dots \\ & =& {\int}_{0}^{\infty} P(L >t, N(t) = 0) dt + \sum\limits_{m = 1}^{\infty} {\int}_{g^{m-1}(\delta_{0})}^{\infty} P(L >t, N(t) = m) dt \end{array} $$

(A9)

On using Eqs. A2 and A4 in the above expression, we get the required result. □

Proof of Corollary 3.2

Since δ(t) = δ₀ + γt, we have g(t) = (1 + γ)t + δ₀. From Eq. A9, we have

$$ E(L) = \sum\limits_{m = 0}^{\infty} {\int}_{\delta_{0} \left( \frac{(1+ \gamma)^{m} - 1}{\gamma} \right) }^{\infty} P(L >t, N(t) = m) dt . $$

(A10)

Note that, for $t > \delta _{0}\left (\frac {(1+ \gamma )^{m} - 1}{\gamma } \right )$,

$$P(L >t, N(t) = m) = \left( \frac{b}{b + \alpha t} \right)^{\frac{\beta}{\alpha}} \frac{\Gamma\left( \frac{\beta}{\alpha} + m\right)}{\Gamma\left( \frac{\beta}{\alpha}\right) m!} \frac{ \alpha^{m} }{ (1+\gamma)^{\frac{m(m-1)}{2}} } \left[ \frac{ t - \left( \frac{ (1 + \gamma )^{m} - 1 }{\gamma} \right) \delta_{0} }{ \alpha t + b} \right]^{m}.$$

On using the above expression in Eq. A10, we get

$$ \begin{array}{@{}rcl@{}} E(L) \!& = &\! \sum\limits_{m = 0}^{\infty} {\int}_{\delta_{0} \left( \frac{(1+ \gamma)^{m} - 1}{\gamma} \right) }^{\infty}\left[ \left( \frac{b}{b + \alpha t} \right)^{\frac{\beta}{\alpha}} \frac{\Gamma\left( \frac{\beta}{\alpha} + m\right)}{\Gamma\left( \frac{\beta}{\alpha}\right) m!} \frac{ \alpha^{m} }{ (1+\gamma)^{\frac{m(m-1)}{2}} } \left\{ \frac{ t - \left( \frac{ (1 + \gamma )^{m} - 1 }{\gamma} \right) \delta_{0} }{ \alpha t + b} \right\}^{m} \right]dt \\ \!& = &\!\sum\limits_{m = 0}^{\infty} \frac{\Gamma\left( \frac{\beta}{\alpha} + m\right)}{\Gamma\left( \frac{\beta}{\alpha}\right) m!} \frac{ \alpha^{m} }{ (1+\gamma)^{\frac{m(m-1)}{2}} } {\int}_{\delta_{0} \left( \frac{(1+ \gamma)^{m} - 1}{\gamma} \right) }^{\infty} \left( \frac{b}{b + \alpha t} \right)^{\frac{\beta}{\alpha}} \left\{ \frac{ t - \left( \frac{ (1 + \gamma )^{m} - 1 }{\gamma} \right) \delta_{0} }{ \alpha t + b} \right\}^{m} dt \\ \!& = &\!\sum\limits_{m = 0}^{\infty} \frac{\Gamma\left( \frac{\beta}{\alpha} + m\right)}{\Gamma\left( \frac{\beta}{\alpha}\right) m!} \frac{ \alpha^{m} b^{\frac{\beta}{\alpha}}}{ (1 + \gamma)^{\frac{m(m-1)}{2}} } {\int}_{\delta_{0} \left( \frac{(1+ \gamma)^{m} - 1}{\gamma} \right) }^{\infty} \frac{1}{(\alpha t + b)^{\frac{\beta}{\alpha} + m}} \left\{ t - \left( \frac{ (1 + \gamma )^{m} - 1 }{\gamma} \right) \delta_{0} \right\}^{m} dt \end{array} $$

$$ \begin{array}{@{}rcl@{}} & = & \sum\limits_{m = 0}^{\infty} \frac{\Gamma\left( \frac{\beta}{\alpha} + m\right)}{\Gamma\left( \frac{\beta}{\alpha}\right) m!} \frac{ \left( \frac{b}{\alpha}\right)^{\frac{\beta}{\alpha}}}{ (1+\gamma)^{\frac{m(m-1)}{2}} } {\int}_{0}^{\infty} u^{m} \left( \frac{1}{u+\frac{b}{\alpha}+\left( \frac{(1+ \gamma)^{m} - 1}{\gamma} \right)\delta_{0}} \right)^{m + \frac{\beta}{\alpha}} du \\ & = & \sum\limits_{m = 0}^{\infty} \frac{\Gamma\left( \frac{\beta}{\alpha} + m\right)}{\Gamma\left( \frac{\beta}{\alpha}\right) m!} \frac{ \left( \frac{b}{\alpha}\right)^{\frac{\beta}{\alpha}}}{ (1+\gamma)^{\frac{m(m-1)}{2}} \left[ \frac{b}{\alpha} + \left( \frac{(1+ \gamma)^{m} - 1}{\gamma} \right)\delta_{0} \right]^{\frac{\beta}{\alpha} - 1} } {\int}_{0}^{\infty} \frac{x^{m}}{(1+x)^{m+\frac{\beta}{\alpha}}} dx \\&=&\frac{b}{\beta-\alpha} \sum\limits_{m=0}^{\infty} \left[ \frac{b}{ b + \left( \frac{(1+\gamma)^{m} - 1}{\gamma} \right) \alpha \delta_{0}} \right]^{\frac{\beta}{\alpha} - 1} \left[ \frac{1}{ (1+\gamma)^{\frac{m(m-1)}{2}}} \right], \end{array} $$

where the last equality follows from the fact that

$$ {\int}_{0}^{\infty} \frac{x^{m}}{(1+x)^{m+\frac{\beta}{\alpha}}} dx = \frac{\Gamma(m+1) {\Gamma}(\frac{\beta}{\alpha} - 1)}{\Gamma(m+ \frac{\beta}{\alpha})} = \frac{\Gamma(\frac{\beta}{\alpha} ) m!}{(\frac{\beta}{\alpha} -1) {\Gamma}(m+ \frac{\beta}{\alpha}) } . $$

□

Proof of Proposition 4.1

Note that

$$ \begin{array}{@{}rcl@{}} P(M = m) &=& P(X_{1} > \delta_{0}, X_{2} > \delta(X_{1}), X_{3} > \delta(X_{1} + X_{2} ), {\dots} , X_{m - 1 } \\&>& \delta(X_{1} + X_{2} + {\dots} + X_{m-2}), \\ && X_{m} \leq \delta(X_{1} + X_{2} + {\dots} + X_{m-1} ) )\\ & = & {\int}_{0}^{\infty} P(X_{1} > \delta_{0}, X_{2} > \delta(X_{1}), X_{3} > \delta(X_{1} + X_{2} ), {\dots} , X_{m - 1 } \\&>& \delta(X_{1} + X_{2} + {\dots} + X_{m-2}), \\ && X_{m} \leq \delta(X_{1} + X_{2} + {\dots} + X_{m-1} )|{\Lambda}=\lambda) dH(\lambda)\\ & = & {\int}_{0}^{\infty} P(X_{1} > \delta_{0}, X_{2} > \delta_{0} + \gamma X_{1} , X_{3} > \delta_{0} + \gamma (X_{1} + X_{2}), {\dots} , X_{m - 1 } \!>\! \delta_{0} \\ &&+ \gamma (X_{1} + X_{2} + {\dots} + X_{m-2}), X_{m} \leq \delta_{0} \\&&+ \gamma (X_{1} + X_{2} + {\dots} + X_{m-1} )| {\Lambda}=\lambda) dH(\lambda), \\ \end{array} $$

(A11)

where Λ is a structure random variable with the probability density given by

$$ dH(\lambda) = \frac{\left( \frac{b}{\alpha}\right)^{\frac{\beta}{\alpha}}}{\Gamma\left( \frac{\beta}{\alpha}\right)} \lambda^{\left( \frac{\beta}{\alpha}\right) -1} \exp\left\{ - \left( \frac{b}{\alpha} \right) \lambda \right\} d \lambda. $$

(A12)

This is because the GPP with a set of parameters {λ(t) = 1/(b + αt), α, β} is actually a Pólya process with the parameters set {β/α, b/α} (Cha and Finkelstein 2018). Furthermore, on condition Λ = λ, the Pólya process is the same as the HPP with intensity λ (see Beichelt (2006), p.130). Thus, the conditional probability density function of $(X_{1},X_{2},{\dots } , X_{m})$ given Λ = λ is given by

$$ f_{(X_{1},X_{2},{\dots} , X_{m}| {\Lambda})}(x_{1},x_{2},\dots, x_{m}|\lambda) = \prod\limits_{i = 1}^{m} \lambda \exp\{ - \lambda x_{i} \}, \quad 0 < x_{1},x_{2},\dots, x_{m} < \infty. $$

(A13)

Now, consider

$$ \begin{array}{@{}rcl@{}} &&P(X_{1} > \delta_{0}, X_{2} > \delta_{0} + \gamma X_{1} , X_{3} > \delta_{0} + \gamma (X_{1} + X_{2}), {\dots} , X_{m - 1 } \\&&> \delta_{0} + \gamma (X_{1} + X_{2} + {\dots} + X_{m-2}), \\ &&X_{m} \leq \delta_{0} + \gamma (X_{1} + X_{2} + {\dots} + X_{m-1} )| {\Lambda}=\lambda) \\ && {}= {\int}_{\delta_{0}}^{\infty} {\int}_{\delta_{0} + \gamma x_{1}}^{\infty} {\dots} {\int}_{\delta_{0} + \gamma \cdot (x_{1} + x_{2} + {\dots} + x_{m-2} )}^{\infty} {\int}_{0}^{\delta_{0} + \gamma \cdot (x_{1} + x_{2} + {\dots} + x_{m-1} )} \\&&f_{(X_{1},X_{2},{\dots} , X_{m}| {\Lambda})}(x_{1},x_{2},\dots, x_{m}|\lambda) \\ && dx_{m} dx_{m-1} {\dots} dx_{2} dx_{1} \\ && {}= {\int}_{\delta_{0}}^{\infty} {\int}_{\delta_{0} + \gamma x_{1}}^{\infty} {\dots} {\int}_{\delta_{0} + \gamma \cdot (x_{1} + x_{2} + {\dots} + x_{m-2} )}^{\infty} {\int}_{0}^{\delta_{0} + \gamma \cdot (x_{1} + x_{2} + {\dots} + x_{m-1} )} \\&&\left( \prod\limits_{i = 1}^{m} \lambda \exp\{ - \lambda x_{i} \} \right) dx_{m} dx_{m-1} {\dots} dx_{2} dx_{1} \\ && {}= \frac{ \exp \left\{ - \lambda \delta_{0} \left( \frac{(1 + \gamma )^{(m-1)} - 1}{\gamma} \right) \right\} }{ (1 + \gamma )^{\frac{(m-1)(m-2)}{2}}} - \frac{ \exp \left\{ - \lambda \delta_{0} \left( \frac{(1 + \gamma )^{m} - 1}{\gamma} \right) \right\} }{ (1 + \gamma )^{\frac{(m-1)(m)}{2}}} , \end{array} $$

(A14)

where the second equality follows from Eq. A13. Now, on using Eqs. A12 and A14 in Eq. A11, we get

$$ \begin{array}{@{}rcl@{}} P(M = m) &=& {\int}_{0}^{\infty} \left\{ \frac{ \exp \left\{ - \lambda \delta_{0} \left( \frac{(1 + \gamma )^{(m-1)} - 1}{\gamma} \right) \right\} }{ (1 + \gamma )^{\frac{(m-1)(m-2)}{2}}} - \frac{ \exp \left\{ - \lambda \delta_{0} \left( \frac{(1 + \gamma )^{m} - 1}{\gamma} \right) \right\} }{ (1 + \gamma )^{\frac{(m-1)(m)}{2}}} \right\} \\ && \times \frac{\left( \frac{b}{\alpha}\right)^{\frac{\beta}{\alpha}}}{\Gamma\left( \frac{\beta}{\alpha}\right)} \lambda^{\left( \frac{\beta}{\alpha}\right) -1} \exp\left\{ - \left( \frac{b}{\alpha} \right) \lambda \right\} d \lambda \\ & = & {\int}_{0}^{\infty} \left[ \frac{ \exp \left\{ - \lambda \delta_{0} \left( \frac{(1 + \gamma )^{(m-1)} - 1}{\gamma} \right) \right\} }{ (1 + \gamma )^{\frac{(m-1)(m-2)}{2}}} \right] \frac{\left( \frac{b}{\alpha}\right)^{\frac{\beta}{\alpha}}}{\Gamma\left( \frac{\beta}{\alpha}\right)} \lambda^{\left( \frac{\beta}{\alpha}\right) -1} \exp\left\{ - \left( \frac{b}{\alpha} \right) \lambda \right\} d \lambda \\ && - {\int}_{0}^{\infty} \left[ \frac{ \exp \left\{ - \lambda \delta_{0} \left( \frac{(1 + \gamma )^{m} - 1}{\gamma} \right) \right\} }{ (1 + \gamma )^{\frac{(m-1)(m)}{2}}} \right] \frac{\left( \frac{b}{\alpha}\right)^{\frac{\beta}{\alpha}}}{\Gamma\left( \frac{\beta}{\alpha}\right)} \lambda^{\left( \frac{\beta}{\alpha}\right) -1} \exp\left\{ - \left( \frac{b}{\alpha} \right) \lambda \right\} d \lambda \\ & = & \frac{\left( \frac{b}{\alpha}\right)^{\frac{\beta}{\alpha}}}{\Gamma\left( \frac{\beta}{\alpha}\right)} \frac{1}{(1 + \gamma )^{\frac{(m-1)(m-2)}{2}}} {\int}_{0}^{\infty} \lambda^{\left( \frac{\beta}{\alpha}\right) -1} \exp \left[- \lambda \left\{\frac{b}{\alpha} + \delta_{0} \left( \frac{(1 + \gamma )^{(m-1)} - 1}{\gamma} \right) \right\} \right] d \lambda \\ && - \frac{\left( \frac{b}{\alpha}\right)^{\frac{\beta}{\alpha}}}{\Gamma\left( \frac{\beta}{\alpha}\right)} \frac{1}{(1 + \gamma )^{\frac{(m-1)(m)}{2}}} {\int}_{0}^{\infty} \lambda^{\left( \frac{\beta}{\alpha}\right) -1} \exp \left[ - \lambda \left\{\frac{b}{\alpha} + \delta_{0} \left( \frac{(1 + \gamma )^{(m)} - 1}{\gamma} \right) \right\} \right] d \lambda \\ & = & \frac{1}{(1+\gamma)^{\frac{(m-1)(m-2)}{2}}} \left[\frac{b}{b + \alpha \delta_{0} \left( \frac{(1+\gamma)^{(m-1)} - 1}{\gamma}\right)}\right]^{\frac{\beta}{\alpha}} \\&& - \frac{1}{(1+\gamma)^{\frac{m(m-1)}{2}}} \left[\frac{b}{b + \alpha \delta_{0} \left( \frac{(1+\gamma)^{m} - 1}{\gamma}\right)}\right]^{\frac{\beta}{\alpha}}, \end{array} $$

where the last equality follows from the fact that

$$ {\int}_{0}^{\infty} x^{k-1} \exp\{-c x\} dx = \frac{ {\Gamma}(k) }{ c^{k} }. $$

□

Proof of Theorem 4.2

Let g_i(t) = δ_i(t) + t, i = 1, 2, for t ≥ 0. Note that δ₁(t) ≤ δ₂(t) is equivalent to the fact that g₁(t) ≤ g₂(t), for all t ≥ 0. This means that $\gamma _{1} \leq \gamma _{2} \text { and } \delta _{0}^{(1)} \leq \delta _{0}^{(2)}$. From Corollary 3.1, we have

$$ \begin{array}{@{}rcl@{}} \bar F_{L_{i}}(t) &=& \left( \frac{b}{b + \alpha t} \right)^{\frac{\beta}{ \alpha}} \sum\limits_{m = 0}^{M_{0}^{(i)}(t)} {{\frac{\beta}{\alpha} + m - 1}\choose{m}} \frac{ \alpha^{m} }{ (1+\gamma_{i})^{\frac{m(m-1)}{2}} } \left[ \frac{ t - \left( \frac{ (1 + \gamma_{i} )^{m} - 1 }{\gamma_{i}} \right) \delta_{0}^{(i)} }{ b+\alpha t } \right]^{m}, \\ i&=&1,2, \end{array} $$

where $M_{0}^{(i)}(t) =\max \limits \{ n \geq 1 | g_{i}^{n-1}(\delta _{0}^{(i)}) < t \}$. Since g₁(t) ≤ g₂(t) and both g₁(⋅) and g₂(⋅) are increasing functions, we get $g_{1}^{n-1}(\delta _{0}^{(1)}) \leq g_{2}^{n-1}(\delta _{0}^{(2)})$. Thus, if there exists an n such that $g_{2}^{n-1}(\delta _{0}^{(2)}) < t$, then $g_{1}^{n-1}(\delta _{0}^{(1)}) < t$. Consequently, $M_{0}^{(2)}(t) \leq M_{0}^{(1)}(t)$. Again, $\gamma _{1} \leq \gamma _{2} \text { and } \delta _{0}^{(1)} \leq \delta _{0}^{(2)}$ together imply that

$$ \frac{\left( t - \left( \frac{ [(1 + \gamma_{2} )^{m} - 1] }{\gamma_{2}} \right) \delta_{0}^{(2)} \right)^{m}}{ (1+\gamma_{2})^{\frac{m(m-1)}{2}}} \leq \frac{\left( t - \left( \frac{ [(1 + \gamma_{1} )^{m} - 1] }{\gamma_{1}} \right) \delta_{0}^{(1)} \right)^{m}}{ (1+\gamma_{1})^{\frac{m(m-1)}{2}}}, \text{ for all } t\geq 0.$$

On combining these two facts, we get $\bar F_{L_{2}}(t) \leq \bar F_{L_{1}}(t)$, for all t > 0, and hence the result follows. □

Proof of Theorem 4.3

Case-I: Let 0 ≤ t < δ₀. Then, from Eq. 3.1, we have $\bar F_{L}(t) = \exp \{ - \lambda t \}$. Thus, r_L(t) = λ, and hence the result follows.

Case-II: Let δ₀ ≤ t < (2 + γ)δ₀. Then, from Eq. 3.1, we have

$$\bar F_{L}(t) = \exp \{ - \lambda t \} + \lambda \exp \{ - \lambda t \}(t- \delta_{0} ),$$

which gives

$$ \begin{array}{@{}rcl@{}} f_{L}(t)& =& \lambda \exp \{ - \lambda t \} + \lambda^{2} \exp \{ - \lambda t \} (t - \delta_{0} ) - \lambda \exp \{ - \lambda t \} \\&=&\lambda \bar F_{L}(t) - \lambda \exp \left\{- \lambda \left( \frac{\gamma t + \delta_{0}}{1 + \gamma} \right) \right\} \bar F_{L} \left( \frac{t - \delta_{0}}{ 1 + \gamma } \right), \end{array} $$

where the last equality holds because 0 ≤ (t − δ₀)/(1 + γ) < δ₀ and $\bar F_{L}(t) = \exp \{ - \lambda t \}$ for 0 ≤ t < δ₀. Thus,

$$ \begin{array}{@{}rcl@{}} r_{L}(t) = \lambda - \lambda \exp \left\{- \lambda \left( \frac{\gamma t + \delta_{0}}{1 + \gamma} \right) \right\} \left[ \frac{ \bar F_{L} \left( \frac{t - \delta_{0}}{ 1 + \gamma } \right) }{ \bar F_{L}(t) } \right], \text{ for } \delta_{0} \leq t < (2 + \gamma) \delta_{0}. \end{array} $$

Case-III: Let (2 + γ)δ₀ ≤ t < (1 + (1 + γ) + (1 + γ)²)δ₀. Then, from Eq. 3.1, we get

$$ \begin{array}{ll} & \bar F_{L}(t) = \exp \{ - \lambda t \} + \lambda \exp \{ - \lambda t \} (t - \delta_{0} ) + \lambda^{2} \exp \{ - \lambda t \} \frac{(t - (2 + \gamma) \delta_{0} )^{2}}{ (2!)(1 + \gamma )}, \end{array} $$

which gives

$$ \begin{array}{@{}rcl@{}} f_{L}(t) & = &\lambda \left( \exp \{ - \lambda t \} + \lambda \exp \{ - \lambda t \} (t - \delta_{0} ) + \lambda^{2} \exp \{ - \lambda t \} \frac{(t - (2 + \gamma) \delta_{0} )^{2}}{ 2!(1 + \gamma )} \right) \\ && - \left( \lambda \exp \{ - \lambda t \} + \lambda^{2} \exp \{ - \lambda t \} \frac{(t - (2 + \gamma) \delta_{0} )}{(1 + \gamma )} \right) \\ &=& \lambda \bar F_{L}(t) - \lambda \left( \frac{\exp \{ - \lambda t \}}{\exp \left\{ - \lambda \left( \frac{t - \delta_{0}}{1+\gamma}\right) \right\}} \right) \\ & &\times\left[ \exp \left\{ - \lambda \left( \frac{t - \delta_{0}}{1+\gamma}\right) \right\} + \lambda \exp \left\{ - \lambda \left( \frac{t - \delta_{0}}{1+\gamma}\right) \right\} \left( \left( \frac{t - \delta_{0}}{ 1 + \gamma }\right) - \delta_{0} \right) \right] \\ & =& \lambda \bar F_{L}(t) - \lambda \exp \left\{- \lambda \left( \frac{\gamma t + \delta_{0}}{1 + \gamma} \right) \right\} \bar F_{L} \left( \frac{t - \delta_{0}}{ 1 + \gamma } \right), \end{array} $$

where the last equality holds because δ₀ ≤ (t − δ₀)/(1 + γ) < (2 + γ)δ₀ and

$$\bar F_{L}(t) = \exp \{ - \lambda t \} + \lambda \exp \{ - \lambda t \}(t- \delta_{0} ),\text{ for }\delta_{0} \leq t < (2 + \gamma) \delta_{0}.$$

Thus,

$$ \begin{array}{@{}rcl@{}} r_{L}(t) &=& \lambda - \lambda \exp \left\{- \lambda \left( \frac{\gamma t + \delta_{0}}{1 + \gamma} \right) \right\} \left[ \frac{ \bar F_{L} \left( \frac{t - \delta_{0}}{ 1 + \gamma } \right) }{ \bar F_{L}(t) } \right], \text{ for } (2 +\gamma) \delta_{0} \\&\leq& t < (1+ (1+ \gamma) + (1 + \gamma)^{2} ) \delta_{0} . \end{array} $$

By proceeding in a similar manner as done in above three cases, we get the required result. □

Proof of Theorem 4.4

On using Eq. 3.1, we have

$$ \exp \left\{- \lambda \left( \frac{\gamma \cdot t + \delta_{0}}{1 + \gamma} \right) \right\} \left[ \frac{ \bar F \left( \frac{t - \delta_{0}}{ 1 + \gamma } \right) }{ \bar F(t) } \right] = \frac{\sum\limits\limits_{m = 0}^{N_{0}(t)-1} \frac{ \lambda^{m} }{ m! } \frac{\left( t - \left( \frac{ [(1 + \gamma )^{m+1} - 1] }{\gamma} \right) \delta_{0} \right)^{m}}{ (1+\gamma)^{\frac{m(m+1)}{2}}}}{\sum\limits\limits_{m = 0}^{N_{0}(t)} \frac{ \lambda^{m}}{ m! } \frac{\left( t - \left( \frac{ [(1 + \gamma )^{m} - 1] }{\gamma} \right) \delta_{0} \right)^{m}}{ (1+\gamma)^{\frac{m(m-1)}{2}}}}. $$

Note that the above expression is a ratio of two polynomials where the degree of the polynomial given in the numerator is strictly less than that in the denominator. Thus,

$$ \lim_{t \to \infty} \exp \left\{- \lambda \left( \frac{\gamma t + \delta_{0}}{1 + \gamma} \right) \right\} \left[ \frac{ \bar F_{L} \left( \frac{t - \delta_{0}}{ 1 + \gamma } \right) }{ \bar F_{L}(t) } \right] = 0,$$

and hence the result follows from Theorem 4.3. □