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On the general \(\delta \)-shock model

Abstract

The \(\delta \)-shock model is one of the basic shock models which has a wide range of applications in reliability, finance and related fields. In existing literature, it is assumed that the recovery time of a system from the damage induced by a shock is constant as well as the shocks magnitude. However, as technical systems gradually deteriorate with time, it takes more time to recover from this damage, whereas the larger magnitude of a shock also results in the same effect. Therefore, in this paper, we introduce a general \(\delta \)-shock model when the recovery time depends on both the arrival times and the magnitudes of shocks. Moreover, we also consider a more general and flexible shock process, namely, the Poisson generalized gamma process. It includes the homogeneous Poisson process, the non-homogeneous Poisson process, the Pólya process and the generalized Pólya process as the particular cases. For the defined survival model, we derive the relationships for the survival function and the mean lifetime and study some relevant stochastic properties. As an application, an example of the corresponding optimal replacement policy is discussed.

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Acknowledgements

The authors are thankful to the Editor-in-Chief, the Associate Editor and the anonymous Reviewers for their valuable constructive comments/suggestions which lead to an improved version of the manuscript. The first author sincerely acknowledges the financial support received from UGC, Govt. of India. The work of the second author was supported by a SRG Project (File Number: SRG/2021/000678), SERB, India.

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Correspondence to Maxim Finkelstein.

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Appendix

Appendix

Proof of Theorem 3.1:

We have

$$\begin{aligned} {\bar{F}}_L(t) =P(L>t)= & {} P(L>t,N(t) = 0) + \sum _{n=1}^{\infty } P(L > t,N(t) = n). \end{aligned}$$
(A1)

From Lemma 3.1, we have

$$\begin{aligned} P(L>t,N(t) = 0)=\frac{\alpha ^{k-\nu }\Gamma _{\nu }(k,(\alpha + \Lambda (t))l)}{\Gamma _{\nu }(k,\alpha l) (\alpha + \Lambda (t))^{k - \nu }}. \end{aligned}$$
(A2)

Further, for \(n \in {\mathbb {N}}\), we have

$$\begin{aligned}&P(L> t|N(t) = n ) \\&\quad = P(T_{1}> \delta _{0}, Y_{1} \le \gamma , T_{2}> T_{1} + \delta (T_{1},Y_{1}), \dots , T_{n}> T_{n-1}\\&\qquad + \delta (T_{n-1},Y_{n-1}),xY_{n}< \gamma | N(t) = n) \\&\quad = P(T_{1}> \delta _{0}, Y_{1} \le \gamma , T_{2}> g(T_{1},Y_{1}), Y_{2} \le \gamma , \dots , T_{n} >\\&\qquad g(T_{n-1},Y_{n-1}),Y_{n} < \gamma | N(t) = n). \end{aligned}$$

Note that the system survives n shocks in [0, t) provided \( t> T_{n}> g(T_{n-1},Y_{n-1}) > g^{2}(T_{n-2},Y_{n-2},Y_{n-1})\) \(> \dots > g^{n-1}(\delta _{0},Y_{1},Y_{2},\dots ,Y_{n-1}) \ge g^{n-1}(\delta _{0},\underbrace{0,0,\dots ,0}_{n-1\text { times}}) \). This implies that, if \(t \le g^{n-1}(\delta _{0},\underbrace{0,0,\dots ,0}_{n-1\text { times}}) \), then the probability of the event “the system survives n shocks till time t” is equal to zero, i.e.,

$$\begin{aligned} P(L>t|N(t) = n) = 0, \text { for } t \le g^{n-1}(\delta _{0},\underbrace{0,0,\dots ,0}_{n-1\text { times}}). \end{aligned}$$

For \(1 \le n \le K_{0}(t) \), we have

$$\begin{aligned}&P(L>t|N(t) =n) \nonumber \\&\quad = \underbrace{\overbrace{\int \int \dots \int }^{n \text {times}}}_{A_0(t,n)} P(T_{1}> \delta _{0}, Y_{1} \le \gamma , \dots , T_{n}> g(T_{n-1},Y_{n-1}),Y_{n} < \gamma | N(t) \nonumber \\&\quad = n,Y_1=y_1,\dots ,Y_n=y_n) f_{Y_{1},Y_{2}, \dots , Y_{n}}(y_{1},y_{2},\dots , y_{n}|N(t)=n) \hbox {d}y_{1} \dots \hbox {d}y_{n}\nonumber \\&\quad = \underbrace{\overbrace{\int \int \dots \int }^{n\text { times}}}_{A_0(t,n)} P(T_{1}> \delta _{0}, T_{2}> g(T_{1},y_{1}),\dots ,T_{n}> g(T_{n-1},y_{n-1})| N(t) = n) \nonumber \\&\qquad \times f_{Y_{1},Y_{2}, \dots , Y_{n}}(y_{1},y_{2},\dots , y_{n}) \hbox {d}y_{1} \dots \hbox {d}y_{n}\nonumber \\&\quad = \underbrace{\overbrace{\int \int \dots \int }^{n\text { times}}}_{A_0(t,n)} P(T_{1}> \delta _{0}, T_{2}> g(T_{1},y_{1}),\dots ,T_{n} > g(T_{n-1},y_{n-1})| N(t) = n) \nonumber \\&\qquad \times \prod _{i=1}^{n} f_{Y_{1}}(y_{i}) \hbox {d}y_{1} \dots \hbox {d}y_{n}, \end{aligned}$$
(A3)

where the second equality follows from Assumptions 2 and 3, and the third equality holds due to Assumption 4. Now, for \(0 \le y_{i} \le \gamma ,\; 1 \le i \le n\), consider

$$\begin{aligned}&P(T_{1}> \delta _{0}, T_{2}> g(T_{1},y_{1}), \dots ,T_{n} > g(T_{n-1},y_{n-1}) | N(t) = n) \\&\quad = \int _{g^{n-1}(\delta _{0},y_{1},y_{2},\dots ,y_{n-1})}^{t} \int _{g^{n-2}(\delta _{0},y_{1},y_{2},\dots ,y_{n-2})}^{h(t_{n},y_{n-1})} \dots \int _{g(\delta _{0},y_{1})}^{h(t_{3},y_{2})} \int _{\delta _{0}}^{h(t_{2},y_{1})}\\&\qquad \times f_{(T_{1},\cdot \cdot \cdot , T_{N(t)} | N(t))}(t_{1},\cdot \cdot \cdot ,t_{n} | n) \\&\qquad \times \hbox {d}t_{1} \hbox {d}t_{2} \dots \hbox {d}t_{n} \\&\quad = \int _{g^{n-1}(\delta _{0},y_{1},y_{2},\dots ,y_{n-1})}^{t} \int _{g^{n-2}(\delta _{0},y_{1},y_{2},\dots ,y_{n-2})}^{h(t_{n},y_{n-1})} \dots \int _{g(\delta _{0},y_{1})}^{h(t_{3},y_{2})} \int _{\delta _{0}}^{h(t_{2},y_{1})} \left( n! \prod _{i = 1}^{n} \frac{\lambda (t_{i})}{\Lambda (t)} \right) \hbox {d}t_{1} \dots \hbox {d}t_{n}, \end{aligned}$$

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

$$\begin{aligned} P(L>t|N(t) = n)=\frac{n!}{(\Lambda (t))^n}u(t,n), \end{aligned}$$

which, by Lemma 3.1(a), gives

$$\begin{aligned} P(L>t,N(t) = n)= & {} P(L>t|N(t) = n)P(N(t)=n)\nonumber \\= & {} \frac{\alpha ^{k - \nu }}{( \alpha + \Lambda (t))^{k + n - \nu }} \frac{\Gamma _{\nu }(k+n, (\alpha + \Lambda (t))l)}{\Gamma _{\nu }(k,\alpha l)} u(t,n).\qquad \quad \end{aligned}$$
(A4)

Finally, on using (A2) and (A4) in (A1), we get the required result. \(\square \)

Proof of Theorem 3.2:

We have

$$\begin{aligned} E(L)= & {} \int _{0}^{\infty } \bar{F}_{L}(t) \hbox {d}t\\= & {} \int _{0}^{\delta _{0}} \bar{F}_{L}(t) \hbox {d}t + \int _{\delta _{0}}^{g(\delta _{0},0)} \bar{F}_{L}(t) \hbox {d}t \\&+ \int _{g(\delta _{0},0)}^{g^{2}(\delta _{0},0,0)} \bar{F}_{L}(t) \hbox {d}t + \dots \\= & {} \int _{0}^{\delta _{0}} P(L> t, N(t) = 0) \hbox {d}t + \int _{\delta _{0}}^{g(\delta _{0},0)} \left[ P(L> t, N(t) = 0)\right. \\&\left. + P(L> t, N(t) = 1) \right] \hbox {d}t + \int _{g(\delta _{0},0)}^{g^{2}(\delta _{0},0,0)} \\&\left[ P(L> t, N(t) = 0) + P(L> t, N(t) = 1) + P(L> t, N(t) = 2) \right] \hbox {d}t + \dots \\= & {} \int _{0}^{\infty } P(L> t, N(t) = 0) \hbox {d}t + \sum _{n = 1}^{\infty } \int _{g^{n-1}(\delta _{0},\underbrace{0,0,\dots ,0}_{n-1 \text {times}})}^{\infty } P(L > t, N(t) = n) \hbox {d}t \end{aligned}$$

On using (A2) and (A4) in the above expression, we get the required result. \(\square \)

Proof of Theorem 3.5:

Since \(\delta _{1}(t,y) \le \delta _{2}(t,y)\), we get \(g_{1}(t,y) \le g_{2}(t,y)\), for all \((t,y) \in [0,\infty )\times [0,\infty )\). We know that both \(g_{1}(\cdot ,y)\) and \(g_{2}(\cdot ,y)\) are bijective and strictly increasing functions on \([{0},\infty )\), for fixed \(y\in [0,\infty )\). Let \(h_{1}(\cdot ,y)\) and \(h_{2}(\cdot ,y)\) be the inverse functions of \(g_{1}(\cdot ,y)\) and \(g_{2}(\cdot ,y)\), respectively, for fixed \(y\in [0,\infty )\). Then, for fixed \(y \in [0,\infty )\), the inequality “\(g_{1}(t,y) \le g_{2}(t,y)\), for all \(t \in [0,\infty )\)” implies \(h_{2}(t,y) \le h_{1}(t,y)\) for all \(t \in [\delta _{0}, \infty )\). Now, from Theorem 3.1, we have

$$\begin{aligned} {\bar{F}}_{L_i}(t)= & {} \frac{\alpha ^{k-\nu }}{\Gamma _{\nu }(k,\alpha l) (\alpha + \Lambda (t))^{k - \nu }} \left[ \Gamma _{\nu }(k,(\alpha + \Lambda (t))l)\right. \\&\left. + \sum _{n=1}^{K_{i0}(t)}\frac{\Gamma _{\nu }(k+n,(\alpha + \Lambda (t))l)}{(\alpha + \Lambda (t))^{n}} u_{i}(t,n) \right] , \end{aligned}$$

where

$$\begin{aligned} K_{i0}(t) = \hbox {max}\{n \ge 1| g_{i}^{n-1}(\delta _{0},\underbrace{0,0,\dots ,0}_{n-1\text { times}}) < t \}, \end{aligned}$$

and

$$\begin{aligned} u_{i}(t,n)= & {} \underbrace{\overbrace{\int \int \dots \int }^{n\text { times}}}_{\{(y_{1},y_{2},\dots , y_{n}): g_{i}^{n-1}(\delta _{0},y_{1},y_{2},\dots ,y_{n-1}) < t, 0 \le y_{j} \le \gamma , 1 \le j \le n \}} \Bigg \{ \\&\int _{g_{i}^{n-1}(\delta _{0},y_{1},y_{2},\dots ,y_{n-1})}^{t} \int _{g_{i}^{n-2}(\delta _{0},y_{i},y_{2},\dots ,y_{n-2})}^{h_{i}(t_{n},y_{n-1})} \dots \int _{g_{i}(\delta _{0},y_{1})}^{h_{i}(t_{3},y_{2})} \int _{\delta _{0}}^{h_{i}(t_{2},y_{1})} \left( \prod _{j=1}^{n} \lambda (t_{j}) \right) \\&\hbox {d}t_{1} \hbox {d}t_{2} \dots \hbox {d}t_{n} \Bigg \} f_{Y_{1},Y_{2}, \dots , Y_{n}}(y_{1},y_{2}, \dots ,y_{n}) \hbox {d}y_{1} \hbox {d}y_{2} \dots \hbox {d}y_{n},\quad i = 1,2. \end{aligned}$$

Since, for fixed \(y\in [0,\infty ),\) \(g_{1}(t,y) \le g_{2}(t,y) \), for all \(t \in [0, \infty )\), and \(h_{2}(t,y) \le h_{1}(t,y)\) for all \(t \in [\delta _{0}, \infty )\), we get

$$\begin{aligned}&\int _{g_{1}^{n-1}(\delta _{0},y_{1},y_{2},\dots ,y_{n-1})}^{t} \dots \int _{g_{1}(\delta _{0},y_{1})}^{h_{1}(t_{3},y_{2})} \int _{\delta _{0}}^{h_{1}(t_{2},y_{1})} \left( \prod _{i=1}^{n} \lambda (t_{i}) \right) \hbox {d}t_{1} \hbox {d}t_{2} \dots \hbox {d}t_{n} \\&\ge \int _{g_{2}^{n-1}(\delta _{0},y_{1},y_{2},\dots ,y_{n-1})}^{t} \dots \int _{g_{2}(\delta _{0},y_{1})}^{h_{2}(t_{3},y_{2})} \int _{\delta _{0}}^{h_{2}(t_{2},y_{1})} \left( \prod _{i=1}^{n} \lambda (t_{i}) \right) \hbox {d}t_{1} \hbox {d}t_{2} \dots \hbox {d}t_{n}, \end{aligned}$$

which further implies that \(u_{1}(t,n) \ge u_{2}(t,n)\), for all \(t \in [0,\infty )\) and \(n \in {\mathbb {N}}\). Again, \(K_{10}(t) \ge K_{20}(t)\), for all \(t \in [0,\infty )\). Thus, \({\bar{F}}_{L_2}(t) \le {\bar{F}}_{L_{1}}( t)\), for all \(t\in [0,\infty )\), and hence, the result is proved. \(\square \)

Proof of Theorem 3.5:

Given that the recovery function is \(\delta (t,y) = \delta _{0} + \sigma _{1} t + \sigma _{2} y \), for all \((t,y)\in [0,\infty )\times [0,\infty )\), which implies \(g(t,y) = \delta _{0} + (1 + \sigma _{1}) t + \sigma _{2} y \), for all \((t,y)\in [0,\infty )\times [0,\infty )\), and \(h(t,y) = ({t - \delta _{0} - \sigma _{2} y})/({1 + \sigma _{1}}) \), for all \((t,y)\in [\delta _0,\infty )\times [0,\infty )\). Consequently,

$$\begin{aligned}&g^{n-1}(\delta _{0},y_{1},y_{2},\dots ,y_{n-1}) = \delta _{0}\left( \frac{ (1+\sigma _{1})^{n} - 1}{\sigma _{1}} \right) \\&\quad + \sigma _{2} \left[ (1+\sigma _{1})^{n-2} y_{1} + (1+\sigma _{1})^{n-3} y_{2} + \dots + y_{n-1} \right] . \end{aligned}$$

Then

$$\begin{aligned} K_{0}(t)= & {} \hbox {max}\left\{ n \ge 1| g^{n-1}(\delta _{0},\underbrace{0,0,\dots ,0}_{n-1\text { times}})< t \right\} \nonumber \\= & {} \hbox {max}\left\{ n \ge 1| \delta _{0}\left( \frac{ (1+\sigma _{1})^{n} - 1}{\sigma _{1}} \right)< t \right\} \nonumber \\= & {} \hbox {max}\left\{ n \ge 1| n < \left( \frac{\ln \left( \frac{\delta _{0} + \sigma _{1} t}{\delta _{0}} \right) }{\ln (1+\sigma _{1})} \right) \right\} \nonumber \\= & {} \left\lfloor \frac{\ln \left( \frac{\delta _{0} + \sigma _{1} t}{\delta _{0}} \right) }{\ln (1+\sigma _{1})} \right\rfloor =S_0(t). \end{aligned}$$
(6.5)

Since \(Y_{i} \sim U[0,\xi ]\), for all \(i \in {\mathbb {N}} \), for some \(\xi > 0\), we have

$$\begin{aligned} f_{Y_{1},Y_{2}, \dots Y_{n}}(y_{1},y_{2},\dots ,y_{n}) = \prod _{i=1}^{n} f_{Y_{i}}(y_{i}) = \frac{1}{\xi ^{n}}, \text { for all } 0 \le y_{1},y_{2},\dots ,y_{n} \le \xi , n \in {\mathbb {N}}.\nonumber \\ \end{aligned}$$
(6.6)

Now, consider the following multiple integration.

$$\begin{aligned} \int _{g^{n-1}(\delta _{0},y_{1},y_{2},\dots ,y_{n-1})}^{t} \int _{g^{n-2}(\delta _{0},y_{1},y_{2},\dots ,y_{n-2})}^{h(t_{n},y_{n-1})} \dots \int _{g(\delta _{0},y_{1})}^{h(t_{3},y_{2})} \int _{\delta _{0}}^{h(t_{2},y_{1})} \hbox {d}t_{1} \dots \hbox {d}t_{n}. \end{aligned}$$

We solve the above integration by iterative process. Let us first consider the following integration.

$$\begin{aligned}&\int _{g(\delta _{0},y_{1})}^{h(t_{3},y_{2})} \int _{\delta _{0}}^{h(t_{2},y_{1})} \hbox {d}t_{1} \hbox {d}t_{2} \\&\quad = \int _{\delta _{0} [ 1 + (1 + \sigma _{1}) ] + \sigma _{2} y_{1} }^{\frac{t_{3} - \sigma _{2} y_{2} - \delta _{0} }{1 + \sigma _{1}}} \int _{\delta _{0}}^{\frac{t_{2} - \sigma _{2} y_{1} - \delta _{0} }{1 + \sigma _{1}}} \hbox {d}t_{1} \hbox {d}t_{2} \\&\quad = \frac{1}{2(1+\sigma _{1})^{3}} [ t_{3} - \delta _{0}(1 + (1+\sigma _{1}) + (1 + \sigma _{1})^{2}) - \sigma _{2}(y_{2} + (1+\sigma _{1}) y_{1}) ]^{2} \\&\quad = \frac{1}{2(1+\sigma _{1})^{3}} [ t_{3} - g^{2}(\delta _{0},y_{1},y_{2}) ]^{2}. \end{aligned}$$

Similarly, we get

$$\begin{aligned}&\int _{g^{2}(\delta _{0},y_{1},y_{2})}^{h(t_{4},y_{3})} \int _{g(\delta _{0},y_{1})}^{h(t_{3},y_{2})} \int _{\delta _{0}}^{h(t_{2},y_{1})} \hbox {d}t_{1} \hbox {d}t_{2} \hbox {d}t_{3} \\&\quad =\int _{\delta _{0} [ 1 + (1 + \sigma _{1}) + (1 + \sigma _{1})^{2} ] + \sigma _{2} (y_{2} + (1 + \sigma _{1})y_{1}) }^{\frac{t_{4} - \sigma _{2} y_{3} - \delta _{0} }{1 + \sigma _{1}}} \int _{\delta _{0} [ 1 + (1 + \sigma _{1}) ] + \sigma _{2} y_{1} }^{\frac{t_{3} - \sigma _{2} y_{2} - \delta _{0} }{1 + \sigma _{1}}} \int _{\delta _{0}}^{\frac{t_{2} - \sigma _{2} y_{1} - \delta _{0} }{1 + \sigma _{1}}} \hbox {d}t_{1} \hbox {d}t_{2} \hbox {d}t_{3} \\&\quad = \frac{1}{2(1+\sigma _{1})^{3}} \int _{\delta _{0} [ 1 + (1 + \sigma _{1}) + (1 + \sigma _{1})^{2} ] + \sigma _{2} (y_{2} + (1 + \sigma _{1})y_{1}) }^{\frac{t_{4} - \sigma _{2} y_{3} - \delta _{0} }{1 + \sigma _{1}}}[ t_{3} - g^{2}(\delta _{0},y_{1},y_{2}) ]^{2} \hbox {d}t_{3} \\&\quad = \frac{1}{3!(1+\sigma _{1})^{6}} [t_{4} - g^{3}(\delta _{0},y_{1},y_{2})]^{3}. \end{aligned}$$

By proceeding in a similar manner, we get

$$\begin{aligned}&\int _{g^{n-2}(\delta _{0},y_{1},y_{2},\dots ,y_{n-2})}^{h(t_{n},y_{n-1})} \dots \int _{g(\delta _{0},y_{1})}^{h(t_{3},y_{2})} \int _{\delta _{0}}^{h(t_{2},y_{1})} \hbox {d}t_{1} \dots \hbox {d}t_{n-1}\\&\quad = \frac{1}{(n-1)! (1 + \sigma _{1})^{\frac{n(n-1)}{2}}} \\&\qquad \times [t_{n} - g^{n-1}(\delta _{0},y_{1},\dots ,y_{n-1}) ]^{n-1}. \end{aligned}$$

Consequently,

$$\begin{aligned}&\int _{g^{n-1}(\delta _{0},y_{1},y_{2},\dots ,y_{n-1})}^{t} \int _{g^{n-2}(\delta _{0},y_{1},y_{2},\dots ,y_{n-2})}^{h(t_{n},y_{n-1})}\nonumber \\&\qquad \dots \int _{g(\delta _{0},y_{1})}^{h(t_{3},y_{2})} \int _{\delta _{0}}^{h(t_{2},y_{1})} \hbox {d}t_{1} \dots \hbox {d}t_{n} \nonumber \\&\quad = \frac{1}{n! (1 + \sigma _{1})^{\frac{n(n-1)}{2}}} [t - g^{n-1}(\delta _{0},y_{1},\dots ,y_{n-1}) ]^{n}. \end{aligned}$$
(6.7)

On using (6.6) and (6.7) in the expression of u(tn), given in Theorem 3.1, we get

$$\begin{aligned} u(t,n)= & {} \left( \frac{\gamma }{\xi ^{n}}\right) \left( \lambda \right) ^{n} \left( \frac{1}{n! (1 + \sigma _{1})^{\frac{n(n-1)}{2}}} \right) v(t,n). \end{aligned}$$

Finally, the result follows from Theorem 3.1 by using the above equality along with (6.5). \(\square \)

Proof of Theorem 3.5:

Let \({\mathcal {X}}\) be a structure random variable with the probability density given by

$$\begin{aligned} \hbox {d}H(\chi )= & {} \frac{\left( \frac{b}{\omega }\right) ^{\frac{\beta }{\omega }}}{\Gamma \left( \frac{\beta }{\omega }\right) }\nonumber \\&\chi ^{\left( \frac{\beta }{\omega }\right) -1} \exp \left\{ - \left( \frac{b}{\omega } \right) \chi \right\} \hbox {d}\chi . \end{aligned}$$
(6.8)

Given that shocks occur according to the HPGGP with the set of parameters \(\{1/b,0, \beta /\omega ,\) \(1/\omega ,l\}\), which is indeed the Pólya process with the set of parameters \(\left\{ \frac{\beta }{\omega }, \frac{b}{\omega } \right\} \). Moreover, on condition \({\mathcal {X}} = \chi \), the Pólya process is the same as the HPP with intensity \(\chi \) [see Beichelt (2006, p. 130)]. Consequently, the inter-arrival times of the given HPGGP are i.i.d. random variables and follow the exponential distribution with parameter \(\chi \). Then, the conditional probability density function of the random vector \((X_{1},X_{2}, \dots , X_{m})\), given \({\mathcal {X}} = \chi \), is given by

$$\begin{aligned} f_{X_{1},X_{2},\dots , X_{m}|{\mathcal {X}}}(x_{1},x_{2},\dots ,x_{m}|\chi ) = \prod _{i=1}^{m} \chi \exp \{ -\chi x_{i} \}, \;\; \; \; 0< x_{1},x_{2}, \dots , x_{m} < \infty .\nonumber \\ \end{aligned}$$
(6.9)

Now,

$$\begin{aligned} P(M =1)= & {} 1 - P(X_{1} > \delta _{0},Y_{1} \le \gamma ) \end{aligned}$$
(6.10)
$$\begin{aligned}= & {} 1 - F_{Y_{1}}(\gamma ) \int _{0}^{\infty } P(X_{1} > \delta _{0} |{\mathcal {X}} = \chi ) \hbox {d}H(\chi ) \nonumber \\= & {} 1 - F_{Y_{1}}(\gamma ) \int _{0}^{\infty } \exp \{ - \chi \delta _{0} \}\frac{ \left( \frac{b}{\omega }\right) ^{\frac{\beta }{\omega }}}{\Gamma (\frac{\beta }{\omega })} \chi ^{\frac{\beta }{\omega } -1} \exp \left\{ - \left( \frac{b}{\omega }\right) \chi \right\} \hbox {d}\chi \nonumber \\ {}= & {} 1 - \frac{\gamma }{\xi } \left( \frac{b}{b + \omega \delta _{0}} \right) ^{\frac{\beta }{\omega }}=\Upsilon _0-\Upsilon _1. \end{aligned}$$
(6.11)

Further, for \(m = 2,3,\dots \), we have

$$\begin{aligned} P(M = m)= & {} P(X_{1}> \delta _{0}, Y_{1} \le \gamma ,\dots , X_{m-1}> g(T_{m-2},Y_{m-2}), Y_{m-1} \le \gamma )\nonumber \\&- P(X_{1}> \delta _{0}, Y_{1} \le \gamma ,\dots , X_{m} > g(T_{m-1},Y_{m-1}), Y_{m} \le \gamma )\nonumber \\= & {} \zeta _{m-1}-\zeta _{m} \text { (say) }, \end{aligned}$$
(6.12)

where

$$\begin{aligned} \zeta _1=P(X_{1} > \delta _{0},Y_{1} \le \gamma )= \frac{\gamma }{\xi } \left( \frac{b}{b + \omega \delta _{0}} \right) ^{\frac{\beta }{\omega }}=\Upsilon _1. \end{aligned}$$
(6.13)

and

$$\begin{aligned} \zeta _m= & {} P(X_{1}> \delta _{0}, Y_{1} \le \gamma , X_{2}> g(T_{1},Y_{1}),\dots , X_{m}> g(T_{m-1},Y_{m-1}), Y_{m} \le \gamma )\nonumber \\= & {} \underbrace{\int _{0}^{\gamma } \int _{0}^{\gamma } \dots \int _{0}^{\gamma }}_{m\text { times}} \Big [P(X_{1}> \delta _{0}, X_{2}> g(T_{1},y_{1}),\dots ,X_{m} > g(T_{m-1},y_{m-1})) \nonumber \\&\times f_{Y_{1},Y_{2},\dots ,Y_{m}}(y_{1},y_{2},\dots ,y_{m}) \Big ] \hbox {d}y_{1} \hbox {d}y_{2} \dots \hbox {d}y_{m}, \quad m=2,3,\dots . \end{aligned}$$
(6.14)

Now, we proceed to find the value of \(\zeta _m\). Consider

$$\begin{aligned}&P(X_{1}> \delta _{0}, X_{2}> g(T_{1},y_{1}),\dots , X_{m}> g(T_{m-1},y_{m-1})) \nonumber \\&\quad = \int _{0}^{\infty } P(X_{1}> \delta _{0}, X_{2}> g(T_{1},y_{1}),\dots , X_{m}> g(T_{m-1},y_{m-1})| {\mathcal {X}} = \chi ) \hbox {d}H(\chi ) \nonumber \\&\quad = \frac{ \left( \frac{b}{\omega }\right) ^{\frac{\beta }{\omega }}}{\Gamma (\frac{\beta }{\omega })} \int _{0}^{\infty } P(X_{1}> \delta _{0}, X_{2}> \delta _{0} + \sigma _{1} X_{1} + \sigma _{2} y_{1}, \dots , X_{m} > \delta _{0} \nonumber \\&\qquad + \sigma _{1} ( X_{1} + \dots + X_{m-1} ) + \sigma _{2} y_{m-1} | {\mathcal {X}} = \chi ) \chi ^{\frac{\beta }{\omega } -1} \exp \left\{ - \left( \frac{b}{\omega }\right) \chi \right\} \hbox {d}\chi , \end{aligned}$$
(6.15)

where the second equality follows from (6.8). Again, consider

$$\begin{aligned}&P(X_{1}> \delta _{0}, X_{2}> \delta _{0}\nonumber \\&\qquad + \sigma _{1} X_{1} + \sigma _{2} y_{1},\dots , X_{m} > \delta _{0} + \sigma _{1} ( X_{1} + \dots + X_{m-1} ) + \sigma _{2} y_{m-1} | {\mathcal {X}} = \chi ) \nonumber \\&\quad = \int _{\delta _{0}}^{\infty } \int _{\delta _{0} + \sigma _{1} x_{1} + \sigma _{2} y_{1}}^{\infty } \dots \int _{\delta _{0} + \sigma _{1} (x_{1} + \dots + x_{m-1}) + \sigma _{2} y_{m-1}}^{\infty }\nonumber \\&\qquad \left( \prod _{i=1}^{m} \chi \exp \{ - \chi x_{i} \}\right) \hbox {d}x_{m} \hbox {d}x_{m-1} \dots \hbox {d}x_{2} \hbox {d}x_{1} \nonumber \\&\quad = \frac{1}{(1+\sigma _{1})^\frac{m(m-1)}{2}}\exp \left\{ - \chi \delta _{0} \left[ 1 + (1 + \sigma _{1}) + \dots + (1 + \sigma _{1})^{m-1} \right] \right\} \nonumber \\&\qquad \times \exp \left\{ - \chi \sigma _{2} \left[ y_{m-1} + (1 + \sigma _{1}) y_{m-2} + \dots + (1+\sigma _{1})^{m-2}y_{1} \right] \right\} \nonumber \\&\quad = \frac{\exp \left\{ -\chi \left( \delta _{0} \left( \frac{(1+\sigma _{1})^{m} - 1}{\sigma _{1}} \right) + \sigma _{2} (y_{m-1} + (1+\sigma _{1})y_{m-2} + \dots + (1 + \sigma _{1})^{m-2}y_{1}) \right) \right\} }{(1+\sigma _{1})^\frac{m(m-1)}{2}},\nonumber \\ \end{aligned}$$
(6.16)

where the first equality follows from (6.9). Then, on using (6.16) in (6.15), we get

$$\begin{aligned}&P(X_{1}> \delta _{0}, X_{2}> g(T_{1},y_{1}),\dots , X_{m} > g(T_{m-1},y_{m-1}))\nonumber \\&\quad =\frac{ \left( \frac{b}{\omega }\right) ^{\frac{\beta }{\omega }}}{\Gamma (\frac{\beta }{\omega })} \int _{0}^{\infty }\left[ \frac{\exp \left\{ - \chi \left( \delta _{0} \left( \frac{(1+\sigma _{1})^{m} - 1}{\sigma _{1}} \right) + \sigma _{2} (y_{m-1} + (1+\sigma _{1})y_{m-2} + \dots + (1 + \sigma _{1})^{m-2}y_{1}) \right) \right\} }{(1+\sigma _{1})^\frac{m(m-1)}{2}}\right. \nonumber \\&\qquad \times \left. \chi ^{\frac{\beta }{\omega } -1} \exp \left\{ - \left( \frac{b}{\omega }\right) \chi \right\} \right] \hbox {d}\chi \nonumber \\&\quad = \frac{1}{(1+\sigma _{1})^\frac{m(m-1)}{2}} \left( \frac{b}{ \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{m} - 1}{\sigma _{1}} \right) + \omega \sigma _{2} y_{m-1} + \omega \sigma _{2} (1 + \sigma _{1} ) y_{m-2} + \dots + \omega \sigma _{2} (1 + \sigma _{1})^{m-2} y_{1} + b } \right) ^{\frac{\beta }{\omega }}. \nonumber \\ \end{aligned}$$
(6.17)

Since \(Y_{i} \sim U[0,\xi ]\), \(i\in {\mathbb {N}}\), for some \(\xi > 0\), we have

$$\begin{aligned}&f_{Y_{1},Y_{2},\dots ,Y_{m}}(y_{1},y_{2},\dots ,y_{m}) \nonumber \\&\quad = \left( \frac{1}{\xi }\right) ^{m}, \quad 0< y_{1},y_{2},\dots , y_{m} < \infty . \end{aligned}$$
(6.18)

On using (6.17) and (6.18) in (6.14), we get

$$\begin{aligned} \zeta _m= & {} \left( \frac{ \gamma b^{\frac{\beta }{\omega }}}{ \xi ^{m} (1+\sigma _{1})^\frac{m(m-1)}{2}} \right) I_m, \end{aligned}$$
(6.19)

where

$$\begin{aligned} I_m= & {} \underbrace{\int _{0}^{\gamma } \int _{0}^{\gamma } \dots \int _{0}^{\gamma }}_{m-1\text { times}} \Bigg \{ \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{m} - 1}{\sigma _{1}} \right) + \omega \sigma _{2} y_{m-1} \nonumber \\&+ \omega \sigma _{2} (1 + \sigma _{1} ) y_{m-2} + \dots + \omega \sigma _{2} (1 + \sigma _{1})^{m-2} y_{1} + b \Bigg \}^{-\frac{\beta }{\omega }} \hbox {d}y_{1} \hbox {d}y_{2} \dots \hbox {d}y_{m-1}. \end{aligned}$$

Now, we solve \(I_m\) by iterative process. Consider

$$\begin{aligned} I_{2}= & {} \int _{0}^{\gamma } \left\{ \omega \delta _{0} \left( 1 + ( 1 + \sigma _{1}) \right) + \omega \sigma _{2} y_{1} + b \right\} ^\frac{-\beta }{\omega } \hbox {d}y_{1} \\= & {} \left[ \frac{ \{ \omega \delta _{0} (1 + (1 + \sigma _{1})) + \omega \sigma _{2} \gamma + b \}^ {\left( - \frac{\beta }{\omega } + 1 \right) } - \{ \omega \delta _{0} (1 + (1 + \sigma _{1})) + b \}^ {\left( - \frac{\beta }{\omega } + 1\right) } }{ \sigma _{2} (\omega - \beta ) } \right] \\= & {} \frac{1}{\sigma _{2} (\omega - \beta ) }\left[ (-1) \left\{ \omega \delta _{0} (1 + (1 + \sigma _{1})) + \sum _{s \in S_{A_{1},0}^{1}} s + b \right\} ^{ \left( -\frac{\beta }{\omega } +1 \right) } \right. \\&\quad \left. + (-1)^{0} \left\{ \omega \delta _{0} (1 + (1 + \sigma _{1})) + \sum _{s \in S_{A_{1},1}^{1}} s + b \right\} ^{ \left( -\frac{\beta }{\omega } +1 \right) } \right] , \end{aligned}$$

where \(A_1=\{ \omega \sigma _{2} \gamma \}\). Similarly,

$$\begin{aligned} I_{3}= & {} \int _{0}^{\gamma } \int _{0}^{\gamma } \{ \omega \delta _{0} (1 + (1 + \sigma _{1}) + (1 + \sigma )^{2} ) + \omega \sigma _{2} y_{2} + \omega \sigma _{2} (1 + \sigma _{1}) y_{1} + b \}^{- \frac{\beta }{\omega } } \hbox {d}y_{1} \hbox {d}y_{2} \\ \\= & {} \frac{1}{\sigma _{2}^{2} (1 + \sigma _{1})(2 \omega - \beta )(\omega - \beta )} \Bigg [ \Bigg \{ \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{3} -1}{\sigma _{1}}\right) + \sum _{s \in S_{A_{2},2}^{1}} s + b \Bigg \}^{ \left( - \frac{\beta }{\omega } + 2 \right) } \\&- \sum _{l =1}^{2} \Bigg \{ \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{3} -1}{\sigma _{1}}\right) + \sum _{s \in S_{A_{2},1}^{l}} s + b \Bigg \}^{ \left( - \frac{\beta }{\omega } + 2 \right) } \\&+ \Bigg \{ \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{3} -1}{\sigma _{1}}\right) + \sum _{s \in S_{A_{2},0}^{1}} s + b \Bigg \}^{ \left( - \frac{\beta }{\omega } + 2 \right) } \Bigg ] \end{aligned}$$

and

$$\begin{aligned} I_{4}= & {} \int _{0}^{\gamma } \int _{0}^{\gamma } \int _{0}^{\gamma } \Bigg \{ \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{4} -1}{\sigma _{1}}\right) + \omega \sigma _{2} y_{3} + \omega \sigma _{2} (1 + \sigma _{1}) y_{2} \\&+ \omega \sigma _{2} (1 + \sigma _{1})^{2} y_{1} \Bigg \}^{- \frac{\beta }{\omega }} \hbox {d}y_{3} \hbox {d}y_{2} \hbox {d}y_{1} \\= & {} \frac{1}{\sigma _{2}^{3}(1 + \sigma _{1})^{1+2} (3\omega - \beta )(2 \omega - \beta )(\omega - \beta )} \Bigg [ \Bigg \{ \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{4} -1}{\sigma _{1}}\right) \\&+ \sum _{s \in S_{A_{3},3}^{1}} s + b \Bigg \}^{\left( -\frac{\beta }{\omega } + 3\right) } \\&- \sum _{l=1}^{3} \Bigg \{ \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{4} -1}{\sigma _{1}}\right) + \sum _{s \in S_{A_{3},2}^{l}} s + b \Bigg \}^{\left( -\frac{\beta }{\omega } + 3\right) } \\&+ \sum _{l=1}^{3}\Bigg \{ \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{4} -1}{\sigma _{1}}\right) + \sum _{s \in S_{A_{3},1}^{l}} s + b \Bigg \}^{\left( -\frac{\beta }{\omega } + 3\right) } \\&- \Bigg \{ \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{4} -1}{\sigma _{1}}\right) + \sum _{s \in S_{A_{3},0}^{1}} s + b \Bigg \}^{\left( -\frac{\beta }{\omega } + 3\right) } \Bigg ], \end{aligned}$$

where \(A_{2} := \{ \omega \sigma _{2} \gamma , \omega \sigma _{2} (1 + \sigma _{1}) \gamma \}\) and \(A_{3} = \{ \omega \sigma _{2} \gamma , \omega \sigma _{2} \gamma (1 + \sigma _{1}), \omega \sigma _{2} \gamma (1 + \sigma _{1})^{2} \}\). By proceeding in the same line, we get

$$\begin{aligned} I_{m}= & {} \frac{1}{\sigma _{2}^{m-1} (1 + \sigma _{1})^{\frac{(m-1)(m-2)}{2}} \left( \prod _{i=1}^{m-1} \{i\omega - \beta \}\right) } \\&\times \left[ \sum _{i=0}^{m-1} (-1)^{m-1-i} \sum _{l = 1}^{{m-1 \atopwithdelims ()i}} \left( \omega \delta _{0} \left( \frac{(1+\sigma _{1})^{m} - 1}{\sigma _{1}} \right) + \left( \sum _{s \in S_{A_{m-1},i}^{l}} s \right) + b \right) ^{-\frac{\beta }{\omega } + m - 1} \right] , \end{aligned}$$

where \(A_{m-1} = \{ \omega \sigma _{2} \gamma , \omega \sigma _{2} \gamma (1 + \sigma _{1}), \dots , \omega \sigma _{2} \gamma (1 + \sigma _{1})^{m-2} \} \). On using the above value of \(I_m\) in (6.19), we get \(\zeta _m=\Upsilon _m\) and hence, the required result follows from (6.11), (6.12) and (6.13). \(\square \)

Proof of Theorem 4.3:

Since \(Y_{i} \sim \exp (\theta ) \), for all \( i \in {\mathbb {N}} \), for some \(\theta >0\), we have

$$\begin{aligned} f_{Y_{1},Y_{2},\dots ,Y_{n}}(y_{1},y_{2},\dots ,y_{m}) = \prod _{i=1}^{m} f_{Y_{i}}(y_{i}) = \prod _{i=1}^{m} \theta \exp \{ - \theta y_{i} \},\; \; \; \; 0< y_{1},y_{2},\dots ,y_{n} < \infty ,\nonumber \\ \end{aligned}$$
(6.20)

and \(F_{Y_{m}}(\gamma ) = (1 - \exp \{ - \theta \gamma \})\) for all \( m \in {\mathbb {N}} \). Then, from (6.10), we have

$$\begin{aligned} P(M =1 ) = 1 - (1 - \exp \{ -\theta \gamma \}) \exp \{ - \lambda \delta _{0} \}=\Delta _0-\Delta _1 . \end{aligned}$$
(6.21)

By proceeding in the same line as done in the proof of Theorem 4.2, we get, from (6.14) and (6.20), that

$$\begin{aligned} \zeta _m= & {} \left( \frac{(1 - \exp \{ -\theta \gamma \})}{(1 + \sigma _{1})^{\frac{m(m-1)}{2}}} \right) \exp \left\{ - \lambda \delta _{0}\left( \frac{(1+\sigma _{1})^{m} -1}{\sigma _{1}} \right) \right\} \theta ^{m-1}J_m, \end{aligned}$$

where

$$\begin{aligned} J_{m}= & {} \underbrace{\int _{0}^{\gamma } \int _{0}^{\gamma } \dots \int _{0}^{\gamma }}_{m-1\text { times}} \prod _{i=1}^{m-1} \exp \left\{ - y_{i}\left( \lambda \sigma _{2} (1 + \sigma _{1})^{m-1-i} + \theta \right) \right\} \hbox {d}y_{1} \hbox {d}y_{2} \dots \hbox {d}y_{m-1} \\= & {} \prod _{i=1}^{m-1} \left( \int _{0}^{\gamma } \exp \left\{ - y_{i}\left( \lambda \sigma _{2} (1 + \sigma _{1})^{m-1-i} + \theta \right) \right\} \hbox {d}y_{i} \right) \\= & {} \prod _{i=1}^{m-1} \left( \frac{1 - \exp \left\{ - \gamma ( \lambda \sigma _{2} ( 1 + \sigma _{1})^{m-1-i} + \theta ) \right\} }{ \lambda \sigma _{2} (1 + \sigma _{1})^{m-1-i} + \theta } \right) . \end{aligned}$$

On using the above value of \(\zeta _m\) in (6.12), we get

$$\begin{aligned} P(M = m)= & {} \theta ^{m-2} \left( \frac{(1 - \exp \{ -\theta \gamma \})}{(1 + \sigma _{1})^{\frac{(m-1)(m-2)}{2}}} \right) \exp \left\{ - \lambda \delta _{0}\left( \frac{(1+\sigma _{1})^{m-1} -1}{\sigma _{1}} \right) \right\} J_{m-1}\nonumber \\&- \theta ^{m-1} \left( \frac{(1 - \exp \{ -\theta \gamma \})}{(1 + \sigma _{1})^{\frac{m(m-1)}{2}}} \right) \exp \left\{ - \lambda \delta _{0}\left( \frac{(1+\sigma _{1})^{m} -1}{\sigma _{1}} \right) \right\} J_{m}\nonumber \\= & {} \Delta _{m-1}-\Delta _m, \quad m=2,3,\dots . \end{aligned}$$
(6.22)

Finally, the result follows from (6.21) and (6.22). \(\square \)

Proof of Theorem 4.3:

From Theorem 4.4, we have

$$\begin{aligned} E(L)= & {} \frac{\alpha ^{k - \nu }}{ \Gamma _{\nu }(k,\alpha l)} \sum _{n =0}^{\infty } \int _{ \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \delta _{0}}^{\infty }\\&\quad \frac{\lambda ^{n}}{ n!} \frac{\Gamma _{\nu }(k+n, (\alpha + \lambda t)l)}{(\alpha + \lambda t)^{k + n - \nu }} \frac{1}{(1 + \sigma _{1})^{\frac{n(n-1)}{2}}} \left( \frac{\gamma }{\xi }\right) ^{n}\\&\times \left[ t - \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \delta _{0} \right] ^{n} \hbox {d}t \\= & {} \frac{\alpha ^{k - \nu }}{ \Gamma _{\nu }(k,\alpha l)} \sum _{n=0}^{\infty } \left( \frac{\gamma }{\xi }\right) ^{n} \frac{\lambda ^{n}}{ n!} \frac{1}{(1 + \sigma _{1})^{\frac{n(n-1)}{2}}} \int _{ \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \delta _{0}}^{\infty }\\&\quad \frac{\Gamma _{\nu }(k+n, (\alpha + \lambda t)l)}{(\alpha + \lambda t)^{k + n - \nu }} \\&\times \left[ t - \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \delta _{0} \right] ^{n} \hbox {d}t \\= & {} \frac{\alpha ^{k - \nu }}{ \Gamma _{\nu }(k,\alpha l)} \sum _{n=0}^{\infty } \left( \frac{\gamma }{\xi }\right) ^{n} \frac{\lambda ^{n}}{ n!} \frac{1}{(1 + \sigma _{1})^{\frac{n(n-1)}{2}}} \int _{ \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \delta _{0}}^{\infty } \int _{0}^{\infty } \\&\frac{y^{n+k-1} \exp \{ - (\alpha + \lambda t) y\}}{ (y + l)^{\nu }} \\&\times \left[ t - \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \delta _{0} \right] ^{n} \hbox {d}y \hbox {d}t\\= & {} \frac{\alpha ^{k - \nu }}{ \Gamma _{\nu }(k,\alpha l)} \sum _{n=0}^{\infty } \left( \frac{\gamma }{\xi }\right) ^{n} \frac{\lambda ^{n}}{ n!} \frac{1}{(1 + \sigma _{1})^{\frac{n(n-1)}{2}}} \int _{0}^{\infty } \frac{y^{n+k-1} \exp \{ - \alpha y\}}{ (y + l)^{\nu }} \\&\Bigg \{\int _{ \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \delta _{0}}^{\infty } \exp \{ - \lambda t y \} \\&\times \left[ t - \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \delta _{0} \right] ^{n} \hbox {d}t \Bigg \} \hbox {d}y \\= & {} \frac{\alpha ^{k - \nu }}{ \Gamma _{\nu }(k,\alpha l)} \sum _{n=0}^{\infty } \left( \frac{\gamma }{\xi }\right) ^{n} \frac{\lambda ^{n}}{ n!} \frac{\Gamma (n+1)}{(1 + \sigma _{1})^{\frac{n(n-1)}{2}}} \int _{0}^{\infty } \frac{y^{n+k-1} \exp \{ - \alpha y\}}{ (y + l)^{\nu }}\\&\frac{ \exp \left\{ - \lambda y \delta _{0} \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \right\} }{ (\lambda y)^{n+1}} \hbox {d}y \\= & {} \frac{\alpha ^{k - \nu }}{ \lambda \Gamma _{\nu }(k,\alpha l)} \sum _{n=0}^{\infty } \left( \frac{\gamma }{\xi }\right) ^{n}\\&\frac{1}{(1 + \sigma _{1})^{\frac{n(n-1)}{2}}} \int _{0}^{\infty } \frac{y^{k-2} \exp \left\{ - y \left[ \alpha + \lambda \delta _{0} \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \right] \right\} }{ (y + l)^{\nu }} \hbox {d}y \\= & {} \frac{\alpha ^{k - \nu }}{ \lambda \Gamma _{\nu }(k,\alpha l)} \sum _{n=0}^{\infty } \left( \frac{\gamma }{\xi }\right) ^{n} \frac{1}{(1 + \sigma _{1})^{\frac{n(n-1)}{2}}} \frac{\Gamma _{\nu }\left( k-1, \alpha l + \lambda \delta _{0} l \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \right) }{ \left( \alpha + \lambda \delta _{0} \left( \frac{(1 + \sigma _{1})^{n} - 1}{\sigma _{1}} \right) \right) ^{k-\nu -1}}, \end{aligned}$$

where the third and the last equalities follow from equation (2.1). Thus the result is proved. \(\square \)

Proof of Theorem 4.3:

From Theorem 4.4, we have

$$\begin{aligned} E(L)= & {} \int _{0}^{\infty } \frac{\alpha ^{k-\nu }}{ (\alpha + \lambda t)^{k - \nu }} \frac{\Gamma _{\nu }(k,(\alpha + \lambda t)l)}{\Gamma _{\nu }(k,\alpha l)} \hbox {d}t \\&+ \frac{\alpha ^{k-\nu } \gamma }{ \Gamma _{\nu }(k,\alpha l) (\alpha + \lambda t)^{k - \nu }} \sum _{n=1}^{\infty } \int _{n \delta _{0}}^{\infty } \left( \frac{\lambda }{\xi }\right) ^{n} \frac{\Gamma _{\nu }(k+n,(\alpha + \lambda t)l)}{n! (\alpha + \lambda t)^{n}} v(t,n) \hbox {d}t \\= & {} \frac{\alpha ^{k-\nu }}{\Gamma _{\nu }(k,\alpha l)} \int _{0}^{\infty } \int _{0}^{\infty } \frac{\exp \{ - (\alpha + \lambda t) y\} y^{k-1}}{ (y + l)^{\nu }} \hbox {d}y \hbox {d}t \\&+ \frac{\alpha ^{k-\nu } \gamma }{\Gamma _{\nu }(k,\alpha l)} \sum _{n=1}^{\infty } \int _{n \delta _{0}}^{\infty } \int _{0}^{\infty } \left( \frac{\lambda }{\xi }\right) ^{n} \frac{\exp \{ - (\alpha + \lambda t) y\} y^{k+n-1}}{ (y + l)^{\nu }} \frac{v(t,n)}{n!} \hbox {d}y \hbox {d}t \\= & {} \frac{\alpha ^{k-\nu }}{\Gamma _{\nu }(k,\alpha l)} \int _{0}^{\infty } \frac{\exp \{- \alpha y\} y^{k-1}}{ (y + l)^{\nu }} \left( \int _{0}^{\infty } \exp \{ - \lambda t y\} \hbox {d}t \right) \hbox {d}y \\&+ \frac{\alpha ^{k-\nu } \gamma }{\Gamma _{\nu }(k,\alpha l)} \sum _{n=1}^{\infty } \int _{0}^{\infty } \left( \frac{\lambda }{\xi }\right) ^{n} \frac{\exp \{ - \alpha y\} y^{k+n-1}}{ n! (y + l)^{\nu }} \left( \int _{n \delta _{0}}^{\infty } \exp \{ - \lambda t y\} v(t,n) \hbox {d}t \right) \hbox {d}y\\= & {} \frac{\alpha ^{k-\nu }}{\Gamma _{\nu }(k,\alpha l)} \int _{0}^{\infty } \frac{\exp \{- \alpha y\} y^{k-1}}{ (y + l)^{\nu }} \left( \int _{0}^{\infty } \exp \{ - \lambda t y\} \hbox {d}t \right) \hbox {d}y \\&+ \frac{\alpha ^{k-\nu } \gamma }{\Gamma _{\nu }(k,\alpha l)} \sum _{n=1}^{\infty } \sum _{i=0}^{n-1} \int _{0}^{\infty }(-1)^{i} { n-1 \atopwithdelims ()i} \left( \frac{\lambda }{\xi }\right) ^{n} \frac{\exp \{ - \alpha y\} y^{k+n-1}}{ n! (y + l)^{\nu }} \\&\times \left( \int _{n \delta _{0} + i \gamma \sigma _{2}}^{\infty } \frac{\exp \{ -\lambda y t \} \left( t - n \delta _{0} - i \gamma \sigma _{2} \right) ^{2n -1}}{ (2n-1)!} \hbox {d}t \right) \hbox {d}y \\= & {} \frac{\alpha ^{k-\nu }}{\Gamma _{\nu }(k,\alpha l)} \int _{0}^{\infty } \frac{\exp \{- \alpha y\} y^{k-1}}{ (y + l)^{\nu }} \left( \int _{0}^{\infty } \exp \{ - \lambda t y\} \hbox {d}t \right) \hbox {d}y \\&+ \frac{\alpha ^{k-\nu } \gamma }{\Gamma _{\nu }(k,\alpha l)} \sum _{n=1}^{\infty } \sum _{i=0}^{n-1} \int _{0}^{\infty }(-1)^{i} { n-1 \atopwithdelims ()i} \\&\left( \frac{\lambda }{\xi }\right) ^{n} \frac{\exp \{ - (\alpha + n \lambda \delta _{0} + i \gamma \sigma _{2} \lambda ) y\} y^{k+n-1}}{ n! (y + l)^{\nu }} \\&\times \left( \int _{0}^{\infty } \frac{\exp \{ -\lambda y x \} x^{2n-1}}{(2n-1)!} \hbox {d}x \right) \hbox {d}y \\= & {} \frac{\alpha ^{k-\nu }}{\Gamma _{\nu }(k,\alpha l)} \int _{0}^{\infty } \frac{\exp \{- \alpha y\} y^{k-2}}{ \lambda (y + l)^{\nu }} \hbox {d}y \\&+ \frac{\alpha ^{k-\nu } \gamma }{\Gamma _{\nu }(k,\alpha l)} \sum _{n=1}^{\infty } \sum _{i=0}^{n-1} \int _{0}^{\infty }(-1)^{i} { n-1 \atopwithdelims ()i} \\&\left( \frac{1}{ \lambda \xi }\right) ^{n} \frac{\exp \{ - (\alpha + n \lambda \delta _{0} + i \gamma \sigma _{2} \lambda ) y\} y^{k-n-1}}{ n! (y + l)^{\nu }} \hbox {d}y \\= & {} \frac{\alpha ^{k-\nu }}{\Gamma _{\nu }(k,\alpha l)} \left[ \frac{\Gamma _{\nu }(k-1, \alpha l)}{\lambda \alpha ^{k-\nu -1}} + \sum _{n=1}^{\infty } \sum _{i=0}^{n-1} (-1)^{i} { n-1 \atopwithdelims ()i}\right. \\&\left. \left( \frac{1}{ \lambda \xi }\right) ^{n} \frac{\Gamma _{\nu }(k-n, (\alpha + n \lambda \delta _{0} + i \gamma \sigma _{2} \lambda ) l)}{(\alpha + n \lambda \delta _{0} + i \gamma \sigma _{2} \lambda )^{k-\nu -n}} \right] , \end{aligned}$$

where the second and the last equalities follow from (2.1), for \(k > n,\; \text {for all }n \in {\mathbb {N}}\). Therefore,

$$\begin{aligned} E(L)= & {} \lim _{k \rightarrow \infty } \frac{\alpha ^{k-\nu }}{\Gamma _{\nu }(k,\alpha l)} \left[ \frac{\Gamma _{\nu }(k-1, \alpha l)}{\lambda \alpha ^{k-\nu -1}}\right. \\&\left. + \sum _{n=1}^{\infty } \sum _{i=0}^{n-1} (-1)^{i} { n-1 \atopwithdelims ()i} \left( \frac{1}{ \lambda \xi }\right) ^{n} \frac{\Gamma _{\nu }(k-n, (\alpha + n \lambda \delta _{0} + i \gamma \sigma _{2} \lambda ) l)}{(\alpha + n \lambda \delta _{0} + i \gamma \sigma _{2} \lambda )^{k-\nu -n}} \right] \end{aligned}$$

and hence, the result is proved. \(\square \)

Proof of Theorem 4.5:

We can write

$$\begin{aligned} L = X_{1} + X_{2} + \dots + X_{M}, \end{aligned}$$

where M is the random variable representing the number of shocks that have occurred before the failure of the system. As the shocks occur according to the HPP with intensity \(\lambda >0\), \(\{X_{1}, X_{2}, \dots , X_{n}\}\) are i.i.d. random variables with the common mean \({1}/{\lambda }\). Thus, from the Wald’s equation, we get

$$\begin{aligned} E(L) = E(X_{1})E(M) = \frac{1}{\lambda } E(M), \end{aligned}$$

where E(M) is the same as in Theorem 4.3. Hence, the result is proved. \(\square \)

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Goyal, D., Hazra, N.K. & Finkelstein, M. On the general \(\delta \)-shock model. TEST (2022). https://doi.org/10.1007/s11749-022-00810-5

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  • DOI: https://doi.org/10.1007/s11749-022-00810-5

Keywords

  • Reliability
  • \(\delta \)-shock model
  • Poisson generalized gamma process
  • Homogeneous Poisson process

Mathematics Subject Classification

  • 60E15
  • 60K10