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(t, n), 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. \(\square \)
