Appendix
Proof of Theorem 1
By the definition of T, we have
$$\begin{aligned} P(T>t)=P\left( \sum _{i=1}^{N}X_{i}>t\right) =\sum _{n=1}^{\infty }P\left( \sum _{i=1}^{N}X_{i}>t,N=n\right) . \end{aligned}$$
Let
$$\begin{aligned} \sum _{n=1}^{\infty }P\left( \sum _{i=1}^{N}X_{i}>t,N=n\right) =A+B+C+D, \end{aligned}$$
for \(n=1\) we have
$$\begin{aligned}&P(X_1>t,X_1<\delta ,Z_1>\gamma )+P(X_1>t,X_1>\delta ,Z_1>\gamma )\\&\quad =P(X_1>t,Z_1>\gamma )\\ \end{aligned}$$
and for \(n>1\)
$$\begin{aligned}&A=\sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) P\left( \sum _{i=1}^{n}X_{i}>t,X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{n-k}>\delta ,Z_{n-k}<\gamma ,\right. \\&\left. X_{n-k+1}<\delta ,Z_{n-k+1}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma ,X_{n}<\delta \right) , \\&B=\sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) P\left( \sum _{i=1}^{n}X_{i}>t,X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{n-k}>\delta ,Z_{n-k}<\gamma ,\right. \\&\left. X_{n-k+1}<\delta ,Z_{n-k+1}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma ,X_{n}>\delta ,Z_{n}>\gamma \right) , \\&C=\sum _{n=2}^{\infty }\sum _{j=0}^{k-2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) P\left( \sum _{i=1}^{n}X_{i}>t,X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{n-j-1}>\delta ,\right. \\&\left. Z_{n-j-1}<\gamma ,X_{n-j}<\delta ,Z_{n-j}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma ,X_{n}>\delta ,Z_{n}>\gamma \right) , \\&D=\sum _{n=2}^{\infty }\sum _{j=0}^{k-2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) P\left( \sum _{i=1}^{n}X_{i}>t,X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{n-j-1}>\delta , \right. \\&\left. Z_{n-j-1}<\gamma ,X_{n-j}<\delta ,Z_{n-j}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma ,X_{n}<\delta ,Z_{n}>\gamma \right) . \end{aligned}$$
In the following, we evaluate A, B, C and D. First note that
$$\begin{aligned} A&= \sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) P\left( \sum _{i=0}^{n}X_{i}>t,X_{1}> \delta ,Z_{1}<\gamma ,\ldots ,X_{n-k}>\delta ,Z_{n-k}<\gamma , \right. \\&\left. X_{n-k+1}<\delta ,Z_{n-k+1}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma ,X_{n}<\delta \right) \\& = \sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) P\left( \sum _{i=1}^{n-1}X_{i}>t-X_{n},X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{n-k}>\delta , \right. \\&\left. Z_{n-k}<\gamma ,X_{n-k+1}<\delta ,Z_{n-k+1}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma ,X_{n}<\delta \right) \\&= \sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) \int _{0}^{\delta }P\left( \sum _{i=1}^{n-1}X_{i}>t-x,X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{n-k}>\delta ,\right. \\&\left. Z_{n-k}<\gamma ,X_{n-k+1}<\delta ,Z_{n-k+1}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma \right) dF(x) \\&= \sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) P(X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{n-k}>\delta ,Z_{n-k}<\gamma ,X_{n-k+1}<\delta , \\&Z_{n-k+1}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma ) \int _{0}^{\delta }P\left( \sum _{i=1}^{n-1}X_{i}>t-x|X_{1}>\delta , \right. \\&\left. Z_{1}<\gamma ,\ldots ,X_{n-k}>\delta ,Z_{n-k}<\gamma ,X_{n-k+1}\right. \\&\left.<\delta ,Z_{n-k+1}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma \right) dF(x). \end{aligned}$$
Thus
$$\begin{aligned} A=\sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) p_{1}^{n-k}p_{2}^{k-1}\int _{0}^{\delta }P(S_{n-k}^{*}+S_{k-1}^{**}>t-x)dF(x), \end{aligned}$$
and
$$\begin{aligned} B& = \sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) P\left( \sum _{i=1}^{n}X_{i}>t,X_{1}> \delta ,Z_{1}<\gamma ,\ldots ,X_{n-k}>\delta ,Z_{n-k}<\gamma , \right. \\&\left. X_{n-k+1}<\delta ,Z_{n-k+1}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma ,X_{n}>\delta ,Z_{n}>\gamma \right) \\& = \sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) P\left( \sum _{i=1}^{n}X_{i}>t,X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{n-k}>\delta ,Z_{n-k}<\gamma , \right. \\&\left. X_{n-k+1}<\delta ,Z_{n-k+1}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma ,X_{n}>\delta \right) \\&-\sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) P\left( \sum _{i=1}^{n}X_{i}>t,X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{n-k}>\delta ,Z_{n-k}<\gamma , \right. \\&\left. X_{n-k+1}<\delta ,Z_{n-k+1}<\gamma ,\ldots ,X_{n-1}<\delta ,Z_{n-1}<\gamma ,X_{n}>\delta ,Z_{n}<\gamma \right) . \end{aligned}$$
Thus
$$\begin{aligned} B&= \sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) p_1^{n-k}p_2^{k-1}\\&\int _{\delta }^{\infty }P(S_{n-k}^{*}+S_{k-1}^{**}>t-x)dF(x) \\&-\sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) p_1^{n-k+1}p_2^{k-1} P(S_{n-k+1}^{*}+S_{k-1}^{**}>t), \\ \end{aligned}$$
and similar to above, we have
$$\begin{aligned} C& = \sum _{n=2}^{\infty }\sum _{j=0}^{k-2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_{1}^{n-j-1}p_{2}^{j}\\&\int _{\delta }^{\infty }P(S_{n-j-1}^{*}+S_{j}^{**}>t-x)dF(x) \\&-\sum _{n=2}^{\infty }\sum _{j=0}^{k-2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_{1}^{n-j}p_{2}^{j}P(S_{n-j}^{*}+S_{j}^{**}>t), \end{aligned}$$
and
$$\begin{aligned} D& = \sum _{n=2}^{\infty }\sum _{j=0}^{k-2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_{1}^{n-j-1}p_{2}^{j}\\&\int _{0}^{\delta }P(S_{n-j-1}^{*}+S_{j}^{**}>t-x)dF(x) \\&-\sum _{n=2}^{\infty }\sum _{j=0}^{k-2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_{1}^{n-j-1}p_{2}^{j+1}P(S_{n-j-1}^{*}+S_{j+1}^{**}>t). \end{aligned}$$
Thus
$$\begin{aligned}&P(T>t) \\&\quad =P(X_1>t,Z_1>\gamma )\\&\qquad +\sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) p_1^{n-k}p_2^{k-1}\int _{0}^{ \infty }P(S_{n-k}^{*}+S_{k-1}^{**}>t-x)dF(x) \\&\qquad -\sum _{n=k}^{\infty }\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) p_1^{n-k+1}p_2^{k-1}P(S_{n-k+1}^{*}+S_{k-1}^{**}>t) \\&\qquad +\sum _{n=2}^{\infty }\sum _{j=0}^{k-2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_1^{n-j-1}p_2^{j} \int _{0}^{\infty }P(S_{n-j-1}^{*}+S_{j}^{**}>t-x)dF(x) \\&\qquad -\sum _{n=2}^{\infty }\sum _{j=0}^{k-2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_1^{n-j}p_2^{j}P(S_{n-j}^{*}+S_{j}^{**}>t) \\&\qquad -\sum _{n=2}^{\infty }\sum _{j=0}^{k-2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_1^{n-j-1}p_2^{j+1}P(S_{n-j-1}^{*}+S_{j+1}^{**}>t). \end{aligned}$$
\(\square\)
Proof of Theorem 2
We have
$$\begin{aligned} P(S>t)& = P\left( \sum _{i=1}^{M}X_{i}>t\right) =\sum _{m=1}^{\infty }P\left( \sum _{i=1}^{m}X_{i}>t,M=m\right) \\ \end{aligned}$$
For \(m=1\) we have
$$\begin{aligned}&P(X_1>t,X_1<\delta )+P(X_1>t,X_1>\delta ,Z_1>\gamma )\\&\quad =(P(X_1<\delta )-P(X_1<t))I_{[0,\delta )}(t)+P(X_1>\max (t,\delta ),Z_1>\gamma ) \end{aligned}$$
and for \(m>1\) we have
$$\begin{aligned}&= \sum _{m=2}^{\infty }P\left( \sum _{i=1}^{m}X_{i}>t,X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{m-1}>\delta ,Z_{m-1}<\gamma ,X_{m}<\delta \right) \\&\quad +\sum _{m=2}^{\infty }P\left( \sum _{i=1}^{m}X_{i}>t,X_{1}>\delta ,Z_{1}<\gamma ,\ldots ,X_{m-1}>\delta ,Z_{m-1}<\gamma ,\right. \\&\quad \left. X_{m}>\delta ,Z_{m}>\gamma \right) \\&= \sum _{m=2}^{\infty }p_1^{m-1}\int _{0}^{\infty }P(S_{m-1}^{*}>t-x)dF(x)-\sum _{m=2}^{\infty }p_1^{m}P(S_{m}^{*}>t). \end{aligned}$$
Proof of Lemma 1
Proof of a) is clear. For b) note that
$$\begin{aligned} P(N-M=a|N>M)=\dfrac{P(N-M=a,N>M)}{P(N>M)}, \end{aligned}$$
by part a,
$$\begin{aligned} P(N-M& = a,N>M)=\sum _{m=1}^{\infty }P(N=m+a,M=m) \\&= \sum _{m=1}^{\infty }\left( {\begin{array}{c}a-1\\ k-2\end{array}}\right) p_{2}^{k-1}p_{1}^{m+a-k}(p_{3}+p_{4}) \\&+\sum _{m=1}^{\infty }\sum _{j=0}^{k-3}\left( {\begin{array}{c}a-1\\ j\end{array}}\right) p_{2}^{j+1}p_{1}^{m+a-j-2}(p_{4}+p_{5}) \\& = \dfrac{p_{2}}{1-p_{1}}\bigg [\left( {\begin{array}{c}a-1\\ k-2\end{array}}\right) p_{2}^{k-2}p_{1}^{a-k+1}(p_{3}+p_{4}) \\&+\sum _{j=0}^{k-3}\left( {\begin{array}{c}a-1\\ j\end{array}}\right) p_{2}^{j}p_{1}^{a-j-1}(p_{4}+p_{5})\bigg ], \end{aligned}$$
and
$$\begin{aligned}&P(N>M)\\&\quad =\sum _{m=1}^{\infty }\sum _{n=m+k-1}^{\infty }\left( {\begin{array}{c}n-m-1\\ k-2\end{array}}\right) p_{2}^{k-1}p_{1}^{n-k}(p_{3}+p_{4}) \\&\qquad +\sum _{m=1}^{\infty }\sum _{n=m+j+1}^{\infty }\sum _{j=0}^{k-3}\left( {\begin{array}{c}n-m-1\\ j\end{array}}\right) p_{2}^{j+1}p_{1}^{n-j-2}(p_{4}+p_{5}) \\&\quad =\dfrac{p_{2}}{1-p_{1}}\bigg (\dfrac{p_{2}^{k-2}}{(1-p_{1})^{k-2}}+1-\dfrac{ p_{2}^{k-2}}{(1-p_{1})^{k-2}}\bigg ) \\&\quad =\dfrac{p_{2}}{1-p_{1}}. \end{aligned}$$
Thus
$$\begin{aligned}&P(N-M=a|N>M) \\&\quad =\left( {\begin{array}{c}a-1\\ k-2\end{array}}\right) p_{2}^{k-2}p_{1}^{a-k+1}(p_{3}+p_{4}) \\&\qquad +\sum _{j=0}^{k-3}\left( {\begin{array}{c}a-1\\ j\end{array}}\right) p_{2}^{j}p_{1}^{a-j-1}(p_{4}+p_{5}). \end{aligned}$$
It is the distribution of N for \(k-1\) instead of k. \(\square\)
Proof of Theorem 3
We have
$$\begin{aligned}&P(T-S=0)\\&\quad =P(T-S=0,N=M)+P(T-S=0,N>M)\\&\quad =P(T-S=0,N=M) \\&\quad = P(T-S=0|N=M)P(N=M)=P(N=M)\\&\quad =\sum _{n=1}^{\infty }P(N=n,M=n)\\&\quad =\sum _{n=1}^{\infty }p_1^{n-1}(p_4+p_5)\\&\quad =\dfrac{p_4+p_5}{1-p_1}, \end{aligned}$$
For \(t>0,\)
$$\begin{aligned} P(T-S>t)& = P\left( \sum _{i=1}^{N}X_{i}-\sum _{i=1}^{M}X_{i}>t|N>M\right) P(N>M) \\& = P\left( \sum _{i=M+1}^{N}X_{i}>t|N>M\right) P(N>M) \\& = P\left( \sum _{i=1}^{N-M}X_{i}>t|N>M\right) P(N>M) \\& = P(N>M)P\left( \sum _{i=1}^{N^{*}}X_{i}>t\right) , \end{aligned}$$
where \(P(\sum _{i=1}^{N^{*}}X_{i}>t)\) is similar to that of Theorem 1 for \(k-1\) instead of k.
Proof of Proposition 1
(a) We have
$$\begin{aligned} E(T)& = E\left( \sum _{i=1}^{N}X_{i}\right) =\sum _{n=1}^{\infty }E\left( \left( \sum _{i=1}^{n}X_{i}\right) |N=n\right) P(N=n) \\& = \sum _{n=1}^{\infty }\sum _{i=1}^{n}E(X_{i}|N=n)P(N=n). \end{aligned}$$
From the definition of N, we find
$$\begin{aligned}&E(X_{i}|N=n)\\&\quad =\left\{ \begin{array}{ll} E_{1}, &{} i=1,\ldots ,n-k \\ E_{2}, &{} i=n-k+1,\ldots ,n-1 \\ \dfrac{p_{3}}{p_{3}+p_{4}}E_{3}+\dfrac{p_{4}}{p_{3}+p_{4}}E_{4}, &{} i=n \end{array} \right. \end{aligned}$$
and for \(j=0,1,\ldots ,k-2,\)
$$\begin{aligned}&E(X_{i}|N=n)\\&\quad =\left\{ \begin{array}{ll} E_{1}, &{} i=1,\ldots ,n-j-1 \\ E_{2}, &{} i=n-j,\ldots ,n-1 \\ \dfrac{p_{4}}{p_{4}+p_{5}}E_{4}+\dfrac{p_{5}}{p_{4}+p_{5}}E_{5}, &{} i=n. \end{array} \right. \end{aligned}$$
Thus
$$\begin{aligned}&\sum _{n=1}^{\infty }\sum _{i=1}^{n}E(X_{i}|N =n)P(N=n) \\&\quad =A_{1}\times \left[ E(N-k)E_{1}+(k-1)E_{2}\right. \\&\qquad \left. +\dfrac{p_{3}}{p_{3}+p_{4}} E_{3}+\dfrac{p_{4}}{p_{3}+p_{4}}E_{4}\right] \\&\qquad +\sum _{j=2}^{k}A_{j}\times \left[ E(N-(j-1))E_{1}+(j-2)E_{2}+\dfrac{p_{5} }{p_{4}+p_{5}}E_{5}\right. \\&\qquad \left. +\dfrac{p_{4}}{p_{4}+p_{5}}E_{4}\right] . \end{aligned}$$
(b) According to Parvardeh and Balakrishnan (2015), and by using the Wald identity, we have
$$\begin{aligned} E\left( \sum _{i=1}^{M}X_{i}\right) =E(M)E(X_{1}). \end{aligned}$$
(c)
$$\begin{aligned}&E(T-S)\\&\quad =E\left( \sum _{i=1}^{N}X_{i}-\sum _{i=1}^{M}X_{i}\right) \\&\quad =E\left( \sum _{i=1}^{N}X_{i}- \sum _{i=1}^{M}X_{i}|N=M\right) P(N=M) \\&\qquad +E\left( \sum _{i=1}^{N}X_{i}-\sum _{i=1}^{M}X_{i}|N>M\right) P(N>M) \\&\quad =E\left( \sum _{i=1}^{N}X_{i}-\sum _{i=1}^{M}X_{i}|N>M\right) P(N>M) \\&\quad =E\left( \sum _{i=M+1}^{N}X_{i}|N>M\right) P(N>M) \\&\quad =E\left( \sum _{i=1}^{N^{*}}X_{i}\right) P(N>M), \end{aligned}$$
where \(E(\sum _{i=1}^{N^{*}}X_{i})\) is similar to E(T) for \(k-1\) instead of k. \(\square\)
Proof of Proposition 2
We prove \(E(T^2)\) only. \(E(S^2)\) and \(E(T-S)^2\) are similar.
$$\begin{aligned} E(T^{2})=E\left( \sum _{i=1}^{N}X_{i}\right) ^{2}=\sum _{n=1}^{\infty }E\left( \left( \sum _{i=1}^{n}X_{i}\right) ^{2}|N=n\right) P(N=n), \end{aligned}$$
where
$$\begin{aligned} E\left( \left( \sum _{i=1}^{n}X_{i}\right) ^{2}|N=n\right)& = \sum _{i=1}^{n}E(X_{i}^{2}|N=n)\\&+2\sum \sum _{1\le i<l\le n}E(X_{i}X_{l}|N=n). \end{aligned}$$
From the definition of N, we find
$$\begin{aligned} E(X_{i}^{2}|N=n)=\left\{ \begin{array}{ll} E_{1}^{^{\prime }}, &{} i=1,\ldots ,n-k \\ E_{2}^{^{\prime }}, &{} i=n-k+1,\ldots ,n-1 \\ \dfrac{p_{3}}{p_{3}+p_{4}}E_{3}^{^{\prime }}+\dfrac{p_{4}}{p_{3}+p_{4}} E_{4}^{^{\prime }}, &{} i=n \end{array} \right. \end{aligned}$$
and for \(j=0,1,\ldots ,k-2,\)
$$\begin{aligned} E(X_{i}^{2}|N=n)=\left\{ \begin{array}{ll} E_{1}^{^{\prime }}, &{} i=1,\ldots ,n-j-1 \\ E_{2}^{^{\prime }}, &{} i=n-j,\ldots ,n-1 \\ \dfrac{p_{4}}{p_{4}+p_{5}}E_{4}^{^{\prime }}+\dfrac{p_{5}}{p_{4}+p_{5}} E_{5}^{^{\prime }}, &{} i=n. \end{array} \right. \end{aligned}$$
Thus
$$\begin{aligned}&\sum _{n=1}^{\infty }\sum _{i=1}^{n}E(X_{i}^{2}|N = n)P(N=n) \\&\quad =A_{1}\times \left[ E(N-k)E_{1}^{^{\prime }}+(k-1)E_{2}^{^{\prime }}+ \dfrac{p_{3}}{p_{3}+p_{4}}E_{3}^{^{\prime }}+\dfrac{p_{4}}{p_{3}+p_{4}} E_{4}^{^{\prime }}\right] \\&\qquad +\sum _{j=2}^{k}A_{j}\times \left[ E(N-(j-1))E_{1}^{^{\prime }}+(j-2)E_{2}^{^{\prime }}+\dfrac{p_{5}}{p_{4}+p_{5}}E_{5}^{^{\prime }}\right. \\&\qquad \left. + \dfrac{p_{4}}{p_{4}+p_{5}}E_{4}^{^{\prime }}\right] . \end{aligned}$$
and
$$\begin{aligned}&E(X_{i}X_{l}|N=n) \\&\quad =\left\{ \begin{array}{ll} E_{1}^{2}, &{} 1\le i<l\le n-k \\ E_{2}^{2}, &{} n-k+1\le i<l\le n-1, \\ E_{1}E_{2}, &{} 1\le i\le n-k,n-k+1\le l\le n-1 \\ \dfrac{p_{3}}{p_{3}+p_{4}}E_{1}E_{3}+\dfrac{p_{4}}{p_{3}+p_{4}}E_{1}E_{4}, &{} 1\le i\le n-k,l=n \\ \dfrac{p_{3}}{p_{3}+p_{4}}E_{2}E_{3}+\dfrac{p_{4}}{p_{3}+p_{4}}E_{2}E_{4}, &{} n-k+1\le i\le n-1<l=n \end{array} \right. \end{aligned}$$
For \(j=0\) we have
$$\begin{aligned} E(X_{i}X_{l}|N=n)= \\ \left\{ \begin{array}{ll} E_{1}^{2}, &{} 1\le i<l\le n-1 \\ \dfrac{p_{4}}{p_{4}+p_{5}}E_{1}E_{4}+\dfrac{p_{5}}{p_{4}+p_{5}}E_{1}E_{5}, &{} 1\le i<l=n, \end{array} \right. \end{aligned}$$
and for \(j=1,2,\ldots ,k-2\) we have
$$\begin{aligned}&E(X_{i}X_{l}|N=n) \\&\quad =\left\{ \begin{array}{ll} E_{1}^{2}, &{} 1\le i<l\le n-j-1 \\ E_{1}E_{2}, &{} 1\le i\le n-j-1,n-j\le l\le n-1 \\ E_{2}^{2}, &{} n-j\le i<l\le n-1 \\ \dfrac{p_{4}}{p_{4}+p_{5}}E_{1}E_{4}+\dfrac{p_{5}}{p_{4}+p_{5}}E_{1}E_{5}, &{} 1\le i\le n-j-1,l=n \\ \dfrac{p_{4}}{p_{4}+p_{5}}E_{2}E_{4}+\dfrac{p_{5}}{p_{4}+p_{5}}E_{2}E_{5}, &{} n-j\le i\le n-1,l=n. \end{array} \right. \end{aligned}$$
Then
$$\begin{aligned}&2\sum _{n=1}^{\infty }\sum \sum _{1\le i<l\le n}E(X_{i}X_{l}|N=n)P(N=n) \\&\quad =2\times \left[ A_{1}\times \left( E\left( \frac{(N-k)(N-k-1)}{2}\right) E_1^{2} +\frac{ (k-1)(k-2)}{2}E_2^{2} \right. \right. \\&\qquad \left. +(k-1)E(N-k)E_1E_2+E(N-k)\right. \\&\qquad \left. \left( \frac{p_3}{p_3+p_4}E_1E_3+\frac{p_4}{p_3+p_4} E_1E_4\right) \right. \\&\qquad \left. \left. +(k-1)\left( \frac{p_3}{p_3+p_4}E_2E_3 +\frac{p_4}{p_3+p_4}E_2E_4\right) \right) \right. \\&\qquad \left. +\sum _{j=2}^{k}A_{j} \left( \frac{E(N-j+1)(N-j)}{2}E_1^{2} +(j-2)E(N-j+1)E_1E_2 \right. \right. \\&\qquad \left. \left. +\frac{(j-2)(j-3)}{2}E_2^{2} +(j-2)\right. \right. \\&\qquad \left. \left. \times \left( \frac{p_4}{p_4+p_5}E_2E_4+ \frac{p_5}{p_4+p_5}E_2E_5\right) \right. \right. \\&\qquad \left. \left. +E(N-j+1)\times \left( \frac{p_4}{p_4+p_5}E_1E_4+\frac{p_5}{p_4+p_5}E_1E_5 \right) \right) \right] . \end{aligned}$$
In this part E(N) and \(E(N^{2})\) are calculated.
$$\begin{aligned} E(N)& = \sum _{n=k}^{\infty }n\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) p_{2}^{k-1}p_{1}^{n-k}(p_{3}+p_{4}) \\&+\sum _{n=1}^{\infty }\sum _{j=0}^{k-2}n\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_{2}^{j}p_{1}^{n-j-1}(p_{4}+p_{5}). \end{aligned}$$
Let
$$\begin{aligned} A^{1}& = \sum _{n=k}^{\infty }n\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) p_{2}^{k-1}p_{1}^{n-k}(p_{3}+p_{4}) \\ B^{1}& = \sum _{n=1}^{\infty }\sum _{j=0}^{k-2}n\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_{2}^{j}p_{1}^{n-j-1}(p_{4}+p_{5}). \end{aligned}$$
Then
$$\begin{aligned} A^{1}& = p_{2}^{k-1}(p_{3}+p_{4}).\sum _{n=k}^{\infty }k\left( {\begin{array}{c}n\\ k\end{array}}\right) p_{1}^{n-k}\\& = p_{2}^{k-1}(p_{3}+p_{4})k\frac{1}{(1-p_{1})^{k+1}} \\& = k\dfrac{p_{2}^{k-1}}{(1-p_{1})^{k}}, \end{aligned}$$
and
$$\begin{aligned} B^{1}& = \sum _{n=1}^{\infty }\sum _{j=0}^{k-2}n\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_{1}^{n-j-1}p_{2}^{j}(p_{4}+p_{5})\\& = (p_{4}+p_{5})\frac{\partial }{\partial p_{2}}\dfrac{\frac{p_{2}}{1-p_{1}}\left( 1-\left( \frac{p_{2}}{1-p_{1}}\right) ^{k-1}\right) }{ 1-p_{2}-p_{1}}. \end{aligned}$$
Then
$$\begin{aligned} E(N)=\dfrac{(1-p_{1})^{k}-p_{2}^{k}}{(1-p_{2}-p_{1})(1-p_{1})^{k}}, \end{aligned}$$
and
$$\begin{aligned} E(N^{2})& = \sum _{n=k}^{\infty }n^{2}\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) p_{2}^{k-1}p_{1}^{n-k}(p_{3}+p_{4}) \\&+\sum _{n=1}^{\infty }\sum _{j=0}^{k-2}n^{2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_{2}^{j}p_{1}^{n-j-1}(p_{4}+p_{5}). \end{aligned}$$
Define \(C^{1}\) and \(D^{1}\) as
$$\begin{aligned} C^{1}& = \sum _{n=k}^{\infty }n^{2}\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) p_{2}^{k-1}p_{1}^{n-k}(p_{3}+p_{4}), \\ D^{1}& = \sum _{n=1}^{\infty }\sum _{j=0}^{k-2}n^{2}\left( {\begin{array}{c}n-1\\ j\end{array}}\right) p_{2}^{j}p_{1}^{n-j-1}(p_{4}+p_{5}). \end{aligned}$$
Then
$$\begin{aligned} C^{1}=\frac{kp_{2}^{k-1}(p_{1}+k)}{(1-p_{1})^{k+1}}, \end{aligned}$$
and
$$\begin{aligned} D^{1}&= \dfrac{p_{4}+p_{5}}{1-p_{1}}\left[ \frac{\partial }{\partial p_{2}}\left( \dfrac{p_{2}-\frac{p_{2}^{k}}{(1-p_{1})^{k-1}}}{1-p_{2}-p_{1}}\right) \right. \\&\quad \left. +(p_{2}+p_{1}) \frac{\partial ^{2}}{\partial p_{2}^{2}}\left( \dfrac{p_{2}-\frac{p_{2}^{k}}{ (1-p_{1})^{k-1}}}{1-p_{2}-p_{1}}\right) \right] . \end{aligned}$$
\(\square\)
Proof of Lemma 2
We know
$$\begin{aligned} P(M=m|N>M)=\dfrac{P(M=m,N>M)}{P(N>M)}, \end{aligned}$$
and by using Lemma 1, we have
$$\begin{aligned} P(N>M)=\dfrac{p_{2}}{1-p_{1}}, \end{aligned}$$
and
$$\begin{aligned}&P(M=m,N>M) \\&\quad = \sum _{n=m+1}^{\infty }\left( {\begin{array}{c}n-m-1\\ k-2\end{array}}\right) p_{2}^{k-1}p_{1}^{n-k}(p_{3}+p_{4}) \\&\qquad +\sum _{n=m+1}^{\infty }\sum _{j=0}^{k-3}\left( {\begin{array}{c}n-m-1\\ j\end{array}}\right) p_{2}^{j+1}p_{1}^{n-j-2}(p_{4}+p_{5}) \\&\quad =p_{1}^{m-1}p_{2}. \end{aligned}$$
Then
$$\begin{aligned}&P(M=m|N>M)=\dfrac{p_{1}^{m-1}p_{2}}{\frac{p_{2}}{1-p_{1}}}\\&\quad =p_{1}^{m-1}(1-p_{1}),\quad m=1,2,\ldots \end{aligned}$$
\(\square\)
