Refined tail asymptotic properties for the $M^X/G/1$ retrial queue

Abstract

In the literature, retrial queues with batch arrivals and heavy-tailed service times have been studied and the so-called equivalence theorem has been established under the condition that the service time is heavier than the batch size. The equivalence theorem provides the distribution (or tail) equivalence between the total number of customers in the system for the retrial queue and the total number of customers in the corresponding standard (non-retrial) queue. In this paper, under the assumption of regularly varying tails, we eliminate this condition by allowing that the service time can be either heavier or lighter than the batch size. The main contribution made in this paper is an asymptotic characterization of the difference between two tail probabilities: the probability of the total number of customers in the system for the $M^X/G/1$ retrial queue and the probability of the total number of customers in the corresponding standard (non-retrial) queue.

Appendices

Collection of concepts and results

Definition A.1

(e.g. see [11], pp. 564–565) A measurable function $U:(0,\infty )\rightarrow (0,\infty )$ is regularly varying at $\infty$ with index $\sigma \in (-\infty ,\infty )$, denoted by $U\in \mathcal R_{\sigma }$, iff $\lim _{x\rightarrow \infty }U(tx)/U(x)=t^{\sigma }$ for all $t>0$. If $\sigma =0$ we call U slowly varying, i.e. $\lim _{x\rightarrow \infty }U(tx)/U(x)=1$ for all $t>0$.

Furthermore, the class of the regularly varying distributions is defined as (see, for example, [11], p. 50)

$$\begin{aligned} \mathcal R=\{F \text{ df } \text{ on } (0,\infty ): 1- F\in \mathcal R_{\sigma } \text{ for } \text{ some } \sigma \le 0\}. \end{aligned}$$

The following lemma is referred to the uniform convergence theorem for regularly varying functions.

Lemma A.1

(Bingham et al. [7], p. 22) If f is regularly varying at $\infty$ with index $\sigma \le 0$,then for $\sigma <0$, $f(tx)/f(x)\rightarrow t^{\sigma }$ as $x\rightarrow \infty$ uniformly in t on each $[a,\infty )$ $(0<a<\infty )$; for $\sigma =0$, $f(tx)/f(x)\rightarrow 1$ as $x\rightarrow \infty$ uniformly in t on each [a, b] $(0<a\le b<\infty )$.

The following lemma is referred to the monotone density theorem for regularly varying functions.

Lemma A.2

([11], p. 568) Let $U(x)=\int _x^{\infty }u(y)\hbox {d}y$ (or $\int _0^x u(y)\hbox {d}y$) where u is ultimately monotone (i.e. u is monotone on $(z,\infty )$ for some $z>0$). If $U(x)\sim c x^{\sigma }L(x)$ as $x\rightarrow \infty$ with $c> 0,\ \sigma \in (-\infty ,\infty )$, then $u(x)\sim c\sigma x^{\sigma -1}L(x)$ as $x\rightarrow \infty$.

Definition A.2

(e.g. see Foss et al. [16], p. 40) A distribution F on $(0,\infty )$ belongs to the class of the subexponential distributions, denoted by $F\in \mathcal S$, if $\lim _{x\rightarrow \infty }(1-F^{(2)}(x))/(1-F(x))=2$, where $F^{(n)}=\underbrace{F*F* \cdots * F}_{n}$ denotes the n-fold convolution of F to itself.

Note that $\mathcal R$ is a subset of $\mathcal S$, i.e. $\mathcal R\subset \mathcal S$ (see, e.g., [11], p. 50).

Definition A.3

(Grandell [18], p. 146) A distribution F on $(0,\infty )$ is called light-tailed, if there exists $s_0>0$ such that $\int _{0}^{\infty }e^{sx}\hbox {d}F(x)<\infty$ for all $s<s_0$.

Lemma A.3

([11], p. 567) Let L be a slowly varying function on $(0,\infty )$. Then, for $b>1$, $\int _{x}^{\infty }t^{-b}L(t)\textrm{d}t\sim (b-1)^{-1} x^{-b+1}L(x)$ as $x\rightarrow \infty$.

Lemma A.4

([6], or Foss and Korshunov [15]) Assume that $N_t$ is a Poisson process with rate $\lambda >0$, and $T>0$ is a rv independent of $N_t$ with tail $P\{T>x\}$ heavier than $e^{-\sqrt{x}}$. Then, $P(N_T>j)\sim P\{T>j/\lambda \},\ j\rightarrow \infty$.

Note that by Assumption A1, both the service time B and the equilibrium service time $B^{(e)}$ have tails heavier than $e^{-\sqrt{x}}$.

Lemma A.5

Let N be a discrete non-negative integer-valued rv, and let $\{Y_k\}_{k=1}^{\infty }$ be a sequence of non-negative, independently and identically distributed rvs. Define $S_0\equiv 0$ and $S_n=\sum _{k=1}^n Y_k$.

• (i) If $P\{Y_k>x\}\sim c_Y x^{-h}L(x)$ as $x\rightarrow \infty$ and $P\{N>n\}\sim c_N n^{-h}L(n)$ as $n\rightarrow \infty$, where $h> 1$, $c_Y\ge 0$ and $c_N\ge 0$, then

  $$\begin{aligned} P\{S_N > x\}\sim & {} \left( c_N \mu _Y^{h}+ c_Y\mu _N\right) x^{-h}L(x),\quad x\rightarrow \infty , \end{aligned}$$

  (A.1)

  where $E(N)=\mu _N<\infty$ and $E(Y_k)=\mu _Y<\infty$.

• (ii) If $P\{N>n\}\sim c_N n^{-h_N}L(n)$ as $n\rightarrow \infty$, where $0\le h_N<1,\ c_N\ge 0$, and $E(Y_k)=\mu _Y<\infty$, then

  $$\begin{aligned} P\{S_N > x\}\sim & {} c_N (x/\mu _Y)^{-h_N}L(x),\quad x\rightarrow \infty . \end{aligned}$$

  (A.2)

• (iii) If $P\{N>n\}\sim c_N n^{-1}L(n)$ as $n\rightarrow \infty$, where $c_N\ge 0$, and $x^b P\{Y_k>x\}\le c<\infty$ for some $b>1$, then

  $$\begin{aligned} P\{S_N > x\}\sim & {} c_N (x/\mu _Y)^{-1}L(x),\quad x\rightarrow \infty , \end{aligned}$$

  (A.3)

  where $E(Y_k)=\mu _Y<\infty$.

In Lemma A.5, Parts (i) and (ii) are directly from Corollary 8.1 and Corollary 8.2 in [18] (pp. 163–165), and Part (iii) is due to Lemma 2.8 in Stam [36], p. 315.

Lemma A.6 is the discrete version of Karamata’s Theorem and Monotone Density Theorem.

Lemma A.6

([11], pp. 567–568) Let $\{q(j)\}_{j=0}^{\infty }$ be a nonnegative sequence, and $b>1$. If $q(j)\sim j^{-b}L(j)$ as $j\rightarrow \infty$, then $\sum _{k=j+1}^{\infty }q(k)\sim \displaystyle \frac{1}{b-1} j^{-b+1}L(j)$ as $j\rightarrow \infty$. Conversely, if $\sum _{k=j+1}^{\infty }q(k)\sim \displaystyle \frac{1}{b-1} j^{-b+1}L(j)$ as $j\rightarrow \infty$ and $\{q(j)\}_{j=0}^{\infty }$ is ultimately monotonic (i.e. q(j) is monotone except for first finite many terms), then $q(j)\sim j^{-b}L(j)$ as $j\rightarrow \infty$.

Lemma A.7

([16], p. 48) Suppose that $F(x)\in \mathcal S$.

• (i) If $1-G(x)=o(1-F(x))$ as $x\rightarrow \infty$, then $F*G\in \mathcal S$ and $1-F*G(x)\sim 1-F(x)$.

• (ii) If $(1-G_i(x))/(1-F(x))\rightarrow c_i$ as $x\rightarrow \infty$ for some $c_i\ge 0$, i=1,2, then $(1-G_1*G_2(x))/(1-F(x))\rightarrow c_1+c_2$ as $x\rightarrow \infty$.

Lemma A.8

([11], pp. 580–581) Suppose $0<b<1$, and distribution functions F and G are related as $G(x)=(1-b)\sum _{n=1}^{\infty }b^nF^{(n)}(x)$. Then, the following statements are equivalent:

• (i) $F\in \mathcal S$;

• (ii) $G\in \mathcal S$;

• (iii) $\lim _{x\rightarrow \infty }(1-G(x))/(1-F(x))=b/(1-b)$.

For proving our key property, Theorem 5.1, we need the following concepts and properties. Let $\{g(j)\}_{j=0}^{\infty }$ be a discrete probability distribution with the GF $G(z)=\sum _{j=0}^{\infty }g(j)z^j$. Denote by $\gamma _n(n\ge 0)$ the nth factorial moment of $\{g(j)\}_{j=0}^{\infty }$, i.e,

$$\begin{aligned} \gamma _0=1\quad \text{ and }\quad \gamma _n=\sum _{k=n}^{\infty }k(k-1)\cdots (k-n+1)g(k),\quad n\ge 1. \end{aligned}$$

(A.4)

It is well known that if $\gamma _n<\infty$, then $\gamma _n=\lim _{z\uparrow 1}\hbox {d}^n G(z)/\hbox {d}z^n$ and

$$\begin{aligned} G(z)=\sum _{k=0}^{n}(-1)^k\frac{\gamma _k}{k!}(1-z)^k +o((1-z)^n)\quad \text{ as } \ z\uparrow 1. \end{aligned}$$

(A.5)

Next, if $\gamma _n<\infty$, we introduce notations $G_n(\cdot )$ and $\widehat{G}_n(\cdot )$ as follows:

$$\begin{aligned}{} & {} G_n(z){\mathop {=}\limits ^\textrm{def}}(-1)^{n+1}\left( G(z)-\sum _{k=0}^{n}(-1)^k\frac{\gamma _k}{k!}(1-z)^k\right) ,\quad n\ge 0, \end{aligned}$$

(A.6)

$$\begin{aligned}{} & {} \widehat{G}_n (z){\mathop {=}\limits ^\textrm{def}}\frac{G_n(z)}{(1-z)^{n+1}},\quad n\ge 0. \end{aligned}$$

(A.7)

So,

$$\begin{aligned} G(z)=\sum _{k=0}^{n}(-1)^k\frac{\gamma _k}{k!}(1-z)^k+(-1)^{n+1}G_n(z). \end{aligned}$$

(A.8)

It follows that if $\gamma _n<\infty$, then for $n\ge 1$,

$$\begin{aligned} G_{n-1}(z)= & {} \frac{\gamma _{n}}{n!}(1-z)^n-G_n(z), \end{aligned}$$

(A.9)

$$\begin{aligned} \widehat{G}_{n-1}(z)= & {} \frac{\gamma _{n}}{n!}-(1-z)\widehat{G}_n (z), \end{aligned}$$

(A.10)

$$\begin{aligned} \widehat{G}_{n-1}(1)= & {} \frac{\gamma _n}{n!}-\lim _{z\uparrow 1}\frac{G_{n}(z)}{(1-z)^n}=\frac{\gamma _n}{n!}. \end{aligned}$$

(A.11)

In the following Lemma, we verify that $\widehat{G}_n (z)$ is the GF of a nonnegative sequence. To this end, we define recursively

$$\begin{aligned} \overline{g}_{0}(j)= & {} g(j),\quad j\ge 0, \end{aligned}$$

(A.12)

$$\begin{aligned} \overline{g}_{n+1}(j)= & {} \sum _{i=j+1}^{\infty }\overline{g}_{n}(i),\quad j\ge 0;\ n\ge 0. \end{aligned}$$

(A.13)

Lemma A.9

Suppose that $\{g(j)\}_{j=0}^{\infty }$ is a discrete probability distribution with $\gamma _{n}<\infty ,\ n\ge 0$. Then, $\widehat{G}_k(z)$ is the GF of sequence $\{\overline{g}_{k+1}(j)\}_{j=0}^{\infty }$ for $0\le k\le n$, that is,

$$\begin{aligned} \sum _{j=0}^{\infty }\overline{g}_{k+1}(j)z^j= & {} \widehat{G}_k(z),\quad 0\le k\le n. \end{aligned}$$

(A.14)

Proof

Notice that

$$\begin{aligned} \sum _{j=0}^{\infty }\overline{g}_{k+1}(j)z^j= & {} \sum _{j=0}^{\infty }\left( \sum _{i=j+1}^{\infty }\overline{g}_{k}(i)\right) z^j =\sum _{i=1}^{\infty }\sum _{j=0}^{i-1}\overline{g}_{k}(i) z^j\nonumber \\= & {} \frac{1}{1-z}\sum _{i=0}^{\infty } \overline{g}_{k}(i) (1-z^i). \end{aligned}$$

(A.15)

Next, we proceed with the mathematical induction on k. For $k=0$,

$$\begin{aligned} \sum _{j=0}^{\infty }\overline{g}_{1}(j)z^j=\frac{1-G(z)}{1-z}=\widehat{G}_0(z)\quad \text{(by } (\text {A.15}), (\text {A.6}) \text{ and } (\text {A.7})\text{) }. \end{aligned}$$

Under the induction hypothesis that (A.14) holds for $k=i-1\in \{0,1,\cdots ,n-1\}$, we have

$$\begin{aligned} \sum _{j=0}^{\infty }\overline{g}_{i+1}(j)z^j= & {} \frac{\widehat{G}_{i-1}(1)-\widehat{G}_{i-1}(z)}{1-z}\quad \text{(by } (\text {A.15}) \text{ and } \text{ the } \text{ induction } \text{ hypothesis } \text{) } \\= & {} \frac{\gamma _{i}/i!-\widehat{G}_{i-1}(z)}{1-z} \quad \text{(by } (\text {A.11})\text{) } \\= & {} \widehat{G}_i(z)\quad \text{(by } (\text {A.10})\text{) }. \end{aligned}$$

Therefore, (A.14) holds for $k=i\in \{1,2,\cdots ,n\}$. $\square$

The following lemma is referred to the Karamata’s Tauberian theorem for power series.

Lemma A.10

([7], pp. 40) Let $\{q(j)\}_{j=0}^{\infty }$ be a non-negative sequence such that $Q(z){\mathop {=}\limits ^\textrm{def}}\sum _{j=0}^{\infty }q(j)z^j$ converges for $0\le z<1$, let $L(\cdot )$ be slowly varying at $\infty$, and $b\ge 0$, then the following two statements are equivalent:

$$\begin{aligned} \text{(i) }&\quad Q(z)\sim (1-z)^{-b}L\left( 1/(1-z)\right) ,\quad z\uparrow 1; \text{ and } \end{aligned}$$

(A.16)

$$\begin{aligned} \text{(ii) }&\quad \sum _{k=0}^{j}q(k)\sim \frac{1}{\Gamma (b+1)}j^{b}L(j), \quad j\rightarrow \infty . \end{aligned}$$

(A.17)

Furthermore, if the sequence $\{q(j)\}_{j=0}^{\infty }$ is ultimately monotonic and $b>0$, then both (i) and (ii) are equivalent to

$$\begin{aligned} \text{(iii) }&\quad q(j)\sim \frac{1}{\Gamma (b)}j^{b-1}L(j),\quad j\rightarrow \infty . \end{aligned}$$

(A.18)

Lemma A.11

Let $\{g(j)\}_{j=0}^{\infty }$ be a discrete probability distribution with the GF G(z). Assume that $n< d<n+1$ for some $n\in \{0,1,2,\cdots \}$. The sequence $\{\overline{g}_{n+1}(j)\}_{j=0}^{\infty }$ is defined by (A.13). Let $L(\cdot )$ be slowly varying. The following two statements are equivalent:

$$\begin{aligned} \text{(i) }&\quad G_n(z)\sim (1-z)^{d}L(1/(1-z)),\quad z\uparrow 1; \text{ and } \end{aligned}$$

(A.19)

$$\begin{aligned} \text{(ii) }&\quad \overline{g}_{1}(j)\sim \frac{\Gamma (d)}{\Gamma (d-n)\Gamma (n+1-d)} j^{-d} L(j), \quad j\rightarrow \infty . \end{aligned}$$

(A.20)

Proof

By the definition of $\widehat{G}_n(z)$ in (A.7), (A.19) is equivalent to

$$\begin{aligned} \widehat{G}_n(z)\sim (1-z)^{-(n+1-d)} L\left( 1/(1-z)\right) . \end{aligned}$$

(A.21)

Note that $0< n+1-d<1$ and the sequence $\{\overline{g}_{n+1}(j)\}_{j=0}^{\infty }$ is decreasing with the GF $\widehat{G}_n(z)$ (by Lemma A.9). Applying Lemma A.10 (taking $b=n+1-d$ in (A.16) and (A.18)), we know that (A.21) is equivalent to

$$\begin{aligned} \overline{g}_{n+1}(j)\sim \frac{1}{\Gamma (n+1-d)} j^{-d+n} L(j),\quad j\rightarrow \infty . \end{aligned}$$

(A.22)

Next, we prove the equivalence of (A.20) and (A.22). Noting the recursive relation (A.13) and repeatedly applying Lemma A.6, (A.22) is equivalent to

$$\begin{aligned} \overline{g}_{1}(j)\sim & {} \frac{(d-1)\cdots (d-n)}{\Gamma (n+1-d)} j^{-d} L(j),\quad j\rightarrow \infty . \end{aligned}$$

(A.23)

Note that $\Gamma (d)=(d-1)\cdots (d-n)\Gamma (d-n)$. $\square$

Lemma A.12

(e.g. [7], p. 172) Suppose that $\{q(j)\}_{j=0}^{\infty }$ is a nonnegative sequence with GF $Q(z)=\sum _{j=0}^{\infty } q(j)z^j$. Let $r(t)=\sum _{0\le j\le t} q(j)$, $t\in [0,\infty )$. The following two statements are equivalent:

$$\begin{aligned} \text{(i) }&\quad \lim _{t\uparrow \infty }\frac{r(xt)-r(t)}{L(t)}=\log x, \quad \text {for all x>0}; \text{ and } \end{aligned}$$

(A.24)

$$\begin{aligned} \text{(ii) }&\quad \lim _{s\downarrow 0}\frac{ Q(e^{-xs}) - Q(e^{-s})}{L(1/s)}= -\log x, \quad \text {for all x>0}. \end{aligned}$$

(A.25)

Lemma A.13

Suppose that $\{q(j)\}_{j=0}^{\infty }$ is a nonnegative sequence with GF $Q(z)=\sum _{j=0}^{\infty } q(j)z^j$. The above (A.25) is equivalent to

$$\begin{aligned} \lim _{u\downarrow 0}\frac{ Q(1-xu) - Q(1-u)}{L(1/u)}= -\log x,\quad \text{ for } \text{ all } x>0. \end{aligned}$$

(A.26)

Proof

By taking $u=1-e^{-s}$ in (A.25), we know that (A.25) is equivalent to

$$\begin{aligned} \lim _{u\downarrow 0}\frac{ Q((1-u)^x ) - Q(1-u)}{L(1/u)}= -\log x,\quad \text{ for } \text{ all } x>0. \end{aligned}$$

(A.27)

So we only need confirm the equivalence of (A.26) and (A.27).

Suppose that (A.26) holds. Note that $(1-u)^x=1-xu+o(u)$. For $0<\epsilon <x$, we have $1-(x+\epsilon )u \le (1-u)^x \le 1-(x-\epsilon )u$ for $u>0$ small enough. Therefore,

$$\begin{aligned} Q(1-(x+\epsilon )u) \le Q((1-u)^x) \le Q(1-(x-\epsilon )u), \end{aligned}$$

(A.28)

which, along with (A.26), implies

$$\begin{aligned}{} & {} -\log (x+\epsilon )\le \liminf _{u\downarrow 0}\frac{ Q((1-u)^x ) - Q(1-u)}{L(1/u)}\\{} & {} \qquad \le \limsup _{u\downarrow 0}\frac{ Q((1-u)^x ) - Q(1-u)}{L(1/u)}= -\log (x-\epsilon ). \end{aligned}$$

Letting $\epsilon \downarrow 0$, we get (A.27).

Conversely, suppose that (A.27) holds. Note that for $0<\epsilon <x$, $(1-u)^{x+\epsilon }\le 1-xu\le (1-u)^{x-\epsilon }$ for $u>0$ small enough, by which (A.26) can be verified similarly. $\square$

Details of some proofs

1.1 Proof of Theorem 4.1

Case 1: $d_X>d_B>1$ in Assumptions A1 and A2

This is the case, in which the batch size X has a tail lighter than the service time B. It is worthwhile to mention that in this case X is not necessarily light-tailed (see Definition A.3).

Lemma B.1

If $d_X>d_B>1$ in Assumptions A1 and A2, then as $j\rightarrow \infty$,

$$\begin{aligned} P\{X>j\}= & {} o(j^{-d_B}L(j)), \end{aligned}$$

(B.1)

$$\begin{aligned} P\{X_0>j\}= & {} o(j^{-d_B}L(j)), \end{aligned}$$

(B.2)

$$\begin{aligned} P\{X^{(de)}>j\}= & {} o(j^{-d_B+1}L(j)). \end{aligned}$$

(B.3)

Proof

Because of $d_X>d_B$, (B.1) and (B.2) directly follow from Assumptions A1 and A2. We now prove (B.3). By Assumption A2, $P\{X>j\}\le c'_X j^{-d_X}L(j)$ for some $c'_X>0$. Since $P\{X^{(de)}=j\}=P\{X>j\}/\chi _1$ (by the definition of the equilibrium distribution),

$$\begin{aligned}{} & {} P\{X^{(de)}>j\} \le (c'_X/\chi _1)\sum _{k=j+1}^{\infty } k^{-d_X}L(k) \ \sim \ \frac{c'_X/\chi _1}{d_X-1} j^{-d_X+1}L(j)\nonumber \\{} & {} \quad \text{(by\;Lemma\;A.6) } \end{aligned}$$

(B.4)

which leads to (B.3) due to $d_X>d_B$. $\square$

By (4.1), (4.2), (B.1), and (B.3), we immediately have $P\{X>j\}=o(P\{N_{B}>j\})$ and $P\{X^{(de)}>j\}=o(P\{N_{B^{(e)}}>j\})$. By the definitions of $N_{B_X}$ and $N_{B^{(e)}_X}$ in Facts A and D, and applying Lemma A.5, we have

$$\begin{aligned} P\{N_{B_X}>j\}\sim & {} (\lambda \chi _1)^{d_B} j^{-d_B}L(j), \end{aligned}$$

(B.5)

$$\begin{aligned} P\{N_{B^{(e)}_X}>j\}\sim & {} \frac{(\lambda \chi _1)^{d_B-1}}{(d_B-1)\beta _1} j^{-d_B+1}L(j). \end{aligned}$$

(B.6)

By the definitions in Facts B and D, $N_{B_XX_0}=N_{B_X}+X_0$ and $N_{B^{(e)}_XX^{(de)}}=N_{B^{(e)}_X}+X^{(de)}$, and (B.5) and (B.6) lead to $P\{X_0>j\}=o(P\{N_{B_X}>j\})$ and $P\{X^{(de)}>j\}=o(P\{N_{B^{(e)}_X}>j\})$ due to $d_X>d_B$. Applying Part (i) of Lemma A.7, we have

$$\begin{aligned} P\{N_{B_XX_0}>j\}\sim & {} P\{N_{B_X}>j\}\ \sim \ (\lambda \chi _1)^{d_B} j^{-d_B}L(j), \end{aligned}$$

(B.7)

$$\begin{aligned} P\{N_{B^{(e)}_XX^{(de)}}>j\}\sim & {} P\{N_{B^{(e)}_X}>j\}\ \sim \ \frac{(\lambda \chi _1)^{d_B-1}}{(d_B-1)\beta _1} j^{-d_B+1}L(j). \end{aligned}$$

(B.8)

Now we are ready to present the asymptotic property for the tail probability of K. By (3.9) and (B.8), and applying Lemma A.8, we get

$$\begin{aligned} P\{K^{\circ }>j\}=\frac{\rho }{1-\rho }P\{N_{B^{(e)}_XX^{(de)}}>j\}\sim & {} \frac{(\lambda \chi _1)^{d_B}}{(d_B-1)(1-\rho )} j^{-d_B+1}L(j), \quad \quad \end{aligned}$$

(B.9)

where in the first equality we have used the fact that $\rho /(1-\rho )$ is the mean of rv J in (3.9). By Facts B and C, and (B.7),

$$\begin{aligned} P\{K^{*}=j\}=P\{N_{B_XX_0}^{(de)}=j\}=\frac{P\{N_{B_XX_0}>j\}}{E(N_{B_XX_0})}\sim \frac{(\lambda \chi _1)^{d_B} }{\rho +\chi _1-1} j^{-d_B}L(j). \end{aligned}$$

Applying Lemma A.6 gives

$$\begin{aligned} P\{K^{*}>j\}\sim & {} \frac{(\lambda \chi _1)^{d_B}}{(d_B-1)(\rho +\chi _1-1)} j^{-d_B+1}L(j). \end{aligned}$$

(B.10)

By (3.10), (B.10), and (B.9) and using Part (ii) of Lemma A.7, we have

$$\begin{aligned} P\{K>j\}\sim & {} \frac{(\lambda \chi _1)^{d_B}\chi _1}{(d_B-1)(1-\rho )(\rho +\chi _1-1)}\cdot j^{-d_B+1}L(j), \end{aligned}$$

(B.11)

which is the conclusion in Theorem 4.1 for Case 1.

Case 2: $1<d_X<d_B$ and $c_X> 0$ in Assumptions A1 and A2

This is the case, in which the batch size X has a tail heavier than the service time B. By the definitions of $N_{B}$, $N_{B_X}$ and $N_{B_XX_0}$ in Facts A and B, and applying Lemma A.5 and Part (ii) of Lemma A.7, we have

$$\begin{aligned} P\{N_{B_X}>j\}\sim & {} \lambda \beta _1\cdot c_X j^{-d_X}L(j), \end{aligned}$$

(B.12)

$$\begin{aligned} P\{N_{B_XX_0}>j\}\sim & {} (1+\lambda \beta _1)\cdot c_X j^{-d_X}L(j), \end{aligned}$$

(B.13)

where we have used the facts $E(N_B)=\lambda \beta _1$ and $P\{X_0>j\}\sim P\{X>j\}$.

By (4.2) and the definition of $N_{B^{(e)}_X}$ in Fact D, and applying Lemma A.5,

$$\begin{aligned} P\{N_{B^{(e)}_X}>j\}\le & {} c''_X \max \left( j^{-d_B+1}L(j),\ j^{-d_X}L(j)\right) \quad \text{ for } \text{ some } c''_X>0.\qquad \end{aligned}$$

(B.14)

By Lemma A.6, we have $P\{X^{(de)}>j\}\sim (\chi _1(d_X-1))^{-1} c_X j^{-d_X+1}L(j)$, which implies $P\{N_{B^{(e)}_X}>j\}=o(P\{X^{(de)}>j\})$. By the definition $N_{B^{(e)}_XX^{(de)}}$ in Fact D, and applying Part (i) of Lemma A.7, we get

$$\begin{aligned} P\{N_{B^{(e)}_XX^{(de)}}>j\}\sim P\{X^{(de)}>j\}\sim \frac{c_X}{\chi _1(d_X-1)} j^{-d_X+1}L(j). \end{aligned}$$

(B.15)

Now we are ready to present the asymptotic property for the tail probability of K. By (3.9) and (B.15), and applying Lemma A.8, we get

$$\begin{aligned} P\{K^{\circ }>j\}=\frac{\rho }{1-\rho }P\{N_{B^{(e)}_XX^{(de)}}>j\}\sim & {} \frac{\lambda \beta _1 c_X}{(1-\rho )(d_X-1)} j^{-d_X+1}L(j).\qquad \end{aligned}$$

(B.16)

By Facts B and C, and (B.13),

$$\begin{aligned} P\{K^{*}=j\}=P\{N_{B_XX_0}^{(de)}=j\}=\frac{P\{N_{B_XX_0}>j\}}{E(N_{B_XX_0})}\sim \frac{(1+\lambda \beta _1)c_X}{\rho +\chi _1-1} j^{-d_X}L(j). \end{aligned}$$

Applying Lemma A.6,

$$\begin{aligned} P\{K^{*}>j\}\sim \frac{(1+\lambda \beta _1)c_X}{(d_X-1)(\rho +\chi _1-1)}j^{-d_X+1}L(j). \end{aligned}$$

(B.17)

By (3.10), (B.17)–(B.16) and using Part (ii) of Lemma A.7,

$$\begin{aligned} P\{K>j\}\sim & {} \frac{c_X}{(d_X-1)(1-\rho )(\rho +\chi _1-1)}\cdot j^{-d_X+1}L(j), \end{aligned}$$

(B.18)

which is the conclusion in Theorem 4.1 for Case 2.

Case 3: $d_X=d_B=a>1$ and $c_X> 0$ in Assumptions A1 and A2

This is the case, in which the batch size X has a tail equivalent to the service time B. Following the same procedure as in Cases 1 and 2, we can prove that

$$\begin{aligned} P\{N_{B_X}>j\}\sim & {} ((\lambda \chi _1)^{a}+\lambda \beta _1c_X )\cdot j^{-a}L(j), \end{aligned}$$

(B.19)

$$\begin{aligned} P\{N_{B^{(e)}_XX^{(de)}}>j\}\sim & {} \frac{(\lambda \chi _1)^{a}+\lambda \beta _1c_X}{(a-1)\rho }\cdot j^{-a+1}L(j), \end{aligned}$$

(B.20)

$$\begin{aligned} P\{K^{\circ }>j\}\sim & {} \frac{(\lambda \chi _1)^a+\lambda \beta _1 c_X}{(a-1)(1-\rho )}\cdot j^{-a+1}L(j), \end{aligned}$$

(B.21)

$$\begin{aligned} P\{K^{*}>j\}\sim & {} \frac{(\lambda \chi _1)^a+(1+\lambda \beta _1) c_X}{(a-1)(\rho +\chi _1-1)}\cdot j^{-a+1}L(j), \end{aligned}$$

(B.22)

$$\begin{aligned} P\{K>j\}\sim & {} \frac{(\lambda \chi _1)^{a}\chi _1+c_X}{(a-1)(1-\rho )(\rho +\chi _1-1)}\cdot j^{-a+1}L(j), \end{aligned}$$

(B.23)

where we have skipped detailed derivations to avoid repetition.

1.2 Proof of Theorem 5.1

First let us rewrite (2.6) as follows:

$$\begin{aligned} D^{(0)}(z) = 1 -\psi \int _z^1 K(u)\hbox {d}u +\sum _{k=2}^{\infty }{(-\psi )^k \over k!}\left( \int _z^1 K(u)\hbox {d}u\right) ^k. \end{aligned}$$

(B.24)

As shown in Facts A–D, K(z) is the GF of the rv K with the discrete probability distribution $k(j) = P\{K=j\}$, $j\ge 0$. In the proof, we use the notation $\kappa _n$ to represent the nth factorial moment of K (see (A.4) in Appendix A for the definition of nth factorial moment).

Proof for the case of non-integer $a>1$

Suppose $m<a<m+1$, $m\in \{1,2,\cdots \}$. By Theorem 4.1, $P\{K>j\}\sim c_K\cdot j^{-a+1}L(j)$. So $\kappa _{m-1}<\infty$ and $\kappa _{m}=\infty$.

Define $K_{m-1}(z)$ in a manner similar to the definition of $G_n(z)$ in (A.6). Corresponding to the sequence $\{k(j)\}_{j=0}^{\infty }$, we also define $\overline{k}_n(j)$, $n\in \{0,1,\cdots m-1\}$ in a way similar to the definition of $\overline{g}_n(j)$ in (A.12) and (A.13). Note that $\overline{k}_1(j)=P\{K>j\}\sim c_K\cdot j^{-a+1}L(j)$. By Lemma A.11,

$$\begin{aligned} K_{m-1}(z)\sim & {} \frac{\Gamma (a-m)\Gamma (m+1-a)}{\Gamma (a-1)} c_K(1-z)^{a-1}L(1/(1-z)),\quad z\uparrow 1.\qquad \quad \end{aligned}$$

(B.25)

By Karamata’s theorem (see, Lemma A.3),

$$\begin{aligned} \int _z^{1}K_{m-1}(u)\hbox {d}u\sim & {} \frac{\Gamma (a-m)\Gamma (m+1-a)}{\Gamma (a-1)a} c_K(1-z)^{a}L(1/(1-z)),\quad z\uparrow 1.\nonumber \\ \end{aligned}$$

(B.26)

Next, we present a relation between $D^{(0)}_m(z)$ and $K_{m-1}(z)$. By the definition of $K_{m-1}(z)$,

$$\begin{aligned} K(z)= & {} \sum _{k=0}^{m-1}(-1)^k\frac{\kappa _k}{k!}(1-z)^k+(-1)^{m}K_{m-1}(z), \end{aligned}$$

(B.27)

$$\begin{aligned} \int _z^1K(u)\hbox {d}u= & {} -\sum _{k=1}^{m}(-1)^k\frac{\kappa _{k-1}}{k!}(1-z)^k+(-1)^{m}\int _z^{1}K_{m-1}(u)\hbox {d}u. \end{aligned}$$

(B.28)

Note that $\int _z^{1}K_{m-1}(u)\hbox {d}u/(1-z)^{m}\rightarrow 0$ and $\int _z^{1}K_{m-1}(u)\hbox {d}u/(1-z)^{m+1}\rightarrow \infty$ as $z\uparrow 1$.

From (B.24) and (B.28), there are constants $\{v_k;\ k=0,1,2,\cdots ,m\}$ satisfying

$$\begin{aligned} D^{(0)}(z)= & {} \sum _{k=0}^{m}(-1)^k v_k (1-z)^k+(-1)^{m+1}\psi \int _z^{1}K_{m-1}(u)\hbox {d}u + O((1-z)^{m+1}),\quad z\uparrow 1. \nonumber \\ \end{aligned}$$

(B.29)

Define $D^{(0)}_m(z)$ in a manner similar to the definition of $G_n(z)$ in (A.6). By (B.29),

$$\begin{aligned} D^{(0)}_m(z)= & {} \psi \int _z^{1}K_{m-1}(u)\hbox {d}u + O((1-z)^{m+1})\nonumber \\\sim & {} \psi \int _z^{1}K_{m-1}(u)\hbox {d}u,\quad z\uparrow 1. \end{aligned}$$

(B.30)

By (B.26) and (B.30),

$$\begin{aligned} D^{(0)}_m(z)\sim & {} \frac{\Gamma (a-m)\Gamma (m+1-a)}{\Gamma (a)}\cdot \frac{(a-1)c_K\psi }{a}(1-z)^{a}L(1/(1-z)),\quad z\uparrow 1.\nonumber \\ \end{aligned}$$

(B.31)

By applying Lemma A.11,

$$\begin{aligned} P\{D^{(0)}>j\}\sim & {} \frac{(a-1)c_K\psi }{a} j^{-a}L(j),\quad j\rightarrow \infty , \end{aligned}$$

(B.32)

which completes the proof of Theorem 5.1 for non-integer $a>1$.

Proof for the case of integer $a>1$

Suppose $a=m\in \{2,3,\cdots \}$. By Theorem 4.1, $P\{K>j\}\sim c_K\cdot j^{-m+1}L(j)$. So, $\kappa _{m-2}<\infty$. Unfortunately, whether $\kappa _{m-1}$ is finite or not remains uncertain, which is determined essentially by whether $\sum _{k=1}^{\infty }k^{-1}L(k)$ is convergent or not. For this reason, we have to sharpen our analytical tool by introducing the following lemma.

Lemma B.2

Suppose that $\{q(j)\}_{j=0}^{\infty }$ is a nonnegative non-increasing sequence with GF Q(z). Let $r(t)=\sum _{0\le j\le t} q(j)$, $t\in [0,\infty )$. The following three statements are equivalent:

$$\begin{aligned} \text{(i) }&\quad q(j)\sim j^{-1}L(j),\quad j\rightarrow \infty ; \end{aligned}$$

(B.33)

$$\begin{aligned} \text{(ii) }&\quad \lim _{t\uparrow \infty }\frac{r(xt)-r(t)}{L(t)}=\log x,\quad \text{ for } \text{ all } x>0; \text{ and } \end{aligned}$$

(B.34)

$$\begin{aligned} \text{(iii) }&\quad \lim _{u\downarrow 0}\frac{ Q(1-xu) - Q(1-u)}{L(1/u)}= -\log x,\quad \text{ for } \text{ all } x>0. \end{aligned}$$

(B.35)

Proof

Set $f(t)=q(k+1)$ for $t\in (k, k+1],\ k= 0,1,\cdots$, and $f(t)=0$ for $t\le 0$. So $f(t)\ge 0$ is a nonincreasing function. For $t\ge 1$, we have

$$\begin{aligned} r(t)= & {} q(0)+\int _0^{\lfloor t \rfloor } f(y)\hbox {d}y, \end{aligned}$$

(B.36)

$$\begin{aligned} r(xt)-r(t)= & {} \int _{\lfloor t \rfloor }^{\lfloor xt \rfloor } f(y) \hbox {d}y, \quad \text{ for } \text{ all } x>0, \end{aligned}$$

(B.37)

where $\lfloor t \rfloor$ represents the greatest integer less than or equal to t. Note that for $\epsilon >0$, we have $(x-\epsilon )t\le \lfloor xt \rfloor \le (x+\epsilon )t$ and $(1-\epsilon )t\le \lfloor t \rfloor \le (1+\epsilon )t$ for t large enough. Therefore,

$$\begin{aligned} \int _{(1+\epsilon )t}^{(x-\epsilon )t} f(y) \hbox {d}y\le & {} r(xt)-r(t)\ \le \ \int _{(1-\epsilon )t}^{(x+\epsilon )t} f(y) \hbox {d}y, \quad \text{ for } x>1,\ 0<2\epsilon < x-1, \nonumber \\ \end{aligned}$$

(B.38)

$$\begin{aligned} \int _{(x+\epsilon )t}^{(1-\epsilon )t} f(y) \hbox {d}y\le & {} r(t)-r(xt)\ \le \ \int _{(x-\epsilon )t}^{(1+\epsilon )t} f(y) \hbox {d}y, \quad \text{ for } 0<x<1, \ 0<2\epsilon < 1-x.\nonumber \\ \end{aligned}$$

(B.39)

Apparently, (B.33) is equivalent to

$$\begin{aligned} f(t)\sim t^{-1}L(t),\quad t\rightarrow \infty . \end{aligned}$$

(B.40)

Next, we will prove the equivalence of (B.40) and (B.34). Suppose that (B.40) holds. It follows from (B.38) that, for $x>1$,

$$\begin{aligned} \limsup _{t\rightarrow \infty }\frac{r(xt)-r(t)}{L(t)}\le & {} \lim _{t\rightarrow \infty }\int _{1-\epsilon }^{x+\epsilon }\frac{tf(ty)}{L(t)}\hbox {d}y = \int _{1-\epsilon }^{x+\epsilon }1/y \hbox {d}y =\log \frac{x+\epsilon }{1-\epsilon },\nonumber \\ \end{aligned}$$

(B.41)

$$\begin{aligned} \liminf _{t\rightarrow \infty }\frac{r(xt)-r(t)}{L(t)}\ge & {} \lim _{t\rightarrow \infty }\int _{1+\epsilon }^{x-\epsilon }\frac{tf(ty)}{L(t)}\hbox {d}y = \int _{1+\epsilon }^{x-\epsilon }1/y \hbox {d}y =\log \frac{x-\epsilon }{1+\epsilon },\nonumber \\ \end{aligned}$$

(B.42)

where we have used the uniform convergence theorem (see Lemma A.1) on slowly varying functions for interchanging the limit and the integral. Letting $\epsilon \downarrow 0$ in (B.41) and (B.42) implies (B.34) for $x>1$. (B.34) for $0<x<1$ can be similarly proved by using (B.39), and the proof of (B.34) for $x=1$ is trivial.

Conversely, suppose that (B.34) holds. By (B.38),

$$\begin{aligned} r(xt)-r(t)\le & {} \int _{(1-\epsilon )t}^{(x+\epsilon )t} f(y) \hbox {d}y\ \le \ f((1-\epsilon )t)(x-1+2\epsilon )t, \quad \text{ for } 0<2\epsilon < x-1,\nonumber \\ \end{aligned}$$

(B.43)

which, along with (B.34), implies

$$\begin{aligned} \log x=\lim _{t\rightarrow \infty }\frac{r(xt)-r(t)}{L(t)} \le \frac{x-1+2\epsilon }{1-\epsilon }\liminf _{t\rightarrow \infty } \frac{t f(t)}{L(t)}. \end{aligned}$$

(B.44)

Taking $\epsilon \downarrow 0$ gives $\frac{\log x}{x-1}\le \liminf _{t\rightarrow \infty } \frac{t f(t)}{L(t)}$. Then, letting $x\downarrow 1$ leads to $1\le \liminf _{t\rightarrow \infty } \frac{t f(t)}{L(t)}$. Similarly, for the case of $0<x<1$, by (B.39) we have

$$\begin{aligned} r(t)-r(xt)\ge & {} \int _{(x+\epsilon )t}^{(1-\epsilon )t} f(y) \hbox {d}y\ \ge \ f((1-\epsilon )t)(1-x-2\epsilon )t, \text{ for } 0<2\epsilon < 1-x, \nonumber \\ \end{aligned}$$

(B.45)

from which, again by (B.34),

$$\begin{aligned} -\log x=-\lim _{t\rightarrow \infty }\frac{r(xt)-r(t)}{L(t)} \ge \frac{1-x-2\epsilon }{1-\epsilon }\limsup _{t\rightarrow \infty } \frac{t f(t)}{L(t)}. \end{aligned}$$

(B.46)

Taking $\epsilon \downarrow 0$ gives $-\frac{\log x}{1-x}\ge \limsup _{t\rightarrow \infty } \frac{t f(t)}{L(t)}$. Then, letting $x\uparrow 1$ leads to $\limsup _{t\rightarrow \infty } \frac{t f(t)}{L(t)}\le 1$. Therefore, $\lim _{t\rightarrow \infty } \frac{t f(t)}{L(t)}= 1$, which is (B.40).

The equivalence of (B.34) and (B.35) is immediate by using Lemma A.12 and Lemma A.13. $\square$

Lemma B.3

Let $\{g(j)\}_{j=0}^{\infty }$ be a discrete probability distribution with GF G(z), and $n\in \{1,2,\cdots \}$. The following two statements are equivalent:

$$\begin{aligned} \text{(i) }&\quad \overline{g}_1(j)\sim j^{-n}L(j)\quad \text{ as } j\rightarrow \infty ; \end{aligned}$$

(B.47)

$$\begin{aligned} \text{(ii) }&\quad \lim _{u\downarrow 0}\frac{\widehat{G}_{n-1}(1-xu)-\widehat{G}_{n-1}(1-u)}{L(1/u)/(n-1)!}=-\log x\quad \text{ for } \text{ all } x>0, \end{aligned}$$

(B.48)

where $\overline{g}_1(j)$ and $\widehat{G}_{n}(x)$ are defined in (A.13) and (A.7), respectively.

Proof

By using Lemma A.6 repeatedly, (B.47) is equivalent to

$$\begin{aligned} \overline{g}_{n}(j)\sim j^{-1}L(j)/(n-1)!\quad \text{ as } j\rightarrow \infty . \end{aligned}$$

(B.49)

Note that the sequence $\{\overline{g}_{n}(j)\}_{j=0}^{\infty }$ has the GF $\widehat{G}_{n-1}(z)$ (by Lemma A.9). The equivalence of (B.48) and (B.49) is proved by applying Lemma B.2. $\square$

Since $\kappa _{m-2}<\infty$, we can define $K_{m-2}(z)$ in a manner similar to the definition of $G_n(z)$ in (A.6). Then, by comparing (A.8), we have

$$\begin{aligned} K(z)= & {} \sum _{k=0}^{m-2}(-1)^k\frac{\kappa _k}{k!}(1-z)^k+(-1)^{m-1}K_{m-2}(z), \end{aligned}$$

(B.50)

where $K_{m-2}(z)=o\left( (1-z)^{m-2}\right)$ as $z\uparrow 1$. Furthermore,

$$\begin{aligned} \int _z^{1}K(u)\hbox {d}u= & {} -\sum _{k=1}^{m-1}(-1)^k\frac{\kappa _{k-1}}{k!}(1-z)^k+(-1)^{m-1}\int _z^{1}K_{m-2}(u)\hbox {d}u,\quad \quad \end{aligned}$$

(B.51)

where $\int _z^{1}K_{m-2}(u)\hbox {d}u=o((1-z)^{m-1})$ as $z\uparrow 1$.

It follows from (B.24) and (B.51) that for some constants $\{v_k;\ k=0,1,2,\cdots ,m\}$,

$$\begin{aligned} D^{(0)}(z)= & {} \sum _{k=0}^{m}(-1)^kv_k (1-z)^k+(-1)^{m} \psi \int _z^{1}K_{m-2}(u)\hbox {d}u +o((1-z)^{m}), \quad z\uparrow 1.\nonumber \\ \end{aligned}$$

(B.52)

Define $\widehat{D}^{(0)}_{m-1}(z)$ in a manner similar to the definition of $\widehat{G}_n (z)$ in (A.7). Then, we have

$$\begin{aligned} \widehat{D}^{(0)}_{m-1}(z)=v_m+ \frac{\psi }{(1-z)^m}\int _z^{1} (1-u)^{m-1}\widehat{K}_{m-2}(u)\hbox {d}u + o(1),\quad z\uparrow 1,\nonumber \\ \end{aligned}$$

(B.53)

which immediately leads to:

$$\begin{aligned}{} & {} \widehat{D}^{(0)}_{m-1}(1-w)=v_m+ \frac{\psi }{w^m}\int _0^{w} u^{m-1}\widehat{K}_{m-2}(1-u)\hbox {d}u + o(1),\quad w\downarrow 0, \nonumber \\ \end{aligned}$$

(B.54)

$$\begin{aligned}{} & {} \widehat{D}^{(0)}_{m-1}(1-xw)=v_m+ \frac{\psi }{(xw)^m}\int _0^{xw} u^{m-1}\widehat{K}_{m-2}(1-u)\hbox {d}u + o(1)\nonumber \\{} & {} = v_m+ \frac{\psi }{w^m}\int _0^{w} u^{m-1}\widehat{K}_{m-2}(1-xu)\hbox {d}u + o(1),\quad w\downarrow 0. \end{aligned}$$

(B.55)

By (B.54) and (B.55)

$$\begin{aligned}{} & {} \widehat{D}^{(0)}_{m-1}(1-xw)-\widehat{D}^{(0)}_{m-1}(1-w)\nonumber \\{} & {} =\frac{\psi }{w^m}\int _0^{w} u^{m-1}\left( \widehat{K}_{m-2}(1-xu)-\widehat{K}_{m-2}(1-u)\right) \hbox {d}u + o(1)\quad w\downarrow 0.\quad \quad \quad \end{aligned}$$

(B.56)

Note that $\overline{k}_1(j)=P\{K>j\}\sim c_K\cdot j^{-m+1}L(j)$. By Lemma B.3, we obtain

$$\begin{aligned} \widehat{K}_{m-2}(1-xu)-\widehat{K}_{m-2}(1-u)\sim -(\log x)c_KL(1/u)/(m-2)! \quad u\downarrow 0.\nonumber \\ \end{aligned}$$

(B.57)

By Karamata’s theorem (see Lemma A.3), we know

$$\begin{aligned}{} & {} \int _0^{w} u^{m-1}\left( \widehat{K}_{m-2}(1-xu)-\widehat{K}_{m-2}(1-u)\right) \hbox {d}u\sim -(\log x)\nonumber \\{} & {} \quad \frac{c_K}{m}w^m L(1/w)/(m-2)! \quad w\downarrow 0.\nonumber \\ \end{aligned}$$

(B.58)

Therefore,

$$\begin{aligned} \lim _{w\downarrow 0}\frac{\widehat{D}^{(0)}_{m-1}(1-xw)-\widehat{D}^{(0)}_{m-1}(1-w)}{L(1/w)/(m-1)!} =-\frac{m-1}{m}c_K\psi \log x \quad \text{(by } (\text {B.56}) \text{ and } (\text {B.58})\text{) }. \nonumber \\ \end{aligned}$$

(B.59)

By applying Lemma B.3, we obtain from (B.59) that

$$\begin{aligned} p\{D^{(0)}>j\}\sim & {} \frac{m-1}{m}c_K\psi j^{-m}L(j)\quad \text{ as }\ j\rightarrow \infty , \end{aligned}$$

(B.60)

which completes the proof of Theorem 5.1 for integer $a=m\in \{2,3,\cdots \}$.

1.3 Proof of Lemma 6.1

The proof method presented here resembles that for Lemma 1.3.1 in [11], p. 37. For $0<\varepsilon <1$ and $t>0$, we have

$$\begin{aligned}{} & {} P\{X_1+X_2>t\} =P\{X_1>(1-\varepsilon )t,X_2>\varepsilon t\}\nonumber \\{} & {} \quad \quad +P\{X_1+X_2>t,X_2\le \varepsilon t\}+P\{X_1+X_2>t,X_1\le (1-\varepsilon ) t\}\nonumber \\{} & {} \quad =\overline{F}_1((1-\varepsilon )t)\overline{F}_2(\varepsilon t)+P\{X_1>t,X_2\le \varepsilon t\} +P\{X_1+X_2>t,X_1\le t, X_2\le \varepsilon t\}\nonumber \\{} & {} \quad \quad +P\{X_1\le (1-\varepsilon ) t,X_2>t\}+P\{X_1+X_2>t, X_1\le (1-\varepsilon )t,X_2\le t\}\nonumber \\{} & {} \quad =\overline{F}_1((1-\varepsilon )t)\overline{F}_2(\varepsilon t)+\overline{F}_1(t)-\overline{F}_1(t)\overline{F}_2(\varepsilon t) +\pi _1(t,\varepsilon )\nonumber \\{} & {} \quad \quad +\overline{F}_2(t)-\overline{F}_2(t)\overline{F}_1((1-\varepsilon )t)+\pi _2(t,\varepsilon ), \end{aligned}$$

(B.61)

where

$$\begin{aligned} \pi _1(t,\varepsilon )= & {} \int _0^{\varepsilon t}(\overline{F}_1(t-y)-\overline{F}_1(t))\hbox {d}F_2(y),\\ \pi _2(t,\varepsilon )= & {} \int _0^{(1-\varepsilon )t}(\overline{F}_2(t-y)-\overline{F}_2(t))\hbox {d}F_1(y). \end{aligned}$$

By the monotone density theorem (Lemma A.2), we know that $f_1(t)\sim (d-1)c_1 t^{-d} L_1(t)\sim (d-1)t^{-1}\overline{F}_1(t)$. By the mean value theorem, there exists $\theta =\theta (t,y)\in (0, 1)$ such that $\overline{F}_1(t-y)-\overline{F}_1(t)=f_1(t-\theta y)\cdot y$ for $0\le y\le \varepsilon t$. The decreasing property of $f_1$ yields $f_1(t)\le f_1(t-\theta y)\le f_1((1-\varepsilon )t)$ for $0\le y\le \varepsilon t$. Therefore,

$$\begin{aligned} f_1(t)\int _0^{\varepsilon t}y \hbox {d}F_2(y)\le \pi _1(t,\varepsilon )\le f_1((1-\varepsilon )t)\int _0^{\varepsilon t}y \hbox {d}F_2(y), \end{aligned}$$

(B.62)

from which it follows that

$$\begin{aligned} (d-1)\mu _{F_2}=\liminf _{t\rightarrow \infty }\frac{\pi _1(t,\varepsilon )}{t^{-1}\overline{F}_1(t)} \le \limsup _{t\rightarrow \infty }\frac{\pi _1(t,\varepsilon )}{t^{-1}\overline{F}_1(t)}=(d-1)(1-\varepsilon )^{-d}\mu _{F_2}\nonumber \\ \end{aligned}$$

(B.63)

Next, we prove the following asymptotic result:

$$\begin{aligned} \frac{\pi _2(t,\varepsilon )}{\overline{F}_2(t)}=\int _0^{(1-\varepsilon )t}\left[ \frac{\overline{F}_2(t-y)}{\overline{F}_2(t)} -1\right] \hbox {d}F_1(y)= & {} o(1). \end{aligned}$$

(B.64)

Note that

$$\begin{aligned}{} & {} \int _0^{(1-\varepsilon )t}\left[ \frac{\overline{F}_2(t-y)}{\overline{F}_2(t)} -1\right] \hbox {d}F_1(y)-\int _0^{(1-\varepsilon )t}\left[ \left( 1-{y\over t}\right) ^{-d}-1\right] \hbox {d}F_1(y)\nonumber \\{} & {} \quad =\int _0^{(1-\varepsilon )t}\left( 1-{y\over t}\right) ^{-d}\left[ \frac{L_2(t(1-y/t))}{L_2(t)}(1+o(1))-1\right] \hbox {d}F_1(y)\nonumber \\{} & {} \quad =o(1) \end{aligned}$$

(B.65)

where the last equality is due to the fact: by the uniform convergence theorem for slowly varying functions, $L_2(t(1-y/t))/L_2(t)\rightarrow 1$ uniformly on $y\in [0,(1-\varepsilon )t]$ (or $1-y/t\in [\varepsilon ,1]$).

Since for $b> 0$ and $0\le w\le 1-\varepsilon$,

$$\begin{aligned} (1-w)^{-b} -1 = b \int _{1-w}^{1} v^{-b-1} dv \le b w(1-w)^{-b-1} \end{aligned}$$

(B.66)

we have

$$\begin{aligned} 0\le \int _0^{(1-\varepsilon )t}\left[ \left( 1-{y\over t}\right) ^{-d}-1\right] \hbox {d}F_1(y)\le \frac{d}{\varepsilon ^{d+1}}\cdot \frac{1}{t} \int _0^{(1-\varepsilon )t}y\hbox {d}F_1(y) =o(1),\nonumber \\ \end{aligned}$$

(B.67)

where the last equality is due to $\int _0^t x \hbox {d}F_1(x)=-t \overline{F}_1(t)+\int _0^t \overline{F}_1(x)\hbox {d}x$. Now, (B.64) follows from (B.65) and (B.67).

Now, let us recall (B.61). Note that for all $0<\varepsilon <1$, $\overline{F}_1((1-\varepsilon )t)\overline{F}_2(\varepsilon t)=o(\overline{F}_2(t))$, $\overline{F}_1(t)\overline{F}_2(\varepsilon t)=o(\overline{F}_2(t))$, $\overline{F}_2(t)\overline{F}_1((1-\varepsilon )t)=o(\overline{F}_2(t))$ and $\pi _2(t,\varepsilon )=o(\overline{F}_2(t))$ (by B.64)). So, (B.61) can be rewritten as

$$\begin{aligned} P\{X_1+X_2>t\}=\overline{F}_1(t)+\pi _1(t,\varepsilon )+\overline{F}_2(t)(1+o(1)). \end{aligned}$$

Using (B.63) and taking $\varepsilon \rightarrow 0^+$, we complete the proof.

Proof of $\Delta$-analyticity of $Ez^{L_{\infty }}$ for Examples 1 and 2

Definition C.1

([14], p. 389) An open domain is called a $\Delta$-domain at 1 if it is in the form $\Delta =\{z|\ |z|<R,\ z\ne 1,\ |\text{ Arg }(z-1)|>\phi \}$ for some $R>1$ and $0<\phi <\frac{\pi }{2}$. A function is called $\Delta$-analytic at 1 if it is analytic in some $\Delta$-domain at 1.

In the following, for Example 1 and Example 2, we will prove, respectively, that $Ez^{L_{\infty }}$ can be analytically continued to $\mathbb {C}\backslash [1,\infty )$, which immediately implies its $\Delta$-analyticity at 1.

1.1 Example 1

Recall that (8.2) and (8.3). Note that $\beta (s)$ and $\beta ^{(e)}(s)$ are analytic in $\mathbb {D}=\mathbb {C}\backslash (-\infty ,0]$, and both X(z) and $X^{(de)}(z)$ can be analytically continued to $\mathbb {C}\backslash [1,\infty )$. For $Ez^{L_{\infty }}$ to be analytic in $\mathbb {C}\backslash [1,\infty )$, we only need to verify two things: (i) $\lambda (1-X(z))\in \mathbb {D}$ for any $z\in \mathbb {C}\backslash [1,\infty )$; (ii) $1-\rho \beta ^{(e)}(\lambda -\lambda X(z))\cdot X^{(de)}(z)$ is nonzero in $\mathbb {C}\backslash [1,\infty )$.

(i) Let $s=(1-z)/(1-q)=x+iy$. By (8.8),

$$\begin{aligned} 1-X(z)=\frac{s}{1+qs} = \frac{(x+iy)(1+qx-iqy)}{(1+qx)^2+(qy)^2}=\frac{x(1+qx)+qy^2+iy}{(1+qx)^2+(qy)^2}. \end{aligned}$$

It follows that if $\text{ Im }(z)\not = 0$ (or say $y\not = 0$), then $\text{ Im }\big (1-X(z)\big )\not = 0$. Additionally, for any real $z\in (-\infty ,1)$ (or say $s>0$), $1-X(z)=s/(1+qs)>0$. Hence, $\lambda (1-X(z))\in \mathbb {D}$ for any $z\in \mathbb {C}\backslash [1,\infty )$.

(ii) By (11) in [33], for $s\in \mathbb {D}$,

$$\begin{aligned} \beta ^{(e)}(s)= & {} \frac{a-1}{\Gamma (a)} \int _0^{\infty }\frac{t^{a-1}e^{-t}}{t+b s} \hbox {d}t. \end{aligned}$$

(C.1)

Set $s=(1-z)/(1-q)=x+iy$. By the expressions of $1-X(z)=s/(1+qs)$ and $X^{(de)}(z)=1/(1+qs)$, and using (C.1), we have

$$\begin{aligned}{} & {} \beta ^{(e)}(\lambda -\lambda X(z))\cdot X^{(de)}(z) =\beta ^{(e)}\Big (\frac{\lambda s}{1+qs}\Big )\cdot \frac{1}{1+qs}\nonumber \\{} & {} \quad =\frac{a-1}{\Gamma (a)} \int _0^{\infty }\frac{t^{a-1}e^{-t}}{(1+qs)t+\lambda b s} \hbox {d}t\nonumber \\{} & {} \quad =\frac{a-1}{\Gamma (a)} \int _0^{\infty }\frac{t^{a-1}e^{-t}}{(1+qx)t+\lambda b x+iy(qt+\lambda b)} \hbox {d}t\nonumber \\{} & {} \quad =\frac{a-1}{\Gamma (a)} \int _0^{\infty }\frac{[(1+qx)t+\lambda b x-iy(qt+\lambda b)]t^{a-1}e^{-t}}{[(1+qx)t+\lambda b x]^2+[y(qt+\lambda b)]^2} \hbox {d}t, \end{aligned}$$

(C.2)

from which we infer that $\text{ Im }(\beta ^{(e)}\left( \lambda -\lambda X(z))\cdot X^{(de)}(z)\right) \ne 0$ for $y\ne 0$ or $\text{ Im }(z)\not =0$. So, $1-\rho \beta ^{(e)}(\lambda -\lambda X(z))\cdot X^{(de)}(z)$ is nonzero in $\mathbb {C}\backslash (-\infty ,\infty )$. In addition, for any real $z\in (-\infty ,1)$, we have $s=x> 0$, then $\beta ^{(e)}(\lambda -\lambda X(z))\cdot X^{(de)}(z)=\beta ^{(e)}\big (\frac{\lambda x}{1+qx}\big )\cdot \frac{1}{1+qx}< 1$. So, $1-\rho \beta ^{(e)}(\lambda -\lambda X(z))\cdot X^{(de)}(z)$ is nonzero for $z\in (-\infty ,1)$.

1.2 Example 2

Note that $\tau (s)$ and $\tau ^{(e)}(s)$, $\beta (s)$ and $\beta ^{(e)}(s)$ can be analytically continued to $\mathbb {D}=\mathbb {C}\backslash (-\infty ,0]$, and both X(z) and $X^{(de)}(z)$ can be analytically continued to $\mathbb {C}\backslash [1,\infty )$. For $Ez^{L_{\infty }}$ to be analytic in $\mathbb {C}\backslash [1,\infty )$, we only need to verify two things: (i) $\lambda (1-X(z))\in \mathbb {D}$ for any $z\in \mathbb {C}\backslash [1,\infty )$; (ii) $1-\rho \beta ^{(e)}(\lambda -\lambda X(z))\cdot X^{(de)}(z)$ is nonzero in $\mathbb {C}\backslash [1,\infty )$.

(i) Similar to (C.1), we know that

$$\begin{aligned} \tau ^{(e)}(s)=\frac{u-1}{\Gamma (u)} \int _0^{\infty }\frac{t^{u-1}e^{-t}}{t+v s} \hbox {d}t, \qquad \text{ for } s\in \mathbb {\mathbb {D}}=\mathbb {C}\backslash (-\infty ,0]. \end{aligned}$$

(C.3)

By the definition of X(z), we can immediately write

$$\begin{aligned} 1-X(z)=1-zX_0(z)=1-z\tau (1-z). \end{aligned}$$

(C.4)

Set $s=1-z=x+iy$ in the above equality. Note that $\tau (s)=1-\tau _1 s\tau ^{(e)}(s)$. We then have

$$\begin{aligned} 1-X(z)=1-(1-s)\tau (s)= & {} s\big [1+\tau _1\cdot (1-s)\tau ^{(e)}(s)\big ]=s\varphi (s), \end{aligned}$$

(C.5)

where

$$\begin{aligned}{} & {} \varphi (s)=1+\tau _1\cdot (1-s)\tau ^{(e)}(s)\nonumber \\{} & {} \qquad =1+ \frac{v}{\Gamma (u)} \int _0^{\infty }\frac{1-s}{t+v s}t^{u-1}e^{-t} \hbox {d}t\nonumber \\{} & {} \qquad =1+ \frac{v}{\Gamma (u)} \int _0^{\infty }\frac{(1-x)(t+vx)-vy^2-(t+v)yi}{(t+v x)^2+(vy)^2}t^{u-1}e^{-t} \hbox {d}t\nonumber \\{} & {} \qquad {\mathop {=}\limits ^\textrm{def}} f(x,y)-iyg(x,y). \end{aligned}$$

(C.6)

It is worthwhile to mention that $g(x,y)=\frac{v}{\Gamma (u)} \int _0^{\infty }\frac{t+v}{(t+v x)^2+(vy)^2}\cdot t^{u-1}e^{-t} \hbox {d}t>0$ for all $x>0$ and $y\not =0$. It follows from (C.5) and (C.6) that

$$\begin{aligned} \text{ Im }\big (1-X(z))= & {} y[f(x,y)-xg(x,y)]\nonumber \\= & {} y\Big [1+\frac{v}{\Gamma (u)} \int _0^{\infty }\frac{t-2xt-vx^2-vy^2}{(t+v x)^2+(vy)^2}\cdot t^{u-1}e^{-t} \hbox {d}t\Big ]\nonumber \\= & {} y\cdot \frac{1}{\Gamma (u)} \int _0^{\infty }\frac{vt+t^2}{(t+v x)^2+(vy)^2}\cdot t^{u-1}e^{-t} \hbox {d}t. \end{aligned}$$

(C.7)

Therefore, if $\text{ Im }(z)\not = 0$ (or $y\not = 0$), then $\text{ Im }\big (1-X(z)\big )\not = 0$. Additionally, for any real $z\in (-\infty ,1)$ (or $s>0$), $1-X(z)\ge 1-(1-s)>0$ (see (C.5)). Hence, $\lambda (1-X(z))\in \mathbb {D}$ for any $z\in \mathbb {C}\backslash [1,\infty )$.

(ii) It follows from (C.5) and (C.6) that

$$\begin{aligned} X^{(de)}(z)= & {} \frac{1-X(z)}{\chi _1(1-z)}=\frac{\varphi (s)}{1+ \tau _1}, \end{aligned}$$

(C.8)

$$\begin{aligned} \frac{1}{\varphi (s)}= & {} \frac{f(x,y)+iyg(x,y)}{(f(x,y))^2+(yg(x,y))^2}{\mathop {=}\limits ^\textrm{def}}f_0(x,y)+iyg_0(x,y), \end{aligned}$$

(C.9)

where $g_0(x,y)=\frac{g(x,y)}{(f(x,y))^2+(yg(x,y))^2}>0$ for all $x>0$ and $y\not =0$. It follows from (C.5), (C.8), and (C.1) that

$$\begin{aligned}{} & {} \beta ^{(e)}(\lambda -\lambda X(z))\cdot X^{(de)}(z)=\beta ^{(e)}\big (\lambda s\varphi (s)\big )\cdot \frac{\varphi (s)}{1+\tau _1}\nonumber \\{} & {} \quad =\frac{a-1}{\Gamma (a)(1+\tau _1)} \int _0^{\infty }\frac{t^{a-1}e^{-t}}{\frac{t}{\varphi (s)}+\lambda b s} \hbox {d}t\nonumber \\{} & {} \quad =\frac{a-1}{\Gamma (a)(1+\tau _1)} \int _0^{\infty }\frac{t^{a-1}e^{-t}}{tf_0(x,y)+\lambda b x +iy(tg_0(x,y)+\lambda b)} \hbox {d}t\quad \text{(by } (\text {C.9})\text{) }\nonumber \\{} & {} \quad =\frac{a-1}{\Gamma (a)(1+\tau _1)} \int _0^{\infty }\frac{[tf_0(x,y)+\lambda b x -iy(tg_0(x,y)+\lambda b)]t^{a-1}e^{-t}}{[tf_0(x,y)+\lambda b x]^2+y^2(tg_0(x,y)+\lambda b)^2} \hbox {d}t, \end{aligned}$$

(C.10)

from which we know that $\text{ Im }(\beta ^{(e)}\left( \lambda -\lambda X(z))\cdot X^{(de)}(z)\right) \ne 0$ for all $y\ne 0$. So, $1-\rho \beta ^{(e)}(\lambda -\lambda X(z))\cdot X^{(de)}(z)$ is nonzero in $\mathbb {C}\backslash (-\infty ,\infty )$. Furthermore, for any real $z\in (-\infty ,1)$, we have $s=x> 0$ and by (C.6),

$$\begin{aligned} \varphi (x)= & {} 1+ \frac{v}{\Gamma (u)} \int _0^{\infty }\frac{1-x}{t+v x}t^{u-1}e^{-t} \hbox {d}t=\frac{1}{\Gamma (u)} \int _0^{\infty }\frac{t+v}{t+v x}t^{u-1}e^{-t} \hbox {d}t>0,\nonumber \\ \varphi (x)< & {} 1+ \frac{v}{\Gamma (u)} \int _0^{\infty }\frac{1}{t}t^{u-1}e^{-t} \hbox {d}t=1+\frac{v}{u-1}=1+\tau _1, \end{aligned}$$

(C.11)

hence $\beta ^{(e)}(\lambda -\lambda X(z))\cdot X^{(de)}(z)=\beta ^{(e)}\big (\lambda x\varphi (x)\big )\cdot \frac{\varphi (x)}{1+\tau _1}< 1$. So, $1-\rho \beta ^{(e)}(\lambda -\lambda X(z))\cdot X^{(de)}(z)$ is nonzero for $z\in (-\infty ,1)$.