Tail asymptotics for delay in a half-loaded GI/GI/2 queue with heavy-tailed job sizes

Abstract

We obtain asymptotic bounds for the tail distribution of steady-state waiting time in a two-server queue where each server processes incoming jobs at a rate equal to the rate of their arrivals (that is, the half-loaded regime). The job sizes are taken to be regularly varying. When the incoming jobs have finite variance, there are basically two types of effects that dominate the tail asymptotics. While the quantitative distinction between these two manifests itself only in the slowly varying components, the two effects arise from qualitatively very different phenomena (arrival of one extremely big job or two big jobs). Then there is a phase transition that occurs when the incoming jobs have infinite variance. In that case, only one of these effects dominates the tail asymptotics; the one involving arrival of one extremely big job.

Appendices

Appendix 1: Lyapunov bound techniques for a uniform bound on ${\mathbb {P}}_\mathbf{w} \{ \tau _b^{(2)} < \tau _0\}$

We use the Lyapunov bound technique that has been employed in [2, 3] and [7]. The strategy is to define a Markov kernel $Q_{\theta }\left( \mathbf{w},\cdot \right)$ (indexed by some parameter $\theta$) and a non-negative function $H_{b}\left( w_{1},w_{2}\right)$ satisfying the following conditions:

• (L1) For every $\mathbf{w} = (w_1,w_2)$ such that $w_2 < b,$

  $$\begin{aligned} {\mathbb {E}}_\mathbf{w}^{\theta }\left[ r_\theta \left( \mathbf{w} ,\mathbf{W}_{1}\right) H_{b}\left( \mathbf{W}_{1}\right) \right] \le H_{b}\left( \mathbf{w} \right) , \end{aligned}$$

  where ${\mathbb {P}}_\mathbf{w}\{ \mathbf{W}_1 \in \cdot \}$ is the nominal transition kernel induced by recursions (5a) and (5b), $r_\theta (\mathbf{w}, \mathbf{x}) := {\mathbb {P}}_\mathbf{w} \{ \mathbf{W}_1 \in d\mathbf{x}\}/Q_\theta ( \mathbf{w}, d\mathbf{x})$ is the corresponding Radon-Nikodym derivative with respect to $Q_\theta (\mathbf{w}, \cdot )$, and ${\mathbb {E}}_{\mathbf{w}}^{\theta }\left[ \cdot \right]$ is the expectation associated with the probability measure in path space for the Markov evolution induced by $Q_{\theta }\left( \mathbf{w},\cdot \right) .$

• (L2) Whenever $\mathbf{w} = (w_1,w_2)$ is such that $w_2 > b,$ $H_{b}\left( w_{1},w_{2}\right) \ge 1.$

If conditions (L1) and (L2) are satisfied, then following the analysis in Part (iii) of [2, Theorem 2], we have that

$$\begin{aligned} {\mathbb {P}}_\mathbf{w} \left\{ \tau _{b}^{\left( 2\right) }<\tau _{0} \right\}\le & {} {\mathbb {E}}_\mathbf{w}^{\theta } \left[ \prod \limits _{n=0}^{\tau _{b}^{\left( 2\right) }-1} r_{\theta }\left( \mathbf{W}_{n},\mathbf{W}_{n+1}\right) H_{b}\left( \mathbf{W}_{\tau _b^{(2)}}\right) I\left( \tau _{b}^{\left( 2\right) }<\tau _{0} \right) \right] \nonumber \\\le & {} H_{b}\left( w_{1},w_{2}\right) . \end{aligned}$$

(32)

The construction of $Q_{\theta }\left( \mathbf{w},\cdot \right)$ and $H_{b}\left( \cdot \right)$ follows the intuition explained in [2, 3]: We wish to select $Q_{\theta }\left( \mathbf{w},\cdot \right)$ as closely as possible to the conditional distribution of the process $\left\{ \mathbf{W}_{n}:n\ge 0\right\}$ given that $\{\tau _{b}^{\left( 2\right) }<\tau _{0}\},$ because in that case, it happens that (32) is automatically satisfied with equality. Additionally, we shall find a suitable non-negative function $G_{b}\left( \cdot \right)$ so that $H_{b}\left( w_{1},w_{2}\right) =G_{b}\left( w_{1}+w_{2}\right)$ satisfies the Lyapunov inequality (L1).

For ease of notation, let us write

$$\begin{aligned} l :=w_{1}+w_{2},\ L := W_1^{(1)} + W_2^{(2)} \text { and } \varDelta := L - l. \end{aligned}$$

In order to construct $Q_{\theta }\left( \mathbf{w},\cdot \right)$ and $G_{b}\left( \cdot \right) ,$ first define the Markov transition kernel

$$\begin{aligned} Q^{\prime }\left( \mathbf{w},A\right)= & {} {\mathbb {P}}_\mathbf{w}\left\{ \mathbf{W}_{1}\in A\text { } |\ X_1 >a\left( b-l\right) \right\} p\left( \mathbf{w}\right) \\&+\,{\mathbb {P}}_\mathbf{w}\left\{ \mathbf{W}_{1}\in A\text { }|\ X_1 \le a\left( b-l\right) ,W^{(2)}_1 >0\right\} \left( 1-p\left( \mathbf{w}\right) \right) , \end{aligned}$$

where $p\left( \mathbf{w}\right)$ will be specified momentarily, and the choice $a \in (0,1)$ is arbitrary. On the set $\{\tau _{b}^{\left( 2\right) }<\tau _{0}\}$, given $\mathbf{w}=\left( w_{1},w_{2}\right)$ with $w_1 \le w_{2} < b$, we have that the nominal kernel ${\mathbb {P}}_\mathbf{w}\{ \mathbf{W}_1 \in \cdot \}$ is absolutely continuous with respect to $Q^{\prime }\left( \mathbf{w},\cdot \right) .$ Now, for $z \ge 0,$ define

$$\begin{aligned} h_{b}\left( z\right) = \int _{0}^{z+\kappa _{0}}{\mathbb {P}}\left\{ X>b-z+t\right\} dt = \int _{b-z}^{b+\kappa _{0}}{\mathbb {P}}\left\{ X>u\right\} du. \end{aligned}$$

Next, write

$$\begin{aligned} G_{b}\left( l\right) =\min (\kappa _{1}h_{b}\left( l\right) ,1) \end{aligned}$$

and set

$$\begin{aligned} p\left( \mathbf{w} \right) = \frac{{\mathbb {P}}\left\{ X>a\left( b-l\right) \right\} }{\kappa _2 h_{b}\left( l\right) } \end{aligned}$$

where $\kappa _2$ is a number larger than

$$\begin{aligned} \sup _{x > 0} \frac{{\mathbb {P}}\left\{ X > a x\right\} }{\int _{x}^{x+l+\kappa _0} {\mathbb {P}}\left\{ X > u\right\} du} < \infty . \end{aligned}$$

Finally, define $\theta =\left( \kappa _{0},\kappa _{1},\kappa _{2}\right)$ and write

$$\begin{aligned} Q_{\theta }\left( \mathbf{w},\cdot \right) =Q^{\prime }\left( \mathbf{w},\cdot \right) I(G_{b}\left( l\right) <1)+K\left( \mathbf{w},\cdot \right) I(G_{b}\left( l\right) =1). \end{aligned}$$

Recall the notation $l = w_1 + w_2 \text { and } L =W_{1}^{\left( 1\right) }+W_{1}^{\left( 2\right) }.$ Condition (L1) is verified via the following proposition:

Proposition 5

For every $\mathbf{w} = (w_1,w_2)$ such that $w_2 < b,$ we have that

$$\begin{aligned} {\mathbb {E}}_\mathbf{w}^\theta \left[ r_\theta \left( \mathbf{w}, \mathbf{W}_1\right) G_b(L) \right] \le G_b(l). \end{aligned}$$

For proving Proposition 5, we consider only the case $G_b(l) < 1.$ When $G_b(l) = 1,$ the inequality is satisfied trivially. The following results are crucial in the proof of Proposition 5.

Lemma 17

There exist positive constants $\mu$ and C such that

$$\begin{aligned} {\mathbb {E}}_{(w_1,w_2)} \left[ \varDelta I\left( W_1^{(2)} > 0 \right) \right] < -\mu \end{aligned}$$

whenever $w_2 > C.$

Proof

First, observe that

$$\begin{aligned} {\mathbb {E}}_\mathbf{w} \left[ \varDelta I\left( W_1^{(2)} > 0 \right) \right] = {\mathbb {E}}_\mathbf{w} \left[ \varDelta \right] - {\mathbb {E}}_\mathbf{w} \left[ \varDelta I\left( W_1^{(2)} = 0 \right) \right] . \end{aligned}$$

Additionally, note that $\varDelta = -(w_1+w_2)$ when $W_1 ^{(2)} = 0.$ Therefore,

$$\begin{aligned} {\mathbb {E}}_{(w_1,w_2)} \left[ \varDelta I\left( W_1^{(2)} = 0 \right) \right] = -\left( w_1+w_2 \right) {\mathbb {P}}\left\{ w_1 + V - T \le 0, \ w_2 - T \le 0 \right\} . \end{aligned}$$

Therefore, due to Lemma 3,

$$\begin{aligned} {\mathbb {E}}_\mathbf{w} \left[ \varDelta I\left( W_1^{(2)} > 0 \right) \right]&\le {\mathbb {E}}_\mathbf{w} \left[ \varDelta \right] + \left( w_1+w_2 \right) {\mathbb {P}}\left\{ w_2 - T \le 0 \right\} \\&\le -\epsilon + 2w_2 {\mathbb {P}}\left\{ T > w_2 \right\} , \end{aligned}$$

where $w_2{\mathbb {P}}\{ T > w_2\}$ can be made arbitrarily small by choosing $C > w_2$ large enough. Hence the claim stands verified. $\square$

Lemma 18

Recall that $l = w_1+w_2.$ The following holds as $(b-l) \rightarrow \infty :$

$$\begin{aligned}&{\mathbb {E}}_\mathbf{w}^{\theta }\left[ r_{\theta }\left( \mathbf{w},\mathbf{W}_{1}\right) \frac{G_{b}\left( L \right) }{G_b(l)} I\left( X_1 \le a\left( b-l\right) \right) \right] \nonumber \\&\quad \le {\mathbb {P}}_\mathbf{w}\left\{ W_{1}^{\left( 2\right) }>0\right\} + \frac{{\mathbb {P}}\left\{ X > b-l \right\} }{h_b(l)}{\mathbb {E}}_\mathbf{w}\left[ \varDelta I\left( W_{1}^{\left( 2\right) }>0\right) \right] \left( 1 + o(1)\right) . \end{aligned}$$

Proof

Since

$$\begin{aligned} G_{b}\left( L_{1}\right) =G_{b}\left( l\right) +\int _{0}^{1}G_{b}^{\prime }\left( l+u\varDelta \left( 1\right) \right) \varDelta \left( 1\right) du, \end{aligned}$$

we introduce a uniform random variable U, independent of everything else, to write

$$\begin{aligned}&{\mathbb {E}}_\mathbf{w}^{\theta }\left[ r_{\theta }\left( \mathbf{w},\mathbf{W}_{1}\right) \frac{G_{b}\left( L \right) }{G_b(l)} I\left( X_1 \le a\left( b-l\right) \right) \right] \nonumber \\&\quad ={\mathbb {E}}_\mathbf{w}\left[ \frac{G_{b}\left( L \right) }{G_b(l)} I\left( X_1 \le a\left( b-l\right) ,W_{1}^{\left( 2\right) }>0 \right) \right] \\&\quad ={\mathbb {E}}_\mathbf{w}\left[ \left( 1 \!+\!\frac{G_{b}^{\prime }\left( l\!+\!U\varDelta \right) }{G_b(l)} \varDelta \right) I\left( X_1 \!\le \! a\left( b\!-\!l\right) ,W_{1}^{\left( 2\right) }\!>\!0\right) \right] \!\le \! {\mathbb {P}}_\mathbf{w}\left\{ W_{1}^{\left( 2\right) }\!>\!0\right\} \nonumber \\&\quad \qquad +\, \frac{{\mathbb {P}}\left\{ X \!>\! b-l \right\} }{h_b(l)}{\mathbb {E}}_\mathbf{w}\left[ \frac{{\mathbb {P}}\left\{ X > b-l-U\varDelta \right\} }{{\mathbb {P}}\left\{ X > b-l \right\} }\varDelta I\left( X_1\!\le \! a\left( b-l\right) ,W_{1}^{\left( 2\right) }\!>\!0\right) \right] . \nonumber \end{aligned}$$

(33)

We have used $G_{b}^{\prime }\left( l+U\varDelta \left( 1\right) \right) =\kappa _1 {\mathbb {P}}\left\{ X > b-l-U\varDelta \left( 1\right) \right\}$ to write the last step. Additionally, whenever $X_1\le a\left( b-l\right) ,$ observe that

$$\begin{aligned} \varDelta =\left( w_{1}+X_1 \right) ^{+}-w_{1}+\left( w_{2}-T_1 \right) ^{+}-w_{2}\le X_1^{+}\le a\left( b-l\right) , \end{aligned}$$

and therefore, ${\mathbb {P}}\left\{ X \!>\! b-l-U\varDelta \right\} \!\le \! {\mathbb {P}}\left\{ X > (1-a)(b-l) \right\} \!\le \! m_{1-a} {\mathbb {P}}\left\{ X \!>\! b - l \right\} ,$ where

$$\begin{aligned} m_t := \sup _{x > 0}\frac{{\mathbb {P}}\{ X > t x\}}{{\mathbb {P}}\{ X > x\}} < \infty , \end{aligned}$$

for every $t > 0.$ Here, the finiteness of $m_t$ follows from the regularly varying nature of the tail distribution of X (recall that ${\mathbb {P}}\{ X > x\} \sim \bar{B}(x)$ as $x \rightarrow \infty$). As a result, we have the following uniform bound for various values of b and l:

$$\begin{aligned} {\mathbb {E}}_\mathbf{w}\left[ \frac{{\mathbb {P}}\left\{ X > b-l-U\varDelta \right\} }{{\mathbb {P}}\left\{ X > b-l \right\} }\varDelta I\left( X_1\le a\left( b-l\right) ,W_{1}^{\left( 2\right) }>0\right) \right] \le m_{1-a} {\mathbb {E}}X^+. \end{aligned}$$

(34)

Consequently, due to the dominated convergence theorem, we obtain that

$$\begin{aligned}&{\mathbb {E}}_\mathbf{w}\left[ \frac{{\mathbb {P}}\left\{ X > b-l-U\varDelta \right\} }{{\mathbb {P}}\left\{ X > b-l \right\} }\varDelta I\left( X_1\le a\left( b-l\right) ,W_{1}^{\left( 2\right) }>0\right) \right] \\&\quad \sim {\mathbb {E}}_\mathbf{w} \left[ \varDelta I \left( W_1^{(2)} > 0 \right) \right] , \end{aligned}$$

as $(b-l) \rightarrow \infty .$ Now, the statement of Lemma 18 is immediate from (33) and the above stated convergence. $\square$

Proof

(Proof of Proposition 5) As mentioned before, we consider $G_{b}\left( l\right) <1.$ First, observe that

$$\begin{aligned} {\mathbb {E}}_\mathbf{w}^{\theta }\left[ r_{\theta } \left( \mathbf{w}, \mathbf{W}_{1}\right) \frac{G_{b}\left( L \right) }{G_b(l)}I\left( X_1>a\left( b-l\right) \right) \right]&= {\mathbb {E}}_\mathbf{w}\left[ \frac{G_{b}\left( L \right) }{G_b(l)}I\left( X_1 >a\left( b-l\right) \right) \right] \nonumber \\&\le \frac{{\mathbb {P}}\left\{ X>a\left( b-l\right) \right\} }{\kappa _{1} h_b(l)} \end{aligned}$$

(35)

because $G_b(\cdot ) \le 1.$ For a respective bound on the complementary event $\{ X_1 \le a(b-l) \},$ it is easy to see that our strategy must use Lemmas 17 and 18 in the following way: Given $\delta > 0,$ there exists a constant $C_\delta$ large enough such that for all initial conditions $\mathbf{w} = (w_1,w_2)$ satisfying $w_2 > C$ and $b-l > C_\delta ,$

$$\begin{aligned}&{\mathbb {E}}_\mathbf{w}^{\theta }\left[ r_{\theta }\left( \mathbf{w},\mathbf{W}_{1}\right) \frac{G_{b}\left( L \right) }{G_b(l)} I\left( X_1 \le a\left( b-l\right) \right) \right] \\&\quad \le {\mathbb {P}}_\mathbf{w}\left\{ W_{1}^{\left( 2\right) }>0\right\} - (1-\delta ) \mu \frac{{\mathbb {P}}\left\{ X > b-l \right\} }{h_b(l)}. \end{aligned}$$

Combining this bound with (35), we obtain

$$\begin{aligned} {\mathbb {E}}_\mathbf{w}^{\theta }\left[ r_{\theta }\left( \mathbf{w},\mathbf{W}_{1}\right) \frac{G_{b}\left( L \right) }{G_b(l)} \right]&\le {\mathbb {P}}_\mathbf{w}\left\{ W_{1}^{\left( 2\right) }>0\right\} + \frac{{\mathbb {P}}\left\{ X > a(b-l) \right\} }{h_b(l)} \left( \frac{1}{\kappa _1} - \frac{(1-\delta ) \mu }{m_a}\right) \\&\le 1 + \kappa _2 p\left( \mathbf{w} \right) \left( \frac{1}{\kappa _1} - \frac{(1-\delta ) \mu }{m_a}\right) \end{aligned}$$

which is, in turn, smaller than 1 for $\kappa _1$ suitably large. In addition to this, in the region $\{(w_1,w_2): w_2 > C, b - l < C_\delta \},$ we simply let $G_b(l) = 1$ by again choosing $\kappa _1$ large enough. This flexibility in the choice of $\kappa _1$ yields

$$\begin{aligned} {\mathbb {E}}_\mathbf{w}^{\theta }\left[ r_{\theta }\left( \mathbf{w},\mathbf{W}_{1}\right) \frac{G_{b}\left( L \right) }{G_b(l)} \right] \le 1 \end{aligned}$$

(36)

for initial conditions $\mathbf{w} = (w_1,w_2)$ satisfying $w_2 > C.$ Now, turning our attention to the values of $\mathbf{w}$ such that $w_2 \le C,$ we see that $l = w_1+ w_2 \le 2C,$ and as a consequence of the regularly varying nature of the tail of X,  we obtain

$$\begin{aligned} \frac{{\mathbb {P}}\left\{ X > b-l\right\} }{h_b(l)} = \left( \int _{0}^{l+\kappa _0} \frac{{\mathbb {P}}\left\{ X > b-l + u\right\} }{{\mathbb {P}}\left\{ X > b-l\right\} } du\right) ^{-1} = \frac{1+o(1)}{l+\kappa _0} \end{aligned}$$

as $b \rightarrow \infty .$ Then, it is immediate from (33) that whenever $w_2 \le C,$

$$\begin{aligned}&{\mathbb {E}}_\mathbf{w}^{\theta }\left[ r_{\theta }\left( \mathbf{w},\mathbf{W}_{1}\right) \frac{G_{b}\left( L \right) }{G_b(l)} I\left( X_1 \le a\left( b-l\right) \right) \right] \le {\mathbb {P}}_{\left( C,C\right) } \left\{ W_1^{(2)} > 0 \right\} \\&\quad + \frac{m_{1-a} {\mathbb {E}}\left[ X_1^+\right] }{\kappa _0} \left( 1+o(1) \right) . \end{aligned}$$

Combining this bound with the one obtained in (35), we get

$$\begin{aligned}&{\mathbb {E}}_\mathbf{w}^{\theta }\left[ r_{\theta }\left( \mathbf{w},\mathbf{W}_{1}\right) \frac{G_{b}\left( L \right) }{G_b(l)} \right] \le \frac{p\left( \mathbf{w}\right) \kappa _2}{\kappa _1} + {\mathbb {P}}_{\left( C,C\right) } \left\{ W_1^{(2)} > 0 \right\} \\&\qquad + \frac{m_{1-a} {\mathbb {E}}\left[ X_1^+\right] }{\kappa _0} \left( 1+o(1) \right) , \end{aligned}$$

which can also be made smaller than 1 by picking $\kappa _0$ and $\kappa _1$ large enough. Thus, for all initial conditions $\mathbf{w},$ we have a consistent choice of parameters $(\kappa _0,\kappa _1,\kappa _2)$ that satisfies (L1). $\square$

Since $G_b(l) = 1$ whenever $w_1+w_2 \ge b-C_\delta ,$ we also have $G_b(l) = 1$ if $w_2 > b.$ This verifies condition (L2). Since both (L1) and (L2) are satisfied, it follows from (32) that if $w_{2}<b\delta _{+}$ for some $\delta _{+}<1/2,$ then

$$\begin{aligned} {\mathbb {P}}_\mathbf{w}\left\{ \tau _{b}^{\left( 2\right) }<\tau _{0}\right\}&\le \kappa _{1}h_{b}\left( l\right) =\kappa _{1}\int _{b-l}^{b+\kappa _{0}}{\mathbb {P}} \left\{ X>u\right\} du \\&\le \kappa _{1}{\mathbb {P}}\left\{ X > b-l \right\} \left( \kappa _{0}+l\right) \le \kappa _{1}\bar{B}\left( b\left( 1-2\delta _{+}\right) \right) \\&\quad \left( \kappa _{0}+2w_{2}\right) \left( 1+o(1)\right) . \end{aligned}$$

The right-hand side of the previous inequality is equivalent to the statement of Lemma 16, so we conclude the proof.

Appendix 2: Proofs for other estimates

Proof

(Proof of Lemma 3) First, observe that

$$\begin{aligned} {\mathbb {E}}\left[ (w_1 + V - T)^+ - w_1\right]= & {} {\mathbb {E}}\left[ V - T \right] - {\mathbb {E}}\left[ (V-T)I\left( w_1 + V - T < 0 \right) \right] \\&- w_1 {\mathbb {P}}\left\{ w_1 + V - T < 0\right\} \\= & {} - {\mathbb {E}}\left[ (V-T)I\left( w_1 + V - T < 0 \right) \right] \\&- w_1 {\mathbb {P}}\left\{ w_1 + V - T < 0\right\} , \text { and }\\ {\mathbb {E}}\left[ (w_2- T)^+ - w_2\right]= & {} -{\mathbb {E}}T + {\mathbb {E}}\left[ T I\left( w_2 -T < 0\right) \right] - w_2 {\mathbb {P}}\left\{ w_2 -T < 0 \right\} . \end{aligned}$$

Then, it follows from the definition of $\mathbf{W}_{1}$ in recursions (5a) and (5b) that

$$\begin{aligned}&{\mathbb {E}}_{(w_1,w_2)}\left[ \left( W_1^{(1)} + W_1^{(2)}\right) - (w_1 + w_2) \right] \\&\quad = {\mathbb {E}}\left[ (w_1 + V - T)^+ - w_1\right] + {\mathbb {E}}\left[ (w_2 + T)^+ - w_2 \right] \\&\quad = - {\mathbb {E}}\left[ V I\left( w_1 + V - T < 0 \right) \right] - w_1 {\mathbb {P}}\left\{ w_1 + V - T < 0\right\} \\&\quad \quad - {\mathbb {E}}\left[ T I\left( w_1 + V - T \ge 0 \right) \right] \\&\quad \quad - w_2 {\mathbb {P}}\{ w_2 - T < 0\} + {\mathbb {E}}\left[ T I\left( w_2 - T < 0 \right) \right] \end{aligned}$$

which is negative if ${\mathbb {E}}[TI(T > w_2)]$ is small enough, and this can be achieved by choosing $C < w_2$ large enough. This completes the proof. $\square$

Proof

(Proof of Lemma 5) From recursions (5a) and (5b), it is evident that for every $1 \le k \le n,$

$$\begin{aligned} W_k^{(i)} \le \left( W_{k-1}^{(i)} + X_{k}\right) ^+, \quad i=1,2. \end{aligned}$$

We repeatedly expand the recursion, as below, to obtain

$$\begin{aligned} W_k^{(i)}&\le \max \left\{ 0, \ W_{k-1}^{(i)} + X_k \right\} \\&\le \max \left\{ 0, \ X_k, \ W_{k-2}^{(i)} + X_{k-1} + X_{k} \right\} \\&\le \! \max \left\{ 0, \ X_k, \ X_{k-1} \!+\! X_k, \ X_{k-2} \!+\! X_{k-1} \!+\! X_k, \ldots , \ W_{0}^{(i)}\right. \\&\quad \left. + X_1+ \ldots + X_{k-1} + X_{k} \right\} \\&\le S_k - \min _{0 \le j \le k}S_j + w_i, \end{aligned}$$

where we have used that $S_0:=0, S_j := X_1+\ldots +X_j$ and $w_i \ge 0.$ Then

$$\begin{aligned} \max _{0 < k \le n} W_k^{(i)} \le \max _{0 \le k \le n}S_k + \max _{0 < k \le n} \max _{0 \le j \le k} \left( -S_j \right) + w_i \le 2\max _{0 \le k \le n} \left| S_k \right| + w_i, \end{aligned}$$

and this proves the result. $\square$

Proof

(Proof of Lemma 7) According to [20, Corollary 1], we have that

$$\begin{aligned} {\mathbb {P}}\left\{ \max _{0\le n\le m}S_{n}>x\right\} =\left( {\mathbb {P}}\left\{ \max _{0\le t\le 1}\sigma B\left( t\right) >\frac{x}{m^{1/2}}\right\} +m {\mathbb {P}}\left\{ X > x\right\} \right) \left( 1+o\left( 1\right) \right) . \end{aligned}$$

(37)

uniformly over $y\ge m^{1/2},$ as $m \rightarrow \infty$ (actually, [20] states that the asymptotic is valid assuming $x/m^{1/2}\rightarrow \infty$ but the case $x/m^{1/2}=O\left( 1\right)$ follows from the Central Limit Theorem). Also, from the development in [20], because ${\mathbb {P}}\{ T>x\} =o\left( \bar{B}\left( x\right) \right) ,$ for each ${\varepsilon }> 0,$ there is a positive integer $m_{{\varepsilon }}$ such that for all $m > m_{{\varepsilon }},$

$$\begin{aligned} {\mathbb {P}}\left\{ \max _{0\le n\le m}\left( -S_{n}\right) > x\right\} \le (1+{\varepsilon })\left( {\mathbb {P}}\left\{ \max _{0\le t\le 1}\sigma B\left( t\right) >\frac{x}{m^{1/2}}\right\} +m{\mathbb {P}}\left\{ -X>x\right\} \right) . \end{aligned}$$

(38)

Additionally, since

$$\begin{aligned} {\mathbb {P}}\left\{ \max _{0\le n\le m}\left| S_{n}\right| >x\right\} \le {\mathbb {P}}\left\{ \max _{0\le n\le m}S_{n}>x\right\} +{\mathbb {P}}\left\{ \max _{0\le n\le m}\left( -S_{n}\right) >x\right\} \end{aligned}$$

the statement of Lemma 7 immediately follows from (37) and (38). $\square$

Proof of Lemma 9

Let

$$\begin{aligned} I_1(b)&:= {\mathbb {E}}\left[ I\left( \max _{0\le n\le N_A(X) + 1} 2\left| S_{n}\right| >\left( \delta -\delta _{-}\right) b, N_A(X) + 1 \le 2X\right) \right. \\&\quad \qquad \left. \max _{0\le n\le N_A(X) + 1} \left| S_{n}\right| \ \Big | X > b\delta _+ \right] ,\\ I_2(b)&:= {\mathbb {E}}\left[ I\left( \max _{0\le n\le N_A(X) + 1} 2\left| S_{n}\right| >\left( \delta -\delta _{-}\right) b, N_A(X) + 1 > 2X\right) \right. \\&\qquad \quad \left. \max _{0\le n\le N_A(X) + 1} \left| S_{n}\right| \ \Big | X > b\delta _+ \right] . \end{aligned}$$

Then our objective is to show that $I_1(b) + I_2(b) = O(b^2\bar{B}(b)).$ This is an immediate consequence of the following two results.

Lemma 19

Under Assumption 1 with $\alpha > 2,$ and Assumption 2,

$$\begin{aligned} I_1(b) = O \left( b^2\bar{B}(b)\right) . \end{aligned}$$

Lemma 20

Under Assumption 1 with $\alpha > 2,$ and Assumption 2,

$$\begin{aligned} I_2(b) = O \left( \exp \left( -\nu b\right) \right) , \end{aligned}$$

for a suitable $\nu > 0.$

Proof

(Proof of Lemma 19) First, observe that

$$\begin{aligned} I_1(b)&\le \int _{b\delta _+}^\infty {\mathbb {E}}\left[ I \left( \max _{0 \le n \le 2t} \left| S_n \right| > \frac{\delta - \delta _-}{2}b\right) \max _{0 \le n \le 2t} \left| S_n \right| \right] \frac{{\mathbb {P}}\left\{ X \in dt \right\} }{{\mathbb {P}}\left\{ X > b\delta _+\right\} }. \end{aligned}$$

Additionally, letting $c = (\delta - \delta _-)/2,$ observe that

$$\begin{aligned}&{\mathbb {E}}\left[ I \left( \max _{0 \le n \le 2t} \left| S_n \right| > cb\right) \max _{0 \le n \le 2t} \left| S_n \right| \right] \\&\quad = cb {\mathbb {P}}\left\{ \max _{0 \le n \le 2t} \left| S_n \right| > cb\right\} \\&\qquad + \int _{cb}^\infty {\mathbb {P}}\left\{ \max _{0 \le n \le 2t} \left| S_n \right| > u \right\} du. \end{aligned}$$

Therefore, due to (4),

$$\begin{aligned} I_1(b)&= O \left( \frac{1}{\bar{B}(b\delta _+)} \int _{b\delta _+}^\infty \left( b {\mathbb {P}}\left\{ \max _{0 \le n \le 2t} \left| S_n \right| > cb\right\} \right. \right. \nonumber \\&\left. \left. \qquad + \int _{cb}^\infty {\mathbb {P}}\left\{ \max _{0 \le n \le 2t} \left| S_n \right| > u \right\} du \right) {\mathbb {P}}\{ X \in dt\} \right) . \end{aligned}$$

(39)

Due to the applicability of the uniform asymptotic presented in Lemma 7 in the region $2t \le c^2b^2,$ and because of the applicability of Central Limit Theorem in the region $2t > c^2b^2,$ we obtain

$$\begin{aligned}&\int _{b\delta _+}^\infty {\mathbb {P}}\left\{ \max _{0 \le n \le 2t} \left| S_n \right| > cb\right\} {\mathbb {P}}\{ X \in dt \} \nonumber \\&\quad = O\left( \int _{b \delta _+}^\infty {\mathbb {P}}\left\{ \max _{0 \le s \le 1}\sigma \left| B(s) \right| > \frac{cb}{\sqrt{2t}}\right\} {\mathbb {P}}\{ X \in dt\}\right. \\&\qquad \left. + \int _{b\delta _+}^{\frac{c^2b^2}{2}} t {\mathbb {P}}\left\{ \left| X \right| > cb\right\} {\mathbb {P}}\{X \in dt\}\right) \\&\quad = O\left( b \int _{b\delta _+}^\infty \frac{{\mathbb {P}}\left\{ X > t\right\} }{\sqrt{t^3}} \exp \left( -\frac{c^2 b^2}{4\sigma ^2t}\right) dt + {\mathbb {P}}\{ \vert X \vert > cb \}\right. \\&\left. \quad {\mathbb {E}}\left[ X I \left( X \in \left[ b\delta _+, \frac{c^2b^2}{2}\right] \right) \right] \right) \end{aligned}$$

due to integration by parts. Now, one can apply Lemma 10 to evaluate the first integration, and Karamata’s theorem for the second integration, together with the observation that ${\mathbb {P}}\{ \vert X \vert > x\} = O(\bar{B}(x)),$ to obtain

$$\begin{aligned} \int _{b\delta _+}^\infty {\mathbb {P}}\left\{ \max _{0 \le n \le 2t} \left| S_n \right| > cb\right\} {\mathbb {P}}\{ X \in dt \} = O\left( \bar{B}\left( b^2 \right) + \bar{B}(b)\times b\bar{B}(b)\right) . \end{aligned}$$

(40)

On similar lines of reasoning using Lemma 7, again via careful integration by parts and subsequent application of Lemma 10 and Karamata’s theorem, one can derive

$$\begin{aligned}&\int _{b\delta _+}^\infty \int _{cb}^\infty {\mathbb {P}}\left\{ \max _{0 \le n \le 2t} \left| S_n \right| > u \right\} du {\mathbb {P}}\{ X \in dt\}\\&\quad = O\left( \int _{b \delta _+}^\infty \int _{cb}^\infty {\mathbb {P}}\left\{ \max _{0 \le s \le 1}\sigma \left| B(s) \right| > \frac{u}{\sqrt{2t}}\right\} du \ {\mathbb {P}}\{ X \in dt\}\right. \\&\left. \qquad + \int _{b\delta _+}^\infty \int _{\sqrt{2t}}^\infty t {\mathbb {P}}\left\{ \left| X \right| > u\right\} du \ {\mathbb {P}}\{X \in dt\} \right) \\&\quad = O\left( \int _{b\delta _+}^\infty \frac{{\mathbb {P}}\{X > t\}}{\sqrt{t}} \exp \left( -\frac{c^2b^2}{4\sigma ^2 t}\right) dt + \int _{\sqrt{2b\delta _+}}^{\infty }\int _{b\delta _+}^{\frac{u^2}{2}} t{\mathbb {P}}\{ X \in dt\} {\mathbb {P}}\{ \vert X \vert > u\}du\right) \\&\quad = O\left( b\bar{B} \left( b^2 \right) + b\bar{B}(b) \times b\bar{B}(b)\right) . \end{aligned}$$

This bound, along with (39), (40) and the observation that $\bar{B}(b\delta _+) = \varTheta (\bar{B}(b)),$ prove Lemma 19. $\square$

Proof

(Proof of Lemma 20) Since $T_1 + \cdots + T_{N_A(t)} \le t$ (which follows from the definition of $N_A(t)$) and $V_0 := 0,$

$$\begin{aligned}&{\mathbb {E}}\left[ I\left( N_A(t) + 1 > 2t\right) \max _{0 \le n \le N_A(t) + 1} \left| S_n \right| \right] \\&\quad \le {\mathbb {E}}\left[ I\left( N_A(t) > 2t -1 \right) \sum _{n=1}^{N_A(t) + 1} \left( V_{n-1} + T_{n} \right) \right] \\&\quad \le {\mathbb {E}}\left[ I\left( N_A(t) > 2t - 1 \right) \left( \sum _{n=1}^{N_A(t)} V_{n} + t + T_{N_A(t) + 1} \right) \right] \\&\quad \le {\mathbb {E}}V \times {\mathbb {E}}\left[ N_A(t) I \left( N_A(t) > 2t - 1\right) \right] + \left( t + {\mathbb {E}}T \right) {\mathbb {P}}\left\{ N_A(t) > 2t - 1\right\} \\&\quad \le C_1 t {\mathbb {P}}\left\{ N_A(t) > 2t-1 \right\} + C_2 \int _{2t-1}^\infty {\mathbb {P}}\left\{ N_A(t) > s\right\} ds \end{aligned}$$

for suitable positive constants $C_1$ and $C_2$ independent of t. Here, note that the penultimate inequality is simply due to the independence between $V_n$ and $T_n$ for $n \ge 1.$ Therefore,

$$\begin{aligned} I_2(b)\le & {} \int _{b\delta _+}^\infty {\mathbb {E}}\left[ I\left( N_A(t) + 1 > 2t\right) \max _{0 \le n \le N_A(t) + 1} \left| S_n \right| \right] \frac{{\mathbb {P}}\left\{ X \in dt \right\} }{{\mathbb {P}}\left\{ X > b \delta _+ \right\} } \nonumber \\\le & {} C_1\int _{b\delta _+}^\infty t {\mathbb {P}}\left\{ N_A(t) > 2t-1 \right\} \frac{{\mathbb {P}}\left\{ X \in dt \right\} }{{\mathbb {P}}\left\{ X > b \delta _+ \right\} } \\&+\, C_2 \int _{b\delta _+}^\infty \int _{2t-1}^\infty {\mathbb {P}}\left\{ N_A(t) > s \right\} ds \frac{{\mathbb {P}}\left\{ X \in dt \right\} }{{\mathbb {P}}\left\{ X > b \delta _+ \right\} } \nonumber \\\le & {} {\mathbb {P}}\left\{ N_A \left( b\delta _+ \right) > 2b \delta _+-1 \right\} \left( C_1\int _{b\delta _+}^\infty t \frac{{\mathbb {P}}\left\{ X \in dt \right\} }{{\mathbb {P}}\left\{ X > b \delta _+ \right\} }\right. \nonumber \\&\quad \left. +\, C_2 \int _{2b\delta _+-1}^\infty \frac{{\mathbb {P}}\left\{ X > \frac{s}{2} \right\} }{{\mathbb {P}}\left\{ X > b \delta _+ \right\} } ds \right) ,\nonumber \end{aligned}$$

(41)

where we have used a simple change of order of integration to arrive at the above conclusion. Since $N_A(x)/x \rightarrow 1$ as $x \rightarrow \infty ,$ the event $\{ N_A(b\delta _+) > 2b\delta _+-1\}$ is a large deviations event with probability exponentially decaying in b,  whereas the sum appearing in the parenthesis in (41) is O(b) due to Karamata’s theorem. This proves the claim that $I_2(b) = O(\exp (-\nu b))$ for a suitable constant $\nu > 0.$ $\square$

As mentioned earlier, Lemmas 19 and 20 together complete the proof of Lemma 9.

Proof

(Proof of Lemma 10) Due to Potter’s bounds (25), given ${\varepsilon }> 0,$ we have

$$\begin{aligned} \int _{b}^\infty \frac{v(t)}{v \left( b^2 \right) } \exp \left( -\frac{cb^2}{t} \right) dt = O \left( \int _b^\infty \left( \frac{b^2}{t}\right) ^{\alpha - {\varepsilon }} \exp \left( -\frac{cb^2}{t} \right) dt \right) \end{aligned}$$

for all suitably large values of b. Changing variables $u = b^2/t,$ we obtain

$$\begin{aligned} \int _{b}^\infty \frac{v(t)}{v \left( b^2 \right) } \exp \left( -\frac{cb^2}{t} \right) dt = O \left( b^2 \int _0^\infty u^{\alpha -2-\epsilon } \exp (-cu) du \right) = O\left( b^2 \right) \end{aligned}$$

for all $\epsilon$ small enough such that $\alpha -2 - \epsilon > 0$, and this verifies the claim. $\square$

Proof

(Proof of Lemma 14) Letting $\bar{c} = (\delta -\delta _-)/2,$ observe that

$$\begin{aligned}&{\mathbb {E}}\left[ I \left( \max _{0 \le n \le 2X} \left| S_n \right| > \bar{c} b \right) \max _{0 \le n \le 2X} \left| S_n \right| \ \Big | X > b\delta _+ \right] \\&\quad = \bar{c}b {\mathbb {P}}\left\{ \max _{0 \le n \le 2X} \left| S_n \right| > \bar{c} b \ \Big \vert X > b\delta _+ \right\} + \int _{\bar{c}b}^\infty {\mathbb {P}}\left\{ \max _{0 \le n \le 2X} \left| S_n \right| > u \ \Big \vert X > b\delta _+ \right\} du \nonumber \\&\quad \le 3\bar{c}b {\mathbb {P}}\left\{ Z_* > \frac{\bar{c}b}{\left( 2cX\right) ^{\frac{1}{\alpha }}} \ \Big \vert X > b\delta _+\right\} + 3\int _{\bar{c}b}^\infty {\mathbb {P}}\left\{ Z_* > \frac{u}{\left( 2cX\right) ^{\frac{1}{\alpha }}} \ \Big \vert X > b\delta _+ \right\} du \nonumber \end{aligned}$$

(42)

because of Lemma 12. Since ${\mathbb {P}}\{ X > x\} \sim cx^{-\alpha }$ as $x \rightarrow \infty ,$ after simple integration using Karamata’s theorem (24), one can show that

$$\begin{aligned}&\bar{c}b {\mathbb {P}}\left\{ Z_* > \frac{\bar{c}b}{\left( 2cX\right) ^{\frac{1}{\alpha }}} \ \Big \vert X > b\delta _+\right\} = \bar{c}b {\mathbb {E}}\left[ \frac{{\mathbb {P}}\left\{ X > \frac{1}{2c} \left( \frac{\bar{c} b}{Z_*} \right) ^{\alpha }\right\} }{{\mathbb {P}}\left\{ X > b\delta _+\right\} } \wedge 1\right] \\&\quad = O\left( b^2\bar{B} (b) \right) , \text { and }\\&\int _{\bar{c}b}^\infty {\mathbb {P}}\left\{ Z_* > \frac{u}{\left( 2cX\right) ^{\frac{1}{\alpha }}} \ \Big \vert X > b\delta _+ \right\} du = \int _{\bar{c} b}^\infty \left( \frac{{\mathbb {P}}\left\{ X > \frac{u^\alpha }{2c Z_*^\alpha } \right\} }{{\mathbb {P}}\left\{ X > b\delta _+\right\} } \wedge 1\right) du \\&\quad = O \left( b^2\bar{B}(b) \right) . \end{aligned}$$

Therefore, due to (42), we obtain

$$\begin{aligned} {\mathbb {E}}\left[ I \left( \max _{0 \!\le \! n \le 2X} \left| S_n \right| > \bar{c} b , \ N_A(X) \!+\! 1 \le 2X \right) \max _{0 \!\le \!n \!\le \! 2X} \left| S_n \right| \ \Big | X \!>\! b\delta _+ \right] \!= \!O\left( b^2 \bar{B}(b)\right) . \end{aligned}$$

On the other hand, the component corresponding to the large deviations event $\{ N_A(X) + 1 \ge 2X \}$ is handled similarly to Lemma 20, and this upper bounding procedure results in

$$\begin{aligned} {\mathbb {E}}\left[ I \left( \max _{0 \!\le \! n \le 2X} \left| S_n \right| \!>\! \bar{c} b , \ N_A(X) \!+\! 1 \!>\! 2X \right) \max _{0 \!\le \! n \!\le \! 2X} \left| S_n \right| \ \Big | X \!>\! b\delta _+ \right] = O\left( \exp (-\nu b) \right) , \end{aligned}$$

for some $\nu > 0.$ The last two upper bounds are enough to conclude the statement of Lemma 14. $\square$