2. Proof of Proposition 1

Before giving the proof we collect some auxiliary results that we will need in this and the following sections.

We will require the following statement, the first part of which follows from Theorem 2 in [Reference Foss, Palmowski and Zachary12] (see also [Reference Denisov4] for a short proof), and the second part from [7, Theorem 3.2].

Proposition 2. Let $\mathbb{E}[X_1]=-a$ and either (a) $\overline{F}(x)\coloneqq \mathbb{P}(X_1>x)=x^{-\alpha}L(x)$ with some $\alpha>1$ or (b) $\overline{F}(x)\sim x^{-\varkappa} {\rm e}^{-g(x)}$ , where g(x) is a monotone continuously differentiable function satisfying $\frac{g(x)}{x^\beta}\downarrow$ for $\beta\in(0,1/2)$ , and $\mathbb{E}|X_1|^\varkappa<\infty$ for some $\varkappa>1/(1-\beta)$ ; then, for any fixed k,

(13) \begin{align}\mathbb{P}(M_k>y)&\sim \mathbb{P}(S_k>y)\sim k\overline F(y), \quad\quad y\to \infty , \end{align}

(14) \begin{align}\mathbb{P}\left(\max_{n\le\tau\wedge k}S_n>y\right)&\sim \mathbb{E}(\tau\wedge k)\overline{F}(y),\quad y\to \infty , \end{align}

(15) \begin{align}\hspace*{-2pt}\mathbb{P}(M_\tau \gt y)&\sim\mathbb{E}\tau\bar{F}(y),\qquad\qquad\qquad y\to\infty\end{align}

and

(16) \begin{equation}\mathbb{P}(\tau \gt n)\sim \mathbb{E}[\tau]\overline F(an), \qquad\qquad\quad\,\, n\to \infty.\end{equation}

In the proof we will need some properties of the function $\overline F(x)\sim x^{-\varkappa} {\rm e}^{-g(x)}$ that we will summarise in the following lemma, which will also be used later in the paper.

Lemma 1. Let the distribution function $\overline F(x)$ be such that $\overline{F}(x)\sim x^{-\varkappa} {\rm e}^{-g(x)}$ , where g(x) is a monotone continuously differentiable function satisfying $\frac{g(x)}{x^\beta}\downarrow$ for $\beta\in(0,1)$ , and $\mathbb{E}|X_1|^\varkappa<\infty$ for some $\varkappa>1/(1-\beta)$ . Then,

(17) \begin{align}g'(x) & \le \beta \frac{g(x)}{x},\qquad\qquad x>0 , \end{align}

(18) \begin{align}g(x) - g(y) & \le \beta g(y)\frac{x-y}{y},\quad x>y>0 , \end{align}

(19) \begin{align}g(x) - g(x-y) & \le \beta g(y),\qquad x\ge 2y>0 , \end{align}

(20) \begin{align}\sup_{y\le x^{1/\varkappa}}\frac{\overline F(x-y)}{\overline F(x)} & \to 1,\qquad x\to \infty . \end{align}

Proof. Since g(x) is continuously differentiable and $\frac{g(x)}{x^\beta}$ is monotone decreasing then, with necessity,

\begin{equation*}0\ge \left(\frac{g(x)}{x^\beta}\right)'=\frac{g'(x)x^\beta-\beta x^{\beta-1}g(x)}{x^{2\beta}} ,\end{equation*}

implying (17). To prove (18), note that

\begin{equation*}g(x)-g(y) =\int_y^x g'(t) \, {\rm d} t \le \beta\int_y^x \frac{g(t)}{t} \, {\rm d} t\le \beta \frac{g(y)}{y^\beta}\int_y^x \frac{1}{t^{1-\beta}} \, {\rm d} t= \beta g(y)\frac{x-y}{y}.\end{equation*}

To prove (18), note that, since $x-y\ge y$ ,

\begin{align*}g(x)-g(x-y) &=\int_{x-y}^x g'(t) \, {\rm d} t \le \beta\int_{x-y}^x \frac{g(t)}{t} \, {\rm d} t\le \beta \frac{g(y)}{y^\beta}\int_{x-y}^x \frac{1}{t^{1-\beta}} \, {\rm d} t\\[3pt] &\le \beta \frac{g(y)}{y^\beta}\int_{y}^{2y} \frac{1}{t^{1-\beta}} \, {\rm d} t\le \beta g(y)\frac{(2y)^\beta-y^\beta}{\beta y^\beta}\le \beta g(y),\end{align*}

since $2^\beta\le 1+\beta$ for $\beta\in[0,1]$ . To show (20), note that, uniformly in $y\le x^{1/\varkappa}$ , as $x\to\infty$ ,

\begin{align*}1&\le \frac{\overline F(x-y)}{\overline F(x)}\le \frac{\overline F(x-x^{1/\varkappa})}{\overline F(x)}=(1+o(1)) \exp\left\{g(x)-g(x-x^{1/\varkappa})\right\} \\[3pt]&\le (1+o(1))\exp\left\{\beta g(x-x^{1/\varkappa})\frac{x^\varkappa}{x-x^\varkappa}\right\}\le (1+o(1))\exp\left\{C\frac{x^{1/\varkappa}}{(x-x^{1/\varkappa})^{1-\beta}}\right\}\to 1,\end{align*}

since $1/\varkappa<1-\beta$ . Here, we have also made use of (18).

Proof of Proposition 2. To prove (13), (14), and (15), by Theorem 2 of [Reference Foss, Palmowski and Zachary12] it is sufficient to show that (a) or (b) implies that $F\in\mathcal S^*$ , that is, $\int_0^\infty \overline F(y)<\infty$ and

\begin{equation*}\int_0^x\overline F(y)\overline F(x-y) \, {\rm d} y \sim 2\overline F(x) \int_0^\infty \overline F(y) \, {\rm d} y , \qquad x\to \infty.\end{equation*}

The fact that (a) implies $F\in\mathcal S^*$ is well known and follows immediately from the dominated convergence theorem, since $ \overline F(x)\sim \overline F(x-y)$ for all fixed y and

\begin{equation*}\int_0^x \frac{\overline F(y)\overline F(x-y)}{\overline F(x) } \, {\rm d} y=2\int_0^{x/2} \frac{\overline F(y)\overline F(x-y)}{\overline F(x) } \, {\rm d} y ,\end{equation*}

and $\overline F(x-y)\le C\overline F(x)$ for some $C>0$ when $y\le x/2$ . Now, assume that (b) holds and show that $F\in\mathcal S^*$ . Consider

\begin{align*}2\int_0^{x/2} \frac{\overline F(y)\overline F(x-y)}{\overline F(x) } \, {\rm d} y.\end{align*}

Uniformly in $y\in [\ln x ,x/2]$ we have, by (19),

(21) \begin{equation}\frac{\overline F(y)\overline F(x-y)}{\overline F(x) }\le C \frac{x^\varkappa}{(x-y)^\varkappa y^\varkappa}{\rm e}^{g(x)-g(x-y)-g(y)}\le Cy^{-\varkappa} {\rm e}^{\beta g(y)-g(y)} ,\end{equation}

and therefore, since $\varkappa>1$ ,

\begin{align*}2\int_{\ln x }^{x/2} \frac{\overline F(y)\overline F(x-y)}{\overline F(x) } \, {\rm d} y \to 0.\end{align*}

Next, applying (20) we see that $\frac{\overline F(x-y)}{\overline F(x)} \to 1 $ uniformly in $y\in [0, \ln x]$ , which implies that $F\in\mathcal S^*$ .

The proof of (16) can be done by verification that (8) and (9) imply that the conditions of Theorem 3.1 (and hence of Theorem 3.2) of [Reference Denisov and Shneer7] hold. We will provide the arguments in the more complicated case (b). First, $X_1+a$ satisfies $\mathbb{E}[|X_1+a]^\varkappa<\infty$ by the assumption of this proposition. Convergence (3.1) in Theorem 3.1 of [Reference Denisov and Shneer7] holds by (20). Let

\begin{align*}\varepsilon(n)\coloneqq \sup_{x\ge 2n^{1/\varkappa}}\frac{\mathbb{P}(\xi_1>n^{1/\varkappa},\xi_2>n^{1/\varkappa}, \xi_1+\xi_2>x)}{\mathbb{P}(\xi_1>x)},\end{align*}

where $\xi_i=X_i+a$ , $i=1,2$ . To show (3.2) in Theorem 3.1 of [Reference Denisov and Shneer7] we need to prove that $\varepsilon(n)=o(1/n)$ . For $x\ge 2n^{1/\varkappa}$ we have

\begin{align*}\frac{\mathbb{P}(\xi_1>n^{1/\varkappa},\xi_2>n^{1/\varkappa}, \xi_1+\xi_2>x)}{\mathbb{P}(\xi_1>x)}&= 2P_1+P_2\\[3pt] &\coloneqq 2\int_{n^{1/\varkappa}}^{x/2}\mathbb{P}(\xi_1\in {\rm d} y)\frac{\mathbb{P}(\xi_2>x-y)}{\mathbb{P}(\xi_1>x)}+\frac{\mathbb{P}(\xi_1>x/2)^2}{\mathbb{P}(\xi_1>x)} .\end{align*}

Then, using (19),

\begin{align*}P_1&\le\int_{n^{1/\varkappa}}^{x/2}\mathbb{P}(\xi_1\in {\rm d} y)\frac{\mathbb{P}(\xi_2>x-y)}{\mathbb{P}(\xi_1>x)}\\[3pt]&\le C \int_{n^{1/\varkappa}}^{x/2}\mathbb{P}(\xi_1\in {\rm d} y) \frac{\overline F(x-y)}{\overline F(x)}\le C \int_{n^{1/\varkappa}}^{x/2}\mathbb{P}(\xi_1\in {\rm d} y) {\rm e}^{\beta g(y)} \end{align*}

for some C. Integrating by parts,

\begin{align*}P_1 & \le C\mathbb{P}(\xi_1>n^{1/\varkappa}){\rm e}^{\beta g(n^{1/\varkappa})}+C\int_{n^{1/\varkappa}}^{x/2} {\rm d} y g'(y) {\rm e}^{\beta g(y)}\mathbb{P}(\xi_1>y)\\[3pt]&\le C n^{-1}{\rm e}^{(\beta-1) g(n^{1/\varkappa})}+C \int_{n^{1/\varkappa}}^{x/2} {\rm d} y g'(y) y^{-\kappa} {\rm e}^{(\beta-1) g(y)}\\[3pt]&\le o(n^{-1})+cn^{-1}\int_{n^{1/\varkappa}}^{x/2} {\rm d} y g'(y) {\rm e}^{(\beta-1) g(y)}=o(n^{-1}) \end{align*}

uniformly in $x\ge 2n^{1/\varkappa}$ . Using (21), $P_2\le Cx^{-\varkappa} {\rm e}^{(\beta-1)g(x)}=o(1/n)$ uniformly in $x\ge 2n^{1/\varkappa}$ , which proves that $\varepsilon(n)=o(1/n)$ .

Define $\sigma_y=\inf\lbrace n<\tau\,{:}\,S_n>y\rbrace.$

Lemma 2. Under the conditions of Proposition 2,

\begin{equation*}\lim_{y\to\infty}\mathbb{P}(\sigma_y=k \mid M_\tau \gt y)\eqqcolon q_k,\quad k\ge1 ; \qquad\sum_{k=1}^\infty q_k=1.\end{equation*}

Proof. For every $k\geq 1$ ,

\begin{align*}\mathbb{P}(\sigma_y=k \mid M_\tau \gt y)&=\frac{\mathbb{P}(\sigma_y=k)}{\mathbb{P}(M_\tau \gt y)}\\[3pt]&=\frac{\mathbb{P}\left(\max_{n\le\tau\wedge k}S_n>y\right)-\mathbb{P}\left(\max_{n\le\tau\wedge(k-1)}S_n>y\right)}{\mathbb{P}(M_\tau \gt y)}.\end{align*}

It follows from (14) and (15) that

(22) \begin{align}\nonumber\lim_{y\rightarrow\infty}\mathbb{P}(\sigma_y=k \mid M_\tau \gt y)&=\frac{\mathbb{E}\tau\wedge k-\mathbb{E}\tau\wedge(k-1)}{\mathbb{E}\tau}\\[3pt]&=\frac{\mathbb{P}(\tau \gt k-1)}{\mathbb{E}\tau}\eqqcolon q_k, \qquad k\geq 1.\end{align}

It is clear that

\begin{align*}\sum_{k=1}^\infty q_k=\frac{1}{\mathbb{E}\tau}\sum_{k=0}^\infty\mathbb{P}(\tau \gt k-1)=1. \\[-40pt]\end{align*}

Lemma 3. For every fixed k,

\begin{align*}\sup_{v>y}\Bigg\vert\frac{\mathbb{P}(S_k>v,\sigma_y=k)}{\overline{F}(v)}-\mathbb{P}(\tau \gt k-1)\Bigg\vert\rightarrow 0 \text{ as } y\rightarrow\infty .\end{align*}

Proof. Fix some $N>0$ and define the events

\begin{equation*}D_{k,N}=\cup_{j=1}^k\left\lbrace X_j>v+kN, \vert X_l\vert \leq N \text{ for all } l\neq j, l\le k\right\rbrace.\end{equation*}

It is clear that $D_{k,N}\subseteq \lbrace S_k>v\rbrace.$ Therefore,

\begin{align*}\mathbb{P}(S_k>v,\sigma_y=k)&=\mathbb{P}(D_{k,N},\sigma_y=k)+\mathbb{P}(S_k>v, D^c_{k,N},\sigma_y=k)\\[3pt]&=\mathbb{P}(X_k>v+kN,\vert X_l\vert\leq N, \text{for all } l<k, \sigma_y>k-1)\\[3pt]& \quad + \mathbb{P}(S_k>v, D^c_{k,N},\sigma_y=k).\end{align*}

For the first term we have $(y>(k-1)N)$

(23) \begin{equation}\begin{split}\mathbb{P}(X_k>v+kN,& \vert X_l\vert\leq N, \text{ for all } l<k, \sigma_y>k-1)\\[3pt]&=\mathbb{P}(\tau \gt k-1,\vert X_l\vert\leq N, l<k)\overline{F}(v+kN)\\[3pt]&=\mathbb{P}(\tau \gt k-1)\overline{F}(v)-\varepsilon_N^{(1)}\overline{F}(v)+o(\overline{F}(v)) \end{split}\end{equation}

uniformly in $v > y$ , where $\varepsilon_N^{(1)}\coloneqq \mathbb{P}(\tau \gt k-1,\vert X_l\vert> N \text{ for some }l<k)\to0$ as $N \to \infty$ .

Furthermore,

(24) \begin{equation}\begin{split}\mathbb{P}(S_k>v, & D_{k,N}^c, \sigma_y=k)\leq\mathbb{P}(S_k>v, D_{k,N}^c)=\mathbb{P}(S_k>v)-\mathbb{P}(D_{k,N})\\[3pt]&=\mathbb{P}(S_k>v)-k\mathbb{P}(X_1>v+kN)(\mathbb{P}(\vert X_1\vert\leq N))^{k-1}\\[3pt]&=\varepsilon_N^{(2)}\overline{F}(v)+o(\overline{F}(v)),\end{split}\end{equation}

where $\varepsilon_N^{(2)}\coloneqq k\left(1-(\mathbb{P}(\vert X_1\vert\leq N))^{k-1}\right)\to0,$ as $N \to \infty$ . Combining (23) and (24) and letting $N\rightarrow\infty$ , we get the desired relation.

We now turn to the study of the tail behaviour of $A_\tau$ on the event $\{\sigma_y=k\}$ . For the corresponding result we need the following property of $\overline{F}$ .

Lemma 4. Assume that the conditions of Proposition 1 are fulfilled. Then

(25) \begin{equation}\lim_{y\to\infty}\frac{\overline{F}(y+h)}{\overline{F}(y)}=1\end{equation}

for any $h=o(y/g(y))$ . Furthermore, for every $R>0$ there exists a constant C such that

(26) \begin{equation}\overline{F}(y-h)\le C\overline{F}(y) \text{ for all } h\le R\frac{y}{g(y)}.\end{equation}

Proof. Since $\overline{F}(x)\sim x^{-\varkappa}{\rm e}^{-g(x)}$ and $\frac{y}{y+h}\to1$ , (25) will follow from

(27) \begin{equation}g(y+h)-g(y)\to0.\end{equation}

Since $\frac{g(x)}{x^\beta}$ is monotone decreasing and g is differentiable,

\begin{equation*}g'(x)\le \beta \frac{g(x)}{x}.\end{equation*}

Then, for $h<0$ ,

(28) \begin{align}g(y)-g(y-h)=\int^{y-h}_{y} g'(t) \, {\rm d} t\le \beta \int^{y-h}_{y} \frac{g(t)}{t} \, {\rm d} t\le \beta \frac{g(y-h)}{y-h}h.\end{align}

In the last step we have used the fact that $\frac{g(t)}{t}$ is decreasing. Similarly, for $h>0$ ,

\begin{equation*}g(y+h)-g(y)\le \beta \frac{g(y)}{y}h.\end{equation*}

These estimates yield (27).

To prove the second claim we note that, by (28),

\begin{align*}\frac{\overline{F}(y-h)}{\overline{F}(y)}\le C\left(\frac{y}{y-h}\right)^\varkappa {\rm e}^{g(y)-g(y-h)}\le C\exp\left\{\beta \frac{g(y-h)}{y-h}h\right\}.\end{align*}

If $h\le R\frac{y}{g(y)}$ then

\begin{align*}\frac{\overline{F}(y-h)}{\overline{F}(y)}\le C\exp\left\{\beta R \frac{yg(y-h)}{(y-h)g(y)}\right\}\le C\exp\left\{\beta R \frac{1}{(1-R/g(y))}\right\}.\end{align*}

This completes the proof.

Lemma 5. Assume that the conditions of Proposition 1 hold. Then

\begin{equation*}\mathbb{P}\left(\sum_{j=k}^{\tau-1}S_j>z,\sigma_y=k\right)\sim q_k\mathbb{E}\tau\overline{F}(\sqrt{2az}),\qquad k\ge1,\end{equation*}

uniformly in $y\in[\varepsilon\sqrt{z},\sqrt{2az}]$ for regularly varying tails $\overline{F}$ and in $y\in\big[\sqrt{2ax}-\frac{R\sqrt{2ax}}{g(\sqrt{2ax})},\sqrt{2ax}\big]$ for tails satisfying the conditions of part (b) in Proposition 1.

Proof. By the Markov property, for every $z>0$ ,

\begin{align*}\mathbb{P}\left(\sum_{j=k}^{\tau-1}S_j>z, \sigma_y=k\right)=\int_y^\infty\mathbb{P}(S_k\in {\rm d} v, \sigma_y=k)\mathbb{P}(A_\tau \gt z \mid S_0=v).\end{align*}

Let $\varkappa\in(1/(1-\beta),2)$ if $\overline{F}$ satisfies the conditions of part (b), and let $\varkappa=1$ in the case when $\overline{F}$ is regularly varying. Fix some $\delta>0$ and consider the set

\begin{equation*}B_v\coloneqq \left\{v-\delta v^{1/\varkappa}\le S_n+na\le v+\delta v^{1/\varkappa}\mbox{ for all }n\le\frac{v+\delta v^{1/\varkappa}}{a}\right\}.\end{equation*}

Since $\mathbb{E}|X_1|^\varkappa<\infty$ , it follows from the Marcinkiewicz–Zygmund law of large numbers that

\begin{equation*}\lim_{R\to\infty}\mathbb{P}\left(-R-\frac{\delta}{2} n^{1/\varkappa}<S_n+na<R+\frac{\delta}{2} n^{1/\varkappa}\text{ for all }n\ge1\right)=1.\end{equation*}

Consequently, $\mathbb{P}(B_v \mid S_0=v)\to 1$ as $v\to\infty$ . This implies that, as $y\rightarrow\infty$ ,

\begin{align*}\mathbb{P}\Bigg(\sum_{j=k}^{\tau-1}S_j>z & , \sigma_y=k\Bigg)\\[3pt]&=\int_y^\infty\mathbb{P}\left(S_k\in {\rm d} v,\sigma_y=k\right)\mathbb{P}(A_\tau \gt z \mid S_0=v)\\[3pt]&=\int_y^\infty\mathbb{P}\left(S_k\in {\rm d} v,\sigma_y=k\right)\mathbb{P}\left(\lbrace A_\tau \gt z\rbrace\cap B_v \mid S_0=v\right)+o\left(\mathbb{P}(\sigma_y=k)\right).\end{align*}

On the event $B_v\cap\{S_0=v\}$ one has

\begin{equation*}\frac{v-\delta v^{1/\varkappa}}{a}<\tau<\frac{v+\delta v^{1/\varkappa}}{a}.\end{equation*}

Consequently,

\begin{equation*}A_\tau=\sum_{n=0}^{\tau-1}S_n\ge \sum_{n=0}^{\tau-1}(v-\delta v^{1/\varkappa}-na)\ge \tau\left(v-\delta v^{1/\varkappa}-\frac{a\tau}{2}\right)\ge \frac{(v-\delta v^{1/\varkappa})^2}{2a}\end{equation*}

and

\begin{equation*}A_\tau=\sum_{n=0}^{\tau-1}S_n\le v+ \sum_{n=1}^{\tau-1}(v+\delta v^{1/\varkappa}-na)\le \tau\left(v+\delta v^{1/\varkappa}-\frac{a\tau}{2}\right)\le \frac{(v+\delta v^{1/\varkappa})^2}{2a}\end{equation*}

on the same event. In other words, $\mathbb{P}\left(\lbrace A_\tau \gt z\rbrace\cap B_v \mid S_0=v\right)=\mathbb{P}(B_v \mid S_0=v)$ if $v-\delta v^{1/\varkappa}\ge\sqrt{2az}$ , and $\mathbb{P}\left(\lbrace A_\tau \gt z\rbrace\cap B_v \mid S_0=v\right)=0$ if $v+\delta v^{1/\varkappa}<\sqrt{2az}$ . Therefore, for all v large enough,

\begin{align*}\mathbb{P}\left(\sum_{j=k}^{\tau-1}S_j>z, \sigma_y=k\right)&\leq\int_{\sqrt{2az}-\delta(2az)^{1/2\varkappa}}^\infty \mathbb{P}(S_k\in {\rm d} v, \sigma_y=k)+o(\mathbb{P}(\sigma_y=k))\\[3pt]&=\mathbb{P}\left(S_{\sigma_y}>\sqrt{2az}-\delta(2az)^{1/2\varkappa}, \sigma_y=k\right)+o(\mathbb{P}(\sigma_y=k))\end{align*}

and

\begin{align*}\mathbb{P}\left(\sum_{j=k}^{\tau-1}S_j>z, \sigma_y=k\right)&\geq\int_{\sqrt{2az}+2\delta(2az)^{1/2\varkappa}}^\infty \mathbb{P}(S_k\in {\rm d} v, \sigma_y=k)\mathbb{P}(B_v)+o(\mathbb{P}(\sigma_y=k))\\&=\mathbb{P}\left(S_{\sigma_y}>\sqrt{2az}+2\delta(2az)^{1/2\varkappa}, \sigma_y=k\right)+o(\mathbb{P}(\sigma_y=k)).\end{align*}

By Lemma 3, $\mathbb{P}(S_{\sigma_y}>v,\sigma_y=k)\sim\bar{F}(v)\mathbb{P}(\tau \gt k-1)$ as $y\rightarrow\infty$ uniformly in $v\geq y$ and, consequently,

\begin{align*}\mathbb{P}\left(\sum_{j=k}^{\tau-1}S_j>z, \sigma_y=k\right) & \leq \bar{F}\left((\sqrt{2az}-\delta(2az)^{1/2\varkappa})\vee y\right)(\mathbb{P}(\tau \gt k-1)+o(1))\\& \quad +o(\mathbb{P}(\sigma_y=k))\end{align*}

and

\begin{align*}\mathbb{P}\left(\sum_{j=k}^{\tau-1}S_j>z, \sigma_y=k\right) & \geq \overline{F}\left(\sqrt{2az}+2\delta(2az)^{1/2\varkappa}\right)(\mathbb{P}(\tau \gt k-1)+o(1))\\& \quad +o(\mathbb{P}(\sigma_y=k)).\end{align*}

Under our assumptions on $\overline{F}$ , we have

\begin{equation*}\lim_{\delta\to0}\lim_{z\to\infty}\frac{\overline{F}\left(\sqrt{2az}+2\delta(2az)^{1/2\varkappa}\right)}{\overline{F}\left((\sqrt{2az}-\delta(2az)^{1/2\varkappa})\vee y\right)}=1.\end{equation*}

Indeed, this relation is obvious for regularly varying tails, and under the conditions of part (b) it is a particular case of (25). Therefore,

\begin{align*}&\mathbb{P}\left(\sum_{j=k}^{\tau-1}S_j>z, \sigma_y=k\right)=\overline{F}\left(\sqrt{2az}\right)(\mathbb{P}(\tau \gt k-1)+o(1))+o(\mathbb{P}(\sigma_y=k)).\end{align*}

Combining (15) and (22), we get $\mathbb{P}(\sigma_y=k)\sim q_k\mathbb{E}\tau\overline{F}(y).$ Thus, it remains to show that $\overline{F}(y)=O(\overline{F}(\sqrt{2az}))$ . This is obvious for regularly varying tails and $y\ge \varepsilon\sqrt{z}$ . For distributions satisfying the conditions of part (b), it suffices to apply (26) with $y=\sqrt{2az}$ .

Proof of Proposition 1. For every fixed $N\geq1$ we have

(29) \begin{align}&\mathbb{P}(A_\tau > x, M_\tau \gt y) \nonumber\\[3pt] &\quad =\sum_{k=1}^N\mathbb{P}(A_\tau \gt x, \sigma_y=k, M_\tau \gt y)+\mathbb{P}(A_\tau \gt x, \sigma_y>N, M_\tau \gt y).\end{align}

For the last term on the right-hand side we have

\begin{align*}\mathbb{P}(A_\tau \gt x, \sigma_y>N, M_\tau \gt y)&\leq\mathbb{P}(\sigma_y>N, M_\tau \gt y)\\[3pt]&=\mathbb{P}(M_\tau \gt y)\mathbb{P}(\sigma_y>N \mid M_\tau \gt y).\end{align*}

It follows from (22) that $\mathbb{P}(\sigma_y>N \mid M_\tau \gt y)\rightarrow\sum_{j=N+1}^\infty q_j$ as $y\rightarrow\infty$ . Then, using (15), we get

(30) \begin{align}\mathbb{P}(A_\tau \gt x,\sigma_y>N, M_\tau \gt y)\leq \varepsilon_N\bar{F}(y),\end{align}

where $\varepsilon_N\rightarrow 0$ as $N\rightarrow\infty$ .

For every fixed k we have $\mathbb{P}(A_\tau \gt x, \sigma_y=k, M_\tau \gt y)=\mathbb{P}(A_\tau \gt x, \sigma_y=k)$ . Since $S_j\in(0,y)$ for all $j<k$ , we obtain

\begin{align*}\mathbb{P}(A_\tau \gt x, \sigma_y=k)\leq\mathbb{P}\left(\sum_{j=k}^{\tau-1}S_j>x-(k-1)y, \sigma_y=k\right)\end{align*}

and

\begin{align*}\mathbb{P}(A_\tau \gt x, \sigma_y=k)\geq\mathbb{P}\left(\sum_{j=k}^{\tau-1}S_j>x, \sigma_y=k\right).\end{align*}

Using Lemma 5 with $z=x$ and with $z=x-ky$ , we conclude that $\mathbb{P}(A_\tau \gt x,\sigma_y=k)\sim q_k\mathbb{E}\tau\overline{F}(\sqrt{2ax})$ . Consequently,

(31) \begin{equation}\sum_{k=1}^N\mathbb{P}(A_\tau \gt x, \sigma_y=k, M_\tau \gt y) \sim\overline{F}(\sqrt{2ax})\mathbb{E}\tau\sum_{k=1}^Nq_k.\end{equation}

Plugging (30) and (31) into (29) and letting $N\rightarrow\infty$ , we obtain

\begin{align*}\mathbb{P}(A_\tau \gt x, M_\tau \gt y)=(\mathbb{E}\tau+o(1))\overline{F}(\sqrt{2ax})+o(\overline{F}(y)).\end{align*}

Recalling that $\overline{F}(y)=O(\overline{F}(\sqrt{2ax}))$ , we finish the proof.

3. Proof of Theorem 1

We start by proving an exponential estimate for the area $A_n$ when random variables $X_j$ are truncated. Let $\overline X_n=\max(X_1,\ldots,X_n)$ . The next result is our main technical tool to investigate trajectories without big jumps.

Lemma 6. Let $\mathbb{E}[X_1]=-a$ and $\sigma^2\coloneqq {\rm Var}(X_1)<\infty$ . Assume that the distribution function F of $X_j$ satisfies (8) and that (9) holds with $\gamma_0=1$ . Then there exists a constant $C_0>0$ such that

\begin{equation*}\mathbb{P}(A_n>x,\overline X_n\le y) \le \exp\left\{-\lambda \frac{x}{n} - \lambda\frac{a n}{2} +C_0\lambda^2 n\right\},\end{equation*}

where $\lambda = \frac{g(y)}{y}$ .

Proof. We will prove this lemma by using the exponential Chebyshev inequality. For that, we need to obtain estimates for the moment-generating function of $A_n$ . First,

\begin{equation*}\mathbb{E}\left[ {\rm e}^{\frac{\lambda}{n}A_n};\ \overline X_n\le y\right]=\mathbb{E} \left[{\rm e}^{\frac{\lambda}{n}\sum_1^n(n-j+1)X_j};\ \overline X_n\le y\right]=\prod_{j=1}^n\varphi_y\left(\lambda_{n,j}\right),\end{equation*}

where $\varphi_y(t) \coloneqq \mathbb{E}[{\rm e}^{t X_j};\ X_j\le y]$ and $\lambda_{n,j} \coloneqq \lambda\frac{(n-j+1)}{n}$ . Then,

\begin{equation*}\varphi_y(\lambda_{n,j}) = \mathbb{E}[{\rm e}^{\lambda_{n,j}X_j};\ X_j\le 1/\lambda_{n,j}]+\mathbb{E}[{\rm e}^{\lambda_{n,j}X_j};\ 1/\lambda_{n,j} < X_j\le y] \eqqcolon E_1+E_2.\end{equation*}

Using the elementary bound ${\rm e}^x\le 1+x+x^2$ for $x\le 1$ , we obtain

\begin{align*}E_1\le 1+\lambda_{n,j}\mathbb{E}[X_j]+\lambda_{n,j}^2 \mathbb{E}[X_j^2]=1-a\lambda_{n,j}+(a^2+\sigma^2)\lambda_{n,j}^2.\end{align*}

Next, using integration by parts and the assumption in (8),

\begin{align*}E_2&=\int_{1/\lambda_{n,j}}^{y} {\rm e}^{\lambda_{n,j}t} \, {\rm d} F(t)=-\overline F(t){\rm e}^{\lambda_{n,j}t}\biggl |_{t=1/\lambda_{n,j}}^{t=y}+\lambda_{n,j}\int_{1/\lambda_{n,j}}^{y} {\rm e}^{\lambda_{n,j}t}\overline F(t) \, {\rm d} t \\[3pt]&\le {\rm e} \overline F(1/\lambda_{n,j})+C \lambda_{n,j}\int_{1/\lambda_{n,j}}^{y} {\rm e}^{\lambda_{n,j}t-g(t)}t^{-2} \, {\rm d} t.\end{align*}

Now note that, for $t\le y$ ,

\begin{equation*}\lambda_{n,j} t - g(t) = t \left(\lambda_{n,j}-\frac{g(t)}{t}\right)\le t \left(\lambda_{n,j}-\frac{g(y)}{y}\right),\end{equation*}

due to the condition in (9). Then,

\begin{equation*}\lambda_{n,j}-\frac{g(y)}{y} \le\lambda - \frac{g(y)}{y}=0\end{equation*}

and, therefore,

\begin{equation*}E_2\le {\rm e} \overline F(1/\lambda_{n,j})+C \lambda_{n,j}\int_{1/\lambda_{n,j}}^{y} t^{-2} \, {\rm d} t \le (C+e)\lambda_{n,j}^2,\end{equation*}

where we also used the Chebyshev inequality. As a result, for some constant C,

\begin{equation*}\varphi_y(\lambda_{n,j})= E_1+E_2\le 1-a\lambda_{n,j} +C\lambda_{n,j}^2.\end{equation*}

Consequently,

\begin{align*}\mathbb{E}\left[{\rm e}^{\frac{\lambda}{n}A_n};\ \overline X_n\le y\right]&\le \prod_{j=1}^n \left(1-a\lambda_{n,j} +C\lambda_{n,j}^2\right) \\[3pt]&=\exp\left\{\sum_{j=1}^n\ln \left(1-a\lambda_{n,j} +C\lambda_{n,j}^2\right)\right\}\\[3pt]&\le\exp\left\{\sum_{j=1}^n\left(-a\lambda_{n,j} +C\lambda_{n,j}^2\right)\right\}\\[3pt]&=\exp\left\{\sum_{j=1}^n\left(-a\lambda\frac{n-j+1}{n} +C\left(\lambda\frac{n-j+1}{n} \right)^2\right)\right\}\\[3pt]&\le \exp\left\{-\frac{a \lambda}{2} n +C\lambda^2 n\right\}.\end{align*}

Finally,

\begin{align*}\mathbb{P}(A_n>x,\overline X_n\le y)\le {\rm e}^{-\lambda\frac{x}{n}}\mathbb{E}\left[{\rm e}^{\frac{\lambda}{n}A_n};\ \overline X_n\le y\right]\le\exp\left\{-\lambda \frac{x}{n} -\frac{a \lambda}{2} n +C\lambda^2 n\right\}.\\[-40pt]\end{align*}

We can now obtain upper bounds for the tail of $A_\tau$ using the exponential bound in Lemma 6.

Lemma 7. Let $\mathbb{E}[X_1]=-a<0$ and ${\rm Var}(X_1)<\infty$ . Assume that the distribution function F of $X_j$ satisfies (8), and that (9) holds with $\gamma_0=1$ . Then there exists a constant $C>0$ such that

(32) \begin{align}\mathbb{P}(A_\tau \gt x, \overline X_\tau\le y)\le Cx^{1/4} \exp\left\{-2\frac{g(y)}{y}\sqrt{\left(\frac{a}{2}-\frac{2C_0g(y)}{y}\right)x}\right\}\end{align}

for all y satisfying $C_0 g(y)\le ay/4$ , where $C_0$ is the constant given by Lemma 6. Moreover,

\begin{equation*}\mathbb{P}(A_\tau \gt x)\le Cx^{1/4} \exp\left\{-g(\sqrt{2ax})\sqrt{\left(1-\frac{2C_0g(\sqrt{2ax})}{a\sqrt{2ax}}\right)^+}\right\}\ x\ge1,\end{equation*}

Proof. Using Lemma 6 with $y=\sqrt{2ax}$ we obtain

\begin{align*}\mathbb{P}(A_\tau \gt x, \overline X_\tau\le y)&\le \sum_{n=0}^\infty\mathbb{P}(A_n\geq x, \overline X_n \le \sqrt{2ax},\tau=n+1)\\&\le \sum_{n=1}^\infty\exp\left\{-\lambda \frac{x}{n} -\frac{a \lambda}{2} n +C\lambda^2 n\right\}=\sum_{n=1}^\infty\exp\left\{-\lambda \frac{x}{n} - \lambda I n\right\},\end{align*}

where $\lambda = \frac{g(\sqrt {2ax} )}{\sqrt {2ax}}$ and $I=\frac{a}{2}-C\lambda.$ The assumption $C_0g(y)\le y\frac{a}{4}$ implies that $I>\frac{a}{4}$ . Since I is positive, we have the inequality

\begin{equation*}\int_{n-1}^n\exp\left\lbrace-\lambda \frac{x}{y}-\lambda I(y+1)\right\rbrace {\rm d} y\ge\exp\left\lbrace-\lambda \frac{x}{n}-\lambda In\right\rbrace,\qquad n\ge1.\end{equation*}

With formula (25) on page 146 of [Reference Bateman1], we have

\begin{align*}\sum_{n=1}^\infty\exp\left\lbrace-\lambda\frac{x}{n}-\lambda I n\right\rbrace&\leq \int_0^\infty\exp\left\lbrace-\lambda \frac{x}{y}-\lambda I(y+1)\right\rbrace {\rm d} y \\&={\rm e}^{-\lambda I}\sqrt{\frac{4x}{I}}K_1(2\lambda\sqrt{Ix}).\end{align*}

Now, using the asymptotics for the modified Bessel function

\begin{equation*}K_1(z)\sim \sqrt{\frac{\pi}{2z}}{\rm e}^{-z} ,\end{equation*}

we obtain

\begin{align*}\sum_{n=1}^\infty\exp\left\lbrace-\lambda\frac{x}{n}-\lambda I n\right\rbrace\leq Cx^{1/4}\exp\{-2\lambda\sqrt{Ix}\}.\end{align*}

Therefore, (32) is proven.

The second claim in the lemma obviously holds for all x such that $C_0g(\sqrt{2ax})\ge a\sqrt{2ax}$ . Assume that x is so large that $C_0g(\sqrt{2ax})< a\sqrt{2ax}$ . Clearly,

\begin{equation*}\mathbb{P}(A_\tau \gt x)\le \mathbb{P}(A_\tau \gt x, \overline X_\tau\le \sqrt{2ax})+\mathbb{P}(A_\tau \gt x, \overline X_\tau \gt \sqrt{2ax})\eqqcolon P_1+P_2.\end{equation*}

By (32) with $y=\sqrt{2ax}$ ,

\begin{equation*}P_1\le Cx^{1/4} \exp\left\{-g(\sqrt{2ax})\sqrt{1-\frac{2C_0g(\sqrt{2ax})}{a\sqrt{2ax}}}\right\}.\end{equation*}

Next,

\begin{align*}P_2&\le \sum_{n=0}^\infty\mathbb{P}(A_\tau\geq x, M_n \le \sqrt{2ax}, X_{n+1}>\sqrt{2ax} ,\tau \gt n)\\&\le \sum_{n=0}^\infty\mathbb{P}( X_{n+1}>\sqrt{2ax}) \mathbb{P}(\tau \gt n) \le\mathbb{E}[\tau] \overline F(\sqrt {2a x}) = o(P_1).\end{align*}

Then, the claim follows.

Now we will give a lower bound.

Lemma 8. Let $\mathbb{E}[X_1]=-a<0$ and ${\rm Var}(X_1)<\infty$ . Then, for any $\varepsilon>0$ there exists $C>0$ such that

\begin{equation*}\liminf_{x\to\infty}\frac{\mathbb{P}(A_\tau \gt x)}{\overline F(\sqrt{2ax}+Cx^{1/4+\varepsilon/2})}\ge \mathbb{E}\tau.\end{equation*}

Proof. Fix $N\ge 1$ . Put $y^+ =\sqrt{2ax}+Cx^{1/4+\varepsilon/2},$ where C will be picked later. Since $\mathbb{E}[X_1^2]<\infty$ , by the strong law of large numbers,

\begin{equation*}\frac{S_{l}+al}{l^{1/2+\varepsilon}}\to 0, \qquad l\to \infty \mbox{ almost surely}.\end{equation*}

Hence, for any $\delta>0$ we can pick $R>0$ such that

\begin{equation*}\mathbb{P}\left( \min_{l\le \sqrt{2x/a}} (S_{l}+al +R + l^{1/2+\varepsilon})>0\right)>(1-\delta).\end{equation*}

Define

\begin{equation*}E^+_k\coloneqq \left\{\min_{l\le \sqrt{2x/a}} (S_{k+l}-S_k+al +R + l^{1/2+\varepsilon})>0,\tau \gt k, S_k>y^+\right\}.\end{equation*}

If $C>1+(2/a)$ then, for all x large enough, $al+l^{1/2+\varepsilon}+R\le\sqrt{2ax}+(2x/a)^{1/4+\varepsilon/2}+R\le y^+$ for all $l\le\sqrt{2x/a}$ . Therefore, for every $k\le N$ , $E^+_k\subset \{\tau \gt k+\sqrt{2x/a}\}$ . Furthermore, if $\tau \gt k+\sqrt{2x/a}$ then, on the event $E_k^+$ ,

\begin{align*}A_\tau&>\sum_{l=0}^{k+\sqrt{2x/a}}S_{k+l}>y^+\sqrt{2x/a}+\sum_{l=0}^{k+\sqrt{2x/a}}(S_{k+l}-S_l)\\&>y^+\sqrt{2x/a}-\sum_{l=0}^{k+\sqrt{2x/a}}(al+l^{1/2+\varepsilon}+R) .\end{align*}

Now, we can choose C so large that, for every $k\le N$ , $E_k^+\subset \{A_\tau \gt x\}$ . Hence,

\begin{align*}&\mathbb{P}(A_{\tau}>x) \ge\sum_{k=1}^N\mathbb{P}(A_\tau \gt x, \overline X_{k-1}\le y^+, X_k>y^+, \tau \gt k)\\&\quad\ge\sum_{k=1}^N\mathbb{P}(E_k^+, \overline X_{k-1}\le y^+, X_k>y^+, \tau \gt k)\\&\quad\ge\sum_{k=1}^N\mathbb{P}\left(\overline X_{k-1}\le y^+,\tau \gt k-1, X_k>y^+, \min_{l\le \sqrt{2x/a}} (S_{l+k}-S_k +R + l^{1/2+\varepsilon})>0\right)\\&\quad\ge (1-\delta)\sum_{k=1}^N \mathbb{P}\left(\overline X_{k-1}\le y^+,\tau \gt k-1\right)\overline F(y^+).\end{align*}

For every fixed k we have $\mathbb{P}\left(\overline X_{k-1}\le y^+,\tau \gt k-1\right) \to \mathbb{P}\left(\tau \gt k-1\right)$ as $x\to\infty$ . Furthermore, $\sum_{k=0}^N \mathbb{P}(\tau \gt k)\to\mathbb{E}\tau$ as $N\to\infty$ . Therefore, we can pick sufficiently large N such that

\begin{equation*}\liminf_{x\to\infty}\sum_{k=1}^N \mathbb{P}\left(\overline X_{k-1}\le y^+,\tau \gt k-1\right)\ge (1-\delta)\mathbb{E}\tau.\end{equation*}

Then, for all x sufficiently large, $\mathbb{P}(A_{\tau}>x)\ge (1-\delta)^2\mathbb{E}\tau\overline F(y^+)$ . As $\delta>0$ is arbitrarily small, we arrive at the conclusion.

Proof of Theorem 1. The upper bound follows from Lemma 7. The lower bound follows from Lemma 8. The rough asymptotics follow immediately from the lower and upper bounds, and from the observation that

(33) \begin{equation}\sup_{|y|\le x\rho(x)}\left|\frac{\log \overline F(x)}{\log \overline F(x+y)}-1\right|\to0,\end{equation}

where $\rho(x)\to0$ . To prove (33) we note that by (9) and (10)

\begin{align*} g(x+y)-g(x)&=\int_x^{x+y} g'(t) \, {\rm d} t \le\gamma_0 \int_x^{x+y} \frac{g(t)}{t} \, {\rm d} t\le \gamma_0 \frac{g(x)}{x^{\gamma_0}}\int_x^{x+y} \frac{1}{t^{1-\gamma_0}} \, {\rm d} t \\[3pt]\nonumber&\le \gamma_0\frac{g(x)}{x^{\gamma_0}}\frac{y}{x^{1-\gamma_0}}=\gamma_0g(x)\frac{y}{x},\qquad y>0.\end{align*}

This implies that, as $x\to\infty$ ,

(34) \begin{equation}\sup_{|y|\le x\rho(x)}\left|\frac{g(x+y)}{g(x)}-1\right|\to0.\end{equation}

Recalling that $\log\overline{F}(x)\sim-g(x)-2\log x$ , one easily obtains (33).

4. Proof of Theorem 2

Set

\begin{equation*}h(x)\coloneqq \frac{\sqrt{2ax}}{g(\sqrt{2ax})}, \qquad y=\sqrt{2ax} - Ch(x) \log x ,\end{equation*}

where $C>\frac{5/4}{1-\gamma_0}$ . We first split the probability $\mathbb{P}(A_{\tau}>x)$ as follows:

\begin{align*}\mathbb{P}(A_{\tau}>x)&=\mathbb{P}(A_{\tau}>x, \overline X_\tau\le y)+\mathbb{P}\left(A_{\tau}>x, \overline X_\tau \gt \sqrt{2ax}-\frac{1}{\log x}h(x)\right)\\[3pt]& \quad + \mathbb{P}\left(A_{\tau}>x, \overline X_\tau\in \left[y, \sqrt{2ax}-\frac{1}{\log x}h(x))\right]\right)\eqqcolon P_1+P_2+P_3.\end{align*}

The first term will be estimated using the exponential bound proved in Lemma 6.

Proof. According to (32),

\begin{align*}P_1&\le Cx^{1/4}\exp\left\{-2\frac{g(y)}{y}\sqrt{\left(\frac{a}{2}-2C_0\frac{g(y)}{y}\right)x}\right\}.\end{align*}

Since (9) holds for some $\gamma_0<1/2$ , $g^2(y)/y\to 0$ , and hence

\begin{equation*}P_1 \le Cx^{1/4}\exp\left\{-\frac{g(y)}{y}\sqrt{2ax}\right\}.\end{equation*}

Then,

\begin{equation*}\frac{P_1}{\overline F(\sqrt{2ax})}\le C x^{5/4}\exp\left\{g(\sqrt{2ax})-\frac{g(y)}{y}\sqrt{2ax}\right\}.\end{equation*}

To finish the proof, it is sufficient to show that

(35) \begin{equation}g(\sqrt{2ax})-\frac{g(y)}{y}\sqrt{2ax} + \frac{5}{4}\log x \to -\infty,\qquad x\to \infty.\end{equation}

We first note that

\begin{align*}d(x)&\coloneqq g(\sqrt{2ax})-\frac{g(y)}{y}\sqrt{2ax} =g(\sqrt{2ax}) - \frac{g(y)}{1-C\frac{\log x}{g(\sqrt{2ax})}}\\[3pt]&= g(\sqrt{2ax}) - g(y) - (C+o(1)) \log x \frac{g(y)}{g(\sqrt{2ax})}.\end{align*}

Using (18), we can see that

\begin{equation*}g(\sqrt{2ax}) - g(y) \le \gamma_0 \frac{g(y)}{y} (\sqrt{2ax}-y)=\gamma_0 C \frac{g(y)}{y}\log x \frac{\sqrt{2ax}}{g(\sqrt{2ax})}.\end{equation*}

Hence,

\begin{equation*}d(x)\le \left(\gamma_0\frac{\sqrt{2ax}}{y}-1 \right) (C+o(1)) \frac{g(y)}{g(\sqrt{2ax})}\log x.\end{equation*}

According to (34), $g(y)\sim g(\sqrt{2ax})$ . Therefore, (35) is valid for any C satisfying $C(\gamma_0-1)+\frac{5}{4}<0$ .

The next lemma gives the term that dominates in $\mathbb{P}(A_\tau \gt x)$ .

Lemma 10. Under the assumptions of Lemma 9 we have the following estimate:

\begin{equation*}P_2\le (\mathbb{E}\tau+o(1))\overline F(\sqrt{2ax}), \qquad x\to \infty.\end{equation*}

Proof. Put

\begin{equation*}y^*=\sqrt{2ax}-\frac{h(x)}{\log x}.\end{equation*}

By the total probability formula,

\begin{align*}P_2&\le \sum_{n=0}^\infty\mathbb{P}(A_\tau\geq x, \overline{X}_n \le y^*, X_{n+1}>y^* ,\tau \gt n)\\&\le \sum_{n=0}^\infty\mathbb{P}( X_{n+1}>y^*) \mathbb{P}(\tau \gt n)= \mathbb{E}[\tau] \overline F(y^*).\end{align*}

Now, note that, by (18) and (34),

\begin{align*}\frac{\overline F(y^*)}{\overline F(\sqrt{2ax})}&\le (1+o(1)) {\rm e}^{g(\sqrt{2ax})-g(y^*)}\le (1+o(1)) \exp\left\{\frac{\gamma_0 g(y^*)}{y^*}(\sqrt{2ax}-y^*)\right\} \\&\le (1+o(1)) \exp\left\{\frac{\gamma_0 g(y^*)}{y^*}\frac{1}{\log x}\frac{\sqrt{2ax}}{g(\sqrt{2ax})}\right\} = 1+o(1).\end{align*}

Then the statement immediately follows.

We proceed to the analysis of $P_3$ . Fix some $\delta>0$ and set $z=\frac{1}{a}\left(\sqrt{2ax} + \delta\sqrt{x} \right)$ . We split $P_3$ further as follows:

\begin{align*}P_3\le P_{31}+P_{32}+P_{33} & \coloneqq \mathbb{P}\left(A_{\tau}>x, \overline X_\tau\in \left[y, \sqrt{2ax}-\frac{1}{\log x}h(x)\right];\ J_1;\ \tau\le z\right )\\& \quad +\mathbb{P}\left(A_{\tau}>x, \overline X_\tau\in \left[y, \sqrt{2ax}-\frac{1}{\log x}h(x)\right];\ J_{\ge 2}, \tau\le z\right)\\& \quad + \mathbb{P}(\tau \gt z),\end{align*}

where $J_1 =\{ \mbox{there exists} $ k (1, $\tau$ ) such that $X_k>y \mbox{ and }\max_{1\le i\le \tau, i\neq k} X_i \le y\}$ and, correspondingly, $J_{\ge 2} =\{\mbox{there exist} $ k, l $\in(1,\tau$ ) such that $X_k>y \mbox{ and } X_l>y\}$ .

We start with the easier terms $P_{32}$ and $P_{33}$ . To deal with these terms we will use Proposition 2.

Proof. We have, by Proposition 2, $P_{33}=\mathbb{P}(\tau \gt z)\le (\mathbb{E}\tau+o(1))\overline F(az) =O\big(\overline F(\sqrt{2ax}+\delta \sqrt{x})\big)$ . Therefore,

\begin{align*}\frac{P_{33}}{\overline F(\sqrt{2ax})}&\le C\exp\left\{g(\sqrt{2ax})-g(\sqrt{2ax}+\delta \sqrt{x})\right\}.\end{align*}

By the mean value theorem and by the assumption (11), $g(cx)-g(x)\to\infty$ as $x\to\infty$ for every $c>1$ . This completes the proof.

Proof. We can use the formula of total probability to write

\begin{equation*}P_{32} \le \sum_{k=1}^z \mathbb{P}(\tau=k, J_{\ge 2})\le \sum_{k=1}^z \frac{k^2}{2}\overline F(y)^2.\end{equation*}

Then,

\begin{equation*}\frac{P_{32}}{\overline F(\sqrt{2ax})}\le C x^{3/2}\frac{\overline F(y)^2}{\overline F(\sqrt{2ax})}\le C x^{1/2} {\rm e}^{g(\sqrt{2ax})-2g(y)}.\end{equation*}

Using (18) we can see that, in view of (12),

\begin{align*}\frac{P_{32}}{\overline F(\sqrt{2ax})} \le C x^{1/2} {\rm e}^{C\ln x -g(y)}\to 0 . \\[-40pt]\end{align*}

$P_{31}$ remains to be analysed. For that, introduce $\mu(y)\coloneqq \min\{n\ge 1\,{:}\, X_k>y\}$ . Now we will complete the proof with the following lemma.

Proof. First, represent event $J_1$ as $J_1=J_{11}\cup J_{12}$ , where

\begin{align*}J_{11}&\coloneqq \{\mbox{$X_k>y$ for exactly one $k\in(0,\tau)$ and $X_i\le x^\varepsilon$ for all other $i<\tau$} ,\}\\[3pt]J_{12}&\coloneqq \{\mbox{$X_k>y$ for exactly one $k\in(0,\tau)$ and $X_i>x^\varepsilon $ for some $i\neq k, i<\tau$}\}.\end{align*}

Then,

\begin{align*}Q_{2}&\coloneqq \mathbb{P}\left(A_{\tau}>x, \overline X_\tau\in \left[y, \sqrt{2ax}-\frac{1}{\log x}h(x)\right];\ J_{12},\tau\le z\right )\\[3pt]&\le \sum_{j=1}^z \mathbb{P}(\tau=j, J_{12})\le \sum_{j=1}^z \frac{j^2}{2}\overline F(y)\overline F(x^\varepsilon)\le z^3 \overline F(y)\overline F(x^\varepsilon) , \end{align*}

so

\begin{align*}\frac{Q_{2}}{\overline F(\sqrt{2ax})}\le Cx^{3/2-2\varepsilon} {\rm e}^{g(\sqrt{2ax})-g(y)-g(x^{\varepsilon}))}\end{align*}

By (18), $g(\sqrt{2ax})-g(y)\le C\ln x$ . Then, in view of the relation (12), we have $g(\sqrt{2ax})-g(y)-g(x^{\varepsilon})\le -4\ln x$ , which implies that $Q_{2} = o(\overline F(\sqrt{2ax}))$ .

To estimate

\begin{equation*}Q_{1}\coloneqq \mathbb{P}\left(A_{\tau}>x, \overline X_\tau\in \left[y, \sqrt{2ax}-\frac{1}{\log x}h(x)\right];\ J_{11},\tau\le z\right )\end{equation*}

we make use of the exponential bound given in Lemma 6. Put

\begin{equation*}x^+(k)=x-k\left(\sqrt{2ax}-\frac{h(x)}{\log x}\right).\end{equation*}

Then, we have

\begin{align*}Q_{1}&=\sum_{k=0}^{z-1} \sum_{j=1}^k \mathbb{P}\left(A_k>x, \max_{i\neq j, i\le k} X_i\le x^\varepsilon, X_j\in\left[y, \sqrt{2ax}-\frac{h(x)}{\log x}\right] , \tau=k+1\right)\\[3pt]&\le\sum_{k=1}^z (k+1)\mathbb{P}(A_k>x^+(k), \overline X_k\le x^\varepsilon)\overline F(y)\\[3pt]&\le Cx^{1/2}\overline F(y)\sum_{k=1}^z\exp\left\{-\lambda \frac{x^+(k)}{k} -\frac{a \lambda}{2} k +C\lambda^2 k\right\},\end{align*}

where $\lambda = \frac{g(x^\varepsilon)}{x^\varepsilon}$ . Now note that

\begin{equation*}-\lambda \frac{x^+(k)}{k} -\frac{a \lambda}{2} k=-\lambda\left(-\sqrt{2ax}+\frac{h(x)}{\log x} +\frac{x}{k}+\frac{ak}{2}\right).\end{equation*}

Since

\begin{equation*}\frac{x}{k}+\frac{ak}{2}\ge \sqrt{2ax},\qquad k\ge1,\end{equation*}

we obtain

\begin{equation*}-\lambda \frac{x^+(k)}{k} -\frac{a \lambda}{2} k\le -\lambda \frac{h(x)}{\log x},\qquad k\ge1.\end{equation*}

Thus, $Q_{1} \le Cx {\rm e}^{-\lambda h(x)/\log x+\lambda^2 z}\overline{F}(y)$ . Next, we can pick $\varepsilon = \frac{1}{4(1-\gamma_0)} $ to achieve

\begin{align*}\lambda^2 z &\le C \left(\frac{g(x^\varepsilon)}{x^\varepsilon}\right)^2x^{1/2}=C\left(\frac{g(x^\varepsilon)}{x^{\varepsilon(1-1/(4\varepsilon))}}\right)^2=C\left(\frac{g(x^\varepsilon)}{x^{\gamma_0\varepsilon}}\right)^2\\[3pt]&<C\sup_t\left(\frac{g(t)}{t^{\gamma_0}}\right)^2<\infty \end{align*}

by the condition (9). Note that the assumption $\gamma_0<1/2$ implies that $\varepsilon=\frac{1}{4(1-\gamma_0)}<1/2$ . Then, using (8), we obtain

\begin{align*}\frac{Q_{1}}{\overline F(\sqrt{2ax})}\le Cx {\rm e}^{g(\sqrt{2ax})-g(y)-\lambda h(x)/\log x},\end{align*}

and, using (18),

\begin{align*}\frac{Q_{1}}{\overline F(\sqrt{2ax})}\le Cx^{C} {\rm e}^{-\lambda h(x)/\log x}.\end{align*}

Finally, noting that

\begin{align*}\lambda h(x) = \frac{g(x^\varepsilon)}{x^{\varepsilon}} \frac{\sqrt{2ax}}{g(\sqrt{2ax})}\end{align*}

grows polynomially, we obtain the required convergence to 0. The polynomial growth can be immediately seen for $g(x)=x^{\gamma_0}$ . However, a proper proof goes as follows:

\begin{align*}g(C\sqrt{x}) &= g(x^\varepsilon)+\int_{x^\varepsilon}^{C\sqrt{x}} g'(t) \, {\rm d} t\le g(x^\varepsilon)+\gamma_0\int_{x^\varepsilon}^{C\sqrt{x}} \frac{g(t)}{t} \, {\rm d} t \\[3pt]&\le g(x^\varepsilon)+\gamma_0\int_{x^\varepsilon}^{C\sqrt{x}} \frac{g(t)}{t^{\gamma_0}} t^{\gamma_0-1} \, {\rm d} t\le g(x^\varepsilon)+\frac{g(x^\varepsilon)}{x^{\varepsilon\gamma_0}}\int_{x^\varepsilon}^{C\sqrt{x}} t^{\gamma_0-1} \, {\rm d} t \\[3pt]&\le g(x^\varepsilon)+C\frac{g(x^\varepsilon)}{x^{\varepsilon \gamma_0}} x^{\gamma_0/2}\le C g(x^\varepsilon)x^{\gamma_0(1/2-\varepsilon)}.\end{align*}

Therefore $\lambda h(x)\ge x^{1/2-\varepsilon} x^{-\gamma_0(1/2-\varepsilon)}=x^{(1-\gamma_0)/2-1/4}$ , where we have used the equality $\varepsilon=\frac{1}{4(1-\gamma_0)}$ .

Proof of Theorem 2. Combining the preceding lemmas give us the upper bound. The lower bound has been shown in (5) under even weaker conditions.