Proof. We first consider the case $p_1=0$ . Let d be the greatest common divisor of $\{j-k\,{:}\,j\neq k,\ p_j p_k>0\}$ . According the proof of [Reference Fleischmann and Wachtel20, Theorem 6], for any fixed $\varepsilon\in(0,1)$ there exist constants $c_3\geq c_4>0$ such that, for any $b^n\leq k\leq \lfloor m^{(1-\varepsilon)n}\rfloor$ , $k=b^n\ (\textrm{mod}\ d)$ , and $n\geq 1$ ,

(4) \begin{equation} \exp\left[ -c_3\left(\frac{k}{m^n}\right)^{-\frac{\rho}{1-\rho}}\right]\leq m^n\mathbb{P}(|Z_n|=k)\leq\exp\left[ -c_4 \left(\frac{k}{m^n}\right)^{-\frac{\rho}{1-\rho}}\right].\end{equation}

Hence, for the upper bound, we have

(5) \begin{align} \mathbb{E}\left[\textrm{e}^{-l|Z_n|^{\beta}}\right] & \leq \mathbb{E}\left[\textrm{e}^{-l|Z_n|^{\beta}}\textbf{1}_{\{|Z_n|\leq m^{(1-\varepsilon)n}\}}\right]+\textrm{e}^{-lm^{\beta(1-\varepsilon)n}}\nonumber\\ & \leq m^{-n} \!\!\!\!\! \sum_{\scriptsize\begin{array}{c}b^n\leq k\leq \lfloor m^{(1-\varepsilon)n}\rfloor \\ k=b^n(\textrm{mod}\ d)\end{array}} \!\!\!\!\! \exp(-lk^{\beta}) \exp\left[ -c_4 \left(\frac{k}{m^n}\right)^{-\frac{\rho}{1-\rho}}\right]+\textrm{e}^{-lm^{\beta(1-\varepsilon)n}}.\end{align}

Note that there exists a positive constant T depending only on l, $c_4$ , $\rho$ , and $\beta$ such that, for n large enough,

\begin{equation*}\min\limits_{b^n\leq x\leq\lfloor m^{(1-\varepsilon)n}\rfloor}\left\{lx^{\beta}+c_4 \left(\frac{x}{m^n}\right)^{-\frac{\rho}{1-\rho}}\right\}\geq Tm^\frac{\beta\rho n}{\beta+\rho-\beta\rho}.\end{equation*}

As a consequence, if $1-\varepsilon>\frac{\rho }{\beta+\rho-\beta\rho}$ then

\begin{align*} \mathbb{E}\left[\textrm{e}^{-l|Z_n|^{\beta}}\right]& \leq m^{\varepsilon n}\exp\left(-Tm^\frac{\beta\rho n}{\beta+\rho-\beta\rho}\right)+\textrm{e}^{-lm^{\beta(1-\varepsilon)n}}\\ &\leq2m^{\varepsilon n}\exp\left(-(T\wedge l)m^\frac{\beta\rho n}{\beta+\rho-\beta\rho}\right),\nonumber \end{align*}

which yields

\begin{equation*}\liminf_{n\rightarrow\infty}\frac{1}{n}\log\left[-\log \mathbb{E}\left[\textrm{e}^{-l|Z_n|^{\beta}}\right]\right]\geq \frac{\beta\rho}{\beta+\rho-\beta\rho}\log m.\end{equation*}

For the lower bound, similarly, using the left-hand side of (4) we have, for n large enough,

\begin{align*} \mathbb{E}\left[\textrm{e}^{-l|Z_n|^{\beta}}\right]&\geq m^{-n}\sum_{\scriptsize\begin{array}{c}b^n\leq k\leq \lfloor m^{(1-\varepsilon)n}\rfloor \\ k=b^n(\textrm{mod}\ d)\end{array}}\exp(-lk^{\beta}) \exp\left[ -c_3 \left(\frac{k}{m^n}\right)^{-\frac{\rho}{1-\rho}}\right]\\ &\geq m^{-n}\exp\left(-T'm^\frac{\beta\rho n}{\beta+\rho-\beta\rho}\right),\nonumber \end{align*}

where $T'>0$ depends on l, $c_3$ , $\rho$ , $\beta$ , and d. Taking limits yields

\begin{equation*}\limsup_{n\rightarrow\infty}\frac{1}{n}\log\left[-\log \mathbb{E}\left[\textrm{e}^{-l|Z_n|^{\beta}}\right]\right]\leq \frac{\beta\rho}{\beta+\rho-\beta\rho}\log m.\end{equation*}

For $p_1>0$ , by [Reference Fleischmann and Wachtel21, Lemma 13] there exists a universal constant $c>0$ such that, for any $k,n\geq1$ , $\mathbb{P}(|Z_n|=k)\leq cp^n_1k^{-(\log p_1/\log m)-1}$ . Thus, $p^n_1e^{-l}\leq\mathbb{E}\big[\textrm{e}^{-l|Z_n|^{\beta}}\big]\leq cp^n_1\sum_{k\geq1} k^{-(\log p_1/\log m)-1}\textrm{e}^{-lk^{\beta}}$ , which implies the desired result.

Proof. We first consider the case $\mathbb{E}\big[|Z_1|^{\beta}\big]<\infty$ for some $\beta>1$ . By the branching property, for any $n\geq 1$ and $\epsilon\in(0,\frac{1}{3})$ , we have

(7) \begin{align} \mathbb{P}\Bigg(\bar{Z}^{\xi}_n(A) & \geq \frac{1}{|\xi|}\sum_{x\in\xi}\nu_n\left(A-S_x\right)+\epsilon\Bigg)\nonumber\\ & = \mathbb{P}\Bigg(\frac{\frac{1}{|\xi|}\sum_{x\in\xi}{Z}^{x}_n(A)/m^n}{\frac{1}{|\xi|}\sum_{x\in\xi}{Z}^{x}_n/m^n}\geq\frac{1}{|\xi|}\sum_{x\in\xi}\nu_n(A-S_x)+\epsilon\Bigg)\nonumber\\ & \leq \mathbb{P}\Bigg(\sum_{x\in\xi}{W}^{x}_n<|\xi|(1-\epsilon/2)\Bigg)+\mathbb{P}\Bigg(\sum_{x\in\xi}\left({W}^{x}_n(A)-\nu_n\left(A-S_x\right)\right)>|\xi|\epsilon/3\Bigg)\nonumber\\ & \leq \mathbb{P}\Bigg(\sum_{x\in\xi}(1-W^x_n)>\frac{\epsilon}{3}{|\xi|}\Bigg)+\mathbb{P}\Bigg(\sum_{x\in\xi}\left({W}^{x}_n(A)-\nu_n\left(A-S_x\right)\right)>|\xi|\epsilon/3\Bigg)\nonumber\\ &\,{=\!:}\,I_1+I_2,\end{align}

where $\epsilon\in\big(0,\frac{1}{3}\big)$ is used in the first inequality.

We first consider $I_2$ . By [Reference Nagaev34, Corollary 1.6], we know that if $Y_x$ , $x\in\xi$ , are independent random variables with zero mean, and $A^+_t=\sum_{x\in\xi}\mathbb{E}\big[Y^t_x \textbf{1}_{\{Y_x\geq 0\}}\big]<\infty$ for some $1\leq t\leq 2$ , then, for $y^t\geq 4A^+_t$ and $z>y$ ,

(8) \begin{equation} \mathbb{P}\Bigg(\sum\limits_{x\in\xi} Y_x\geq z\Bigg)\leq\sum\limits_{x\in\xi}\mathbb{P}(Y_x>y)+\left(\textrm{e}^2A^+_t/zy^{t-1}\right)^{z/2y}.\end{equation}

So, if we choose $t=1+\frac{\beta-1}{2\beta}$ , $y=\frac{1}{4\beta}z$ , and $z=|\xi|\epsilon/3$ , then $t\in(1,2\wedge\beta]$ . Let $Y_x={W}^{x}_n(A)-\nu_n(A-S_x)$ . Note that by (3), there exists a constant $C_{\beta}>0$ such that

(9) \begin{equation}\sup_{n}\mathbb{E}\big[W^{t}_n\big]<C_{\beta}<\infty.\end{equation}

Therefore, $A^+_t<\infty$ . Furthermore, by (9), there exists a constant $C_{\epsilon,\beta}>0$ such that, for all $|\xi|>C_{\epsilon,\beta}$ and $n\geq1$ ,

\begin{equation*}y^t=\left(\frac{\epsilon|\xi|}{12\beta}\right)^t\geq 4|\xi|C_{\beta}\geq 4|\xi|\mathbb{E}[W^t_n]\geq 4\sum\limits_{x\in\xi}\mathbb{E}[Y^t_x \textbf{1}_{\{Y_x\geq 0\}}].\end{equation*}

So, by (8), it follows that, for all $|\xi|>C_{\epsilon,\beta}$ , $n\geq 1$ , and $A\subset\mathbb{R}$ ,

(10) \begin{align}I_2&\leq|\xi|\mathbb{P}\bigg({W}^{x}_n(A)-\nu_n(A-S_x)>\frac{\epsilon|\xi|}{12\beta}\bigg)+\bigg(\frac{\textrm{e}^2C_{\beta}3^{t}}{(4\beta)^{1-t}\epsilon^t}\bigg)^{2\beta}|\xi|^{(1-t)2\beta}\nonumber\\ &\leq|\xi|\mathbb{P}\bigg({W}_n>\frac{\epsilon|\xi|}{12\beta}\bigg)+\bigg(\frac{\textrm{e}^2C_{\beta}3^{t}}{(4\beta)^{1-t}\epsilon^t}\bigg)^{2\beta}|\xi|^{1-\beta}.\end{align}

By (3) and the Markov inequality, there exists a constant $C^{\prime}_{\epsilon,\beta}>0$ such that, for all $n\geq1$ ,

(11) \begin{equation} \mathbb{P}\bigg({W}_n>\frac{\epsilon|\xi|}{12\beta}\bigg)\leq \bigg(\frac{12\beta}{\epsilon|\xi|}\bigg)^{\beta}\mathbb{E}\bigg[\sup_nW^{\beta}_n\bigg]\leq C^{\prime}_{\epsilon,\beta}|\xi|^{-\beta}.\end{equation}

Plugging (11) into (10) yields, for all $|\xi|>C_{\epsilon,\beta}$ , $n\geq 1$ , and $A\subset\mathbb{R}$ ,

\begin{equation*}I_2\leq C^{\prime}_{\epsilon,\beta}|\xi|^{1-\beta}+\bigg(\frac{\textrm{e}^2C_{\beta}3^{t}}{(4\beta)^{1-t}\epsilon^t}\bigg)^{2\beta}|\xi|^{1-\beta}.\end{equation*}

For $I_1$ , using (8) again and letting $Y_x=1-W^x_n$ , we have, for all $|\xi|>\frac{12\beta}{\epsilon}$ and $n\geq1$ ,

\begin{equation*}I_1\leq|\xi|\mathbb{P}\bigg(1-W^x_n>\frac{\epsilon|\xi|}{12\beta}\bigg)+\bigg(\frac{\textrm{e}^23^{t}}{(4\beta)^{1-t}\epsilon^t}\bigg)^{2\beta}|\xi|^{(1-t)2\beta}= \bigg(\frac{\textrm{e}^23^{t}}{(4\beta)^{1-t}\epsilon^t}\bigg)^{2\beta}|\xi|^{1-\beta}.\end{equation*}

Plugging the above two inequalities into (7) means that there exists a constant $T(\epsilon,\beta)>0$ such that, for $|\xi|>C_{\epsilon,\beta}\vee\frac{12\beta}{\epsilon}$ ,

\begin{equation*}\mathbb{P}\Bigg(\bar{Z}^{\xi}_n(A)\geq \frac{1}{|\xi|}\sum_{x\in\xi}\nu_n\left(A-S_x\right)+\epsilon\Bigg)\leq T(\epsilon,\beta)|\xi|^{1-\beta}.\end{equation*}

Hence, to obtain (6), we can take $C_1\,{:\!=}\,\big(C_{\epsilon,\beta}\vee\frac{12\beta}{\epsilon}\big)^{\beta-1}\vee T(\epsilon,\beta)$ . For the case $\mathbb{P}(|Z_1|>x)\sim \Theta(1)x^{-\beta}$ , as a consequence of Lemma 1 we only need to replace (11) by

\begin{equation*}\mathbb{P}\bigg({W}_n>\frac{\epsilon|\xi|}{12\beta}\bigg)\leq c_2\bigg(\frac{\epsilon}{12\beta}\bigg)^{-\beta}|\xi|^{-\beta}.\end{equation*}

Replacing A with $A^\textrm{c}$ , we obtain (6) with $>$ and $+\epsilon$ replaced by $<$ , $-\epsilon$ .

Proof. From (7), for every $\epsilon\in\left(0,\frac{1}{3}\right)$ we have

(13) \begin{align} \mathbb{P}\Bigg(\bar{Z}^{\xi}_n(A) & \geq \frac{1}{|\xi|}\sum_{x\in\xi}\nu_n(A-S_x)+\epsilon\Bigg)\nonumber\\ & \leq \mathbb{P}\Bigg(\sum_{x\in\xi}(1-W^x_n)>\frac{\epsilon}{3}{|\xi|}\Bigg) + \mathbb{P}\Bigg(\sum_{x\in\xi}{W}^{x}_n(A)>\sum_{x\in\xi}\nu_n(A-S_x)+|\xi|\epsilon/3\Bigg)\nonumber\\ & \,{=\!:}\, I_1 + I_2. \end{align}

For $I_1$ , note that $\sup\limits_{x,n}\mathbb{E}\left[\textrm{e}^{1-W^x_n}\right]<\infty$ . Furthermore, by Lemma 1, $\mathbb{E}\left[((1-W^x_n)^-)^2\right]\leq 3+\sup_n\mathbb{E}\left[W^2_n\right]<\infty$ . Hence, by [Reference Louidor and Perkins32, Lemma 2.3], there exists a constant $c_5>0$ such that, for all n, $\xi$ , and $\epsilon$ small enough,

(14) \begin{equation} I_1=\mathbb{P}\Bigg(\sum_{x\in\xi}(1-W^x_n)>\frac{\epsilon}{3}{|\xi|}\Bigg) \leq \textrm{e}^{-c_5\epsilon^2|\xi|}. \end{equation}

For $I_2$ , by Lemma 1 there exists a constant $c_6>0$ such that, for all $y>0$ and $n\geq1$ , $\mathbb{P}({W}^{x}_n(A)>y)\leq\mathbb{P}(W_n>y)\leq c_6\exp[-l(m-1)^{\beta}y^\beta]$ . Let $X_i$ , $i\geq 1$ , be i.i.d. copies of ${W}^{x}_n(A)$ , and

\begin{equation*} a_i(|\xi|)= \begin{cases} \Big(|\xi|\epsilon/3+\sum_{x\in\xi}\nu_n(A-S_x)\Big)^{-1}, & 1\leq i\leq |\xi|; \\ \\[-7pt] 0 , & i > |\xi|. \end{cases} \end{equation*}

Then, by slight modifications of the proof of the upper bound [Reference Aurzada5, Theorem 2.1], there exist positive constants T, T $^{\prime}$ depending on $\epsilon$ and $\beta$ such that, for any $|\xi|>T$ , $n\geq 1$ , and any $A\subset\mathbb{R}$ ,

(15) \begin{equation} I_2=\mathbb{P}\Bigg(\sum_{x\in\xi}{W}^{x}_n(A)>\sum_{x\in\xi}\nu_n(A-S_x)+|\xi|\epsilon/3\Bigg)\leq \textrm{e}^{-T'|\xi|^{\beta}}. \end{equation}

Plugging (14) and (15) into (13) concludes the proof of this lemma with $C_2\,{:\!=}\,\textrm{e}^{T'T^{\beta}}$ and $C_3\,{:\!=}\,T'\wedge(c_5\epsilon^2)$ .

Proof. The upper bound can be found in [Reference Chen and He13, Lemma B.1]. For the lower bound, since $C\textrm{e}^{-\lambda x^{\alpha}}\leq\mathbb{P}(X>x)\leq C'\textrm{e}^{-\lambda x^{\alpha}}$ for some constants $C'>C>0$ , we have, for n large enough,

\begin{align*} \mathbb{P}\left(\sum^{t_n}_{i=1}X_i\in(a\sqrt n,b\sqrt n)\right) & \geq \mathbb{P}\left(X_i\in\left(\frac{a\sqrt n}{t_n},\frac{b\sqrt n}{t_n} \right), \textrm{ for all } 1\leq i\leq t_n\right) \\[4pt] & \geq \left(C\exp\left[-\lambda\left(\frac{a\sqrt n}{t_n}\right)^{\alpha}\right]-C'\exp\left[-\lambda\left(\frac{b\sqrt n}{t_n}\right)^{\alpha}\right]\right)^{t_n} \\[4pt] & \geq C^{t_n}\exp\left[-\lambda a^{\alpha}\frac{n^{\alpha/2}}{t^{\alpha-1}_n}\right]. \end{align*}

Then, the desired lower bound follows, provided $t_n=o(\sqrt n)$ .

Proof. We start with the lower bound. Fix $\epsilon>0$ . By the continuity of $I_A(\!\cdot\!)$ at p, there exist some $\delta>0$ and $|x|<I_A(p)+\epsilon$ such that, for any small $\eta>0$ , $\inf_{y\in[x-\eta,x+\eta]}\nu(A-y)\geq p+\delta$ . Consequently, we can choose $M>1$ large enough that $\frac{1}{1+M^{-1}}\inf_{y\in[x-\eta,x+\eta]}\nu(A-y)\geq p+\frac{\delta}{2}$ . Let $t_n\,{:\!=}\,\lfloor h\sqrt n\rfloor$ with some $h>0$ , and $|v|$ be the generation of particle v. Set $Z^v_{k} \,{:\!=}\, \sum_ {u\in Z_{|v|+k},u\textrm{ is a descent of }v}\delta_{S_u-S_v}$ ; $H(u) \,{:\!=}\, \{w\in Z_{t_n+1} : w\notin Z^u_{1}\}$ for $u\in Z_{t_n}$ ; $\mathcal{E} \,{:\!=}\, \{(\xi,k,r)\in \mathcal{M}\times\mathbb{N^+}\times\mathbb{N^+} \,{:}\, \xi$ has exactly one atom z such that $S_z \in [(x-\eta)\sqrt n,(x+\eta)\sqrt n];\ k>2Mr\}$ ; $E \,{:\!=}\, \big\{Z_{t_n}$ has exactly one particle u such that $S_u \in [(x-\eta)\sqrt n,(x+\eta)\sqrt n]$ and $|Z^u_1|>2M\sum_{v\neq u,v\in Z_{t_n}}|Z^v_1|\big\}$ . Namely, $Z^v_{k}$ is the kth generation of the sub-BRW emanating from particle v, and H(u) represents a collection of particles at time $t_n+1$ who are not the children of u. The proof of the lower bound is divided into two main steps.

For the first step, we will show that there exists a constant $C_M>0$ such that, for n large enough, $\mathbb{P}(\bar{Z}_n(\sqrt nA)\geq p)\geq C_M\mathbb{P}(E)$ . Let $\{Z^{i}_n\}_{n\geq 0}$ , $i\geq 1$ , be i.i.d. copies of $\{Z_n\}_{n\geq 0}$ and $W^{v}_{n-t_n-1}\,{:\!=}\,|Z^{v}_{n-t_n-1}|m^{-(n-t_n-1)}$ , and let $x_v$ be the displacement of particle v. By the Markov property, we have

\begin{align*} & \mathbb{P}\big(\bar{Z}_n(\sqrt nA)\geq p\big)\nonumber \\[4pt] & \geq \mathbb{P}\left(\vphantom{\frac{\sum\limits_{v\in Z^u_{1}}|Z^{v}_{n-t_n-1}|}{\sum\limits_{w\in H(u)}|Z^{w}_{n-t_n-1}|}}\textrm{there exists } u\in Z_{t_n} \textrm{ such that} \right. \nonumber \\[4pt] & \qquad \qquad \frac{\sum\limits_{v\in Z^u_{1}}Z^{v}_{n-t_n-1}(\sqrt nA-x_v-S_u)+\sum\limits_{w\in H(u)}Z^{w}_{n-t_n-1}(\sqrt nA-S_w)}{\sum\limits_{v\in Z^u_{1}}|Z^{v}_{n-t_n-1}|+\sum\limits_{w\in H(u)}|Z^{w}_{n-t_n-1}|}\geq p \textrm{ and} \nonumber \\[4pt] & \left. \qquad \qquad \frac{\sum\limits_{v\in Z^u_{1}}|Z^{v}_{n-t_n-1}|}{\sum\limits_{w\in H(u)}|Z^{w}_{n-t_n-1}|}>M \right) \nonumber \\ \end{align*}

(16) \begin{align} \nonumber\\[-30pt] & \geq \int_{\mathcal{E}}\mathbb{P}\left(\frac{\sum^k\limits_{i=1}Z^{i}_{n-t_n-1}(\sqrt nA-x_v-S_z)}{\sum^k\limits_{i=1}|Z^{i}_{n-t_n-1}|+\sum^r\limits_{j=1}|Z^{j}_{n-t_n-1}|}\geq p \textrm{ and} \right. \nonumber \\ & \left. \qquad \qquad \frac{\sum^k\limits_{i=1}|Z^{i}_{n-t_n-1}|}{\sum^r\limits_{j=1}|Z^{j}_{n-t_n-1}|}>M \right) \mathbb{P}(Z_{t_n}\in \textrm{d}\xi, \, |Z^{z}_{1}|\in \textrm{d} k, \, |H(z)|\in \textrm{d} r)\nonumber \\ & \geq \int_{\mathcal{E}}\mathbb{P}\left(\frac{\sum^k\limits_{i=1}Z^{i}_{n-t_n-1}(\sqrt nA-x_v-S_z)}{(1+M^{-1})\sum^k\limits_{i=1}|Z^{i}_{n-t_n-1}|}\geq p \textrm{ and} \right. \nonumber \\ & \left. \qquad \qquad \frac{\sum^k\limits_{i=1}{W^{i}_{n-t_n-1}}}{\sum^r\limits_{j=1}W^{j}_{n-t_n-1}}>M \right) \mathbb{P}\left(Z_{t_n}\in \textrm{d}\xi, \, |Z^{z}_{1}|\in \textrm{d} k, \, |H(z)|\in \textrm{d} r\right)\nonumber \\ & \,{=\!:}\, \int_{\mathcal{E}} P_n(\xi,k,r)\mathbb{P}(Z_{t_n}\in \textrm{d}\xi, \, |Z^{z}_{1}|\in \textrm{d} k, \, |H(z)|\in \textrm{d} r),\end{align}

where $\mathbb{P}(Z_{t_n}\in \textrm{d}\xi)$ represents the distribution of the point process $Z_{t_n}$ (for a serious definition of a point process’s distribution, see [Reference Bovier8, Section 2.1]). To finish the first step, it suffices to show that

(17) \begin{equation}\lim\limits_{n\rightarrow\infty}\inf\limits_{(\xi,k,r)\in \mathcal{E}}P_n(\xi,k,r)>0.\end{equation}

Since $A\in\mathcal{A}$ (see Assumption 1), we can write $A=\sum^l_{i=1}(a_i,b_i]$ for some natural number l, where $-\infty\leq a_i<b_i\leq+\infty$ . Let $A(x,\eta)\,{:\!=}\,\sum^l_{i=1}\left(a_i-x+\eta,b_i-x-\eta\right)$ . Since $S_z\in[(x-\eta)\sqrt n,(x+\eta)\sqrt n]$ , we can choose $\eta$ small enough such that

(18) \begin{equation}\frac{1}{1+M^{-1}}\nu(A(x,\eta))>p+\frac{\delta}{4} \quad \textrm{and} \quad \sqrt nA(x,\eta)\subset \sqrt nA-S_z.\end{equation}

Thus, by the central limit theorem, there exists a constant $C(M,h,x,\eta,\delta)>0$ such that, for $n>C(M,h,x,\eta,\delta)$ ,

(19) \begin{equation}\frac{1}{1+M^{-1}}\nu_{n-t_n}(\sqrt nA(x,\eta))>p+\frac{\delta}{8}.\end{equation}

Recall that $Z^{i}_{n-t_n-1}$ , $W^{i}_{n-t_n-1}$ , and $X_i$ , $1\leq i\leq k$ , are respectively i.i.d. copies of $Z_{n-t_n-1}$ , $W_{n-t_n-1}$ , and X. Since $\sqrt nA(x,\eta)\subset \sqrt nA-S_z$ , by (16), we have

\begin{align*}P_n(\xi,k,m)&\geq \mathbb{P}\left(\frac{\sum\limits^k_{i=1}Z^{i}_{n-t_n-1}(\sqrt nA(x,\eta)-X_i)}{(1+M^{-1})\sum\limits^k_{i=1}|Z^{i}_{n-t_n-1}|}\geq p \textrm{ and } \frac{\sum\limits^k_{i=1}{W^{i}_{n-t_n-1}}}{\sum\limits^{\lfloor\frac{k}{2M}\rfloor}_{j=1}W^{j}_{n-t_n-1}}>M\right)\\[3pt] &\,{=\!:}\,\mathbb{E}\left[{\bf 1}_{\{A_{n,k}\geq p\}}{\bf 1}_{\{B_{n,k}\geq M\}}\right].\nonumber\end{align*}

Fix $k> 2M$ (since $(\xi,k,r)\in\mathcal{E}$ ) and $\epsilon'\in(0,\eta)$ . A random variable $N(\epsilon', k)>0$ exists such that, for $n>N(\epsilon', k)$ and $1\leq i\leq k$ , $\sqrt nA(x,\eta+\epsilon')\subset\sqrt nA(x,\eta)-X_i\subset \sqrt nA(x,\eta-\epsilon')$ . Thus,

\begin{align*} &\frac{\sum\limits^k_{i=1}m^{-(n-t_n-1)}Z^{i}_{n-t_n-1}(\sqrt nA(x,\eta+\epsilon'))}{(1+M^{-1})\sum\limits^k_{i=1}m^{-(n-t_n-1)}|Z^{i}_{n-t_n-1}|}\leq A_{n,k} \\[3pt] &\quad \leq \frac{\sum\limits^k_{i=1}m^{-(n-t_n-1)}Z_{n-t_n-1}^{i}(\sqrt nA(x,\eta-\epsilon'))}{(1+M^{-1})\sum\limits^k_{i=1}m^{-(n-t_n-1)}|Z^{i}_{n-t_n-1}|}.\end{align*}

Since $\bar{Z}_n(\sqrt nA)\rightarrow\nu(A)$ and $W_n\rightarrow W$ , the above implies that $\lim_{n\rightarrow \infty}A_{n,k}= \frac{1}{1+M^{-1}}\nu(A(x,\eta))$ , $\mathbb{P}$ -a.s. On the other hand, it is easy to see that

\begin{equation*}\lim\limits_{n\rightarrow\infty}B_{n,k}= {\sum\limits^k_{i=1}{W^{i}}} \Big / {\sum\limits^{\lfloor\frac{k}{2M}\rfloor}_{j=1}W^{j}} \,{=\!:}\,B_k \qquad \mathbb{P}\textrm{{-}a.s.},\end{equation*}

where $W^i$ and $W^j$ are i.i.d. copies of $W\,{:\!=}\,\lim_{n\rightarrow\infty}|Z_n|/m^n$ . Hence, by the dominated convergence theorem and (18), for any fixed $k>2M$ we have

(20) \begin{equation}\lim\limits_{n\rightarrow \infty}\mathbb{E}\left[{\bf 1}_{\{A_{n,k}\geq p\}}{\bf 1}_{\{B_{n,k}\geq M\}}\right]=\mathbb{P}\left(B_k\geq M\right).\end{equation}

So, to achieve (17), it suffices to show that

(21) \begin{equation} \lim\limits_{n\rightarrow\infty}\sup_{k> 2M}\left|\mathbb{E}\left[{\bf 1}_{\{A_{n,k}\geq p\}}{\bf 1}_{\{B_{n,k}\geq M\}}\right]-\mathbb{P}\left(B_k\geq M\right)\right|=0 \end{equation}

and

(22) \begin{equation} \inf_{k>2M}\mathbb{P}\left(B_k>M\right)>0.\end{equation}

By the strong law of large numbers, we can easily obtain $\inf_{k>2M}\mathbb{P}\left(B_k>M\right)>0$ . For (20), observe that

(23) \begin{align} & \left|\mathbb{E}\left[{\bf 1}_{\{A_{n,k}\geq p\}}{\bf 1}_{\{B_{n,k}\geq M\}}\right]-\mathbb{P}\left(B_k\geq M\right)\right|\nonumber\\[1pt]& \leq\mathbb{P}(A_{n,k}<p)+\mathbb{P}(B_{n,k}\geq M,B_k<M)+\mathbb{P}(B_{n,k}<M,B_k\geq M).\end{align}

For the first term on the right-hand side of (23), let $\xi=\sum^k_{i=1}\delta_{X_i}$ be a point process (recall that the $X_i$ , $i\geq 1$ , are i.i.d. copies of X). By Remark 7 and (19), there exists a constant $C_1>0$ depending on $\delta$ and $\beta$ such that, for $n>C(M,h,x,\eta,\delta)$ ,

\begin{align*}\mathbb{P}(A_{n,k}<p) &=\mathbb{P}\left(\frac{\sum\limits^k_{i=1}Z^{i}_{n-t_n-1}(\sqrt nA(x,\eta)-X_i)}{(1+M^{-1})\sum\limits^k_{i=1}|Z^{i}_{n-t_n-1}|}<p \right)\nonumber \\ &=\mathbb{P}\left(\bar{Z}^{\xi}_{n-t_n-1}(\sqrt nA(x,\eta))<p\big(1+M^{-1}\big)\right)\nonumber \\ &\leq \mathbb{P}\left(\bar{Z}^{\xi}_{n-t_n-1}(\sqrt nA(x,\eta))<\nu_{n-t_n}(\sqrt nA(x,\eta))-\frac{\delta}{8}\right)\nonumber \end{align*}

(24) \begin{align} \nonumber\\[-40pt] &=\mathbb{P}\left(\bar{Z}^{\xi}_{n-t_n-1}(\sqrt nA(x,\eta))<\mathbb{E}\left[\frac{1}{|\xi|}\sum_{z\in\xi}\nu_{n-t_n-1}(\sqrt nA(x,\eta)-S_z)\right]-\frac{\delta}{8}\right)\nonumber \\ &\leq C_1k^{1-\beta},\end{align}

where the third equality holds since

\begin{align*} \nu_{n-t_n}\left(\sqrt nA(x,\eta)\right)&=\mathbb{P}\left(X_1+X_2 + \cdots + X_{n-t_n}\in\sqrt nA(x,\eta)\right)\\ &=\mathbb{E}[\nu_{n-t_n-1}(\sqrt nA(x,\eta)-S_z)].\nonumber\end{align*}

For the second term on the right-hand side of (23), by the strong law of large numbers we have

(25) \begin{equation} \mathbb{P}(B_{n,k}\geq M,B_k<M)\leq\mathbb{P}(B_k<M)\rightarrow 0 \qquad \textrm{as}\ k\rightarrow\infty.\end{equation}

For the third term on the right-hand side of (23), there exists a constant $C(M,\beta)>0$ such that, for any $k>2M$ and $n\geq1$ ,

(26) \begin{align}\mathbb{P}(B_{n,k}<M & , B_k\geq M)\nonumber \\ &\leq\mathbb{P}(B_{n,k}<M)\nonumber \\ &=\mathbb{P}\left(\frac{\sum\limits^k_{i=1}{W^{i}_{n-t_n-1}}}{\sum\limits^{\lfloor\frac{k}{2M}\rfloor}_{j=1}W^{j}_{n-t_n-1}}<M\right)\nonumber \\ &\leq\mathbb{P}\left(\sum\limits^k_{i=1}W^{i}_{n-t_n-1}<M\sum\limits^{\lfloor\frac{k}{2M}\rfloor}_{j=1}W^{j}_{n-t_n-1},\sum\limits^{\lfloor\frac{k}{2M}\rfloor}_{j=1}W^{j}_{n-t_n-1}<\frac{k}{2M}\frac{3}{2}\right) \nonumber \\ & \quad + \mathbb{P}\left(\sum\limits^{\lfloor\frac{k}{2M}\rfloor}_{j=1}W^{j}_{n-t_n-1}\geq\frac{k}{2M}\frac{3}{2}\right)\nonumber \\ &\leq\mathbb{P}\left(\sum\limits^k_{i=1}W^{i}_{n-t_n-1}<\frac{3k}{4}\right)+\mathbb{P}\left(\sum\limits^{\lfloor\frac{k}{2M}\rfloor}_{j=1}W^{j}_{n-t_n-1}\geq\frac{k}{2M}\frac{3}{2}\right)\nonumber \\ &\leq\mathbb{P}\left(\sum\limits^k_{i=1}\left(1-W^{i}_{n-t_n-1}\right)>\frac{k}{4}\right)+\mathbb{P}\left(\sum\limits^{\lfloor\frac{k}{2M}\rfloor}_{j=1}\left(W^{j}_{n-t_n-1}-1\right)\geq\frac{k}{2M}\frac{1}{2}\right)\nonumber \\ &\leq C(M,\beta)k^{1-\beta},\end{align}

where, for the last inequality, we used exactly the same arguments as for bounding $I_1$ and $I_2$ in Lemma 4. Plugging (25), (26), and (27) into (23), we obtain that for every $\epsilon>0$ there exists a constant $C(\epsilon,M,\beta,\delta)$ such that, for $k>C(\epsilon,M,\beta,\delta)$ and $n>C(M,h,x,\eta,\delta)$ , $\left|\mathbb{E}\left[{\bf 1}_{\{A_{n,k}\geq p\}}{\bf 1}_{\{B_{n,k}\geq M\}}\right]-\mathbb{P}\left(B_k\geq M\right)\right|<\epsilon$ . Combining this with (20), there exists some large constant $C(\epsilon,M,\beta,h,x,\eta,\delta,p)$ such that, for $n>C(\epsilon,M,\beta,h,x,\eta,\delta,p)$ and $k>2M$ , $\left|\mathbb{E}\left[{\bf 1}_{\{A_{n,k}\geq p\}}{\bf 1}_{\{B_{n,k}\geq M\}}\right]-\mathbb{P}\left(B_k\geq M\right)\right|<\epsilon$ . Thus, (21) holds. This, combined with (22) and (16), implies that there exist positive constants $C_M$ and $C(M,\beta,h,x,\eta,\delta,p)$ such that, for $n>C(M,\beta,h,x,\eta,\delta,p)$ , $\mathbb{P}\big(\bar{Z}_n(\sqrt nA)\geq p\big)\geq C_M\mathbb{P}(E)$ , which completes the first step.

For the second step, we will give a lower bound of $\mathbb{P}(E)$ . Define $\mathcal{F}_{t_n}\,{:\!=}\,\sigma(Z_{i}$ , $1\leq i\leq t_n)$ . By the first step, for $n>C(M,\beta,h,x,\eta,\delta,p)$ ,

(27) \begin{align}&\mathbb{P}\left(\bar{Z}_n(\sqrt nA)\geq p\right)\nonumber \\ &\geq C_M\mathbb{P}\left(\!\textrm{there exists }u\in Z_{t_n} \textrm{ such that } S_u\in[(x-\eta)\sqrt n,(x+\eta)\sqrt n],|Z^{u}_{1}|>2M\!\!\sum_{\scriptsize\begin{array}{c}v\neq u\, v\in Z_{t_n}\end{array}}\!\!\!|Z^{v}_{1}|\right)\nonumber \\ &=C_M\mathbb{E}\Bigg[\sum_{u\in Z_{t_n}}{\bf 1}_{\{S_u\in[(x-\eta)\sqrt n,(x+\eta)\sqrt n]\}}{\bf 1}_{\{|Z^{u}_{1}|>2M\sum\limits_{v\neq u, v\in Z_{t_n}}|Z^{v}_{1}|\}}\Bigg]\nonumber \\ &=C_M\mathbb{E}\Bigg[\sum_{u\in Z_{t_n}}{\bf 1}_{\{S_u\in[(x-\eta)\sqrt n,(x+\eta)\sqrt n]\}}\mathbb{E}\left[{\bf 1}_{\{|Z^{u}_{1}|>2M\sum\limits_{v\neq u, v\in Z_{t_n}}|Z^{v}_{1}|\}} \ \Big| \ \mathcal{F}_{t_n}\right]\Bigg],\end{align}

where the first equality follows from the fact that the random variable inside the expectation can only be 0 or 1 (since there exists at most one individual satisfying $|Z^{u}_{1}|>2M\sum_{v\neq u, v\in Z_{t_n}}|Z^{v}_{1}|$ for $u\in Z_{t_n}$ ). Let $k_i$ , $i\geq 0$ , be i.i.d. copies of $|Z_1|$ , and independent of $Z_{t_n}$ . Since $|Z_1|\sim \textrm{Pareto}(\beta)$ , there exists a constant $C_4>0$ such that $\mathbb{P}(|Z_1|>x)>C_4 x^{-\beta}$ for all $x>1$ . Recall the well-known fact that if U and V are independent random variables then, for any bounded measurable function F(x,y), we have $\mathbb{E}[F(U,V) \mid \sigma(V)]=\mathbb{E}[F(U,v)]\big|_{v=V}$ . Using this fact, we have

(28) \begin{align}\mathbb{E}\left[{\bf 1}_{\{|Z^{u}_{1}|>2M\sum\limits_{\scriptsize\begin{array}{c}v\neq u \\ v\in Z_{t_n}\end{array}}|Z^{v}_{1}|\}} \ \big| \ \mathcal{F}_{t_n}\right]&=\mathbb{E}\left[{\bf 1}_{\{k_0>2M\sum^{|Z_{t_n}|-1}_{i=1}k_i\}} \ \Big | \ \mathcal{F}_{t_n}\right]\nonumber \\ &=\mathbb{E}\left[{\bf 1}_{\{k_0>2M\sum^{j}_{i=1}k_i\}}\right]\Big{|}_{j=|Z_{t_n}|-1}\nonumber \\ &=\mathbb{E}\left[\mathbb{E}\left[{\bf 1}_{\{k_0>2M\sum^{j}_{i=1}k_i\}}\Big{|}\sigma(k_1,\ldots,k_j)\right]\right]\Big{|}_{j=|Z_{t_n}|-1}\nonumber \\ &\geq C_4\mathbb{E}\left[\left(2M\sum^{j}_{i=1}k_i\right)^{-\beta}\right]\Bigg{|}_{j=|Z_{t_n}|-1}\nonumber \\ &= C_4\mathbb{E}\left[\left(2M\sum^{|Z_{t_n}|-1}_{i=1}k_i\right)^{-\beta} \ \Big | \ \sigma(|Z_{t_n}|)\right]\nonumber \\ &\geq C_4(2M)^{-\beta}\mathbb{E}\left[ |Z_{t_n+1}|^{-\beta} \ \big | \ \mathcal{F}_{t_n}\right].\end{align}

Plugging (28) into (27) yields, for $n>C(M,\beta,h,x,\eta,\delta,p)$ ,

(29) \begin{align}&\mathbb{P}\left(\bar{Z}_n(\sqrt nA)\geq p\right)\nonumber \\ &\geq C_MC_4(2M)^{-\beta}\mathbb{E}\left[\mathbb{E}\left[\sum_{u\in Z_{t_n}}{\bf 1}_{\{S_u\in[(x-\eta)\sqrt n,(x+\eta)\sqrt n]\}} |Z_{t_n+1}|^{-\beta} \ \Big | \ \mathcal{F}_{t_n}\right]\right]\nonumber \\ &= C_MC_4(2M)^{-\beta}\mathbb{E}\left[\frac{|Z_{t_n}|}{|Z_{t_n+1}|^{\beta}}\right]\nu_{t_n}\left([(x-\eta)\sqrt n,(x+\eta)\sqrt n]\right),\end{align}

where the last equality follows from the fact that the branching and motion are independent. By Fatou’s lemma, for n large enough,

(30) \begin{align}\mathbb{E}\left[\frac{|Z_{t_n}|}{|Z_{t_n+1}|^{\beta}}\right]=m^{-\beta}m^{-(\beta-1)t_n}\mathbb{E}\left[\frac{W_{t_n}}{(W_{t_{n+1}})^{\beta}}\right]\geq 0.9 \, \mathbb{E}\left[W^{1-\beta}\right] m^{-\beta}m^{-(\beta-1)t_n}.\end{align}

Plugging (30) into (29) yields the existence of a constant $C'(\epsilon,M,\beta,h,x,\eta,\delta,p)$ such that, for $n>C'(\epsilon,M,\beta,h,x,\eta,\delta,p)$ ,

\begin{align*}\mathbb{P}(\bar{Z}_n&(\sqrt nA)\geq p) \\[3pt] &\geq C_MC_4(2M)^{-\beta}0.9m^{-\beta}\mathbb{E}\left[W^{1-\beta}\right]m^{-(\beta-1)t_n}\nu_{t_n}\big([(x-\eta)\sqrt n,(x+\eta)\sqrt n]\big)\\[3pt] &\geq C_MC_4(2M)^{-\beta}0.9m^{-\beta}\mathbb{E}\left[W^{1-\beta}\right] m^{-(\beta-1)h\sqrt n}\exp\bigg[{-}\bigg(\Lambda\bigg(\frac{x-\eta}{h}\bigg)+\epsilon\bigg)h\sqrt n\bigg],\nonumber\end{align*}

where the last inequality follows by Cramér’s theorem. Hence, for every $\epsilon$ , $\eta$ small enough and $h>0$ , the above yields

\begin{equation*}\liminf_{n\rightarrow\infty}\frac{1}{\sqrt{n}\,}\log\mathbb{P}(\bar{Z}_n(\sqrt nA)\geq p)\geq-\left\{h(\beta-1)\log m+\left(\Lambda\left(\frac{x-\eta}{h}\right)+\epsilon \right)h\right\}.\end{equation*}

First, let $\epsilon\rightarrow 0$ , and then maximize the lower bound with h, to finally obtain

\begin{equation*}\liminf_{n\rightarrow\infty}\frac{1}{\sqrt{n}\,}\log\mathbb{P}(\bar{Z}_n(\sqrt nA)\geq p)\geq-\inf_{h>0}\left\{h(\beta-1)\log m+\Lambda\left(\frac{I_A(p)}{h}\right)h\right\},\end{equation*}

which concludes the desired lower bound for the case of $\beta-1<\frac{-\log{p_1}}{\log m}$ . For $\beta-1\geq \frac{-\log{p_1}}{\log m}$ , [Reference Chen and He13, Lemma 3.1] gives

\begin{equation*} \liminf_{n\rightarrow\infty}\frac{1}{\sqrt{n}\,}\log\mathbb{P}(\bar{Z}_n(\sqrt nA)\geq p)\geq -\inf_{h>0}\left\{h\log \frac{1}{p_1}+\Lambda\left(\frac{I_A(p)}{h}\right)h\right\}. \end{equation*}

We next consider the upper bound in Theorem 1. By the definition of $I_A(p)$ , for every $\eta\in(0,I_A(p))$ there exists $\delta>0$ such that

(31) \begin{equation}\sup\limits_{|y|\leq I_A(p)-\eta}\nu(A-y)\leq p-\delta.\end{equation}

Set $B_n\,{:\!=}\,[(-I_A(p)+\eta)\sqrt n,(I_A(p)-\eta)\sqrt n]$ , $\mathcal{M}_1\,{:\!=}\,\big\{\xi\in \mathcal{M}:\frac{\xi(B_n^\textrm{c})}{|\xi|}\leq \frac{\delta}{2}\big\}$ , and $t_n\,{:\!=}\,\lfloor h\sqrt n\rfloor$ for some $h>0$ . For every $\xi\in\mathcal{M}_1$ and n large enough, we have

\begin{align*} &\frac{1}{|\xi|}\sum\limits_{z\in\xi}\nu_{n-{t_n}}(\sqrt n A-S_z)+\frac{\delta}{4} \\[2pt]& \leq \frac{1}{|\xi|}\sum\limits_{\scriptsize\begin{array}{c}z\in\xi \\ z\in B^\textrm{c}_n \end{array}}\nu_{n-{t_n}}(\sqrt n A-S_z) + \frac{1}{|\xi|}\sum\limits_{\scriptsize\begin{array}{c}z\in\xi \\ z\in B_n \end{array}}\nu_{n-{t_n}}(\sqrt n A-S_z)+\frac{\delta}{4}\\[2pt] & \leq \frac{\xi(B_n^\textrm{c})}{|\xi|}+\sup_{|z|\leq I_A(p)-\eta}\nu_{n-{t_n}}(\sqrt {n} (A-z))+\frac{\delta}{4}\\[2pt] & \leq \frac{\delta}{2}+p-\frac{3\delta}{4}+\frac{\delta}{4}\\[2pt] & = p,\nonumber\end{align*}

where the third inequality follows by the generalized central limit theorem (see [Reference Louidor and Perkins32, Lemma 2.2]) and (31). Hence, by Lemma 4 and the Markov inequality, for n large enough,

(32) \begin{align}&\mathbb{P}\left(\bar{Z}_{n}(\sqrt n A)\geq p\right)\nonumber \\ &\leq \mathbb{P}\left(\bar{Z}_{t_n}(B_n^\textrm{c})\geq \frac{\delta}{2}\right)+\mathbb{P}\left(\bar{Z}_{t_n}(B_n^\textrm{c})\leq \frac{\delta}{2},\bar{Z}_{n}(\sqrt n A)\geq p\right)\nonumber \\ &\leq \mathbb{E}\left[\frac{2}{\delta}\frac{Z_{t_n}(B_n^\textrm{c})}{|Z_{t_n}|}\right]+\int_{\mathcal{M}_1}\!\!\!\!\mathbb{P}\Bigg(\bar{Z}^\xi_{n-t_n}(\sqrt nA)\geq \frac{1}{|\xi|}\sum_{z\in\xi}\nu_{n-t_n}(\sqrt nA-S_z)+\frac{\delta}{4}\Bigg)\mathbb{P}(Z_{t_n}\in \textrm{d}\xi)\nonumber \\ &\leq \mathbb{E}\left[\frac{2}{\delta}\frac{Z_{t_n}(B_n^\textrm{c})}{|Z_{t_n}|}\right]+C_1\mathbb{E}\big[|Z_{t_n}|^{-(\beta-1)}\big].\end{align}

For the first term on the right-hand side of (32), define $\mathcal{G}_{t_n}\,{:\!=}\,\sigma(|Z_i|, 1\leq i\leq t_n)$ . Since the branching and motion are independent we have, for n large enough,

(33) \begin{align}\mathbb{E}\left[\frac{2}{\delta}\frac{Z_{t_n}(B_n^\textrm{c})}{|Z_{t_n}|}\right]&=\mathbb{E}\left[\mathbb{E}\left[\frac{2}{\delta}\frac{Z_{t_n}(B_n^\textrm{c})}{|Z_{t_n}|}\ \big| \ \mathcal{G}_{t_n}\right]\right] \nonumber \\ &= \frac{4}{\delta}\nu_{t_n}\left((I_A(p)-\eta)\sqrt n\right)\nonumber \\ &\leq\frac{4}{\delta} \exp\bigg[{-}\left(\Lambda\left(\frac{I_A(p)-\eta}{h}\right)-\epsilon\right)h\sqrt n\bigg],\end{align}

where the second equality comes from the symmetry of the step size, and the last inequality follows by Cramér’s theorem. For the second term on the right-hand side of (32), by Lemma 2, for n large enough,

(34) \begin{equation}\mathbb{E}\left[|Z_{t_n}|^{-(\beta-1)}\right]\leq\begin{cases}2C_0 m^{-(\beta-1)h\sqrt n}, &\beta-1\geq \dfrac{-\log{p_1}}{\log m}; \\ \\[-7pt] 2C_0 p_1^{h\sqrt n}, &0<\beta-1<\dfrac{-\log{p_1}}{\log m}.\end{cases}\end{equation}

Plugging (33) and (34) into (32) yields that if $\beta-1\geq \frac{-\log{p_1}}{\log m}$ then

\begin{equation*}\mathbb{P}\left(\bar{Z}_{n}(\sqrt n A)\geq p\right)\leq \frac{4}{\delta}\exp\bigg[-\left(\Lambda\left(\frac{I_A(p)-\eta}{h}\right)-\epsilon\right)h\sqrt n\bigg] + 2C_0 C_1(p,\delta,h)m^{-(\beta-1)h\sqrt n},\end{equation*}

and if $0<\beta-1<\frac{\log \frac{1}{p_1}}{\log m}$ then

\begin{equation*}\mathbb{P}\left(\bar{Z}_{n}(\sqrt n A)\geq p\right)\leq \frac{4}{\delta}\exp\bigg[-\left(\Lambda\left(\frac{I_A(p)-\eta}{h}\right)-\epsilon\right)h\sqrt n\bigg] + 2C_0 C_1(p,\delta,h)p_1^{h\sqrt n}.\end{equation*}

So, the upper bound follows by optimizing h on $(0,+\infty)$ .

Proof. We start with the lower bound. By the continuity of $I_A(\!\cdot\!)$ at p, for every $\epsilon>0$ there exist $\eta, \delta>0$ such that, for some $|x|\leq I_A(p)+\epsilon$ , $\inf_{y\in[x-\eta,x+\eta]}\nu(A-y)\geq p+\delta$ . Without loss of generality, we write $A=\sum^h_{j=1}(a_j,b_j]$ . Set $A(x,\eta)\,{:\!=}\,\sum^h_{j=1}\left(a_j-(x-\eta),b_j-(x+\eta)\right)$ , $A(\epsilon)\,{:\!=}\,\sum^h_{j=1}\left(a_j+\epsilon,b_j-\epsilon\right)$ . Obviously, we can choose $\epsilon, \eta$ small enough such that $\nu(A(x,\eta))>p+\frac{\delta}{2}$ , $\nu(A(\epsilon))>\nu(A)-\frac{\delta}{2}$ . Set $\mathcal{E}\,{:\!=}\,\big\{\xi\in\mathcal{M} \,{:}\, \xi=\sum^b_{i=1}\delta_{{x_i}\sqrt{n}}$ , where $x_1\in(x-\eta,x+\eta)$ , $x_i\in(-\epsilon,\epsilon)$ , $i=2, \ldots, b\big\}$ . By the Markov property, $\mathbb{P}(\bar{Z}_{n}(\sqrt n A)\geq p)\geq\int_\mathcal{E} \mathbb{P}\big(\bar{Z}^{\xi}_{n-1}(\sqrt n A)\geq p\big)\mathbb{P}(Z_1\in \textrm{d}\xi)$ . For every $\xi\in \mathcal{E}$ , it is easy to see that $A-x_1\supset A(x,\eta)$ , $A-x_i\supset A(\epsilon)$ , $i=2, \ldots, b$ . Hence,

(36) \begin{align} \mathbb{P}(\bar{Z}_{n}&(\sqrt n A) \geq p) \geq \int_\mathcal{E}\mathbb{P}\left(\bar{Z}^{\xi}_{n-1}(\sqrt nA)\geq p\right)\mathbb{P}(Z_1\in \textrm{d}\xi)\nonumber\\[2pt] & \geq \int_\mathcal{E} \mathbb{P}\left(\frac{Z^{1}_{n-1}(\sqrt n A(x,\eta))+\sum^b_{i=2}Z^{i}_{n-1}(\sqrt n A(\epsilon))}{\sum^b_{i=1}|Z^{i}_{n-1}|}\geq p\right)\mathbb{P}(Z_1\in \textrm{d}\xi)\nonumber\\[2pt] & = \mathbb{P}\left(\frac{Z^{1}_{n-1}(\sqrt n A(x,\eta))+\sum^b_{i=2}Z^{i}_{n-1}(\sqrt n A(\epsilon))}{\sum^b_{i=1}|Z^{i}_{n-1}|}\geq p\right)\mathbb{P}(Z_1\in \mathcal{E}), \end{align}

where $Z^{i}_{n-1}(\!\cdot\!)$ , $1\leq i\leq b$ , are i.i.d. copies of $Z_{n-1}(\!\cdot\!)$ . Since $\bar{Z}_{n}(\sqrt n A)\rightarrow \nu(A)$ and $|Z_n|m^{-n}\rightarrow W$ almost surely, we have, as $n\to\infty$ ,

(37) \begin{align} & \frac{Z^{1}_{n-1}(\sqrt n A(x,\eta))+\sum^b_{i=2}Z^{i}_{n-1}(\sqrt n A(\epsilon))}{\sum^b_{i=1}|Z^{i}_{n-1}|}\nonumber\\[2pt] &\qquad\qquad\qquad\qquad\qquad \longrightarrow \frac{\sum^b_{i=2}\nu(A(\epsilon))W_i+\nu(A(x,\eta))W_1}{\sum^b_{i=1}W_i} \qquad \mathbb{P}\textrm{-a.s.}, \end{align}

where $W_i$ , $1\leq i\leq b$ , are i.i.d. copies of W. If $b<B$ , then $W_i$ has a continuous density on $(0,+\infty)$ (see [Reference Athreya and Ney4, Chapter II, Lemma 2]). In this case, by the dominated convergence theorem and (37),

(38) \begin{equation} \lim\limits_{n\rightarrow\infty}\mathbb{P}\left(\frac{Z^{1}_{n-1}(\sqrt n A(x,\eta))+\sum^b_{i=2}Z^{i}_{n-1}(\sqrt n A(\epsilon))}{\sum^b_{i=1}|Z^{i}_{n-1}|}\geq p\right)\,{=\!:}\,C(\epsilon,\eta,x)>0. \end{equation}

Since $\mathbb{P}(X>x)\sim\kappa x^{-\alpha}$ , there exists a constant $C(\kappa,x,\epsilon,\eta)\,{=\!:}\, C_7>0$ such that, for n large enough,

(39) \begin{equation} \mathbb{P}(Z_1\in\mathcal{E})\geq C_7p_bn^{-\alpha/2}. \end{equation}

Plugging (38) and (39) into (36) yields, for n large enough, $\mathbb{P}(\bar{Z}_{n}(\sqrt n A)\geq p)\geq 0.9C(\epsilon,\eta,x)C_7p_bn^{-\alpha/2}$ , which implies the desired lower bound if $b<B$ .

If $b=B$ and A is unbounded, without loss of generality we assume $A=(a,+\infty)$ . Set $\mathcal{E'}\,{:\!=}\,\big\{\xi\in\mathcal{M} \,{:}\, \xi=\sum^b_{i=1}\delta_{{x_i}\sqrt{n}}$ , where $x_1\in[t\sqrt n,+\infty)$ , $x_i\in(-\epsilon,\epsilon)$ , $i=2, \ldots, b\big\}$ , where t is some positive constant. Using similar arguments to above, we obtain

\begin{equation*} \mathbb{P}(\bar{Z}_{n}(\sqrt n A)\geq p)\geq\mathbb{P}\left(\frac{Z^{1}_{n-1}(\sqrt n(a-t,+\infty))+\sum^b_{i=2}Z^{i}_{n-1}(\sqrt n A(\epsilon))}{\sum^b_{i=1}Z^{i}_{n-1}}\geq p\right)\mathbb{P}(Z_1\in \mathcal{E'}). \end{equation*}

For any $p\in\big(\nu(A),\nu(A)+\frac{1-\nu(A)}{b}\big)$ , there exists $\delta'>0$ such that

(40) \begin{equation} p<\frac{(b-1)(\nu(A)-\delta')+1-\delta'}{b}. \end{equation}

If we choose $\epsilon$ small enough and t large enough, then $\nu(A(\epsilon))\geq \nu(A)-\delta'$ , $\nu((a-t,+\infty))>1-\delta'$ . This, together with (37) and (40), shows that, almost surely,

\begin{equation*}\frac{Z^{1}_{n-1}(\sqrt n(a-t,+\infty))+\sum^b_{i=2}Z^{i}_{n-1}(\sqrt n A(\epsilon))}{\sum^b_{i=1}|Z^{i}_{n-1}|}\rightarrow \frac{(b-1)\nu(A(\epsilon))+\nu((a-t,+\infty))}{b}>p,\end{equation*}

which implies the desired lower bound.

For the upper bound, set $t_n\,{:\!=}\,\lfloor c\log n\rfloor$ for some $c>0$ . By copying the arguments from (31) to (32), there exist constants $\delta, \eta >0$ such that, for n large enough, $\mathbb{P}\left(\bar{Z}_{n}(\sqrt n A)\geq p\right)\leq \frac{4}{\delta}\mathbb{P}\left(S_{t_n}\geq(I_A(p)-\eta)\sqrt n\right)+ C_1\mathbb{E}\big[|Z_{t_n}|^{-(\beta-1)}\big]$ , where the last inequality follows from Lemma 4. For the first term of the right-hand side of the latter inequality, by (8), there exists a constant $C_8>0$ depending on c, $\kappa$ , $\alpha$ , and $\eta$ such that, for n large enough, $\frac{4}{\delta}\mathbb{P}\left(S_{t_n}\geq(I_A(p)-\eta)\sqrt n\right) \leq C_8 n^{-\frac{\alpha}{2}}\log n$ . For the second term, by Lemma 2, for n large enough, $\mathbb{E}\left[|Z_{t_n}|^{-(\beta-1)}\right]\leq 2C_0\left(m^{-t_n(\beta-1)}+p_1^{t_n}\right)$ . Hence, for n large enough, $\mathbb{P}\left(\bar{Z}_{n}(\sqrt n A)\geq p\right)\leq C_8n^{-\frac{\alpha}{2}}\log n+2C_0\left(m^{-t_n(\beta-1)}+p_1^{t_n}\right)$ . Since $t_n=\lfloor c\log n\rfloor$ , if we choose $c>\frac{\alpha}{2(\beta-1)\log m}\vee\frac{\alpha}{-2\log p_1}$ then, for large n, $\mathbb{P}\left(\bar{Z}_{n}(\sqrt n A)\geq p\right)\leq 2C_8n^{-\frac{\alpha}{2}}\log n$ . Thus, we have

\begin{align*} \limsup\limits_{n\rightarrow\infty}\frac{1}{\log n}\log\mathbb{P}(\bar{Z}_n(\sqrt nA)\geq p)\leq-\frac{\alpha}{2}.\\[-40pt] \end{align*}

Proof. Considering the lower bound, we first examine $|Z_1|$ with a Pareto tail. If $\beta-1\geq \frac{-\log{p_1}}{\log m}$ then from [Reference Chen and He13, Lemma 3.4] we have $\liminf_{n\rightarrow\infty}\frac{1}{n}\log\mathbb{P}(\bar{Z}_n(\sqrt nA)\geq p)\geq J_A(p)\log{p_1}$ . Now we shall consider the case of $0<\beta-1<\frac{-\log{p_1}}{\log m}$ . Since $J_A(p)$ is continuous at p, for any $\epsilon>0$ , there exists $\delta>0$ such that, for $r\,{:\!=}\,J_A(p+2\delta)\in(0,1)$ , we have $|r-J_A(p)|<\epsilon$ . Moreover, by the definition of $J_A(p)$ , there exists $x\in\mathbb{R}$ such that

\begin{equation*}\nu\left(\frac{A-x}{\sqrt{1-r}\,}\right)\geq p+\frac{3}{2}\delta.\end{equation*}

So, for any small $\eta>0$ ,

(41) \begin{equation} \nu\left(\bigcap\limits_{y\in[x-\eta,x+\eta]}\frac{A-y}{\sqrt{1-r}\,}\right)\geq p+\delta. \end{equation}

Set $t_n\,{:\!=}\,\lfloor rn\rfloor$ , and for large M set $E\,{:\!=}\,\big\{$ there exists $u\in Z_{t_n}$ such that $S_u\in[(x-\eta)\sqrt n,(x+\eta)\sqrt n]$ , $|Z^{u}_{1}|>2M\sum_{v\neq u,v\in Z_{t_n}}|Z^{v}_{1}|\big\}$ . Similar to (18), from (41) and $S_u\in[(x-\eta)\sqrt n,(x+\eta)\sqrt n]$ , we can choose $\eta$ small enough such that

(42) \begin{align} \frac{1}{1+M^{-1}}\nu\left(\frac{A(x,\eta)}{\sqrt{1-r}\,}\right)>p+\frac{\delta}{4} , \qquad \sqrt nA(x,\eta)\subset \sqrt nA-S_u, \end{align}

where $A(x,\eta)\,{:\!=}\,\sum^l_{i=1}\left(a_i-x+\eta,b_i-x-\eta\right).$ By [Reference Louidor and Perkins32, Lemma 2.2], it follows that

\begin{equation*}\lim_{n\to\infty}\nu_{n-t_n}(\sqrt nA(x,\eta))=\nu\left(\frac{A(x,\eta)}{\sqrt{1-r}\,}\right).\end{equation*}

This, combined with (42), shows that there exists a constant $C(M,r,x,\eta,\delta)>0$ such that, for $n>C(M,r,x,\eta,\delta)>0$ , $\frac{1}{1+M^{-1}}\nu_{n-t_n}(\sqrt nA(x,\eta))>p+\frac{\delta}{8}$ , which plays the same role in this proof as (19) in Theorem 1. Using similar arguments to the proof of the lower bound in Theorem 1, there exists a constant $C_M>0$ such that, for n large enough,

(43) \begin{align} \mathbb{P}(\bar{Z}_n(\sqrt nA)\geq p) &\geq C_M\mathbb{P}\left(E\right) \nonumber\\[3pt] &\geq C_M 0.9C_0m^{-t_n(\beta-1)}\nu_{t_n}\left([(x-\eta)\sqrt n,(x+\eta)\sqrt n]\right)\nonumber\\[3pt] &\geq 0.9^2C_MC_0C_9m^{-t_n(\beta-1)}, \end{align}

where the last inequality holds since $\lim_{n\rightarrow\infty}\nu_{t_n}\left([(x-\eta)\sqrt n,(x+\eta)\sqrt n]\right){=}C(r,x,\eta)\,{=\!:}\,C_9{>}0$ . Taking limits in (43) yields $\liminf_{n\rightarrow\infty}\frac{1}{n}\log\mathbb{P}(\bar{Z}_n(\sqrt nA)\geq p)\geq-(J_A(p)+\epsilon)(\beta-1)\log m$ . Then, the desired lower bound follows by letting $\epsilon \rightarrow 0$ .

We now consider the upper bound. For $\epsilon \in(0,J_A(p))$ , set $t_n\,{:\!=}\,\lfloor(J_A(p)-\epsilon)n\rfloor$ . By the definition of $J_A(p)$ there exists $\delta>0$ such that, for $\epsilon'\in[\epsilon,2\epsilon]$ ,

(44) \begin{equation} \sup_{y\in\mathbb{R}}\nu\left(\frac{A-y}{\sqrt{1-J_A(p)+\epsilon'}\,}\right)\leq p-\delta. \end{equation}

Thus, for any $\xi\in\mathcal{M}$ ,

\begin{align*} \frac{1}{|\xi|}\sum_{y\in\xi}\nu_{n-t_n}(\sqrt nA-S_y)+\frac{\delta}{2}&\leq\frac{1}{|\xi|}\sum_{y\in\xi}\nu\left(\frac{\sqrt n}{\sqrt{n-t_n}\,}A-\frac{S_y}{\sqrt{n-t_n}\,}\right)+\frac{3\delta}{4}\\ &\leq p-\frac{3\delta}{4}+\frac{3\delta}{4} =p,\nonumber \end{align*}

where the first inequality follows from the generalized central limit theorem, and the second inequality follows from (44) and the fact that $g(u,v)=\nu(uA+v)$ is a continuous function on $\mathbb{R}^2$ . Thus, for n large enough,

\begin{equation*} \mathbb{P}(\bar{Z}_n(\sqrt nA)\geq p) \leq\int_{\mathcal{M}}\mathbb{P}\bigg(\bar{Z}^\xi_{n-t_n}(\sqrt nA)\geq \frac{1}{|\xi|}\sum_{y\in\xi}\nu_{n-t_n}(\sqrt nA-S_y)+\frac{\delta}{2}\bigg)\mathbb{P}(Z_{t_n}\in \textrm{d}\xi). \end{equation*}

Hence, by Lemma 4, there exists $C_1>0$ such that, for large n, $\mathbb{P}(\bar{Z}_n(\sqrt nA)\geq p)\leq C_1\mathbb{E}\left[|Z_{t_n}|^{-(\beta-1)}\right]$ . As a consequence, the upper bound follows by Lemma 2.

If $|Z_1|$ has a Weibull tail, similar arguments to those above, together with Lemmas 3 and 5, get the desired results.