2.1. Heuristics for Theorem 1

Let $p\in (0, \, 1)$ . Write

$$a_n(k) \;:\!=\; {{\mathbb P}}(D_n =k), \quad n\ge 0, \ k\ge 1.$$

By (1.1), we get, for $k\ge 1$ ,

(2.1) \begin{equation} a_{n+1} (k)=p \sum_{1\leq i<k} a_n(i) a_n(k-i) + (1-p) \Bigg( 2a_n(k) \Bigg( 1-\sum_{1\leq i<k} a_n(i) \Bigg) - a_n(k)^2 \Bigg) . \end{equation}

Let $p=p_c+\varepsilon = \frac12 + \varepsilon$ and assume that for large n and for k such that $\log k = \mathcal{O} (\sqrt{n})$ that $a_n(k)$ takes the following scaling form:

(2.2) \begin{equation}a_n(k) = \frac{1}{k\sqrt{n}} f\bigg( n, \, \frac{\log k}{\sqrt{n}} \bigg),\end{equation}

with an appropriate bivariate function $(t, \, x) \mapsto f(t, \, x)$ . This scaling was already used by Auffinger and Cable [Reference Auffinger and Cable4] in the case $p=p_c$ . Using the fact that

$$\frac{1}{i(k-i)}= \frac{1}{k} \left( \frac{1}{i} + \frac{1}{k-i}\right),$$

we show that (2.1) implies heuristically the following equation (the details of the derivation of (2.3) from (2.1) are given in Appendix A)

(2.3) \begin{equation} t \frac{\partial f}{\partial t} = \frac{x}{2} \frac{\partial f}{\partial x} + \frac{f}{2} - K f \frac{\partial f}{\partial x} + \varepsilon t \bigg( \! - 2f + 4f \int_0^x f(t, \, y) \, \textrm{d} y \bigg) , \end{equation}

where

$$K \;:\!=\; \int_0^1 \frac{\log \left(\frac{1}{1-u}\right)}{u} \, \textrm{d} u = \frac{\pi^2}{6} .$$

For $\varepsilon=0$ one can solve (2.3) exactly for an arbitrary initial condition, $F_0(x) \equiv f(t_0,x)$ at time $t_0>0$ . It is in fact easy to write the solution in a parametric form at arbitrary time $t \ge t_0$ :

(2.4) \begin{equation}f(t, \, x)= \sqrt{\frac{t}{t_0}} F_0 (y) ; \quad x= K \left( \sqrt{\frac{t}{t_0}} -\sqrt{\frac{t_0}{t}} \right) F_0(y) + y \sqrt{\frac{t_0}{t}} .\end{equation}

If one writes

(2.5) \begin{equation}\textrm{d} x = \Bigg[ K \Bigg(\sqrt{\frac{t}{t_0}} -\sqrt{\frac{t_0}{t}}\, \Bigg) F_0'(y) + \sqrt{\frac{t_0}{t}} \Bigg] \, \textrm{d} y ,\end{equation}

and if $F_0(y)$ vanishes at the boundaries, then one can see that

(2.6) \begin{equation}\int f(t,x) \, \textrm{d} x = \int F_0(y) \, \textrm{d} y ,\end{equation}

so that the normalization of $F_0$ is conserved. However, after a finite time $t^*>t_0$ , given by the first time for which ${\textrm{d} x}/{\textrm{d} y}$ as in (2.5) vanishes for some y i.e.

$$t^*= t_0 \Bigg[ 1- \frac1K \max_y \frac{1}{F_0'(y)} \Bigg] ,$$

the solution F of (2.4) becomes multi-valued (see Figure 3 below). Then the solution is still given by (2.4) for some ranges of x with one of several shocks as on the right of Figure 3. In the example of Figure 3, if $f_a(t,x) > f_b(t,x) > f_c(t,x)$ are the three expressions of f(t, x) given by the parametric form (2.4) in the region $(x_1(t), \, x_2(t))$ where $f(t, \, x)$ is multi-valued, the true solution is

$$ f(t,x) = \begin{cases}f_a(t,x) & \quad\text{for $x_1< x < x_*$,}\\[5pt] f_c(t,x) & \quad\text{for $x_* < x <x_2$,}\end{cases}$$

with $x_*(t)$ determined by

$$\int_{x_1(t)}^{x_*(t)}[ f_a(t,x)-f_b(t,x)] \, \textrm{d} x =\int_{x_*(t)}^{x_2(t)}[ f_b(t,x)-f_c(t,x)] \, \textrm{d} x .$$

With this choice of $x_*(t)$ the normalization (2.6) of f is preserved.

Figure 3.

On the left, solution $F(t,\, x)$ using the parametric form (2.4) at times $t_0$ , $3 t_0$ , $5 t_0$ , and $7t_0$ for $F_0(z)= z \, {\textrm{e}}^{-z}$ . The same with the shock on the right.

In the long time limit, in that way from (2.4) one obtains that

(2.7) \begin{equation} f(t,x) = \begin{cases}\frac{x}{K} & \quad \text{for } 0 < x < x_*(\infty),\\[5pt] 0 & \quad \text{for } x \ge x_*(\infty),\end{cases} \quad \text{with} \ x_*(\infty) \;:\!=\; \sqrt{2 K}. \end{equation}

For $\varepsilon=0$ , this is in agreement with the description in [Reference Auffinger and Cable4].

For $\varepsilon \ne 0$ , we did not succeed in solving (2.3) for an arbitrary initial condition. However, if one starts at $t=0$ with the $\varepsilon=0$ solution (2.7), one can solve (2.3) explicitly:

(2.8) \begin{equation} f(t, \, x) = \frac{1}{{\textrm{e}}^{2\varepsilon t}-1} \sqrt{\frac{\varepsilon t}{K}} \sinh \Bigg( 2x \sqrt{\frac{\varepsilon t}{K}} \, \Bigg) . \end{equation}

Since $\sum_{k=1}^\infty a_n(k)=1$ for all n, the function $x\mapsto f(t, \, x)$ is a probability density function for all t. By analogy with the case $\varepsilon=0$ , we assume that (2.8) is valid only for $x\in [0, \, x_*(t)]$ , where $x_*(t)$ is such that

$$\int_0^{x_*(t)} \frac{1}{{\textrm{e}}^{2\varepsilon t}-1} \sqrt{\frac{\varepsilon t}{K}} \sinh \Bigg( 2x \sqrt{\frac{\varepsilon t}{K}} \, \Bigg) \, \textrm{d} x = 1$$

and we take $f(t, \, x) =0$ for $x> x_*(t)$ . When $t\to \infty$ , we have

$$x_*(t) \approx \sqrt{K \varepsilon t} .$$

This implies that for $n\to \infty$ ,

$$\log {{\mathbb{E}}} [D_n (p_c+ {\varepsilon})]\approx\sqrt{K \varepsilon} n.$$

Consequently,

$$\alpha(p_c+\varepsilon)\;:\!=\;\lim_{n\to \infty} \frac{1}{n} \log {{\mathbb{E}}} [D_n (p_c+\varepsilon)]\approx\sqrt{K \varepsilon} ,\quad\varepsilon\to 0^+ ,$$

which leads to the statement of Theorem 1.

3.2. Proof of Lemma 2

We fix $\eta\in (0, \, \eta_0)$ and $\varepsilon \in (0, \, \frac12)$ . Writing $\textrm{LHS}_{(3.9)}$ for the expression on the left-hand side of (3.9), we need to check that $\textrm{LHS}_{(3.9)} \ge 0$ for all large n, uniformly in $y\in [1, \, \lambda_{n+1}]$ because this inequality is obvious if $y\in [0, 1)$ and $y\in (\lambda_{n+1} , \infty)$ . Let $ n_{3} (\eta, \varepsilon)$ be such that $\eta^{1/\delta} \lambda_n \geq 2$ for all $n\geq n_{3}(\eta, \varepsilon)$ . Thus, to prove (3.9), we suppose that $n\geq n_{3} (\eta, \varepsilon)$ and we consider three situations:

$$\textbf{Case 1:} \ y\in [1, \, \eta^{1/\delta} \lambda_n], \quad\textbf{Case 2:} \ y\in (\eta^{1/\delta} \lambda_n, \, \lambda_n], \quad \textrm{and} \quad\textbf{Case 3:} \ y\in (\lambda_n, \, \lambda_{n+1}].$$

For the sake of clarity, these situations are dealt with in three distinct parts.

Proof of Lemma 2: first case $y\in [1, \, \eta^{1/\delta} \lambda_n]$ . In this case, we first note that

$${{\mathbb P}} \big( Y_n + \widehat{Y}_n > y\big) \ge1 - {{\mathbb P}} \big(Y_n \le y ; \, \widehat{Y}_n \le y \big) =1 - {{\mathbb P}} \big(Y_n \le y \big)^2.$$

Therefore, writing $F_n(y) \;:\!=\; {{\mathbb P}}(Y_n\le y)$ , we obtain

For $y\in [1, \, \eta^{1/\delta} \lambda_n]$ , we have $y\le \lambda_n < \lambda_{n+1}$ , thus $F_n(y) = 2a_n \varphi_\delta (y\wedge \lambda_n) = 2a_n \varphi_\delta (y)$ , and $F_{n+1}(y) = 2a_{n+1} \varphi_\delta (y) = (1 - 2\varepsilon + \eta \delta^2) 2a_n \varphi_\delta (y)$ . Accordingly,

Observe that $\delta^2/(2\varepsilon) = 12 (1+\eta)^2/ \pi^2 >1$ , thus $\frac{\eta \delta^2}{2\varepsilon} > \eta$ , which equals $\lim_{n\rightarrow \infty} 2a_n \varphi_\delta (\eta^{1/\delta} \lambda_n)$ by Lemma 1(ii). Thus, there exists $n_{4}(\eta, \varepsilon)\geq n_{3} (\eta, \varepsilon)$ such that for all $n\geq n_{4}(\eta, \varepsilon)$ and all $y\in [1, \eta^{1/\delta} \lambda_n ]$ , we have

$$2a_n \varphi_{\delta} (y) \leq \frac{\eta \delta^2}{2\varepsilon} ,$$

which implies that

as desired.

Proof of Lemma 2: second case $y\in [\eta^{1/\delta} \lambda_n, \, \lambda_n]$ . In this case and in the next case, the convolution term $Y_n+ \widehat{Y_n}$ matters more specifically; we use the following lemma to get estimates for the law of $Y_n+ \widehat{Y_n}$ . This is a key lemma where the constant $\zeta(2)$ appears.

Lemma 3. We keep the previous notation and we recall, in particular, $\delta$ from (3.2). Let $r\in (0, 1]$ . We define

(3.10) \begin{equation} \mathtt{cvl}_{\delta , r} (x) \;:\!=\; \int_1^{rx-1} \varphi_\delta'(t) \, \big( \varphi_\delta(x) - \varphi_\delta(x - t)\big)\, \textrm{d} t , \quad x \in \Big[\frac{2}{r} , \, \infty \Big) . \end{equation}

For $r \in (0, 1]$ , we also set $\kappa(r) \;:\!=\; \sum_{k=1}^\infty \frac{r^k}{k^2} $ and for $ x \in \big[\frac{2}{r} , \infty \big) $ ,

$$ \kappa_{\delta, r} (x) \;:\!=\; \kappa \bigg( r - \frac{1}{x} \bigg) - \frac{2}{x} -c_0 \, \delta \coth (\delta \log x), $$

with

$$c_0 \;:\!=\; \int_0^1\,\bigg( \log \frac{1}{s(1-s)} \bigg) \bigg( \log \frac{1}{1-s}\bigg) \, \frac{\textrm{d} s}{s} \in (0, \, \infty).$$

Then, for $r \in (0, 1]$ and $x \in [{2}/{r} , \, \infty )$ ,

(3.11) \begin{equation}\kappa_{\delta, r} (x) \big(\delta \sinh(\delta \log x)\big)^2 \leq \mathtt{cvl}_{\delta, r} (x) \le \kappa(r) \, \big( \delta \sinh(\delta \log x)\big)^2 . \end{equation}

Proof. We first prove the upper bound in (3.11). Let $\widetilde{\varphi}_\delta(u) \;:\!=\; \varphi_\delta({\textrm{e}}^u) = \cosh(\delta u) - 1$ , for $u \in \mathbb{R}_+$ . We fix $r\in (0, 1]$ and $ x \in [{2}/{r} , \infty ) $ . Since $\widetilde{\varphi}_\delta$ is convex on $\mathbb{R}_+$ , we get $\widetilde{\varphi}_\delta( \log x) - \widetilde{\varphi}_\delta (\!\log (x - t)) \le (\!\log x - \log (x - t)) \widetilde{\varphi}_\delta'(\!\log x)$ for $t\ge 0$ and $x\ge t+1$ . Thus,

$$\mathtt{cvl}_{\delta, r} (x) =\int_1^{rx-1} \varphi_\delta'(t) \big( \widetilde{\varphi}_\delta( \log x) - \widetilde{\varphi}_\delta (\!\log (x - t)) \big) \, \textrm{d} t \leq\widetilde{\varphi}{\,} '_{ \delta}(\!\log x) \int_1^{rx-1} \varphi_\delta'(t) \log \frac{x}{x-t} \, \textrm{d} t .$$

By definition, $\widetilde{\varphi}{\,} '_{ \delta}(\!\log x) = \delta \sinh(\delta \log x)$ , and $\varphi_\delta'(t) = \frac{\delta}{t} \sinh(\delta\log t) \le \frac{\delta}{t} \sinh(\delta\log x)$ if $t \leq rx-1$ . Thus,

\begin{eqnarray*}\mathtt{cvl}_{\delta, r} (x) & \,\leq\, &\big( \delta \sinh(\delta \log x)\big)^2 \int_1^{rx-1} \frac{1 }{t} \log \frac{x}{x-t} \,\textrm{d} t\le \big( \delta \sinh(\delta \log x)\big)^2 \int_0^{rx} \frac{1 }{t} \log \frac{x}{x-t} \, \textrm{d} t \\[5pt] &\,=\,& \big( \delta \sinh(\delta \log x)\big)^2 \int_0^r \frac{\log \frac{1}{1 - u} }{u} \, \textrm{d} u.\end{eqnarray*}

By Fubini–Tonelli,

(3.12) \begin{equation} \int_0^r \frac{\log \frac{1}{1 - u} }{u} \, \textrm{d} u = \sum_{k\ge 1} \int_0^r \frac1u \frac{u^k}{k} \,\textrm{d} u = \kappa (r) , \end{equation}

which yields the upper bound in (3.11).

We turn to the proof of the lower bound in (3.11). Since on $\mathbb{R}_+$ , all the derivatives of $\widetilde{\varphi}_\delta$ are positive, $\widetilde{\varphi}_\delta$ and $\widetilde{\varphi}_\delta'$ are convex which implies for all $b \geq a \geq 0$ that

$$\frac{\widetilde{\varphi}_\delta (b) - \widetilde{\varphi}_\delta (a) }{b-a} \geq \widetilde{\varphi}_\delta'(a) = \widetilde{\varphi}_\delta'(b) - (\widetilde{\varphi}_\delta'(b) - \widetilde{\varphi}_\delta'(a)) \geq \widetilde{\varphi}_\delta'(b) -(b - a) \widetilde{\varphi}_\delta''(b) .$$

We suppose that $1\leq t\leq rx-1$ . Taking $b = \log x$ and $a = \log (x - t)$ , we get that

(3.13)

where we have set

We first look at $\textrm{RHS}^{(2)}_{(3.13)}$ . Since $\varphi'_\delta(t) = \frac{\delta}{t} \sinh(\delta\log t) \le \frac{\delta}{t} \sinh(\delta\log x)$ , we have

$$\int_1^{rx-1} \varphi'_\delta(t) \bigg(\!\log \frac{x}{x-t} \bigg)^2 \, \textrm{d} t \le\delta \sinh(\delta\log x) \int_1^{rx-1} \frac1t \, \bigg(\!\log \frac{x}{x-t} \bigg)^2 \, \textrm{d} t .$$

We observe that

$$\int_1^{rx-1} \frac1t \bigg(\!\log \frac{x}{x-t} \bigg)^2 \, \textrm{d} t\le\int_0^x \frac1t \bigg(\!\log \frac{x}{x-t} \bigg)^2 \, \textrm{d} t=\int_0^1 \frac1s \bigg(\!\log \frac{1}{1-s} \bigg)^2 \, \textrm{d} s\;=\!:\;c_1 .$$

Therefore,

(3.14)

We now turn to

and look for a lower bound. We still assume that $1\leq t\leq rx-1$ . Since the function $\sinh$ is convex on $\mathbb{R}_+$ , we have $ \sinh(\delta\log t) \ge \sinh(\delta\log x) - \delta(\!\log x - \log t) \cosh(\delta \log x)$ . Therefore,

\begin{eqnarray*} &&\int_1^{rx-1} \delta \sinh(\delta \log t) \frac{1}{t} \big(\!\log \frac{x}{x-t} \big) \, \textrm{d} t \\[5pt] &\,\ge\,& \delta\int_1^{rx-1} \big(\!\sinh(\delta\log x) - \delta \left(\!\log \frac{x}{t}\right) \cosh(\delta \log x) \big) \frac{1}{t} \Big(\!\log \frac{x}{x-t} \Big) \, \textrm{d} t \\[5pt] &\,=\,& \delta \sinh(\delta\log x) \int_1^{rx-1} \frac{\log \frac{x}{x-t}}{t} \, \textrm{d} t - \delta^2 \cosh(\delta \log x) \int_1^{rx-1} \frac{ \big(\!\log \frac{x}{t} \big) \log \frac{x}{x-t}}{t} \, \textrm{d} t .\end{eqnarray*}

Let us look at the two integrals on the right-hand side. The second integral is easy to deal with:

$$\int_1^{rx-1} \frac{ \big(\!\log \frac{x}{t} \big) \log \frac{x}{x-t}}{t} \, \textrm{d} t\le\int_0^x \frac{ \big(\!\log \frac{x}{t} \big) \log \frac{x}{x-t}}{t} \, \textrm{d} t=\int_0^1 \frac{ \big(\!\log \frac{1}{s} \big) \log \frac{1}{1-s}}{s} \, \textrm{d} s\;=\!:\;c_2 \in (0, \, \infty) .$$

The first integral is handled as follows: let $r_1 \;:\!=\; r-\frac1x \in (0, \, r)$ . Then $rx-1 = r_1x$ , so that

$$\int_1^{rx-1} \frac{\log \frac{x}{x-t}}{t} \, \textrm{d} t=\int_0^{r_1 x} \frac{\log \frac{x}{x-t}}{t} \, \textrm{d} t-\int_0^1 \frac{\log \frac{x}{x-t}}{t} \, \textrm{d} t=\kappa(r_1) - \kappa \left( \frac1x \right) ,$$

since $\int_0^r \frac1u \log (1 - u)^{-1} \, \textrm{d} u = \kappa (r)$ by (3.12). Consequently,

$$\int_1^{rx-1} \delta \sinh(\delta \log t) \, \frac{1}{t} \, \Big(\!\log \frac{x}{x-t}\Big) \, \textrm{d} t \ge\bigg(\kappa(r_1) - \kappa \bigg( \frac1x \bigg) \bigg)\delta \sinh(\delta\log x)-c_2 \delta^2 \cosh(\delta \log x) .$$

This implies that

Now observe that $ x\kappa \big(\frac1x \big) \leq \zeta(2) \leq 2$ . Together with (3.13) and (3.14), this yields the lower bound in (3.11) with $c_0 = c_1+ c_2$ , and completes the proof of Lemma 3.

Let us now proceed to the proof of (3.9) (i.e. Lemma 2) in the second case $y\in (\eta^{1/\delta} \lambda_n, \, \lambda_n]$ . We write

(3.15)

where we have set

\begin{eqnarray*} \textrm{I}_n(y) & \;:\!=\; & F_n(y)^2 - {{\mathbb P}}(Y_n + \widehat{Y}_n \le y), \\[5pt] \textrm{I I}_n(y) & \;:\!=\; & F_{n+1}(y) - F_n(y) \; \, \textrm{(which is negative),} \\[5pt] \textrm{I I I}_n(y) & \;:\!=\; & 2F_n(y) -F_n(y)^2 - {{\mathbb P}}(Y_n + \widehat{Y}_n \le y) .\end{eqnarray*}

We first look for a lower bound for $\textrm{I}_n(y)$ . Note that $F_n(y)^2 \ge F_n(y)\, F_n(y-1) = F_n(y)\, \int_1^{y-1} F'_n(t) \, \textrm{d} t$ and that ${{\mathbb P}}(Y_n + \widehat{Y}_n \le y) = \int_1^{y-1} F'_n(t) F_n(y-t) \, \textrm{d} t$ . Therefore,

\begin{align*} \textrm{I}_n(y)& \ge\int_1^{y-1} F'_n(t) \big( F_n(y) - F_n(y - t)\big) \, \textrm{d} t\\& = 4a_n^2 \int_1^{y-1} \varphi_\delta^\prime (t)\big( \varphi_\delta(y) - \varphi_\delta (y - t) \big) \, \textrm{d} t = 4a_n^2 \mathtt{cvl}_{\delta, 1} (y) ,\end{align*}

where $\mathtt{cvl}_{\delta, 1} $ is defined in Lemma 3. By the first inequality in (3.11) of Lemma 3, $\mathtt{cvl}_{\delta, 1}(y) \geq \kappa_{\delta, 1} (y) (\delta \sinh (\delta \log y))^2$ for $y \in [2, \infty)$ . Since

(3.16) \begin{equation}\mbox{for all } t \in \mathbb{R}_+, \quad \cosh (t) \geq \sinh (t) \geq \cosh (t)-1\geq 0,\end{equation}

we have $\mathtt{cvl}_{\delta, 1} (y) \geq \kappa_{\delta, 1} (y) \delta^2 \varphi_\delta (y)^2$ . Thus, $\textrm{I}_n(y) \geq \kappa_{\delta, 1} (y) \delta^2 F_n (y)^2$ . Since the function $\coth$ decreases to 1 and since $ \kappa_{\delta, 1}$ is increasing on $[2, \, \infty)$ , we get, for $y\in [\eta^{1/\delta} \lambda_n, \, \lambda_n]$ ,

\begin{align*} & \kappa_{\delta, 1}(y) \geq \kappa_{\delta, 1} (\eta^{1/\delta} \lambda_n)\\ &\quad = \kappa \bigg( 1 - \frac{1}{\eta^{1/\delta} \lambda_n} \bigg) -\frac{2}{\eta^{1/\delta} \lambda_n} -c_0 \delta \coth (\delta \log (\eta^{1/\delta} \lambda_n)) \underset{n\to \infty}{-\!\!\! -\!\!\! -\!\!\!\longrightarrow} \zeta(2) -c_0\delta ,\end{align*}

because $\kappa(1)=\zeta(2)$ . By definition, $\delta = 2\zeta(2)^{-1/2}(1+\eta)\sqrt{\varepsilon}$ , thus

$$\lim_{n\to \infty }\delta^2 \kappa_{\delta, 1} (\eta^{1/\delta} \lambda_n)= \delta^2 \zeta(2) -c_0 \delta^3= 4(1+\eta)^2\varepsilon - 8c_0\zeta(2)^{-3/2} (1+\eta)^3 \varepsilon^{3/2} .$$

Let $\varepsilon_{3} (\eta) \in (0,\, \frac12)$ be such that for $\varepsilon \in (0, \,\varepsilon_{3} (\eta))$ ,

$$4(1+\eta)^2\varepsilon - 8c_0\zeta(2)^{-3/2} (1+\eta)^3 \varepsilon^{3/2} >4\varepsilon.$$

Therefore, there exists an integer $n_{5} (\eta, \varepsilon) \geq n_{4} (\eta, \varepsilon) $ such that

(3.17) \begin{equation} \mbox{for all }n \geq n_{5} (\eta, \varepsilon), \mbox{ for all } y\in [\eta^{1/\delta} \lambda_n, \, \lambda_n] , \quad \textrm{I}_n(y) \ge 4 \varepsilon F_n(y)^2 . \end{equation}

On the other hand, since $a_{n+1}/a_n = 1-2 {\varepsilon} (1-\widetilde{\eta} )$ , we get

\begin{eqnarray*} \textrm{I I}_n(y) & \,=\, & 2(a_{n+1} - a_n) \varphi_{\delta} (y) = - 2(1 - \widetilde{\eta})\varepsilon F_n(y), \\[5pt] \textrm{I I I}_n(y) & \,=\, & 2(F_n(y) - F_n(y)^2) + \textrm{I}_n(y) \ge 2(F_n(y) - F_n(y)^2) .\end{eqnarray*}

Thus, for all $\varepsilon \in (0, \varepsilon_{3} (\eta))$ , all integers $n\geq n_{5}(\eta, \varepsilon) $ and all $y\in [\eta^{1/\delta} \lambda_n, \, \lambda_n]$ ,

This completes the proof of Lemma 2 in the second case $y\in [\eta^{1/\delta} \lambda_n, \, \lambda_n]$ .

Proof of Lemma 2: third and last case $y\in (\lambda_n, \, \lambda_{n+1}]$ . Here again the law of $Y_n+ \widehat{Y}_n$ matters specifically and we use Lemma 3 to handle it. Let us fix $\varepsilon \in (0, \varepsilon_{3} (\eta))$ and $n \geq n_{5}(\eta, \varepsilon) $ . We first observe that

Since $F_{n+1}(\lambda_n) =\big(1 - 2(1 - \widetilde{\eta})\varepsilon \big) F_n(\lambda_n) = 1 - 2(1 - \widetilde{\eta})\varepsilon$ , we get for all $y\in (\lambda_n, \, \lambda_{n+1}]$ that

(3.18)

We now look for a lower bound for ${{\mathbb P}}(Y_n + \widehat{Y}_n > \lambda_{n+1})$ . By Lemma 1(iii), there is $\varepsilon_{4} (\eta) \in (0, \varepsilon_{3} (\eta))$ such that for $\varepsilon \in (0, \varepsilon_{4} (\eta))$ , there exists an integer $n_{6} (\eta, \varepsilon) \geq n_{5} (\eta, \varepsilon)$ which satisfies $\lambda_{n+1} - \lambda_n+1\le \delta \lambda_n$ , for all $n\geq n_{6}(\eta, \varepsilon)$ . We have

\begin{eqnarray*} {{\mathbb P}} \big( Y_n + \widehat{Y}_n > \lambda_{n+1} \big) &\ge & {{\mathbb P}} \big( \lambda_n - 1 > Y_n > \lambda_{n+1} - \lambda_n ; \, Y_n + \widehat{Y}_n > \lambda_{n+1} \big) \\[5pt] &\,=\,& \int_{\lambda_{n+1}-\lambda_n}^{\lambda_n-1} F'_n(t) \big( 1 - F_n(\lambda_{n+1} - t)\big) \, \textrm{d} t .\end{eqnarray*}

Writing $1-F_n(\lambda_{n+1}-t) = F_n(\lambda_n) -F_n(\lambda_{n+1}-t)$ , this leads to

(3.19)

where:

• • ;

• • ;

• • .

We apply Lemma 3 to

to get that

Note that $\lim_{n\to \infty} \kappa_{\delta, 1} (\lambda_n)\delta^2 = 4(1+\eta)^2\varepsilon - 8c_0\zeta(2)^{-3/2} (1+\eta)^2 \varepsilon^{3/2}$ . Therefore there exists $\varepsilon_{5} (\eta) \in (0, \varepsilon_{4} (\eta))$ such that for all $\varepsilon \in (0, \varepsilon_{5} (\eta))$ there is an integer $n_{7} (\eta, \varepsilon) \geq n_{6} (\eta, \varepsilon)$ satisfying

(3.20)

Let us next provide an upper bound for

. For $n\geq n_{7} (\eta, \varepsilon)$ , we have $\lambda_{n+1} - \lambda_n+1\le \delta \lambda_n$ ; thus

Therefore, there exists $\varepsilon_{6} (\eta) \in (0, \varepsilon_{5} (\eta))$ such that for all $\varepsilon \in (0, \varepsilon_{6} (\eta))$ , there is an integer $n_{8} (\eta, \varepsilon)\geq n_{7} (\eta, \varepsilon)$ satisfying

(3.21)

We finally look for an upper bound for

. Since $\varphi'_\delta(t)=\frac{\delta}{t}\sinh(\delta \log t)$ , we have

Since $\lambda_{n+1} - \lambda_n +1 \leq \delta \lambda_n$ , we get $\lambda_{n+1} \leq (1+ \delta) \lambda_n $ , and thus

Thus, there exists $\varepsilon_{7} (\eta) \in (0, \varepsilon_{6} (\eta))$ such that for all $\varepsilon \in (0, \varepsilon_{7} (\eta))$ , there is an integer $n_{9} (\eta, \varepsilon)\geq n_{8} (\eta, \varepsilon)$ such that

Combined with (3.20) and (3.21), it entails that

Returning to (3.18), we obtain $\textrm{LHS}_{(3.9)}\ge2\varepsilon - 2(1 - \widetilde{\eta})\varepsilon= 2\widetilde{\eta}\varepsilon> 0 ,$ which proves Lemma 2 in the third and last case $y\in (\lambda_n, \, \lambda_{n+1}]$ .

4.2. Proof of Lemma 6

This section is devoted to the proof of Lemma 6. We fix $\eta \in (0, \, 1)$ and $\varepsilon \in (0, \, \varepsilon_{1} (\eta))$ . Recall from (4.1) the definition of $\gamma$ and $b_n$ and from (4.2) the definition of $\sigma_n$ and $\tau_n$ . By (4.3), $\lim_{n\to \infty} \sigma_n = 1 < 1+ \eta \gamma = \lim_{n\to \infty} (1 + \eta \gamma ) \sigma_{n+1} < \lim_{n\to \infty} (\eta \varepsilon)^{1/\gamma} \tau_n = \infty$ . Therefore, there is an integer $n_{3} (\eta, \, \varepsilon) \geq n_{1} (\eta,\, \varepsilon) $ such that

(4.7) \begin{equation} 2\sigma_n < 2(1 + \eta \gamma) \sigma_{n+1} < (\eta \varepsilon)^{1/\gamma} \tau_n < \gamma^2\tau_n < {\textrm{e}}^{1/\sqrt{\gamma}} \tau_n , \quad n\geq n_{3} (\eta, \varepsilon) .\end{equation}

Writing $\textrm{LHS}_{(4.5)}$ for the expression on the left-hand side of (4.5), we need to check that $\textrm{LHS}_{(4.5)} \le 0$ for all small enough $\varepsilon >0$ , all sufficiently large integer n and all $z \ge (1+\eta \gamma) \sigma_{n+1}$ . This is done by distinguishing four cases:

\begin{eqnarray*}\textbf{Case 1:} \ (1+\eta \gamma) \sigma_{n+1} \leq z <(\eta \varepsilon)^{1/\gamma} \tau_n , & \quad & \textbf{Case 2:} \ (\eta \varepsilon)^{1/\gamma} \tau_n \leq z < \gamma^2 \tau_n, \\[5pt] \textbf{Case 3:} \ \gamma^2 \tau_n \le z < {\textrm{e}}^{1/\sqrt{\gamma}} \tau_n, & \quad\textrm{and}\quad & \textbf{Case 4:} \ z \geq {\textrm{e}}^{1/\sqrt{\gamma}} \tau_n. \end{eqnarray*}

Proof of Lemma 6: first case $(1+\eta \gamma) \sigma_{n+1} \le z < (\eta \varepsilon)^{1/\gamma} \tau_n$ . Here $\varepsilon \in (0, \varepsilon_{1} (\eta))$ and $n\geq n_{3} (\eta, \varepsilon)$ . We use the trivial fact ${{\mathbb P}} \big( Z_n+\widehat{Z}_n > z\big) \le 1$ , so that

(4.8)

Recall the notation $G_n(z) \;:\!=\; {{\mathbb P}} (Z_n \leq z)$ . Then

(4.9)

We have used the fact (by (4.7)) that $z\ge \sigma_n$ in the last inequality. Since $\varphi_\gamma(z) \le \cosh (\gamma \log z)$ and $z < (\eta \varepsilon)^{1/\gamma} \tau_n$ , we get that

Thus, there exists $n_{4} (\eta, \, \varepsilon) \geq n_{3} (\eta, \, \varepsilon)$ such that $G_n (z) < 2\eta\varepsilon$ for all $n\geq n_{4} (\eta, \, \varepsilon)$ and for all $z\in [(1+\eta \gamma) \sigma_{n+1}, (\eta \varepsilon)^{1/\gamma} \tau_n]$ . Thus, we get

$$\frac12 G_n(z)^2 - (1- 2\varepsilon) G_n(z)\le- \big(1 - (2 + \eta)\varepsilon \big) G_n(z)=- 2b_{n+1} \frac{G_n(z)}{2b_n} .$$

By (4.8), this leads to

We claim for all sufficiently small $\varepsilon $ and large n, and all $z\in [(1+\eta \gamma) \sigma_{n+1}, (\eta \varepsilon)^{1/\gamma} \tau_n]$ , that

(4.10) \begin{align} \varphi_\gamma(z+\tau_n) - \varphi_\gamma(\sigma_n+\tau_n) &\leq \frac{1}{2}\bigg(\varphi_\gamma(z) - \varphi_\gamma\bigg( \frac{z}{1+\eta \gamma}\bigg) \bigg) \quad \textrm{and}\nonumber \\\varphi_\gamma(\sigma_n) & \leq \frac{1}{2}\bigg(\varphi_\gamma(z) - \varphi_\gamma\bigg( \frac{z}{1+\eta \gamma}\bigg) \bigg) ,\end{align}

which will readily imply Lemma 6 in the first case.

To check the second inequality in (4.10), we look for a suitable upper bound for $\varphi_\gamma (\sigma_n) \;:\!=\; \cosh(\gamma \log \sigma_n)-1$ . We note that $\cosh (\lambda) -1 \leq 2 \lambda^2$ for $\lambda \in [0, \, 2]$ (indeed, since $\cosh (2) < 4$ , the second derivative of $\lambda \in [0, \, 2] \mapsto \cosh (\lambda )-1 - 2 \lambda^2 $ is negative, so is the derivative, and the function itself is decreasing). Since $\sigma_n \to 1^+$ as $n \to \infty$ , there exists $n_{5} (\eta, \, \varepsilon) \geq n_{4}(\eta, \, \varepsilon)$ such that $\gamma \log \sigma_n \in [0, \, 2]$ for $n\geq n_{5} (\eta, \, \varepsilon)$ ; accordingly,

(4.11) \begin{equation} \varphi_\gamma(\sigma_n) \le 2 (\gamma \log \sigma_n)^2 \le 2\gamma^2 (\sigma_n - 1)^2 ,\end{equation}

since $\log x\leq x-1$ for $x\ge 1$ .

We now look for lower bounds for $\varphi_\gamma(z) - \varphi_\gamma\left(\frac{z}{1+\eta \gamma}\right)$ . Using the formula $\cosh (a) - \cosh (b) = 2 \sinh\left( \frac{a + b}{2}\right) \sinh \left(\frac{a - b}{2}\right)$ , we get that

$$\varphi_\gamma(z) - \varphi_\gamma \bigg( \frac{z}{1+\eta \gamma} \bigg) =2 \sinh \bigg( \frac{\gamma}{2}\log (1 + \eta \gamma) \bigg) \, \sinh \bigg( \gamma \log z - \frac{\gamma}{2}\log (1 + \eta \gamma) \bigg) .$$

Since $z\in [(1+\eta \gamma) \sigma_{n+1}, (\eta \varepsilon)^{1/\gamma} \tau_n]$ , observe that

(4.12) \begin{equation} \gamma \log z - \frac{\gamma}{2}\log (1 + \eta \gamma) \geq \gamma \log (1 + \eta \gamma) + \gamma \log \sigma_{n+1} - \frac{\gamma}{2}\log (1 + \eta \gamma) \geq 0.\end{equation}

Since $\sinh (x) \geq x$ (for $x\in \mathbb{R}_+$ ) and $\log (1+ x)\geq \frac{1}{2} x$ (for $x\in [0, \, 1]$ ), we have

(4.13) \begin{eqnarray} \varphi_\gamma(z) - \varphi_\gamma \bigg( \frac{z}{1+\eta \gamma} \bigg) &\,\ge\,& \gamma \, [\!\log (1 + \eta \gamma)] \sinh \bigg( \gamma \log z - \frac{\gamma}{2}\log (1 + \eta \gamma) \bigg) \nonumber \\[5pt] &\,\ge\,& \frac{1}{2}\eta \gamma^2 \sinh \bigg( \gamma \log z - \frac{\gamma}{2}\log (1 + \eta \gamma) \bigg) \end{eqnarray}

(4.14)

Since $2\gamma^2 (\sigma_n - 1)^2 \to 0$ as $n \to \infty$ , there exists $n_{6} (\eta, \, \varepsilon) \ge n_{5} (\eta,\, \varepsilon)$ such that $2\gamma^2 (\sigma_n - 1)^2 < \frac{1}{8}\eta \gamma^3\log (1 + \eta \gamma)$ for $n \geq n_{6} (\eta, \, \varepsilon)$ ; in view of (4.11), this implies the desired second inequality in (4.10):

$$\varphi_\gamma(\sigma_n) \leq \frac{1}{2} \bigg(\varphi_\gamma(z) - \varphi_\gamma \bigg( \frac{z}{1+\eta \gamma} \bigg) \bigg) , \quad n\geq n_{6} (\eta, \varepsilon), \ z\in \big[ (1 + \eta \gamma)\sigma_{n+1} , (\eta \varepsilon)^{1/\gamma} \tau_n \big).$$

We now turn to the proof of the first inequality in (4.10). Observe that

(4.15) \begin{eqnarray}\varphi_\gamma(z+\tau_n) - \varphi_\gamma(\sigma_n+\tau_n) &\,=\,& \gamma \int_{\sigma_n+ \tau_n}^{z+ \tau_n} \sinh (\gamma \log x) \frac{\textrm{d} x}{x} \nonumber \\[5pt] &\,\leq\, &\frac{\gamma}{2} \int_{\sigma_n+ \tau_n}^{z+ \tau_n} \frac{\textrm{d} x}{x^{1-\gamma}} \leq \frac{\gamma (z- \sigma_n)}{2(\sigma_n+ \tau_n)^{1-\gamma}} \leq\frac{\gamma (z- 1)}{2\tau_n^{1-\gamma}}, \end{eqnarray}

since $\sigma_n \geq 1$ . Consequently, for $z\in \big[ (1 + \eta \gamma)\sigma_{n+1}, \, {\textrm{e}}^{2/\gamma} \big)$ ,

$$\varphi_\gamma(z+\tau_n) - \varphi_\gamma(\sigma_n+\tau_n)\leq \frac{\gamma ({\textrm{e}}^{2/\gamma} - 1)}{2\tau_n^{1-\gamma}} \underset{n\rightarrow \infty}{-\!\!\! -\!\!\! -\!\!\!\longrightarrow} 0 < \frac{1}{8}\eta \gamma^3\log (1 + \eta \gamma) .$$

Thus, by (4.14), there is $n_{7} (\eta,\, \varepsilon) \geq n_{6} (\eta,\, \varepsilon)$ such that for $n\geq n_{7} (\eta,\, \varepsilon)$ and $z\in \big[ (1 + \eta \gamma)\sigma_{n+1}, \, {\textrm{e}}^{2/\gamma} \big)$ ,

$$\varphi_\gamma(z+\tau_n) - \varphi_\gamma(\sigma_n+\tau_n)\leq \frac{1}{2}\bigg( \varphi_\gamma(z) - \varphi_\gamma \bigg( \frac{z}{1+\eta \gamma} \bigg)\bigg) .$$

To complete the proof of the first inequality in (4.10), we still need to consider $z\in [{\textrm{e}}^{2/\gamma}, \, (\eta\varepsilon)^{1/\gamma} \tau_n)$ (with $n\geq n_{7}(\eta,\varepsilon )$ ). Noting that $\frac{\gamma}{2}\log (1 + \eta \gamma) < 1$ , we have $\gamma \log z - \frac{\gamma}{2}\log (1 + \eta \gamma) > \gamma \log z - 1 \geq 1$ . By (4.13), and writing $c_0 \;:\!=\; \inf_{x\in [2, \, \infty) } [{\textrm{e}}^{-x} \sinh(x-1)]$ ,

$$\varphi_\gamma(z) - \varphi_\gamma \bigg( \frac{z}{1+\eta \gamma} \bigg)\ge\frac{1}{2}\eta \gamma^2 \sinh ( \gamma \log z - 1)\ge\frac{1}{2} c_0\eta \gamma^2 {\textrm{e}}^{\gamma \log z} .$$

Since ${\textrm{e}}^{\gamma \log z} = \frac{z}{z^{1-\gamma}} \ge \frac{z-1}{((\eta\varepsilon)^{1/\gamma}\tau_n)^{1-\gamma}} = \frac{1}{\gamma (\eta\varepsilon)^{(1-\gamma)/\gamma}} \, \frac{\gamma (z-1)}{\tau_n^{1-\gamma}}$ , this implies that

$$\varphi_\gamma(z) - \varphi_\gamma \bigg( \frac{z}{1+\eta \gamma} \bigg)\ge\frac{1}{2} c_0 \eta \gamma^2 \frac{z - 1}{((\eta\varepsilon)^{1/\gamma}\tau_n)^{1-\gamma}}=\frac{c_1 (1 - \eta)}{\eta^{\frac{1}{\gamma}-2} \, \varepsilon^{\frac{1}{\gamma} - \frac{3}{2}}} \, \frac{\gamma (z - 1)}{2\tau_n^{1-\gamma}} ,$$

with $c_1 \;:\!=\; \frac{2}{\sqrt{\zeta(2)}} c_0$ . Since $\frac{c_1 (1 - \eta)}{\eta^{\frac{1}{\gamma}-2} \, \varepsilon^{\frac{1}{\gamma} - \frac{3}{2}}} \to \infty$ as $\varepsilon \to 0^+$ , there exists $\varepsilon_{3} (\eta) \in (0, \varepsilon_{1} (\eta))$ such that $\frac{c_1 (1 - \eta)}{\eta^{(1/\gamma)-2} \, \varepsilon^{1/\gamma - 3/2}} >2$ for $\varepsilon \in (0, \, \varepsilon_{3} (\eta))$ and there exists $n_{8} (\eta,\, \varepsilon)\geq n_{7} (\eta, \, \varepsilon)$ such that for $n \geq n_{8} (\eta, \, \varepsilon)$ and $z\in [{\textrm{e}}^{2/\gamma} , \, (\eta\varepsilon)^{1/\gamma} \tau_n)$ ,

This yields (4.10) and, thus, implies Lemma 6 in the first case $(1+\eta \gamma) \sigma_{n+1} \le z < (\eta \varepsilon)^{1/\gamma} \tau_n$ .

Proof of Lemma 6: second case $(\eta \varepsilon)^{1/\gamma} \tau_n \le z < \gamma^2 \tau_n$ . We first prove the following lemma.

Lemma 7. We keep the previous notation. For $0 < a < b < \infty$ , we have

(4.16) \begin{equation} \limsup_{n\rightarrow \infty} \sup_{\theta \in [a, \, b]} | G_n (\theta \tau_n)\theta^{-\gamma}-1| = \frac{(1+ b)^{\gamma} -1}{b^\gamma} .\end{equation}

Proof. We remind the reader that $\lim_{n\to \infty} b_n\tau_n^\gamma=1$ and that $\lim_{n\to \infty} b_n\tau_n^{-\gamma}=0$ . Then, for $\theta \in [a, \, b]$ , we have

\begin{align*} G_n (\theta \tau_n) &= 2b_n \big( \varphi_\gamma(\tau_n \theta) - \varphi_\gamma(\sigma_n) - \varphi_\gamma(\tau_n (1+\theta)) + \varphi_\gamma(\sigma_n+\tau_n) \big) \\[5pt] & = b_n\tau_n^{\gamma} \theta^{\gamma} + b_n\tau_n^{-\gamma} \theta^{-\gamma} -2b_n -2b_n \varphi_{\gamma} (\sigma_n) + b_n\tau_n^{\gamma} \big( 1 - (1+ \theta)^{\gamma} \big)- b_n\tau_n^{-\gamma} (1+ \theta)^{-\gamma} \\[5pt] & \quad + b_n\tau_n^{\gamma} \big((1+ \sigma_n \tau_n^{-1})^{\gamma}-1 \big) + b_n\tau_n^{-\gamma} (1+ \sigma_n\tau^{-1}_n)^{-\gamma} \\[5pt] & = b_n\tau_n^{\gamma} \theta^{\gamma}+ b_n\tau_n^{\gamma} \big( 1 - (1+ \theta)^{\gamma} \big) + b_n\tau_n^{-\gamma} (\theta^{-\gamma} -(1+ \theta)^{-\gamma} ) + u_n (\eta, \varepsilon),\end{align*}

where $u_n (\eta, \, \varepsilon)$ does not depend on $\theta$ and satisfies $\lim_{n\to \infty} u_n (\eta, \varepsilon)= 0$ . Thus,

$$G_n (\theta \tau_n)\theta^{-\gamma} - 1 =-\frac{(1 + \theta)^{\gamma} - 1}{ \theta^{\gamma}} +R_n( \eta, \, \gamma, \, \theta) ,$$

where

$$R_n( \eta, \, \gamma, \, \theta) \;:\!=\;(b_n\tau_n^{\gamma} - 1) \bigg( 1 - \frac{(1 + \theta)^{\gamma} - 1}{ \theta^{\gamma}} \bigg) +\frac{b_n\tau_n^{-\gamma} (\theta^{-\gamma} -(1+ \theta)^{-\gamma} ) + u_n (\eta, \varepsilon)}{\theta^{\gamma}} .$$

Since $\lim_{n\to \infty} b_n\tau_n^{\gamma} =1$ by (4.3), we have $\lim_{n\to \infty}\sup_{\theta \in [a, \, b]}|R_n( \eta, \gamma, \theta) | = 0$ , which implies that

$$\limsup_{n\rightarrow \infty} \sup_{\theta \in [a,\, b]} | G_n (\theta \tau_n)\theta^{-\gamma}-1| = \sup_{\theta \in [a, \, b]} \frac{(1+ \theta)^{\gamma} -1}{\theta^\gamma} .$$

Note that $\frac{(1+ \theta)^{\gamma} -1}{\theta^\gamma} = \gamma \int_0^1 \frac{\textrm{d} v}{(\theta^{-1} +v)^{1-\gamma}}$ , which is increasing in $\theta$ . This entails (4.16).

We now turn to the proof of Lemma 6 in the second case. Applying Lemma 7 to $a=\frac12(\eta \varepsilon)^{1/\gamma}$ and $b=\gamma^2$ , and since $\lim_{\varepsilon \to 0^+}\gamma^{-3} \gamma^{-2\gamma} ((1+ \gamma^2)^{\gamma} - 1)= 1$ , it follows that there exists $\varepsilon_{4}(\eta) \in (0, \, \varepsilon_{3} (\eta))$ such that for $\varepsilon \in (0,\, \varepsilon_{4} (\eta))$ there is an integer $n_{9} (\eta, \, \varepsilon) \geq n_{8} (\eta, \, \varepsilon)$ such that for $n\geq n_{9} (\eta, \, \varepsilon)$ and $z\in \big( \frac12 (\eta \varepsilon)^{1/\gamma} \tau_n, \, \gamma^2 \tau_n \big)$ ,

(4.17) \begin{equation}(1 - 2 \gamma^3) \Big( \frac{z}{\tau_n}\Big)^{\gamma} \leq G_n (z) \leq (1 + 2 \gamma^3) \Big( \frac{z}{\tau_n}\Big)^{\gamma} .\end{equation}

We recall that $\textrm{LHS}_{(4.5)}$ stands for the expression on the left-hand side of (4.5). Then

where:

• • $\widetilde{\textrm{I}}_n(z) \;:\!=\; G_n(z)^2 - {{\mathbb P}}(Z_n + \widehat{Z}_n \le z)$ ;

• • $\widetilde{\textrm{I I}}_n(z) \;:\!=\; G_{n+1}\left(\frac{z}{1+\eta \gamma}\right) - G_n(z)$ ;

• • $\widetilde{\textrm{I I I}}_n(z) \;:\!=\; 2G_n(z) - G_n(z)^2 - {{\mathbb P}}(Z_n + \widehat{Z}_n \le z)=2G_n(z) - 2G_n(z)^2 + \widetilde{\textrm{I}}_n(z)$ .

We claim that there exists $\varepsilon_{5} (\eta) \in (0, \, \varepsilon_{4} (\eta))$ such that for $\varepsilon \in (0, \, \varepsilon_{5} (\eta))$ there is an integer $n_{10} (\eta, \, \varepsilon)$ such that for $n\geq n_{10} (\eta, \, \varepsilon)$ ,

(4.18) \begin{equation} \widetilde{\textrm{I}}_n(z) \le 4 (1 - \eta) \varepsilon \, G_n(z)^2, \quad z\in [(\eta \varepsilon)^{1/\gamma} \tau_n, \, \gamma^2 \tau_n). \end{equation}

To see why (4.18) is true, we recall that $G_n(z) = \int_{\sigma_n}^z G_n'(t) \, \textrm{d} t$ and that ${{\mathbb P}}(Z_n + \widehat{Z}_n \le z) = \int_{\sigma_n}^{z-\sigma_n} G_n'(t) G_n(z-t) \, \textrm{d} t$ . Since $G_n(\sigma_n)= 0$ , by definition of $G_n$ , we thus get

$$\widetilde{\textrm{I}}_n (z)=G_n(z) \big( G_n(z) - G_n(z - \sigma_n) \big) +\, \int_{\sigma_n}^{z-\sigma_n} G_n'(t) \big( G_n(z) - G_n(z - t)\big) \, \textrm{d} t .$$

To get an upper bound for the first term on the right-hand side, we observe that

$$G_n(z) - G_n(z - \sigma_n)=\int_{z-\sigma_n}^{z} G_n'(t) \, \textrm{d} t = \int_{z-\sigma_n}^{z} 2b_n \big(\varphi_\gamma'(t) - \varphi_\gamma'(t+\tau_n) \big) \, \textrm{d} t \leq \int_{z-\sigma_n}^{z} 2b_n \varphi_\gamma'(t) \, \textrm{d} t .$$

For $t\in (z-\sigma_n, \, z)$ , we have

$$\varphi_\gamma'(t)=\frac{\gamma}{t} \sinh( \gamma \log t) \le\frac{\gamma}{z - \sigma_n} \sinh( \gamma \log z) \le\frac{\gamma}{2(z - \sigma_n)} z^\gamma ,$$

so that for $z\in [(\eta \varepsilon)^{1/\gamma} \tau_n, \, \gamma^2 \tau_n)$ ,

$$G_n(z) - G_n(z - \sigma_n)\le\frac{b_n \sigma_n \gamma z^\gamma}{z - \sigma_n}\le\frac{b_n \sigma_n \gamma z^\gamma}{(\eta \varepsilon)^{1/\gamma} \tau_n - \sigma_n}=\frac{\gamma}{\varepsilon \tau_n} \frac{b_n \tau_n^\gamma \sigma_n}{(\eta \varepsilon)^{1/\gamma} - \frac{\sigma_n}{\tau_n}} \varepsilon \Bigg( \frac{z}{\tau_n}\Bigg)^{ \gamma} .$$

By (4.17), this yields that

$$G_n(z) - G_n(z - \sigma_n)\le\frac{\gamma}{(1 - 2\gamma^3) \varepsilon \tau_n} \, \frac{b_n \tau_n^\gamma \sigma_n}{(\eta \varepsilon)^{1/\gamma} - \frac{\sigma_n}{\tau_n}} \, \varepsilon G_n(z) .$$

Let $\varepsilon_{5} (\eta) \in (0, \, \varepsilon_{4} (\eta))$ be such that $6 \gamma^3 < \eta$ for $\varepsilon \in (0, \, \varepsilon_{5} (\eta))$ . Note that

(4.19) \begin{equation}4 (1 - \eta) \left(\eta - 6 \gamma^3\right) + \left(1+ 6\gamma^3\right) 4 (1 - \eta)^2 < 4 (1 - \eta) .\end{equation}

Let $\varepsilon \in (0, \, \varepsilon_{5} (\eta))$ . Since

$$\lim_{n\to \infty} \frac{1}{\varepsilon (1 - 2\gamma^3)\tau_n} = 0\quad \mbox{and} \quad \lim_{n\to \infty} \frac{b_n\tau^{\gamma}_n \gamma \sigma_n}{(\eta \varepsilon)^{1/\gamma} - \frac{\sigma_n}{\tau_n}} = \gamma (\eta \varepsilon)^{-\frac{1}{\gamma}}$$

by (4.3), whereas $4 (1 - \eta) (\eta - 6 \gamma^3) >0$ , there exists $n_{11} (\eta, \, \varepsilon) \geq n_{9} (\eta, \, \varepsilon)$ such that for $n\ge n_{11} (\eta, \, \varepsilon)$ ,

(4.20) \begin{equation}G_n(z) \big( G_n(z) - G_n(z - \sigma_n)\big) \leq 4 (1 - \eta) (\eta - 6\gamma^3) \, \varepsilon G_n(z)^2 , \quad z\in [(\eta \varepsilon)^{1/\gamma} \tau_n, \, \gamma^2 \tau_n) .\end{equation}

The specific choice of the factor in front of $ \varepsilon G_n(z)^2$ in the right-hand side member of the last inequality is justified further.

To deal with $\int_{\sigma_n}^{z-\sigma_n} G_n'(t) (G_n(z) - G_n(z - t)) \, \textrm{d} t$ , we recall that $G'_n(t) \leq 2b_n \varphi_\gamma'(t)$ and we note that $G_n(z) - G_n(z-t) \le \int_{z-t}^z 2b_n \varphi_\gamma'(s)\, \textrm{d} s = 2b_n (\varphi_\gamma(z) - \varphi_\gamma (z - t))$ , so

\begin{eqnarray*} \int_{\sigma_n}^{z-\sigma_n} G_n'(t) \big( G_n(z) - G_n(z - t) \big) \, \textrm{d} t &\,\le\,& 4b_n^2 \int_{\sigma_n}^{z-\sigma_n} \varphi_\gamma'(t) \big( \varphi_\gamma(z) - \varphi_\gamma (z - t) \big) \, \textrm{d} t \\[5pt] &\,\le\,& 4b_n^2 \int_{1}^{z-1} \varphi_\gamma'(t) \big( \varphi_\gamma(z) - \varphi_\gamma (z - t) \big) \, \textrm{d} t .\end{eqnarray*}

(Indeed, (4.7) implies that if $z\geq (\eta \varepsilon)^{1/\gamma} \tau_n $ , then $z> 2\sigma_n >2$ .) The integral on the right-hand side is $\mathtt{cvl}_{\gamma, 1} (z)$ by our definition in (3.10). By Lemma 3, we get that

$$\int_{\sigma_n}^{z-\sigma_n} G_n'(t) \big( G_n(z) - G_n(z - t) \big) \, \textrm{d} t\le4b_n^2 \zeta(2) \gamma^2 \big(\sinh(\gamma \log z) \big)^2 .$$

Observe that

$$4 \big(\!\sinh(\gamma \log z) \big)^2 \le z^{2\gamma} = \tau_n^{2\gamma} \left(\frac{z}{\tau_n}\right)^{2\gamma},$$

which is bounded by $(\tau_n^{2\gamma}/(1-2\gamma^3)^2) G_n(z)^2$ (see (4.17)). Therefore,

$$\int_{\sigma_n}^{z-\sigma_n}G_n'(t) \big( G_n(z) - G_n(z - t) \big) \, \textrm{d} t\le\zeta(2) \gamma^2 \, \frac{(b_n \tau_n^\gamma)^2}{(1-2\gamma^3 )^2} \, G_n(z)^2 .$$

Recall that $\zeta(2) \gamma^2 = 4(1 - \eta)^2 \varepsilon$ , and that $\lim_{n\to \infty} b_n\tau_n^{\gamma}= 1$ (see (4.3)). Since ${1}/{(1-2\gamma^3)^2} < 1+ 6 \gamma^3$ (because $\gamma < \tfrac13$ ), there exists $n_{10} (\eta, \, \varepsilon) \geq n_{11} (\eta, \, \varepsilon) $ such that for $n\ge n_{10} (\eta, \, \varepsilon)$ , we have

(4.21) \begin{equation} \int_{\sigma_n}^{z-\sigma_n} G_n'(t) \big( G_n(z) - G_n(z - t) \big) \, \textrm{d} t \leq (1+ 6\gamma^3) 4 (1 - \eta)^2 \, \varepsilon G_n(z)^2 , \quad z\in [(\eta \varepsilon)^{1/\gamma} \tau_n, \, \gamma^2 \tau_n) .\end{equation}

We obtain (4.18) by (4.20), (4.21), and (4.19).

Let us consider $\widetilde{\textrm{I I}}_n(z)$ . Let $\varepsilon \in (0, \, \varepsilon_{5} (\eta))$ , $n\geq n_{10} (\eta, \, \varepsilon)$ , and $z\in [(\eta \varepsilon)^{1/\gamma} \tau_n, \, \gamma^2 \tau_n)$ . We start with the trivial inequality $\widetilde{\textrm{I I}}_n(z) \le G_{n+1}(z) - G_n(z)$ , and look for an upper bound for $G_{n+1}(z) - G_n(z)$ . Since $\lim_{\varepsilon \to 0^+} (1 - (2+\eta)\varepsilon)^{1/\gamma} = 1$ and $\lim_{\varepsilon \to 0^+} (\gamma^3/\varepsilon) = 0$ , there exists $\varepsilon_{5} (\eta)\in (0, \, \min\{ \varepsilon_{5} (\eta), \, \eta/2\} )$ such that

(4.22) \begin{equation} \frac{1}{2} <(1 - (2+\eta)\varepsilon)^{1/\gamma}\quad \textrm{and} \quad 5\gamma^3 < \frac{1}{2}\eta \varepsilon, \quad \varepsilon \in (0, \, \varepsilon_{6} (\eta)).\end{equation}

We fix $\varepsilon \in (0, \, \varepsilon_{6} (\eta))$ . Since $\lim_{n\to \infty} \frac{\tau_n}{\tau_{n+1}} = (1 - (2+\eta)\varepsilon)^{1/\gamma}$ (by (4.3)), there exists an integer $\varepsilon_{6} (\eta, \, \varepsilon) \geq n_{11} (\eta, \,\varepsilon) $ such that

$$\frac{\tau_n}{\tau_{n+1}} >\frac{1}{2}$$

for $n\geq n_{12} (\eta, \, \varepsilon)$ , which implies that $z\in \left(\frac{1}{2} (\eta \varepsilon)^{1/\gamma} \tau_{n+1}, \, \gamma^2 \tau_{n+1}\right)$ ; so applying (4.17) twice leads to

$$G_{n+1} (z)\le\big(1 + 2 \gamma^3\big) \left( \frac{z}{\tau_{n+1}}\right)^{\gamma}\le\frac{1 + 2 \gamma^3 }{1 - 2 \gamma^3 } \frac{\tau_n^\gamma}{\tau_{n+1}^\gamma} G_n (z) .$$

Note that

$$\frac{1 + 2 \gamma^3 }{1 - 2 \gamma^3 } < 1 + 5 \gamma^3$$

(because $\gamma < \frac13$ ), and that

$$\lim_{n\to \infty} \frac{\tau_n^\gamma}{\tau_{n+1}^\gamma} = \lim_{n\to \infty} \frac{b_{n+1}}{b_n} = 1 - (2 + \eta) \varepsilon$$

(see (4.3)). Thus, there exists an integer $\varepsilon_{13} (\eta, \, \varepsilon) \geq n_{12} (\eta, \, \varepsilon)$ such that $G_{n+1} (z) \leq (1 + 5 \gamma^3 ) \big( 1 - (2 + \eta) \varepsilon\big) G_n (z)$ for $n\ge n_{13} (\eta, \, \varepsilon)$ and $z\in [(\eta \varepsilon)^{1/\gamma} \tau_n, \, \gamma^2 \tau_n)$ ; consequently,

$$\widetilde{\textrm{I I}}_n(z)\le G_{n+1} (z) - G_n (z)\le\big[ (1 + 5 \gamma^3 ) ( 1 - (2 + \eta) \varepsilon) -1 \big] G_n (z) .$$

Since $\big(1 + 5 \gamma^3 \big) ( 1 - (2 + \eta) \varepsilon) -1 = 5\gamma^3 - (2 + \eta) \varepsilon - 5\gamma^3 (2 + \eta)\varepsilon < 5\gamma^3 - (2 + \eta) \varepsilon$ , which is smaller than $-(2 + \tfrac12 \eta) \varepsilon$ (by (4.22)), we deduce that for $n\ge n_{13} (\eta, \, \varepsilon)$ and $z\in [(\eta \varepsilon)^{1/\gamma} \tau_n, \, \gamma^2 \tau_n)$ ,

(4.23) \begin{equation} \widetilde{\textrm{I I}}_n(z) \le -\left(2 + \frac12\eta\right) \, \varepsilon G_n (z) . \end{equation}

Finally, we turn to $\widetilde{\textrm{I I I}}_n(z)$ , which is easy to estimate:

(4.24) \begin{eqnarray} \widetilde{\textrm{I I I}}_n(z) &\,=\,& 2G_n(z) - 2G_n(z)^2 + \widetilde{\textrm{I}}_n(z) \nonumber \\[5pt] &\,\le\,& 2G_n(z) - 2G_n(z)^2 + 4 (1 - \eta)\varepsilon G_n(z)^2 \nonumber \\[5pt] &\,\le\,& 2G_n(z) - 2G_n(z)^2 + 4 \varepsilon G_n(z)^2 .\end{eqnarray}

Assembling (4.18), (4.23), and (4.24) yields that for $\varepsilon \in (0, \, \varepsilon_{6} (\eta))$ , $n\geq n_{13} (\eta,\, \varepsilon) $ , and $z\in [(\eta \varepsilon)^{1/\gamma} \tau_n, \, \gamma^2 \tau_n)$ ,

which is non-positive since $\varepsilon< \varepsilon_{6} (\eta) < \frac{\eta}{2}$ . This completes the proof of Lemma 6 when $z\in [(\eta \varepsilon)^{1/\gamma} \tau_n, \, \gamma^2 \tau_n)$ (i.e. in the second case).

Proof of Lemma 6: third case $\gamma^2 \tau_n \le z < {\textrm{e}}^{1/\sqrt{\gamma}} \tau_n$ . We easily check that there exists ${q}_0 \in (0, \, \frac19)$ such that for all ${q} \in (0, q_0)$ the following holds true.

(4.25) \begin{eqnarray} &&\left({\textrm{e}}^{1/\sqrt{{q}}} + 1\right)^{2{q}}\leq 1 + 3 \sqrt{{q}} < 2, \end{eqnarray}

(4.26) \begin{eqnarray} && {q}^2 - {\textrm{e}}^{-4/\sqrt{{q}}} > {q}^3 > {\textrm{e}}^{-1/\sqrt{{q}}}, \end{eqnarray}

(4.27) \begin{eqnarray} && ( 1 - {q}) \bigg( \frac{{q}^3}{2} \bigg)^{q} > 1 - \sqrt{{q}}. \end{eqnarray}

For any ${q} \in (0, \, {q}_0]$ , we define the function

(4.28) \begin{equation} g_{q} (\theta) \;:\!=\; (\theta+1)^{q} - \theta^{q}, \quad \theta \in \mathbb{R}_+^* \;:\!=\; (0, \, \infty) . \end{equation}

Since $\lim_{\theta\to 0^+} g_{q} (\theta) = 1$ , $\lim_{\theta\to \infty} g_{q} (\theta) = 0$ , and $g^\prime_{q} (\theta) = {q} ((\theta+1)^{{q}-1} - \theta^{{q} -1} ) < 0$ for $\theta\in \mathbb{R}_+^*$ , it follows that $g_{q}$ is a decreasing bijection from $\mathbb{R}_+^*$ onto $(0, \, 1)$ .

We collect some elementary properties of $g_{q}$ .

We proceed to the proof of Lemma 6 in the third case. Let ${q}_0\in (0, \, \frac19)$ be the small constant ensuring (4.25)–(4.27) hold. Fix $\eta \in (0, \, 1)$ . We set

(4.32) \begin{equation}c_* = c_* (\eta) \;:\!=\; \eta(5 -2\eta) + \frac{4\eta(1 -\eta)^2}{\zeta(2)} >0.\end{equation}

Let $\varepsilon\in (0, \, \varepsilon_{7}(\eta))$ , where $\varepsilon_{7}(\eta) \in (0, \, \min\{\varepsilon_{6}(\eta), \, \frac12\})$ is such that

(4.33) \begin{eqnarray} &&\gamma = \gamma(\varepsilon, \, \eta) \;:\!=\; \frac{2}{\sqrt{\zeta(2)}} (1 - \eta)\sqrt{\varepsilon} \in (0, \, {q}_0) , \quad (1 + \eta \gamma) \big( 1 - (2 + \eta) \varepsilon \big)^{ -\frac{1}{\gamma}} <2 , \end{eqnarray}

(4.34) \begin{eqnarray} &&\bigg(\frac{1}{2} + \varepsilon\bigg)\left(1 + 3\sqrt{\gamma}\,\right) \le \frac{1}{2} + 2 \sqrt{\gamma} , \end{eqnarray}

(4.35) \begin{eqnarray} &&c_* \varepsilon > \bigg(\eta \gamma + \frac{(2 + \eta) \varepsilon}{\gamma}\bigg) (\gamma^{3/2} + \eta \gamma^2 + (2+ \eta) \varepsilon) + 2 \gamma^{5/2} \zeta(2) + 12\varepsilon \gamma^{1/2} + 16{\textrm{e}}^{-1/\sqrt{\gamma}} .\nonumber \\[-2pt] && \end{eqnarray}

The seemingly complicated form of the right-hand member of (4.35) is justified in the following: it is the sum of several explicit inequalities. We could have used slightly simpler terms, but the price would have been an extra layer of numbered constants, which we wanted to avoid for the sake of clarity.

We want to check (4.5) in Lemma 6 when $z= \tau_n x$ , with $x\in [\gamma^2, \, {\textrm{e}}^{1/\sqrt{\gamma}})$ . To this end we first need to find a lower bound for ${{\mathbb P}} (Z_{n+1}> x\tau_n /(1+ \eta \gamma) )$ . More precisely, in the third case (4.5) is equivalent to prove that there exists ${\varepsilon}'\in (0, \infty)$ (depending on $\eta$ ) such that for all ${\varepsilon}\in (0, {\varepsilon}')$ there is $n({\varepsilon}')$ such that for all $n\geq n({\varepsilon}')$ and all $ x\in [\gamma^2, \, {\textrm{e}}^{1/\sqrt{\gamma}})$ :

(4.36) \begin{equation} [\Psi_{ \frac12 + \varepsilon}(\nu_n')] ((x \tau_n, \, \infty)) \le {{\mathbb P}} \left( Z_{n+1} > \frac{x \tau_n}{1+\eta \gamma} \right) . \end{equation}

where $\Psi_{ p}$ is the transformation in (2.9).

To this end, we first study the tail probability of $\frac{Z_n}{\tau_n}$ . Let $a>0$ . By (4.4), for all sufficiently large n such that $a \tau_n \ge \sigma_n$ , and all $\theta\ge a$ ,

\begin{eqnarray*} {{\mathbb P}} (Z_n > \theta \tau_n) &\,=\, & b_n\tau_n^{\gamma} \big( (\theta + 1)^\gamma - \theta^\gamma \big) - b_n\tau_n^{-\gamma} \big( \theta^{-\gamma} - (\theta + 1)^{-\gamma}\big) \\[5pt] &\,=\,& g_\gamma (\theta) \big( b_n\tau_n^{\gamma} - b_n\tau_n^{-\gamma} \theta^{-\gamma} (\theta+1)^{-\gamma} \big) ,\end{eqnarray*}

where $g_\gamma$ is the function in (4.28). Hence,

$$\sup_{\theta\in [a, \, \infty)}\big| g_\gamma (\theta)^{-1} {{\mathbb P}} (Z_n > \theta \tau_n) - 1\big|\le|b_n\tau_n^{\gamma} - 1| + b_n\tau_n^{-\gamma} a^{-\gamma} .$$

Since $\lim_{n\to \infty} b_n\tau_n^{\gamma} = 1$ and $\lim_{n\to \infty} b_n\tau_n^{-\gamma} = 0$ (by (4.3)), this implies that

(4.37) \begin{equation} \lim_{n\to \infty} \sup_{\theta \in [a, \, \infty )} \big| g_\gamma (\theta)^{-1} {{\mathbb P}} (Z_n > \theta \tau_n) - 1\big| = 0 . \end{equation}

Note that $\sup_{\theta \in [a, \, \infty )} g_\gamma (\theta) = g_\gamma (a) < \infty$ ; thus, we also have $\lim_{n\to \infty} \sup_{\theta \in [a, \, \infty )} \big| {{\mathbb P}} (Z_n > \theta \tau_n) - g_\gamma (\theta)\big| = 0$ . Taking $a= \frac12 {\textrm{e}}^{-4/\sqrt{\gamma}}$ yields the existence of a positive integer $n_{14} (\eta, \, \varepsilon)$ such that for $n\ge n_{14}( \eta, \, \varepsilon)$ and $\theta \in [\frac12 {\textrm{e}}^{-4/\sqrt{\gamma}}, \, \infty)$ ,

(4.38) \begin{eqnarray} \big| g_\gamma (\theta)^{-1} {{\mathbb P}} (Z_n > \theta \tau_n) - 1\big| & \,\le\, &{\textrm{e}}^{-2/\sqrt{\gamma}}, \end{eqnarray}

(4.39) \begin{eqnarray} \big| {{\mathbb P}} (Z_n > \theta \tau_n) - g_\gamma (\theta) \big| & \,\le\, &{\textrm{e}}^{-2/\sqrt{\gamma}} . \end{eqnarray}

In order to deal with ${{\mathbb P}} ( Z_{n+1} > x \tau_n/({1+\eta \gamma}))$ on the right-hand side of (4.36), we write

$$\varrho_n (\gamma) \;:\!=\;(1 + \eta \gamma) \, \frac{\tau_{n+1}}{\tau_n} .$$

By (4.3), $\lim_{n\to \infty} b_n \tau^\gamma_n =1$ , whereas by definition, $\frac{b_{n+1}}{b_n} = 1-(2 + \eta) \varepsilon$ . By using the inequality $(1- y)^{-z} \ge 1+ yz$ for $y \in (0, \, 1)$ and $z \ge 0$ , we therefore get

$$\lim_{n\to \infty} \rho_n (\gamma) =(1 + \eta \gamma) ( 1 - (2 + \eta) \varepsilon )^{ -\frac{1}{\gamma}} \ge(1 + \eta \gamma) \bigg( 1 + \frac{(2 + \eta) \varepsilon}{\gamma} \bigg) >1 + \eta \gamma + \frac{(2 + \eta) \varepsilon}{\gamma} .$$

In addition, by (4.33), $\lim_{n\to \infty} \rho_n (\gamma) < 2$ . Therefore, there exists an integer $n_{15} (\eta, \, \varepsilon) \geq n_{14} (\eta, \, \varepsilon)$ such that for $n\ge n_{15} (\eta, \, \varepsilon)$ ,

$$1 + \eta \gamma + \frac{(2 + \eta) \varepsilon}{\gamma} < \rho_n (\gamma) < 2 .$$

Let $x\in [\gamma^2, \, {\textrm{e}}^{1/\sqrt{\gamma}})$ . We have

$$\frac{x \tau_n}{1+\eta \gamma} = \frac{x\tau_{n+1}}{\rho_n (\gamma)},$$

and

$$\frac{x}{\rho_n (\gamma)} > \frac{\gamma^2}{2} \ge \frac12 {\textrm{e}}^{-4/\sqrt{\gamma}}$$

(by the elementary inequality $z^2 \le {\textrm{e}}^{4\sqrt{z}}$ , for $z \ge 1$ ), so we are entitled to apply (4.39) to $\theta \;:\!=\; x/{\rho_n (\gamma)}$ to see that

(4.40) \begin{equation} {{\mathbb P}} \bigg( Z_{n+1} > \frac{x \tau_n}{1 + \eta \gamma} \bigg) \ge g_\gamma \bigg( \frac{x}{\rho_n (\gamma)} \bigg) - {\textrm{e}}^{-2/\sqrt{\gamma}} \ge g_\gamma \bigg( \frac{x}{1+ \eta \gamma + \frac{(2 + \eta) \varepsilon}{\gamma}} \bigg) - {\textrm{e}}^{-2/\sqrt{\gamma}} . \end{equation}

To prove (4.36), we need to find an appropriate upper bound for $[\Psi_{ \frac12 + \varepsilon}(\nu_n')] ((x \tau_n, \, \infty))$ . Let $Z_n^* \;:\!=\; \frac{Z_n}{\tau_n}$ , and let $\widehat{Z}_n^*$ denote an independent copy of $Z_n^*$ . For $\theta \in \big(\gamma^3, \, {\textrm{e}}^{1/\sqrt{\gamma}}\big)$ , we define the event $B_n(\theta) \;:\!=\; \{ Z^*_n + \widehat{Z}_n^* \ge \theta \}$ , and note that

\begin{eqnarray*} {{\mathbb P}} (B_n(\theta)) &\,\le\,& {{\mathbb P}} \big(Z^*_n \geq \theta \big) + {{\mathbb P}} \big( \widehat{Z}_n^* \geq \theta ; \, Z^*_n < \theta \big ) + {{\mathbb P}} \big( \widehat{Z}_n^* < \theta ; \, Z^*_n < \theta; \, B_n(\theta)\big) \\[5pt] & \,=\, & 2{{\mathbb P}} \big(Z^*_n \geq \theta \big) - [{{\mathbb P}} \big(Z^*_n \geq \theta\big)]^2 + {{\mathbb P}} \big( \widehat{Z}_n^* < \theta ; \, Z^*_n < \theta; \, B_n(\theta)\big) .\end{eqnarray*}

Therefore,

(4.41) \begin{eqnarray} \big[\Psi_{ \frac12 + \varepsilon}(\nu_n')\big] ((x \tau_n, \, \infty)) & \,=\, & \bigg(\frac{1}{2} + \varepsilon \bigg) {{\mathbb P}} (B_n(x)) + \bigg(\frac{1}{2} - \varepsilon \bigg)\, [{{\mathbb P}}( Z^*_n \ge x)]^2 \nonumber \\[5pt] &\,\le\,& (1 + 2\varepsilon) {{\mathbb P}} (Z^*_n \geq x) + \bigg(\frac{1}{2} + \varepsilon \bigg) {{\mathbb P}} ( \widehat{Z}_n^* < x ; \, Z^*_n < x; \, B_n(x)) .\end{eqnarray}

We first get an upper bound for the second term in the right-hand side member of (4.41). To this end, we write $x_1 \;:\!=\; x - {\textrm{e}}^{-4/\sqrt{\gamma}}$ , so by (4.26), $x_1 \ge \gamma^2 - {\textrm{e}}^{-4/\sqrt{\gamma}} > \gamma^3 > {\textrm{e}}^{-1/\sqrt{\gamma}}$ . We consider $\{ \widehat{Z}_n^* < x\}$ as the union of $\{ \widehat{Z}_n^* < {\textrm{e}}^{-4/\sqrt{\gamma}} \}$ , $\{ \widehat{Z}_n^* \in [{\textrm{e}}^{-4/\sqrt{\gamma}}, \, x_1]\}$ , and $\{ \widehat{Z}_n^* \in (x_1, \, x)\}$ . On $\{ \widehat{Z}_n^* < {\textrm{e}}^{-4/\sqrt{\gamma}} \} \cap B_n(x)$ , we have $Z_n > x_1$ . This implies that

(4.42) \begin{eqnarray} &&\big[\Psi_{ \frac12 + \varepsilon}(\nu_n')\big] ((x \tau_n, \, \infty)) \nonumber \\[5pt] & \,\le\, & (1 + 2\varepsilon) {{\mathbb P}} (Z^*_n \geq x) + (1 + 2\varepsilon) {{\mathbb P}} (Z^*_n \in (x_1, \, x) ) \nonumber \\[5pt] && +\, \bigg(\frac{1}{2} + \varepsilon \bigg) {{\mathbb P}} \big( \widehat{Z}_n^* \in \big[{\textrm{e}}^{-4/\sqrt{\gamma}}, \, x_1\big] ; \, Z^*_n \in \big(x - \widehat{Z}_n^*, \, x\big)\big) \nonumber \\[5pt] & \,=\, & (1 + 2\varepsilon) {{\mathbb P}} (Z^*_n > x_1) + \bigg(\frac{1}{2} + \varepsilon \bigg) {{\mathbb P}} \big( \widehat{Z}_n^* \in \big[{\textrm{e}}^{-4/\sqrt{\gamma}}, \, x_1\big] ; \, Z^*_n \in \big(x - \widehat{Z}_n^*, \, x\big)\big) . \end{eqnarray}

We then estimate the two probability expressions on the right-hand side. Since we have proved that $x_1 > e^{-1/\sqrt{\gamma}}$ , (4.39) applies with $\theta= x_1$ and we get ${{\mathbb P}} (Z^*_n \ge x_1) \le g_\gamma (x_1) + {\textrm{e}}^{-2/\sqrt{\gamma}}$ . By the mean-value theorem, $g_\gamma (x_1) - g_\gamma (x) = -(x - x_1)g_\gamma'(y) = - {\textrm{e}}^{-4/\sqrt{\gamma}}g_\gamma'(y)$ for some $y\in [x_1, \, x]$ . Since $-g_\gamma'(y) = \gamma [y^{-(1-\gamma)} - (y + 1)^{-(1-\gamma)}] \le \gamma y^{-(1-\gamma)} \le \gamma (x_1)^{-(1-\gamma)} < \gamma^{1-3(1-\gamma)} \le \gamma^{-2}$ (we have used the fact that $x_1 > \gamma^3$ ), we obtain that

(4.43) \begin{equation} g_\gamma (x_1) - g_\gamma (x) \le \gamma^{-2}\, {\textrm{e}}^{-4/\sqrt{\gamma}} \le {\textrm{e}}^{-2/\sqrt{\gamma}} \, \end{equation}

(here, we have used also the elementary inequality $z^2 \le {\textrm{e}}^{2\sqrt{z}}$ for $z\ge 1$ ). Therefore,

$$ {{\mathbb P}} (Z^*_n > x_1) \le g_\gamma (x) + 2{\textrm{e}}^{-2/\sqrt{\gamma}}.$$

By (4.29) (applied to $a= {\textrm{e}}^{-1/\sqrt{\gamma}}$ and ${q} = \gamma$ , which is possible since $\gamma < {q}_0$ ), $g_\gamma (x) \le \frac{6 \gamma^{1/2}}{x+1}$ . Since $2(1 + 2\varepsilon) \le 3$ , this implies that

(4.44) \begin{eqnarray} (1 + 2\varepsilon) {{\mathbb P}} (Z^*_n > x_1) &\,\le\, & g_\gamma (x)+ 2{\varepsilon} g_\gamma (x) + 2(1+2{\varepsilon}) {\textrm{e}}^{-2/\sqrt{\gamma}} \nonumber\\[5pt] & \,\leq\, & g_\gamma (x) + \frac{12 \varepsilon \gamma^{1/2}}{x + 1} + 3{\textrm{e}}^{-2/\sqrt{\gamma}} . \end{eqnarray}

We next prove an upper bound for ${{\mathbb P}} (\widehat{Z}_n^* \in [{\textrm{e}}^{-4/\sqrt{\gamma}}, \, x_1] ; \, Z^*_n \in (x - \widehat{Z}_n^*, \, x))$ thanks to (4.39). We fix $t\in (0, \, x_1)$ and we first observe that $x\geq x-t \geq x-x_1= e^{-4/\sqrt{\gamma}}$ . Then (4.39) applies with $\theta$ equal to x and $x-t$ and we get

$${{\mathbb P}}( Z^*_n \in (x - t, \, x))= {{\mathbb P}} ( Z^*_n > \; x-t) - {{\mathbb P}} (Z^*_n >x) \leq g_\gamma(x - t) - g_\gamma(x) + 2{\textrm{e}}^{-2/\sqrt{\gamma}} .$$

To simplify, we next denote by $f_{Z^*_n}({\cdot})$ the density of $Z^*_n$ and we set $\gamma_1 \;:\!=\; {\textrm{e}}^{-4/\sqrt{\gamma}} = x - x_1$ . Then

(4.45) \begin{align} {{\mathbb P}} ( \widehat{Z}_n^* \in [\gamma_1, \, x_1] ; \, Z^*_n \in (x - \widehat{Z}_n^*, \, x)) &= \int_{\mathbb R} f_{Z^*_n} (t) \textbf 1_{[\gamma_1 , x_1]} (t) {{\mathbb P}} \big( Z^*_n \in (x-t, x] \big) \, \textrm{d} t \nonumber \\[5pt] &\le \int_{\mathbb R}\textbf 1_{[\gamma_1 , x_1]} (t) f_{Z^*_n}(t) \big( g_\gamma(x - t) - g_\gamma(x) \big) \, \textrm{d} t + 2{\textrm{e}}^{-2/\sqrt{\gamma}}. \end{align}

Then, by Fubini, observe that

\begin{align*} \int_{\mathbb R} \textbf 1_{[\gamma_1 , x_1]} (t) f_{Z^*_n}(t) \big( g_\gamma(x - t) - g_\gamma(x) \big) \, \textrm{d} t &= \int_{\mathbb R} \textrm{d} t \int_{\mathbb R} \textrm{d} s \, f_{Z^*_n} (t) \textbf 1_{[\gamma_1 , x_1]} (t) \textbf 1_{[0, t]} (x-s) \big( - g'_\gamma(s) \big)\\[5pt] & = \int_{0}^{x_1} \textrm{d} u \int_{\mathbb R} \textrm{d} t \, f_{Z^*_n} (t) \textbf 1_{[\gamma_1 , x_1]} (t) \textbf 1_{[u, \infty)} (t) \big|g'_\gamma(x-u) \big| \\[5pt] &= \int_{0}^{x_1} \big|g'_\gamma(x-u) \big| {{\mathbb P}} \big( Z^*_n \in [\gamma_1 \vee u, x_1]\big) \, \textrm{d} u. \end{align*}

Note that if $u\in [0, x_1]$ , then $x_1 \geq u\vee \gamma_1 \geq \gamma_1 > \frac12 {\textrm{e}}^{-4/\sqrt{\gamma}}$ . Thus, (4.39) applies and we get

(4.46) \begin{align}& \int_{0}^{x_1} \big|g'_\gamma(x-u) \big| {{\mathbb P}} ( Z^*_n \in [\gamma_1 \vee u, x_1]) \textrm{d} u \leq \int_{0}^{x_1} \big|g'_\gamma(x - u) \big| ( g_\gamma (u \vee \gamma_1) - g_\gamma (x_1) + 2e^{-2/\sqrt{\gamma}}) \textrm{d} u\nonumber \\[5pt] &\leq \int_{0}^{x} \big|g'_\gamma(x- u) \big| \big( g_\gamma (u) - g_\gamma (x) \big) \, \textrm{d} u + 2e^{-2/\sqrt{\gamma}} \int_{0}^{x_1} \big|g'_\gamma(x- u) \big| \, \textrm{d} u ,\end{align}

since $g_\gamma$ is decreasing. Then

(4.47) \begin{eqnarray} \int_{0}^{x_1} \big|g'_\gamma(x-u) \big| {{\mathbb P}} \big( Z^*_n & \,\in\, & [\gamma_1 \vee u, x_1]\big) \textrm{d} u \nonumber \\[5pt] &\,\leq \, & \int_{0}^{x} \big|g'_\gamma(t) \big| \big( g_\gamma (x - t) - g_\gamma (x) \big) \textrm{d} t + 2e^{-2/\sqrt{\gamma}} \big( g_{\gamma} (\gamma_1) - g_\gamma (x_1) \big)\nonumber \\[5pt] &\,\leq \, & (1 + 3\sqrt{\gamma}\,) \gamma^2 \, \frac{\zeta(2)}{x + 1} + 2e^{-2/\sqrt{\gamma}}\end{eqnarray}

by (4.31), Lemma 8(iii) (with ${q}=\gamma< {q}_0$ and $ \theta= x < {\textrm{e}}^{1/\sqrt{\gamma}}$ ), and since $x - x_1= \gamma_1 $ and $ g_{\gamma} (\gamma_1) - g_\gamma (x_1)\leq g_{\gamma} (\gamma_1) < 1$ . By (4.45) and (4.47) we then get

$${{\mathbb P}} ( \widehat{Z}_n^* \in [\gamma_1, \, x_1] ; \, Z^*_n \in (x - \widehat{Z}_n^*, \, x))\le(1 + 3\sqrt{\gamma}\,) \gamma^2 \, \frac{\zeta(2)}{x + 1} +4{\textrm{e}}^{-2/\sqrt{\gamma}} .$$

Since $(\frac{1}{2} + \varepsilon)(1 + 3\sqrt{\gamma}\,) \le \frac{1}{2} + 2 \sqrt{\gamma}$ by (4.34) and $\frac{1}{2} + \varepsilon \le 1$ , it follows that

$$\bigg(\frac{1}{2} + \varepsilon \bigg) {{\mathbb P}} ( \widehat{Z}_n^* \in [\gamma_1, \, x_1] ; \, Z^*_n \in (x - \widehat{Z}_n^*, \, x))\le\bigg(\frac{1}{2} + 2\sqrt{\gamma}\bigg) \gamma^2 \frac{\zeta(2)}{x + 1} +4{\textrm{e}}^{-2/\sqrt{\gamma}}.$$

Combined with (4.42) and (4.44), we obtain that

\begin{align*} [\Psi_{ \frac12 + \varepsilon}(\nu_n')] ((x \tau_n, \, \infty)) & \le g_\gamma (x) + 3{\textrm{e}}^{-2/\sqrt{\gamma}} + \frac{12 \varepsilon \gamma^{1/2}}{x+1} + \bigg(\frac{1}{2} + 2\sqrt{\gamma}\bigg) \gamma^2 \frac{\zeta(2)}{x + 1} +4{\textrm{e}}^{-2/\sqrt{\gamma}} \\[5pt] & = g_\gamma (x) + \frac{12\varepsilon \gamma^{1/2} + \bigg(\frac{1}{2} + 2\sqrt{\gamma}\bigg) \gamma^2 \zeta(2)}{x + 1} + 7{\textrm{e}}^{-2/\sqrt{\gamma}}.\end{align*}

In view of (4.40), this yields that

\begin{eqnarray*} && {{\mathbb P}} \bigg( Z_{n+1} > \frac{x \tau_n}{1+ \eta \gamma} \bigg) - [\Psi_{ \frac12 + \varepsilon}(\nu_n')] ((x \tau_n, \, \infty)) \\[5pt] &\,\ge\,& g_\gamma \bigg( \frac{x}{1 + \eta \gamma + \frac{(2+ \eta) \varepsilon}{\gamma}} \bigg) - g_\gamma (x) - {\textrm{e}}^{-2/\sqrt{\gamma}} - \frac{12\varepsilon \gamma^{1/2} + (\frac{1}{2} + 2\sqrt{\gamma}\,) \gamma^2 \zeta(2)}{x + 1} - 7{\textrm{e}}^{-2/\sqrt{\gamma}} \\[5pt] &\,=\,& g_\gamma \bigg( \frac{x}{1 + \eta \gamma + \frac{(2+ \eta) \varepsilon}{\gamma}} \bigg) - g_\gamma (x) - \frac{\frac{\zeta(2)}{2} \gamma^2 + (2 \gamma^{5/2} \zeta(2) + 12\varepsilon \gamma^{1/2})}{x + 1} - 8{\textrm{e}}^{-2/\sqrt{\gamma}} .\end{eqnarray*}

We easily check that we can apply (4.30) with ${q}= \gamma <{q}_0$ , $\theta = x \in [\gamma^3, {\textrm{e}}^{1/\sqrt{\gamma}} ]$ , and $r= 1 + \eta \gamma + \frac{(2 + \eta) \varepsilon}{\gamma} \in [1,2]$ . Then we get

\begin{eqnarray*} g_\gamma \bigg( \frac{x}{1 + \eta \gamma + \frac{(2 + \eta) \varepsilon}{\gamma}} \bigg) - g_\gamma (x) &\,\ge\,& \frac{\gamma(1 - \gamma^{1/2})\left(\eta \gamma + \frac{(2 + \eta) \varepsilon}{\gamma}\right) - \gamma \left(\eta \gamma + \frac{(2 + \eta) \varepsilon}{\gamma}\right)^2}{x + 1} \\[5pt] &\,=\,& \frac{\eta \gamma^2 + (2 + \eta) \varepsilon- \left(\eta \gamma + \frac{(2 + \eta) \varepsilon}{\gamma}\right) (\gamma^{3/2} + \eta \gamma^2 + (2+ \eta) \varepsilon)}{x + 1} .\end{eqnarray*}

As a consequence,

$${{\mathbb P}} \left( Z_{n+1} > \frac{x \tau_n}{1+\eta \gamma} \right) -[\Psi_{ \frac12 + \varepsilon}(\nu_n')] ((x \tau_n, \, \infty))\ge\frac{\eta \gamma^2 + (2 + \eta) \varepsilon - \frac{\zeta(2)}{2} \gamma^2 - R_\varepsilon}{x + 1}-8 {\textrm{e}}^{-2/\sqrt{\gamma}},$$

where

$$R_\varepsilon \;:\!=\; \left(\eta \gamma + \frac{(2 + \eta) \varepsilon}{\gamma}\right) (\gamma^{3/2} + \eta \gamma^2 + (2+ \eta) \varepsilon) + 2 \gamma^{5/2} \zeta(2) + 12\varepsilon \gamma^{1/2}.$$

Recall that $\eta\in (0, \, 1)$ is a constant and recall from (4.32), (4.33), (4.34), and (4.35) that $\gamma = ({2}/{\sqrt{\zeta(2)}}) (1 - \eta)\sqrt{\varepsilon}$ . Thus,

$$\eta \gamma^2 + (2 + \eta) \varepsilon - \frac{\zeta(2)}{2} \gamma^2=c_* \, \varepsilon ,$$

where

$$c_*= \eta(5 -2\eta) + \frac{4\eta(1 -\eta)^2}{\zeta(2)} > 0.$$

Since $8 (x +1) {\textrm{e}}^{-2/\sqrt{\gamma}} \le 8 ({\textrm{e}}^{1/\sqrt{\gamma}} + 1) {\textrm{e}}^{-2/\sqrt{\gamma}} \le 16 {\textrm{e}}^{-1/\sqrt{\gamma}}$ , it follows from (4.35) that

$$\frac{\eta \gamma^2 + (2 + \eta) \varepsilon - \frac{\zeta(2)}{2} \gamma^2) - R_\varepsilon}{x + 1} - 8{\textrm{e}}^{-2/\sqrt{\gamma}} >0.$$

Therefore, we have found $\varepsilon_{7}(\eta) \in (0, 1)$ such that for all $\varepsilon\in (0, \, \varepsilon_{7}(\eta))$ , there is $ n_{15} (\eta, \, \varepsilon)$ such that (4.36) holds true for all $n\ge n_{15} (\eta, \, \varepsilon)$ and all $x\in [\gamma^2 , {\textrm{e}}^{1/\sqrt{\gamma}} ]$ , which completes the proof of Lemma 6 in the third case $\gamma^2 \tau_n \leq z < {\textrm{e}}^{1/\sqrt{\gamma}} \tau_n$ .

Proof of Lemma 6: fourth (and last) case $z \ge {\textrm{e}}^{1/\sqrt{\gamma}} \tau_n$ . Let $\eta, {\varepsilon} \in (0, \, 1)$ and recall the definition of $\gamma$ from (4.1): namely, $\gamma = \gamma(\varepsilon, \, \eta) \;:\!=\; 2\zeta(2)^{-1/2} (1 - \eta)\sqrt{\varepsilon}$ . We fix $\eta$ and we easily check that we can find $\varepsilon_{8}(\eta) \in (0, \, \varepsilon_{7}(\eta))$ such for all ${\varepsilon} \in (0, \varepsilon_{8}(\eta))$ , the following inequalities hold true:

(4.48) \begin{equation} \gamma\le \frac{1}{400}, \quad (1 + 2\varepsilon) \, \frac{\big(1 + 2{\textrm{e}}^{-1/(5\sqrt{\gamma})}\big)^4}{1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}} <\big( 1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}\big)^2 \, (1 - (2 + \eta)\varepsilon )^{-(1-\gamma) /\gamma} . \end{equation}

The following lemma gives an estimate of ${{\mathbb P}} (Z_n \ge \tau_n \theta)$ when $\theta$ is sufficiently large.

Lemma 9. Let $\eta \in (0, \, 1)$ . Let ${\varepsilon} \in (0, \, \varepsilon_{8} (\eta))$ and $n \geq n_{15} (\eta, \, {\varepsilon})$ . Then

(4.49) \begin{equation} 1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})} \le \frac{\theta^{1-\gamma}}{\gamma} \, {{\mathbb P}} (Z_n \geq \tau_n \theta) \le 1 + {\textrm{e}}^{-2/\sqrt{\gamma}}, \quad \theta \in [{\textrm{e}}^{1/(2\sqrt{\gamma})}, \, \infty) .\end{equation}

Furthermore,

(4.50) \begin{equation}{{\mathbb P}} (Z_n \geq {\textrm{e}}^{3/(4\sqrt{\gamma})} \tau_n) \le {\textrm{e}}^{-3/(4\sqrt{\gamma})} .\end{equation}

Proof. For $\theta \in \mathbb{R}^*_+$ , write as before

$$g_\gamma (\theta) \;:\!=\; (\theta +1)^\gamma - \theta^{\gamma}= \frac{\gamma}{\theta^{1-\gamma}} \int_0^1 \frac{\textrm{d} u}{( 1+ \frac{u}{\theta})^{1-\gamma}}.$$

Thus,

$$\frac{\theta^{1-\gamma}}{\gamma} g_\gamma (\theta) \le 1.$$

On the other hand,

$$\frac{\theta^{1-\gamma}}{\gamma} g_\gamma (\theta) \ge \int_0^1 \frac{\textrm{d} u}{1+ \frac{u}{\theta}} \ge \int_0^1 \frac{\textrm{d} u}{1+ \frac{1}{\theta}} = \frac{\theta}{1+\theta}.$$

Thus,

$$\mbox{for all }\theta \in \mathbb{R}^*_+ ,\quad\frac{\theta}{1+\theta} \le \frac{\theta^{1-\gamma}}{\gamma} g_\gamma (\theta) \le 1 .$$

By (4.38), for $n\ge n_{15}( \eta, \, \varepsilon)$ and $\theta \in [\frac12 {\textrm{e}}^{-4/\sqrt{\gamma}}, \, \infty)$ , we get

(4.51) \begin{equation} (1 - {\textrm{e}}^{-2/\sqrt{\gamma}})\frac{\theta}{1+\theta} \le \frac{\theta^{1-\gamma}}{\gamma} \, {{\mathbb P}} (Z_n \geq \tau_n \theta) \le 1 + {\textrm{e}}^{-2/\sqrt{\gamma}} . \end{equation}

If $\theta > {\textrm{e}}^{1/(2\sqrt{\gamma})}$ , then $\frac{1}{1+\theta} \le \frac{1}{\theta} \le {\textrm{e}}^{-1/(2\sqrt{\gamma})}$ , so $(1 - {\textrm{e}}^{-2/\sqrt{\gamma}})\frac{\theta}{1+\theta} \ge 1 - \frac{1}{1+\theta} - {\textrm{e}}^{-2/\sqrt{\gamma}} \ge 1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}$ , and (4.49) follows.

To get (4.50), it suffices to take $\theta \;:\!=\; {\textrm{e}}^{3/(4\sqrt{\gamma})}$ , and note that in this case,

$$\frac{\gamma}{\theta^{1-\gamma}} = \frac{\gamma {\textrm{e}}^{3\sqrt{\gamma}\,/4}}{{\textrm{e}}^{3/(4\sqrt{\gamma})}} \le \frac{\gamma {\textrm{e}}}{{\textrm{e}}^{3/(4\sqrt{\gamma})}} \le \frac12 {\textrm{e}}^{-3/(4\sqrt{\gamma})},$$

whereas $1 + {\textrm{e}}^{-2/\sqrt{\gamma}} \le 2$ , so (4.50) follows from the second inequality in (4.51).

We proceed to the proof of Lemma 6 in the fourth case: $z \ge {\textrm{e}}^{1/\sqrt{\gamma}} \tau_n$ . Let ${\varepsilon} \in (0, \, \varepsilon_{8} (\eta))$ , $n \geq n_{15} (\eta, \, {\varepsilon})$ , and $x \in [{\textrm{e}}^{1/\sqrt{\gamma}}, \, \infty)$ . We write as before $Z_n^* \;:\!=\; \frac{Z_n}{\tau_n}$ where $\widehat{Z}_n^*$ denotes an independent copy of $Z_n^*$ . We have, for $x' \in (0, \, x)$ ,

\begin{eqnarray*} {{\mathbb P}}\big(Z^*_n + \widehat{Z}^*_n > x\big) &\,\le\,& {{\mathbb P}}(Z^*_n > x') + {{\mathbb P}}(\widehat{Z}^*_n > x') + {{\mathbb P}}(\widehat{Z}^*_n > x - x', \, Z^*_n > x-x') \\[5pt] & \,=\, &2 {{\mathbb P}}(Z^*_n > x') + \big( {{\mathbb P}}(Z^*_n > x - x') \big)^2 .\end{eqnarray*}

We now take $x' \;:\!=\; (1 - {\textrm{e}}^{-1/(4\sqrt{\gamma}\,)})x$ , so $x' \in [{\textrm{e}}^{1/(2\sqrt{\gamma})}, \, \infty)$ (because $1 - {\textrm{e}}^{-1/(4\sqrt{\gamma}\,)} > {\textrm{e}}^{-1/(2\sqrt{\gamma})}$ as $\sqrt{\gamma}\le \tfrac{1}{20}$ according to (4.48)) and $x - x'\in [{\textrm{e}}^{3/(4\sqrt{\gamma})}, \, \infty)$ . By (4.50),

(4.52) \begin{equation} {{\mathbb P}}(Z^*_n > x - x') \le {{\mathbb P}}\big(Z^*_n > {\textrm{e}}^{3/(4\sqrt{\gamma})}\big) \le {\textrm{e}}^{-3/(4\sqrt{\gamma})} . \end{equation}

Hence,

$${{\mathbb P}}\big(Z^*_n + \widehat{Z}^*_n > x\big)\le2 {{\mathbb P}}(Z^*_n > x') +{\textrm{e}}^{-3/(4\sqrt{\gamma})}\, {{\mathbb P}}(Z^*_n > x - x') .$$

We easily check that we can apply (4.49) to $\theta =x'$ and to $\theta= x - x'$ , which yields

$${{\mathbb P}}\big(Z^*_n + \widehat{Z}^*_n > x\big)\le\left( 2\left(\frac{x}{x'}\right)^{1-\gamma} +{\textrm{e}}^{-3/(4\sqrt{\gamma})}\, \left(\frac{x}{x - x'}\right)^{1-\gamma} \right) \frac{1 + {\textrm{e}}^{-2/\sqrt{\gamma}}}{1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}} \, {{\mathbb P}}(Z^*_n > x) .$$

We have

$$\Big(\frac{x}{x'}\Big)^{1-\gamma} \le \frac{x}{x'} = \frac{1}{1 - {\textrm{e}}^{-1/(4\sqrt{\gamma})}} < 1 + {\textrm{e}}^{-1/(5\sqrt{\gamma})}$$

(noting that $(1 + {\textrm{e}}^{-1/(5\sqrt{\gamma})})(1 - {\textrm{e}}^{-1/(4\sqrt{\gamma})}) > 1$ because $\sqrt{\gamma} \le \tfrac{1}{20}$ by (4.48)),

$$\Big(\frac{x}{x - x'}\Big)^{1-\gamma} \le \frac{x}{x - x'} = {\textrm{e}}^{1/(4\sqrt{\gamma}\,)}.$$

Thus,

$$2\Big(\frac{x}{x'}\Big)^{1-\gamma} +{\textrm{e}}^{-3/(4\sqrt{\gamma})}\, \Big(\frac{x}{x - x'}\Big)^{1-\gamma}\le2(1 + {\textrm{e}}^{-1/(5\sqrt{\gamma})}) +{\textrm{e}}^{-1/(2\sqrt{\gamma})}\le2\big(1 + 2{\textrm{e}}^{-1/(5\sqrt{\gamma})}\big) ,$$

which implies that

\begin{eqnarray*} {{\mathbb P}}\big(Z^*_n + \widehat{Z}^*_n > x\big) & \,\le\, & 2\big(1 + 2{\textrm{e}}^{-1/(5\sqrt{\gamma})}\big)\, \frac{1 + {\textrm{e}}^{-2/\sqrt{\gamma}}}{1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}} \, {{\mathbb P}}(Z^*_n > x) \\[5pt] & \,\le\, & \frac{2\big(1 + 2{\textrm{e}}^{-1/(5\sqrt{\gamma})}\big)^2}{1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}}\, {{\mathbb P}}(Z^*_n > x) .\end{eqnarray*}

Let $V_n \;:\!=\; (Z_n + \widehat{Z}_n) \mathscr{E}_n + (Z_n \wedge \widehat{Z}_n) (1 - \mathscr{E}_n)$ , where $\mathscr{E}_n$ denotes a Bernoulli random variable that is independent of $(Z_n, \widehat{Z}_n)$ and such that ${{\mathbb P}}(\mathscr{E}_n = 1) = \frac{1}{2} + \varepsilon$ . Then

$${{\mathbb P}} (V_n \geq x \tau_n)=\bigg( \frac12 + \varepsilon \bigg) \, {{\mathbb P}}(Z^*_n + \widehat{Z}^*_n > x)+\bigg( \frac12 - \varepsilon \bigg)\, \big( {{\mathbb P}}(Z^*_n > x)\big)^2 .$$

We use the trivial inequality $\frac12 - \varepsilon \le 1$ . By (4.52) ${{\mathbb P}}(Z^*_n > x) \le {{\mathbb P}}(Z^*_n > x - x') \le {\textrm{e}}^{-3/(4\sqrt{\gamma})}$ . Therefore,

\begin{eqnarray*} {{\mathbb P}} (V_n \geq x \tau_n) & \,\le\, & (1 + 2\varepsilon) \, \frac{\big(1 + 2{\textrm{e}}^{-1/(5\sqrt{\gamma})}\big)^2}{1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}} \, {{\mathbb P}}(Z^*_n > x) +{\textrm{e}}^{-3/(4\sqrt{\gamma})} \, {{\mathbb P}}(Z^*_n > x) \\[5pt] & \,\le\, & (1 + 2\varepsilon) \, \frac{\big(1 + 2{\textrm{e}}^{-1/(5\sqrt{\gamma})}\big)^3}{1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}} \, {{\mathbb P}}(Z^*_n > x) .\end{eqnarray*}

To complete the proof of the lemma, we observe that

We can apply (4.49) to $x \geq {\textrm{e}}^{1/\sqrt{\gamma}} > {\textrm{e}}^{1/(2\sqrt{\gamma})}$ and we get ${{\mathbb P}}(Z^*_n > x) \le ({\gamma}/{x^{1-\gamma}}) (1 + {\textrm{e}}^{-2/\sqrt{\gamma}})$ , which is bounded by $({\gamma}/{x^{1-\gamma}}) \big(1 + 2{\textrm{e}}^{-1/(5\sqrt{\gamma})}\big)$ . Thus,

(4.53)

We look for a lower bound for ${{\mathbb P}} ( Z^*_{n+1} > ({\tau_n}/{\tau_{n+1}}) x )$ . By definition, $\lim_{n\to \infty} \tau_{n}/\tau_{n+1}$ $=$ $(1 - (2 + \eta)\varepsilon)^{1/\gamma}$ and $\lim_{{\varepsilon} \to 0^+} (1 - (2 + \eta)\varepsilon)^{1/\gamma} = 1$ . Therefore, there exists $\varepsilon_{9} (\eta) \in (0,\, \varepsilon_{8}(\eta))$ such that for ${\varepsilon} \in (0, \,\varepsilon_{9} (\eta))$ , there is an integer $n_{7} (\eta, {\varepsilon}) \ge n_{15} (\eta, {\varepsilon})$ such that for all $n\geq n_{16} (\eta, \, {\varepsilon})$ ,

$$ \frac{\tau_{n}}{\tau_{n+1}} \geq {\textrm{e}}^{-1/(2 \sqrt{\gamma}) } \quad \textrm{and} \quad \Big( \frac{ \tau_{n+1}}{\tau_{n}} \Big)^{1-\gamma} \ge \big( 1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}\big) (1 - (2 + \eta)\varepsilon )^{-(1-\gamma) /\gamma} .$$

Since $\frac{\tau_{n}}{\tau_{n+1}} x \ge {\textrm{e}}^{1/(2 \sqrt{\gamma}) }$ , (4.49) applies to $\theta \;:\!=\; \frac{\tau_{n}}{\tau_{n+1}} x$ and for all ${\varepsilon} \in (0, \, \varepsilon_{9} (\eta))$ and all $n\ge n_{16} (\eta, \, {\varepsilon})$ , we get

\begin{eqnarray*} {{\mathbb P}} \bigg( Z^*_{n+1} > \frac{\tau_n}{\tau_{n+1}} x \bigg) & \,\ge\, & \big( 1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}\big) \Big( \frac{ \tau_{n+1}}{\tau_{n}} \Big)^{1-\gamma} \frac{\gamma}{x^{1-\gamma}} \\[5pt] & \,\ge\, & \big( 1 - 2{\textrm{e}}^{-1/(2\sqrt{\gamma})}\big)^2 \, (1 - (2 + \eta)\varepsilon )^{-(1-\gamma) /\gamma} \frac{\gamma}{x^{1-\gamma}} .\end{eqnarray*}

Going back to (4.53), for all ${\varepsilon} \in (0, \, \varepsilon_{9} (\eta))$ , all $n\ge n_{16} (\eta, {\varepsilon})$ , and all $x \in [{\textrm{e}}^{1/\sqrt{\gamma}} , \, \infty)$ , we obtain

which is negative according to (4.48). This implies Lemma 6 in the fourth case, and thus completes the proof of the lemma.