Dynamic Phase Diagram of the REM

Abstract

By studying the two-time overlap correlation function, we give a comprehensive analysis of the phase diagram of the Random Hopping Dynamics of the Random Energy Model (REM) on time-scales that are exponential in the volume. These results are derived from the convergence properties of the clock process associated to the dynamics and fine properties of the simple random walk in the n-dimensional discrete cube.

Appendices

Appendix A. Calculations

This appendix contains calculatory results on the moments of $f_{\delta }({\gamma }_n(x))$ and $g_{\delta }({\gamma }_n(x))$ that are needed in several places in the proofs. Our first lemma provides asymptotic bounds on $a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))\right)$ and $a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))^2\right)$ needed in the verification of Condition (A3’).

Lemma A.1

For all $0<\varepsilon \le 1$ and all $0<\beta <\infty$ such that ${\alpha }(\varepsilon )=1$ we have that for all ${\delta }>0$ and large enough n, there exist constants $0<c_0,c_1<\infty$ such that

$$\begin{aligned} a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))\right) \le c_0 {\delta }, \end{aligned}$$

(A.1)

$$\begin{aligned} a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))^2\right) \le {\delta }^4+c_1. \end{aligned}$$

(A.2)

Proof

We observe that

$$\begin{aligned} f_{\delta }(u)\le {\delta }^2 \quad \forall u\in (0,\infty ). \end{aligned}$$

(A.3)

We decompose $a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))\right)$ in the following way

$$\begin{aligned} a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))\right) = a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))\mathbbm {1}_{\{{\gamma }_n(x)>{\delta }\}}\right) +a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))\mathbbm {1}_{\{{\gamma }_n(x)\le {\delta }\}}\right) \equiv (1) + (2) \end{aligned}$$

(A.4)

By (3.2) of Lemma 3.1 and the assumption that ${\alpha }(\varepsilon )=1$,

$$\begin{aligned} (1)\le {\delta }^2a_n{\mathbb P}\left( {\gamma }_n(x)>{\delta }\right) \sim {\delta }. \end{aligned}$$

(A.5)

Turning to (2) we have

$$\begin{aligned} (2)\le & {} a_n{\mathbb E}\left( {\gamma }_n(x)^2\mathbbm {1}_{\{{\gamma }_n(x)\le {\delta }\}}\right) \nonumber \\= & {} \frac{a_n e^{2n{\beta }^2}}{c_n^2}\int _{-\infty }^{\frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}-2\sqrt{n}{\beta }} \frac{e^{-u^2/2}}{\sqrt{2\pi }} \text{ d }u\nonumber \\\sim & {} \frac{a_n e^{2n{\beta }^2}}{c_n^2} \left( \sqrt{2\pi }\left( -\frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}+2\sqrt{n}{\beta }\right) \right) ^{-1}e^{-\frac{1}{2}\left( \frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}-2\sqrt{n}{\beta }\right) ^2}, \end{aligned}$$

(A.6)

where we used that by (3.21), for ${\alpha }(\varepsilon )=1$, $\frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}-2\sqrt{n}{\beta }\rightarrow -\infty$ as $n\rightarrow \infty$. Now we start to expand the exponent of the exponential function and plug in (3.21).

$$\begin{aligned} (A.6) = a_n\left( \sqrt{2\pi }\left( -\frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}+2\sqrt{n}{\beta }\right) \right) ^{-1} {\delta }^2 e^{-\frac{1}{2}\left( \frac{\log c_n{\delta }}{\sqrt{n}{\beta }}\right) ^2} = c_0' {\delta }(1+o(1)), \end{aligned}$$

(A.7)

where $0<c_0'<\infty$. Putting our estimates together we have that for n large enough there exists a constant $0<c_0<\infty$ such that

$$\begin{aligned} a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))\right) \le c_0 {\delta }. \end{aligned}$$

(A.8)

In a similar way we treat $a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))^2\right)$. This time we truncate at one, namely

$$\begin{aligned} a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))^2\right) = a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))^2\mathbbm {1}_{\{{\gamma }_n(x)>1\}}\right) + a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))^2\mathbbm {1}_{\{{\gamma }_n(x)\le 1\}}\right) . \end{aligned}$$

(A.9)

For the first summand we use again the bound on f and the definition of the time-scale to bound it by ${\delta }^4$. And for the second summand we use the same method as for (2): applying Gaussian estimates, expanding the resulting term and plugging in the exact representation of $c_n$. The bound we obtain is a constant. Putting these estimates together we have for n large enough

$$\begin{aligned} \textstyle a_n{\mathbb E}\left( f_{\delta }({\gamma }_n(x))^2\right) \le {\delta }^4+c_1. \end{aligned}$$

(A.10)

   $\square$

In the verification of Condition (A3) a slightly different function $g_{\delta }$ appeared. In the forthcoming lemma we control the first and second moment of $g_{\delta }({\gamma }_n(x))$ when $0<{\alpha }(\varepsilon )<1$.

Lemma A.2

Let $c_n$ be an intermediate time-scale and $0<{\beta }<\infty$ and $0<{\alpha }(\varepsilon )<1$. Then there exists constants $c_3$ and $c_4$ such that the following holds for n large enough

$$\begin{aligned} a_n{\mathbb E}\left( g_{\delta }({\gamma }_n(x))\right) \le c_3 {\delta }^{1-{\alpha }(\varepsilon )}, \end{aligned}$$

(A.11)

$$\begin{aligned} a_n{\mathbb E}\left( g_{\delta }({\gamma }_n(x))^2\right) \le c_4. \end{aligned}$$

(A.12)

Proof

We observe that

$$\begin{aligned} g_{\delta }(u) \le {\delta }\quad \forall u>0. \end{aligned}$$

(A.13)

As in the proof of the previous lemma we write

$$\begin{aligned} a_n{\mathbb E}\left( g_{\delta }({\gamma }_n(x))\right) = a_n{\mathbb E}\left( g_{\delta }({\gamma }_n(x))\mathbbm {1}_{\{{\gamma }_n(x)>{\delta }\}}\right) +a_n{\mathbb E}\left( g_{\delta }({\gamma }_n(x))\mathbbm {1}_{\{{\gamma }_n(x)\le {\delta }\}}\right) \equiv (1) + (2). \nonumber \end{aligned}$$

The first summand (1) we control by

$$\begin{aligned} (1) \le {\delta }a_n{\mathbb P}\left( {\gamma }_n(x)>{\delta }\right) \sim {\delta }^{1-{\alpha }(\varepsilon )}. \end{aligned}$$

(A.14)

For (2) we use Gaussian estimates.

$$\begin{aligned} (2)\le & {} a_n{\mathbb E}\left( {\gamma }_n(x) \mathbbm {1}_{{\gamma }_n(x)\le {\delta }}\right) = \frac{a_ne^{n{\beta }^2/2}}{c_n}\int _{-\infty }^{\frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}-\sqrt{n}{\beta }} \frac{e^{-u^2/2}}{\sqrt{2\pi }}\text{ d }u\nonumber \\\sim & {} \frac{a_ne^{n{\beta }^2/2}}{c_n} \left( \sqrt{2\pi }\left( \sqrt{n}{\beta }-\frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}\right) \right) ^{-1}e^{-\frac{1}{2}\left( \frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}-\sqrt{n}{\beta }\right) ^2}, \end{aligned}$$

(A.15)

where we used that since ${\beta }>{\beta }_c(\varepsilon )$ we have by (3.21) $\frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}-\sqrt{n}{\beta }\rightarrow -\infty$ as $n\rightarrow \infty$. Now, expanding the terms and inserting the exact representation of $c_n$, (A.15) is equal to

$$\begin{aligned} {\delta }a_n\left( \sqrt{2\pi }\left( \sqrt{n}{\beta }-\frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}\right) \right) ^{-1} e^{-\frac{1}{2}\left( \frac{\log (c_n{\delta })}{\sqrt{n}{\beta }}\right) ^2} \le c_3' {\delta }^{1-{\alpha }(\varepsilon )}, \end{aligned}$$

(A.16)

for some constant $0<c_3'<\infty$. Putting the estimates on (1) and (2) together we get that there exists a constant $0<c_3<\infty$ such that

$$\begin{aligned} a_n{\mathbb E}\left( g_{\delta }({\gamma }_n(x))\right) \le c_3 {\delta }^{1-{\alpha }(\varepsilon )}. \end{aligned}$$

(A.17)

To control $a_n{\mathbb E}\left( g_{\delta }({\gamma }_n(x))^2\right)$ one proceeds in exactly the same way.    $\square$

To study the behavior of $M_n(t)$, and in particular to check Condition (B1), we needed a control on the moments of $g_1({\gamma }_n(x))$ when ${\beta }={\beta }_c(\varepsilon )$ which is done in the next lemma.

Lemma A.3

Let $c_n$ be an intermediate scale.

• (i) Let ${\beta }={\beta }_c(\varepsilon )$. Then

  $$\begin{aligned} {\mathbb E}\left( g_1({\gamma }_n(x))\right) \le \frac{e^{n{\beta }^2/2}}{c_n}(1+o(1)). \end{aligned}$$

  (A.18)

  Moreover, if $\lim _{n\rightarrow \infty }\sqrt{n}{\beta }-\frac{\log c_n}{\beta \sqrt{n}} = \theta$ for some $\theta \in (-\infty ,\infty )$. Then,

  $$\begin{aligned} a_n{\mathbb E}\left( g_1({\gamma }_n(x))\right) = \Phi (\theta ) \frac{a_ne^{n{\beta }^2/2}}{c_n}(1+o(1)) = \Phi (\theta )\beta \sqrt{2n\pi } e^{\theta ^2/2}(1+o(1)). \end{aligned}$$

  (A.19)

• (ii) Let ${\beta }={\beta }_c(\varepsilon )$. For n large enough there exists a constant $0<c_2<\infty$ such that

  $$\begin{aligned} a_n {\mathbb E}\left( g_1({\gamma }_n(x))^l\right) \le c_2, \quad 2\le l\le 4. \end{aligned}$$

  (A.20)

• (iii) Let $\beta <\beta _c$. Then

  $$\begin{aligned} {\mathbb E}\left( g_1({\gamma }_n(x))\right) = \frac{e^{n{\beta }^2/2}}{c_n}(1+o(1)). \end{aligned}$$

  (A.21)

  If $\beta >\beta _c/2$, then

  $$\begin{aligned} a_n {\mathbb E}\left( g_1({\gamma }_n(x))^2\right) \le c_2 . \end{aligned}$$

  (A.22)

  Otherwise $a_n {\mathbb E}\left( g_1({\gamma }_n(x))^2\right) \le a_ne^{2n\beta ^2}/c_n^2$ and

  $$\begin{aligned} \log \frac{a_n {\mathbb E}\left( g_1({\gamma }_n(x))^2\right) }{a_n^2 e^{n\beta ^2}/c_n^2}= n(2\beta -\beta _c)/2. \end{aligned}$$

  (A.23)

Proof

Recall that $g_1(u) \le 1$, $\forall u>0$. To prove assertion (i) we rewrite ${\mathbb E}\left( g_1({\gamma }_n(x))\right)$ as

$$\begin{aligned}&\frac{e^{n{\beta }^2/2}}{\sqrt{2\pi }c_n} \int _{-\infty }^{\infty } e^{\sqrt{n}{\beta }z }\left( 1-e^{-c_n e^{-\sqrt{n}{\beta }z}}\right) e^{-z^2/2}\text{ d }z\nonumber \\= & {} \frac{e^{n{\beta }^2/2} }{c_n} - \frac{ e^{n{\beta }^2/2}}{c_n {\beta }\sqrt{2 \pi n}} \int _{-\infty }^{\infty } e^{y+\log c_n}e^{-\left( \frac{y}{{\beta }\sqrt{n}}+\frac{\log c_n}{{\beta }\sqrt{n}}\right) ^2- e^{-y}}\text{ d }y. \end{aligned}$$

(A.24)

Now one can cut the domain of integration into different pieces. Observe that in the region $y>\log n$ the integral is equal to

$$\begin{aligned}&(1+o(1))\frac{e^{n{\beta }^2/2}}{c_n {\beta }\sqrt{2\pi n}} \int ^{\infty }_{\log n} e^{y+\log c_n}e^{-\left( \frac{y}{{\beta }\sqrt{n}}+\frac{\log c_n}{{\beta }\sqrt{n}}\right) ^2/2}\text{ d }y\nonumber \\= & {} (1+o(1))\frac{ e^{n{\beta }^2/2}}{c_n\sqrt{2\pi }}\int ^{\infty }_{\frac{\log n}{\sqrt{n}{\beta }} -\frac{\log c_n}{{\beta }\sqrt{n}}+\sqrt{n}{\beta }}e^{-y^2/2}\text{ d }y. \end{aligned}$$

(A.25)

If $\sqrt{n}{\beta }-\frac{\log c_n}{\beta \sqrt{n}} \rightarrow \theta$ for some constant $\theta$ as $n\rightarrow \infty$ we have that (A.24) is equal to $(1+o(1))\frac{e^{n{\beta }^2/2}}{c_n}(1-\Phi (\theta ))$. Proceeding as in (A.25) one can bound the integral in (A.24) on the domain of integration $|y|<\log n$ by $o(1)\frac{ e^{n{\beta }^2/2}}{c_n}$. For $y<-\log n$, $e^{-y}>n$ which implies that the on that part of the domain of integration the integral in (A.24) is equal to $o(1)\frac{ e^{n{\beta }^2/2}}{c_n}$. This yields the first equality in (A.19), and as the Gaussian integral is always between zero and one, this also implies (A.18). The second inequality in (A.19) follows from the first by (B.5) of Lemma B.2. We now turn to assertion (ii) and consider ${\mathbb E}\left( g_1({\gamma }_n(x))^2\right)$. We will split this term into two terms:

$$\begin{aligned} a_n {\mathbb E}\left( g_1({\gamma }_n(x))^2\right) = a_n{\mathbb E}\left( g_1({\gamma }_n(x))^2\mathbbm {1}_{\{{\gamma }_n(x)>1\}}\right) +a_n {\mathbb E}\left( g_1({\gamma }_n(x))^2\mathbbm {1}_{\{{\gamma }_n(x)\le 1\}}\right) \equiv (1) + (2). \nonumber \end{aligned}$$

For (1) we use the definition of the scaling $a_n$ and $c_n$ and the bound (A.13)

$$\begin{aligned} (1) \le a_n{\mathbb P}\left( {\gamma }_n(x)>1\right) = 1. \end{aligned}$$

(A.26)

For Term (2) we use exact Gaussian estimates to bound

$$\begin{aligned} (2)\le & {} \frac{a_n}{c_n^2} \int _{-\infty }^{\frac{\log c_n}{\sqrt{n}{\beta }}} e^{2{\beta }\sqrt{n} u}\left( 1-e^{- c_ne^{-\sqrt{n}{\beta }u}}\right) ^2 \frac{e^{-u^2/2}}{\sqrt{2\pi }} \text{ d }u\nonumber \\\le & {} \frac{a_n e^{2n{\beta }^2}}{c_n^2}\int _{-\infty }^{\frac{\log c_n}{\sqrt{n}{\beta }}-2\sqrt{n}{\beta }} \frac{e^{-r^2/2}}{\sqrt{2\pi }} \text{ d }r\nonumber \\\sim & {} \frac{a_n e^{2n{\beta }^2}}{c_n^2} \left( \sqrt{2\pi }\left( -\frac{\log c_n}{\sqrt{n}{\beta }}+2\sqrt{n}{\beta }\right) \right) ^{-1}e^{-\left( \frac{\log c_n}{\sqrt{n}{\beta }}-2\sqrt{n}{\beta }\right) ^2/2}, \end{aligned}$$

(A.27)

where we use that by (3.22), $\frac{\log c_n}{\sqrt{n}{\beta }}-2\sqrt{n}{\beta }\rightarrow -\infty$ as $n\rightarrow \infty$. Plugging in (3.22) yields

$$\begin{aligned} (A.27) = a_n \left( \sqrt{2\pi }\left( -\frac{\log c_n}{\sqrt{n}{\beta }}+2\sqrt{n}{\beta }\right) \right) ^{-1} e^{-\left( \frac{\log c_n}{\sqrt{n}{\beta }}\right) ^2/2} = c_2'(1+o(1)), \end{aligned}$$

(A.28)

where $0<c_2'<\infty$. Putting both estimates together we get that for n large there exists a constant $0< c_2<\infty$ such that

$$\begin{aligned} a_n {\mathbb E}\left( g_1({\gamma }_n(x))^2\right) \le c_2 . \end{aligned}$$

(A.29)

Proceeding in exactly the same way with $a_n {\mathbb E}\left( g_1({\gamma }_n(x))^3\right)$ and $a_n {\mathbb E}\left( g_1({\gamma }_n(x))^4\right)$, one readily obtains (A.20) for $l=3$ and $l=4$.

Part (iii) follows from computations similar to those of (i) and (ii). (A.23) follows from (3.20).    $\square$

Appendix B. The Centering Term $M_n(t)$ at Criticality

In this appendix we collect the fine asymptotics needed to control the centering term $M_n(t)$ on the critical line ${\beta }={\beta }_c(\varepsilon )$, $0<\varepsilon \le 1$. Computing ${\mathbb E}\left( {\mathcal E}\left( M_n(t) \right) \right)$ at ${\beta }={\beta }_c(\varepsilon )$ gives

$$\begin{aligned} {\mathbb E}\left( {\mathcal E}\left( M_n(t) \right) \right)= & {} {\mathbb E}\left( \sum _{i=1}^{[a_nt]} {\mathcal E}\left( c_n^{-1}{\tau _n(J_n(i)) e_{n,i}}1_{\left\{ 0<c_n^{-1}{\tau _n(J_n(i)) e_{n,i}}<1 \right\} } \right) \right) \nonumber \\= & {} \lfloor a_nt\rfloor {\mathbb E}\left( g_1({\gamma }_n(x))\right) = c t\frac{a_ne^{n{\beta }^2/2}}{c_n}(1+o(1)), \end{aligned}$$

(B.1)

where by the first equality in (A.19) c is some constant $>0$. The following lemma proves the general diverging behavior of ${\mathbb E}({\mathcal E}(M_n(1))$. Recall the notation (1.3)–(1.11) of Definition 1.1.

Lemma B.1

Given $0<\varepsilon \le 1$, let $a_n$ and $c_n$ be sequences satisfying (1.3) and (1.4) and let ${\beta }={\beta }_c(\varepsilon )$. Then

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{a_ne^{n{\beta }^2/2}}{c_n} = \infty . \end{aligned}$$

(B.2)

Proof

By (3.20) with ${\beta }={\beta }_c(\varepsilon )$, $\log a_n = \frac{1}{2}(n{\beta }^2+f(n))$ for some sequence f(n) such that $\frac{f(n)}{n{\beta }^2}=o(1)$. Furthermore, by (3.21), $\log (\log a_n) = \log (\frac{n{\beta }^2+f(n)}{2})$ and $\sqrt{2\log a_n}= \sqrt{n{\beta }^2+f(n)}$. Note that due to the asymptotic behavior of f(n), $\log (\log a_n)$ is positive for n large enough. Hence it suffices to show that

$$\begin{aligned} \frac{\log (a_n e^{n{\beta }^2/2})}{\sqrt{n}{\beta }} \ge \sqrt{2\log a_n}. \end{aligned}$$

(B.3)

Plugging in the expressions for $\log a_n$, (B.3) reads

$$\begin{aligned} \sqrt{n}{\beta }+ \frac{f(n)}{2\sqrt{n}{\beta }}\ge \sqrt{n{\beta }^2 + f(n)}, \end{aligned}$$

(B.4)

which is always satisfied and equality holds if and only if $f(n)=0$.    $\square$

Lemma B.2

If in addition to the assumptions of Lemma B.1, $\lim _{n\rightarrow \infty }\frac{\log c_n}{\sqrt{n}\beta }-\sqrt{n}\beta =\theta$ for some $\theta \in (-\infty ,\infty )$, then

$$\begin{aligned} \lim _{n\rightarrow \infty }\sqrt{n}\frac{c_n}{a_ne^{n\beta ^2/2}}=\frac{1}{\beta \sqrt{2\pi } } e^{-\theta ^2/2}. \end{aligned}$$

(B.5)

Proof

Using the notation of the proof of Lemma B.1, (B.5) follows from (3.21) with $\lim _{n\rightarrow \infty } \frac{f(n)}{2\sqrt{n}\beta }=\theta$. Namely, under the assumption of the lemma, (3.21) may be written as

$$\begin{aligned} c_n=\frac{1}{\beta \sqrt{2\pi n}} e^{n\beta ^2+\frac{f(n)}{2}-\frac{1}{8}\left( \frac{f(n)}{\beta \sqrt{n}}\right) ^2+o(1)} \end{aligned}$$

(B.6)

by Taylor expansion of the square root. Equation (B.6) also implies that $\lim _{n\rightarrow \infty }\frac{\log c_n}{\sqrt{n}\beta }-\sqrt{n}\beta =\theta$ if and only if $\lim _{n\rightarrow \infty } \frac{f(n)}{2\sqrt{n}\beta }=\theta$.    $\square$

Appendix C. Auxiliary Lemmas Needed in the Proof of Theorem 1.5

Recall that $\widetilde{a}_n$ and $A_n(t)$ are defined in (8.1) and (8.3), respectively.

Lemma C.1

Let $c_n$ be an intermediate scale with $\lim _{n\rightarrow \infty }\frac{\log c_n}{\sqrt{n}\beta }-\sqrt{n}\beta =\theta$ for some $\theta \in (-\infty ,\infty )$ and ${\beta }={\beta }_c(\varepsilon )$ with $0<\varepsilon \le 1$. If $\sum a_n/2^n<\infty$ we have for all $t,s>0$ and for all $\epsilon >0$, ${\mathbb P}$-a.s.

$$\begin{aligned} \lim _{n\rightarrow \infty }\sqrt{n} {\mathcal P}\left( \left| \sum _{k=1}^{\lfloor \widetilde{a}_nt\rfloor } {\mathcal P}\Bigl ({\tau }_n(J_n(k+1))e_{n,k+1}>c_ns|J_n(k)\Bigr ) -\frac{\widetilde{a}_nt}{a_ns}\right| >\frac{\widetilde{a}_n\epsilon }{a_n\sqrt{A_n(t)}} \right) = 0 \end{aligned}$$

(C.1)

If $\sum a_n/2^n=\infty$ the same holds in ${\mathbb P}$-probability.

Proof

Proceeding as in the the proof of Proposition 5.1 one readily establishes that

$$\begin{aligned}&{\mathcal P}\left( \left| \sum _{k=1}^{\lfloor \widetilde{a}_nt\rfloor } {\mathcal P}\Bigl ({\tau }_n(J_n(k+1))e_{n,k+1}>c_ns|J_n(k)\Bigr ) -\frac{\lfloor \widetilde{a}_nt\rfloor }{a_n}\nu _n(s,\infty )\right| >\frac{\widetilde{a}_n\epsilon }{a_n\sqrt{A_n(t)}}\right) \nonumber \\&\le \left( \frac{a_n \sqrt{A_n(t)}}{\widetilde{a}_n \epsilon }\right) ^2 \left( \left( \frac{\lfloor \widetilde{a}_nt\rfloor }{a_n}\right) ^2\frac{\nu _n^2(u,\infty )}{2^{n-1}} +\frac{\lfloor \widetilde{a}_nt\rfloor }{a_n}\Theta ^1_n(u)\right) . \end{aligned}$$

(C.2)

where $\Theta ^1_n(u)$ is defined in (5.7). Using Proposition 6.1 and (8.2) yields the claim of Lemma C.1.    $\square$

Lemma C.2

Let $c_n$ be an intermediate scale with $\lim _{n\rightarrow \infty }\frac{\log c_n}{\sqrt{n}\beta }-\sqrt{n}\beta =\theta$ for some $\theta \in (-\infty ,\infty )$ and ${\beta }={\beta }_c(\varepsilon )$ with $0<\varepsilon \le 1$. Then we have for all $x>0$ that ${\mathbb P}$-a.s.

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{a_n}{\widetilde{a}_n}\sum _{l=1}^{\theta _n}{\mathcal P}\left( {\tau }_n(J_n(k+1)e_{n,k+1}>c_n x\right) =0. \end{aligned}$$

(C.3)

Proof

Using a first order Tchebychev inequality we have

$$\begin{aligned} {\mathbb P}\left( \frac{a_n}{\widetilde{a}_n}\sum _{l=1}^{\theta _n}{\mathcal P}\left( {\tau }_n(J_n(k+1)e_{n,k+1}>c_nx\right) >\epsilon \right) \le \epsilon ^{-1} \frac{a_n}{\widetilde{a}_n}\frac{\theta _n}{a_n}{\mathbb E}\nu _n(x,\infty ). \end{aligned}$$

(C.4)

In view of Lemma 6.2 and since $\sum \frac{\theta _n}{\widetilde{a}_n}<\infty$, the claim of Lemma C.2 follows.    $\square$

Lemma C.3

Let $c_n$ be an intermediate scale with $\lim _{n\rightarrow \infty }\frac{\log c_n}{\sqrt{n}\beta }-\sqrt{n}\beta =\theta$ for some $\theta \in (-\infty ,\infty )$ and ${\beta }={\beta }_c(\varepsilon )$ with $0<\varepsilon \le 1$. If $\sum a_n/2^n<\infty$ we have for all $t,s>0$ and for all $\epsilon '>0$ that ${\mathbb P}$-a.s.

$$\begin{aligned} \lim _{n\rightarrow \infty }\sqrt{n}{\mathcal P}\left( M_n(\lfloor \widetilde{a}_nt(1+\epsilon ')/a_n\rfloor ) < t\right) =0. \end{aligned}$$

(C.5)

If $\sum a_n/2^n=\infty$ the same holds in ${\mathbb P}$-probability.

Proof

Observe first that $M_n(\lfloor \widetilde{a}_nt(1+\epsilon ')/a_n\rfloor )=M_n(ct/\sqrt{n})$ for some constant c. We know from Proposition (5.4) and Lemma 6.5 that it concentrates around ${\mathcal E}M_n(\lfloor \widetilde{a}_nt(1+\epsilon ')/a_n\rfloor )$ either ${\mathbb P}$-a.s. or in ${\mathbb P}$-probability and the bounds are in the worst case linear in t. Moreover by linearity of $yyy{\mathcal E}M_n(\lfloor \widetilde{a}_nt(1+\epsilon ')/a_n\rfloor )$ and Lemma 6.5 the claim of Lemma C.3 follows.    $\square$