High points of a random model of the Riemann-zeta function and Gaussian multiplicative chaos

TL;DR

This paper investigates the behavior of high points in a random model of the Riemann-zeta function, demonstrating convergence of their total mass to Gaussian multiplicative chaos using probabilistic methods.

Contribution

It establishes the almost sure convergence of the total mass of high points to Gaussian multiplicative chaos in a specific random model of the Riemann-zeta function.

Findings

Total mass of high points converges to Gaussian multiplicative chaos

Almost sure convergence established using second moment method

Branching approximation aids in the proof

Abstract

We study the total mass of high points in a random model for the Riemann-Zeta function. We consider the same model as in [8], [2], and build on the convergence to 'Gaussian' multiplicative chaos proved in [14]. We show that the total mass of points which are a linear order below the maximum divided by their expectation converges almost surely to the Gaussian multiplicative chaos of the approximating Gaussian process times a random function. We use the second moment method together with a branching approximation to establish this convergence.

Equations

$$X_N(x)=\sum _{j=1}^N\frac{1}{\sqrt{p_j}}\left(\cos (x\ln p_j)\cos ({\theta }_{p_j})+\sin (x\ln p_j)\sin ({\theta }_{p_j})\right)x\in [0,1].$$

$$G_N(x)=\sum _{j=1}^N\frac{1}{2\sqrt{p_j}}\left(W_j^{(1)}\cos (x\ln p_j)+W_j^{(2)}\sin (x\ln p_j)\right),$$

$$X_N(x)-G_N(x)\equiv E_N(x),x\in [0,1],$$

$$\mathbb{E}\left(e^{\lambda {\sup }_{N\ge 1,x\in [0,1]}{E}_N(x)}\right)<\infty ,$$

$$M_{\alpha ,N}(dx)=\frac{e^{\alpha X_N(x)}}{\mathbb{E}{e}^{\alpha X_N(x)}}dx.$$

$$M_{\alpha }=\underset{N\to \infty }{\lim }{\int }_0^1{M}_{\alpha ,N}(dx)a.s.$$

$$W_{\alpha ,N}=\mathrm{Leb}\{X_N(x)>\frac{\alpha }{2}\ln \ln N\}$$

$$\frac{W_{\alpha ,N}}{\mathbb{E}\left(W_{\alpha ,N}\right)}\to M_{\alpha },$$

$$\underset{T\to \infty }{\lim }\frac{W_{\alpha ,T}}{\mathbb{E}[W_{\alpha ,T}]}=\underset{T\to \infty }{\lim }\frac{{\int }_0^1|\zeta (1/2+i(\tau +h)){|}^{\alpha }}{\mathbb{E}[|\zeta (1/2+i\tau {|}^{\alpha }]}a.s.$$

$${\sigma }_R^2(N)\equiv \mathrm{Var}(G_N(x)-G_R(x))=\frac{1}{2}\sum _{R<p\le N}{p}^{-1}$$

$${\sigma }_R^2(N)=\left|\sum _{R<p\le N}^N{p}^{-1}-\frac{1}{2}(\ln \ln N-\ln \ln R)\right|=o(1)\text{as N→∞ and R→∞. }$$

$$\underset{R\to \infty }{\lim }\underset{N\to \infty }{\lim }\frac{\mathbb{E}\left(W_{\alpha ,N}|{\mathcal{F}}_R\right)}{\mathbb{E}\left(W_{\alpha ,N}\right)}=M_{\alpha }\text{mbox}a.s.$$

$$\displaystyle\int_{0}^{1}{\mathbb{P}}\left(X_{N}(x)>\frac{\alpha}{2}\ln\ln N\Big{|}{\mathcal{F}}_{R}\right)dx$$

$$\int_{0}^{1}{\mathbb{P}}\left(G_{N}(x)-G_{R}(x)>\frac{\alpha}{2}\ln\ln N-G_{R}(x)-E_{R}(x)-\epsilon)\Big{|}{\mathcal{F}}_{R}\right)dx,$$

$$\frac{{\sigma }_R(N)}{\sqrt{2\pi }(\alpha \ln \ln N-G_R(x)-E_R(x)-ϵ)}\exp \left(-\frac{(\frac{\alpha }{2}\ln \ln N-G_R(x)-E_R(x)-ϵ{)}^2}{2{\sigma }_R^2(N)}\right)$$

$$=\frac{{\sigma }_R(N)}{\sqrt{2\pi }(\alpha \ln \ln N)}\exp \left(-\frac{{\alpha }^2(\ln \ln N{)}^2}{8{\sigma }_R^2(N)}+\alpha (E_R(x)+G_R(x)+ϵ)\right)(1+o(1)),$$

$$\mathbb{E}\left(W_{\alpha ,N}\right)=\mathbb{E}\left(\mathbb{E}\left(W_{\alpha ,N}|{\mathcal{F}}_R\right)\right)\le \frac{{\sigma }_r(N)}{\sqrt{2\pi }(\frac{\alpha }{2}\ln \ln N)}\exp \left(-\frac{{\alpha }^2(\ln \ln N{)}^2}{8{\sigma }_R^2(N)}\right)$$

$$\times {\int }_0^1\mathbb{E}\left(\exp \left(\alpha (E_R(x)+G_R(x)-ϵ)\right)\right)dx(1+o(1))$$

$$\frac{{\int }_0^1\exp \left(\alpha (E_R(x)+G_R(x)+ϵ)\right)}{{\int }_0^1\mathbb{E}\left(\exp \left(\alpha (E_R(x)+G_R(x)-ϵ)\right)\right)dx}(1+o(1))$$

$$\le$$

$$\mathbb{E}\left(\frac{{\int }_0^1\exp \left(\alpha (E_R(x)+G_R(x))\right)dx}{{\int }_0^1\mathbb{E}\left(\exp \left(\alpha (E_R(x)+G_R(x))\right)\right)dx}\right)=1,$$

$$M_{\alpha ,R}{e}^{2\alpha ϵ}(1+o(1))\le \frac{\mathbb{E}\left(W_{\alpha ,N}|{\mathcal{F}}_R\right)}{\mathbb{E}\left(W_{\alpha ,N}\right)}\le M_{\alpha ,R}{e}^{-2\alpha ϵ}(1+o(1)).$$

$$M_{\alpha }{e}^{2\alpha ϵ}(1+o(1))\le \underset{N,r\to \infty }{\mathrm{liminf}}\frac{\mathbb{E}\left(W_{\alpha ,N}|{\mathcal{F}}_R\right)}{\mathbb{E}\left(W_{\alpha ,N}\right)}\le \underset{N,r\to \infty }{\mathrm{limsup}}\frac{\mathbb{E}\left(W_{\alpha ,N}|{\mathcal{F}}_R\right)}{\mathbb{E}\left(W_{\alpha ,N}\right)}\le M_{\alpha }{e}^{-2\alpha ϵ}(1+o(1)).$$

$$\underset{R\to \infty }{\lim }\underset{N\to \infty }{\lim }\frac{\mathbb{E}\left(W_{\alpha ,N}|{\mathcal{F}}_R\right)}{\mathbb{E}\left(W_{\alpha ,N}\right)}=M_{\alpha }.$$

$$W_{\alpha ,N}^{>}=\mathrm{Leb}\{x\in [0,1]:X_N(x)\ge \alpha \ln \ln N;\exists k\in \left[R,N\right]:X_k(x)>(\alpha +ϵ)\ln \ln k\}.$$

$$\mathbb{P}\left(W_{\alpha ,N}^{>}>c\mathbb{E}{W}_{\alpha ,N}\right)\le e^{-ϵr},$$

$$\displaystyle\int_{0}^{1}{\mathbb{P}}\Big{(}\left\{G_{N}(x)-G_{R}(x)\geq\alpha\ln\ln N-X_{R}(x)-\epsilon^{\prime}\right\}$$

$$\displaystyle\hskip 28.45274pt\cap\left\{\exists K\in\left[R,N\right]:\;G_{K}(x)-G_{R}(x)>(\alpha+\epsilon)\ln\ln K-X_{R}(x)-\epsilon^{\prime}\right\}\Big{|}{\mathcal{F}}_{R}\Big{)},$$

$${\sigma }_R^2(K)=\frac{1}{2}\sum _{R<p\le K}{p}^{-1}.$$

$$B_K(x)=G_K(x)-G_R(x)-\frac{{\sigma }_R^2(K)}{{\sigma }_R^2(N)}(G_N(x)-G_R(x)),$$

$${\int }_0^1{\int }_{\frac{\alpha }{2}\ln \ln N-X_R(x)-{ϵ}^{\prime }}^{\infty }\mathbb{P}(G_N(x)-G_R(x)\in dy)$$

$$\displaystyle\qquad\times{\mathbb{P}}\left(\exists K\in\left[R,N\right]:\;B_{K}(x)>\frac{\alpha+\epsilon}{2}\ln\ln K-X_{R}(x)-\epsilon^{\prime}-\frac{\sigma_{R}^{2}(K)}{\sigma_{R}^{2}(N)}y\ \Big{|}{\mathcal{F}}_{R}\right)dx$$

$$\mathbb{P}\left(X_R(x)\ge \frac{ϵr}{3}\right)\le \mathbb{P}\left(G_R(x)\ge \frac{ϵr}{4}\right)+\mathbb{P}\left(E_R(x)\ge \frac{ϵr}{12}\right).$$

Full text

The research of L.-P. A. is supported in part by NSF CAREER DMS-1653602.

1. Introduction

1.1. The model

Let $\mathcal{P}$ denote the set of all prime numbers. Let $(\theta_{p})_{p\in\mathcal{P}}$ be independent identically distributed random variables, being uniformly distributed on $[0,2\pi]$. For $N\in\mathbb{N}$, a good model for the large values of the logarithm of the Riemann-zeta function on a typical interval of length $1$ of the critical line as proposed in [8] is

[TABLE]

By Theorem 7 in [14], the process $X_{N}$ can be well approximated by a log-correlated Gaussian field $G_{N}(x),x\in[0,1]$. Namely, take

[TABLE]

where $(W_{j}^{(i)})_{j\in\mathbb{N},i\in\{1,2\}}$ are i.i.d. standard normal distributed. It is shown in [14] that

[TABLE]

where $E_{N}(x)$ converges almost surely uniformly to a random function $E(x)$. Moreover, the error $E_{N}(x)$ has uniform exponential moments

[TABLE]

where ${\mathbb{E}}$ denotes expectation with respect to the $\theta_{p}$’s.

Some of the behavior of the large values of the process $X_{N}(x)$, $x\in[0,1]$ is captured by the random measure

[TABLE]

By the independence of the $\theta_{p}$’s, it is not hard to see that $M_{\alpha,N}$ converges almost surely as $N\to\infty$. By Theorem 4 in [14], the almost sure weak limit of $M_{\alpha,N}(dx)$ is non-trivial for $0<\alpha<2$. We denote the limit of the total mass by $M_{\alpha}$

[TABLE]

For log-correlated Gaussian field the analogous limiting measure is called Gaussian multiplicative chaos and $M_{\alpha}$ corresponds to the total mass of the limiting measure. For Gaussian multiplicative chaos it was first proven by [9] that the limit is nontrivial for small ${\alpha}$ and was recently revisited (see [13, 12]). Note that in our case the limit of $M_{{\alpha},N}(dx)$ is almost a Gaussian multiplicative measure (see [14]). The connection between the Riemann-zeta function and Gaussian multiplicative chaos has been further analysed in [15].

The fact that the Riemann-zeta function (or a random model of it) can be well approximated by a log-correlated field have recently been used to study the extremes on a random interval [4, 11, 2].

1.2. Main result

Consider the Lebesgue measure of $\alpha$-high points:

[TABLE]

The main result of this note is to relate the limit $M_{\alpha}$ to the Lebesgue measure of high points building on the ideas of [7]:

Theorem 1.1**.**

For any $0<{\alpha}<2$ and $M_{\alpha}$ as in (1.6), we have

[TABLE]

in probability as $N\to\infty$.

It was proved in [2] that the maximum of $X_{N}(x)$ on $[0,1]$ is $\ln\ln N-(3/4\pm\epsilon)\ln\ln\ln N$ with large probability. In view of this and of Theorem 1.1, it is not surprising to see that the $M_{\alpha}$ is non-trivial for $\alpha<2$. The critical case where $\alpha\to 2$ is interesting as it is related to the fluctuations of the maximum of $X_{N}$. It is reasonable to expect that our approach can be adapted to the method of [5] to prove the critical case. Another upshot of the proof is that it highlights the fact that $M_{\alpha}$ depends on small primes, cf. Lemma 2.1.

The problem for the Riemann-zeta function is trickier. We expect that the equivalent of Theorem 1.1 still holds:

Conjecture 1.2**.**

Let $\tau$ be a uniform random variable on $[T,2T]$. Let $W_{\alpha,T}={\rm Leb}\{h\in[0,1]:\ln|\zeta(1/2+i(\tau+h))|>\frac{\alpha}{2}\ln\ln T\}$. Then we have

[TABLE]

This would be consistent with the conjecture of Fyodorov & Keating for the Lebesgue measure of high points, see Section 2.5 in [6] One issue is that it is not obvious that a result akin to Equation (1.3) holds, mainly because of the singularities of $\ln\zeta$ at the zeros. One way around this would be to restrict to Gaussian comparison to one-point and two-point large deviation estimates. This seems doable in view of Lemmas 3.2 and 3.3 and the Gaussian comparison theorem proved for the zeta function in [1].

1.3. Outline of the proof

The proof of Theorem 1.1 is based on a first and a second moment estimate and follow the global strategy proposed in [7] for branching Brownian motion. First, we prove convergence of a conditional first moment to the desired limiting object in Lemma 2.1. Its proof builds on results on the Gaussian comparison and convergence to Gaussian multiplicative chaos established in [14]. Next, a localisation result is established in Lemma 2.2. Finally, we turn to the proof of Proposition 3.4 which is based on a second moment computation. We use a branching approximation similar to the one employed in [2]. Using the obtained first and second moment estimates we are finally in the position to prove Theorem 1.1.

Acknowledgements. Lisa Hartung and Nicola Kistler thank the Rhein-Main Stochastic group for creating an interactive research environment leading to this article.

2. First moment estimates

For $R\leq N$, we define ${\mathcal{F}}_{R}$ to be the ${\sigma}$-algebra generated by $(\theta_{p})_{p\leq R}$. We will often condition on ${\mathcal{F}}_{R}$ to fix the dependence on the small primes. The variance of $G_{N}(x)-G_{R}(x)$, $x\in[0,1]$ is by definition

[TABLE]

The prime number theorem, see e.g. [10], implies that the density of the primes goes like $(\ln p)^{-1}$. More precisely, we have

[TABLE]

It turns out that the non-trivial contribution to Theorem 1.1 comes from the small primes.

Lemma 2.1**.**

For $W_{\alpha,N}$ as in (1.7), we have for $0<\alpha<2$

[TABLE]

Proof.

We start by computing ${\mathbb{E}}\left(W_{\alpha,N}|{\mathcal{F}}_{R}\right)$. Using Fubini’s Theorem we can write the left-hand side of (2.3) as

[TABLE]

where we used (1.3). Moreover, again for each $\epsilon>0$ there is $R_{0}$ such that for all $R\geq R_{0}$ $|E_{R}(x)-E_{N}(x)|<\epsilon$ almost surely and uniformly in $x$. Hence, we can again upper bound (2) by

[TABLE]

and a corresponding lower by replacing $\epsilon$ by $-\epsilon$. Next, observe that by definition of $X_{N}(x)$ and $E_{N}(x)$, $G_{N}(x)-G_{R}(x)$ are independent of ${\mathcal{F}}_{R}$. We have that the probability in (2.5) is bounded from above by

[TABLE]

Next, we turn to ${\mathbb{E}}\left(W_{\alpha,N}\right)$. We have that

[TABLE]

A corresponding lower bound we obtain by replacing $\epsilon$ by $-\epsilon$. Taking the quotient of (2) and and (2) and integrating with respect to $x$ we get

[TABLE]

Pulling the terms involving $\epsilon$ out of the integral and noting the normalization of $M_{{\alpha},R}$ is chosen such that ${\mathbb{E}}M_{{\alpha},R}=1$ and noting that

[TABLE]

we can rewrite (2) as

[TABLE]

Note that (2.10) holds for all $\epsilon>0$. When taking $N,R\uparrow\infty$ $M_{{\alpha},R}$ converges a.s. to $M_{\alpha}$ hence we have a.s.

[TABLE]

As (2.11) does not depend on $r$ and $N$ anymore, we can take the limit as $\epsilon\to 0$ and obtain

[TABLE]

∎

Next, we want to control

[TABLE]

The idea is that, at high points, the value $X_{N}(x)$ is most likely shared equally by the increments as defined in (3.1) below.

Lemma 2.2**.**

For all $\epsilon>0$ there exists $R_{0}$ such that for all $R=o(N)$ and $R,N>R_{0}$ such that for all $c>0$

[TABLE]

where $r=\ln\ln R$.

Proof.

We want to use Markov’s inequality to bound the probability on (2.14). Hence, we need to bound ${\mathbb{E}}W_{{\alpha},N}^{>}$ from above. First, we bound ${\mathbb{E}}\left(W_{{\alpha},N}^{>}|{\mathcal{F}}_{R}\right)$ from above by

[TABLE]

where we used (1.3) and the fact that $E_{R}(x)$ converges a.s. uniformly to a continuous function $E(x)$. Hence, for all $\epsilon^{\prime}>0$ there is $R_{0}$ such that for all $K\geq R_{0}$ and all $x$ we have $|E_{K}(x)-E_{R}(x)|<\epsilon^{\prime}$.

Similarly as in (2.1), the variable $G_{K}(x)-G_{R}(x)$ is Gaussian with mean [math] and variance

[TABLE]

Let

[TABLE]

then $(B_{K}(x))_{K=1}^{N}$ are points on a time-changed brownian bridge from zero to zero in time $\sigma_{R}(N)^{2}$. As a Brownian bridge is independent from its endpoint, Equation (2.15) is equal to

[TABLE]

as $|\sum_{p\leq K}p^{-1}-\ln\ln K|<C$. Let $r=\ln\ln R$ and $n=\ln\ln N$. Next, let us control the probability that $X_{R}(x)$ is too large.

[TABLE]

The second probability in (2.18) is bounded by $Ce^{-\frac{\epsilon r}{12}}$ by (1.4). For the first probability in (2.18) is bounded by $Ce^{-\frac{\epsilon r}{32}}$ by Gaussian tail asymptotics and the variance estimate for $G_{R}(x)$ for $r$ large enough and uniformly in $x$. On the event that $\{X_{R}(x)\leq\frac{\epsilon r}{3}\}$ we can bound the second probability in (2.17) from above by

[TABLE]

where $b(s)$ is a Brownian bridge from zero to zero in time $\sigma_{r}^{2}(N)$. Consider the line $l$ from $(0,\frac{\epsilon}{6}r)$ to $\left(\sigma_{r}^{2}(N),({\alpha}+\epsilon)(n/2-C)-\epsilon^{\prime}-y\right)$. One checks that $l(s)\leq({\alpha}+\epsilon)\left((s+{\sigma}_{0}(R)^{2})-C\right)-\frac{\epsilon r}{3}-\epsilon^{\prime}-\frac{s}{\sigma_{r}^{2}(N)}y$ for all $r$ large enough. The probability of Brownian bridge to stay under a linear function is well known, see e.g., Lemma 2.2 in [3],

[TABLE]

Hence, on the event we can bound the expectation of (2.17) by

[TABLE]

Using the Gaussian tail asymptotics for $G_{N}(x)-G_{R}(x)$ together with (2), Equation (2.21) is bounded above by

[TABLE]

This implies the claim of Lemma 2.2. ∎

3. Branching approximation and second moment estimates

3.1. Definition of the increments

The goal is to use a branching approximation similar to [2] to compute the necessary second moments. To this end, we define for $k\in{\mathbb{N}}$ and $x\in(0,1)$

[TABLE]

By definition, we have

[TABLE]

where for the rest of the section we set $n\equiv\ln\ln N$. The increments $Y_{k}$ are such that

[TABLE]

The covariances can be computed again by the prime number theorem. This is done in Lemma 2.1 in [2]. It is convenient to state the result to introduce branching point of $x,x^{\prime}\in(0,1)$ by

[TABLE]

Lemma 3.1** (Lemma 2.1 in [2]).**

For $k\geq 1$ and $x,x^{\prime}\in(0,1)$ we have

[TABLE]

and

[TABLE]

There is a fast decoupling between the increments after the branching point where the distribution of $Y_{k}(x)$ and $Y_{k}(x^{\prime})$ is very close to independent Gaussians with mean zero and variance $1/2$. We introduce a parameter $\Delta$ that gives some room before and after the branching point to ensure uniform estimates.

Lemma 3.2**.**

Let $\Delta>0$. Let $x,x^{\prime}\in(0,1)$ and $m>x\wedge x^{\prime}+\Delta$. Then we have

[TABLE]

where $(Y_{i})_{i\in{\mathbb{N}}}$ are iid Gaussians with mean zero and variance $\sigma^{2}$.

Proof.

As$\left(\sum_{k=m+1}^{n}Y_{k}(x),\sum_{k=m+1}^{n}Y_{k}(x^{\prime})\right)$ is a Gaussian process and its covariance is controlled in Lemma 3.1 it suffices to compare densities. This follows the same lines starting from Eq. (61) in [2] only that in our setting $\mu=0$. ∎

Before the branching point we want to show that $Y_{k}(x)$ and $Y_{k}(x^{\prime})$ are almost fully correlated. This is specified in the lemma below.

Lemma 3.3**.**

Let $\Delta>0$. Let $x,x^{\prime}\in(0,1)$ and $r<m<x\wedge x^{\prime}-\Delta$. Then we have

[TABLE]

Proof.

As $G$ is a Gaussian process this follows from the density estimates in Lemma 3.1. ∎

3.2. Second moment computation

The main result of this section is:

Proposition 3.4**.**

There exists $\kappa_{\alpha}>0$ such that for $R=o(\ln\ln N)$ as $N\to\infty$ we have

[TABLE]

where $r\equiv\ln\ln R$ and $C>0$ a constant depending on $c$.

To prove Proposition 3.4 we essentially need to control the second moment of

[TABLE]

Remark*.*

Throughout the proof we restrict our computations to $R$ and $N$ such that $r=\ln\ln R$ and $n=\ln\ln N$ are natural numbers. The general case follows in the same way by considering the last resp. first summands in the representation in (3.2) of $G_{N}$ separately. The desired estimates carry over by minor adjustments but would require a more involved notation. To keep the computations that follow as clear as possible and not to burden the reader with heavier notations we restrict ourselves to the case where $r,n\in\mathbb{N}$.

Indeed, Markov’s inequality and Lemma 2.2 imply

[TABLE]

Clearly, we have

[TABLE]

Let $0<{\Delta}<r$. We divide the right side into four terms depending on the branching point:

[TABLE]

The term $(IV)$ is controlled in the following Lemma.

Lemma 3.5**.**

For $R=o(\ln\ln N)$ we have

[TABLE]

Proof of Lemma 3.5.

As $x\wedge x^{\prime}<r-{\Delta}$ and by a similar rewriting in (2.5) we have by Lemma 3.2 that it is bounded from above by

[TABLE]

We now compare (3.2) with $\left({\mathbb{E}}\left(W^{\leq}_{{\alpha},N}|{\mathcal{F}}_{R}\right)\right)^{2}$ which is bounded below by

[TABLE]

for any $\epsilon>0.$ By 2.1 and the Gaussian approximation given in (3.1) the absolute value of the difference of (3.2) and (3.14) is bounded by

[TABLE]

Hence (3.15) divided by ${\mathbb{E}}(W_{{\alpha},N})^{2}$ converges almost surely to

[TABLE]

for all $\epsilon,\Delta>0$. Note that (3.15) converges to zero as $\epsilon\to 0$ and ${\Delta}\to\infty$. ∎

To control the terms $(I)$, $(II)$ and $(III)$, we prove the following lemma.

Lemma 3.6**.**

Let $0<{\alpha}<2$. There exists $\kappa_{\alpha}>0$ such that for $R=o(\ln\ln N)$ as $N\to\infty$ we have

[TABLE]

Proof of Lemma 3.6.

We bound ${\mathbb{E}}((I)|{\mathcal{F}}_{R})$ from above by

[TABLE]

by (2). Hence,

[TABLE]

as ${\mathbb{E}}(W_{{\alpha},N})=cn^{-1/2}e^{-\frac{\alpha^{2}}{4}n}$ and $0<\alpha<2$.

Next, we turn to ${\mathbb{E}}((II)|{\mathcal{F}}_{R})$. Using that uniformly in $y$ for all $R,N$ large enough $|E_{N}(y)-E_{R}(y)|\leq\epsilon$, we can bound ${\mathbb{E}}((II)|{\mathcal{F}}_{R})$ from above by

[TABLE]

Dropping the barrier constraint except at $x\wedge x^{\prime}-\Delta$ and $x\wedge x^{\prime}+\Delta$ we can bound the probability in (3.20) from above by

[TABLE]

We evaluate the probability in the integral at a fixed $x\wedge x^{\prime}=m$, and sum the contributions over $m$ afterwards. We introduce an extra conditioning. Let ${\mathcal{F}}^{Y}_{k}=\sigma(Y_{j},j\leq k)$. We condition on ${\mathcal{F}}^{Y}_{m+{\Delta}}$, slightly after the branching point. Lemma 3.2 applied to (3.21) then yields

[TABLE]

We distinguish two cases. First, consider the case when for $y=x$ or $y=x^{\prime}$,

[TABLE]

Note that due to the barrier in (3.2) this can only happen jointly with the barrier event if $m\geq\frac{{\alpha}}{{\alpha}+\epsilon}n-C^{\prime}\epsilon$ for some constant $C^{\prime}>0$ independent of $\epsilon$. In this case we bound the probabilities above by one and bound (3.2) from above by

[TABLE]

As for an upper bound we can drop all constraints in the expectation with respect $x^{\prime}$ (if $y=x$) and $x$ otherwise, let us assume without loss of generality that $y=x$. We need to distinguish whether $\frac{{\alpha}}{2}n-X_{R}(x)-\epsilon>0$ or not. On the event $\frac{{\alpha}}{2}n-X_{R}(x)-\epsilon\leq 0$ we bound the expectation in (3.24) by one and obtain that the expectation of (3.24) from above by

[TABLE]

by the exponential Chebyshev inequality. Hence, integrating over $x,x^{\prime}$ in (3.2) we get

[TABLE]

by (2). When $\frac{{\alpha}}{2}n-X_{R}(x)-\epsilon>0$, we bound (3.24) from above using Gaussian tail asymptotics by

[TABLE]

The integral of (3.27) with respect to $x$ and $x^{\prime}$ can be bounded from above by

[TABLE]

Using that in the range of summation in (3.2) ${\sigma}_{r}(m+{\Delta})$ is bounded form above and below by $\frac{1}{2}(m-r)+C$ resp. $\frac{1}{2}(m-r)-C$ , for some constant large enough, we can bound (3.2) from above by

[TABLE]

As $m\geq\frac{{\alpha}}{{\alpha}+\epsilon}n-C^{\prime}\epsilon$, exponantial term in $\epsilon$ bounded by $e^{C\epsilon}$ and as $0<{\alpha}<2$ we have that on the one hand $\frac{{\alpha}^{2}}{4}-1<0$ and on the other hand we can choose together with (2) we can bound the corresponding expectation in (3.29) from above by

[TABLE]

for some $c>0$.

Finally, we turn to bound (3.2) for $\frac{{\alpha}}{2}n-X_{r}(y)-\epsilon-\sum_{r<j\leq m+{\Delta}}Y_{j}(y)\geq 0$ we can bound (3.2) from above by a Gaussian tail bound and obtain

[TABLE]

Next, we condition on ${\mathcal{F}}^{Y}_{m-\Delta}$. The terms depending on $\sum_{m-{\Delta}<j\leq m+{\Delta}}Y_{j}$ can be bounded by the moment generating function:

[TABLE]

Hence, Equation (3.31) is bounded above by

[TABLE]

Using the fact that the variables $Y_{j}(x)$ and $Y_{j}(x^{\prime})$ almost coincide for $j\leq m-{\Delta}$ by Lemma 3.3, we have that (3.33) is bounded above by

[TABLE]

The expectation in (3.34) is equal to

[TABLE]

The integrand with respect to $z$ is minimal for

[TABLE]

When $\frac{{\alpha}+\epsilon}{2}m-z^{*}\ll 0$ which is the case when $m<(1-{\delta})n$ for some ${\delta}>0$, we can use Gaussian tail asymptotics to bound (3.35) from above by

[TABLE]

Plugging this bound into (3.34), summing over $m<(1-{\delta})n$, and computing the squares in the exponential, we obtain that (3.34) is bounded from above by

[TABLE]

If $x\wedge x^{\prime}<(1-{\delta})n$ we can bound the Gaussian integral by one and get that (3.8) is bounded from above by

[TABLE]

Using (3.36) we can bound the expectation of (3.39) for $m>(1-{\delta})n$ by

[TABLE]

Plugging this into (3.34) we can bound the contribution from above

[TABLE]

Noting that $2n-n{\delta}\leq n+m\leq 2n$ the above term can be bounded from above by

[TABLE]

Note that the exponent in (3.42) is negative for ${\delta}$ sufficiently small.

Finally, we want to bound ${\mathbb{E}}((III))$. By Lemma 3.2 we have similar to (3.2) that ${\mathbb{E}}((III))$ is bounded from above by $\left(1+e^{-c{\Delta}}\right)$ times

[TABLE]

If $\frac{{\alpha}}{2}n-X_{R}(y)-\epsilon-\sum_{j=r+1}^{m+{\Delta}}Y_{j}(y)>0$ for $y\in\{x,x^{\prime}\}$ we can use Gaussian tail asymptotics for the probabilities in (3.2) to bound the expectation in (3.2) from above by

[TABLE]

Noticing that the polynomial prefactor is bounded by $C/n$ and otherwise proceeding as in (3.32) we can bound (3.2) from above by

[TABLE]

by (2) for any $\epsilon>0$ and ${\Delta}>0$. Note first that due to the barrier event in (3.2) the case $\frac{{\alpha}}{2}n-X_{R}(y)-\epsilon-\sum_{j=r+1}^{m+{\Delta}}Y_{j}(y)\leq 0$ for at least one $y\in\{x,x^{\prime}\}$ can be excluded for $m\in\{r-{\Delta},r+{\Delta}\}$.

This completes the control of $(III)$ and hence also the proof of Theorem 3.6.

∎

Proof of Proposition 3.4.

We bound (3.10) from above by

[TABLE]

where we used Chebyshev’s inequality. By Lemma 3.6 we can bound (3.10) from above by

[TABLE]

which yields Proposition 3.4 by possibly modifying the constants and noting that $\epsilon$ in (3.47) is arbitrary (but fixed) as the claim of Lemma 3.5 holds almost surely in the $N\to\infty$ limit. ∎

4. Proof of Theorem 1.1

Finally, we are in the position to prove Theorem 1.1 using Lemma 2.1 and Proposition 3.4.

Proof of Theorem 1.1.

First, we rewrite

[TABLE]

By Proposition 3.4 the second summand on the right hand side of (4.1) converges to zero in probability when first $N\to\infty$ and then $R\to\infty$. By Lemma 2.1 the term the first summand on the right hand side of (4.1) converges almost surely to $M_{\alpha}$ defined in (1.6). This completes the proof of Theorem 1.1. ∎