# The Fyodorov-Hiary-Keating Conjecture. II.

Louis-Pierre Arguin  Department of Mathematics, Baruch College and Graduate
Center, City University of New York, USA<EMAIL_ADDRESS>,  Paul Bourgade  Courant Institute, New York University, USA
<EMAIL_ADDRESS>and  Maksym Radziwiłł  Department of Mathematics,
Caltech, Department of Mathematics, U.T. Austin, USA
<EMAIL_ADDRESS>
###### Abstract.

We prove a lower bound on the maximum of the Riemann zeta function in a
typical short interval on the critical line. Together with the upper bound
from [ArgBouRad2020], this implies tightness of

$\max_{|h|\leq
1}|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|\cdot\frac{(\log\log
T)^{3/4}}{\log T},$

for large $T$, where $\tau$ is uniformly distributed on $[T,2T]$. The
techniques are also applied to bound the right tail of the maximum, proving
the distributional decay $\asymp ye^{-2y}$ for $y$ positive. This confirms the
Fyodorov-Hiary-Keating conjecture, which states that the maximum of $\zeta$ in
short intervals lies in the universality class of logarithmically correlated
fields.

###### Contents

  1. 1 Introduction
  2. 2 Proof of Theorem 1
  3. 3 Approximations by Gaussian Random Walks
  4. 4 Proof of Proposition 3
  5. 5 Proof of Proposition 4
  6. 6 Proof of Proposition 6
  7. 7 Proof of Theorem 2
  8. A Some Auxiliary Results
  9. B Ballot Theorem

##  1\. Introduction

The distribution of the Riemann zeta function on the critical line is
conjecturally related to random matrices, a fact discovered by Montgomery
[Mon1972] for local statistics of the zeros. It was extended in many
directions including to distributions for families of $L$-functions
[KatSar1999] and their moments [KeaSna2000]. Fyodorov, Hiary & Keating
[FyoHiaKea2012] and Fyodorov & Keating [FyoKea2014] proposed to further expand
the scope of this analogy at the level of extreme values. Based on a similar
conjecture for random unitary matrices, they put forward the very precise
asymptotics

$\frac{1}{T}\cdot\text{\rm meas}\Big{\\{}T\leq t\leq 2T:\max_{|h|\leq
1}|\zeta(\tfrac{1}{2}+\mathrm{i}t+\mathrm{i}h)|>e^{y}\cdot\frac{\log
T}{(\log\log T)^{3/4}}\Big{\\}}\rightarrow F(y),$

as $T\rightarrow\infty$, where the limiting distribution function $F$
satisfies $F(y)\sim Cye^{-2y}$ for large $y$. While the explicit form of $F$
is not expected to be universal, the exponent $3/4$ and the tail asymptotics
$ye^{-2y}$ characterize the universality class of logarithmically correlated
fields.

In the first part of this series [ArgBouRad2020], we showed the upper bound of
this conjecture, $F(y)\ll ye^{-2y}$. The main goal of this paper is to
complete this work and establish tightness in the Fyodorov-Hiary-Keating
conjecture, by showing $F(y)\to 1$ as $y\to-\infty$. The following is the main
result.

######  Theorem 1.

There exists $c>0$ such that for any $T\geq 100$ and $0\leq y\leq(\log\log
T)^{1/10}$ we have

$\frac{1}{T}\cdot\text{\rm meas}\Big{\\{}T\leq t\leq 2T:\max_{|h|\leq
1}|\zeta(1/2+\mathrm{i}t+\mathrm{i}h)|<e^{-y}\frac{\log T}{(\log\log
T)^{3/4}}\Big{\\}}\leq c^{-1}y^{-c}.$

A direct consequence of the above result and [ArgBouRad2020, Theorem 1] is the
expected tightness of maxima on short intervals, and existence of
subsequential limits.

######  Corollary 1.

For every $\varepsilon>0$ there exists $C>0$ such that for any $T\geq 100$,
for $t\in[T,2T]$ in a set of measure larger than $(1-\varepsilon)T$ we have

$\left|\max_{|h|\leq
1}\log|\zeta(\tfrac{1}{2}+\mathrm{i}t+\mathrm{i}h)|-(\log\log
T-\frac{3}{4}\log\log\log T)\right|\leq C.$

In particular, there exists a subsequence $T_{\ell}\to\infty$ and a
distribution function $F$ such that

$\frac{1}{T_{\ell}}\cdot\text{\rm
meas}\Big{\\{}t\in[T_{\ell},2T_{\ell}]:\max_{|h|\leq
1}|\zeta(\tfrac{1}{2}+\mathrm{i}t+\mathrm{i}h)|>e^{y}\cdot\frac{\log
T_{\ell}}{(\log\log T_{\ell})^{3/4}}\Big{\\}}\rightarrow F(y),$

uniformly in $y\in\mathbb{R}$ outside of a countable set.

Previous results in the direction of Theorem 1 were limited to the first order
$\log T$, conditionally on the Riemann Hypothesis by Najnudel [Naj2018] and
unconditionally by the authors with Belius and Soundararajan
[ArgBelBouRadSou2019]. This contrasts with the developments on the upper
bound, starting with the first order $\log T$ proved in [Naj2018,
ArgBelBouRadSou2019], then the second order by Harper [Har2019], and finally
the optimal upper bound with the tail distribution [ArgBouRad2020]. In fact,
the present work builds on many techniques developed for the upper bound in
[ArgBouRad2020], as well as new inputs as we now explain.

Progress towards the Fyodorov-Hiary-Keating conjecture has relied on the
observation that the maxima of $|\zeta|$ on a short interval are related to
extremes of branching processes. Indeed, the emergence of large values of
$|\zeta|$ follows a scenario first identified by Bramson [Bra1978] in the
setting of branching Brownian motion. As explained in the introduction of
[ArgBouRad2020], the explicit branching structure behind $\zeta$ comes from
the Dirichlet polynomials $(S_{k}(h),k\geq 1)$, $|h|\leq 1$, defined in (3).
These polynomials behave similarly to correlated random walks, the time index
$k$ corresponding to primes in the loglog scale. Bramson’s scenario translates
into the ballistic behavior of $(S_{k}(h),k\leq n_{\mathcal{L}})$ conditioned
not to cross an upper barrier. Estimating the maximum of $\zeta$ with a
precision of order one is a delicate task because the final index
$n_{\mathcal{L}}$ needs to be $y$-dependent and very large, i.e., the sum must
include primes very close to $T$.

The proof of Theorem 1 is decomposed into two parts. First, it is shown that
large values of $S_{n_{\mathcal{L}}}$ indeed imply large values of
$\log|\zeta|$, cf. Proposition 1. Second, we prove that large values of
$S_{n_{\mathcal{L}}}$ of the claimed size are achieved, see Proposition 2.
Proposition 2 builds on two techniques from [ArgBouRad2020], namely the
introduction of a lower barrier ensuring that large deviations of the
increments of $S_{k}$ can be obtained even for large primes, and the precise
encoding through Dirichlet sums of the event that $S$ remains in the corridor
defined by an upper barrier and lower barrier.

To justify that large values of $S_{n_{\mathcal{L}}}$ imply large values of
$\log|\zeta|$, the first order asymptotics from [ArgBelBouRadSou2019] relied
on working on the right of the critical line. However, implementing this
method for the much finer tightness would be considerably more involved.
Instead, Proposition 1 uses a new, simpler argument allowing to work directly
on the critical line, through an integral approximation of $\zeta$ by a finite
Euler product (Lemma 5), and a control of the regularity in $h$ of
$S_{n_{\mathcal{L}}}$ on high points (Proposition 4).

With these methods developed for Theorem 1, we can also complement the upper
bound $F(y)\ll ye^{-2y}$ from [ArgBouRad2020, Theorem 1], and show that
$F(y)\asymp ye^{-2y}$ for positive $y$.

######  Theorem 2.

For any $C>0$ there exists $c>0$ such that for any $10\leq y\leq
C\frac{\log\log T}{\log\log\log T}$, we have

$\mathbb{P}\Big{(}\max_{|h|\leq
1}|\zeta(1/2+\mathrm{i}\tau+\mathrm{i}h)|>e^{y}\frac{\log T}{(\log\log
T)^{3/4}}\Big{)}\geq c\,ye^{-2y}e^{-y^{2}/\log\log T}.$ (1)

This proves the matching lower bound of the upper tail not only in the in the
exponential regime $y\leq\sqrt{\log\log T}$ but also in the Gaussian regime
$\sqrt{\log\log T}\leq y\leq C\frac{\log\log T}{\log\log\log T}$, because the
proof of [ArgBouRad2020, Theorem 1] implies the Gaussian decay in this range.

The estimate (1) essentially matches the range $y={\rm o}(t)$ proved by
Bramson [Bra1983] for the branching Brownian motion up to time $t$. (The time
$t$ corresponds to $\log\log T$ in our problem.) It is weaker by a logarithmic
factor as it would corresponds to $y\leq C\,t/\log t$ in the branching
Brownian motion case. We are not aware of other examples of log-correlated
processes where the order of the right tail of the maximum is known to this
level of precision. In fact, any form of decay has only been proved for a few
models in this universality class. Notably for the branching random walk, the
best known range is $y={\rm O}(\sqrt{t})$ [BraDinZei2016bis], which matches
the known precision for the two dimensional discrete Gaussian free field on
the $N\times N$ square grid, $y={\rm O}(\sqrt{\log N})$ [Din2013, DinZei2014,
BraDinZei2016]. A finer control of the contributions from small primes in the
random walk would improve this range of $y$ in Theorem 2 to match Bramson’s.

The distributional limit obtained in Corollary 1 is presumably unique but we
believe this is out of reach with current number theory techniques. Moreover,
no explicit formula for $F$ was conjectured. Indeed, denoting $U_{n}$ a Haar-
distributed $n\times n$ unitary matrix, [FyoHiaKea2012] proposed a very
precise limiting distribution for

$\sup_{|z|=1}\big{(}\log|{\rm det}(z-U_{n})|-\log n+\frac{3}{4}\log\log
n\big{)},$ (2)

but as explained in [FyoKea2014] this limit is not expected to coincide with
$F$: It primarily suggested the characteristic exponent $3/4$ and the tail
distribution $ye^{-2y}$ for $\zeta$, which are the prominent signatures of
extremal statistics in log-correlated fields [CarLed01]. Progress on a limit
for (2) culminated in the breakthrough proofs of tightness [ChhMadNaj2018] and
uniqueness [PaqZei2022] of a limiting distribution for the more general
circular beta ensembles, after initial steps verifying the first
[ArgBelBou2017] and second order terms [PaqZei2018].

The exact form of the limiting distribution of (2), and universality of its
right tail, remain open.

Acknowledgment. L.-P. A. is supported by the grants NSF CAREER 1653602 and NSF
DMS 2153803, P. B. is supported by the NSF grant DMS 2054851, and M. R. is
supported by the NSF grant DMS 1902063.

Notation. Throughout the paper, $\tau$ will denote a random variable uniformly
distributed in $[T,2T]$, and $T$ will be some large parameter that is usually
taken to go to infinity. With this notation, for any measurable function $f$
on $[T,2T]$ and event $A$, we have

$\mathbb{P}(f(\tau)\in A):=\frac{1}{T}\cdot\text{\rm meas}\Big{\\{}T\leq t\leq
2T:f(t)\in A\Big{\\}}.$

##  2\. Proof of Theorem 1

Let

$n_{0}:=\lfloor y\rfloor\text{ and }n:=\lfloor\log\log T\rfloor\text{ and
}n_{\mathcal{L}}:=n-n_{0}.$

For $n_{0}\leq k\leq n_{\mathcal{L}}$ and $|h|\leq 1$, we consider the partial
sums

$S_{k}(h)=\sum_{n_{0}<\log\log p\leq k}{\rm
Re}\Big{(}p^{-(1/2+\mathrm{i}\tau+\mathrm{i}h)}+\frac{1}{2}\cdot
p^{-2(1/2+\mathrm{i}\tau+\mathrm{i}h)}\Big{)}.$ (3)

Essentially one can think of $S_{k}(h)$ as an approximation to

$\int_{\mathbb{R}}\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h+\mathrm{i}x)|f(e^{k}x)\
e^{k}{\rm d}x.$

for some choice of smoothing with $\widehat{f}$ compactly supported.

We will show that with high probability the local maxima of
$S_{n_{\mathcal{L}}}(h)$ arise at those $h$ at which the partial sums
$S_{k}(h)$ evolve in a predictable manner as $k$ runs from $n_{0}$ to
$n_{\mathcal{L}}$. More precisely, the partial sums $S_{k}(h)$ of maximizing
$h$’s are constrained between $L_{k}$ and $U_{k}$ (defined below) for all
$n_{0}\leq k\leq n_{\mathcal{L}}$. Once $k$ reaches $n_{\mathcal{L}}$ there
are only ${\rm O}(1)_{y}$ well-spaced (i.e, $1/\log T$ spaced) values of $h$
that can satisfy all those constraints, thus identifying the maximum almost
uniquely.

In order to define $L_{k}$ and $U_{k}$ we introduce the slope,

$\alpha=1-\frac{3}{4}\frac{\log n}{n},$ (4)

Furthermore given a function $f$, we define a symmetrized version,

$\mathcal{S}_{\mathcal{L}}(f)(k):=\begin{cases}f(k-n_{0})&\text{ for
}n_{0}<k\leq\frac{n}{2},\\\ f(n_{\mathcal{L}}-k)&\text{ for
}\frac{n}{2}<k<n_{\mathcal{L}},\\\ 0&\text{ for }k\geq n_{\mathcal{L}}\text{
or }k\leq n_{0}.\end{cases}$

Then, the so-called barriers (i.e., values $L_{k}$ and $U_{k}$) are defined as

$\displaystyle U_{k}$
$\displaystyle=\frac{y}{10}+\alpha(k-n_{0})-10\mathcal{S}_{\mathcal{L}}(x\mapsto\log(x))(k),$
(5) $\displaystyle L_{k}$
$\displaystyle=-10y+\alpha(k-n_{0})-\mathcal{S}_{\mathcal{L}}(x\mapsto
x^{3/4})(k).$ (6)

We now introduce the set of good points $G_{\mathcal{L}}$, defined more
generally for $n_{0}\leq\ell\leq n_{\mathcal{L}}$ as

$\displaystyle G_{0}$ $\displaystyle=[-\tfrac{1}{2},\tfrac{1}{2}]\cap
e^{-(n_{\mathcal{L}}-n_{0})}\mathbb{Z},$ (7) $\displaystyle G_{\ell}$
$\displaystyle=\\{h\in G_{0}:S_{k}(h)\in[L_{k},U_{k}]\text{ for all
}k\leq\ell\\}.$

We will show that with high probability the local maximum belongs to
$G_{\mathcal{L}}$.

We first comment on the above choices of barriers and discrete sets. The
interval $[-\tfrac{1}{2},\tfrac{1}{2}]$ defining $G_{0}$ needs to be strictly
included in the original interval $[-1,1]$, as it will be apparent in the
proof of Proposition 3. Moreover, the discretization step
$e^{-(n_{\mathcal{L}}-n_{0})}$ will be convenient for the proof of Proposition
6 as it corresponds to the number of steps of the random walk (3), but it is
not essential and any step in
$[e^{-n_{\mathcal{L}}},e^{-(n_{\mathcal{L}}-n_{0})}]$ would work. However,
contrary to [ArgBouRad2020], it is essential that the upper barrier is convex
and not concave, as we will see in the proof of Proposition 6.

The proof of the main theorem reduces now to two main propositions. In the
first proposition, we show how the local maxima of the zeta function arise
from the good points $h\in G_{\mathcal{L}}$.

######  Proposition 1.

There exists an absolute constant $C>0$ such that uniformly in $T\geq 100$ and
$0\leq y\leq(\log\log T)^{1/10}$ we have

$\displaystyle\mathbb{P}\Big{(}\max_{|h|\leq 1}$
$\displaystyle\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|\geq
n-\frac{3}{4}\log n-100y-C\Big{)}\geq\mathbb{P}\Big{(}\exists h\in
G_{\mathcal{L}}\Big{)}+{\rm O}(e^{-y}).$

In the second proposition, we then show that good points exist with high
probability.

######  Proposition 2.

There exists $c>0$ such that uniformly in $T\geq 100$ and $0\leq
y\leq(\log\log T)^{1/10}$ we have

$\displaystyle\mathbb{P}\Big{(}\exists h\in G_{\mathcal{L}}\Big{)}=1+{\rm
O}(y^{-c}).$

Combining Proposition 1 and Proposition 2 yields Theorem 1. We now describe
the proofs of Proposition 2 and Proposition 1

###  2.1. Proof of Proposition 1

The proof of Proposition 1 breaks down into two propositions.

######  Proposition 3.

There exists $C>0$ such that for any $1000<y<n^{1/10}$

$\mathbb{P}\Big{(}\max_{|h|\leq
1}\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|\geq\max_{h\in
G_{0}}\min_{|u|\leq
1}(S_{n_{\mathcal{L}}}(h+u)+\sqrt{|u|e^{n_{\mathcal{L}}}})-2C-20y\Big{)}\geq
1-{\rm O}(e^{-y}).$

We then show that with high probability for all $h\in G_{\mathcal{L}}$ and all
$|u|\leq 1$,

$\displaystyle|S_{n_{\mathcal{L}}}(h+u)-S_{n_{\mathcal{L}}}(h)|\leq
20y+\sqrt{|u|e^{n_{\mathcal{L}}}}.$

######  Proposition 4.

For any $1000<y<n^{1/10}$ we have

$\mathbb{P}\Big{(}\forall h\in G_{\mathcal{L}}\ \forall|u|\leq
1:|S_{n_{\mathcal{L}}}(h+u)-S_{n_{\mathcal{L}}}(h)|\leq
20y+\sqrt{|u|e^{n_{\mathcal{L}}}}\Big{)}=1-{\rm O}(e^{-y}).$

On the event that there exists a $h\in G_{\mathcal{L}}$, Proposition 4 now
implies

$\displaystyle\max_{v\in G_{0}}\min_{|u|\leq
1}(S_{n_{\mathcal{L}}}(v+u)+\sqrt{|u|e^{n_{\mathcal{L}}}})$
$\displaystyle\geq\min_{|u|\leq
1}(S_{n_{\mathcal{L}}}(h+u)+\sqrt{|u|e^{n_{\mathcal{L}}}})$ $\displaystyle\geq
S_{n_{\mathcal{L}}}(h)-20y\geq n-\frac{3}{4}\log n-50y$

outside of a set of probability ${\rm O}(e^{-y})$. Proposition 3 then yields
that outside of a set of $\tau$ of probability $\ll e^{-y}$,

$\max_{|h|\leq
1}\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|>n-\frac{3}{4}\log
n-100y-2C.$

In other words,

$\mathbb{P}(\exists h\in G_{\mathcal{L}})\leq\mathbb{P}\Big{(}\max_{|h|\leq
1}\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|>n-\frac{3}{4}\log
n-100y-2C\Big{)}+{\rm O}(e^{-y})$

and Proposition 1 follows.

###  2.2. Proof of Proposition 2

Cauchy-Schwarz inequality readily implies

$\mathbb{P}\Big{(}\exists h\in
G_{\mathcal{L}}\Big{)}\geq\frac{\mathbb{E}[\\#G_{\mathcal{L}}]^{2}}{\mathbb{E}[(\\#G_{\mathcal{L}})^{2}]}.$
(8)

For fixed $k$ and $h,h^{\prime}\in G_{0}$, we posit that the random variables
$(S_{k}(h),S_{k}(h^{\prime}))$ can be well approximated by two correlated
Gaussian random variables $(\mathcal{G}_{k}(h),\mathcal{G}_{k}(h^{\prime}))$
with,

$\displaystyle\mathcal{G}_{k}(h):=\sum_{n_{0}\leq j\leq
k}\mathcal{N}_{j}\text{ and }\mathcal{G}_{k}(h^{\prime}):=\sum_{n_{0}\leq
j\leq k}\mathcal{N}_{j}^{\prime},$ (9)

where the increments $\mathcal{N}_{j}$ and $\mathcal{N}_{j}^{\prime}$ are
Gaussian random variables with mean $0$, equal variance

$\mathbb{E}[\mathcal{N}_{k}^{2}]=\mathbb{E}[{\mathcal{N}^{\prime}_{k}}^{2}]=\mathfrak{s}_{k}^{2}:=\sum_{e^{k-1}<\log
p\leq e^{k}}\Big{(}\frac{1}{2p}+\frac{1}{8p^{2}}\Big{)},$ (10)

and covariance

$\mathbb{E}[\mathcal{N}_{k}\mathcal{N}^{\prime}_{k}]=\rho_{k}:=\sum_{e^{k-1}<\log
p\leq e^{k}}\Big{(}\frac{\cos(|h-h^{\prime}|\log
p)}{2p}+\frac{\cos(2|h-h^{\prime}|\log p)}{8p^{2}}\Big{)}.$ (11)

The analog of the good sets (7) for the Gaussian random variables is

$\mathfrak{G}_{\mathcal{L}}^{\pm}:=\\#\Big{\\{}h\in
G_{0}:\mathcal{G}_{k}(h)\in[L_{k}\mp 1,U_{k}\pm 1]\text{ for all }n_{0}\leq
k\leq n_{\mathcal{L}}\Big{\\}}.$

We then show that in (8) we can replace the arithmetic good set
$G_{\mathcal{L}}$ by the purely probabilistic good sets
$\mathfrak{G}_{\mathcal{L}}^{\pm}$.

######  Proposition 5.

Uniformly in $T\geq 100$ and $100\leq y\leq n^{1/10}$, we have

$\frac{\mathbb{E}[\\#G_{\mathcal{L}}]^{2}}{\mathbb{E}[(\\#G_{\mathcal{L}})^{2}]}\geq\Big{(}1+{\rm
O}(y^{-10})\Big{)}\frac{\mathbb{E}[\\#\mathfrak{G}_{\mathcal{L}}^{+}]^{2}}{\mathbb{E}[(\\#\mathfrak{G}_{\mathcal{L}}^{-})^{2}]}.$

###### Proof.

This result is an immediate consequence of the comparison with Gaussian random
walks as stated in Propositions 7 and 8 in the next Section 3. ∎

The problem is now reduced to a purely probabilistic computation. The proof of
Proposition 2 is concluded by the next proposition building on ideas of
Bramson.

######  Proposition 6.

There is an absolute constant $c>0$ such that for any $T\geq 100$ and
$c^{-1}\leq y\leq n^{1/10}$,

$\frac{\mathbb{E}[\\#\mathfrak{G}_{\mathcal{L}}^{+}]^{2}}{\mathbb{E}[(\\#\mathfrak{G}_{\mathcal{L}}^{-})^{2}]}\geq
1-y^{-c}.$

Combining the two above propositions with the lower bound from (8) yield
Proposition 2.

##  3\. Approximations by Gaussian Random Walks

The proof of Proposition 5 relies on approximating one-point and two-point
correlations in terms of correlations of Gaussian random variables, see
Propositions 7 and 8 below. Note that the one-point estimate contains an
additional twist by a Dirichlet polynomial. This will be needed in the proof
of Proposition 4. The proofs of Propositions 7 and 8 are independent of the
rest of the paper and can be skipped on a first reading.

######  Proposition 7.

Let $h\in[-1,1]$. Let $n_{0}\leq\ell\leq n_{\mathcal{L}}$. Let
$(S_{k}(h),n_{0}\leq k\leq n_{\mathcal{L}})$ and
$(\mathcal{G}_{k}(h),n_{0}\leq k\leq n_{\mathcal{L}})$ be as in Equations (3)
and (9). Let $\mathcal{Q}$ be a Dirichlet polynomial of length
$\leq\exp(\tfrac{1}{100}e^{n})$ and supported on integers such that all their
prime factors are greater than $\exp(e^{\ell})$. Then, we have for $n_{0}$
large enough,

$\displaystyle\mathbb{E}\Big{[}$
$\displaystyle|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|^{2}\mathbf{1}\Big{(}S_{k}(h)\in[L_{k},U_{k}],k\leq\ell\Big{)}\Big{]}$
(12)
$\displaystyle\geq(1+n_{0}^{-10})\mathbb{E}\Big{[}|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|^{2}\Big{]}\cdot\mathbb{P}\Big{(}\mathcal{G}_{k}(h)\in[L_{k}+1,U_{k}-1],n_{0}\leq
k\leq\ell\Big{)}$

and

$\displaystyle\mathbb{E}\Big{[}$
$\displaystyle|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|^{2}\mathbf{1}\Big{(}S_{k}(h)\in[L_{k},U_{k}],k\leq\ell\Big{)}\Big{]}$
(13)
$\displaystyle\leq(1+n_{0}^{-10})\mathbb{E}\Big{[}|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|^{2}\Big{]}\cdot\mathbb{P}(\mathcal{G}_{k}(h)\in[L_{k}-1,U_{k}+1],n_{0}<k\leq\ell).$

######  Proposition 8.

Let $h,h^{\prime}\in[-1,1]$. Consider $(S_{k}(h),S_{k}(h))$ and
$(\mathcal{G}_{k}(h),\mathcal{G}_{k}^{\prime}(h))$ for $n_{0}<k\leq
n_{\mathcal{L}}$ as defined in Equations (3) and (9). We have for $n_{0}$
large enough

$\displaystyle\mathbb{P}\big{(}(S_{k}(h),S_{k}(h^{\prime}))\in[L_{k},U_{k}]^{2},n_{0}<k\leq
n_{\mathcal{L}}\big{)}$ (14) $\displaystyle\hskip
56.9055pt\leq(1+n_{0}^{-10})\cdot\mathbb{P}\big{(}(\mathcal{G}_{k}(h),\mathcal{G}_{k}(h^{\prime}))\in[L_{k}-1,U_{k}+1]^{2},n_{0}<k\leq
n_{\mathcal{L}}\big{)}.$

A similar lower bound can be proved, but is actually not needed in the proofs
of Theorem 1 and 2.

The proof of both propositions rely on an extension of the techniques of
[ArgBouRad2020] to estimate the probability of events involving the partial
sums (3) in terms of random walk estimates. The first step is to approximate
indicator functions in terms of explicit polynomials in Section 3.1. The
relations between the partial sums and the random walks are then established
in Section 3.2 via Dirichlet polynomials.

###  3.1. Approximation of Indicator Functions by Polynomials

First, we state a slight modification of [ArgBouRad2020, Lemma 6] that is more
convenient when working with lower bounds. Throughout the paper, the
normalization for the Fourier transform is

$\widehat{f}(u)=\int_{\mathbb{R}}e^{-\mathrm{i}2\pi ux}f(x){\rm d}x.$

######  Lemma 1.

There exists an absolute constant $C>0$ such that for any $\Delta,A\geq 3$,
there exist entire functions $G^{-}_{\Delta,A}$ and $G^{+}_{\Delta,A}(x)\in
L^{2}(\mathbb{R})$ such that:

  1. (1)

The Fourier transforms $\widehat{G^{\pm}_{\Delta,A}}$ are supported on
$[-\Delta^{2A},\Delta^{2A}]$.

  2. (2)

We have, $0\leq G^{-}_{\Delta,A}(x)\leq G^{+}_{\Delta,A}(x)\leq 1$ for all
$x\in\mathbb{R}$.

  3. (3)

We have

$\displaystyle\mathbf{1}(x\in[0,\Delta^{-1}])$ $\displaystyle\leq
G^{+}_{\Delta,A}(x)\cdot(1+Ce^{-\Delta^{A-1}}),$
$\displaystyle\mathbf{1}(x\in[0,\Delta^{-1}])$ $\displaystyle\geq
G^{-}_{\Delta,A}(x)-Ce^{-\Delta^{A-1}}.$

  4. (4)

We have

$\displaystyle G^{+}_{\Delta,A}(x)$
$\displaystyle\leq\mathbf{1}(x\in[-\Delta^{-A/2},\Delta^{-1}+\Delta^{-A/2}])+Ce^{-\Delta^{A-1}},$
$\displaystyle G^{-}_{\Delta,A}(x)$
$\displaystyle\geq\mathbf{1}(x\in[\Delta^{-A/2},\Delta^{-1}-\Delta^{-A/2}])\cdot(1-Ce^{-\Delta^{A-1}}).$

  5. (5)

We have $\int_{\mathbb{R}}|\widehat{G^{\pm}_{\Delta,A}}(x)|{\rm d}x\leq
2\Delta^{2A}.$

###### Proof.

This is proved the same way as [ArgBouRad2020, Lemma 6] with

$G^{-}_{\Delta,A}(x)=\int_{\Delta^{-A/2}-\Delta^{-A}}^{\Delta^{-1}-\Delta^{-A/2}+\Delta^{-A}}\Delta^{2A}F(\Delta^{2A}(x-t)){\rm
d}t$

and

$G^{+}_{\Delta,A}(x)=\int_{-\Delta^{-A}}^{\Delta^{-1}+\Delta^{-A}}\Delta^{2A}F(\Delta^{2A}(x-t)){\rm
d}t$

with the approximate identity $F=F_{0}/\|F_{0}\|_{1}$, where the existence of
$F_{0}$ is given by the following lemma.∎

######  Lemma 2.

[ArgBouRad2020, Lemma 5] There exists a smooth function $F_{0}$ such that

  1. (1)

For all $x\in\mathbb{R}$, we have $0\leq F_{0}(x)\leq 1$ and
$\widehat{F}_{0}(x)\geq 0$.

  2. (2)

$\widehat{F}_{0}$ is compactly supported on $[-1,1]$.

  3. (3)

Uniformly in $x\in\mathbb{R}$, we have

$F_{0}(x)\ll e^{-|x|/\log^{2}(|x|+10)}.$

With Lemma 1, we get the following estimate of indicator functions expressed
in terms of polynomials.

######  Lemma 3.

Let $A\geq 3$ and $\Delta$ large enough. There exist polynomials
$\mathcal{D}^{-}_{\Delta,A}(x)$ and $\mathcal{D}^{+}_{\Delta,A}(x)$ of degree
at most $\Delta^{10A}$ with $\ell$-th coefficient bounded by
$2\Delta^{2A(\ell+1)}$ such that for all $|x|\leq\Delta^{6A}$

$\displaystyle{\mathbf{1}}(x\in[0,\Delta^{-1}])$
$\displaystyle\leq(1+Ce^{-\Delta^{A-1}})|\mathcal{D}^{+}_{\Delta,A}(x)|^{2}$
(15) $\displaystyle|\mathcal{D}^{+}_{\Delta,A}(x)|^{2}$
$\displaystyle\leq\mathbf{1}(x\in[-\Delta^{-A/2},\Delta^{-1}+\Delta^{-A/2}])+Ce^{-\Delta^{A-1}},$

and

$\displaystyle{\mathbf{1}}(x\in[0,\Delta^{-1}])$
$\displaystyle\geq|\mathcal{D}^{-}_{\Delta,A}(x)|^{2}-Ce^{-\Delta^{A-1}}$ (16)
$\displaystyle|\mathcal{D}^{-}_{\Delta,A}(x)|^{2}$
$\displaystyle\geq(1-Ce^{-\Delta^{A-1}})\mathbf{1}(x\in[\Delta^{-A/2},\Delta^{-1}-\Delta^{-A/2}]),$

for some absolute constant $C>0$.

###### Proof.

We prove the inequalities (16) for $\mathcal{D}^{-}_{\Delta,A}$. The ones for
$\mathcal{D}^{+}_{\Delta,A}(x)$ were proved in [ArgBouRad2020] using the
function $G^{+}_{\Delta,A}$, cf. Equations (32), (33) and (41), (42) there.
The treatment is very similar to the one below.

For the first inequality in (16), item (3) of Lemma 1 ensures the existence of
a function $G^{-}_{\Delta,A}(x)$ in $L^{2}$ such that

${\mathbf{1}}(x\in[0,\Delta^{-1}])\geq
G^{-}_{\Delta,A}(x)-Ce^{-\Delta^{A-1}}.$ (17)

For $\nu=\Delta^{10A}$, we write $G^{-}_{\Delta,A}(x)$ as

$G^{-}_{\Delta,A}(x)=\int_{\mathbb{R}}e^{2\pi\mathrm{i}\xi
x}\widehat{G^{-}_{\Delta,A}}(\xi){\rm
d}\xi=\mathcal{D}^{-}_{\Delta,A}(x)+\sum_{\ell>\nu}\frac{(2\pi\mathrm{i}x)^{\ell}}{\ell!}\int_{\mathbb{R}}\xi^{\ell}\widehat{G^{-}_{\Delta,A}}(\xi){\rm
d}\xi,$ (18)

where

$\mathcal{D}^{-}_{\Delta,A}(x)=\sum_{\ell\leq\nu}\frac{(2\pi\mathrm{i}x)^{\ell}}{\ell!}\int_{\mathbb{R}}\xi^{\ell}\widehat{G^{-}_{\Delta,A}}(\xi){\rm
d}\xi.$ (19)

Clearly, the degree of $\mathcal{D}^{-}_{\Delta,A}$ is $\nu=\Delta^{10A}$,and

$\int_{\mathbb{R}}|\xi|^{\ell}|\widehat{G^{-}_{\Delta,A}}(\xi)|{\rm
d}\xi\leq\Delta^{2A\ell}\int_{\mathbb{R}}|\widehat{G^{-}_{\Delta,A}}(\xi)|{\rm
d}\xi\leq 2\Delta^{2A(\ell+1)},$ (20)

by properties (1) and (5) of Lemma 1. Thus, the coefficients of
$\mathcal{D}^{-}_{\Delta,A}(x)$ are bounded by $\ll\Delta^{2A(\ell+1)}$.

Assuming that $|x|\leq\Delta^{6A}$, then the error term in Equation (18) is
smaller than

$\frac{(2\pi)^{\nu}}{\nu!}|x|^{\nu}\int_{\mathbb{R}}|\xi^{\nu}||\widehat{G^{-}_{\Delta,A}}(\xi)|{\rm
d}\xi\leq\frac{10^{\nu}}{\nu!}\Delta^{6A\nu}\,\Delta^{2A(\nu+1)}\leq\frac{10^{\nu}}{\nu!}\Delta^{9A\nu}.$
(21)

This is $\leq e^{-\Delta^{A}}$ for the choice $\nu=\Delta^{10A}$. This shows
that whenever $|x|\leq\Delta^{6A}$

$G^{-}_{\Delta,A}(x)=\mathcal{D}^{-}_{\Delta,A}(x)+{\rm
O}^{*}(e^{-\Delta^{A}}),$ (22)

where the ${\rm O}^{\star}$ means that the implicit constant is smaller than
$1$. If $x\notin[0,\Delta^{-1}]$, then Equations (17) and (22) with the fact
that $G^{-}_{\Delta,A}\geq 0$ imply

$-e^{-\Delta^{A}}\leq\mathcal{D}^{-}_{\Delta,A}(x)\leq 2Ce^{-\Delta^{A-1}},$

for $\Delta$ large enough (depending on $C$). Therefore, in this case, the
following holds

$\displaystyle{\mathbf{1}}(x\in[0,\Delta^{-1}])$
$\displaystyle\geq|\mathcal{D}^{-}_{\Delta,A}(x)|^{2}+(\mathcal{D}^{-}_{\Delta,A}(x)-|\mathcal{D}^{-}_{\Delta,A}(x)|^{2})-2Ce^{-\Delta^{A-1}}$
$\displaystyle\geq|\mathcal{D}^{-}_{\Delta,A}(x)|^{2}-2|\mathcal{D}^{-}_{\Delta,A}(x)|-2Ce^{-\Delta^{A-1}}$
$\displaystyle\geq|\mathcal{D}^{-}_{\Delta,A}(x)|^{2}-6Ce^{-\Delta^{A-1}}.$

If $x\in[0,\Delta^{-1}]$, then the fact that $G^{-}_{\Delta,A}\leq 1$ implies
instead.

$-e^{-\Delta^{A}}\leq\mathcal{D}^{-}_{\Delta,A}(x)\leq 1+2Ce^{-\Delta^{A-1}}.$

We deduce that:

$\displaystyle{\mathbf{1}}(x\in[0,\Delta^{-1}])$
$\displaystyle\geq|\mathcal{D}^{-}_{\Delta,A}(x)|^{2}+(\mathcal{D}^{-}_{\Delta,A}(x)-|\mathcal{D}^{-}_{\Delta,A}(x)|^{2})-2Ce^{-\Delta^{A-1}}$
$\displaystyle=|\mathcal{D}^{-}_{\Delta,A}(x)|^{2}+|\mathcal{D}^{-}_{\Delta,A}(x)|\Big{(}\text{sgn}\mathcal{D}^{-}_{\Delta,A}(x)-|\mathcal{D}^{-}_{\Delta,A}(x)|\Big{)}-2Ce^{-\Delta^{A-1}}$
$\displaystyle\geq|\mathcal{D}^{-}_{\Delta,A}(x)|^{2}-6Ce^{-\Delta^{A-1}}.$

This establishes the first inequality in (16) by redefining $C$.

For the second inequality in (16), item (4) of Lemma 1 and Equation (22) give

$|\mathcal{D}^{-}_{\Delta,A}(x)+{\rm
O}^{*}(e^{-\Delta^{A-1}})|^{2}\geq(1-Ce^{-\Delta^{A-1}})\cdot\mathbf{1}(x\in[\Delta^{-A/2},\Delta^{-1}-\Delta^{-A/2}]).$

Since the constant in ${\rm O}^{*}$ is $\leq 1$, the dominant term on the
left-hand side is $\mathcal{D}^{-}_{\Delta,A}(x)$, and we can absorb the
additive error in a multiplicative factor to get the second inequality in
(16). ∎

###  3.2. Proof of Propositions 7 and 8.

For these proofs, we need two preliminary steps. First, the constraints for
the random walk $(S_{k})_{k}$ (7) are re-expressed in terms of its increments.
Second, this allows to write the probabilities for the Dirichlet sums $S_{k}$
in terms of a probabilistic model.

Constraints and increments. First, the polynomial approximation of indicator
functions from Lemma 3 will be related to events involving the partial sums
$S_{k}$, $n_{0}<k\leq n_{\mathcal{L}}$. Fix $h\in[-1,1]$. Consider the
increments

$Y_{j}(h)=S_{j}(h)-S_{j-1}(h),\quad n_{0}<j\leq n_{\mathcal{L}}.$

To shorten the notation, we consider the set of times

$\mathcal{J}_{\ell}=\\{n_{0}+1,n_{0}+2,\dots,\ell-1,\ell\\}.$ (23)

with $n_{0}\leq\ell\leq n_{\mathcal{L}}$. We will partition the intervals of
values taken by $Y_{j}$, $j\in\mathcal{J}_{\ell}$ into sub-intervals of length
$\Delta_{j}^{-1}$ where

$\Delta_{j}=(j\wedge(n-j))^{4}.$

The exponent $4$ is chosen to ensure summability. In particular we will simply
use that for $y$ chosen large enough we have

$\sum_{j\in\mathcal{J}_{\ell}}\Delta_{j}^{-1}\leq\sum_{j\geq
n_{0}}\Delta_{j}^{-1}\leq 1.$ (24)

We consider events for the partial sums of the form

$\\{S_{j}(h)\in[L_{j},U_{j}],j\in\mathcal{J}_{\ell}\\},\quad h\in[-1,1].$

We would like to decompose the above in terms of events for the increments

$\\{Y_{j}(h)\in[u_{j},u_{j}+\Delta_{j}^{-1}],j\in\mathcal{J}_{\ell}\\},\quad
h\in[-1,1],$ (25)

for a given tuple $(u_{j},j\in\mathcal{J}_{\ell})$. Note that such events are
disjoint for two distinct tuples. On an event of the form (25), from (24) we
have

$\displaystyle\sum_{i\leq j}u_{i}\leq S_{j}(h)$ $\displaystyle\leq\sum_{i\leq
j}(u_{i}+\Delta_{i}^{-1})$ $\displaystyle\leq\sum_{i\leq j}u_{i}+1,\quad\text{
for all $j\in\mathcal{J}_{\ell}$,}$ (26)

This means that we have the following inclusions

$\displaystyle\\{S_{j}(h)\in[L_{j},U_{j}+1],j\in\mathcal{J}_{\ell}\\}$
$\displaystyle\supset\bigcup_{\mathbf{u}\in\mathcal{I}}\\{Y_{j}(h)\in[u_{j},u_{j}+\Delta_{j}^{-1}],j\in\mathcal{J}_{\ell}\\},$
(27) $\displaystyle\\{S_{j}(h)\in[L_{j}+1,U_{j}],j\in\mathcal{J}_{\ell}\\}$
$\displaystyle\subset\bigcup_{\mathbf{u}\in\mathcal{I}}\\{Y_{j}(h)\in[u_{j},u_{j}+\Delta_{j}^{-1}],j\in\mathcal{J}_{\ell}\\},$
(28)

where $\mathcal{I}$ is the set of tuples
$\mathbf{u}=(u_{j},j\in\mathcal{J}_{\ell})$,
$u_{j}\in\Delta_{j}^{-1}\mathbb{Z}$, such that $\sum_{i\leq
j}u_{i}\in[L_{i},U_{i}]$ for all $j\in\mathcal{J}_{\ell}$. The definition of
$\mathcal{I}$ imposes restrictions on the $u_{j}$’s. Indeed, we must have

$u_{j}\leq U_{j}-L_{j-1}\leq 10\Delta_{j}^{1/4}\qquad u_{j}\geq
L_{j}-U_{j-1}\geq-10\Delta_{j}^{1/4}.$

In all cases, we have the following bound which will be repeatedly used:

$|u_{j}|\leq 100\Delta_{j}^{1/4},\quad j\in\mathcal{J}_{\ell}.$ (29)

Probabilistic model for the increments. Additionally to the original random
walk (3) and its Gaussian counterpart (9), as an intermediate we now consider
another probabilistic model needed for the proofs of Propositions 7 and 8. For
$h\in[-1,1]$, let

$\mathcal{S}_{k}(h)=\sum_{n_{0}\leq\log\log p\leq k}{\rm
Re}\Big{(}e^{\mathrm{i}\theta_{p}}\,p^{-(1/2+\mathrm{i}h)}+\tfrac{1}{2}\,e^{2\mathrm{i}\theta_{p}}\,p^{-(1+2\mathrm{i}h)}\Big{)},\quad
k\leq n_{\mathcal{L}},$ (30)

where $(\theta_{p},p\text{ prime})$ are i.i.d. random variables distributed
uniformly on $[0,2\pi]$, and define the corresponding increments

$\mathcal{Y}_{k}(h)=\mathcal{S}_{k}(h)-\mathcal{S}_{k-1}(h),\quad k\leq
n_{\mathcal{L}}.$ (31)

It is easy to see that $\mathcal{S}_{k}$ and $\mathcal{Y}_{k}$ have mean $0$.
The variance of the increments $\mathcal{Y}_{k}$ coincides with (10) and by a
quantitative version of the Prime Number Theorem (see [ArgBouRad2020, Equation
(74)]) they satisfy

$\mathfrak{s}_{j}^{2}=\frac{1}{2}+{\rm O}(e^{-{c\sqrt{j}}}).$ (32)

for some universal $c>0$. These precise asymptotics are not used in the
comparison with the Gaussian model, i.e. in the proof of Proposition 7 below,
and they will be used only for convenience in the first and second moment for
the Gaussian model, Proposition 6. In fact to apply the Ballot theorem from
Proposition 14 we will only rely on
$\mathfrak{s}_{j}^{2}\in[\kappa,\kappa^{-1}]$ for some fixed $\kappa>0$.

###### Proof of Proposition 7.

We prove (12). The upper bound (13) is proved in a similar way, see
Proposition 8. We define the weighted expectation,

$\mathbb{E}_{\mathcal{Q}}[X]:=\mathbb{E}\Big{[}|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau)|^{2}\cdot
X(\tau)\Big{]}\cdot\mathbb{E}\Big{[}|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau)|^{2}\Big{]}^{-1},$

and the corresponding measure
$\mathbb{P}_{\mathcal{Q}}(A):=\mathbb{E}_{\mathcal{Q}}[\mathbf{1}(\tau\in
A)]$.

In what follows, we drop the dependence on $h$ as it plays no role. Equation
(27) directly implies (by taking $U_{k}$ instead of $U_{k}+\epsilon$):

$\mathbb{P}_{\mathcal{Q}}(S_{k}\in[L_{k},U_{k}],k\in\mathcal{J}_{\ell})\geq\sum_{\mathbf{u}\in\mathcal{I}}\mathbb{P}_{\mathcal{Q}}(Y_{k}-u_{k}\in[0,\Delta_{k}^{-1}],k\in\mathcal{J}_{\ell}),$
(33)

where $\mathcal{I}$ is now the set of tuples
$\mathbf{u}=(u_{j},j\in\mathcal{J}_{\ell})$,
$u_{j}\in\Delta_{j}^{-1}\mathbb{Z}$, such that $\sum_{i\leq
j}u_{i}\in[L_{i},U_{i}-1]$ for all $j\in\mathcal{J}_{\ell}$. By introducing
the indicator functions
$\prod_{k}{\mathbf{1}}(|Y_{k}-u_{k}|\leq\Delta_{k}^{6A})$, Equation (16) of
Lemma 3 can be applied with $A=10$ (say), thanks to the bound (29). This
yields

$\mathbb{P}_{\mathcal{Q}}(Y_{k}-u_{k}\in[0,\Delta_{k}^{-1}],k\in\mathcal{J}_{\ell})\geq\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k}\Big{(}|\mathcal{D}^{-}_{\Delta_{k},A}(Y_{k}-u_{k})|^{2}-Ce^{-\Delta_{k}^{A-1}}\Big{)}{\mathbf{1}}(|Y_{k}-u_{k}|\leq\Delta_{k}^{6A})\Big{]}.$
(34)

The tricky part is to get rid of the indicator function. For simplicity, let’s
write $\mathcal{D}_{k}$ for
$|\mathcal{D}^{-}_{\Delta_{k},A}(Y_{k}-u_{k})|^{2}-Ce^{-\Delta_{k}^{A-1}}$.
Since
${\mathbf{1}}(|Y_{k}-u_{k}|\leq\Delta_{k}^{6A})=1-{\mathbf{1}}(|Y_{k}-u_{k}|>\Delta_{k}^{6A})$,
we can rewrite the above as

$\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in\mathcal{J}_{\ell}}\mathcal{D}_{k}\Big{]}+\sum_{J\subseteq\mathcal{J}_{\ell},J\neq\emptyset}(-1)^{|J|}\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in\mathcal{J}_{\ell}}\mathcal{D}_{k}\prod_{j\in
J}{\mathbf{1}}(|Y_{j}-u_{j}|>\Delta_{j}^{6A})\Big{]}.$ (35)

We start with the first term, which will be dominant. Each $Y_{j}$ is a
Dirichlet polynomial of length at most $\exp(2e^{j})$. Therefore, from Lemma
3, for any subset $\mathcal{M}\subset\mathcal{J}_{\ell}$ the Dirichlet
polynomial
$\prod_{j\in\mathcal{M}}\mathcal{D}^{-}_{\Delta_{j},A}(Y_{j}-u_{j})$ is of
length at most

$\exp(2e^{n_{\mathcal{L}}}\Delta_{n_{\mathcal{L}}}^{100})\leq\exp\Big{(}\frac{1}{100}e^{n}\Big{)}$
(36)

for $y$ large enough. Therefore, Lemma 9 applies to compare with the random
model with increments $\mathcal{Y}_{k}$ given in (31):

$\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in\mathcal{M}}|\mathcal{D}^{-}_{\Delta_{k},A}(Y_{k}-u_{k})|^{2}\Big{]}=(1+{\rm
O}(T^{-99/100}))\prod_{k\in\mathcal{M}}\mathbb{E}\Big{[}|\mathcal{D}^{-}_{\Delta_{k},A}(\mathcal{Y}_{k}-u_{k})|^{2}\Big{]},$
(37)

where we have split the expectation $\mathbb{E}_{\mathcal{Q}}$ and used
$\mathbb{E}=\mathbb{E}_{\mathcal{Q}}$ for the probabilistic model, thanks to
the independence of the $\mathcal{Y}_{k}$’s. Moreover, for each $k$, we have

$\displaystyle\mathbb{E}\Big{[}|\mathcal{D}^{-}_{\Delta_{k},A}(\mathcal{Y}_{k}-u_{k})|^{2}\Big{]}$
$\displaystyle\geq\mathbb{E}\Big{[}|\mathcal{D}^{-}_{\Delta_{k},A}(\mathcal{Y}_{k}-u_{k})|^{2}{\mathbf{1}}(|\mathcal{Y}_{k}-u_{k}|\leq\Delta_{k}^{6A})\Big{]}$
(38)
$\displaystyle\geq(1-Ce^{-\Delta_{k}^{A-1}})\cdot\mathbb{P}(\mathcal{Y}_{k}-u_{k}\in[\Delta_{k}^{-A/2},\Delta_{k}^{-1}-\Delta_{k}^{-A/2}]),$

where the second inequality follows from (16), noting that the condition
$|\mathcal{Y}_{k}-u_{k}|\leq\Delta_{k}^{6A}$ is implied by
$\mathcal{Y}_{k}-u_{k}\in[\Delta_{k}^{-A/2},\Delta_{k}^{-1}-\Delta_{k}^{-A/2}]$,
and thus can be dropped. We now rewrite this probability in terms of Gaussian
increments. Lemma 13 in Appendix A gives

$\mathbb{P}(\mathcal{Y}_{k}-u_{k}\in[\Delta_{k}^{-A/2},\Delta_{k}^{-1}-\Delta_{k}^{-A/2}])=\mathbb{P}(\mathcal{N}_{k}-u_{k}\in[\Delta_{k}^{-A/2},\Delta_{k}^{-1}-\Delta_{k}^{-A/2}])+{\rm
O}(e^{-ce^{k/2}}).$ (39)

The overspill $\Delta_{j}^{-A/2}$ can be removed at no cost: from (29) and
$\mathfrak{s}_{k}\asymp 1$, uniformly in $x,y\in
u_{k}+[\Delta_{k}^{-A/2},\Delta_{k}^{-1}-\Delta_{k}^{-A/2}]$ the density
$f_{k}$ of $\mathcal{N}_{k}$ satisfies $f_{k}(x)\asymp f_{k}(y)$, so

$\mathbb{P}(\mathcal{N}_{k}-u_{k}\in[\Delta_{k}^{-A/2},\Delta_{k}^{-1}-\Delta_{k}^{1-A/2}])=(1+{\rm
O}(\Delta_{k}^{-A/2}))\cdot\mathbb{P}(\mathcal{N}_{k}-u_{k}\in[0,\Delta_{k}^{-1}]).$
(40)

Moreover,

$\mathbb{P}(\mathcal{N}_{k}-u_{k}\in[0,\Delta_{k}^{-1}])\gg\Delta_{k}^{-1}e^{-2u_{k}^{2}}\gg\Delta_{k}^{-1}e^{-100^{2}\Delta_{k}^{1/2}}.$
(41)

This is much larger than the additive error term ${\rm O}(e^{-ce^{k/2}})$ in
(39), which can therefore be replaced by a multiplicative error. Both
multiplicative errors together give for $k\leq n_{\mathcal{L}}$

$\mathbb{P}(\mathcal{Y}_{k}-u_{k}\in[\Delta_{k}^{-A/2},\Delta_{k}^{-1}-\Delta_{k}^{-A/2}])=\Big{(}1+{\rm
O}\Big{(}(k\wedge(n-k))^{-2A}\Big{)}\cdot\mathbb{P}(\mathcal{N}_{k}-u_{k}\in[0,\Delta_{k}^{-1}]).$
(42)

The product over $k\in\mathcal{J}_{\ell}$ of the error terms above is $(1+{\rm
O}(n_{0}^{-A}))$. Going back to Equations (37) and (38), we have established
that

$\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in\mathcal{M}}|\mathcal{D}_{\Delta_{k},A}^{-}(Y_{k}-u_{k})|^{2}\Big{]}\geq(1+{\rm
O}(n_{0}^{-A}))\prod_{k\in\mathcal{M}}\mathbb{P}(\mathcal{N}_{k}-u_{k}\in[0,\Delta_{k}^{-1}]).$
(43)

Remember that we aim at a similar estimate for
$\mathcal{D}_{k}=|\mathcal{D}^{-}_{\Delta_{k},A}(Y_{j}-u_{j})|^{2}-Ce^{-\Delta_{k}^{A-1}}$.
From (41), $\mathbb{P}(\mathcal{N}_{k}-u_{k}\in[0,\Delta_{k}^{-1}])\gg
e^{-\Delta_{k}}$ and (43) holds for arbitrary
$\mathcal{M}\subset\mathcal{J}_{\ell}$, so that by a simple expansion we have

$\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in\mathcal{J}_{\ell}}\mathcal{D}_{k}\Big{]}\geq(1+{\rm
O}(n_{0}^{-A}))\prod_{k\in\mathcal{J}_{\ell}}\mathbb{P}(\mathcal{N}_{k}-u_{k}\in[0,\Delta_{k}^{-1}]).$
(44)

We now bound the second term in (35). Let’s fix the non-empty subset
$J\subseteq\mathcal{J}_{\ell}$ in the sum. Since
${\mathbf{1}}(|X|>\lambda)\leq\frac{|X|^{2q}}{\lambda^{2q}},$ we have

$\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in\mathcal{J}_{\ell}}\mathcal{D}_{k}\prod_{j\in
J}{\mathbf{1}}(|Y_{j}-u_{j}|>\Delta_{j}^{6A})\Big{]}\leq\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in\mathcal{J}_{\ell}}\mathcal{D}_{k}\prod_{j\in
J}\frac{|Y_{j}-u_{j}|^{2q_{j}}}{\Delta_{j}^{12Aq_{j}}}\Big{]},$ (45)

where we pick $q_{j}=\lfloor\Delta_{j}^{6A}\rfloor$, $A=10$. As for the first
term, we need to handle the error $Ce^{-\Delta_{k}^{A-1}}$ in
$\mathcal{D}_{k}$. For this we abbreviate
$d_{k}(x)=D^{-}_{\Delta_{k},A}(x-u_{k})$,
$\varepsilon_{k}=Ce^{-\Delta_{k}^{A-1}}$, and expand

$\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in\mathcal{J}_{\ell}}\mathcal{D}_{k}\prod_{j\in
J}\frac{|Y_{j}-u_{j}|^{2q_{j}}}{\Delta_{j}^{12Aq_{j}}}\Big{]}\leq\sum_{B\subset\mathcal{J}_{\ell}}\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in
B}|d_{k}(Y_{k})|^{2}\prod_{k\in\mathcal{J}_{\ell}\setminus
B}\varepsilon_{k}\prod_{j\in
J}\frac{|Y_{j}-u_{j}|^{2q_{j}}}{\Delta_{j}^{12Aq_{j}}}\Big{]}.$ (46)

From Lemma 3, the Dirichlet polynomial $d_{j}$ is of length at most
$\exp(2e^{j}\Delta_{j}^{100})$. The choice of $q_{j}$ implies that the
Dirichlet polynomial $\prod_{k\in A}d_{k}\prod_{j\in J}(Y_{j}-u_{j})^{q_{j}}$
has length at most
$\exp(2e^{n_{\mathcal{L}}}\Delta_{n_{\mathcal{L}}}^{100})\leq\exp(\tfrac{1}{100}e^{n})$
as in (36). Therefore, we can use Lemma 9 again, and work with the random
model term by term. Again, the fact that $\mathcal{Q}$ is supported on
integers with primes $p$ with $\log p>e^{\ell}$ means that for the random
model the expectation with respect to $\mathbb{E}_{\mathcal{Q}}$ is equal to
the expectation with respect to $\mathbb{E}$. We start with the case $j\in
B\cap J$. We have

$\mathbb{E}\Big{[}|d_{j}(\mathcal{Y}_{j})|^{2}|\mathcal{Y}_{j}-u_{j}|^{2q_{j}}\Big{]}\ll\mathbb{E}\Big{[}|d_{j}(\mathcal{Y}_{j})|^{4}\Big{]}^{1/2}\cdot\mathbb{E}\Big{[}|\mathcal{Y}_{j}-u_{j}|^{4q_{j}}\Big{]}^{1/2}.$
(47)

The definition of $\mathcal{D}^{-}_{\Delta_{j},A}$ in Equations (19) and (20)
implies the following bound on all $2k$-moments, $k\in\mathbb{N}$,

$\mathbb{E}[|d_{j}(\mathcal{Y}_{j})|^{2k}]\leq\mathbb{E}\Big{[}\Big{(}\sum_{\ell\leq\Delta_{j}^{10A}}\frac{(2\pi)^{\ell}}{\ell!}2\Delta_{j}^{2A(\ell+1)}(|\mathcal{Y}_{j}|+100\Delta_{j}^{1/4})^{\ell}\Big{)}^{2k}\Big{]}\\\
\ll\Delta_{j}^{4kA}\,\mathbb{E}[\exp(4\pi
k\Delta_{j}^{2A}(|\mathcal{Y}_{j}|+100\Delta_{j}^{1/4}))]\ll_{k}e^{\Delta_{j}^{5A}},$
(48)

where the third inequality follows from Lemma 10. By Lemma 10 and the
inequality $\frac{x^{4q}}{q^{4q}}\leq\frac{(4q)!}{(\lambda
q)^{4q}}\cdot(e^{\lambda x}+e^{-\lambda x})$ with the choice
$q=q_{j}=\lfloor\Delta_{j}^{6A}\rfloor$, $\lambda=10$, we have for any $j\leq
n_{\mathcal{L}}$

$\mathbb{E}\Big{[}\frac{|\mathcal{Y}_{j}-u_{j}|^{4q_{j}}}{\Delta_{j}^{24Aq_{j}}}\Big{]}\ll
e^{-2\Delta_{j}^{6A}},$ (49)

by Stirling’s formula and the fact that $|u_{j}|\leq 100\Delta_{j}^{1/4}$.
From equations (45) (46) (47) (48) and (49) we have proved

$\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in\mathcal{J}_{\ell}}\mathcal{D}_{k}\prod_{j\in
J}{\mathbf{1}}(|Y_{j}-u_{j}|>\Delta_{j}^{6A})\Big{]}\ll\sum_{B\subset\mathcal{J}_{\ell}}\prod_{J}e^{-\Delta_{j}^{6A}}\prod_{B\backslash
J}\mathbb{E}[|d_{j}({\mathcal{Y}}_{j})|^{2}]\prod_{\mathcal{J}_{\ell}\backslash
B}\varepsilon_{j}\\\
=\prod_{j\in{\mathcal{J}}_{\ell}}\mathbb{E}[|d_{j}({\mathcal{Y}}_{j})|^{2}]\sum_{B\subset\mathcal{J}_{\ell}}\prod_{J}\frac{e^{-\Delta_{j}^{6A}}}{\mathbb{E}[|d_{j}({\mathcal{Y}}_{j})|^{2}]}\prod_{\mathcal{J}_{\ell}\backslash(B\cup
J)}\frac{\varepsilon_{j}}{\mathbb{E}[|d_{j}({\mathcal{Y}}_{j})|^{2}]}\prod_{J\backslash
B}\varepsilon_{j}.$

Moreover, from (38) with the estimates (41), (42), we have
$\mathbb{E}[|d_{j}({\mathcal{Y}}_{j})|^{2}]\gg\Delta_{j}^{-1}e^{-100^{2}\Delta_{j}^{1/2}}$.
We have obtained

$\left|\sum_{J\subseteq\mathcal{J}_{\ell},J\neq\emptyset}(-1)^{|J|}\mathbb{E}_{\mathcal{Q}}\Big{[}\prod_{k\in\mathcal{J}_{\ell}}\mathcal{D}_{k}\prod_{j\in
J}{\mathbf{1}}(|Y_{j}-u_{j}|>\Delta_{j}^{6A})\Big{]}\right|\\\
\ll\prod_{j\in{\mathcal{J}}_{\ell}}\mathbb{E}[|d_{j}({\mathcal{Y}}_{j})|^{2}]\sum_{J\subset\mathcal{J}_{\ell},J\neq\emptyset}\sum_{B\subset\mathcal{J}_{\ell}}\prod_{J}e^{-\tfrac{1}{2}\Delta_{j}^{6A}}\prod_{\mathcal{J}_{\ell}\backslash
B}\varepsilon_{j}^{1/2}\\\
=\prod_{j\in{\mathcal{J}}_{\ell}}\mathbb{E}[|d_{j}({\mathcal{Y}}_{j})|^{2}]\big{(}\prod_{\mathcal{J}_{\ell}}(1+e^{-\tfrac{1}{2}\Delta_{j}^{6A}})-1\big{)}\prod_{\mathcal{J}_{\ell}}(1+\sqrt{\varepsilon_{j}})\ll
e^{-n_{0}^{100}}\prod_{j\in{\mathcal{J}}_{\ell}}\mathbb{E}[|d_{j}({\mathcal{Y}}_{j})|^{2}].$

The above product is
$\ll\prod_{j\in\mathcal{J}_{\ell}}\mathbb{P}(\mathcal{N}_{j}-u_{j}\in[0,\Delta_{j}^{-1}])$
as easily proved by combining (16) and (49). (A similar bound in the more
general case of joint increments is detailed in (54).)

Equations (34),(35) and (44) with the above finally yield

$\mathbb{P}_{\mathcal{Q}}(Y_{j}-u_{j}\in[0,\Delta_{j}^{-1}],j\in\mathcal{J}_{\ell})\geq(1+{\rm
O}(n_{0}^{-10}))\prod_{j\in\mathcal{J}_{\ell}}\mathbb{P}(\mathcal{N}_{j}-u_{j}\in[0,\Delta_{j}^{-1}]).$

The claim (12) follows by summing over $\mathbf{u}\in\mathcal{I}$ as in
Equation (33), and by applying the inclusion (28) for the Gaussian random walk
with increments $\mathcal{N}_{j}$. ∎

For the proof Proposition 8 below, we will also consider the partial sums at
$h$ and $h^{\prime}$ jointly, i.e., $S_{k}(h)$ and $S_{k}(h^{\prime})$,
$n_{0}<k\leq n_{\mathcal{L}}$, as well as the joint increments
$\mathcal{Y}_{j}(h)$ and $\mathcal{Y}_{j}(h^{\prime})$. These increments have
covariance and correlations identical to those of $\mathcal{N}_{j}$ and
$\mathcal{N}_{j}^{\prime}$, i.e., they are given by (11), which satisfies the
asymptotics

$\displaystyle\rho_{j}$ $\displaystyle=\begin{cases}\mathfrak{s}_{j}^{2}+{\rm
O}((e^{j}|h-h^{\prime}|)^{2})\ &\text{if $j\leq\log|h-h^{\prime}|^{-1}$},\\\
{\rm O}((e^{j}|h-h^{\prime}|)^{-1})\ &\text{if
$j\geq\log|h-h^{\prime}|^{-1}$},\end{cases}$ (50)

as is easily proved using the Prime Number Theorem as in [ArgBelHar2017, Lemma
2.1]. We also define $\varepsilon_{j}=\varepsilon_{j}(h,h^{\prime})$ by

$\displaystyle\rho_{j}=\begin{cases}\mathfrak{s}_{j}^{2}-\varepsilon_{j}&\text{
if }j\leq\log|h-h^{\prime}|^{-1},\\\ \varepsilon_{j}&\text{ if
}j>\log|h-h^{\prime}|^{-1}.\end{cases}$ (51)

The precise asymptotics of the covariances in 51 will not play a role in the
proof of Proposition 8 below. However, it will be crucial in the proof of
Proposition 6.

###### Proof of Proposition 8.

We write $(S_{k},S_{k}^{\prime})$ for $(S_{k}(h),S_{k}(h^{\prime}))$ for
conciseness, and similarly for the increments. The event on the left-hand side
of (13) is decomposed using the increments as in Equation (28). Then, Equation
(15) can be used to bound the indicator functions for both points. We take
$A=10$ (say). This gives that the left-hand side of (14) is

$\leq(1+{\rm
O}(e^{-n_{0}^{10}}))\sum_{\mathbf{u},\mathbf{u}^{\prime}\in\mathcal{I}}\mathbb{E}\Big{[}\prod_{j\in\mathcal{J}_{\mathcal{L}}}|\mathcal{D}^{+}_{\Delta_{j},A}(Y_{j}-u_{j})\mathcal{D}^{+}_{\Delta_{j},A}(Y^{\prime}_{j}-u^{\prime}_{j})|^{2}\Big{]},$
(52)

where we write $\mathcal{J}_{\ell}$ as in (23). We proceed as in Equation
(37). From Lemma 3, the Dirichlet polynomial
$\prod_{j}\mathcal{D}^{+}_{\Delta_{j},A}(Y_{j}-u_{j})$ is of length at most
$\exp(2e^{n_{\mathcal{L}}}\Delta_{n_{\mathcal{L}}}^{100})$. So the product of
the polynomials for $h$ and $h^{\prime}$ has length smaller than
$\exp(4e^{n_{\mathcal{L}}}\Delta_{n_{\mathcal{L}}}^{100})\leq T^{1/100}$, as
in (36). Lemma 9 then implies

$\displaystyle\mathbb{E}\Big{[}\prod_{j\in\mathcal{J}_{\mathcal{L}}}|\mathcal{D}^{+}_{\Delta_{j},A}(Y_{j}-u_{j})\mathcal{D}^{+}_{\Delta_{j},A}(Y^{\prime}_{j}-u^{\prime}_{j})|^{2}\Big{]}$
(53) $\displaystyle=(1+{\rm
O}(T^{-99/100}))\prod_{j\in\mathcal{J}_{\mathcal{L}}}\mathbb{E}\Big{[}|\mathcal{D}^{+}_{\Delta_{j},A}(\mathcal{Y}_{j}-u_{j})\mathcal{D}^{+}_{\Delta_{j},A}(\mathcal{Y}^{\prime}_{j}-u^{\prime}_{j})|^{2}\Big{]}.$

We estimate the expectation for each $j$. Write for short
$\mathcal{D}^{+}_{\Delta_{j},A}(\mathcal{Y}_{j}-u_{j})=\mathcal{D}_{j}$ and
similarly for $\mathcal{D}_{j}^{\prime}$. We would like to introduce the
indicator functions ${\mathbf{1}}(|\mathcal{Y}_{j}-u_{j}|\leq\Delta_{j}^{6A})$
and
${\mathbf{1}}(|\mathcal{Y}^{\prime}_{j}-u^{\prime}_{j}|\leq\Delta_{j}^{6A})$.
For this, note first that

$\mathbb{E}\Big{[}|\mathcal{D}_{j}\mathcal{D}^{\prime}_{j}|^{2}{\mathbf{1}}(|\mathcal{Y}_{j}-u_{j}|>\Delta_{j}^{6A})\Big{]}\leq\mathbb{E}\Big{[}|\mathcal{D}_{j}|^{6}\Big{]}^{1/3}\cdot\mathbb{E}\Big{[}|\mathcal{D}^{\prime}_{j}|^{6}\Big{]}^{1/3}\cdot\mathbb{P}\Big{(}|\mathcal{Y}_{j}-u_{j}|>\Delta_{j}^{6A}\Big{)}^{1/3}\ll
e^{-\Delta_{j}^{6A}},$

by Equation (48) (with $a=3$) and Markov’s inequality using (49). This
observation implies that

$\displaystyle\mathbb{E}\Big{[}|\mathcal{D}_{j}\mathcal{D}^{\prime}_{j}|^{2}\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}|\mathcal{D}_{j}\mathcal{D}^{\prime}_{j}|^{2}{\mathbf{1}}(|\mathcal{Y}_{j}-u_{j}|\leq\Delta_{j}^{6A},|\mathcal{Y}^{\prime}_{j}-u^{\prime}_{j}|\leq\Delta_{j}^{6A})\Big{]}+{\rm
O}\Big{(}e^{-\Delta_{j}^{6A}}\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}(\mathcal{Y}_{j}-u_{j},\mathcal{Y}_{j}^{\prime}-u_{j}^{\prime})\in[-\Delta_{j}^{-A/2},\Delta_{j}^{-1}+\Delta_{j}^{-A/2}]^{2}\Big{)}+{\rm
O}\Big{(}e^{-\Delta_{j}^{A-1}}\Big{)},$ (54)

by Equation (15) applied to both $\mathcal{D}_{j}$ and
$\mathcal{D}_{j}^{\prime}$. The Berry-Esseen approximation of Lemma 13 can now
be applied:

$\displaystyle\mathbb{E}\Big{[}|\mathcal{D}_{j}\mathcal{D}^{\prime}_{j}|^{2}\Big{]}$
$\displaystyle\leq(1+\Delta_{j}^{-A/2})\mathbb{P}\Big{(}(\mathcal{N}_{j}-u_{j},\mathcal{N}_{j}^{\prime}-u_{j}^{\prime})\in[-\Delta_{j}^{-A/2},\Delta_{j}^{-1}+\Delta_{j}^{-A/2}]^{2}\Big{)}+{\rm
O}\Big{(}e^{-\Delta_{j}^{A-1}}\Big{)}.$

The overspill $\Delta_{j}^{-A/2}$ can be also removed as in (40). We conclude
that the above is

$\displaystyle=(1+{\rm
O}(\Delta_{j}^{-A/2}))\mathbb{P}\Big{(}(\mathcal{N}_{j}-u_{j},\mathcal{N}_{j}^{\prime}-u_{j}^{\prime})\in[0,\Delta_{j}^{-1}]^{2}\Big{)}+{\rm
O}\Big{(}e^{-\Delta_{j}^{A-1}}\Big{)}$ $\displaystyle=(1+{\rm
O}(\Delta_{j}^{-A/2}))\mathbb{P}\Big{(}(\mathcal{N}_{j}-u_{j},\mathcal{N}_{j}^{\prime}-u_{j}^{\prime})\in[0,\Delta_{j}^{-1}]^{2}\Big{)},$

since
$\mathbb{P}((\mathcal{N}_{j}-u_{j},\mathcal{N}_{j}^{\prime}-u_{j}^{\prime})\in[0,\Delta_{j}^{-1}]^{2})\gg
e^{-cu_{j}^{2}-c{u_{j}^{\prime}}^{2}}\gg e^{-2c\Delta_{j}^{1/2}}$ by the bound
on $u_{j}$ and $u_{j}^{\prime}$. It remains to use the above bound in (53) and
then (52). The claim then follows from Equation (28) for the Gaussian random
walks. ∎

##  4\. Proof of Proposition 3

We first need preliminary bounds on the size of $\zeta$ and Dirichlet sums. We
will use the notation

$P_{n_{0}}(h)=\sum_{\log\log p\leq n_{0}}{\rm
Re}\Big{(}p^{-(1/2+\mathrm{i}\tau+\mathrm{i}h)}+\frac{1}{2}\cdot
p^{-2(1/2+\mathrm{i}\tau+\mathrm{i}h)}\Big{)}.$ (55)

######  Lemma 4.

We have, for $1000<y<n/10$,

$\displaystyle\mathbb{P}\Big{(}\forall m\geq 1:\max_{|u|\leq
2^{m}}|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}u)|\leq
2^{2m}e^{2n_{\mathcal{L}}}\Big{)}=1-{\rm O}(e^{-n}),$ (56)
$\displaystyle\mathbb{P}\Big{(}\forall m\geq 1:\max_{|u|\leq
2^{m}}|S_{n_{\mathcal{L}}}(u)|\leq
2^{m/100}e^{n_{\mathcal{L}}/100}\Big{)}=1-{\rm O}(e^{-n}),$ (57)
$\displaystyle\mathbb{P}\Big{(}\forall m\geq 1:\max_{|u|\leq
2^{m}}|P_{n_{0}}(u)|\leq 2^{m/100}\cdot 10y\Big{)}=1-{\rm O}(e^{-y}).$ (58)

###### Proof.

By a union bound, the probability of the complement of the first event is

$\sum_{m\geq 1}2^{-4m}e^{-4n_{\mathcal{L}}}\mathbb{E}\Big{[}\max_{|u|\leq
2^{m}}|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}u)|^{2}\Big{]}\ll\sum_{m\geq
1}2^{-4m}e^{-4n_{\mathcal{L}}}\cdot 2^{m}e^{2n}\ll e^{-n},$

as claimed, where the first inequality above relies on the same subharmonicity
argument as [ArgBouRad2020, Lemma 28]. For the second claim, we similarly have
that the probability of the complement is bounded by,

$\sum_{m\geq 1}2^{-4m}e^{-4n_{\mathcal{L}}}\cdot\mathbb{E}\Big{[}\max_{|u|\leq
2^{m}}|S_{n_{\mathcal{L}}}(u)|^{400}\Big{]}\ll\sum_{m\geq
1}2^{-4m}e^{-4n_{\mathcal{L}}}\cdot 2^{m}e^{n}n^{200}\ll e^{-2n},$

where we used (85) in Lemma 11. Finally, the last bound is proved in exactly
the same way, using that, for $v=\lfloor 100y\rfloor$,

$\sum_{m\geq 1}2^{-2vm/100}(10y)^{-2v}\cdot\mathbb{E}\Big{[}\max_{|u|\leq
2^{m}}|P_{n_{0}}(u)|^{2v}\Big{]}\\\ \ll\sum_{m\geq
1}2^{-2vm/100}(10y)^{-2v}\cdot 2^{m}e^{n_{0}}\cdot
v^{1/2}\frac{(2v)!}{2^{v}v!}\cdot(Cy)^{v}\ll e^{-90y},$

where the moments calculation is now based on (86) in Lemma 11. ∎

The main analytic input is the next lemma.

######  Lemma 5.

Let $100\leq T\leq t\leq 2T$ and $|h|\leq 1$. Let $f$ be a smooth function
with $\widehat{f}$ compactly supported in $[-\frac{1}{2\pi},\frac{1}{2\pi}]$
and such that $\widehat{f}(0)=1$. Then,

$\displaystyle\log
X\int_{\mathbb{R}}\zeta(\tfrac{1}{2}+\mathrm{i}t+\mathrm{i}h+\mathrm{i}x)\prod_{p\leq
X}\Big{(}1-\frac{1}{p^{1/2+\mathrm{i}t+\mathrm{i}h+\mathrm{i}x}}\Big{)}f(x\log
X){\rm d}x=1+{\rm O}(T^{-1}).$ (59)

###### Proof.

For $z\in\mathbb{R}$, we have
$f(z)=\int_{\mathbb{R}}\widehat{f}(u)e^{\mathrm{i}2\pi zu}{\rm d}u$. As
$\widehat{f}$ is compactly supported, by Paley-Wiener this defines for
$z\in\mathbb{C}$ an entire function of rapid (faster than polynomial) decay as
$|{\rm Re}\ z|\rightarrow\infty$ inside any fixed strip. We can therefore
shift the contour of integration in (59) and see that it is equal to

$\log
X\int_{2-i\infty}^{2+i\infty}\zeta(s+\mathrm{i}t+\mathrm{i}h)\prod_{p\leq
X}\Big{(}1-\frac{1}{p^{s+\mathrm{i}t+\mathrm{i}h}}\Big{)}f\Big{(}\frac{s-\tfrac{1}{2}}{\mathrm{i}}\cdot\log
X\Big{)}\frac{{\rm d}s}{\mathrm{i}}+{\rm O}(T^{-1})$

where $T^{-1}$ is the contribution of the pole at $s=1-it-ih$ of $\zeta$. On
the line ${\rm Re}\ s=2$ we can write pointwise

$\zeta(s+\mathrm{i}t+\mathrm{i}h)\prod_{p\leq
X}\Big{(}1-\frac{1}{p^{s+\mathrm{i}t+\mathrm{i}h}}\Big{)}=1+\sum_{\begin{subarray}{c}n>1\\\
p|n\Rightarrow p>X\end{subarray}}\frac{1}{n^{s+\mathrm{i}t+\mathrm{i}h}}.$

After nterchanging the sum and integral, the task reduces to estimating

$\frac{\log
X}{\mathrm{i}}\int_{2-\mathrm{i}\infty}^{2+\mathrm{i}\infty}n^{-s-\mathrm{i}t-\mathrm{i}h}f\Big{(}\frac{s-\tfrac{1}{2}}{\mathrm{i}}\cdot\log
X\Big{)}{\rm d}s.$

Shifting the contour back to the line ${\rm Re}\ s=\tfrac{1}{2}$, this is
equal to

$\log
X\int_{\mathbb{R}}\frac{1}{n^{1/2+\mathrm{i}t+\mathrm{i}h+\mathrm{i}x}}\cdot
f(x\log X){\rm d}x=\widehat{f}\Big{(}-\frac{\log n}{2\pi\log
X}\Big{)}\cdot\frac{1}{n^{1/2+\mathrm{i}t+\mathrm{i}h}}.$

If $n=1$, then this is equal to $\widehat{f}(0)=1$. On the other hand if
$n\neq 1$ then $n>X$ and then by assumption $\widehat{f}(-\log n/(2\pi\log
X))=0$. This gives the claim. ∎

We are now ready to prove Proposition 3.

###### Proof of Proposition 3.

From Lemma 2, there exists a smooth function $f\geq 0$ such that
$\widehat{f}(0)=1$, $\widehat{f}$ is compactly supported in
$[-\frac{1}{2\pi},\frac{1}{2\pi}]$ and

$|f(x)|\ll e^{-2\sqrt{|x|}}.$

Applying Lemma 5 with this choice for $f$ and $X=\exp(e^{n_{\mathcal{L}}})$,
we find by the mean-value theorem that for every $\tau$ and $h\in G_{0}$,
there exists a $k\geq 0$ and
$\frac{1}{4}\cdot(2^{k}-1)\leq|u|\leq\tfrac{1}{4}\cdot(2^{k+1}-1)$ such that

$\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h+\mathrm{i}u)|-S_{n_{\mathcal{L}}}(h+u)-P_{n_{0}}(h+u)-\sqrt{|u|e^{n_{\mathcal{L}}}}\geq-C$
(60)

with $C>0$ an absolute constant and where we remind the definition (55).

By (56) and (57) in Lemma 4, the probability (in $\tau$) that there exists an
$|h|\leq 1$ and $k\geq 1$ for which (60) holds is $\ll e^{-n}$. Moreover, by
(58) in Lemma 4, we also know that

$\max_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
1/4\end{subarray}}|P_{n_{0}}(h+u)|\leq 20y$

for all $\tau$ outside of a set of probability $\ll e^{-y}$. Therefore, for
all $\tau$ outside of a set of probability $\ll e^{-y}$ we find that for all
$h\in G_{0}$ there exists a $|u|\leq 1/4$ such that

$\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h+\mathrm{i}u)|-S_{n_{\mathcal{L}}}(h+u)-\sqrt{|u|e^{n_{\mathcal{L}}}}\geq-C-20y.$

Since $G_{0}\subset[-\tfrac{1}{2},\tfrac{1}{2}]$, it follows that for all
$\tau$ outside of a set of measure $\ll e^{-y}$, for all $h\in G_{0}$, there
exists an $|u|\leq 1/4$ such that

$\displaystyle\max_{|v|\leq
1}\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}v)|$
$\displaystyle>S_{n_{\mathcal{L}}}(h+u)+\sqrt{|u|e^{n_{\mathcal{L}}}}-2C-20y$
$\displaystyle\geq\min_{|u|\leq
1}(S_{n_{\mathcal{L}}}(h+u)+\sqrt{|u|e^{n_{\mathcal{L}}}})-2C-20y.$

We now take an $h\in G_{0}$ that maximizes the right-hand side, and the claim
follows. ∎

##  5\. Proof of Proposition 4

The following lemma will be important.

######  Lemma 6.

Let $n_{0}\leq\ell\leq n_{\mathcal{L}}$. Let $v\geq 1$ and $0\leq k\leq n$ be
given. Let $\mathcal{Q}$ be a Dirichlet polynomial supported on primes $p$ or
their squares $p^{2}$, such that $e^{\ell}\leq\log p\leq e^{n_{\mathcal{L}}}$
and of length $\leq\exp(\tfrac{1}{200v}e^{n})$:

$\mathcal{Q}(s)=\sum_{e^{\ell}\leq\log p\leq
e^{n_{\mathcal{L}}}}\left(\frac{a(p)}{p^{s}}+\frac{b(p)}{p^{2s}}\right),$ (61)

where we also assume $|b(p)|\leq 1$. Then

$\displaystyle\mathbb{E}\Big{[}$
$\displaystyle\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h+\mathrm{i}u)-\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|^{2v}\cdot\mathbf{1}_{h\in
G_{\ell}}\Big{]}$ (62) $\displaystyle\ll
e^{n_{\mathcal{L}}-n_{0}-\ell+10((\ell-
n_{0})\wedge(n_{\mathcal{L}}-\ell))^{3/4}+20y}$ $\displaystyle\ \ \ \ \times
100^{v}v!\cdot\Big{(}\Big{(}e^{-2k+4}\sum_{e^{\ell}\leq\log p\leq
e^{k}}\frac{|a(p)|^{2}\log^{2}p}{p}\Big{)}^{v}+\Big{(}16\sum_{e^{k}\leq\log
p}\frac{|a(p)|^{2}}{p}\Big{)}^{v}\cdot e^{n_{\mathcal{L}}-k}+1\Big{)}.$ (63)

###### Proof.

To simplify the exposition we first assume that $b(p)=0$ for all $p$. Since
$G_{\mathcal{\ell}}\subset
G_{0}=e^{-(n_{\mathcal{L}}-n_{0})}\mathbb{Z}\cap[-1,1]$ we have,

$\displaystyle\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}$
$\displaystyle|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h+\mathrm{i}u)-\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|^{2v}\cdot\mathbf{1}_{h\in
G_{\ell}}$ $\displaystyle\leq\sum_{h\in G_{0}}\sup_{|u|\leq
e^{-k+1}}|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h+\mathrm{i}u)-\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|^{2v}\cdot\mathbf{1}_{h\in
G_{\ell}}.$

Taking the expectation we find that (62) is

$\displaystyle\leq
e^{n_{\mathcal{L}}-n_{0}}\cdot\mathbb{E}\Big{[}\sup_{|u|\leq
e^{-k+1}}|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}u)-\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau)|^{2v}\cdot\mathbf{1}_{0\in
G_{\ell}}\Big{]}.$ (64)

We now split the Dirichlet polynomial
$\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}u)-\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau)$
into two parts. One part $\mathcal{Q}_{\leq
k}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}u)-\mathcal{Q}_{\leq
k}(\tfrac{1}{2}+\mathrm{i}\tau)$ composed of primes $p$ with $\log p\leq
e^{k}$ and another part supported on primes $p$ with $\log p>e^{k}$, denoted
$\mathcal{Q}_{>k}(\tfrac{1}{2}+i\tau+iu)-\mathcal{Q}_{>k}(\tfrac{1}{2}+i\tau)$.
For the first part, for $|u|\leq e^{-k+1}$,

$\displaystyle|\mathcal{Q}_{\leq
k}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}u)-\mathcal{Q}_{\leq
k}(\tfrac{1}{2}+\mathrm{i}\tau)|^{2v}$
$\displaystyle\leq\Big{(}\int_{0}^{e^{-k+1}}|\mathcal{Q}_{\leq
k}^{\prime}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}x)|{\rm d}x\Big{)}^{2v}$
$\displaystyle\leq e^{-(2v-1)(k-1)}\int_{0}^{e^{-k+1}}|\mathcal{Q}_{\leq
k}^{\prime}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}x)|^{2v}{\rm d}x.$

Then

$\mathbb{E}\Big{[}|\mathcal{Q}_{\leq
k}^{\prime}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}x)|^{2v}\cdot\mathbf{1}_{0\in
G_{\ell}}\Big{]}\ll v!\cdot\Big{(}\sum_{\log p\leq
e^{k}}\frac{|a(p)|^{2}\log^{2}p}{p}\Big{)}^{v}\cdot e^{-\ell+20y+10((\ell-
n_{0})\wedge(n_{\mathcal{L}}-\ell))^{3/4})},$

using Proposition 7, Lemma 12 and the Ballot theorem from Proposition 14.
Therefore

$\displaystyle e^{n_{\mathcal{L}}-n_{0}}\cdot\mathbb{E}\Big{[}\sup_{|u|\leq
e^{-k+1}}|$ $\displaystyle\mathcal{Q}_{\leq
k}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}u)-\mathcal{Q}_{\leq
k}(\tfrac{1}{2}+\mathrm{i}\tau)|^{2v}\cdot\mathbf{1}_{0\in G_{\ell}}\Big{]}$
$\displaystyle\ll e^{n_{\mathcal{L}}-n_{0}-\ell+20y+10((\ell-
n_{0})\wedge(n_{\mathcal{L}}-\ell))^{3/4}}\cdot
v!\cdot\Big{(}e^{-2k+2}\sum_{\log p\leq
e^{k}}\frac{|a(p)|^{2}\log^{2}p}{p}\Big{)}^{v}.$

For the second part, we bound the contribution of $\mathcal{Q}_{\geq
k}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}u)-\mathcal{Q}_{\geq
k}(\tfrac{1}{2}+\mathrm{i}\tau)$ simply by the triangle inequality and the
discretization Lemma (14) applied to $D=\mathcal{Q}_{\geq k}^{v}$, followed by
Proposition 7. This gives

$\displaystyle e^{n_{\mathcal{L}}-n_{0}}\cdot\mathbb{E}\Big{[}\sup_{|u|\leq
e^{-k+1}}|$ $\displaystyle\mathcal{Q}_{\geq
k}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}u)-\mathcal{Q}_{\geq
k}(\tfrac{1}{2}+\mathrm{i}\tau)|^{2v}\cdot\mathbf{1}_{0\in G_{\ell}}\Big{]}$
$\displaystyle\ll e^{n_{\mathcal{L}}-n_{0}-\ell+20y+10((\ell-
n_{0})\wedge(n_{\mathcal{L}}-\ell))^{3/4}}\cdot e^{n_{\mathcal{L}}-k}\cdot
2^{2v}\,v!\cdot\Big{(}4\sum_{\log p>e^{k}}\frac{|a(p)|^{2}}{p}\Big{)}^{v}.$

Combining everything we obtain the claim when $b(p)=0$. When $b$ is non-
trivial, the only difference is that we cannot dirrectly apply Lemma 12 to
bound the moments of $\mathcal{Q}$: instead, we just use $|X+Y|^{2v}\leq
2^{2v}(|X|^{2v}+|Y|^{2v})$ for $X=\sum\frac{a(p)}{p^{s}}$,
$Y=\sum\frac{b(p)}{p^{2s}}$, and apply Lemma 12 separately to each term. The
assumption $|b(p)|\leq 1$ allows to absorb the contribution of $|Y|^{2v}$ into
the $+1$ in (63). ∎

We are now ready to prove Proposition 4.

###### Proof of Proposition 4.

If there exists an $h\in G_{\mathcal{L}}$ and $|u|\leq 1$ such that

$|S_{n_{\mathcal{L}}}(h+u)-S_{n_{\mathcal{L}}}(h)|>20y+\sqrt{|u|e^{n_{\mathcal{L}}}},$
(65)

then there exists a $0\leq k<n_{\mathcal{L}}^{\prime}:=n_{\mathcal{L}}-\lfloor
2\log y\rfloor$ such that

$\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}|S_{n_{\mathcal{L}}}(h+u)-S_{n_{\mathcal{L}}}(h)|\cdot\mathbf{1}_{h\in
G_{\mathcal{L}}}\geq e^{(n_{\mathcal{L}}-k)/2}.$

Notice that we can stop at
$k=n_{\mathcal{L}}^{\prime}:=n_{\mathcal{L}}-\lfloor 2\log y\rfloor$ thanks to
the term $20y$. Therefore it suffices to bound

$\displaystyle\sum_{0\leq
k<n^{\prime}_{\mathcal{L}}}\mathbb{P}\Big{(}\sup_{\begin{subarray}{c}|h|\leq
1\\\ e^{-k}\leq|u|\leq
e^{-k+1}\end{subarray}}|S_{n_{\mathcal{L}}}(h+u)-S_{n_{\mathcal{L}}}(h)|\cdot\mathbf{1}_{h\in
G_{\mathcal{L}}}\geq e^{(n_{\mathcal{L}}-k)/2}\Big{)}.$ (66)

Suppose now that $|u|\leq e^{-k+1}$ for some $0\leq
k<n^{\prime}_{\mathcal{L}}$. Notice that

$\displaystyle|S_{n_{\mathcal{L}}}(h+u)-S_{n_{\mathcal{L}}}(h)|\mathbf{1}_{h\in
G_{\mathcal{L}}}\leq$ $\displaystyle\sum_{n_{0}\leq
j<k}|(S_{j+1}-S_{j})(h+u)-(S_{j+1}-S_{j})(h)|\mathbf{1}_{h\in G_{j}}$
$\displaystyle\ \ \ \
+|(S_{n_{\mathcal{L}}}-S_{k})(h+u)-(S_{n_{\mathcal{L}}}-S_{k})(h)|\mathbf{1}_{h\in
G_{k}},$

because $S_{n_{0}}=0$. Therefore, by the union bound, for each $0\leq k\leq
n_{\mathcal{L}}^{\prime}$,

$\displaystyle\mathbb{P}$
$\displaystyle\Big{(}\sup_{\begin{subarray}{c}|h|\leq 1\\\ e^{-k}\leq|u|\leq
e^{-k+1}\end{subarray}}|S_{n_{\mathcal{L}}}(h+u)-S_{n_{\mathcal{L}}}(h)|\cdot\mathbf{1}_{h\in
G_{\mathcal{L}}}\geq e^{(n_{\mathcal{L}}-k)/2}\Big{)}$
$\displaystyle\leq\sum_{0\leq
j<k}\mathbb{P}\Big{(}\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}|(S_{j+1}-S_{j})(h+u)-(S_{j+1}-S_{j})(h)|\mathbf{1}_{h\in
G_{j}}\geq\frac{e^{(n_{\mathcal{L}}-k)/2}}{4(k-j)^{2}}\Big{)}$ $\displaystyle\
\ \ \ \ \ \ \ \ +\mathbb{P}\Big{(}\sup_{\begin{subarray}{c}|h|\leq 1\\\
|u|\leq
e^{-k+1}\end{subarray}}|(S_{n_{\mathcal{L}}}-S_{k})(h+u)-(S_{n_{\mathcal{L}}}-S_{k})(h)|\mathbf{1}_{h\in
G_{k}}\geq\frac{e^{(n_{\mathcal{L}}-k)/2}}{4}\Big{)}.$ (67)

We now estimate each of the above probabilities using Chernoff’s bound.
According to Lemma 6 for $0\leq j<k$, for $v\geq 1$, we have

$\displaystyle\mathbb{P}$
$\displaystyle\Big{(}\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}|(S_{j+1}-S_{j})(h+u)-(S_{j+1}-S_{j})(h)|\mathbf{1}_{h\in
G_{j}}\geq\frac{e^{(n_{\mathcal{L}}-k)/2}}{4(k-j)^{2}}\Big{)}$
$\displaystyle\ll(4(k-j))^{4v}\cdot\mathbb{E}\Big{[}\sup_{\begin{subarray}{c}|h|\leq
1\\\ |u|\leq
e^{-k+1}\end{subarray}}\frac{|(S_{j+1}-S_{j})(h+u)-(S_{j+1}-S_{j})(h)|^{2v}}{e^{v(n_{\mathcal{L}}-k)}}\cdot\mathbf{1}_{h\in
G_{j}}\Big{]}$ $\displaystyle\ll(k-j)^{4v}\cdot
e^{n_{\mathcal{L}}-n_{0}-j+20y+10((j-n_{0})\wedge(n_{\mathcal{L}}-j)^{3/4})}\cdot
e^{-v(n_{\mathcal{L}}-k)}\cdot v!\cdot e^{\tilde{C}v}\cdot e^{2v(j-k)}.$ (68)

The above $e^{2v(j-k)}$ factor is due to the contribution of
$\big{(}e^{-2k+4}\sum_{e^{j}\leq\log p\leq
e^{j+1}}\frac{|a(p)|^{2}\log^{2}p}{p}\big{)}^{v}$ in Lemma 6. We choose
$v=\lfloor e^{n_{\mathcal{L}}-j-C}\rfloor$ for fixed $C>0$. Then the Dirichlet
sum $S_{j+1}^{v}$ has length $\exp(e^{j}\cdot
e^{n_{\mathcal{L}}-j-C})\leq\exp(e^{n}/200)$ for large enough $C$, so Lemma 6
can be applied. The above bound becomes, for some absolute positive constant
$\tilde{C}$,

$\displaystyle\ll
e^{n_{\mathcal{L}}-n_{0}-j+20y+10((j-n_{0})\wedge(n_{\mathcal{L}}-j))^{3/4}}\exp\Big{(}v\log
v-(n_{\mathcal{L}}+k-2j-4\log(k-j)-\tilde{C})v\Big{)}$ $\displaystyle\ll
e^{n_{\mathcal{L}}-n_{0}-j+20y+10((j-n_{0})\wedge(n_{\mathcal{L}}-j))^{3/4}}\exp\Big{(}-v(k-j-4\log(k-j)+C-\tilde{C})\Big{)}$
$\displaystyle\ll
e^{n_{\mathcal{L}}-n_{0}-j+20y+10((j-n_{0})\wedge(n_{\mathcal{L}}-j))^{3/4}}\exp\Big{(}-ce^{n_{\mathcal{L}}-j}(k-j)\Big{)},$

for some small constant $c>0$, by choosing $C$ large enough. Summing over
$n_{0}\leq j<k$ we see that the sum is dominated by the contribution of the
last term $j=k-1$. The full sum (over $j$ and $k$) is therefore bounded with

$\sum_{0\leq
k<n^{\prime}_{\mathcal{L}}}e^{n_{\mathcal{L}}-n_{0}-k+20y+10((k-n_{0})\wedge(n_{\mathcal{L}}-k))^{3/4}}\exp(-ce^{n_{\mathcal{L}}-k}),$

which is dominated by $k=n^{\prime}_{\mathcal{L}}-1$ and gives a global bound
$c^{c_{1}y-c_{2}y^{2}}$ for some absolute $c_{1},c_{2}>0$.

The second probability in (67) is again by a Chernoff bound,

$\displaystyle\mathbb{P}$
$\displaystyle\Big{(}\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}|(S_{n_{\mathcal{L}}}-S_{k})(h+u)-(S_{n_{\mathcal{L}}}-S_{k})(h)|\mathbf{1}_{h\in
G_{k}}\geq\frac{e^{(n_{\mathcal{L}}-k)/2}}{4}\Big{)}$ $\displaystyle\ll
4^{4v}\cdot\mathbb{E}\Big{[}\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}\frac{|(S_{n_{\mathcal{L}}}-S_{k})(h+u)-(S_{n_{\mathcal{L}}}-S_{k})(h)|^{2v}}{e^{v(n_{\mathcal{L}}-k)}}\cdot\mathbf{1}_{h\in
G_{k}}\Big{]}$ $\displaystyle\ll
e^{n_{\mathcal{L}}-n_{0}-k+10((k-n_{0})\wedge(n_{\mathcal{L}}-k))^{3/4}}\cdot
e^{-v(n_{\mathcal{L}}-k)}\cdot v!\cdot
e^{\tilde{C}v}\cdot(n_{\mathcal{L}}-k)^{v}\cdot e^{n_{\mathcal{L}}-k},$ (69)

for some absolute $\tilde{C}$. Choosing $v=\lfloor
e^{n_{\mathcal{L}}-k-C}/(n_{\mathcal{L}}-k)^{4}\rfloor$, we see that this is
also $\ll e^{v(\tilde{C}-C)}$. Therefore, for alarge enough absolute constant
$C>0$, the full contribution of (67) after summation over $k$ is

$\ll\sum_{0\leq
k<n_{\mathcal{L}^{\prime}}}e^{n_{\mathcal{L}}-n_{0}-k+10((k-n_{0})\wedge(n_{\mathcal{L}}-k))^{3/4}}\cdot\exp(-e^{n_{\mathcal{L}}-k-C}/(n_{\mathcal{L}}-k)^{4})\ll
e^{\tilde{c}_{1}y-\tilde{c}_{2}\frac{y^{2}}{(\log y)^{4}}},$

for some absolute $\tilde{c}_{1},\tilde{c}_{2}>0$, where we used that the main
contribution comes from $k=n^{\prime}_{\mathcal{L}}$. This concludes the
proof. ∎

##  6\. Proof of Proposition 6

We first need a lemma which precisely captures the coupling/decoupling of the
Gaussian walks $\mathcal{G}_{k}(h)$ defined in (9) as a function of the
distance $|h-h^{\prime}|$. For this, the following elementary lemma will be
key in the decoupling regime $|h-h^{\prime}|>e^{-j}$.

######  Lemma 7.

Let $|\rho|<\mathfrak{s}^{2}$. Consider the following Gaussian vectors and
their covariance matrices:

$\displaystyle(\mathcal{N}_{1},\mathcal{N}_{1}^{\prime}),$
$\displaystyle\mathcal{C}_{1}=\left(\begin{matrix}\mathfrak{s}^{2}&\rho\\\
\rho&\mathfrak{s}^{2}\end{matrix}\right),$
$\displaystyle(\mathcal{N}_{2},\mathcal{N}_{2}^{\prime}),$
$\displaystyle\mathcal{C}_{2}=\left(\begin{matrix}\mathfrak{s}^{2}+|\rho|&0\\\
0&\mathfrak{s}^{2}+|\rho|\end{matrix}\right).$

Then for any measurable set $A\subset\mathbb{R}^{2}$ we have

$\mathbb{P}((\mathcal{N}_{1},\mathcal{N}^{\prime}_{1})\in
A)\leq\sqrt{\frac{\mathfrak{s}^{2}+|\rho|}{\mathfrak{s}^{2}-|\rho|}}\cdot\mathbb{P}((\mathcal{N}_{2},\mathcal{N}^{\prime}_{2})\in
A).$

###### Proof.

The proof is simply by expanding the density of
$(\mathcal{N}_{1},\mathcal{N}^{\prime}_{1})$, which is

$\frac{1}{2\pi\sqrt{\mathfrak{s}^{4}-\rho^{2}}}\exp\left(-\frac{\mathfrak{s}^{2}w^{2}+\mathfrak{s}^{2}z^{2}-2\rho
wz}{2(\mathfrak{s}^{4}-\rho^{2})}\right).$ (70)

If $\rho\geq 0$ then for any $w,z\in\mathbb{R}$ we have
$\mathfrak{s}^{2}w^{2}+\mathfrak{s}^{2}z^{2}-2\rho
wz\geq(\mathfrak{s}^{2}-\rho)(w^{2}+z^{2})$ so that

$\frac{\mathfrak{s}^{2}w^{2}+\mathfrak{s}^{2}z^{2}-2\rho
wz}{2(\mathfrak{s}^{4}-\rho^{2})}\geq\frac{w^{2}+z^{2}}{2(\mathfrak{s}^{2}+\rho)},$

and the conclusion follows. If $\rho\leq 0$ then from the previous case for
any $B\subset\mathbb{R}^{2}$

$\mathbb{P}((\mathcal{N}_{1},-\mathcal{N}^{\prime}_{1})\in
B)\leq\sqrt{\frac{\mathfrak{s}^{2}-\rho}{\mathfrak{s}^{2}+\rho}}\cdot\mathbb{P}((\mathcal{N}_{2},-\mathcal{N}^{\prime}_{2})\in
B),$

which concludes the proof by choosing $B=\\{(x,-y):(x,y)\in A\\}$. ∎

###### Proof of Proposition 6.

We have
$\mathbb{E}[(\\#\mathfrak{G}^{+}_{\mathcal{L}})^{2}]=\sum_{h,h^{\prime}}\mathbb{P}\Big{(}\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime})\Big{)}$
where

$\mathfrak{S}(h)=\\{\mathcal{G}_{k}(h)\in[L_{k}-1,U_{k}+1],\forall n_{0}<k\leq
n_{\mathcal{L}}\\},\quad h\in[-1,1].$ (71)

In what follows, we fix $h$, $h^{\prime}$ and simply write
$(\mathcal{G}_{k},\mathcal{G}_{k}^{\prime})$ for
$(\mathcal{G}_{k}(h),\mathcal{G}_{k}(h^{\prime}))$. We divide the above sum
over pairs in three ranges of $|h-h^{\prime}|$; this is necessary to achieve
the precision $1+{\rm o}(1)$ required by Proposition 6.

###  6.1. Case $|h-h^{\prime}|>e^{-n_{0}/2}$

This is the dominant term. We can express the events $\mathfrak{S}(h)$ in
terms of the increments using 28, and then in terms of independent increments
using Lemma 7. Under the product over $j$, the multiplicative error from Lemma
7 is

$\prod_{n_{0}<j\leq
n_{\mathcal{L}}}\sqrt{\frac{\mathfrak{s}_{j}^{2}+|\rho_{j}|}{\mathfrak{s}_{j}^{2}-|\rho_{j}|}}=\exp\Big{(}{\rm
O}(\sum_{n_{0}\leq j\leq n_{\mathcal{L}}}\rho_{j})\Big{)}=\exp\Big{(}{\rm
O}(\sum_{n_{0}\leq j\leq
n_{\mathcal{L}}}\frac{1}{e^{j}|h-h^{\prime}|}\Big{)}\leq 1+{\rm
O}(e^{-n_{0}/2}),$ (72)

therefore we obtain

$\sum_{|h-h^{\prime}|>e^{-n_{0}/2}}\mathbb{P}\Big{(}\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime})\Big{)}\leq(1+{\rm
O}(n_{0}^{-10}))\cdot\Big{(}\mathbb{P}(\widetilde{\mathcal{G}}_{j}\in[L_{j}-1,U_{j}+2],n_{0}<j\leq
n_{\mathcal{L}})\Big{)}^{2},$

where $\widetilde{\mathcal{G}}_{j}=\sum_{i\leq j}\widetilde{\mathcal{N}}_{j}$
and the independent Gaussian centered $\widetilde{\mathcal{N}}_{j}$’s have
variance $\mathfrak{s}_{j}^{2}+|\rho_{j}|$. Moreover the change from the
original interval $[L_{j}-1,U_{j}+1]$ to $[L_{j}-1,U_{j}+2]$ is due to (27)
when transferring the constraint on increments back to the random walk itself.
From the Ballot theorem in Proposition 14 the barrier can be changed into
$[L_{j}+1,U_{j}-1]$, and the $\widetilde{\mathcal{G}}_{j}$ can be replaced by
${\mathcal{G}}_{j}$ at a combined multiplicative cost of $1+{\rm O}(y^{-c})$,
so that in particular

$\sum_{|h-h^{\prime}|>e^{-n_{0}/2}}\mathbb{P}\Big{(}\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime})\Big{)}\leq(1+{\rm
O}(y^{-c}))(\mathbb{E}[\\#\mathfrak{G}_{\mathcal{L}}^{-}])^{2}.$

All the other cases will be much smaller than
$(\mathbb{E}[\\#\mathfrak{G}_{\mathcal{L}}^{-}])^{2}$.

###  6.2. Case $e^{-n_{0}}<|h-h^{\prime}|\leq e^{-n_{0}/2}$

The same reasoning as above applies in this case. The multiplicative error
term analogue to (72) is now ${\rm O}(1)$, and the precise estimate of this
error is not necessary since there are only $\ll
e^{2(n_{\mathcal{L}}-n_{0})}e^{-n_{0}/2}$ pairs $(h,h^{\prime})$ to consider.
Therefore, we obtain

$\sum_{e^{-n_{0}}<|h-h^{\prime}|\leq
e^{-n_{0}/2}}\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime}))\ll
e^{-\frac{n_{0}}{2}}(\mathbb{E}[\\#\mathfrak{G}_{\mathcal{L}}^{-}])^{2}.$

###  6.3. Case $e^{-(n_{\mathcal{L}}-n_{0})}\leq|h-h^{\prime}|\leq e^{-n_{0}}$

We start by writing,

$\sum_{\begin{subarray}{c}h,h^{\prime}\in\mathfrak{G}^{+}_{\mathcal{L}}\\\
e^{-(n_{\mathcal{L}}-n_{0})}<|h-h^{\prime}|\leq
e^{-n_{0}}\end{subarray}}\mathbb{P}\Big{(}\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime})\Big{)}=\sum_{n_{0}\leq
j^{\star}\leq
n_{\mathcal{L}}}\sum_{\begin{subarray}{c}h,h^{\prime}\in\mathfrak{G}^{+}_{\mathcal{L}}\\\
j^{\star}=\lfloor\log|h-h^{\prime}|^{-1}\rfloor\end{subarray}}\mathbb{P}\Big{(}\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime})\Big{)}.$

In order to evaluate
$\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime}))$, we apply the
Gaussian decorrelation Lemma 7 for the increments $j\geq j^{*}$. For the
increments before $j^{*}$, it will be useful to consider the random variables

$\overline{\mathcal{G}}_{j}=\frac{{\mathcal{G}}_{j}+\mathcal{G}^{\prime}_{j}}{2},\qquad\mathcal{G}^{\perp}_{j}=\frac{{\mathcal{G}}_{j}-\mathcal{G}^{\prime}_{j}}{2},\quad
n_{0}<j\leq n_{\mathcal{L}}.$ (73)

Note that $(\overline{\mathcal{G}}_{j})_{j}$ and
$(\mathcal{G}_{j}^{\perp})_{j}$ are independent and
$\mathcal{G}_{j}=\overline{\mathcal{G}}_{j}+\mathcal{G}_{j}^{\perp}$,
$\mathcal{G}_{j}^{\prime}=\overline{\mathcal{G}}_{j}-\mathcal{G}_{j}^{\perp}.$

As before we can express the events $\mathfrak{S}(h)$ in terms of the
increments using (28); here we only use such a decomposition for the process
$\mathcal{G}_{j^{\star},j}:=\mathcal{G}_{j}-\mathcal{G}_{j^{\star}}$,
approximating its increments with independent ones through Lemma 7, up to a
multiplicative error equal to

$\prod_{j^{*}<j\leq
n_{\mathcal{L}}}\sqrt{\frac{\mathfrak{s}_{j}^{2}+|\rho_{j}|}{\mathfrak{s}_{j}^{2}-|\rho_{j}|}}={\rm
O}(1).$

For $h,h^{\prime}$ such that
$\lfloor\log|h-h^{\prime}|^{-1}\rfloor=j^{\star}$, this gives

$\displaystyle\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime}))\ll\sum_{L_{j^{\star}}-1\leq
v-q,v+q\leq U_{j^{\star}}}\text{C}_{j^{\star}}(h,h^{\prime},v,q)\
\text{D}_{j^{\star}}(h,v-q)\text{D}_{j^{\star}}(h^{\prime},v+q),$ (74)

where

$\displaystyle\text{C}_{j^{\star}}(h,h^{\prime},v,q)$
$\displaystyle:=\mathbb{P}\Big{(}\mathcal{G}_{j},\mathcal{G}^{\prime}_{j}\in[L_{j}-1,U_{j}+1]\text{
for all
}j<j^{\star},\overline{\mathcal{G}_{j^{\star}}}\in[v,v+1],\mathcal{G}_{j^{\star}}^{\perp}\in[q,q+1]\Big{)},$
$\displaystyle\text{D}_{j^{\star}}(h,v)$
$\displaystyle:=\mathbb{P}\Big{(}\widetilde{\mathcal{G}}_{j^{\star},j}(h)+v\in[L_{j}-2,U_{j}+2]\text{
for all }j>j^{\star}\Big{)},$

and
$\widetilde{\mathcal{G}}_{j^{\star},j}=\widetilde{\mathcal{G}}_{j}-\widetilde{\mathcal{G}}_{j^{*}}$.
The proof now reduces to bounding the correlated (C) and decorrelated (D)
terms.

####  6.3.1. The Correlated term

Note that if $\mathcal{G}_{j},\mathcal{G}_{j}^{\prime}\in[L_{j}-1,U_{j}+1]$
for all $j<j^{\star}$ then also
$\overline{\mathcal{G}}_{j}\in[L_{j}-1,U_{j}+1]$ for all $j<j^{\star}$.
Moreover, $\mathcal{G}_{j^{*}}^{\perp}$ is independent of
$(\overline{\mathcal{G}}_{j})_{j\leq j^{*}}$. We can therefore bound

$\text{C}_{j^{\star}}(h,h^{\prime},v,q)\leq\mathbb{P}\Big{(}\overline{\mathcal{G}}_{j}\in[L_{j}-1,U_{j}+1]\text{
for all
}j<j^{\star},\overline{\mathcal{G}_{j^{\star}}}\in[v,v+1]\Big{)}\cdot\mathbb{P}\big{(}\mathcal{G}_{j^{\star}}^{\perp}\in[q,q+1)\big{)}.$

The Gaussian $\mathcal{G}^{\perp}_{j^{\star}}$ is centered with variance
$\ll\sum_{j\leq j^{\star}}\varepsilon_{j}^{2}\ll 1$ from (50) and (51). We
thus have

$\mathbb{P}(\mathcal{G}^{\perp}_{j}\in[q,q+1))\ll e^{-cq^{2}},\quad\text{for
some $c>0$.}$

Moreover, $(\overline{\mathcal{G}}_{j})_{j\leq j^{*}}$ satisfies the
assumptions of Proposition 14, and $\overline{\mathcal{G}}_{j^{*}}$ has
variance $\frac{1}{2}\sum_{j\leq
j^{*}}(\mathfrak{s}_{j}^{2}+\rho_{j})=\frac{j^{*}-n_{0}}{2}+{\rm O}(1)$ from
(32) and (50). Thus, uniformly in $|v|\leq 100(j^{*}-n_{0})$, we have

$\text{C}_{j^{\star}}(h,h^{\prime},v,q)\ll\frac{U_{n_{0}}\cdot(U_{j^{\star}}-v+1)}{(j^{\star}-n_{0})^{3/2}}\cdot
e^{-\frac{v^{2}}{j^{\star}-n_{0}}-cq^{2}}.$

####  6.3.2. The Decorrelated term

We condition on $\mathcal{G}_{j,j^{\star}}\in[v_{2},v_{2}+1]$ which implies
that $v_{1}+v_{2}\in[U_{n_{\mathcal{L}}},L_{n_{\mathcal{L}}}]$. Then by the
Ballot theorem stated in Proposition 14, $\text{D}_{j^{\star}}(h,v_{1}-q)$ is

$\ll\sum_{\begin{subarray}{c}L_{n_{\mathcal{L}}}-2\leq v_{1}+v_{2}\leq
U_{n_{\mathcal{L}}}+2\end{subarray}}\frac{(U_{j^{\star}}-v_{1}+q+1)(U_{n_{\mathcal{L}}}-v_{1}-v_{2}+q+1)}{(n_{\mathcal{L}}-j^{\star})^{3/2}}\cdot
e^{-\frac{(v_{2}-q)^{2}}{n_{\mathcal{L}}-j^{\star}}},$

where we have used from (32) and (50) to obtain
$\mathbb{E}[(\widetilde{\mathcal{G}}_{j^{\star},n_{\mathcal{L}}})^{2}]=\sum_{j^{*}<j\leq
n_{\mathcal{L}}}(\mathfrak{s}_{j}^{2}+|\rho_{j}|)=n_{\mathcal{L}}-j^{*}+{\rm
O}(1)$, and $|v_{2}-q|\leq 100(n_{\mathcal{L}}-j^{*})$. Likewise,
$\text{D}_{j^{\star}}(h,v_{1}+q)$ is

$\ll\sum_{\begin{subarray}{c}L_{n_{\mathcal{L}}}-2\leq v_{1}+v_{3}\leq
U_{n_{\mathcal{L}}}+2\end{subarray}}\frac{(U_{j^{\star}}-v_{1}-q+1)(U_{n_{\mathcal{L}}}-v_{1}-v_{3}-q+1)}{(n_{\mathcal{L}}-j^{\star})^{3/2}}\cdot
e^{-\frac{(v_{3}+q)^{2}}{n_{\mathcal{L}}-j^{\star}}}$

####  6.3.3. Putting it together

The above estimates give, after summing over $q\in\mathbb{Z}$,

$\mathbb{P}\Big{(}\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime})\Big{)}\ll\sum_{\begin{subarray}{c}L_{j^{\star}}-1\leq
v_{1}\leq U_{j^{\star}}+1\\\ L_{n_{\mathcal{L}}}-2\leq
v_{1}+v_{2},v_{1}+v_{3}\leq
U_{n_{\mathcal{L}}}+2\end{subarray}}e^{-\frac{v_{1}^{2}}{j^{\star}-n_{0}}-\frac{v_{2}^{2}}{n_{\mathcal{L}}-j^{\star}}-\frac{v_{3}^{2}}{n_{\mathcal{L}}-j^{\star}}}\\\
\times\frac{U_{n_{0}}(U_{j^{\star}}-v_{1}+1)^{3}(U_{n_{\mathcal{L}}}-v_{1}-v_{2}+1)(U_{n_{\mathcal{L}}}-v_{1}-v_{3}+1)}{(n_{\mathcal{L}}-j^{\star})^{3}\cdot(j^{\star}-n_{0})^{3/2}}.$
(75)

We change the variables to $\overline{v_{1}}=v_{1}-\alpha(j^{\star}-n_{0})$,
$\overline{v_{2}}=v_{2}-\alpha(n_{\mathcal{L}}-j^{\star})$ and
$\overline{v_{3}}=v_{3}-\alpha(n_{\mathcal{L}}-j^{\star})$ so that
$\overline{v_{1}}+\overline{v_{2}}\in[L_{n_{0}},U_{n_{0}}]$ and
$\overline{v_{1}}+\overline{v_{3}}\in[L_{n_{0}},U_{n_{0}}]$, giving

$\frac{e^{-\frac{v_{1}^{2}}{j^{\star}-n_{0}}-\frac{v_{2}^{2}}{n_{\mathcal{L}}-j^{\star}}-\frac{v_{3}^{2}}{n_{\mathcal{L}}-j^{\star}}}}{(n_{\mathcal{L}}-j^{\star})^{3}(j^{\star}-n_{0})^{3/2}}\ll
e^{-2(n_{\mathcal{L}}-j^{\star})-(j^{\star}-n_{0})+2\overline{v_{1}}-2(\overline{v_{1}}+\overline{v_{2}})-2(\overline{v_{1}}+\overline{v_{3}})}\cdot\frac{n^{\frac{3}{2}\frac{j^{\star}}{n}}n^{3(1-\frac{j^{\star}}{n})}}{(j^{\star}-n_{0})^{3/2}(n_{\mathcal{L}}-j^{\star})^{3}}.$

The contribution of the integral over
$\overline{v_{1}}+\overline{v_{2}}\in[L_{n_{0}},U_{n_{0}}]$ is

$\int_{[L_{n_{0}},U_{n_{0}}]}(U_{n_{0}}-z+1)e^{-2z}{\rm
d}z\ll|L_{n_{0}}|e^{2|L_{n_{0}}|}.$

The same bound holds for the integral over
$\overline{v_{1}}+\overline{v_{3}}$. The integral over $\overline{v_{1}}$ is
for
$\mathcal{B}_{j^{\star}}=U_{n_{0}}-10\log((j^{\star}-n_{0})\wedge(n_{\mathcal{L}}-j^{\star}))$

$\ll\int_{-\infty}^{\mathcal{B}_{j^{\star}}}(\mathcal{B}_{j^{\star}}-\overline{v_{1}}+1)^{3}e^{2\overline{v_{1}}}{\rm
d}\overline{v_{1}}\ll e^{2\mathcal{B}_{j^{\star}}}.$

Combining these estimates for the ${\rm
O}(e^{2(n_{\mathcal{L}}-n_{0})-j^{\star}})$ pairs with
$\log|h-h^{\prime}|^{-1}\geq j^{\star}$, we obtain

$\sum_{e^{-(n_{\mathcal{L}}-n_{0})}\leq|h-h^{\prime}|\leq
e^{-n_{0}}}\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime}))\ll
e^{-n_{0}}U_{n_{0}}L_{n_{0}}^{2}e^{4|L_{n_{0}}|}\sum_{j^{\star}}e^{2\mathcal{B}_{j^{\star}}}\frac{n^{\frac{3}{2}\frac{j^{\star}}{n}}n^{3(1-\frac{j^{\star}}{n})}}{(j^{\star}-n_{0})^{3/2}(n_{\mathcal{L}}-j^{\star})^{3}}\\\
\ll e^{-n_{0}}U_{n_{0}}L_{n_{0}}^{2}e^{4|L_{n_{0}}|}\cdot e^{2U_{n_{0}}}.$
(76)

On the other hand, from Proposition 14 we have a simple lower bound for
$\mathbb{E}[\\#\mathfrak{G}^{-}_{\mathcal{L}}]$:

$\mathbb{E}[\\#\mathfrak{G}^{-}_{\mathcal{L}}]=e^{n_{\mathcal{L}}-n_{0}}\cdot\mathbb{P}(\mathcal{G}_{k}\in[L_{k}+1,U_{k}-1],n_{0}<k\leq
n_{\mathcal{L}})\gg U_{n_{0}}|L_{n_{0}}|e^{2|L_{n_{0}}|}.$ (77)

We conclude from (76) and (77) that

$\sum_{e^{-(n_{\mathcal{L}}-n_{0})}\leq|h-h^{\prime}|\leq
e^{-n_{0}}}\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime}))\ll
U_{n_{0}}^{-1}e^{2U_{n_{0}}-n_{0}}(\mathbb{E}[\\#\mathfrak{G}^{-}_{\mathcal{L}}])^{2}\ll
e^{-y/10}(\mathbb{E}[\\#\mathfrak{G}^{-}_{\mathcal{L}}])^{2}$

by the choice of $n_{0}$ and $U_{n_{0}}$.

###  6.4. Conclusion.

When $|h-h^{\prime}|\leq e^{-(n_{\mathcal{L}}-n_{0})}$, because of the spacing
constraint we necessarily have $h=h^{\prime}$, and the contribution from such
trivial pairs admits the same upper bound as for $j^{*}=n_{\mathcal{L}}$
above. All together, we have obtained

$\mathbb{E}[(\\#\mathfrak{G}^{+}_{\mathcal{L}})^{2}]\leq(1+{\rm
O}(y^{-c}))(\mathbb{E}[\\#\mathfrak{G}_{\mathcal{L}}^{-}])^{2},$

which concludes the proof of Proposition 6. ∎

##  7\. Proof of Theorem 2

The proof of the theorem follows the same structure as the one of Theorem 1.
The parameters need to be picked differently. We take for the times

$n_{0}=\lfloor y/100\rfloor,\qquad n_{\mathcal{L}}=\log\log(T^{1/100})=n-\log
100.$

The partial sums on primes are now starting from $p=2$ and not $\exp
e^{n_{0}}$

$S_{j}(h)=\sum_{\log\log p\leq j}{\rm
Re}\Big{(}p^{-(1/2+\mathrm{i}\tau+\mathrm{i}h)}+\frac{1}{2}p^{-2(1/2+\mathrm{i}\tau+\mathrm{i}h)}\Big{)},\quad
j\in\mathbb{N}.$ (78)

The set of good points are

$G_{0}=[-\tfrac{1}{2},\tfrac{1}{2}]\cap e^{-n_{\mathcal{L}}}\mathbb{Z},\qquad
G_{j}=\\{h\in G_{0}:S_{j}\in[L_{j},U_{j}],n_{0}\leq j\leq n_{\mathcal{L}}\\},$

where the barriers are now for $j\geq n_{0}$

$\displaystyle U_{j}$ $\displaystyle=y+\alpha j-10\log(j\wedge(n-j)),$ (79)
$\displaystyle L_{j}$
$\displaystyle=-10+(\alpha+\frac{y}{n_{\mathcal{L}}})j-(j\wedge(n-j))^{3/4}.$

The slope $\alpha$ is $1-\frac{3}{4}\frac{\log n}{n}$ as before. Both barriers
are convex, which is crucial. Note that the final interval for
$S_{n_{\mathcal{L}}}$ is $[L_{n_{\mathcal{L}}},U_{n_{\mathcal{L}}}]$ where

$U_{n_{\mathcal{L}}}=n-\frac{3}{4}\log n+y\qquad
L_{n_{\mathcal{L}}}=n-\frac{3}{4}\log n+y-10.$

The reason for the slightly larger slope in $L_{j}$, i.e.,
$(\alpha+\frac{y}{n_{\mathcal{L}}})$ instead of $\alpha$, is to ensure that
the width of the final interval is order one. The factor $y/n_{\mathcal{L}}$
will not affect the proof.

It is necessary to take $U_{n_{0}}=y+\alpha n_{0}$, as this is the origin of
the factor $y$ in front of the exponential decay in Theorem 2. For $y$ of
order one, it would be possible to take $n_{0}={\rm O}(1)$. However, for
larger $y$, the spread $U_{j}-L_{j}$ could be quite large for small $j$. This
prevents a Gaussian comparison for small primes. For this reason, the barrier
starts at $n_{0}$, a multiple of $y$. For these times, the spread is
proportional to variance and the Gaussian comparison goes through.

Unlike the left tail, we do need to include the small primes in the partial
sums. Dropping the first $\exp e^{n_{0}}$ primes would give a lower bound
$ye^{-2y}e^{-n_{0}}e^{-y^{2}/n}$, which is suboptimal for $n_{0}\asymp y$. A
more involved analysis of the small primes would probably allow to improve the
result range of Theorem 2 to $y={\rm o}(n)$, matching the branching Brownian
motion estimate.

For the proof of Theorem 2, we first need the analogue of Proposition 1.

######  Proposition 9.

We have, for any fixed $C>10$, uniformly in $1\leq y={\rm o}(n)$

$\mathbb{P}\Big{(}\max_{|h|\leq
1}\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|>n-\frac{3}{4}\log
n+y-10C\Big{)}\geq\mathbb{P}\Big{(}\exists h\in G_{\mathcal{L}}\Big{)}-{\rm
O}(e^{-50C}ye^{-2y}e^{-y^{2}/n}).$

Therefore, upon taking $C$ large enough but fixed, the estimate

$\mathbb{P}\Big{(}\\#G_{n_{\mathcal{L}}}\geq 1\Big{)}\gg ye^{-2y}e^{-y^{2}/n}$
(80)

will imply Theorem 2. Equation (80) follows directly from the Paley-Zygmund
inequality from the propositions below.

######  Proposition 10.

Uniformly in $10\leq y\leq C\frac{\log\log T}{\log\log\log T}$,

$\mathbb{E}[\\#G_{n_{\mathcal{L}}}]\gg ye^{-2y}e^{-y^{2}/n}.$

######  Proposition 11.

Uniformly in $10\leq y\leq C\frac{\log\log T}{\log\log\log T}$,

$\mathbb{E}[(\\#G_{n_{\mathcal{L}}})^{2}]\ll ye^{-2y}e^{-y^{2}/n}.$

Unlike the left tail, the dominant term in the second moment will come from
the pairs $h,h^{\prime}$ that are very close, i.e., $|h-h^{\prime}|\ll
e^{-n_{\mathcal{L}}}$.

###  7.1. Proof of Proposition 9

First we have the following easy variant of Proposition 3

######  Proposition 12.

We have, for $1000<y<n^{1/10}$,

$\mathbb{P}\Big{(}\max_{|h|\leq
1}\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|\geq\max_{h\in
G_{0}}\min_{|u|\leq
1}(S_{n_{\mathcal{L}}}(h+u)+\sqrt{|u|e^{n_{\mathcal{L}}}})-2C\Big{)}\geq
1-{\rm O}(e^{-n}),$

with $C>0$ an absolute constant.

###### Proof.

This is the same proof as Proposition 3, the only difference is that this time
we do not need to bound the contribution of the primes $p$ with $\log p\leq
e^{n_{0}}$ and therefore there is no additional term $-20y$. Because of this,
the exceptional set is also better, i.e., $e^{-n}$ instead of $e^{-y}$. ∎

We highlight the changes needed in Lemma 6 and Proposition 4, with the
following two variants.

######  Lemma 8.

Let $1\leq\ell\leq n_{\mathcal{L}}$. Let $v\geq 1$ and $0\leq k\leq n$ be
given. Let $\mathcal{Q}$ be a Dirichlet polynomial as defined in (61), such
that $e^{\ell}\leq\log p\leq e^{n_{\mathcal{L}}}$ and of length
$\leq\exp(\tfrac{1}{200v}e^{n})$. Denote by $a(p)$ the coefficients of
$\mathcal{Q}$. Then

$\displaystyle\mathbb{E}\Big{[}$
$\displaystyle\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}|\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h+\mathrm{i}u)-\mathcal{Q}(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|^{2v}\cdot\mathbf{1}_{h\in
G_{\ell}}\Big{]}$ (81) $\displaystyle\ll e^{n_{\mathcal{L}}}\
\mathbb{P}(\mathcal{G}_{\ell})\cdot
2^{2v}v!\cdot\Big{(}\Big{(}e^{-2k+4}\sum_{e^{\ell}\leq\log p\leq
e^{k}}\frac{|a(p)|^{2}\log^{2}p}{p}\Big{)}^{v}+\Big{(}16\sum_{e^{k}\leq\log
p}\frac{|a(p)|^{2}}{p}\Big{)}^{v}\cdot e^{n_{\mathcal{L}}-k}\Big{)}.$

Moreover, we simply bound $\mathbb{P}(\mathcal{G}_{\ell})\leq 1$ if
$\ell<n_{0}$, and otherwise

$\mathbb{P}(\mathcal{G}_{\ell})\ll\frac{y}{\ell^{3/2}}\exp\Big{(}-\ell+\frac{3}{2}\cdot\frac{\ell\log
n_{\mathcal{L}}}{n_{\mathcal{L}}}-\frac{2y\ell}{n_{\mathcal{L}}}-\frac{y^{2}\ell}{n_{\mathcal{L}}^{2}}+10(\ell\wedge(n-\ell))^{3/4}\Big{)}.$

###### Proof.

The proof is the same as for Lemma 6, the only differences being the different
bound for $\mathbb{P}(\mathcal{G}_{\ell})$ (when $\ell\geq n_{0}$) that arises
from a different barrier. Note that for $1\leq\ell\leq n_{0}$ there is no
barrier so the proof does not rely on Proposition 7, which requires $\ell\geq
n_{0}$. ∎

######  Proposition 13.

We have, for any $C>10$, and for $y=o(n)$,

$\mathbb{P}\Big{(}\forall h\in G_{\mathcal{L}}\ \forall|u|\leq
1:|S_{n_{\mathcal{L}}}(h+u)-S_{n_{\mathcal{L}}}(h)|\leq
C+\sqrt{|u|e^{n_{\mathcal{L}}}}\Big{)}=1+{\rm
O}\Big{(}e^{-50C}ye^{-2y}e^{-y^{2}/n}\Big{)}.$

###### Proof.

The proof is very similar to Proposition 4 but we still find it worthwhile to
include the details. If there exists an $h\in G_{\mathcal{L}}$ and $|u|\leq 1$
such that

$|S_{n_{\mathcal{L}}}(h+u)-S_{n_{\mathcal{L}}}(h)|>C+\sqrt{|u|e^{n_{\mathcal{L}}}}$
(82)

then there exists a $0\leq k<n_{\mathcal{L}}^{\prime}:=n_{\mathcal{L}}-\lfloor
2\log C\rfloor$ such that,

$\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}|S_{n_{\mathcal{L}}}(h+u)-S_{n_{\mathcal{L}}}(h)|\cdot\mathbf{1}_{h\in
G_{\mathcal{L}}}\geq e^{(n_{\mathcal{L}}-k)/2},$

where considering the case $k\leq n_{\mathcal{L}}^{\prime}$ is enough thanks
to the term $C$ in (82). It now suffices to bound (66) through a bound for the
right-hand side of (67), but with our new definitions for $(S_{j})_{j\geq 1}$,
$n_{\mathcal{L}}$, $n^{\prime}_{\mathcal{L}}$ and $G_{k}$. For any $0\leq
j<k$, we have the following analogue of 68, which is also obtained by Lemma 6:

$\displaystyle\mathbb{P}$
$\displaystyle\Big{(}\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}|(S_{j+1}-S_{j})(h+u)-(S_{j+1}-S_{j})(h)|\mathbf{1}_{h\in
G_{j}}\geq\frac{e^{(n_{\mathcal{L}}-k)/2}}{4(k-j)^{2}}\Big{)}$
$\displaystyle\ll(k-j)^{4v}\cdot
e^{n_{\mathcal{L}}-j+10(j\wedge(n-j)^{3/4}}e^{-v(n_{\mathcal{L}}-k)}\cdot
v!\cdot C^{v}\cdot e^{2v(j-k)}\cdot\frac{y}{j^{3/2}}\cdot
e^{\frac{3}{2}\frac{j\log n}{n}-\frac{2yj}{n}-\frac{y^{2}j}{n^{2}}}.$

Pick $v=100$. Summing over $0\leq j<k$ we see that the sum is dominated by the
contribution of the last term $j=k-1$, indeed, the sum is

$\ll
e^{-(v-1)(n_{\mathcal{L}}-k)+10(k\wedge(n-k))^{3/4}}\cdot\frac{y}{k^{3/2}}\exp\Big{(}\frac{3}{2}\cdot\frac{k\log
n}{n}-\frac{2yk}{n}-\frac{y^{2}k}{n^{2}}\Big{)}.$

The contribution of the second term in (67) is bounded similarly to (69), and
we obtain

$\displaystyle\mathbb{P}$
$\displaystyle\Big{(}\sup_{\begin{subarray}{c}|h|\leq 1\\\ |u|\leq
e^{-k+1}\end{subarray}}|(S_{n_{\mathcal{L}}}-S_{k})(h+u)-(S_{n_{\mathcal{L}}}-S_{k})(h)|\mathbf{1}_{h\in
G_{k}}\geq\frac{e^{(n_{\mathcal{L}}-k)/2}}{4}\Big{)}$ $\displaystyle\ll
4^{4v}\cdot e^{n_{\mathcal{L}}-k+10(k\wedge(n-k))^{3/4}}\cdot
e^{-v(n_{\mathcal{L}}-k)}\cdot v!\cdot
C^{v}\cdot(n_{\mathcal{L}}-k)^{v}\cdot\frac{y}{j^{3/2}}\cdot\exp\Big{(}\frac{3}{2}\frac{j\log
n}{n}-\frac{2yj}{n}-\frac{y^{2}j}{n^{2}}\Big{)}.$

Choosing $v=100$ we see that this is also

$\ll
e^{-(v-2)(n_{\mathcal{L}}-k)+10(k\wedge(n-k))^{3/4}}\cdot\frac{y}{k^{3/2}}\exp\Big{(}\frac{3}{2}\cdot\frac{k\log
n}{n}-\frac{2yk}{n}-\frac{y^{2}k}{n^{2}}\Big{)}.$

Therefore with the new definitions for $(S_{j})_{j\geq 1}$, $n_{\mathcal{L}}$,
$n^{\prime}_{\mathcal{L}}$ and $G_{k}$, (66) is bounded with

$\ll\sum_{0\leq k\leq
n_{\mathcal{L}^{\prime}}}e^{-100(n_{\mathcal{L}}-k)}\cdot\frac{y}{k^{3/2}}\exp\Big{(}\frac{3}{2}\cdot\frac{k\log
n}{n}-\frac{2yk}{n}-\frac{y^{2}k}{n^{2}}\Big{)}\ll e^{-50C}\cdot
ye^{-2y}e^{-y^{2}/n}$

as needed, and where the final gain $e^{-50C}$ comes from $k\leq
n_{\mathcal{L}}^{\prime}=n_{\mathcal{L}}-\lfloor 2\log C\rfloor$. ∎

###### Proof of Proposition 9.

If there exists an $h\in G_{\mathcal{L}}$, then from Proposition 13

$\displaystyle\max_{v\in G_{0}}\min_{|u|\leq
1}(S_{n_{\mathcal{L}}}(v+u)+\sqrt{|u|e^{n_{\mathcal{L}}}})$
$\displaystyle\geq\min_{|u|\leq
1}(S_{n_{\mathcal{L}}}(h+u)+\sqrt{|u|e^{n_{\mathcal{L}}}})$ $\displaystyle\geq
S_{n_{\mathcal{L}}}(h)-1\geq n-\frac{3}{4}\log n-10C,$

outside of a set of probability $\ll e^{-50C}ye^{-2y}e^{-y^{2}/n}$.
Proposition 12 then implies that outside of a set of $\tau$ of probability
$\ll e^{-n}$,

$\max_{|h|\leq
1}\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|>n-\frac{3}{4}\log
n+y-10C.$

In other words,

$\mathbb{P}(\exists h\in G_{\mathcal{L}})-{\rm
O}\Big{(}e^{-50C}ye^{-2y}e^{-y^{2}/n}\Big{)}\leq\mathbb{P}\Big{(}\max_{|h|\leq
1}\log|\zeta(\tfrac{1}{2}+\mathrm{i}\tau+\mathrm{i}h)|>n-\frac{3}{4}\log
n+y-10C\Big{)},$

and Proposition 9 follows. ∎

###  7.2. Proof of Proposition 10 and 11

###### Proof of Propositions 10.

Clearly, we have

$\mathbb{E}[\\#G_{n_{\mathcal{L}}}]\gg
e^{n_{\mathcal{L}}}\cdot\mathbb{P}(S_{j}\in[L_{j},U_{j}],n_{0}\leq j\leq
n_{\mathcal{L}}),$

where we write $S_{j}(0)=S_{j}$ for simplicity. By the definition of
$U_{j},L_{j}$, we have for $j\geq n_{0}$

$U_{j}-L_{j}\ll(y-10)-\frac{y}{n}j+(j\wedge(n-j))^{3/4}\ll\Delta_{j}^{1/4}.$

Therefore, the proof of Proposition 7 applies verbatim for all increments
$j\geq n_{0}$. For the first $n_{0}$ increments, the approximation in terms of
Dirichlet polynomials still holds up to a multiplicative constant (as in
[ArgBouRad2020, Equations (31) and (40)], for example). These considerations
yield

$\mathbb{P}(S_{j}\in[L_{j},U_{j}],n_{0}\leq j\leq n_{\mathcal{L}})\\\
\gg\mathbb{P}(\mathcal{S}_{n_{0}}\in[L_{n_{0}}+1,U_{n_{0}}-1],\mathcal{S}_{n_{0}}+\mathcal{G}_{j}\in[L_{j}+1,U_{j}-1],n_{0}<j\leq
n_{\mathcal{L}}),$ (83)

where $(\mathcal{G}_{j})_{j}$ is defined in (9) and is independent of
$\mathcal{S}_{n_{0}}$, now defined as

$\mathcal{S}_{n_{0}}(h)=\sum_{\log\log p\leq n_{0}}{\rm
Re}\Big{(}e^{\mathrm{i}\theta_{p}}\,p^{-(1/2+\mathrm{i}h)}+\tfrac{1}{2}\,e^{2\mathrm{i}\theta_{p}}\,p^{-(1+2\mathrm{i}h)}\Big{)}.$

Note that it differs from (30) as it consists in the first $n_{0}$ increments.
The $\pm 1$ in the barriers will not contribute to the estimate, we henceforth
drop them to lighten the notations. We now write $f(z)$ for the density of
$\mathcal{S}_{n_{0}}$, we condition on $\mathcal{G}_{n_{\mathcal{L}}}$ and
apply the Ballot theorem from Proposition 14: The right-hand side of (83) is
lower bounded with

$\int_{L_{n_{0}}}^{U_{n_{0}}}\mathbb{P}(\mathcal{G}_{j}\in[L_{j}-z,U_{j}-z],n_{0}<j\leq
n_{\mathcal{L}})\ f(z){\rm d}z\\\
\gg\int_{L_{n_{0}}}^{U_{n_{0}}}\int_{L_{n_{\mathcal{L}}}-z}^{U_{n_{\mathcal{L}}}-z}\frac{(U_{n_{0}}-z)(U_{n_{\mathcal{L}}}-z-w)}{(n_{\mathcal{L}}-n_{0})^{3/2}}e^{-\frac{w^{2}}{n_{\mathcal{L}}-n_{0}}}f(z){\rm
d}w{\rm d}z.$

Writing $\bar{w}=w-\alpha(n_{\mathcal{L}}-n_{0})$ and $\bar{z}=z-\alpha
n_{0}$, this becomes

$\gg\int_{\bar{L}_{n_{0}}}^{\bar{U}_{n_{0}}}\int_{y-10-\bar{z}}^{y-\bar{z}}\frac{(\bar{U}_{n_{0}}-\bar{z})(y-\bar{z}-\bar{w})}{(n_{\mathcal{L}}-n_{0})^{3/2}}e^{-(\bar{w}+\alpha(n_{\mathcal{L}}-n_{0}))^{2}/(n_{\mathcal{L}}-n_{0})}f(\bar{z}+\alpha
n_{0}){\rm d}\bar{w}{\rm d}\bar{z}.$

for $\bar{L}_{n_{0}}=L_{n_{0}}+\frac{y}{n_{\mathcal{L}}}n_{0}-n_{0}^{3/4}$ and
$\bar{U}_{n_{0}}=U_{n_{0}}-10\log n_{0}^{3/4}$. Expanding the square gives

$\frac{(\bar{w}+\alpha(n_{\mathcal{L}}-n_{0}))^{2}}{n_{\mathcal{L}}-n_{0}}=\alpha^{2}(n_{\mathcal{L}}-n_{0})+2\alpha\bar{w}+\frac{\bar{w}^{2}}{n_{\mathcal{L}}-n_{0}}=(n_{\mathcal{L}}-n_{0})-\frac{3}{2}\log
t+2\alpha\bar{w}+\frac{\bar{w}^{2}}{n_{\mathcal{L}}-n_{0}}+{\rm o}(1).$ (84)

The integral in $\bar{w}$ becomes

$\int_{y-10-\bar{z}}^{y-\bar{z}}(y-\bar{z}-\bar{w})e^{-2\alpha\bar{w}}e^{-\frac{\bar{w}^{2}}{n_{\mathcal{L}}-n_{0}}}{\rm
d}\bar{w}\gg e^{-2\alpha y}e^{2\alpha\bar{z}}e^{-y^{2}/n}\gg
e^{-2y}e^{-y^{2}/n}e^{2\alpha\bar{z}},$

by the assumption on $y$. So far, we have shown

$\mathbb{E}[\\#G_{n_{\mathcal{L}}}]\gg e^{n_{0}}\cdot
e^{-2y}e^{-y^{2}/n}\cdot\int_{\bar{L}_{n_{0}}}^{\bar{U}_{n_{0}}}(\bar{U}_{n_{0}}-\bar{z})e^{2\alpha\bar{z}}f(\bar{z}+\alpha
n_{0}){\rm d}\bar{z}.$

From the proof of [ArgBouRad2020, Lemma 18], we have $f(u)\ll
e^{-u^{2}/n_{0}}/\sqrt{n_{0}}$ uniformly in $|u|<100n_{0}$. This implies

$\mathbb{E}[\\#G_{n_{\mathcal{L}}}]\gg e^{n_{0}}\cdot
e^{-2y}e^{-y^{2}/n}\cdot\int_{\bar{L}_{n_{0}}}^{\bar{U}_{n_{0}}}(\bar{U}_{n_{0}}-\bar{z})e^{2\alpha\bar{z}}\frac{e^{(-\bar{z}+\alpha
n_{0})^{2}/n_{0}}}{\sqrt{n_{0}}}{\rm d}\bar{z}\\\ \gg
e^{-2y}e^{-y^{2}/n}\int_{\bar{L}_{n_{0}}}^{\bar{U}_{n_{0}}}(\bar{U}_{n_{0}}-\bar{z})\frac{e^{-\bar{z}^{2}/n_{0}}}{\sqrt{n_{0}}}{\rm
d}\bar{z}\gg ye^{-2y}e^{-y^{2}/n}$

since the standard deviation of $\bar{z}$ is $\sqrt{n_{0}}\asymp\sqrt{y}$ and
$\bar{U}_{n_{0}}=y-10\log n_{0}^{3/4}$. ∎

###### Proof of Proposition 11.

Proceeding as in Proposition 10, the estimate is reduced to

$\mathbb{E}[(\\#G_{n_{\mathcal{L}}})^{2}]\ll\sum_{h,h^{\prime}\in
G_{0}}\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime})),$

where

$\mathfrak{S}(h)=\\{\mathcal{S}_{n_{0}}(h)\in[L_{n_{0}}-1,U_{n_{0}}+1],\mathcal{S}_{n_{0}}(h)+\mathcal{G}_{j}(h)\in[L_{j}-1,U_{j}+1],n_{0}<j\leq
n_{\mathcal{L}}\\}.$

Again, since the $\pm 1$ will not contribute to the estimates, we omit them
from the notations. We write $\mathcal{S}_{n_{0}}(h)=\mathcal{S}_{n_{0}}$,
$\mathcal{S}_{n_{0}}(h^{\prime})=\mathcal{S}_{n_{0}}^{\prime}$ and similarly
for $\mathcal{G}$. We condition on the pair
$(\mathcal{S}_{n_{0}},\mathcal{S}_{n_{0}}^{\prime})$ to get

$\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime}))=\int_{[L_{n_{0}},U_{n_{0}}]^{2}}\mathbb{P}(\mathcal{G}_{j}\leq
U_{j}-z,\mathcal{G}_{j}^{\prime}\leq U_{j}-z^{\prime},n_{0}\leq j\leq
n_{\mathcal{L}})f(z,z^{\prime}){\rm d}z{\rm d}z^{\prime},$

where $f$ now stands for the density of
$(\mathcal{S}_{n_{0}},\mathcal{S}_{n_{0}}^{\prime})$. The estimate depends on
the branching time
$j^{\star}=j^{\star}(h,h^{\prime})=\lfloor\log|h-h^{\prime}|^{-1}\rfloor$. We
split into two cases ($j^{\star}\leq n_{0}$ and $j^{\star}>n_{0}$), contrary
to the proof of Proposition (6) which needs three cases as it requires
matching of first and second moments up to $1+{\rm o}(1)$ precision.

Case $j^{\star}\leq n_{0}$. In this case, the decoupling Lemma 7 can be
applied to all increments. The probability in the integral is then

$\ll\int_{[L_{n_{0}},U_{n_{0}}]^{2}}\mathbb{P}(\widetilde{\mathcal{G}}_{j}\leq
U_{j}-z,n_{0}\leq j\leq
n_{\mathcal{L}})\mathbb{P}(\widetilde{\mathcal{G}}^{\prime}_{j}\leq
U_{j}-z^{\prime},n_{0}\leq j\leq n_{\mathcal{L}})f(z,z^{\prime}){\rm d}z{\rm
d}z^{\prime}$

where we recall that $\widetilde{\mathcal{G}}_{j}=\sum_{i\leq
j}\widetilde{\mathcal{N}}_{j}$ and the independent Gaussian centered
$\widetilde{\mathcal{N}}_{j}$’s have variance
$\mathfrak{s}_{j}^{2}+|\rho_{j}|$.

After conditioning on
$(\widetilde{\mathcal{G}}_{n_{\mathcal{L}}},\widetilde{\mathcal{G}}^{\prime}_{n_{\mathcal{L}}})$,
the Ballot theorem from Proposition 14 can be applied to each term. The above
becomes

$\ll\int_{[L_{n_{0}},U_{n_{0}}]^{2}}\int_{[L_{n_{\mathcal{L}}},U_{n_{\mathcal{L}}}]^{2}}\tfrac{(U_{n_{0}}-z)(U_{n_{\mathcal{L}}}-z-w)(U_{n_{0}}-z^{\prime})(U_{n_{\mathcal{L}}}-z^{\prime}-w^{\prime})}{(n_{\mathcal{L}}-n_{0})^{3}}e^{-\frac{w^{2}+w^{\prime
2}}{n_{\mathcal{L}}-n_{0}}}f(z,z^{\prime}){\rm d}w{\rm d}w^{\prime}{\rm
d}z{\rm d}z^{\prime}.$

The Gaussian density can be expanded as in (84). The integral in
$w,w^{\prime}$ gives a contribution ${\rm
O}(e^{-2n_{\mathcal{L}}}e^{2n_{0}}e^{-4y}e^{-2y^{2}/n}).$ There are ${\rm
O}(e^{2n_{\mathcal{L}}})$ pairs $h,h^{\prime}$ with $j^{\star}\leq n_{0}$, so

$\displaystyle\sum_{h,h^{\prime}:j^{\star}\leq
n_{0}}\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h))$ $\displaystyle\hskip
28.45274pt\ll
e^{2n_{0}}e^{-4y}\int_{[\bar{L}_{n_{0}},\bar{U}_{n_{0}}]^{2}}(\bar{U}_{n_{0}}-\bar{z})(\bar{U}_{n_{0}}-\bar{z}^{\prime})e^{2\alpha(\bar{z}+\bar{z}^{\prime})}f(\bar{z}+\alpha
n_{0},\bar{z}^{\prime}+\alpha n_{0}){\rm d}\bar{z}{\rm d}\bar{z}^{\prime}$
$\displaystyle\hskip 28.45274pt\ll
y^{2}e^{2n_{0}}e^{-4y}\int_{[\bar{L}_{n_{0}},\bar{U}_{n_{0}}]^{2}}e^{2\alpha(\bar{z}+\bar{z}^{\prime})}f(\bar{z}+\alpha
n_{0},\bar{z}^{\prime}+\alpha n_{0}){\rm d}\bar{z}{\rm d}\bar{z}^{\prime},$

where we used the barrier range to bound $|\bar{U}_{n_{0}}-\bar{z}|\leq y$.
Using the Cauchy-Schwarz inequality and recalling that $f(u)\ll
e^{-u^{2}/n_{0}}/\sqrt{n_{0}}$, as $n_{0}=y/10$, this is

$\ll
y^{2}e^{2n_{0}}e^{-4y}\cdot\int_{-\infty}^{\infty}e^{4\bar{z}}\frac{e^{-\bar{z}^{2}/n_{0}}}{\sqrt{n_{0}}}\ll
y^{2}e^{2n_{0}}e^{-4y}\cdot e^{8n_{0}}\ll ye^{-2y-y^{2}/n}.$

Case $j^{\star}>n_{0}$. We proceed similarly to the proof of the left tail and
consider the center of mass and the difference between the two Gaussian walks
as in (73). We index the value of $\overline{\mathcal{G}}_{j^{\star}}$ by
$v_{1}$, the values $\mathcal{G}_{j^{\star}}^{\perp}$ by $q$, and the values
of two independent copies of
$\widetilde{\mathcal{G}}_{j^{\star},n_{\mathcal{L}}}$ by $v_{2}$ and $v_{3}$.
Proceeding exactly as for Equation (74), i.e., using Lemma 7 for the
increments after $j^{*}$, we obtain that
$\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h^{\prime}))$ is

$\displaystyle\sum_{q\in\mathbb{Z}}\int_{[L_{n_{0}},U_{n_{0}}]^{2}}\int_{L_{j^{\star}}}^{U_{j^{\star}}}\frac{(U_{n_{0}}-z_{m})(U_{j^{\star}}-z_{m}-v_{1})}{(j^{\star}-n_{0})^{3/2}}\cdot
e^{-cq^{2}}f(z,z^{\prime})e^{-\frac{v_{1}^{2}}{j^{\star}-n_{0}}}$
$\displaystyle\times\mathbb{P}(\widetilde{\mathcal{G}}_{j^{\star},j}+v_{1}+q\leq
U_{j}-z,j>j^{\star})\cdot\mathbb{P}(\widetilde{\mathcal{G}}_{j^{\star},j}+v_{1}-q\leq
U_{j}-z^{\prime},j>j^{\star}){\rm d}v_{1}{\rm d}z{\rm d}z^{\prime}$

where we applied Proposition 14 for $\bar{\mathcal{G}}$ between $n_{0}$ and
$j^{\star}$ and where we denoted
$z_{m}=\frac{z+z^{\prime}}{2}=\bar{\mathcal{G}}_{n_{0}}$. This Ballot theorem
can also be applied to the two probabilities in the integral giving, after
summing over $q$,

$\ll\frac{(U_{j^{\star}}-z_{m}-v_{1})^{2}(U_{n_{\mathcal{L}}}-z-v_{1}-v_{2})(U_{n_{\mathcal{L}}}-z^{\prime}-v_{1}-v_{3})}{(n_{\mathcal{L}}-j^{\star})^{3}}e^{-\frac{v_{2}^{2}+v_{3}^{2}}{n_{\mathcal{L}}-j^{\star}}}.$

After expanding the squares, the densities of $v_{1}$, $v_{2}$, $v_{3}$ become

$e^{-2n_{\mathcal{L}}+j^{\star}}e^{n_{0}}\frac{n^{\frac{3}{2}\frac{j^{\star}}{t}}n^{3(1-\frac{j^{\star}}{n})}}{(j^{\star}-n_{0})^{3/2}(n_{\mathcal{L}}-j^{\star})^{3}}e^{2\alpha\overline{v_{1}}-\frac{\bar{v}_{1}^{2}}{j^{\star}-n_{0}}-2\alpha(\overline{v_{1}}+\overline{v_{2}})-2\alpha(\overline{v_{1}}+\overline{v_{3}})},$

for $\overline{v_{1}}=v_{1}-\alpha(j^{\star}-n_{0})$,
$\overline{v_{i}}=v_{i}-\alpha(n_{\mathcal{L}}-j^{\star})$, $i=2,3$. The
integral over $\overline{v_{1}}+\overline{v_{2}}\in[y-10-\bar{z},y-\bar{z}]$
is

$\int_{y-10-\bar{z}}^{y-\bar{z}}(y-\bar{z}-\overline{v_{1}}-\overline{v_{2}})e^{-2\alpha(\overline{v_{1}}+\overline{v_{2}})}{\rm
d}\overline{v_{1}}{\rm d}\overline{v_{2}}\ll e^{2\alpha\bar{z}-2y}.$

The integral over
$\overline{v_{1}}+\overline{v_{3}}\in[y-10-\bar{z}^{\prime},y-\bar{z}^{\prime}]$
is the same and contributes $\ll e^{2\alpha\bar{z}^{\prime}}e^{-2y}$. The
integral over $\overline{v_{1}}$ is, for
$\bar{z}_{m}=\frac{\bar{z}+\bar{z}^{\prime}}{2}$,

$\ll\int_{\bar{L}_{j}^{\star}}^{\bar{U}_{j^{\star}}-\bar{z}_{m}}(\bar{U}_{j^{\star}}-\bar{z}_{m}-\overline{v_{1}})^{2}e^{2\alpha\overline{v_{1}}}e^{-\frac{\bar{v}_{1}^{2}}{j^{\star}-n_{0}}}{\rm
d}\overline{v_{1}}\ll\frac{e^{2y-2\alpha\bar{z}_{m}-y^{2}/n}}{(j^{\star}\wedge(n-j^{\star}))^{10}},$

where $\bar{U}_{j^{\star}}=y-10\log(j^{\star}\wedge(n-j^{\star}))$ and
$\bar{L}_{j^{\star}}=-10-(j^{\star}\wedge(n-j^{\star}))^{3/4}$.

We now sum over all $j^{*}>n_{0}$ and the ${\rm
O}(e^{-2n_{\mathcal{L}}+j^{\star}})$ pairs with a given $j^{\star}$, so that

$\sum_{h,h^{\prime}:n_{0}<j^{\star}<n_{\mathcal{L}}}\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h))\ll
e^{-2y-y^{2}/n+n_{0}}\int_{\bar{L}_{n_{0}}}^{\bar{U}_{n_{0}}}(\bar{U}_{n_{0}}-\bar{z}_{m})e^{2\alpha\bar{z}_{m}}f(\bar{z}+\alpha
n_{0},\bar{z}^{\prime}+\alpha n_{0}){\rm d}z{\rm d}z^{\prime}\\\
\times\sum_{n_{0}<j^{\star}<n_{\mathcal{L}}}\frac{1}{(j^{\star}\wedge(n-j^{\star}))^{10}}\frac{n^{\frac{3}{2}\frac{j^{\star}}{n}}n^{3(1-\frac{j^{\star}}{n})}}{(j^{\star}-n_{0})^{3/2}(n_{\mathcal{L}}-j^{\star})^{3}}.$

The integral over $z,z^{\prime}$ is over a function of $\bar{z}_{m}$ only,
which has density $\ll e^{-u^{2}/n_{0}}/\sqrt{n_{0}}$ uniformly in
$|u|<100n_{0}$. Moreover, we can simply bound
$|\bar{U}_{n_{0}}-\bar{z}_{m}|\leq y$, hence

$\sum_{h,h^{\prime}:n_{0}<j^{\star}<n_{\mathcal{L}}}\mathbb{P}(\mathfrak{S}(h)\cap\mathfrak{S}(h))\ll
ye^{-2y-y^{2}/n+n_{0}}\int_{\bar{L}_{n_{0}}}^{\bar{U}_{n_{0}}}e^{2\bar{z}_{m}}\frac{e^{-(\bar{z}_{m}+\alpha
n_{0})^{2}/n_{0}}}{\sqrt{n_{0}}}{\rm d}{\bar{z}}_{m}\ll ye^{-2y}e^{-y^{2}/n},$

which concludes the proof of Proposition 11. ∎

##  Appendix A Some Auxiliary Results

Let $(Z_{p},p\text{ prime})$ a sequence of independent and identically
distributed random variables, uniformly distributed on the unit circle
$|z|=1$. For an integer $n$ with prime factorization
$n=p_{1}^{\alpha_{1}}\ldots p_{k}^{\alpha_{k}}$ with $p_{1},\ldots,p_{k}$ all
distinct, consider

$Z_{n}:=\prod_{i=1}^{k}Z_{p_{i}}^{\alpha_{i}}.$

Then we have $\mathbb{E}[Z_{n}\overline{Z}_{m}]=\mathbf{1}_{n=m}$, and
therefore, for an arbitrary sequence $a(n)$ of complex numbers, the following
holds

$\sum_{n\leq N}|a(n)|^{2}=\mathbb{E}\Big{[}\Big{|}\sum_{n\leq
N}a(n)Z_{n}\Big{|}^{2}\Big{]}.$

The next lemma shows that the mean value of Dirichlet polynomial is close to
the one of the above random model. It follows directly from [MonVau1974,
Corollary 3].

######  Lemma 9 (Mean-value theorem for Dirichlet polynomials).

We have,

$\mathbb{E}\Big{[}\Big{|}\sum_{n\leq
N}a(n)n^{\mathrm{i}\tau}\Big{|}^{2}\Big{]}=\Big{(}1+{\rm
O}\Big{(}\frac{N}{T}\Big{)}\Big{)}\sum_{n\leq N}|a(n)|^{2}=\Big{(}1+{\rm
O}\Big{(}\frac{N}{T}\Big{)}\Big{)}\mathbb{E}\Big{[}\Big{|}\sum_{n\leq
N}a(n)Z_{n}\Big{|}^{2}\Big{]}.$

######  Lemma 10 (Exponential moments for the probabilistic model, Lemma 15 in
[ArgBouRad2020]).

Remember the definition (30). There exists an absolute $C>0$ such that for any
$\lambda\in\mathbb{R}$ and $n_{0}\leq j\leq k$ we have

$\mathbb{E}\Big{[}\exp\big{(}\lambda(\mathcal{S}_{k}(h)-\mathcal{S}_{j}(h))\big{)}\Big{]}\leq\exp((k-j+C)\lambda^{2}/4).$

######  Lemma 11 (Gaussian moments of Dirichlet polynomials, Lemma 16 in
[ArgBouRad2020]).

For any $h\in[-1,1]$ and integers $k,j,q$ satisfying $n_{0}\leq j\leq k$,
$2q\leq e^{n-k}$, we have

$\displaystyle\mathbb{E}[|S_{k}(h)-S_{j}(h)|^{2q}]$
$\displaystyle\ll\frac{(2q)!}{2^{q}q!}\,\Big{(}\frac{k-j}{2}\Big{)}^{q}.$ (85)

Moreover, there exists $C>0$ such that for any $0\leq j\leq k$, $2q\leq
e^{n-k}$, we have

$\mathbb{E}[|S_{k}(h)-S_{j}(h)|^{2q}]\ll
q^{1/2}\frac{(2q)!}{2^{q}q!}\,\Big{(}\frac{k-j+C}{2}\Big{)}^{q}.$ (86)

######  Lemma 12 (Gaussian moments of Dirichlet polynomials, Lemma 3 of
[Sou2009]).

Let $2\leq x\leq T$ and $q\in\mathbb{N}$ with $x^{q}\leq T/\log T$. For any
complex numbers $a(p)$, we have

$\mathbb{E}\Big{[}\Big{|}\sum_{p\leq
x}\frac{a(p)}{p^{1/2+\mathrm{i}\tau}}\Big{|}^{2q}\Big{]}\ll
q!\Big{(}\sum_{p\leq x}\frac{|a(p)|^{2}}{p}\Big{)}^{q}.$

######  Lemma 13.

Let $h,h^{\prime}\in[-1,1]$. Consider the increments
$(\mathcal{Y}_{k}(h),\mathcal{Y}_{k}(h^{\prime}))$ for $1\leq k\leq
n_{\mathcal{L}}$, and the corresponding Gaussian vector
$(\mathcal{N}_{k}(h),\mathcal{N}_{k}(h^{\prime}))$, of mean $0$ and with the
covariance given by (10), (11). There exists a constant $c>0$ such that, for
any intervals $A,B$ and $k\geq 1$,

${\mathbb{P}}\Big{(}(\mathcal{Y}_{k}(h),\mathcal{Y}_{k}(h^{\prime}))\in
A\times
B\Big{)}=\mathbb{P}\Big{(}(\mathcal{N}_{k}(h),\mathcal{N}_{k}(h^{\prime}))\in
A\times B\Big{)}+{\rm O}(e^{-ce^{k/2}}).$

###### Proof.

This follows similaly to [ArgBouRad2020, Lemma 20], based on the Berry-Esseen
estimate as stated in [ArgBouRad2020, Lemma 19]. The proof is actually more
immediate because the covariances of $(\mathcal{Y},\mathcal{Y}^{\prime})$ and
$(\mathcal{N},\mathcal{N}^{\prime})$ exactly coincide. ∎

######  Lemma 14.

Let $D$ be a Dirichlet polynomial of length $\leq N$. Then, for any $1\leq
k\leq\log\log N$, we have

$\displaystyle\max_{|h|\leq
e^{-k}}|D(\tfrac{1}{2}+\mathrm{i}t+\mathrm{i}h)|^{2}\ll$
$\displaystyle\sum_{|j|\leq 16e^{-k}\log
N}\Big{|}D\Big{(}\frac{1}{2}+\mathrm{i}t+\frac{2\pi\mathrm{i}j}{8\log
N}\Big{)}\Big{|}^{2}$ (87) $\displaystyle+\sum_{|j|>16e^{-k}\log
N}\frac{1}{1+|j|^{100}}\Big{|}D\Big{(}\frac{1}{2}+\mathrm{i}t+\frac{2\pi\mathrm{i}j}{8\log
N}\Big{)}\Big{|}^{2}.$

###### Proof.

We proceed similarly to the proof of [ArgBouRad2020, Lemma 27], but now with
maxima on intervals of general length $e^{-k}$. With the notations from
[ArgBouRad2020, Lemma 25], we have

$D(\tfrac{1}{2}+\mathrm{i}t+\mathrm{i}h_{0})^{2}=\frac{1}{2+\varepsilon}\sum_{h\in\frac{2\pi\mathbb{Z}}{(2+\varepsilon)\log
N}}D(\tfrac{1}{2}+\mathrm{i}t+\mathrm{i}h)^{2}\widehat{V}\big{(}\frac{(h-h_{0})\log
N}{2\pi}\big{)}.$

Using the triangle inequality and the decay
$\widehat{V}(x)\ll_{A}(1+|x|)^{-A}$ we obtain the result. ∎

##  Appendix B Ballot Theorem

###  B.1. Results

Most ideas for the results in this section are due to Bramson. As we could not
find the exact barrier estimates needed in our setting, this section gives a
self-contained and quantitative analogues of some technical results in
[Bra1978, Bra1983] in the setting of Gaussian random walk with arbitrary,
comparable, variance of the increments.

Let $\kappa>0$ be fixed in all this section, and $(X_{i})_{i\geq 1}$ be
independent, real, centered Gaussian randon variables such that
$\kappa<\mathbb{E}[|X_{i}|^{2}]<\kappa^{-1}$ for all $i$. For $k\in\mathbb{N}$
we denote $S_{k}=\sum_{i\leq k}X_{i}$.

We denote $\mathbb{P}_{(s,x)}$ for the distribution of the process
$(S_{k})_{k}$ starting at time $s$ from $x$,
$\mathbb{P}_{x}=\mathbb{P}_{(0,x)}$, $\mathbb{P}=\mathbb{P}_{0}$, and
$\mathbb{P}_{(s,x)}^{(t,y)}$ for the distribution for $(S_{k})_{k}$ starting
at time $s$ from $x$, and conditioned to end at time $t$ at point $y$.

######  Proposition 14.

Let $\delta>1/2>\alpha>0$. Then there exists $c=c(\alpha,\delta,\kappa)$ such
that uniformly in the time $t\geq 1$, $10\leq y\leq t^{1/10}$, $a,b\in[1,y-1]$
and uniformly in the functions $v_{s}\geq y+\min(s,t-s)^{\delta}$,
$|u_{s}|\leq\min(s,t-s)^{\alpha}$, we have

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq k\leq t}\\{u_{k}\leq S_{k}\leq
v_{k}\\}\right)=\frac{2ab}{\sigma}\cdot\left(1+{\rm
O}_{\alpha,\delta,\kappa}(d^{-c})\right)$

where $d=\min(|y-a|,|y-b|,|a|,|b|)$ and $\sigma=\sum_{k\leq
t}\mathbb{E}[X_{k}^{2}]$.

###  B.2. Preliminaries on Brownian motion

We denote $\mathbb{P}_{(s,x)}$ for the distribution of the Brownian motion
starting at time $s$ from $x$, $\mathbb{P}_{x}=\mathbb{P}_{(0,x)}$,
$\mathbb{P}=\mathbb{P}_{0}$, and $\mathbb{P}_{(s,x)}^{(t,y)}$ for the
distribution for the Brownian bridge starting at time $s$ from $x$, ending at
time $t$ at point $y$. Context will avoid confusion with the notation
$\mathbb{P}$ from Proposition 14 as the Gaussian random walk will always be
denoted $S$, and the Brownian motion $B$.

For such a trajectory $B$, let $M_{t}=\max_{0\leq s\leq t}B_{s}$,
$m_{t}=\min_{0\leq s\leq t}B_{s}$.

######  Lemma 15.

Let $x,y>0$. Then

$\mathbb{P}_{(0,x)}^{(t,y)}\left(m_{t}\geq 0\right)=1-e^{-\frac{2xy}{t}}.$

###### Proof.

From the reflection principle, for any measurable $A\subset(0,\infty)$,

$\mathbb{P}_{x}\left(m_{t}\geq 0,B_{t}\in
A\right)=\mathbb{P}_{x}\left(B_{t}\in
A\right)-\mathbb{P}_{x}\left(B_{t}\in-A\right).$

This implies

$\mathbb{P}_{(0,x)}^{(t,y)}\left(m_{t}\geq 0\right)=1-\lim_{\varepsilon\to
0}\frac{\mathbb{P}_{x}\left(B_{t}\in-[y,y+\varepsilon]\right)}{\mathbb{P}_{x}\left(B_{t}\in[y,y+\varepsilon]\right)}=1-e^{-\frac{2xy}{t}},$

concluding the proof. ∎

######  Lemma 16.

Let $a,c>0$ and $A\subset[-c,a]$ be measurable. Then

$\mathbb{P}\left(M_{t}\leq a,m_{t}\geq-c,B_{t}\in
A\right)\geq\mathbb{P}\left(m_{t}\geq-c,B_{t}\in
A\right)-\mathbb{P}\left(B_{t}\in A-2a\right).$

###### Proof.

The above left-hand side is

$\mathbb{P}\left(m_{t}\geq-c,B_{t}\in A\right)-\mathbb{P}\left(M_{t}\geq
a,m_{t}\geq-c,B_{t}\in A\right)\geq\mathbb{P}\left(m_{t}\geq-c,B_{t}\in
A\right)-\mathbb{P}\left(M_{t}\geq a,B_{t}\in A\right).$

From the reflection principle, this last probability is
$\mathbb{P}\left(B_{t}\in 2a-A\right)$. ∎

######  Lemma 17.

Let $\delta>1/2$, $v_{s}\geq y+\min(s,t-s)^{\delta}$. Let
$c\in(0,2-\frac{1}{\delta})$. Then uniformly in $t\geq 0$, $2\leq y\leq
t^{1/10}$, $a,b\in[1,y-1]$ we have

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq s\leq t}\\{0\leq B_{s}\leq
v_{s}\\}\right)=\mathbb{P}_{(0,a)}^{(t,b)}\left(m_{t}\geq
0\right)\cdot\left(1+{\rm O}(e^{-\min(|y-a|,|y-b|)^{c}})\right).$

###### Proof.

In this proof we abbreviate $B\geq 0$ for $m_{t}\geq 0$ and start with

$\displaystyle\mathbb{P}_{(0,a)}^{(t,b)}\left(B\geq 0,\exists
s:B_{s}>v_{s}\right)\leq$
$\displaystyle\sum_{k=0}^{t/2}\left(\mathbb{P}_{(0,a)}^{(t,b)}\left(B\geq
0,\exists s\in[k,k+1]:B_{s}>v_{k}\right)+\mathbb{P}_{(0,b)}^{(t,a)}\left(B\geq
0,\exists s\in[k,k+1]:B_{s}>v_{k}\right).\right)$

The first probability above is smaller than

$\mathbb{P}_{(0,a)}^{(t,b)}\left(B\geq 0,\exists
s\in[0,k+1]:B_{s}>v_{k}\right)=\mathbb{P}_{(0,a)}^{(t,b)}\left(B\geq
0\right)-\mathbb{P}_{(0,a)}^{(t,b)}\left(B\geq
0,\max_{[0,k+1]}B<v_{k}\right).$

We write

$\mathbb{P}_{(0,a)}^{(t,b)}\left(B\geq 0,\max_{[0,k+1]}B<v_{k}\right)\\\
=\int_{0}^{v_{k}}\mathbb{P}_{(0,a)}^{(k+1,x)}\left(B\geq
0,\max_{[0,k+1]}B<v_{k}\right)\mathbb{P}_{(k+1,x)}^{(t,b)}\left(B\geq
0\right)\mathbb{P}_{(0,a)}^{(t,b)}(B_{k+1}\in{\rm d}x).$

The first probability in this integral is estimated with Lemma 16:

$\mathbb{P}_{(0,a)}^{(k+1,x)}\left(B\geq
0,\max_{[0,k+1]}B<v_{k}\right)\geq\mathbb{P}_{(0,a)}^{(k+1,x)}\left(B\geq
0\right)-e^{-\frac{2(v_{k}-a)(v_{k}-x)}{k+1}}.$

This gives

$\mathbb{P}_{(0,a)}^{(t,b)}\left(B\geq 0,\exists
s\in[k,k+1]:B_{s}>v_{k}\right)\\\
\leq\int_{0}^{v_{k}}e^{-\frac{2(v_{k}-a)(v_{k}-x)}{k+1}}\mathbb{P}_{(k+1,x)}^{(t,b)}\left(B\geq
0\right)\mathbb{P}_{(0,a)}^{(t,b)}(B_{k+1}\in{\rm d}x).$

From Lemma 15, we have $\mathbb{P}_{(k+1,x)}^{(t,b)}\left(B\geq 0\right)\leq
5\frac{xb}{t}$. Moreover, $\mathbb{P}_{(0,a)}^{(t,b)}(B_{s}\in{\rm
d}x)=\mathbb{P}_{(0,0)}^{(t,0)}(B_{s}\in{\rm
d}x+x_{s})=\frac{e^{-(x-x_{s})^{2}/(2w_{s})}}{\sqrt{2\pi w_{s}}}{\rm d}x$
where $w_{s}=s(t-s)/t$, $x_{s}=(1-\frac{s}{t})a+\frac{s}{t}b$. This gives

$\mathbb{P}_{(0,a)}^{(t,b)}\left(B\geq 0,\exists
s\in[k,k+1]:B_{s}>v_{k}\right)\leq
C\frac{b}{t}\int_{0}^{v_{k}}e^{-\frac{2(v_{k}-a)(v_{k}-x)}{k+1}}x\frac{e^{-\frac{(x-x_{k+1})^{2}}{2w_{k+1}}}}{\sqrt{w_{k+1}}}{\rm
d}x.$

In the above integral, the contribution from $|x-x_{k+1}|>v_{k}/3$ is bounded
with

$\int_{|x-x_{k+1}|>v_{k}/3}(x_{k+1}+|x-x_{k+1}|)\frac{e^{-\frac{(x-x_{k+1})^{2}}{2w_{k+1}}}}{\sqrt{w_{k+1}}}{\rm
d}x\leq Cx_{k+1}e^{-\frac{v_{k}^{2}}{100w_{k+1}}}.$ (88)

The regime $|x-x_{k+1}|<v_{k}/3$ gives the error

$\int_{|x-x_{k+1}|<v_{k}/3}e^{-\frac{(k^{\delta}+|y-a|)k^{\delta}}{k+1}}x\frac{e^{-\frac{(x-x_{k+1})^{2}}{2w_{k+1}}}}{\sqrt{w_{k+1}}}{\rm
d}x\leq Cx_{k+1}e^{-\frac{(k^{\delta}+|y-a|)k^{\delta}}{k+1}}.$ (89)

We first bound the sum of the error terms from (89) as $0\leq k\leq t/2$. For
$k^{\delta}<|y-a|$, from the hypothesis $y<t^{1/10}$ we have $x_{k+1}<a+1<2a$,
so that

$\sum_{k^{\delta}<|y-a|}x_{k+1}e^{-\frac{(k^{\delta}+|y-a|)k^{\delta}}{k+1}}\leq
2a\sum_{k^{\delta}<|y-a|}e^{-\frac{|y-a|k^{\delta}}{k+1}}\leq
2a|y-a|^{1/\delta}e^{-|y-a|^{2-\frac{1}{\delta}}}=a\,{\rm O}(e^{-|y-a|^{c}}).$

For $k^{\delta}>|y-a|$, we obtain

$\sum_{k\geq|y-a|^{1/\delta}}x_{k}e^{-k^{2\delta-1}}\leq\sum_{k\geq|y-a|^{1/\delta}}(a+\frac{kb}{t})e^{-k^{2\delta-1}}\leq(a+1)\sum_{k\geq|y-a|^{1/\delta}}ke^{-k^{2\delta-1}}\\\
\leq
C_{\delta}(a+1)\int_{v>|y-a|^{2-\frac{1}{\delta}}}v^{\frac{3-2\delta}{2\delta-1}}e^{-v}{\rm
d}v=a\,{\rm O}(e^{-|y-a|^{c}}).$

The same estimate can be obtained for the sum over $k$ from (88). We have thus
obtained

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq s\leq t}\\{0\leq B_{s}\leq
v_{s}\\}\right)=\mathbb{P}_{(0,a)}^{(t,b)}\left(m_{t}\geq 0\right)+{\rm
O}\left(\frac{ab}{t}e^{-\min(|y-a|,|y-b|)^{c}}\right).$

The result follows from the above estimate and Lemma 15. ∎

######  Lemma 18.

Let $\delta>1/2>\alpha>0$. Then there exists $c=c(\alpha,\delta)$, such that,
uniformly in $t\geq 0$, $y\geq 10$, $a,b\in[1,y-1]$, and uniformly in the
functions $v_{s}\geq y+\min(s,t-s)^{\delta}$,
$|u_{s}|\leq\min(s,t-s)^{\alpha}$, we have

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq s\leq t}\\{u_{s}\leq B_{s}\leq
v_{s}\\}\right)\geq\mathbb{P}_{(0,a)}^{(t,b)}\left(m_{t}\geq
0\right)\cdot\left(1+{\rm O}(d^{-c})\right),$

where $d=\min(|y-a|,|y-b|,|a|,|b|)$.

###### Proof.

Without loss of generality we can assume $\alpha+\delta<1$. We also pick
$\varepsilon\in(0,1)$ and write $d_{0}=d^{1-\varepsilon}$. Let $s_{1},s_{2}$
be the solutions of $s_{1}^{\alpha}=d_{0}$, $(t-s_{2})^{\alpha}=d_{0}$. Let

$\bar{u}_{s}=(d_{0}+(s-s_{1})\alpha
s_{1}^{\alpha-1})\mathbf{1}_{s^{\alpha}<d_{0}}+(d_{0}-(s-s_{2})\alpha(t-s_{2})^{\alpha-1})\mathbf{1}_{(t-s)^{\alpha}<d_{0}}+\min(s,t-s)^{\alpha}\mathbf{1}_{s_{1}<s<s_{2}}.$
(90)

In other words, $\bar{u}$ is the function coinciding with
$\min(s,t-s)^{\alpha}$ on $(s_{0},s_{1})$ and linearly expanded on the
complement, with continuous derivative. Note that
$\bar{u}_{0}=\bar{u}_{t}=(1-\alpha)d_{0}$. We also denote
$\bar{v}_{s}=(1-\frac{s}{t})a+\frac{s}{t}b+\min(s,t-s)^{\delta}+d$. We have

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq s\leq t}\\{u_{s}\leq B_{s}\leq
v_{s}\\}\right)\geq\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq s\leq
t}\\{\bar{u}_{s}\leq B_{s}\leq\bar{v}_{s}\\}\right).$

The Cameron-Martin formula gives

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq s\leq t}\\{\bar{u}_{s}\leq
B_{s}\leq\bar{v}_{s}\\}\right)=\mathbb{E}_{(0,a)}^{(t,b)}\left[e^{-\int_{0}^{t}\dot{\bar{u}}_{s}{\rm
d}B_{s}-\frac{1}{2}\int_{0}^{t}\dot{\bar{u}}_{s}^{2}{\rm
d}s}\mathbf{1}_{\cap_{0\leq s\leq t}\\{d^{1-\varepsilon}\leq
B_{s}\leq\bar{v}_{s}-\int_{0}^{u}\dot{\bar{u}}_{u}{\rm d}u\\}}\right],$

where we denote the derivative in $s$ of $f$ by $\dot{f}$. We now bound both
terms in the measure bias, deterministically. First,

$\int_{0}^{t}\dot{\bar{u}}_{s}^{2}{\rm d}s\leq
C\int_{s>s_{1}}s^{2\alpha-2}{\rm d}s+C\int_{s<s_{1}}s_{1}^{2\alpha-2}{\rm
d}s\leq Cd^{-(\frac{1}{\alpha}-2)(1-\varepsilon)}.$

Moreover, by integration by parts we have (using the fact that $\bar{u}$ has
continuous derivative)

$-\int_{0}^{t}\dot{\bar{u}}_{s}{\rm
d}B_{s}=\int_{0}^{t}B_{s}\ddot{\bar{u}}_{s}{\rm
d}s-B_{t}\dot{\bar{u}}_{t}+B_{0}\dot{\bar{u}}_{0}=\int_{0}^{t}\big{(}B_{s}-((1-\frac{s}{t})B_{0}+\frac{s}{t}B_{t})\big{)}\ddot{\bar{u}}_{s}{\rm
d}s.$

On the set $\cap_{0\leq s\leq t}\\{B_{s}\leq{\bar{v}}_{s}\\}$, we therefore
have the deterministic bound

$-\int_{0}^{t}\dot{\bar{u}}_{s}{\rm
d}B_{s}\geq-\int_{s^{\alpha}>d^{1-\varepsilon}}(d^{1-\varepsilon}+s^{\delta})s^{\alpha-2}{\rm
d}s\geq-C(d^{-(\frac{1}{\alpha}-2)(1-\varepsilon)}+d^{(\frac{\delta}{\alpha}+1-\frac{1}{\alpha})(1-\varepsilon)}).$

We have therefore proved

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq s\leq t}\\{\bar{u}_{s}\leq
B_{s}\leq\bar{v}_{s}\\}\right)\\\ \geq\exp\left({\rm
O}(d^{-(\frac{1}{\alpha}-2)(1-\varepsilon)}+d^{-\frac{1-\alpha-\delta}{\alpha}(1-\varepsilon)})\right)\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq
s\leq t}\\{d^{1-\varepsilon}\leq B_{s}\leq
v_{s}-\int_{0}^{s}\dot{\bar{u}}_{u}{\rm d}u\\}\right).$

The desired lower bound follows by using Lemma 17, noting that
$\frac{(a-d^{1-\varepsilon})(b-d^{1-\varepsilon})}{t}=\frac{ab}{t}(1+{\rm
O}(d^{-\varepsilon}))$. ∎

###  B.3. Proofs of barrier estimates for the random walk, Proposition 14.

The lower bound is a direct consequence of Lemma 18 and Lemma 15.

For the upper bound, we only need to bound
$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq k\leq t}\\{-\bar{u}_{k}\leq
S_{k}\\}\right)$, where $\bar{u}$ is defined in (90). By the change of
variables $S_{k}=\tilde{S}_{k}-\sum_{0\leq i\leq
k-1}(\bar{u}_{i+1}-\bar{u}_{i})$, and denoting
$d_{1}=(1-\alpha)d^{1-\varepsilon}$, we obtain

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq k\leq t}\\{-\bar{u}_{k}\leq
S_{k}\\}\right)=\mathbb{E}_{(0,a)}^{(t,b)}\left[e^{-\frac{1}{2}\sum_{k}(\bar{u}_{k+1}-\bar{u}_{k}))^{2}+\sum_{k}(S_{k+1}-S_{k})(\bar{u}_{k+1}-\bar{u}_{k})}\mathbf{1}_{\cap_{0\leq
k\leq t}\\{-d_{1}\leq S_{k}\\}}\right]\\\
\leq\mathbb{E}_{(0,a)}^{(t,b)}\left[e^{\sum_{k}(S_{k+1}-S_{k})(\bar{u}_{k+1}-\bar{u}_{k})}\mathbf{1}_{\cap_{0\leq
k\leq t}\\{-d_{1}\leq S_{k}\\}}\right].$

Let $\bar{S}_{s}=S_{s}-(a\frac{t-s}{t}+b\frac{s}{t})$. We have

$\sum_{k}(S_{k+1}-S_{k})(\bar{u}_{k+1}-\bar{u}_{k})=\sum_{k}(\bar{S}_{k+1}-\bar{S}_{k})(\bar{u}_{k+1}-\bar{u}_{k})=\sum_{k}a_{k}\bar{S}_{k},$

where the constants $a_{k}$ satisfy $0\leq a_{k}\leq
10\min(k,t-k+1)^{\alpha-2}$ and vanish outside $[d_{2},t-d_{2}]$ where we
define $d_{2}=d^{\frac{1-\varepsilon}{\alpha}}$. As $ab\leq\max(a^{2},b^{2})$,
we have obtained

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq k\leq t}\\{-\bar{u}_{k}\leq
S_{k}\\}\right)\leq\mathbb{E}_{(0,a)}^{(t,b)}\left[e^{2\sum_{k\leq
t/2}a_{k}\bar{S}_{k}}\mathbf{1}_{\cap_{0\leq k\leq t}\\{-d_{1}\leq
S_{k}\\}}\right].$

Let $\varepsilon_{0}\in(0,\frac{1}{2}-\alpha)$ and define
$\kappa=\frac{1-\varepsilon}{2\alpha}(\frac{1}{2}-\alpha-\varepsilon_{0})$.
Note that for any integer $v\geq 1$, $\sum_{k\leq
t/2}a_{k}\bar{S}_{k}>vd^{-\kappa}$ implies that there exists $k$ such that
$d_{2}\leq k\leq t/2$ such that
$\bar{S}_{k}>vk^{\frac{1}{2}+\varepsilon_{0}}$. This observation together with
the union bound gives

$\displaystyle\mathbb{E}_{(0,a)}^{(t,b)}\left[e^{2\sum_{k\leq
t/2}a_{k}\bar{S}_{k}}\mathbf{1}_{\cap_{0\leq k\leq t}\\{-d_{1}\leq
S_{k}\\}}\right]-\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq k\leq
t}\\{-d_{1}\leq S_{k}\\}\right)(1+d^{-\kappa})$ $\displaystyle\leq$
$\displaystyle\sum_{v\geq 1,d_{0}\leq k\leq t/2,w\geq
vk^{1/2+\varepsilon_{0}}}e^{vd^{-\kappa}}\,\mathbb{P}_{(0,a)}^{(t,b)}\left(\\{\bar{S}_{k}\in[w,w+1]\\}\cap_{j\leq
k}\\{S_{j}\geq-d_{1})\\}\right)$ $\displaystyle\ll$ $\displaystyle\sum_{v\geq
1,d_{0}\leq k\leq t/2,w\geq
vk^{1/2+\varepsilon_{0}}}e^{vd^{-\kappa}}\,\mathbb{P}_{(0,a)}^{(t,b)}\left(\bar{S}_{k}\in[w,w+1]\right)\times$
(91) $\displaystyle\hskip
142.26378pt\Big{(}\sup_{c\in[w,w+1]+a\frac{t-k}{t}+b\frac{k}{t}}\mathbb{P}_{(0,c)}^{(t-k,b)}\left(\cap_{1\leq
j\leq t-k}\\{S_{j}\geq-d_{1}\\}\right)\Big{)}$ (92)

where we used the Markov property for the second inequality. To bound the
first probability above, note that under $\mathbb{P}_{(0,a)}^{(t,b)}$, the
random variable $\bar{S}_{k}$ is centered, Gaussian with variance
$k-\frac{k^{2}}{t}\asymp k$. For the second probability, from Lemma 20, we
have uniformly in all parameters

$\mathbb{P}_{(0,c)}^{(t-k,b)}\left(\cap_{1\leq j\leq t-k}\\{S_{j}\geq-
d_{1}\\}\right)\ll\frac{(w+a\frac{t-k}{t}+b\frac{k}{t})b}{t}.$ (93)

This allows to bound the left-hand side of (92) with

$\sum_{v\geq 1,d_{0}\leq k\leq t/2,w\geq
vk^{1/2+\varepsilon_{0}}}e^{vd^{-\kappa}-c\frac{w^{2}}{k}}\cdot\frac{(w+a\frac{t-k}{t}+b\frac{k}{t})b}{t}$

for some absolute $c>0$. The above sum over $w$ and then $v$ is $\ll
e^{-c^{\prime}k^{2\varepsilon_{0}}}$ for some absolute $c^{\prime}>0$. We
conclude that uniformly in our parameters, the left-hand side of (92) is
bounded with

$\frac{b}{t}\sum_{d_{0}\leq k\leq
t/2}e^{-c^{\prime}k^{2\varepsilon_{0}}}(1+a\frac{t-k}{t}+b\frac{k}{t})\ll\frac{ab}{t}e^{-c^{\prime}d_{0}^{2\varepsilon_{0}}}\ll
d^{-\kappa}\frac{ab}{t}.$

Finally, Lemma 21 yields

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq k\leq t}\\{-d_{1}\leq
S_{k}\\}\right)=2\frac{ab}{\sigma}(1+{\rm O}(d^{-c})).$

This concludes the proof.

######  Lemma 19.

There exists $C=C(\kappa)>0$ such that for any $t\geq 10$, $a\geq 1$, we have

$\mathbb{P}_{(0,a)}\left(\cap_{0\leq k\leq t}\\{S_{k}\geq 0\\}\right)\leq
C\frac{a}{\sqrt{t}}.$ (94)

###### Proof.

By monotonicity in $a$, we can consider $a\in\mathbb{N}$ without loss of
generality . Moreover, we have

$\mathbb{P}_{(0,a)}\left(\cap_{0\leq k\leq t}\\{S_{k}\geq
0\\}\right)\leq\mathbb{P}_{(0,a)}\left(\cap_{0\leq
k\leq\frac{t}{a^{2}}}\\{\frac{S_{ka^{2}}}{a}\geq
0\\}\right)=\mathbb{P}_{(0,1)}\left(\cap_{0\leq
k\leq\frac{t}{a^{2}}}\\{\tilde{S}_{k}\geq 0\\}\right),$

where $\tilde{S}_{k}:=S_{ka^{2}}/a$ has independent Gaussian increments with
variance in $[\kappa,\kappa^{-1}]$, like $S$. This proves that (94) only needs
to be proved for $a=1$.

Consider $T=\min(t,\min\\{k\leq t\mid S_{k}\leq 0\\})$. By the stopping time
theorem, $\mathbb{E}[S_{T}]=1,$ so that

$\mathbb{E}[|S_{t}|\mathbf{1}_{T\geq
t}]=1+\mathbb{E}[-S_{T}\mathbf{1}_{T<t}].$

Moreover, we have the correlation inequality

$\mathbb{E}[|S_{t}|\mathbf{1}_{T>t}]\geq\mathbb{E}[\max(0,S_{t})]\mathbb{P}[T>t],$
(95)

which is a simple consequence of the Harris inequality. Indeed, consider the
random walk $Z^{(n)}_{k}=1+\frac{1}{\sqrt{n}}\sum_{1\leq j\leq
kn}\varepsilon_{j}$ with independent Bernoulli random variables
$\varepsilon_{k}$, and $I^{(n)}=\\{\lfloor n{\rm Var}S_{k}\rfloor,k\geq 1\\}$;
denote $U^{(n)}=\min\\{k\in I^{(n)}:Z^{(n)}_{k}\leq 0\\}$. Then the functions
$\mathbf{1}_{U^{(n)}>t}$ and $\max(0,Z^{(n)}_{t})$ are non-decreasing
functions of $(\varepsilon_{k})_{k\leq nt}$, so that
$\mathbb{E}[\max(0,Z^{(n)}_{t})\mathbf{1}_{U^{(n)}>t}]\geq\mathbb{E}[\max(0,Z^{(n)}_{t})]\cdot\mathbb{P}[U^{(n)}>t]$.
This implies (95) by taking $n\to\infty$.

We have obtained

$\mathbb{P}[T>t]\leq C\frac{1+\mathbb{E}[-S_{T}\mathbf{1}_{T<t}]}{\sqrt{t}},$
(96)

and we will now prove that

$\mathbb{E}[-S_{T}\mathbf{1}_{T<t}]\leq C_{1},$ (97)

for some $C_{1}>0$ uniform in $t$, which together with (96) will conclude the
proof of (94).

To prove (97), first note that
$\mathbb{E}[-S_{T}\mathbf{1}_{T<t}]\leq\mathbb{E}[-S_{T_{0}}]$ where
$T_{0}=\min\\{k\geq 0\mid S_{k}\leq 0\\}$. We now consider

$Z_{n}=\sum_{k\geq 0}\mathbf{1}_{S_{k}\in[n,n+1),k<T_{0}},\ n\geq 0,$

the time spent by $S$ in $[n,n+1)$ before it hits $0$. Define $U_{0}=0$ and
$U_{i+1}=\min\\{u\geq U_{i}+n^{2}:S_{u}\in[n,n+1)\\}$. For any $\lambda\geq
n^{2}$ we have the inclusion

$\\{Z_{n}\geq\lambda\\}\subset\cap_{i\leq\lambda/n^{2}}\left\\{S_{U_{i}+n^{2}}-S_{U_{i}}\geq-(n+1)\right\\}.$

By the strong Markov property the events on the right-hand side are
independent, and there exists $\alpha=\alpha(\kappa)$ such that each such
event has probability at most $1-\alpha$, uniformly in $n$. This implies
$\mathbb{E}(Z_{n})\leq Cn^{2}$ for some $C=C(\kappa)$, a key estimate in the
last inequality below: for any $\ell\geq 1$ we have

$\mathbb{P}[|S_{T_{0}}|\geq\ell]\leq\sum_{k,n\geq
0}\mathbb{P}[S_{k}\in[n,n+1),|S_{k+1}-S_{k}|\geq\ell+(n+1),k<T_{0}]\\\
=\sum_{k,n\geq
0}\mathbb{P}[S_{k}\in[n,n+1),k<T_{0}]\cdot\mathbb{P}[|S_{k+1}-S_{k}|\geq\ell+(n+1)]\leq\sum_{n\geq
0}e^{-c(\ell+n)^{2}}\mathbb{E}[Z_{n}]\leq Ce^{-c\ell^{2}},$

which immediately implies $\mathbb{E}[-S_{T_{0}}]\leq C_{1}$ and concludes the
proof. ∎

######  Lemma 20.

With the same notations as Proposition 14, there exists $C=C(\kappa)>0$ such
that for any $t\geq 10$, $a,b\geq 1$ we have

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq k\leq t}\\{S_{k}\geq
0\\}\right)\leq C\frac{ab}{t}.$ (98)

###### Proof.

We first assume that $a,b\leq\sqrt{t}$. Let $n_{1}=\lfloor t/3\rfloor$ and
$n_{2}=\lfloor 2t/3\rfloor$, and $p_{t}(x)=e^{-x^{2}/(2t)}/\sqrt{2\pi t}$, and
abbreviate $\mathbb{P}_{(0,a)}^{(t,b)}=\mathbb{P}_{(0,a)}^{(t,b)}(S>0)$. Then
$\mathbb{P}_{(0,a)}^{(t,b)}(S>0)$ is equal to

$\frac{1}{p_{t}(a-b)}\iint_{x_{1},x_{2}>0}p_{n_{1}}(x_{1}-a)\mathbb{P}_{(0,a)}^{(n_{1},x_{1})}p_{n_{2}-n_{1}}(x_{2}-x_{1})\mathbb{P}_{(n_{1},x_{1})}^{(n_{2},x_{2})}p_{t-n_{2}}(b-x_{2})\mathbb{P}_{(n_{2},x_{2})}^{(t,b)}{\rm
d}x_{1}{\rm d}x_{2}\\\ \leq
C\int_{x_{1}}p_{n_{1}}(x_{1}-a)\mathbb{P}_{(0,a)}^{(n_{1},x_{1})}{\rm
d}x_{1}\int_{x_{2}>0}p_{t-n_{2}}(b-x_{2})\mathbb{P}_{(n_{2},x_{2})}^{(t,b)}{\rm
d}x_{2}\leq C\frac{ab}{t},$

where we have used the trivial bounds
$\mathbb{P}_{(n_{1},x_{1})}^{(n_{2},x_{2})}\leq 1$,
$p_{n_{2}-n_{1}}(x_{2}-x_{1})\leq Ct^{-1/2}$, the estimate (valid for
$a,b\leq\sqrt{t}$) $(p_{t}(a-b))^{-1}\leq C\sqrt{t}$, and Lemma 19.

For the general case, we can assume $a<\sqrt{t}<b$ and $ab<t$. Let $B$ be a
Brownian bridge from $a$ ($s=0$) to $b$ ($s=\sigma$). There exists
$s_{1}<\dots<s_{t}=\sigma$ such that $(S_{k})_{k\leq t}$ and
$(B_{s_{k}})_{k\leq t}$ have the same distribution. Moreover, from [Bra1983,
pages 21, 22], by a simple coupling argument the function

$b\mapsto\mathbb{P}_{(a,0)}^{(b,t)}(B_{s}>0,0\leq s\leq\sigma\mid
B_{s_{i}}>0,0\leq i\leq t)\ \mbox{ is non-decreasing}.$ (99)

This implies

$\mathbb{P}_{(a,0)}^{(b,t)}(B_{s_{i}}\geq 0,1\leq i\leq
t)=\frac{\mathbb{P}_{(a,0)}^{(b,t)}(B_{s_{i}}\geq 0,1\leq i\leq
t)}{\mathbb{P}_{(a,0)}^{(b,t)}(B_{s}\geq 0,0\leq
s\leq\sigma)}\mathbb{P}_{(a,0)}^{(b,t)}(B_{s}\geq 0,0\leq s\leq\sigma)\\\
\leq\frac{\mathbb{P}_{(a,0)}^{(\sqrt{t},t)}(B_{s_{i}}\geq 0,1\leq i\leq
t)}{\mathbb{P}_{(a,0)}^{(\sqrt{t},t)}(B_{s}\geq 0,0\leq
s\leq\sigma)}\mathbb{P}_{(a,0)}^{(b,t)}(B_{s}\geq 0,0\leq s\leq\sigma).$

From Lemma 15, the denominator is lower-bounded with $c\frac{a\sqrt{t}}{t}$
and the last probability is upper-bounded with $\frac{ab}{t}$. And from the
previously discussed case, the numerator is at most $\frac{a\sqrt{t}}{t}$.
This concludes the proof. ∎

######  Lemma 21.

With the same notations as Proposition 14, there exists $C=C(\kappa)>0$ such
that for any $t\geq 10$, $y\leq t^{1/10}$, $1\leq a,b\leq y-1$ we have

$\mathbb{P}_{(0,a)}^{(t,b)}\left(\cap_{0\leq k\leq t}\\{S_{k}\geq
0\\}\right)=\frac{2ab}{\sigma}\cdot\left(1+{\rm
O}_{\alpha,\delta,\kappa}(d^{-c})\right).$

###### Proof.

The lower bound follows directly from Lemma 15.

For the upper bound, consider first the case $a=b=d$. Note that for any $k\leq
t-1$ and $u,v>0$, under $\mathbb{P}_{(k,u)}^{(k+1,v)}$ we can decompose