\[ \lim_{n\to \infty} \frac{1}{n^{1/3}} \log Z_{n,h}^{\omega,\beta} = -\frac{3}{2} (\pi h)^{2/3}, \quad \forall \eps > 0, \mathbf{P}_{n,h}^{\omega,\beta}\left( \Big| n^{-1/3}|\mathcal{R}_n| - c_h \Big| > \eps \right) \xrightarrow[n \to \infty]{} 0 \, . \]

Our result here is to obtain the second order asymptotic for the convergence of $\log Z_{n,h}^{\omega,\beta}$; we deduce a result on the position of the range under $\mathbf{P}_{n,h}^{\omega,\beta}$.

Suppose that $(\omega_z)_{z\in \mathbb Z}$ verifies (<ref>).
Then, for any $h,\beta > 0$, we have the following $\PP$-a.s. convergence

\[ \lim_{n\to \infty} \frac{1}{\beta n^{1/3\alpha}} \left( \log Z_{n,h}^{\omega,\beta} + \frac{3}{2}hc_hn^{1/3} \right) = \sup_{0 \leq u \leq c_h} \left\{ X^{(1)}_u + X^{(2)}_{c_h-u} \right\} \, , \]

where $X^{(1)}$ and $X^{(2)}$ are two independent $\alpha$-stable Lévy processes.

Furthermore, $u_* \defeq \argmax_{u \in [0,c_h]} \left\{ X^{(1)}_u + X^{(2)}_{c_h-u} \right\}$ exists $\PP$-almost surely and

\[ \forall \eps > 0, \mathbf{P}_{n,h}^{\omega, \beta} \left( \Big|\frac{1}{n^{1/3}} (M_n^-,M_n^+) - (-u_*,c_h-u_*) \Big| > \eps \right) \xrightarrow[n \to \infty]{} 0 \qquad \text{$\PP$-a.s.}  \]

The proof is essentially the same as the one of Theorem <ref>.
As in (<ref>), we can write

\[ \log Z_{n,h}^{\omega,\beta} + \frac{3}{2} h c_hn^{1/3} = \log \big(1 + \bar{o}(1)\big)\psi_h + \log \sum_{k_1 = 0}^{c_h/\delta} \sum_{k_2 = \frac{c_h}{\delta} - k_1 - 1}^{\frac{c_h}{\delta} - k_1} Z_{n,h}^{\omega,\beta} (k_1,k_2,\delta) \, , \]

with $Z_{n,h}^{\omega,\beta} (k_1,k_2,\delta)$ defined as in (<ref>). Once again we have

\begin{equation}
\Big| \sum_{z = -x}^y \omega_z - \left( \Sigma^+_{k_2\delta n^{1/3}} + \Sigma^-_{k_1\delta n^{1/3}} \right) \Big| \leq R_n^\delta(k_1\delta, k_2\delta) \, ,
\end{equation}

where the error remainer $R_n^\delta$ is defined for $u,v \geq 0$ by

\[ R_n^\delta(u,v) \defeq \max_{un^{1/3} + 1 \leq j \leq (u+\delta)n^{1/3}-1} \big| \Sigma^-_j - \Sigma^-_{un^{1/3}} \big| + \max_{vn^{1/3} + 1 \leq j \leq (v+\delta)n^{1/3}-1} \big| \Sigma^+_j - \Sigma^+_{vn^{1/3}} \big| \, .
\]

Using the coupling $\hat{\omega}$ and Lemma A.5 of <cit.>, we have $\PP-a.s.$ $\forall \eps > 0$, $\exists n_0 = n_0(\eps,\delta,\omega)$ such that $\forall n \geq n_0$,

\[ \frac{1}{n^{1/3\alpha}} R_n^\delta(u,v) \leq \eps + \sup_{u \leq u' \leq u + \eps + \delta} \big| X^{(1)}_{u'}-X^{(1)}_u \big| + \sup_{v \leq v' \leq v + \eps + \delta} \big| X^{(2)}_{v'}-X^{(2)}_v \big| \]

\[ \left( \left| \frac{1}{n^{1/3\alpha}}\Sigma^+_{vn^{1/3}} - X^{(2)}_v \right| \vee \left| \frac{1}{n^{1/3\alpha}}\Sigma^-_{un^{1/3}} - X^{(1)}_u \right| \right) \leq \eps\, ,
\]

uniformly in $u,v \in U_\delta$ as $U_\delta$ is a finite set (recall the definition (<ref>) of $U_\delta$).
Letting $N \to \infty$ then $\eps \to 0$ we obtain that $\PP$-almost surely,

\[
\begin{split}
\limsup_{n\to \infty} \frac{1}{\beta n^{1/6}} \left( \log Z_{n,h}^{\omega,\beta} + \frac{3}{2}hc_hn^{1/3} \right) \leq  \sup_{\substack{u,v \in U_\delta\\ u+v \in \{c_h,c_h-\delta\}}} \mathcal{W}^+(u,v,\delta) \, , \\
\liminf_{n\to \infty} \frac{1}{\beta n^{1/6}} \left( \log Z_{n,h}^{\omega,\beta} + \frac{3}{2}hc_hn^{1/3} \right)
\geq \sup_{\substack{u,v \in U_\delta\\ u+v \in \{c_h,c_h-\delta\}}} \mathcal{W}^-(u,v,\delta)
\end{split}
\]

in which we wrote

\[ \mathcal{W}^{\pm}(u,v,\delta) =  X^{(1)}_u + X^{(2)}_v \pm \sup_{u \leq u' \leq u + \delta} \big| X^{(1)}_{u'}-X^{(1)}_u \big| \pm \sup_{v \leq v' \leq v + \delta} \big| X^{(2)}_{v'}-X^{(2)}_v \big| \, . \]

Using the càdlàg structure of Lévy processes $X^{(1)}$ and $X^{(2)}$ we push $\delta$ to $0$ and get the desired convergence.

Afterwards, we can use <cit.> and <cit.> to prove that the variational problem is positive and finite (in the sense that $\sup_{0 \leq u \leq c_h} \big\{ X^{(1)}_u + X^{(2)}_{c_h-u} \big\}$ is almost surely positive and finite), which relies on the same reasoning as Lemma <ref>. Then, <cit.> proves the existence and unicity of the maximizer $u_*$. The proof of the second part of Theorem <ref> is exactly the proof of Lemma <ref>.

§ TECHNICAL RESULTS FOR THE BROWNIAN MEANDER

Let $W$ be a standard Brownian motion on $[0,1]$ and denote $\tau \defeq \sup \mathset{t \in [0,1] \; : \, W_t = 0}$.
The Brownian meander on $[0,1]$ is defined as the rescaled trajectory of $W$ between $\tau$ and $1$. More precisely it is the process $M$ defined on $[0,1]$ by

\[ M_t \defeq \frac{1}{\sqrt{1-\tau}} |W_{\tau + t(1-\tau)}| \, . \]

Note that we could define the meander to be on any interval $[0,T]$ by changing how we rescale the trajectory, leading to define a Brownian meander of duration $T$ as the rescaled process $\sqrt{\frac{T}{1-\tau}} |W_{\tau + \frac{t}{T}(1-\tau)}|$ on $[0,T]$.

The Brownian meander also appears when studying a Brownian motion seen from its maximum over an interval. More precisely, we have the following result.

Let $W$ be a Brownian motion on $[0,1]$ and let $\sigma$ be the time at which $W$ reaches its maximum on $[0,1]$. Conditional on the event $\mathset{\sigma = u}$, the processes $(W_{u+s} - W_u, 0 \leq s \leq 1-u)$ and $(W_{u-s} - W_u, 0 \leq s \leq u)$ are independent Brownian meanders of respective duration $1-u$ and $u$.

Recall the notation $\varphi_t(x) \defeq \frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2t}}$ and $\Phi_t(y) \defeq \int_0^y \varphi_t(x) dx$.
The Brownian meander on $[0,1]$ is a continuous, non-homogeneous Markov process starting at $0$, with transition kernel given by

\begin{equation}\label{meandre-transition}
\proba{M_t \in dy \, | \, M_s = x} = p^+(s,x,t,y) dy = \left[\varphi_{t-s}(x-y) - \varphi_{t-s}(x+y)\right] \frac{\Phi_{1-t}(y)}{\Phi_{1-s}(x)} dy
\end{equation}

\begin{equation}\label{meandre-transition-0}
\proba{M_t \in dy} =  p^+(0,0,t,y) dy = 2y t^{-3/2} e^{-\frac{y^2}{2t}} \Phi_{1-t}(y) dy \, .
\end{equation}

For the proofs of these facts, we refer to <cit.> and its references.

Using $\proba{M_t \in dy} \leq 2y t^{-3/2} e^{-\frac{y^2}{2t}} dy$ and $\Phi_{1-t}(y) \leq y/\sqrt{2\pi(1-t)}$, we have the following estimates: for any $a<r/2$,

\begin{equation}
\label{meander-expmoment}
\begin{split}
\esp{e^{a M_r^2}} & \leq \frac{2}{r^{3/2}\sqrt{2\pi }} \int_0^\infty y e^{-\frac{1-2ar}{2r} y^2} \, \dd y =  \left(1-2ra\right)^{-3/2} \, , \\
\esp{(M_r)^{-1} e^{a \mathcal{M}_r^2}} &\leq \frac{1}{r^{3/2}\pi\sqrt{(1-r)}} \int_0^\infty y e^{-\frac{1-2ar}{2r} y^2} \, \dd y = \frac{\sqrt{2\pi}}{\sqrt{r(1-r)}} \left(1-2ra\right)^{-1} \,,
\end{split}
\end{equation}

The asymmetry of the meander can be used to prove the following “reflection principle”.

Let $M$ be a Brownian meander, then for all $b > 0$ and all $0 \leq s < t \leq 1$,

\[  \PP \Big( \sup_{0 \leq r \leq t} M_r \geq b \Big) \leq 2 \proba{M_t \geq b} \, , \quad \PP \Big(\sup_{s \leq r \leq t} |M_r - M_s| \geq b \Big) \leq 4 \proba{M_t - M_s \geq b} \, . \]

If we denote by $T_b$ the hitting time of $b$, we have

\[ \proba{\sup_{0 \leq s \leq t} M_s \geq b} = \int_0^t \proba{T_b \in ds} = \int_0^t \proba{T_b \in ds, M_t < b}  + \int_0^t \proba{T_b \in ds, M_t \geq b} \, .  \]

Now, write $L_b$ the lime of last visit to $b$ before time $t$, on $[L_b,t]$ the process $M_r - b$ is a Brownian bridge conditioned to be above $-b$. We only need to see that any trajectory of $M$ from $b$ to $(0,b]$ which stays above $0$ can thus be transformed into a trajectory from $b$ to $[b,2b)$ that stays above $0$ by reflecting the trajectory between the last visit $L_b$ to $b$ and $t$ (see Figure <ref>).

Reflection of the trajectory $b \to (0,b]$ with respect to the horizontal line at $b$

Since these two Brownian bridges have the same probability and $[b,2b) \subset [b, +\infty)$ it shows that this operation is injective and thus $\proba{T_b \in ds, M_t < b} \leq \proba{T_b \in ds, M_t \geq b}$ for all $s \leq t$ (note that this is a consequence of the Brownian reflection principle). Therefore, we proved

\[ \proba{\sup_{0 \leq s \leq t} M_s \geq b} \leq 2 \int_0^t \proba{T_b \in ds, M_t \geq b} = 2 \proba{T_b \leq t, M_t \geq b} = 2 \proba{M_t \geq b} \, . \]

If we study the supremum of an increment $M_r - M_s$, $s \leq r \leq t$ we only need to repeat the proof for a starting point $M_s = x$ and integrate over all the positions $x$. Since the meander is a Markov process, we get
$\PP \big(\sup_{s \leq r \leq t} M_r - M_s \geq b \big) \leq 2 \proba{M_t - M_s \geq b}.$
Afterwards, we only need to see that again using the asymmetry of $M$, we have that

\[
\PP \Big(\sup_{s \leq r \leq t} |M_r - M_s| \geq b \Big) \leq 2 \PP \Big(\sup_{s \leq r \leq t} M_r - M_s \geq b \Big)\,,
\]

hence the result.

For any $\lambda > 1, a > 0$ and $0 \leq s < t < \frac12$, we have

\[ \PP \Big( \inf_{s \leq r \leq t} M_r \leq a \Big) \leq \proba{M_s \leq \lambda a} + \proba{M_t \leq \lambda a} + \frac{4 a \sqrt{2t}}{t-s} \frac{e^{-\frac{2}{t-s} a^2 (\lambda-1)^2}}{1 - e^{-\frac{2}{t-s} a^2 \lambda^2}} \, , \]

as well as
$ \proba{M_t \leq a}  \leq \frac{4a}{\sqrt{\pi t}}\left( 1 \wedge \frac{a^2}{2t} \right).$

We decompose the probability on whether $M_s, M_t \leq \lambda a$, meaning we only have to consider $\PP \big( \inf_{s \leq r \leq t} M_r \leq a, M_s > \lambda a, M_t > \lambda a \big)$.
For this, we first use Brownian bridge estimates: see that for any $z,w, T > 0$, we have

\[ \begin{split} \probaM{z}{W_T \in dw, \inf_{t \in [0,T]} W_t > 0} &= \frac{1}{\sqrt{2\pi T}} \left( e^{-\frac{1}{2T}(z-w)^2} - e^{-\frac{1}{2T}(z+w)^2} \right) dw \\ \probaM{z}{W_T \in dw} &= \frac{1}{\sqrt{2\pi T}} e^{-\frac{1}{2T}(z-w)^2} dw
\end{split} \]

thus we have

\begin{equation}\label{eq:pont-brownien-pos}
\proba{\inf_{t \in [0,T]} W_t^{z \to w} > 0} = 1 - e^{\frac{1}{2T}(z-w)^2 - \frac{1}{2T}(z+w)^2} = 1 - e^{-\frac{2}{T}zw} \, .
\end{equation}

For any $\alpha > 0$ and $z,w > \alpha$, we define

\[ P_T^\alpha(z,w) \defeq \proba{\inf_{t \in [0,T]} W_t^{z \to w} \leq \alpha \, | \, \inf_{t \in [0,T]} W_t^{z \to w} > 0} = 1 - \frac{\proba{\inf_{t \in [0,T]} W_t^{z \to w} > \alpha}}{\proba{\inf_{t \in [0,T]} W_t^{z \to w} > 0}} \, . \]

Then, using (<ref>) with $z,w,z-\alpha, w - \alpha > 0$, we can deduce

\begin{equation}\label{eq:pont-brownien-fenetre}
P_T^\alpha(z,w) = 1 - \frac{1 - e^{-\frac{2}{T}(z-\alpha)(w-\alpha)}}{1 - e^{-\frac{2}{T}zw}} = \frac{e^{-\frac{2}{T}(z-\alpha)(w-\alpha)} - e^{-\frac{2}{T}zw}}{1 - e^{-\frac{2}{T}zw}} \, .
\end{equation}

Consider the mapping $f_T : (x,y) \mapsto e^{-\frac{2}{T}xy}$. Using the mean value theorem, there is a $c \in [0,1]$ such that

\begin{equation}\label{eq:TAF-2-var}
\begin{split}
f_T(z,w) - f_T(z-\alpha, w-\alpha) &= \nabla f_T \Big( (1-c) \begin{pmatrix} z \\ w \end{pmatrix} + c \begin{pmatrix} z-\alpha \\ w-\alpha \end{pmatrix} \Big) \cdot \Big( \begin{pmatrix} z \\ w \end{pmatrix} - \begin{pmatrix} z-\alpha \\ w-\alpha \end{pmatrix} \Big) \\
&= - \frac{2 \alpha}{T} (z+w-2c\alpha)e^{-\frac{2}{T}(z-c\alpha)(w-c\alpha)} \, .
\end{split}
\end{equation}

Injecting in (<ref>), this yields

\begin{equation}\label{eq:pont-brownien-fenetre-final}
P_T^\alpha(z,w) = \frac{2 \alpha}{T} (z+w-2c\alpha)\frac{e^{-\frac{2}{T}(z-c\alpha)(w-c\alpha)}}{1 - e^{-\frac{2}{T}zw}} \leq \frac{2 \alpha}{T} (z + w) \frac{e^{-\frac{2}{T}(z-c\alpha)(w-c\alpha)}}{1 - e^{-\frac{2}{T}zw}} \, .
\end{equation}

In particular, if we assume $z,w \geq \lambda \alpha$ for some $\lambda > 1$, then $f_T(z,w) \leq f_T(\lambda a, \lambda a)$ and we obtain

\[  P_T^\alpha(z,w) \leq \frac{2 \alpha}{T} (z + w) \frac{e^{-\frac{2}{T}\alpha^2 (\lambda-c)^2}}{1 - e^{-\frac{2}{T}\alpha^2 \lambda^2}} \, . \]

Therefore, for any $\lambda > 1$ and $a > 0$,

\begin{equation}\label{eq:pont-meandre}
\begin{split}
\proba{\inf_{s \leq r \leq t} M_r \leq a, M_s > \lambda a, M_t > \lambda a} &= \esp{P_{t-s}^a(M_s,M_t) \indic{M_s,M_t \geq \lambda a}} \\
&\leq \frac{2 a}{t-s} \frac{e^{-\frac{2}{t-s} a^2 (\lambda-c)^2}}{1 - e^{-\frac{2}{t-s} a^2 \lambda^2}} \esp{M_s + M_t} \, ,
\end{split}
\end{equation}

and we compute $\esp{M_t} \leq \frac{2\sqrt{2}}{\pi} \sqrt{t} \leq \sqrt{2t}$ for $t < 1/2$ to get the desired result.
On the other hand, using (<ref>), we write for $0<t<\frac12$

\begin{align*}
\proba{\mathcal{M}_t \leq a} &= \frac{2}{t^{3/2}} \int_0^a y e^{-\frac{y^2}{2t}} \int_0^y \frac{e^{-\frac{u^2}{2(1-t)}}du}{\sqrt{2\pi(1-t)}} dy \leq \frac{2}{t^{3/2}} \int_0^a y e^{-\frac{y^2}{2t}} \int_0^y \frac{du}{\sqrt{2\pi(1-t)}} dy  \\
&\leq \frac{2 a t^{-3/2}}{\sqrt{2\pi(1-t)}} \int_0^a y e^{-\frac{y^2}{2t}} dy = \frac{4 a (1 - e^{-\frac{a^2}{2t}})}{\sqrt{2 \pi t(1-t)}} \leq \frac{4a}{\sqrt{\pi t}} \left( 1 \wedge \frac{a^2}{2t} \right) \,.
\qedhere
\end{align*}

Let us mention that a process related to the meander is the $3$-dimensional Bessel process $B$. It can be defined as the solution of the SDE $\dd B_t = \dd W_t + B_t^{-1} \dd t$, or as the sum $B_t = |W_t| + L_t$ where $L$ is the local time of $W$ at $0$; it is a homogeneous Markov process that has the Brownian scaling property $(B_{\alpha t})_t \overset{d}{=} (\sqrt{\alpha} B_t)_t$. We refer to <cit.> for those results. The link between the Bessel process and the meander is given by the following result.

The law $\PP^{+,T}$ of the Brownian meander on $[0,T]$ has a density with respect to $\PP^B$ the law of the three-dimensional Bessel process: if $X$ is the canonical process, we have

\[ \PP^{+,T} (A, X_T \in dx) = \frac{1}{x} \sqrt{\frac{\pi T}{2}} \, \PP^B(A,X_T \in dx) \, . \]

In particular, $\forall \alpha > 0, \forall s \leq T, \PP^{+,\alpha T}(X_{\alpha s} \in dx) = \PP^{+,T}(\sqrt{\alpha} X_s \in dx)$.

The formula for the density can be found in <cit.>. Afterwards, for any positive measurable function $f$ and any $\alpha > 0$, we have

\[ \EE^{+,\alpha T} \left[ f\Big(\frac{X_{\alpha s}}{\sqrt{\alpha}} \Big) \right] = \EE^B \left[ \frac{1}{X_{\alpha T}} \sqrt{\frac{\pi \alpha T}{2}} f\Big(\frac{X_{\alpha s}}{\sqrt{\alpha}} \Big) \right] = \sqrt{\frac{\pi}{2}} \EE^B \left[ \frac{\sqrt{T}}{X_T} f(X_s) \right] = \EE^{+,T} \left[ f(X_s) \right] \, . \qedhere \]

§ COUPLING OF BROWNIAN MEANDER, A THREE-DIMENSIONAL BESSEL PROCESS AND A BROWNIAN EXCURSION

In this section we will expand on the way we can construct our different processes to have the almost sure results of Theorems <ref> and <ref>. In particular we want the following result:

\[ \frac{1}{n^{1/6}} \sum_{-u n^{1/3}}^{vn^{1/3}} \omega_z \xrightarrow[n \to \infty]{a.s.} X^{(1)}_u + X^{(2)}_v \quad \text{and} \quad n^{1/18} (X_{u_* + \frac{u}{n^{1/9}}} - X_{u_*}) \xrightarrow[n \to \infty]{a.s.} \mathbf{B}_u \, . \]

Skorokhod's embedding theorem (Theorem <ref>) allows us to sample the Brownian motions $X^{(i)}, i = 1,2$ to get a new environment $\hat\omega^{(n)}$ to obtain the first convergence. Thus we must find how we can couple both processes $X^{(i)}$ to the processes $\mathbf{B},\mathbf{Y}$ in Theorem <ref>,
that is we need to prove Proposition <ref>.
This is based on two intermediate results, Lemmas <ref> and <ref> below, which couple a meander, resp. a Bessel-$3$ process, to a Brownian excursion.

Let $\mathbf{e}$ be a standard Brownian excursion and $U$ a uniform variable on $[0,1]$. Then, the process $M_t = \mathbf{e}_t \indic{t \leq U} + (\mathbf{e}_U + \mathbf{e}_{1-(t-U)}) \indic{t > U}$ is a Brownian meander on $[0,1]$. In particular, there exists a coupling of the Brownian meander $M$ and the Brownian excursion $\mathbf{e}$ on $[0,1]$ such that $M_t = \mathbf{e}_t$ if $t \leq U$.

For any $T \in [0,1]$,
There exists a coupling of the Brownian excursion $\mathbf{e}$ on $[0,1]$ and the three-dimensional Bessel process $\mathbf{B}$ such that there is a positive $\eps(\omega)$ for which we have $\mathbf{B}_t =  \mathbf{e}_{t}$ for any $t \in [0,\eps(\omega)]$.

It is known (see for example <cit.>) that the Brownian excursion can be decomposed into two Bessel bridges of duration $\frac12$ joining at a point $V$ whose law has density $\frac{16}{\sqrt{2\pi}} v^2 e^{-2v^2}$. Thus we only need to define a coupling between a $3d$-Bessel process $\mathbf{B}$ and a $3d$-Bessel bridge $\mathbf{B}'$ with duration $\frac12$ and endpoint $V$. We use the fact that both processes can be realized by the modulus of a three-dimensional Brownian motion.

Consider two independent, three-dimensional Brownian bridges $X$ and $Y$ of duration $1/2$, such that $X_0 = x \in \RR^3$ (resp. $Y_0 = y \in \RR^3$) and $X_\frac12 = Y_\frac12 = 0$. Denote by $\tau$ the first time $X$ and $Y$ have the same modulus: $\tau \defeq \inf \mathset{0 \leq t \leq \frac12 \, : \, |X_t| = |Y_t|}$.
We have the following result.

Almost surely, there exists $\eps(\omega) > 0$ such that $\tau \leq \frac12 - \eps(\omega)$.

Using this lemma, we can conclude the construction of the coupling.
After time $\tau$, we define a coupling by taking the trajectory of $X$ between $\tau$ and $\tfrac12$ and plugging it at $Y_\tau$ after a rotation:

\[ \text{write } X_t = |X_t| e^{i \theta^X_t}, Y_t = |Y_t| e^{i \theta^Y_t} \text{ and define } \hat Y_t = \begin{dcases} Y_t \quad \text{if} \quad t \leq \tau \,,\\ |X_t| e^{i \theta^X_t + i (\theta^Y_\tau - \theta^X_\tau)} \quad \text{if} \quad \tau < t \leq \tfrac12 \,.\end{dcases} \]

The new process $\hat{Y}$ is such that for every $t \in [\tau, \frac12]$, we have $|X_t| = |\hat{Y}_t|$.
Recall that the Brownian bridge is a diffusion process (as the solution to an SDE), thus is Markovian, and $\tau$ is a stopping time for both processes $X$ and $Y$. It follows that $\hat Y$ is a Brownian bridge between $y$ and $0$.

To create the coupling between the two Bessel processes $\mathbf{B}$ and $\mathbf{B}'$, we choose the starting points $x$ and $y$ so that they respectively correspond to $W_\frac12$ (with $W$ a $3d$-Brownian motion) and a uniform variable on the sphere centered at $0$ of radius $V$. Then the processes $\mathbf{B}_t = |X_{\frac12 - t}|$ and $\mathbf{B}'_t = |\hat{Y}_{\frac12 - t}|$ are Bessel processes starting at $0$ that coincide on $[0,\frac12 - \tau]$ and such that $\mathbf{B}'_\frac12 = V$. In particular, the Bessel process $\mathbf{B}$ and the Brownian excursion $\mathbf{e}$ coincide on $[0,\frac12 - \tau]$.

On $[0,\tfrac12]$, consider $\mathbf{B}$ a 3-dimensional Bessel process starting at $0$ and $\mathbf{e}$ the Brownian excursion, which is a Bessel bridge of duration $\tfrac12$ starting at $0$ and ending at $V$. We define $I_{s,t} \defeq \mathset{\forall r \in (s,t), \mathbf{e}_r \neq \mathbf{B}_r}$ the event on which $\mathbf{e}$ and $\mathbf{B}$ never intersect between $0$ and $t$ (with the exception of $0$).
From <cit.>, we have $\probaM{x}{A, \mathbf{B}_t \in dz} = \frac{z}{x} \probaM{x}{A, W_t \in dz, H_0 > t}$, where $W$ is a Brownian motion and $H_0$ its first hitting time of $0$. Then for any $\eps > 0$, conditioning on the values of $(\mathbf{e}_\eps,\mathbf{B}_\eps)$ and $(V,\mathbf{B}_t)$, we can write

\[ \proba{I_{0,t}} \leq \esp{ \proba{I_{\eps,t} \, | \, \mathbf{e}_\eps, \mathbf{B}_\eps} } \leq \esp{ \frac{\mathbf{e}_t \mathbf{B}_t}{\mathbf{e}_\eps \mathbf{B}_\eps} \proba{\mathscr{I}_{\mathbf{e}_\eps \to \mathbf{e}_t}^{\mathbf{B}_\eps \to \mathbf{B}_t}(t-\eps)} } \, , \]

where we have defined

\[ \mathscr{I}_{x \to y}^{a \to b}(T) \defeq \mathset{\forall r \in (0,T), W^{x \to y}_{T}(r) > 0, W^{a \to b}_{T}(r) > 0,  W^{x \to y}_{T}(r) \neq W^{a \to b}_{T}(r)} \, , \]

in which $W^{a \to b}_T$ is a Brownian bridge $a \to b$ of duration $T$ (resp. for $x \to y$). We are interested in taking $t = 1/2$, but this result could be used for any fixed $t > 0$, in the sense that the Bessel process and the Brownian excursion almost surely cross each-other on $]0,t]$ for any fixed $t$.
Take a positive $C > 0$ to be chosen later (we will choose $C = \eps^{-1/8}$). Then, we first get a bound using Cauchy-Schwartz inequality twice:

\[ \esp{ \frac{| \mathbf{e}_t \mathbf{B}_t|}{\mathbf{e}_\eps \mathbf{B}_\eps} \proba{\mathscr{I}_{\mathbf{e}_\eps \to V}^{\mathbf{B}_\eps \to \mathbf{B}_t}(t-\eps)} \indic{\mathbf{B}_t \vee V > C}} \leq \esp{\frac{1}{(\mathbf{e}_\eps \mathbf{B}_\eps)^2}}^{\frac12} \esp{(\mathbf{e}_t \mathbf{B}_t)^4}^{\frac14} \proba{\mathbf{B}_t \vee V > C}^{\frac14} \, . \]

Since $\eps < t$ and $\mathbf{e},\mathbf{B}$ are independent, we have $\esp{(\mathbf{e}_t \mathbf{B}_t)^4} \leq c(t)$ and

\[ \esp{\frac{1}{(\mathbf{e}_\eps \mathbf{B}_\eps)^2}} \leq \frac{2}{\sqrt{\pi}} \Gamma(\frac32) (1-\eps)^{-\frac32} \eps^{-3} \int_{\RR_+^2} e^{-\frac{x^2}{2\eps}} e^{-\frac{y^2}{2\eps(1-\eps)}} dx dy \leq (1-\eps)^{-1} \eps^{-2} \, , \]

where we used the transition probabilities for the Bessel process <cit.>, the Brownian excursion <cit.> and $\Gamma(\frac32) = \sqrt{\pi}/2$. Finally we compute $\proba{\mathbf{B}_t \vee V > C} \leq e^{-C^2/t^{1/6}}$ to get

\begin{equation}\label{eq:majore-intersection-inf-C}
\esp{ \frac{| \mathbf{e}_t \mathbf{B}_t|}{\mathbf{e}_\eps \mathbf{B}_\eps} \proba{\mathscr{I}_{\mathbf{e}_\eps \to V}^{\mathbf{B}_\eps \to \mathbf{B}_t}(t-\eps)} \indic{\mathbf{B}_t \vee \mathbf{e}_t > C}} \leq c_t (1-\eps)^{-\frac12} \eps^{-2} e^{-\frac{C^2}{t^{1/6}}} \, .
\end{equation}

On the other hand,

\begin{equation}\label{eq:separation-intersection-leq}
\esp{ \frac{| \mathbf{e}_t \mathbf{B}_t|}{\mathbf{e}_\eps \mathbf{B}_\eps} \proba{\mathscr{I}_{\mathbf{e}_\eps \to \mathbf{e}_t}^{\mathbf{B}_\eps \to \mathbf{B}_t}(t-\eps)} \indic{\mathbf{B}_t \vee \mathbf{e}_t \leq C}} \leq \esp{ \frac{C^2}{\mathbf{e}_\eps \mathbf{B}_\eps} \proba{\mathscr{I}_{\mathbf{e}_\eps \to \mathbf{e}_t}^{\mathbf{B}_\eps \to \mathbf{B}_t}(t-\eps)}} \, .
\end{equation}

We will use the following lemma to get a bound on $\proba{\mathscr{I}_{\mathbf{e}_\eps \to \mathbf{e}_t}^{\mathbf{B}_\eps \to \mathbf{B}_t}(t-\eps)}$.

For any $T > 0$, there is a $C_T > 0$ such that for any $x,y,a,b > 0$,

\begin{equation}
\proba{\mathscr{I}_{x \to y}^{a \to b}(T)} \leq C_T (x^2 + a^2)^2 (y^2 + b^2)^2 \, .
\end{equation}

Thus, using Lemma <ref> in (<ref>), we have the upper bound

\[ \esp{ \frac{| \mathbf{e}_t \mathbf{B}_t|}{\mathbf{e}_\eps \mathbf{B}_\eps} \proba{\mathscr{I}_{\mathbf{e}_\eps \to \mathbf{e}_t}^{\mathbf{B}_\eps \to \mathbf{B}_t}(t-\eps)} \indic{\mathbf{B}_t \vee \mathbf{e}_t \leq C}} \leq C^6 C_{t,\eps} \esp{ \frac{1}{\mathbf{e}_\eps \mathbf{B}_\eps}\Big( (\mathbf{e}_\eps)^2 + (\mathbf{B}_\eps)^2 \Big)^2} \, , \]

and we compute

\[ \begin{split}
\esp{\frac{\Big( (\mathbf{e}_\eps)^2 + (\mathbf{B}_\eps)^2 \Big)^2}{\mathbf{e}_\eps \mathbf{B}_\eps}} &= \frac{2}{\sqrt{\pi}} \Gamma(\frac32) (1-\eps)^{-\frac32} \eps^{-3} \int_{\RR_+^2} (x^2 + y^2)^2 xy e^{-\frac{x^2}{2\eps}} e^{-\frac{y^2}{2\eps(1-\eps)}} dx dy\\
&\leq (cst.) \eps^{-3} \int_{\RR_+^2} \eps^2 (u^2 + v^2)^2 uv \eps e^{-\frac{u^2}{2}} e^{-\frac{v^2}{2}} \eps du dv \leq (cst.) \eps \, .
\end{split} \]

to get

\begin{equation}\label{eq:majore-intersection-sup-C}
\esp{ \frac{| \mathbf{e}_t \mathbf{B}_t|}{\mathbf{e}_\eps \mathbf{B}_\eps} \proba{\mathscr{I}_{\mathbf{e}_\eps \to \mathbf{e}_t}^{\mathbf{B}_\eps \to \mathbf{B}_t}(t-\eps)} \indic{\mathbf{B}_t \vee \mathbf{e}_t \leq C}} \leq C^6 C_t \eps \, .
\end{equation}

Thus, assembling (<ref>) and (<ref>) while taking $C = \eps^{-1/8}$, for any $t > 0$ we then have

\[ \proba{I_{0,t}} \leq C_t \left( \eps^{-1} e^{-\frac{C^2}{4(t-\eps)^{1/6}}} + C^6 \eps \right) \leq C_t \left( \eps^{-1} \exp \Big(-\frac{\eps^{-1/4}}{2}\Big) + \eps^{1/4} \right) \xrightarrow[\eps \to 0]{} 0 \, . \]

This means that $\proba{I_{0,t}} = 0$ and in particular, taking $t = \frac12$, one can almost-surely find a positive $\eps$ such that Lemma <ref> is true.

We can assume $0 < x < a$ and $0 < y < b$ (otherwise the probability is zero), then we have

\begin{equation}\label{eq:probaBB}
\proba{\mathscr{I}_{x \to y}^{a \to b}(T)}  = \proba{\forall r \in [0,T], 0 < W_T^{x \rightarrow y}(r) < W_T^{a \rightarrow b}(r)}.
\end{equation}

Observe that (<ref>) is exactly the probability for the Brownian bridge $W_T^{x \to y, a \to b} \defeq (W_T^{x \rightarrow y}, W_T^{a \rightarrow b})$ to stay in the cone $\mathscr{C} \defeq \mathset{(x,y) \in \RR^2 \, : \, 0 \leq x \leq y}$ for a time $T$, meaning

\[ \proba{\mathscr{I}_{x \to y}^{a \to b}(T)} = \proba{\forall t \in [0,T], W_T^{x \to y, a \to b}(t) \in \mathscr{C}}. \]

The isotropy of Brownian motion allows us to consider instead $\hat {\mathscr{C}} \defeq \mathset{r e^{i\theta}, 0 \leq \theta \leq \frac{\pi}{4}}$.

Let $W^{z \to z'}$ be a two dimensional Brownian bridge from $z$ to $z'$. Then, there is a positive $C_T$ such that uniformly as $|z| \to 0$ we have

\[ \proba{\forall t \in [0,T], W^{z \to z'}_t \in \hat{\mathscr{C}}} = (1+\bar{o}(1)) C_T |z|^4 |z'|^4 \sin \left( 4  \arg z \right)\sin \left( 4 \arg z' \right) \, . \]

Recall that we identify $\RR^2$ with $\CC$, by writing $W$ for a standard two-dimensional Brownian motion, we have

\[ \begin{split}
\proba{\forall t \in [0,T], W^{z \to z'}_t \in \hat{\mathscr{C}}} &= \lim_{\eta \to 0} \frac{\probaM{x}{\forall t \in [0,T], W_t \in \hat{\mathscr{C}}, W_T \in B(z',\eta)}}{\probaM{z}{W_T \in B(z',\eta)}} \\
&= \lim_{\eta \to 0} \left(C(T) \eta^2 e^{-|z'|^2/2T} \right)^{-1} \int_{B(z',\eta)} K_T^{\hat{\mathscr{C}}}(z,w) dw \, ,
\end{split} \]

where $K_T^{\hat {\mathscr{C}}}(z,w)$ is the heat kernel killed on exiting $\hat{\mathscr{C}}$ and $B(z,r)$ is the ball of radius $r$ centered at $z$.

The key ingredient is the following statement, which is a consequence of <cit.>: as $\delta \to 0$, uniformly in $|z| \leq \delta \sqrt{T}, |w| \leq \sqrt{T/\delta}$, we have

\[ K_T^{\hat{\mathscr{C}}}(z,w) \sim \frac{\chi_0}{T^{5}} e^{-|w|^2/2T} u(w) u(z) \quad \text{for some $\chi_0 > 0$.} \]

where $u(re^{i\theta}) \defeq r^4 \sin(4\theta)$ (this expression is given in <cit.>). This result is also stated in <cit.>. In particular, as $|z| \to 0$,

\[ \begin{split}
\proba{\forall t \in [0,T], W^{z \to z'}_t \in \hat{\mathscr{C}}} &\sim \lim_{\eta \to 0} \left(C(T) \eta^2 e^{-|z'|^2/2T} \right)^{-1} \int_{B(z',\eta)} \frac{\chi_0}{T^{5}} e^{-|w|^2/2T} u(w) u(z) dw \\
&= \lim_{\eta \to 0} e^{|z'|^2/2T} \frac{\chi_0}{T^{5}} e^{-|z'|^2/2T}  u(z') u(z) \frac{\text{Vol}(B(z',\eta))}{C(T) \eta^2} (1 + h(\eta)) \, ,
\end{split} \]

with $h(\eta) \to 0$ and $\text{Vol}(B(z',\eta)) = \pi \eta^2$, leading us to the formula of Lemma <ref>.

We could also use the fact that $\hat{\mathscr{C}}$ is the Weyl chamber $B_2$, thus we can use results from <cit.> after a time scaling by $\eps$ to have that the probability in (<ref>) is of order $(\eps/t)^2$, which is ultimately what we proved.

Thus, we proved Lemma <ref> by injecting $z = (x,a)$ and $z' = (y,b)$.

Assembling Lemmas <ref> and <ref> yields that one can do a coupling of the Brownian meander $\mathcal{M}$ and the three-dimensional Bessel process $\mathbf{B}$ such that almost surely, there is a positive time $\sigma$ for which $\mathcal{M}_t = \mathbf{B}_t$ on $[0,\sigma]$, thus proving Proposition <ref> using (<ref>).

§.§ Acknowledgements

The author would like to thank his PhD advisors Quentin Berger and Julien Poisat for their continual help, as well as Pierre Tarrago for his proof of Lemma <ref>.

§ CALCULATIONS FOR SECTION <REF>

§.§ Trajectories restrictions

Uniformly in $n \geq 1$, we have under Conjecture <ref>

\begin{equation}
\varlimsup_{M \to \infty} \proba{\tilde{\mathcal{Z}}_{n,h}^{\omega,\beta} (M < |\Delta_n - s_* n^{2/9}| < \eta n^{2/9}) \geq e^{-M^{1/4}}} = 0
\end{equation}

We write $\mathscr{E}_{n,M}^\eta$ the event in (REF) and we define $J_{n,M}^\eta \defeq \sup \mathset{j \in \NN \, : \, 2^j M n^{-2/9} \leq \eta}$. For any integer $j \leq J_{n,M}^\eta$ we consider the intervals

\[ \mathscr{C}_j^{M,\pm} \defeq \mathset{s \in \RR \, : \, 2^j M < \Delta_n \mp s_* n^{2/9} < 2^{j+1} M} \]

the $\pm$ representing whether $\Delta_n$ is greater (+) or less (-) than $s_*$ which is assumed to be positive.
If we write $\PP_\eta^\pm \defeq \proba{ \cdot \, | \inf_{[0,\eta]} R^\pm_s > 0}$ the law of $R^\pm$ in the neighbourhood of $s_*$ (see Conjecture <ref>), we can decompose the partition function and use an union bound, similarly to the proof of Proposal <ref>, to get

\begin{equation}\label{eq:decomp-part-simpl}
\begin{split}
\proba{\mathscr{E}_{n,M}^\eta} &\leq \sum_{j = 0}^{J_{n,M}^\eta} \left[\PP_\eta^- \left( e^{-\beta n^{1/9} \inf_{s \in \mathscr{C}_j^{M,-}} R^-_s} \geq \frac{e^{-M^{1/4}}}{2^{j-1}} \right) + \PP_\eta^+ \left( e^{-\beta n^{1/9} \inf_{s \in \mathscr{C}_j^{M,+}} R^+_s} \geq \frac{e^{-M^{1/4}}}{2^{j-1}} \right) \right]\\
&\leq \sum_{j = 0}^{J_{n,M}^\eta} \left[ \PP_\eta^- \left( \inf_{s \in \mathscr{C}_j^{M,-}} R^-_s \leq \frac{M^{1/4}}{\beta n^{1/9}} \right) +\PP_\eta^+ \left( \inf_{s \in \mathscr{C}_j^{M,+}} R^+_s \leq \frac{M^{1/4}}{\beta n^{1/9}} \right) \right]
\end{split}
\end{equation}

We can focus the proof on the “left side” of $s_* > 0$, meaning on $\mathscr{C}_j^{M,-}$ and $R^-_s$, where the condition that $R > 0$ conflicts with the negative drift : the “right side” ($+$) is immediately deduced from this result. Additionally the following proof for the “left side” can easily be adapted to the “right side” process. In the proof we write instead $\mathscr{C}_j^{M}$ and $R_s$ to ease the notations.
Since we condition $R$ to be positive, we will use the following lower bound

\[
\probaM{z}{\inf_{0 \leq s \leq \delta} R_s > 0} = \espM{z}{\indic{\inf_{0 \leq s \leq \delta} B_s > 0} e^{-\alpha B_\delta}} e^{\alpha z - \frac12 \alpha^2 \delta} = \espM{z}{\indic{B_\delta > 0} (1-e^{- \frac{2 z B_\delta}{\delta}}) e^{-\alpha B_\delta}} e^{\alpha z - \frac12 \alpha^2 \delta}
\]

Since $z$ will be taken with $\frac{z}{\sqrt{\delta}} \ll 1$, we have the lower bound

\[ \begin{split}
\espM{z}{\indic{B_\delta > 0} (1-e^{- \frac{2 z B_\delta}{\delta}}) e^{-\alpha B_\delta}} &\geq (cst.) \espM{z}{\indic{B_\delta > 0} \frac{z B_\delta}{\delta}e^{-\alpha B_\delta}} = (cst.) \frac{z}{\sqrt{\delta}}  \espM{\frac{z}{\sqrt{\delta}}}{\indic{B_1 > 0} B_1 e^{-\alpha \sqrt{\delta} B_1}}  \\
&\geq (cst.) \frac{z}{\sqrt{\delta}} \int_0^\infty y e^{-\alpha \sqrt{\delta} y - \frac12 (y -  \frac{z}{\sqrt{\delta}})^2} dy \\
&\geq (cst.) \frac{z}{\sqrt{\delta}} \int_0^{\sigma(z,\delta)^{-1}} y e^{- \frac12 (y^2 - \sigma(z,\delta)^2) - 1} dy
\end{split} \]

where we wrote $\sigma(z,\delta) \defeq 2 \frac{z}{\sqrt{\delta}} + \alpha \sqrt{\delta}$. Finally we get

\[ \probaM{z}{\inf_{0 \leq s \leq \delta} R_s > 0} \geq (cst.) \frac{z}{\sqrt{\delta}} (1-e^{- \frac12 \sigma(z,\delta)^{-2}}) e^{- \sigma(z,\delta)^2 + \alpha z - \frac12 \alpha^2 \delta}
\]

Take an arbitrary $\lambda > 0$ and $x = R_\lambda$, we also write $x_{j,n} \defeq \frac{2^{j/4} M^{3/8}}{\beta n^{1/9}}$ and $j_+ = \frac{2^j M}{n^{2/9}}$.
We can bound the probabilities in (<ref>) using

\[ \mathset{\inf_{s \in \mathscr{C}_j^M} R_s \leq \frac{M^{1/4}}{\beta n^{1/9}}} \subset \mathset{R_{j_+} \leq x_{j,n}} \cup \mathset{R_{j_+} > x_{j,n}, \inf_{j_+ \leq r \leq 2j+} R_r \leq \frac{M^{1/4}}{\beta n^{1/9}}} \]

which stays true after intersecting with $\mathset{\inf_{0 \leq s \leq \eta} R_s > 0}$. Thus, we only need to find an upper bound on

\begin{equation}\label{eq:split-haut-bas}
\probaM{x}{R_{j_+ - \lambda} \leq x_{j,n}, \inf_{0 \leq s \leq \eta} R_s > 0} + \probaM{x}{\inf_{j_+ -\lambda \leq r \leq 2j+ - \lambda} R_r \leq \frac{M^{1/4}}{\beta n^{1/9}}, \inf_{0 \leq s \leq \eta} R_s > 0}
\end{equation}

then divide by $\proba{\inf_{0 \leq s \leq \eta} R_s > 0}$ and finally take the expectation for $x = R_\lambda$. Note that to simplify the notation we wrote $\eta$ instead of $\eta - \lambda$, which doesn't change the results.

Control of the first term of (<ref>)

For any $j \leq J_{n,M}^\eta - 1$, we have

\[ \begin{split}
\probaM{x}{R_{j_+ - \lambda} \leq x_{j,n}, \inf_{0 \leq s \leq \eta} R_s > 0} &= \espM{x}{\indic{R_{j_+} \leq x_{j,n}, \inf_{0 \leq s \leq j_+} R_s > 0} \probaM{R_{j_+}}{\inf_{0 \leq s \leq \eta-j_+} R_s > 0}} \\
&\leq \espM{x}{\indic{W_{j_+} \leq x_{j,n}, \inf_{0 \leq s \leq j_+} W_s > 0} \frac{W_{j_+}}{\sqrt{\eta-j_+}}} e^{\alpha x - \frac12 \alpha^2 (\eta - \lambda)}
\end{split} \]

Denote by $p^-(x,t,y)$ the transition kernel of the Brownian motion killed at $0$, we have

\[ \begin{split}
\probaM{x}{R_{j_+} \leq x_{j,n}, \inf_{0 \leq s \leq \eta} R_s > 0} &\leq \int_0^{x_{j,n}} p^-(x,j_+ - \lambda,y) \frac{y}{\sqrt{\eta-j_+}} \dd y e^{\alpha x - \frac{\alpha^2}{2}(\eta - \lambda)} \\
&\leq \int_0^{x_{j,n}} \frac{2 xy dy}{\sqrt{2\pi (j_+ - \lambda)^3}} \frac{y}{\sqrt{\eta-j_+}} \dd y e^{\alpha x - \frac{\alpha^2}{2}(\eta - \lambda)}\\
&\leq \frac{2 x (x_{j,n})^3 dy}{\sqrt{2\pi (j_+ - \lambda)^3}} \frac{1}{3\sqrt{\eta-j_+}} e^{\alpha x - \frac{\alpha^2}{2}(\eta - \lambda)}
\end{split} \]

Thus, we get the upper bound

\[ \frac{\probaM{x}{R_{j_+ - \lambda} \leq x_{j,n}, \inf_{0 \leq s \leq \eta} R_s > 0}}{\probaM{x}{\inf_{0 \leq s \leq \eta} R_s > 0 }} \leq \frac{2 x (x_{j,n})^3}{\sqrt{2\pi (j_+ - \lambda)^3}} \frac{1}{3\sqrt{\eta-j_+}} \frac{\sqrt{2\pi \eta}}{x} \frac{e^{\sigma(x,\eta)^2 + \frac12 \alpha^2 (\eta - j_+ + \eps) - \frac{x^2}{\eta}}}{1 -e^{-\frac12 \sigma(x,\eta)^{-2}}} \]

Meaning that

\[ \proba{R_{j_+ - \lambda} \leq x_{j,n} \, | \, \inf_{0 \leq s \leq \eta} R_s > 0} \leq (cst.) \frac{(x_{j,n})^3}{\sqrt{(j_+ - \lambda)^3}} \frac{1}{\sqrt{1-\frac{j_+}{\eta}}} \esp{\frac{e^{\sigma(R_\lambda,\eta)^2}}{1 -e^{-\frac12 \sigma(R_\lambda,\eta)^{-2}}}} \]

taking $\lambda = \eps j_+$ for a small $\eps > 0$ and observing that $j \leq J_{n,M}^\eta - 1$ implies $\eta \geq 2 j_+$ and $\forall z > 0, z + \frac{1}{2z} \geq \sqrt{2}$ finally yields

\[ \proba{R_{j_+} \leq x_{j,n} \, | \, \inf_{0 \leq s \leq j_+} R_s > 0} \leq (cst.) \frac{(x_{j,n})^3}{(j_+)^{3/2}} \esp{\frac{1}{e^{-\sigma(R_\lambda,\eta)^2} - e^{-\sqrt{2}}}} \leq (cst.) \frac{2^{-\frac34 j}}{M^{3/8}} \]

where we used in particular that $\sigma(R_\lambda,\eta) \leq 2 R_{\eps j_+} (j_+)^{-1/2} + \alpha \sqrt{\eta} = 2 \mathcal{W_\eps} + \alpha \sqrt{\eta}$ (using Proposition <ref> provided $\eps > 0$ is sufficiently small) is close to $0$, thus $e^{-\sigma(R_\lambda,\eta)^2} \approx 1$.

When $j = J_{n,M}$ we instead use the upper bound

\[ \begin{split}
\probaM{x}{R_{j_+} \leq x_{j,n}, \inf_{0 \leq s \leq \eta} R_s > 0} &\leq \probaM{x}{W_{j_+} \leq x_{j,n}, \inf_{0 \leq s \leq j_+} W_s > 0} = \int_0^{x_{j,n}} p^-(x,j_+ - \lambda,y) \dd y\\
&\leq \int_0^{x_{j,n}} \frac{2 xy dy}{\sqrt{2\pi (j_+ - \lambda)^3}} e^{\alpha x - \frac{\alpha^2}{2}(j_+ - \lambda)} =  (cst.) \frac{x (x_{j,n})^2}{\sqrt{(j_+)^3}} e^{-\alpha x + \frac{\alpha^2}{2}j_+(1 - \eps)}
\end{split} \]

We get the upper bound by using $2j_+ \geq \eta$

\[ \proba{R_{j_+ - \lambda} \leq x_{j,n} \, | \, \inf_{0 \leq s \leq \eta} R_s > 0} \leq \frac{(cst.) x \frac{(x_{j,n})^2}{j_+ \sqrt{\eta}}}{\frac{x}{\sqrt{\eta}} (1 -e^{-\frac12 \lambda(x,\eta)^{-2}}) e^{- \frac{x^2}{2j_+} + \alpha x - \frac12 \alpha^2 j_+}} = (cst.) \frac{(x_{j,n})^2}{j_+} e^{\frac{x^2}{2j_+}} \]

which is greater than our previous bound $2^{-\frac34 j} M^{-3/8}$ but still has a finite sum over $j$.

Control of the second term of (<ref>)

On the other hand we want to bound the quantity

\begin{equation}\label{eq:simp-borne-sec}
\espM{x}{\indic{\inf_{0 \leq s \leq j_+ - \lambda} R_s > 0, R_{j_+ - \lambda} \geq x_{j,n}} \espM{R_{j_+ - \lambda}}{\indic{0 < \inf_{0 \leq s \leq j_+} \bar R_s \leq \frac{M^{1/4}}{n^{1/9}}} \probaM{\bar R_{j_+}}{\inf_{[0, \eta-2j_+]} \tilde R_s > 0}}}
\end{equation}

To do so, first consider any $j \leq J_{n,M}^\eta - 2$, we have

\[ \begin{split}
\espM{R_{j_+ - \lambda}}{\indic{0 < \inf_{0 \leq s \leq j_+} \bar R_s \leq \frac{M^{1/4}}{n^{1/9}}} \probaM{\bar R_{j_+}}{\inf_{[0, \eta-2j_+]} \tilde R_s > 0}} \leq \espM{R_{j_+ - \lambda}}{\indic{0 < \inf_{0 \leq s \leq j_+} \bar R_s \leq \frac{M^{1/4}}{n^{1/9}}} \frac{\bar R_{j_+}}{\sqrt{\eta-2j_+}}}\\
\leq \probaM{R_{j_+ - \lambda}}{0 < \inf_{0 \leq s \leq j_+} \bar R_s \leq \frac{M^{1/4}}{n^{1/9}}}^{1/2} \espM{R_{j_+ - \lambda}}{\frac{\bar R_{j_+}^2}{\eta-2j_+}}^{1/2}
\end{split} \]

We then use the upper bound for $R_{j_+-\lambda} > x_{j,n}$

\[ \begin{split}
\probaM{R_{j_+ - \lambda}}{0 < \inf_{0 \leq s \leq j_+ - \lambda} \bar R_s \leq \frac{M^{1/4}}{n^{1/9}}} \leq \frac{2}{\sqrt{2\pi (j_+ - \lambda)}} \frac{M^{1/4}}{n^{1/9}} e^{- (x_{j,n} - \frac{M^{1/4}}{n^{1/9}})^2/2(j_+ - \lambda)}
\end{split} \]

Using Girsanov theorem with $j_+ \leq \eta$, we can see that it suffices to get an upper bound on

\[ \begin{split}
\espM{x}{\indic{\inf_{0 \leq s \leq j_+ - \lambda} W_s > 0} \espM{W_{j_+ - \lambda}}{\frac{\bar W_{j_+}^2}{\eta-2j_+}}^{1/2}} \leq \espM{x}{\indic{\inf_{0 \leq s \leq j_+ - \lambda} W_s > 0} \espM{\frac{W_{j_+ - \lambda}}{\sqrt{j_+ - \lambda}}}{W_1^2}^{1/2}} \sqrt{\frac{j_+ - \lambda}{\eta-2j_+}}
\end{split} \]

See that

\[ \espM{\frac{W_{j_+ - \lambda}}{\sqrt{j_+ - \lambda}}}{W_1^2}^{1/2} = \esp{\left( W_1 + \frac{W_{j_+ - \lambda}}{\sqrt{j_+ - \lambda}} \right)^2}^{1/2} = \sqrt{1 + \frac{W_{j_+ - \lambda}^2}{j_+ - \lambda}} \]

And then

\[ \begin{split}
\int_0^\infty p^-(x,j_+ - \lambda,y) \sqrt{1 + \frac{y^2}{j_+ - \lambda}} dy &\leq \int_0^\infty \frac{2xy}{\sqrt{(j_+ - \lambda)^3}} e^{- \frac{(y-x)}{2(j_+ - \lambda)}} \sqrt{1 + \frac{y^2}{j_+ - \lambda}} dy \\
&\leq \int_0^\infty 2xz e^{- \frac12 (z-\frac{x}{j_+ - \lambda})^2} \sqrt{\frac{1 + z^2}{j_+ - \lambda}} dz
\end{split}   \]

Thus we obtain the upper bound with $\lambda = \eps j_+$

\[ \begin{split}
\frac{4x}{\sqrt{2\pi j_+}} \frac{M^{1/4}}{n^{1/9}} e^{- (x_{j,n} - \frac{M^{1/4}}{n^{1/9}})^2/2j_+} \sqrt{\frac{j_+(1 - \eps)}{\eta-2j_+}} \frac{e^{\lambda(x,\eta)^2 + \frac12 \alpha^2 (\eta - j_+ + \eps) - \frac{x^2}{\eta}}}{\frac{x}{\sqrt{\eta}}(1 - e^{-\frac12 \lambda(x,\eta)^{-2}})} \int_0^\infty z e^{- \frac12 (z-\frac{x}{j_+(1 - \eps)})^2} \sqrt{\frac{1 + z^2}{j_+(1 - \eps)}} dz\\
\leq (cst.) \frac{2^{-j/2}}{M^{1/4}} \frac{\eta}{\sqrt{\eta-2j_+}} \frac{e^{\lambda(x,\eta)^2 + \frac12 \alpha^2 (\eta - j_+ + \eps) - \frac{x^2}{\eta}}}{1 - e^{-\frac12 \lambda(x,\eta)^{-2}}} \int_0^\infty z e^{- \frac12 (z-\frac{x}{j_+(1 - \eps)})^2} \sqrt{1 + z^2} dz
\end{split} \]

Since the part that depends in $x = R_\lambda$ have finite expectation, with a uniform bound over $(j,M,n)$, we thus proved that

\begin{equation}\label{eq:simp-borne-2-debut}
\begin{split}
\frac{\probaM{x_{j,n}}{0 < \inf_{0 \leq s \leq j_+} R_s \leq \frac{M^{1/4}}{n^{1/9}}}}{\probaM{x_{j,n}}{\inf_{0 \leq s \leq j_+} R_s > 0}} \leq (cst.) \frac{2^{-j/2}}{M^{1/4}} \frac{\eta}{\sqrt{\eta-2j_+}} \leq (cst.) \frac{2^{-j/2}}{M^{1/4}}
\end{split}
\end{equation}

where we used that $j \leq J_{n,M}^\eta$ thus $4j_+ \leq \eta$.

When $j \in \mathset{J_{n,M}^\eta -1, J_{n,M}^\eta}$ we instead bound the probability in (<ref>) by $1$, thus we only need to bound

\[
\espM{x}{\indic{\inf_{0 \leq s \leq j_+ - \lambda} R_s > 0, R_{j_+ - \lambda} \geq x_{j,n}} \probaM{R_{j_+ - \lambda}}{0 < \inf_{0 \leq s \leq j_+} \bar R_s \leq \frac{M^{1/4}}{n^{1/9}}}}
\]

We reuse previous calculations to get the upper bound

\[ \frac{2}{\sqrt{2\pi j_+}} \frac{M^{1/4}}{n^{1/9}} e^{- (x_{j,n} - \frac{M^{1/4}}{n^{1/9}})^2/2j_+} \probaM{x}{\inf_{0 \leq s \leq j_+ - \lambda} R_s > 0} \leq \frac{2}{\sqrt{2\pi j_+}} \frac{M^{1/4}}{n^{1/9}} \frac{x}{\sqrt{j_+ - \lambda}} \]

leading to

\[ \frac{2^{-j/2}}{M^{1/4}} \frac{\sqrt{2\pi \eta}}{\sqrt{j_+ - \lambda}} \frac{e^{\lambda(x,\eta)^2 + \frac12 \alpha^2 (\eta - j_+ + \eps) - \frac{x^2}{\eta}}}{1 -e^{-\frac12 \lambda(x,\eta)^{-2}}} \]

and since $\eta \leq 4j_+$ we get the same upper bound than in (<ref>) but with a different multiplicative constant.

Overall, we proved that we have for any $n \geq 1$,

\[ \proba{\mathscr{E}_{n,M}^\eta} \leq (cst.) \sum_{j = 0}^{J_{n,M}^\eta} \left( \frac{(x_{j,n})^2}{j_+} + \frac{2^{-j/2}}{M^{1/4}} \right) \leq (cst.) \sum_{j = 0}^{+\infty} \left( \frac{(x_{j,n})^2}{j_+} + \frac{2^{-j/2}}{M^{1/4}} \right) \xrightarrow[M \to \infty]{} 0 \]

which proves the theorem as the constant does not depend on $j$, $M$ or $n$.

§.§ Convergence of the partition function

For any $M \geq 1$, there exists a process $W$ such that

\[ \varlimsup_{n \to \infty} \left| \tilde{\mathcal{Z}}_{n,h}^{\omega,\beta} (|\Delta_n - s_* n^{2/9}| \leq M ) - \psi_h \sum_{k = -M}^M e^{-\beta W_{s_*^n + k}} \right| = 0 \]

In particular, one have the following asymptotic development

\[ Z_{n,h}^{\omega,\beta} = \psi_h (1+\bar{o}(1)) \exp \left( -\frac{3}{2} h c_h n^{1/3} + \beta n^{1/6} X_{c_h} + \beta n^{1/9} Y_{s_*} \right) \sum_{k \in \ZZ} e^{-\beta \mathcal{W}_{s_*^n + k}} \]

\[ \tilde{\mathcal{Z}}_{n,h}^{\omega,\beta} = \psi_h \sum_T e^{\beta n^{1/9} \left( Y_{\Delta^T_n n^{-2/9}} - Y_{s_*} \right)} = \psi_h \sum_{T \geq 0} e^{\beta n^{1/9} R_{\Delta^T_n n^{-2/9} - s_*}} \]

We change the summation indexes by $k = \Delta_n - \lfloor s_* n^{2/9} \rfloor \in \ZZ$ and we use Conjecture <ref> to write for any $n \geq n_0$:

\[ \mathcal{Z}_{n,h}^{\omega,\beta} (|\Delta^T_n - s_* n^{2/9}| \leq M ) = \psi_h \sum_{k = 1 - M}^M  e^{-\beta n^{1/9} R_{(k + s_*^n) n^{-2/9}}} = \psi_h \sum_{k = 1 - M}^M  e^{-\beta \mathcal{W}_{k + s_*^n}} \]

proving the desired convergence.

\[ \mathbf{P}_{n,h}^{\omega, \beta} \left( \Delta_n - \lfloor s_* n^{2/9} \rfloor = k \right) = \frac{1}{\theta_\omega (n)} e^{-\beta W_{s_*^n + k}} \quad , \quad \theta_\omega (n) \defeq \sum_{k \in \ZZ} e^{-\beta W_{s_*^n + k}(\omega)} \]

Upper edge of the polymer and the related process $e^{-\beta \mathcal{W}}$ for $s_* < 0$, with a “high probability zone” for the location of the edge.

Missing 'biblatex' package
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Institute of Mathematical Statistics

journaltitleThe Annals of Probability
titleThe intermediate disorder regime for directed polymers in dimension $1+ 1$

issn0178-8051, 1432-2064
journaltitleProbability Theory and Related Fields
titleBrownian motion in cones

titleLévy processes. Theory and applications

titleThe stochastic heat equation with multiplicative Lévy noise: Existence, moments, and intermittency

Institute of Mathematical Statistics
Bernoulli Society

journaltitleElectronic Journal of Probability
titleOne-dimensional polymers in random environments: stretching vs. folding

journaltitleBulletin Des Sciences Mathematiques
titlePath transformations connecting Brownian bridge, excursion and meander

titleScaling limits for the random walk penalized by its range in dimension one

journaltitleJournal of the European Mathematical Society
titlePolynomial chaos and scaling limits of disordered systems

journaltitleInventiones mathematicae
titleThe critical 2d stochastic heat flow

titleDirected polymers in random environments

Academic press

titleStrong approximations in probability and statistics

Institute of Mathematical Statistics

journaltitleThe Annals of Probability
titleRandom walks in cones

Institute of Mathematical Statistics

journaltitleThe Annals of Probability
titlePoly-logarithmic localization for random walks among random obstacles

Institute of Mathematical Statistics

journaltitleThe Annals of Probability
titleWeak Convergence to Brownian Meander and Brownian Excursion
Brownian bridge,Brownian excursion,Brownian meander,Brownian motion,distribution of supremum,weak convergence

John Wiley & Sons

titleAn Introduction to Probability Theory and its Applications

journaltitleAnnales De L Institut Henri Poincare-probabilites Et Statistiques
titleBrownian Motion in a Weyl Chamber, Non-Colliding Particles, and Random Matrices

Cambridge University Press

journaltitleJournal of Applied Probability
titleDensity factorizations for Brownian motion, meander and the three-dimensional Bessel process, and applications

Springer Science & Business Media

titleDiffusion processes and their sample paths: Reprint of the 1974 edition

Institute of Mathematical Statistics
Bernoulli Society

journaltitleElectronic Communications in Probability
titleMoments of the location of the maximum of Brownian motion with parabolic drift

journaltitleThe Annals of Statistics
titleCube root asymptotics

Cambridge University Press

journaltitleJournal of applied probability
titleQuasi-stationary distributions for a Brownian motion with drift and associated limit laws

Institute of Mathematical Statistics

journaltitleThe Annals of Probability
titleThe Growth of Random Walks and Levy Processes

Springer Science & Business Media

titleContinuous martingales and Brownian motion

Courier Dover Publications

titleStudies in the theory of random processes