In particular, for $x\in (\ell_j,r_j)$, we have

\begin{align*}
\mathring{\textup{T}}\textup{ent}_{j+1}(x)\geq   \ug(x)-2(k+1-j)sT.
\end{align*}

\begin{equation*}
\sup_{x\neq y\in [\ell_{j },r_{j }]}\left| \frac{\mathring{\textup{T}}\textup{ent}_{j +1}(x)-\mathring{\textup{T}}\textup{ent}_{j+1 }(y)}{x-y} \right| \leq (k+1-j)T.
\end{equation*}

The strategy of proving (<ref>) and (<ref>) is to compare $ {\textup{T}}\textup{ent}_{j +1}$ with $\mathring{\textup{T}}\textup{ent}_{j +1}$ using Lemma <ref>.

Suppose $p\in \mathring{P}_{j+1}\cap (\ell_i,\ell_i+2d']$ for some $j\in [1,k]_{\mathbb{Z}}$ and $i\in [j+1,k]_{\mathbb{Z}}$. We have

\begin{equation*}
|\ug (\ell_i)-\ug (p)| \leq 2(k+1-j)T^{-1/2}.
\end{equation*}

Similarly, suppose  $p\in \mathring{P}_j\cap [r_i-2d',r_i)$ for some $j\in [1,k]_{\mathbb{Z}}$ and $i\in [j+1,k]_{\mathbb{Z}}$. Then

\begin{equation*}
|\ug(r_i)-\ug (p)| \leq  2(k+1-j)T^{-1/2}.
\end{equation*}

We postpone the proof of Lemma <ref> and show (<ref>) and (<ref>) first. For (<ref>), it remains to prove

\begin{align*}
\textup{Tent}_{j+1}(x)\geq \mathring{\textup{T}}\textup{ent}_{j+1}(x) - (k+1-j)sT.
\end{align*}

Note that $\textup{Tent}_{j+1}(x) $ agrees with $ \mathring{\textup{T}}\textup{ent}_{j+1}(x)$ except for $x$ near $\ell_i$ and $r_i $.

We give the proof for the case $x$ close to $\ell_i$. We may assume that $\mathring{P}_j\cap (\ell_i,\ell_i+2d')\neq \phi$. Otherwise, $\textup{Tent}_{j+1}(x)=\mathring{\textup{T}}\textup{ent}_{j+1}(x)$ near $\ell_i$. Let $p=\mathring{P}_j\cap (\ell_i,\ell_i+2d')$, $q=\ell_i$,  $p_-<p<p_+$ be the elements in $\mathring{P}_{j+1}$ next to $p $. In the interval $[p_-,p_+]$, the tent maps $\mathring{\textup{T}}\textup{ent}_{j+1}(x)$ and $\mathring{\textup{T}}\textup{ent}_{j+1}(x)$ are piecewise linear functions given by

\begin{equation*}
\mathring{\textup{T}}\textup{ent}_{j+1}(x)= \left\{ \begin{array}{cc}
\ug(x) & x=p\ \textup{or}\ x=p_\pm ,\\
\textup{linear inteperlation} & x\in (p_-,p)\cup(p,p_+) .
\end{array} \right.
\end{equation*}

\begin{equation*}
{\textup{T}}\textup{ent}_{j+1}(x)= \left\{ \begin{array}{cc}
\ug(x) & x=q\ \textup{or}\ x=p_\pm ,\\
\textup{linear inteperlation} & x\in (p_-,q)\cup(q,p_+) .
\end{array} \right.
\end{equation*}

To compare $\mathring{\textup{T}}\textup{ent}_{j+1}(x)$ and ${\textup{T}}\textup{ent}_{j+1}(x)$, we apply Lemma <ref>. From Lemma <ref>,

\begin{align*}
|\ug ( \ell_i)-\ug(p)| \leq 2 (k+1-j)T^{-1/2}.
\end{align*}

Note that $|p-q|\leq 2d'=2T^{-3/2}$ and $|p_\pm-p|\geq s\geq 1$. Applying Lemma <ref> yields

\begin{align*}
\textup{Tent}_{j+1}(x)-\mathring{\textup{T}}\textup{ent}_{j+1}(x)\geq -2 (k+1-j)T^{-1/2}  - 2T^{-3/2}\max\{|\mathring{m}_-|,|\mathring{m}_+|\} .
\end{align*}

Here $\mathring{m}_\pm$ is the slope of $\mathring{\textup{T}}\textup{ent}_{j+1}$ in $[p_-,p]$ and $[p,p_+]$. Because the slope of $\mathring{\textup{T}}\textup{ent}_{j+1}$ is bounded by $(k+1-j)T$, we get

$$\textup{Tent}_{j+1}(x)\geq  \mathring{\textup{T}}\textup{ent}_{j+1}(x) -4 (k+1-j)T^{-1/2}.$$

The proof of (<ref>) near $\ell_i$ is finished.

Next, we turn to (<ref>). We again apply Lemma <ref>. The difference between the slopes of $\mathring{\textup{T}}\textup{ent}_{j+1}$ and $ {\textup{T}}\textup{ent}_{j+1}$ is bounded by $8 (k+1-j)T^{-1/2}.$ Therefore, (<ref>) holds true. The argument for the case near $r_i$ is similar.

By symmetry, it suffices to prove the case $p\in \mathring{P}_{j+1}\cap (\ell_i,\ell_i+2d')$. We start by showing $\ug(\ell_i)\leq \ug(p)$. By the concavity of $\mathfrak{c}_{j+1}$,

\begin{align*}
\ug(\ell_i)=&g_{j+1}(\ell_i)\leq \mathfrak{c}_{j+1}(\ell_i) \leq  \mathfrak{c}_{j+1}(p)+(\mathfrak{c}'_{j+1})_+(p)(\ell_i-p).
\end{align*}

Because $p$ is an extreme point of $\mathfrak{c}_{j+1}$,

\begin{align*}
\mathfrak{c}_{j+1}(p)+(\mathfrak{c}'_{j+1})_+(p)(\ell_i-p)=g_{j+1}(p)-(\mathfrak{c}'_{j+1})_+(p)(p-\ell_i).
\end{align*}

From the definitions of $\ug$ and and $g_{j+1}$, see (<ref>) and (<ref>),

\begin{align*}
g_{j+1}(p)-(\mathfrak{c}'_{j+1})_+(p)(p-\ell_i) =&\ug(p)-(\mathfrak{c}'_{j+1})_+(p)(p-\ell_i) .
\end{align*}

By Lemma <ref>, $(\mathfrak{c}'_{j+1})_+(p) \geq -p-10^{-1}T\geq 0$. We conclude $\ug(\ell_i)\leq \ug(p)$.

Next, we show that $\ug(p)\leq \ug(\ell_i)+2(k+1-j)T^{-1/2}.$ Because $p\in (\ell_i,\ell_i+2d')$,

\begin{align*}
\ug(p)=f_{i+1}(p)=g_{i+1}(p).
\end{align*}

By the concavity of $\mathfrak{c}_{i+1}$,

\begin{align*}
g_{i+1}(p)\leq \mathfrak{c}_{i+1}(p)\leq \mathfrak{c}_{i+1}(\ell_i)+(\mathfrak{c}_{i+1}')_+(\ell_i)(p-\ell_i).
\end{align*}

Because $\ell_i$ is an extreme point of $\mathfrak{c}_{i+1}$, it holds that

$$\mathfrak{c}_{i+1}(\ell_i)=g_{i+1}(\ell_i)=f_{i+1}(\ell_i)\leq \ug(\ell_i).$$

From Lemma <ref>, we have
$$\mathfrak{c}_{i+1}'(\ell_i)\leq -\ell_i+10^{-1}T\leq (k+2-i)T\leq (k+1-j)T.$$

Together with $p-\ell_i\leq 2T^{-3/2}$, we conclude that

\begin{align*}
\ug(p)\leq \ug(\ell_i)+2(k+1-j)T^{-1/2}.
\end{align*}

§.§ Jump ensemble

In this subsection, we use the Pole set $P_{j+1}$ to construct the jump ensemble.  We put all $P_{j+1}$ together to form the joint Pole set.

\begin{equation}\label{def:Pstar}
P^*\coloneqq\bigcup_{j=1}^{k }P_{j+1}.
\end{equation}

Recall that $d=64T^{-4}(\log T)^2$. For $j\in [1,k-1]_\mathbb{Z}$, define

\begin{equation*}
\textup{Int}_j\coloneqq \bigg(\bigcup_{p\in P^*\cap (\ell_j,\ell_{j+1}]} [p-d,p]\bigg) \cup \bigg(\bigcup_{p\in P^*\cap  [r_{j+1},r_j)} [p ,p+d]\bigg)
\end{equation*}

\begin{equation*}
\textup{Int}_k\coloneqq \bigg(\bigcup_{p\in P^*\cap (\ell_k,0) } [p-d,p]\bigg) \cup \bigg(\bigcup_{p\in P^*\cap  [0,r_k)} [p ,p+d]\bigg)
\end{equation*}

From (<ref>), we have $\textup{Int}_j\subset (\ell_j,\ell_{j+1}]\cup [r_{j+1},r_j) $ for $j\in [1,k-1]_{\mathbb{Z}}$ and $\textup{Int}_k\subset (\ell_k,r_k)$. $\textup{Int}_j$ and $\textup{Int}_k$ indicates the region where $J$ interacts with the boundary. The index $j$ in $\textup{Int}_j$ records the layer of the boundary data. For instance, in region $\textup{Int}_j$, the lower boundary is given by $f_{j+1}$.

This picture illustrates $W^j_{\textup{jump}}$ near $\ell_j$. In the thick pillar with width $d'$, we use the Hamiltonian $\textbf{H}_t$ as in the scaled KPZ line ensemble. In the thin pillars with width $d$, a weaker Hamiltonian $\textbf{H}_t^{j,j+1}$ is used. There is no interaction on other places.

Now we are ready to define the jump weight $W_{\textup{jump}}$. Recall that we fix the triple $(\bar{\ell},\bar{r},f_{\textup{J}})\in\mathcal{G}$. Fix $j\in [1,k]_{\mathbb{Z}}$. Let

\begin{equation*}
\begin{split}
W^{j}_{\textup{jump}}(\mathcal{L}_j ;f|_{\textup{J}})\coloneqq &\exp\left( -\int_{[\ell_j,\ell_j+d']\cup [r_j-d' ,r_j] } \mathbf{H}_t(\ug (x)-\mathcal{L}_j(x)) dx \right)\\
& \times \exp\left( -\sum_{i=j}^k \int_{\textup{Int}_i} \mathbf{H}_t^{j,i+1}(\ug (x)-\mathcal{L}_j(x)) dx \right) .
\end{split}
\end{equation*}

Here $\ug(x)$ is the lower boundary defined in (<ref>) and $\mathbf{H}_t^{i,{j+1}}(x)$ is the Hamiltonian defined in (<ref>). $\mathbf{H}_t^{i,{j+1}}(x)$ is chosen to guarantee Lemma <ref>.

For any $\ell<r$, we define the truncated version

\begin{equation}\label{def:Wjump_j}
\begin{split}
W^{j, [\ell,r]}_{\textup{jump}}(\mathcal{L}_j ;f|_{\textup{J}})\coloneqq &\exp\left( -\int_{([\ell_j,\ell_j+d']\cup [r_j-d' ,r_j])\cap[\ell,r] } \mathbf{H}_t(\ug (x)-\mathcal{L}_j(x)) dx \right)\\
& \times \exp\left( -\sum_{i=j}^k \int_{\textup{Int}_i\cap[\ell,r]} \mathbf{H}_t^{j,i+1}(\ug (x)-\mathcal{L}_j (x)) dx \right) .
\end{split}
\end{equation}

The jump weight $W_{\textup{jump}} $ is defined by

\begin{align}
W_{\textup{jump}}(\mathcal{L} ;f_{\textup{J}})\coloneqq &\prod_{j=1}^k W^{j}_{\textup{jump}}(\mathcal{L}_j;f_{\textup{J}}).\label{def:Wjump}
\end{align}

In $W_{\textup{jump}} $, the random curves don't interact with each other. In particular, the jump curves $J_i$ and $J_j$, defined in (<ref>), are independent of each other if $i\neq j$.

The $j$-th jump curve $J_j$ interacts with $\ug$ near $\ell_j$ and $r_j$ through $\mathbf{H}_t$. This is the same Hamiltonian in $W_{\mathbf{H}_t}^{ \textup{dom}(\bar{\ell},\bar{r}), f }$. See (<ref>). The reason to keep the Hamiltonian function comes from the possibility that  $f_{\textup{J}}$ may go out of order. If $f_{j+1}>f_j$ near $\ell_j$, the interaction in $-\mathbf{H}_t(f_{j+1}(x)-\mathcal{L}_j(x))$ near $\ell_j$ becomes so drastic such that any mollification would make a huge difference. Thus we preserve the original Hamiltonian.

The other location where $J_j$ interacts with $\ug(x)$ is determined by the Pole set $P^*$, which depends on $f_{\textup{J}}$.

Having the jump weight $W_{\textup{jump}}$, $W_{\textup{rest}}$ is defined by

\begin{align}
W_{\textup{rest}}(\mathcal{L} ;f_{\textup{J}})\coloneqq &W^{\textup{dom}(\bar{\ell},\bar{r}),f}_{\mathbf{H}_t}(\mathcal{L} )/ W_{\textup{jump}}(\mathcal{L} ;f_{\textup{J}}).\label{def:Wrest}
\end{align}

Here $f=f_{\textup{J}}|_{\textup{bdd}(\bar{\ell},\bar{r})}$.

We finish this section by the following lemma.

It holds that

\begin{align*}
W_{\textup{rest}}(\mathcal{L} ;f_{\textup{J}})\leq 1.
\end{align*}

This is equivalent to $W^{\textup{dom}(\bar{\ell},\bar{r}),f}_{\mathbf{H}_t}(\mathcal{L} )\leq   W_{\textup{jump}}(\mathcal{L} ;f_{\textup{J}})$.
From (<ref>), we can rewrite $W_{\mathbf{H}_t}^{ \textup{dom}(\bar{\ell},\bar{r}), f }$ in (<ref>) as

\begin{equation*}
\begin{split}
W_{\mathbf{H}_t}^{ \textup{dom}(\bar{\ell},\bar{r}), f }(\mathcal{L})= &\prod_{j=1}^{k-1} \exp\left( -\int_{[\ell_j,\ell_{j+1}]\cup [r_{j+1},r_j]}  \mathbf{H}_t(\ug (x)-\mathcal{L}_j(x))\, dx  \right)\\
&\times \prod_{j=1}^{k-1} \exp\left( -\int_{[\ell_{j+1},r_{j+1}]} \mathbf{H}_t(\mathcal{L}_{j+1}(x)-\mathcal{L}_j(x))\, dx \right)\\
&\times \exp\left( -\int_{[\ell_{k},r_k]}  \mathbf{H}_t(\ug (x)-\mathcal{L}_k(x))\, dx \right).
\end{split}
\end{equation*}

Suppose $x_0\in (\ell_j,\ell_{j+1})$ for some $j\in [1,k]_{\mathbb{Z}}$, the integrand of $\log W_{\mathbf{H}_t}^{ \textup{dom}(\bar{\ell},\bar{r}), f }$ at $x_0$ is

\begin{align}\label{compare}
-\mathbf{H}_t(\ug (x_0)-\mathcal{L}_j(x_0))-\sum_{i=1}^{j-1}\mathbf{H}_t(\mathcal{L}_{i+1} (x_0)-\mathcal{L}_i(x_0)).
\end{align}

We claim that it is smaller than the integrand of $\log W_{\textup{jump}}(\mathcal{L} ;f_{\textup{J}})$. Suppose $x_0\in (\ell_j,\ell_j+d']$. Because of (<ref>), for any $i \in [1,k]_{\mathbb{Z}}$,

$$\textup{Int}_i\cap (\ell_j,\ell_j+d']=\phi.$$

Hence the integrand of $\log W_{\textup{jump}}(\mathcal{L} ;f_{\textup{J}})$ at $x_0$ is $-\mathbf{H}_t(\ug(x_0)-\mathcal{L}_j(x_0)))$, which is larger than (<ref>). Suppose $x_0\in \textup{Int}_j $, the integrand of $\log W_{\textup{jump}}(\mathcal{L} ;f_{\textup{J}})$ at $x_0$ is

\begin{align*}
-\sum_{i=1 }^j \mathbf{H}_t^{i,j+1}(\ug  (x_0)-\mathcal{L}_i(x_0)).
\end{align*}

In light of Lemma <ref>, it is larger than (<ref>). Suppose $x_0\in ( \ell_j+d',\ell_{j+1})\setminus \textup{Int}_j$. Then the integrand of $\log W_{\textup{jump}}(\mathcal{L} ;f_{\textup{J}})$ at $x_0$ is zero, which is also larger than (<ref>). This completes the discussion for the case $x_0\in (\ell_j,\ell_{j+1})$. Other cases, $x_0\in (r_{j+1},r_{j })$ and $x_0\in (\ell_{k+1},r_{k+1})$ can be treated similarly and we omit the detail. The remaining end points $x_0= \ell_i$ or $x_0= r_i$ have zero contribution. The proof is finished.

§ PROOF OF PROPOSITION <REF>

In this section, we aim to show that

\begin{align*}
\inf \{\mathbb{E}_{J}[ W_{\textup{rest}}]\, |\,  (\bar{\ell},\bar{r},f_{\textup{J}})\in \mathcal{G}   \}  \geq D^{-1} e^{-DT^{5/2}}.
\end{align*}

Here $\mathcal{G}$ is defined in Definition <ref> and $W_{\textup{rest}}$ is defined in (<ref>). Throughout this section, we fix $(\bar{\ell},\bar{r},f_{\textup{J}})\in\mathcal{G}$. In below we explain the strategy of the proof.

Consider the event $\mathsf{NoTouch}$ that requires the jump ensemble $J$ to stay ordered and to lie above $\ug$ except in neighborhoods near $\ell_j$ or $r_j$. Concretely,

\begin{align*}
\mathsf{NoTouch}\coloneqq \{J_j(x)\geq \ug(x)\ \textup{for}\ j\in [1,k]_{\mathbb{Z}}\ \textup{and}\ x\in [\ell_j+d',r_j-d'],\\
J_j(x)\geq J_{j+1}(x)\ \textup{for}\ j\in [1,k-1]_{\mathbb{Z}}\ \textup{and}\ x\in [\ell_{j+1},r_{j+1}]\}.
\end{align*}

When $\mathsf{NoTouch}$ occurs, the weight $W_{\textup{rest}}$ is bounded from below by $e^{-T}$. See Lemma <ref>. Therefore Proposition <ref> is reduced to showing $\mathbb{P}_J(\mathsf{NoTouch})\geq e^{-T^{5/2}}$.

To begin, we show that each curve $J_j$ can jump over $\ug$ in the interval $[\ell_j+d',r_j-d']$ with probability $e^{-T^{5/2}}$. Moreover,a gap of size $T$ between $J_j$ and $\ug$ can be created in the interval $[\ell_j+1,r_j-1]$. This event is denoted by $\mathsf{A}_j$.

Next, we apply induction on the number of curves. Take $j\in [2,k]_{\mathbb{Z}}$. We write $\mathsf{F}_{j+1}$ for the event that $J_1, J_2,\dots, J_{j-1}$ stay ordered and that $J_{j-1}\geq \ug+T $ in $[\ell_{j-1}+1,r_{j-1}-1]$. Suppose that $\mathbb{P}(\mathsf{F}_{j-1})\geq e^{-T^{5/2}}$. Because the $j$-th curve $J_{j}$ is independent of other curves, $\mathbb{P}(\mathsf{F}_{j-1}\cap\mathsf{A}_j)\geq e^{-T^{5/2}}$. To get bounds on $\mathbb{P}_J(\mathsf{F}_j)$, we need to make sure $J_{j}$ does not raise too high and does not touch $J_{j-1}$.

To do so, we apply a stopping domain argument. Consider the leftest and the rightest locations where $J_{j-1}=J_j+2^{-1}T$. Denote them by $\mathfrak{l}''_j$ and $\mathfrak{r}''_j$ respectively. At these two points, we know $J_1\geq J_2\geq \dots \geq J_{j-1}>J_j>\ug.$ Furthermore, $J_{j-1}-J_j=2^{-1}T$ and $J_j-\ug\geq 2^{-1}T$. Applying the strong $\mathbf{H}_t$-Brownian Gibbs property, we resample $J_1,\dots J_j$ in $[\mathfrak{l}''_j, \mathfrak{r}''_j]$. Because the boundary data is now well spaced, we can show that the $J_1,\dots, J_j$, with probability $e^{-T^{5/2}}$, remain ordered and lie beyond $\ug$ in $[\mathfrak{l}''_j, \mathfrak{r}''_j]$. This finishes the proof.

There exists $D=D(k,s)$ such that for all $t\geq 1$, it holds that

\begin{align*}
\mathbbm{1}_{\mathsf{NoTouch}}\cdot W_{\textup{rest}}(;f_{\textup{J}})\geq D^{-1}e^{-DT }\cdot \mathbbm{1}_{\mathsf{NoTouch}}.
\end{align*}

From the definition of $W_{\textup{rest}}$ in (<ref>) and $\mathbf{H}_t^{i,j}\geq 0$, we have

\begin{equation*}
\begin{split}
W_{\textup{rest}}(\mathcal{L};f_{\textup{J}}) \geq  &\prod_{j=1}^{k-1} \exp\left( -\int_{[\ell_j+d',\ell_{j+1}]\cup [r_{j+1},r_j-d']}  \mathbf{H}_t(\ug (x)-\mathcal{L}_j(x))\, dx  \right)\\
&\times \prod_{j=1}^{k-1} \exp\left( -\int_{[\ell_{j+1},r_{j+1}]} \mathbf{H}_t(\mathcal{L}_{j+1}(x)-\mathcal{L}_j(x))\, dx \right)\\
&\times \exp\left( -\int_{[\ell_{k}+d',r_k-d']}  \mathbf{H}_t(\ug (x)-\mathcal{L}_k(x))\, dx \right).
\end{split}
\end{equation*}

When $\mathsf{NoTouch}$ occurs, the integrands above are all bounded from below by $-1$. Hence

\begin{align*}
W_{\textup{rest}}(\mathcal{L};f_{\textup{J}}) \geq e^{-k(r_1-\ell_1)}\geq e^{-2k(k+1)T}.
\end{align*}

There exists a constant $D=D(k,s)$ such that for $E$ large enough, it holds that

\begin{align*}
\mathbb{P}_{J}(\mathsf{NoTouch})\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

From Lemma <ref> and Proposition <ref>

\begin{align*}
\mathbb{E}_J[W_{\textup{rest}}(J;f_{\textup{J}}) ]\geq  \mathbb{P}_J(\mathsf{NoTouch} )\times D^{-1}e^{-DT } \geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

The rest of the section is devoted to prove Proposition <ref>. In subsection <ref>, we deal with a single curve $J_j$ and show that $\mathsf{A}_j$ occurs with probability at least $e^{-T^{5/2}}$. In subsection <ref>, we show that, provided the boundary data is well separated, $J$ remains ordered and lying beyond $\ug$ under resampling. Lastly, we combine the two and finish the induction argument in subsection <ref>.

§.§ Single curve

In this subsection, we focus on a single curve $J_j$ and prove Proposition <ref>. Roughly speaking, Proposition <ref> says the probability that $J_j$ is larger than $\ug$ in $[\ell_j+d',r_j-d']$ is at least $e^{-T^{5/2}}$.

We start by defining $\mathsf{A}_j$. Fix $j\in [1,k]_{\mathbb{Z}}$ and let $\mathsf{A}_j$ be the event that

* $J_j(x)> \ug (x)\ \textup{for all}\ x\in [\ell_j+d',r_j-d']$,

* $J_j(x)> \textup{Tent}_{j+1}(x)+16k(k+1-j)sT\ \textup{for all}\ x\in [\ell_j+1,r_j-1 ]$.

There exists $D =D (k)$ such that for $E$ large enough, we have

\begin{equation*}
\mathbb{P}_{ J }(\mathsf{A}_j)\geq D^{-1}e^{-DT^{5/2}}.
\end{equation*}

We prove Proposition <ref> by gradually lift the curve $J_j$. In below we define the events which record the lifting process.

Let $\mathsf{A}_{j,1}$ be the event such that

* $J_j (\ell_j+d')> f_j(\ell_j)+kT^{1/2},J_j(r_j-d')> f_j(r_j)+kT^{1/2}$,

* $ J_j (\ell_j+1)> f_j(\ell_j)+20k(k+1-j)sT$,

* $J_j(r_j-1)> f_j(r_j)+20k(k+1-j)sT$.

Let $\mathsf{A}_{j,2}$ be the event such that

* $J_j(x)> f_{j+1}(x)\ \textup{for all}\ x\in [\ell_j+d',\ell_j+1]\cup [r_j-1,r_j-d']$,

* $  J_j(\ell_j+1)> \textup{Tent}_{j+1}(\ell_j+1)+18k(k+1-j)sT $

* $ J_j(r_j-1)> \textup{Tent}_{j+1}(r_j-1)+18k(k+1-j)sT $.

Let $\mathsf{A}_{j,3}$ be the event such that

* $J_j(x)> \textup{Tent}_{j+1}(x)+16k(k+1-j) sT\ \textup{for all}\ x\in [\ell_j+1,r_j-1] $.

Recall that $J_j$ is sampled in $[\ell_j,r_j]$ with boundary given by $J_j(\ell_j)=f(\ell_j)$ and $J_j(r_j)=f(r_j)$. In the event $\mathsf{A}_{j,1}$, $J_j$ goes up by $T^{1/2}$ within distance $d'$ and then goes up by $sT$ within distance $1$. We then resample $J_j$ in $[\ell_j+d',\ell_j+1]$ and $[r_j-1,r_j-d']$. Conditioned on $\mathsf{A}_{j,1}$ the boundary data is high enough to ensure $\mathsf{A}_{j,2}$ occurs with probability $D^{-1}$. After that having $\mathsf{A}_{j,2}$, we resample $J_j$ in $[\ell_j+1,r_j-1]$. Applying Theorem <ref> ensures that conditioned on $\mathsf{A}_{j,2}$,  $\mathsf{A}_{j,3}$ occurs with probability $e^{-T^{5/2}}$. This argument is entailed in the following three lemmas. To simplify the notation, we denote $\mathbb{P}^{\textup{dom}(\bar{\ell},\bar{r}),f}_{\free}$ and $\mathbb{E}^{\textup{dom}(\bar{\ell},\bar{r}),f}_{\free}$ by $\mathbb{P}_{\free}$ and $\mathbb{E}_{\free}$ respectively.

There exists $D =D (k,s)$ such that

\begin{align*}
\mathbb{P}_{\free }(\mathsf{A}_{j,1})\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

There exists $D =D (k,s)$ such that

\begin{align*}
\mathbb{P}_{\free }(\mathsf{A}_{j,2}\,|\, \mathsf{A}_{j,1})\geq D^{-1}.
\end{align*}

There exists $D =D (k,s)$ such that for $E$ large enough, it holds that

\begin{align*}
\mathbb{P}_{J}( \mathsf{A}_{j,3}\,|\,\mathsf{A}_{j,2})\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

From (<ref>), we check $\mathsf{A}_{j,2}\cap\mathsf{A}_{j,3}\subset \mathsf{A}_j$.
By Lemmas <ref> and <ref>,

\begin{align*}
\mathbb{P}_{\free }( \mathsf{A}_{j,2} )\geq \mathbb{P}_{\free }( \mathsf{A}_{j,2}\,|\,\mathsf{A}_{j,1} )\times \mathbb{P}_{\free }( \mathsf{A}_{j,1} ) \geq D^{-1}e^{-D^{5/2}}.
\end{align*}

By the stochastic monotonicity Lemma <ref>,

\begin{align*}
\mathbb{P}_{J}( \mathsf{A}_{j,2} )\geq \mathbb{P}_{\free }( \mathsf{A}_{j,2} )\geq D^{-1}e^{-D^{5/2}}.
\end{align*}

With Lemma <ref>, we conclude that

\begin{align*}
\mathbb{P}_{J}(\mathsf{A}_{j})\geq \mathbb{P}_{J}(\mathsf{A}_{j,2}\cap\mathsf{A}_{j,3})= \mathbb{P}_{J}( \mathsf{A}_{j,3}|\mathsf{A}_{j,2})\times \mathbb{P}_{J}( \mathsf{A}_{j,2} )\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

In the rest of this subsection, we prove Lemmas <ref>, <ref> and  <ref>.

The first condition in $\mathsf{A}_{j,1}$ requires $J_j$ to raise by an amount $T^{1/2}$ within a distance $d'=T^{-3/2}.$ This event occurs with probability about $e^{-T^{5/2}}.$ The second and the third condition requires $J_j$ to raise by an amount $T$ within a distance $1$. This event occurs with probability about $e^{-T^{2}}.$ In below we carry out this calculation.

Under the law $\mathbb{P}_{\free}$, $B_j(\ell_j+d')$ is a Gaussian random variable with mean and variance

\begin{align*}
m=\frac{(r_j-\ell_j-d')f_j(\ell_j)+d' f_j(r_j)}{r_j-\ell_j},\ \sigma^2=\frac{(r_j-\ell_j-d')d'}{r_j-\ell_j}.
\end{align*}

Let $N$ be a Gaussian random variable with mean $0$ and variance $1$. It holds that

\begin{align*}
&\mathbb{P}_{\free }(B_j(\ell_j+d')> f_j(\ell_j)+kT^{1/2})=\mathbb{P}(N>\sigma^{-1}(f_j(\ell_j)-m)+ \sigma^{-1} kT^{1/2} ).
\end{align*}

From Condition C1 and Remark <ref>, it holds that $|\ell_j+(k+j-1)T|,|r_j-(k+j-1)T|\leq 10^{-1}T.$ Recall that $d'=T^{-3/2}$. Hence $\sigma^{-1} \leq 2^{1/2}T^{3/4}$ and then

$$\sigma^{-1}kT\leq DT^{5/4}.$$

From Condition C2 , we have
$\left|\frac{f(\ell_j)-f(r_j)}{\ell_j-r_j}\right|\leq DT.$ Thus

\begin{align*}
\sigma^{-1}| f_j(\ell_j)-m | \leq DT^{1/4}.
\end{align*}

As a result,

\begin{align*}
\mathbb{P}_{\free }(B_j(\ell_j+d')> f_j(\ell_j)+kT^{1/2}) \geq \mathbb{P}(N\geq DT^{5/4})\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

Conditioned on $B_j(\ell_j+d')=y$, $B_j(\ell_j+1)$ is a normal distribution with mean and variance

\begin{align*}
m=\frac{(r_j-\ell_j-1 ) y  +(1-d') f_j(r_j) }{ r_j-\ell_j- d'} ,\ \sigma^2=\frac{(r_j-\ell_j-1 ) (1-d') }{ r_j-\ell_j-d'}.
\end{align*}

\begin{align*}
&\mathbb{P}_{\free }(B_j(\ell_j+1')\geq f_j(\ell_j)+20k(k+1-j) sT\, |\,  B_j(\ell_j+d' )=y )\\
=&\mathbb{P}(N>\sigma^{-1}(f_j(\ell_j)-m)+20\sigma^{-1} k(k+1-j)sT)
\end{align*}

Because $\sigma \approx 1$,

\begin{align*}
20\sigma^{-1} k(k+1-j)sT\leq DT.
\end{align*}

When $y\geq f_j(\ell_j)+kT^{1/2}$,

\begin{align*}
f_j(\ell_j)-m \leq \frac{(1-d')(f_j(r_j)-f_j(\ell_j))}{r_j-\ell_j-d'}-\frac{(r_j-\ell_j-1) kT^{1/2}}{r_j-\ell_j-d'}\leq DT.
\end{align*}

As a result,

\begin{align*}
&\mathbbm{1}\{y\geq f_j(\ell_j)+kT^{1/2}\}\times \mathbb{P}_{\free }(B_j(\ell_j+1')> f_j(\ell_j)+20k(k+1-j) sT\, |\,  B_j(\ell_j+d' )=y )\\
\leq & \mathbbm{1}\{y\geq f_j(\ell_j)+kT^{1/2}\}\times D^{-1}e^{-DT^2}.
\end{align*}

This finishes the proof for $\ell_j+d'$ and $\ell_j+1$. The one for $r_j-d'$ and $r_j-1$ is similar and we omit the detail.

We first bound $f_{j+1}$ by a linear function. By (<ref>) and (<ref>), it holds that for $x\in (\ell_j,\ell_j+1]$,

\begin{align*}
f_{j+1}(x)=\ug(x) \leq  f_{j+1}(\ell_j)+(x-\ell_j)(k+1-j)T.
\end{align*}

By Conditions C4, we obtain

\begin{align*}
f_{j+1}(x)=\ug(x) \leq f_j(\ell_j)+3\Delta+(x-\ell_j)(k+1-j)T=:L(x).
\end{align*}

Next, we show that when $\mathsf{A}_{j,1}$ occurs, it holds that

\begin{align*}
& L(\ell_j+d')+2^{-1}kT^{1/2} < J_j(\ell_j+d'),\\
& L(\ell_j+1)+18 k(k+1-j) sT <    J_j(\ell_j+1).
\end{align*}

They can be deduced by

\begin{align*}
L(\ell_j+d')= f_j(\ell_j)+3\Delta+ (k+1-j)T^{-1/2} \leq & f_j(\ell_j)+2^{-1}kT^{1/2}\\
<  &J_j(\ell_j+d')-2^{-1}kT^{1/2},
\end{align*}

\begin{align*}
L(\ell_j+1)= f_j(\ell_j)+3\Delta+ (k+1-j)T \leq &f_j(\ell_j)+2k(k+1-j)T\\
<  &J_j(\ell_j+1)-18 k(k+1-j) sT.
\end{align*}

In particular, the second condition in $\mathsf{A}_{j,2}$ is ensured by $\mathsf{A}_{j,1}$.

Applying a similar argument near $r_j$, the third condition in $\mathsf{A}_{j,2}$ is also ensured by $\mathsf{A}_{j,1}$. Furthermore,

\begin{align*}
\mathsf{A}_{j,1}\cap\left\{\left|J_j^{[\ell_j+d',\ell_j+1]}\right|\leq 2^{-1}kT^{1/2} \right\}\cap\left\{\left|J_j^{[r_j-1 ,r_j-d']}\right|\leq 2^{-1}kT^{1/2} \right\}\subset \mathsf{A}_{j,2}.
\end{align*}

We conclude that

\begin{align*}
\mathbb{P}_{\free }(\mathsf{A}_{j,2}\,|\,\mathsf{A}_{j,1})\geq D^{-1}.
\end{align*}

We first replace $\mathsf{A}_{j,2}$ by $\mathsf{A}_{j,4}$, which fixes the value of $J_j(x)$ at $x=\ell_j+1$ and $x=r_j-1$. Let

\begin{align*}
\mathsf{A}_{j,4}\coloneqq \{    J_j(x)=\textup{Tent}_{j+1}(x)+18k(k+1-j)sT\ \textup{for}\ x\in\{\ell_j+1,r_j-1\}   \}.
\end{align*}

From the stochastic monotonicity Lemma <ref>, it holds that

\begin{align*}
\mathbb{P}_{J}(\mathsf{A}_{j,3}\,|\,\mathsf{A}_{j,2})\geq\mathbb{P}_{J}(\mathsf{A}_{j,3}\,|\,\mathsf{A}_{j,4}).
\end{align*}

It remains to prove that

\begin{align*}
\mathbb{P}_{J}(\mathsf{A}_{j,3}\,|\,\mathsf{A}_{j,4})\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

Consider the event

\begin{equation*}
\begin{split}
\mathsf{L}_j \coloneqq \{ \mathcal{L}_{j}(p)\leq &\textup{Tent}_{j+1}(p)- \Delta_k\  \textup{for some}\  p\in P_{j+1}\cap [\ell_j+1,r_j-1] \}.
\end{split}
\end{equation*}

This event represents the scenario that $J_j$ goes below $\ug$ by an amount $\Delta_k$ at some point in $P_{j+1}\cap [\ell_j+1,r_j-1]$.

Step 1: We show in this step that conditioned on $\mathsf{A}_{j,4}$, $\mathsf{L}_j$ is unlikely.

\begin{align*}
\mathbb{P}_J(\mathsf{L}_j  |\mathsf{A}_{j,4})\leq 2^{-1}.
\end{align*}

We begin by controlling the local fluctuation of $J_j$. Recall that we write $\mathbb{P}_{\free}$ and $\mathbb{E}_{\free}$ for $\mathbb{P}^{\textup{dom}(\bar{\ell},\bar{r}),f}_{\free}$ and $\mathbb{E}^{\textup{dom}(\bar{\ell},\bar{r}),f}_{\free}$ respectively. From (<ref>) and (<ref>), we can apply Lemma <ref> to get

\begin{align*}
\mathbb{P}_ {\free}(B_{j}(x)\geq \ug(x)\ \textup{for}\ x\in [\ell_j+1,r_j-1] \,|\,\mathsf{A}_{j,4})\geq D^{-1}e^{-DT^3}.
\end{align*}

When such event occurs, $W^{j,[\ell_j+1,r_j-1]}_{\textup{Jump} }$ is bounded from below by some constant $D^{-1}$.  Therefore,

$$\mathbb{E}_{\free}\left[W^{j,[\ell_j+1,r_j-1]}_{\textup{Jump} }\,|\,\mathsf{A}_{j,4} \right]\geq D^{-1}e^{-DT^3}.$$

\begin{equation*}
\mathsf{Flct}_j \coloneqq \{| {J}_{j}(x)- {J}_j(y)|\leq \Delta\ \textup{for all}\ x,y\in [\ell_j+1 ,r_j-1]\ \textup{with}\ |x-y|\leq d \}.
\end{equation*}

When $\mathsf{A}_{j,4}$ occurs, $ J_j(x)=\textup{Tent}_{j+1}(x)+18k(k+1-j)sT $ for $x\in \{\ell_j+1,r_j-1\}$. From (<ref>), we have

\begin{align*}
\left| \frac{J_j(\ell_j+1)-J_j(r_j-1) }{r_j-\ell_j-2}\right|=\left| \frac{\textup{Tent}_{j+1}(\ell_j+1)-\textup{Tent}_{j+1}(r_j-1) }{r_j-\ell_j-2}\right|\leq 2kT.
\end{align*}

\begin{align*}
\mathbb{P}_{\free}  (\mathsf{Flct}^{\textup{c}}_j\,|\, \mathsf{A}_{j,4} )\leq &\mathbb{P}_{\free}(\omega_{d,[\ell,r]}(B^{[\ell,r]})\geq \Delta-2kdT)\\
\leq &\mathbb{P}_{\free}(\omega_{d,[\ell,r]}(B^{[\ell,r]})\geq 2^{-1} \Delta)\leq De^{-D^{-1}T^4}.
\end{align*}

\begin{align*}
\omega_{d,[\ell,r]}(B^{[\ell,r]})\coloneqq \max_{x,y\in [\ell,r],\ |x-y|\leq d}\left|B^{[\ell,r]}(x)-B^{[\ell,r]}(y)  \right|.
\end{align*}

As a result, for $E$ large enough, we have

\begin{align*}
\mathbb{P}_{J}  ( \mathsf{Flct}^{\textup{c}}_j\,|\, \mathsf{A}_{j,4}  )\leq De^{DT^3-D^{-1}T^4}\leq 4^{-1}.
\end{align*}

Next, we show that $\mathsf{L}_j\cap  \mathsf{Flct}_j $ is unlikely. From Condition C3, $\mathsf{L}_j $ and $  \mathsf{Flct}_j $ imply there exists an interval in $\textup{Int}_j\cap [\ell_j+1,r_j-1]$ with length $d$ where

\begin{align*}
J_j(x)\leq \ug(x)-k\Delta-k\log k.
\end{align*}

Together with

\begin{align*}
d\, \textbf{H}_{j,k+1}(k\Delta+k\log k)\geq d e^{\Delta }=64T^4(\log T)^2,
\end{align*}

it holds that

\begin{align*}
\mathbbm{1}_{ \mathsf{L}_j \cap  \mathsf{Flct}_j  } \times W^{j,[\ell_j+1,r_j-1]}_{\textup{jump}}\leq  e^{-64T^4(\log T)^2}\times \mathbbm{1}_{ \mathsf{L}_j \cap  \mathsf{Flct}_j  } .
\end{align*}

As a result, for $E$ large enough, it holds that

\begin{align*}
&\mathbb{P}_{J}  \left(\mathsf{L}_j \cap \mathsf{Flct}_j \,|\,\mathsf{A}_{j,4}\right)
\leq  \mathbb{E}_{\free} \left[ W^{j, [\ell_{j}+1,r_j-1]}_{\textup{jump}} \,|\,\mathsf{A}_{j,4} \right]^{-1}\times e^{-64T^4(\log T)^2}\\
\leq & De^{DT^3-64T^4(\log T)^2}\leq 4^{-1}.
\end{align*}

Putting the above estimates together, we conclude that

\begin{align*}
&\mathbb{P}_{J}  (\mathsf{L}_j\,|\,\mathsf{A}_{j,4} )\leq \mathbb{P}_{J}  (\mathsf{L}_j \cap\mathsf{Flct}_j\,|\,\mathsf{A}_{j,4})+\mathbb{P}_{J}  ( \mathsf{Flct}^{\textup{c}}_j\,|\,\mathsf{A}_{j,4})\leq 2^{-1}.
\end{align*}

Step 2: In this step, we want to prove

\begin{align*}
\mathbb{P}_\free ( \mathsf{A}_{j,3}\,|\,\mathsf{A}_{j,4}\cap \mathsf{L}^{\textup{c}}_j) \geq  D^{-1}e^{-DT^{5/2}}.
\end{align*}

$$\mathsf{A}_{j,3}'\coloneqq \{J_j(p)>\textup{Tent}_{j+1}(p)+17k(k+1-j)sT\ \textup{for}\ p\in  P_{j+1}\cap (\ell_j+1,r_j-1)\}.$$

To apply Theorem <ref>, we rewrite the following objects. Set $\breve{\ell}=\ell_j+1,\breve{r}=r_j-1$,

\begin{align*}
\breve{f}(x)\coloneqq\textup{Tent}_{j+1}(x)-\Delta_k ,\ \breve{\mathcal{L}}(x):={J}_j(x),\ \breve{P}=P_{j+1}\cap (\breve{\ell},\breve{r})
\end{align*}

\begin{align*}
\breve{a}^-=\textup{Tent}_{j+1} (\breve{\ell})+18k(k+1-j)sT,\  \breve{a}^+=\textup{Tent}_{j+1} (\breve{r})+18k(k+1-j)sT.
\end{align*}

The law of $\breve{\mathcal{L}}$ in $[\breve{\ell},\breve{r}]$ under $\mathbb{P}_\free (\,|\,\mathsf{A}_{j,4})$ is given by $\mathbb{P}^{1,1,(\breve{\ell},\breve{r}),\breve{a}^-,\breve{a}^+}_\free$. Furthermore, it can easily verified that

\begin{align*}
\mathsf{L}^{\textup{c}}_j=&\{\breve{\mathcal{L}}(p)>\breve{f}(p)  \ \textup{for}\ p\in  \breve{P}\},\\
\mathsf{A}_{j,3}'=& \{\breve{\mathcal{L}}(p)>\breve{f}(p)+17k(k+1-j)sT+\Delta_k\ \textup{for}\ p\in  \breve{P}\}.
\end{align*}

we can applying Theorem <ref> to conclude that

\begin{align*}
\mathbb{P}_\free ( \mathsf{A}_{j,3}' \,|\,\mathsf{A}_{j,4}\cap \mathsf{L}^{\textup{c}}_j) \geq  D^{-1}e^{-DT^{5/2}}.
\end{align*}

Resample $J_j$ between pole points in $P_{j+1}\cap [\ell_j+1,r_j-1]$, we obtain

\begin{align*}
\mathbb{P}_{\free}  (\mathsf{A}_{j,3}  \,|\,\mathsf{A}_{j,4}\cap\mathsf{L}^{\textup{c}}_j )\geq\mathbb{P}_{\free}  (\mathsf{A}_{j,3}  \,|\,\mathsf{A}_{j,4}\cap\mathsf{L}^{\textup{c}}_j\cap \mathsf{A}'_{j,3} )\times \mathbb{P}_{\free}  (\mathsf{A}'_{j,3}  \,|\,\mathsf{A}_{j,4}\cap\mathsf{L}^{\textup{c}}_j ) \geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

Step 3: We finish the proof in this step. When $\mathsf{A}_{j,3} $ occurs, $W^{j,[\ell_j+1,r_j-1]}_{\textup{Jump} }\geq D^{-1}$. Hence

\begin{align*}
\mathbb{P}_{J}  (\mathsf{A}_{j,3}  \,|\,\mathsf{A}_{j,4}\cap\mathsf{L}^{\textup{c}}_j )\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

In conclusion,

\begin{align*}
\mathbb{P}_{J} ( \mathsf{A}_{j,3} \, |\,\mathsf{A}_{j,4} )\geq&  \mathbb{P}_{J} ( \mathsf{A}_{j,3}   \,|\,\mathsf{A}_{j,4} \cap\mathsf{L }^{\textup{c}}_j )\times  \mathbb{P}_{J}  (\mathsf{L }^{\textup{c}}_j |\mathsf{A}_{j,4} )\geq  D^{-1}e^{-DT^{5/2}}.
\end{align*}

§.§ Curve separation over the resampling interval

In this subsection, we prove Proposition <ref>. Roughly speaking, Proposition <ref> says the following. Suppose at two points $\ell$ and $r$, the jump ensemble $J$ is ordered and lie above $\ug $. Then the probability that $J$ remains ordered and lie above $\ug $ in $[\ell,r]$ is at least $e^{-T^{5/2}}$.

We begin with introducing the notation in order to resample $J$ in the interval $[\ell,r]$.  Fix $j\in [1,k]_{\mathbb{Z}}$. Given $\ell_j\leq \ell<r\leq r_j$ and $a^{\pm} =(a^-_1,\dots,a^-_k, a^+_1,\dots,a^+_k)$, define

\begin{align*}
\frac{{d}\mathbb{P}^{1,j,[\ell,r],a^+,a^-}_{\textup{jump}}}{{d}\mathbb{P}^{1,j,[\ell,r],a^+,a^-}_{\free}}(\mathcal{L})\propto \prod_{i=1}^j W^{i,[\ell,r]}_{\textup{jump}}(\mathcal{L}_i;f_\textup{J}).
\end{align*}

Next, we prescribe the boundary condition which is well-spaced. Let $bS_j(\ell,r)$ be the collection of $(a^+,a^-) \in\mathbb{R}^{j}\times \mathbb{R}^{j}$ which satisfies

\begin{align}\label{equ:bS1}
a^\pm_i-a^\pm_{i+1}\geq   (k+2-j)T^{1/2}\ \textup{   for all}\  i\in [1,j-1]_{\mathbb{Z}},
\end{align}

\begin{align}\label{equ:bS2}
a^-_j\geq  \textup{Tent}_{j+1}(\ell)+8k(2k+3-2j)sT\ \textup{and}\  a^+_j \geq \textup{Tent}_{j+1}(r)+8k(2k+3-2j)sT,
\end{align}

\begin{align}\label{equ:bS3}
a^\pm_1\leq  jT^2.
\end{align}

Let $ \underline{a}^\pm$  be the smallest member is $bS_j(\ell,r)$. That is,

\begin{align*}
\underline{a}^{\pm}_i-\underline{a}^{\pm}_{i+1}=&(k+2-j)T^{1/2},\ i\in [1,j-1]_{\mathbb{Z}},\\
\underline{a}^-_j=&\textup{Tent}_{j+1}(\ell)+8k(2k+3-2j)sT,\\
\underline{a}^+_j=&\textup{Tent}_{j+1}(r)+8k(2k+3-2j)sT.
\end{align*}

The following two events capture the scenario that the curves are well-separated in $[\ell,r]$.

\begin{equation}\label{def:Sj}
\begin{split}
\mathsf{S}_j(\ell,r)\coloneqq \{ \mathcal{L}_i(x)>  \mathcal{L}_{i+1}(x)+(k+1-j)T^{1/2}\ \textup{for all}\ i\in [1,j-1]_{\mathbb{Z}}\ \textup{and}\ x\in [\ell,r]  \}\\
\cap\{\mathcal{L}_j(x)>  \textup{Tent}_{j+1}(x)+16k(k+1-j)T \ \textup{for all} \ x\in [\ell,r] \}.
\end{split}
\end{equation}

\begin{equation}\label{def:Upj}
\begin{split}
\mathsf{Up}_j(\ell,r)\coloneqq   \{ \mathcal{L}_1(x)<  (j+1)T^2 \ \textup{for all} \ x\in [\ell,r]\}.
\end{split}
\end{equation}

There exists a constant $D=D(k,s)$ such that for $E$ large enough, the following statement holds. For all $j\in [1,k]_{\mathbb{Z}}$, $\ell,r$ in $[\ell_j,r_j]$ and $(a^+,a^-)\in bS_j(\ell,r)$, we have

\begin{equation*}
\mathbb{P}^{1,j,[\ell,r],a^-,a^+}_{\textup{jump}} (\mathsf{S}_j(\ell,r)\cap\mathsf{Up}_j(\ell,r))\geq D^{-1}e^{-DT^{5/2}}.
\end{equation*}

The proof involves three steps.

Step 1: We first show that

\begin{equation*}
\mathbb{P}^{1,j,[\ell,r],a^-,a^+}_{\textup{jump}}  ( \mathsf{Up}^{\textup{c}}_j(\ell,r)  )\leq D e^{-D^{-1}T^{3}}.
\end{equation*}

This can be achieved by a simple stopping domain argument. Consider the intervals

\begin{equation*}
\begin{split}
I_-\coloneqq  \{x\in [\ell,r]\ |\ \mathcal{L}_1(y)< (j+2^{-1})T^2\ \textup{for all}\ y\in [\ell,x]\},\\
I_+\coloneqq  \{x\in [\ell,r]\ |\ \mathcal{L}_1(y)< (j+2^{-1})T^2\ \textup{for all}\ y\in [x,r] \}.
\end{split}
\end{equation*}

\begin{equation*}
\mathfrak{l}' \coloneqq\left\{ \begin{array}{cc}
\sup I_- & I_-\neq \phi,\\
r & I_-=\phi.
\end{array}
\right.
\end{equation*}

\begin{equation*}
\mathfrak{r}' \coloneqq\left\{ \begin{array}{cc}
\inf I_+ & I_+\neq \phi,\\
\ell & I_+=\phi.
\end{array}
\right.
\end{equation*}

The last condition in $bS_j(\ell,r)$ ensures $I_\pm$ are non-empty. Moreover,

\begin{align*}
\mathsf{Up}_j^{\textup{c}}(\ell,r) \subset \{\ell< \mathfrak{l}' <\mathfrak{r}' <r \} .
\end{align*}

Because $(j+2^{-1})T^2$ is much larger than $\ug(x)$, we have

\begin{align*}
\mathbb{P}^{1,1,[\mathfrak{l}'  ,\mathfrak{r}' ],(j+2^{-1})T^2,(j+2^{-1})T^2}_{\free }\left(B_1(x)\geq \ug (x)\ \textup{for all}\ x\in [\mathfrak{l}'  ,\mathfrak{r}' ] \right)\geq D^{-1} .
\end{align*}

And then

\begin{align*}
\mathbb{E}^{1,1,[\mathfrak{l}'  ,\mathfrak{r}' ],(j+2^{-1})T^2,(j+2^{-1})T^2}_{\free }\left[W^{1,[\mathfrak{l}'  ,\mathfrak{r}' ]}_{\textup{jump}} \right]\geq D^{-1} .
\end{align*}

\begin{align*}
&\mathbb{P}^{1,j,[\mathfrak{l}' ,\mathfrak{r}' ],(j+2^{-1})T^2,(j+2^{-1})T^2}_{\textup{jump}}  ( \mathsf{Up}^{\textup{c}}_j(\mathfrak{l}',\mathfrak{r}' )  )\\
= & \mathbb{E}^{1,1,[\mathfrak{l}'  ,\mathfrak{r}' ],(j+2^{-1})T^2,(j+2^{-1})T^2}_{\free }\left[W^{1,[\mathfrak{l}'  ,\mathfrak{r}' ]}_{\textup{jump}} \right]^{-1}\\
&\times \mathbb{P}^{1,j,[\mathfrak{l}' ,\mathfrak{r}' ],(j+2^{-1})T^2,(j+2^{-1})T^2}_{\free}  ( \mathsf{Up}^{\textup{c}}_j(\mathfrak{l}',\mathfrak{r}' )  )\leq  De^{-D^{-1}T^3}.
\end{align*}

Through the strong Markov property,

\begin{align*}
\mathbb{P}^{1,j,[\ell,r],a^-,a^+}_{\textup{jump}}   ( \mathsf{Up}^{\textup{c}}_j(\ell,r)  )\leq& \mathbb{P}^{1,j,[\ell,r],a^-,a^+}_{\textup{jump}}   ( \mathsf{Up}^{\textup{c}}_j (\mathfrak{l}',\mathfrak{r}')\cap\{ \ell< \mathfrak{l}' <\mathfrak{r}' <r  \}   )\\
\leq& De^{-D^{-1}T^3}.
\end{align*}

This finishes Step 1.

Step 2:

Define the event

\begin{equation*}
\begin{split}
\mathsf{Low}_j(\ell,r) \coloneqq\bigcup_{i=1}^j \{ \mathcal{L}_{i}(p)\leq &\textup{Tent}_{j+1}(p)- \Delta_k\  \textup{for some}\  p\in P_{j+1}\cap [\ell,r] \}.
\end{split}
\end{equation*}

In Step 2, we show that

\begin{align*}
\mathbb{P}^{1,j,[\ell,r],a^-,a^+}_{\textup{jump}}  (\mathsf{Low}_j(\ell,r) )\leq 2^{-1}.
\end{align*}

By the stochastic monotonicity Lemma <ref>, it suffices to prove

\begin{align*}
\mathbb{P}^{1,j,[\ell,r],\underline{a}^-,\underline{a}^+ }_{\textup{jump}}  (\mathsf{Low}_j(\ell,r) )\leq 2^{-1}.
\end{align*}

Fix $i\in [1,j]_{\mathbb{Z}}$. Because $\underline{a}^-_i\geq \textup{Tent}_{j+1}(\ell)+8k(2k+3-2j)sT$, $\underline{a}^+_i\geq \textup{Tent}_{j+1}(r)+8k(2k+3-2j)sT$ and (<ref>), it's straightforward to show that

\begin{align*}
\mathbb{P}^{i,i,[\ell,r],\underline{a}^-_i ,\underline{a}^+_i}_{\free} ( B_i(x)\geq \textup{Tent}_{j+1}(x)+7k(2k+3-2j)sT\ \textup{for all}\ x\in [\ell,r]  )\geq D^{-1}e^{-DT^3}.
\end{align*}

This can be done by applying Lemma <ref> with $x_1=\ell$, $x_2=r$ and $m=2kT$. Together with (<ref>),

\begin{align*}
\mathbb{P}^{1,j,[\ell,r],\underline{a}^-  ,\underline{a}^- }_{\free}  (B_i(x)\geq \ug(x) \ \textup{for all}\ x\in [\ell,r]\ \textup{and}\ i\in [1,j]_{\mathbb{Z}} )\geq D^{-1}e^{-DT^3}.
\end{align*}

\begin{align*}
\mathbb{E}^{1,j,[\ell,r],\underline{a}^-  ,\underline{a}^- }_{\free} \left[\prod_{i=1}^j W^{i,[\ell,r]}_{\textup{jump}} \right]\geq D^{-1}e^{-DT^{3}}.
\end{align*}

Recall that $\Delta=8\log T$ and $d=64T^{-4}(\log T)^2$. Let

\begin{equation}\label{def:Flct}
\mathsf{Flct}_i(\ell,r)\coloneqq \{|\mathcal{L}_{i}(x)-\mathcal{L}_i(y)|\leq \Delta\ \textup{for all}\ x,y\in [\ell,r]\ \textup{with}\ |x-y|\leq d \}.
\end{equation}

Because $\mathcal{L}_i(\ell)=\underline{a}^-_i$, $\mathcal{L}_i(r)=\underline{a}^+_i$ and (<ref>),

\begin{align*}
\left| \frac{\mathcal{L}_i(r)-\mathcal{L}_i(\ell) }{r-\ell}\right|=\left| \frac{\textup{Tent}_{j+1}(r)-\textup{Tent}_{j+1}(\ell) }{r-\ell}\right|\leq 2kT.
\end{align*}

\begin{align*}
\mathbb{P}^{1,j,[\ell,r],\underline{a}^-  ,\underline{a}^- }_{\free}  (\mathsf{Flct}^{\textup{c}}_i(\ell,r) )\leq &\mathbb{P}_{\free}(\omega_{d,[\ell,r]}(B^{[\ell,r]})\geq \Delta-2kdT)\\
\leq &\mathbb{P}_{\free}(\omega_{d,[\ell,r]}(B^{[\ell,r]})\geq 2^{-1} \Delta)\leq De^{-D^{-1}T^4}.
\end{align*}

As a result,

\begin{align*}
\mathbb{P}^{1,j,[\ell,r],\underline{a}^-  ,\underline{a}^+ }_{\textup{jump}}  (\bigcup_{i=1}^j \mathsf{Flct}^{\textup{c}}_i(\ell,r)  )\leq De^{DT^3-D^{-1}T^4}\leq 4^{-1}.
\end{align*}

From Condition C3, $\mathsf{Low}_j(\ell,r) $ and $ \bigcap_{i=1}^j\mathsf{Flct}_i(\ell,r)$ imply there exists $i\in [1,j]_{\mathbb{Z}}$ and an interval in $  \textup{Int}_j\cap [\ell,r]$ with length $d$ in which

\begin{align*}
\mathcal{L}_i(x)\leq \ug(x)-k\Delta-k\log k.
\end{align*}

Together with

\begin{align*}
d \textbf{H}_{i,j+1}(k\Delta+k\log k)\geq d e^{\Delta }=64T^4(\log T)^2,
\end{align*}

we deduce that when $\mathsf{Low}_j(\ell,r)\cap \bigcap_{i=1}^j\mathsf{Flct}_i(\ell,r)$ occurs,

\begin{align*}
\prod_{i=1}^j W^{i,[\ell,r]}_{\textup{jump}}\leq e^{-64T^4(\log T)^2}.
\end{align*}

As a result,

\begin{align*}
&\mathbb{P}^{1,j,[\ell,r],\underline{a}^-  ,\underline{a}^+ }_{\textup{jump}}  \left(\mathsf{Low}_j(\ell,r)\cap \bigcap_{i=1}^j\mathsf{Flct}_i(\ell,r) \right)
\leq  \mathbb{E}^{1,j,[\ell,r],\underline{a}^-  ,\underline{a}^+ }_{\free} \left[\prod_{i=1}^j W^{i,[\ell,r]}_{\textup{jump}} \right]^{-1}\times e^{-64T^4(\log T)^2}\\
\leq & De^{DT^3-64T^4(\log T)^2}\leq 4^{-1}.
\end{align*}

Putting the above estimates together, we conclude that

\begin{align*}
&\mathbb{P}^{1,j,[\ell,r],\underline{a}^-  ,\underline{a}^+ }_{\textup{jump}}  (\mathsf{Low}_j(\ell,r) )\\
\leq&  \mathbb{P}^{1,j,[\ell,r],\underline{a}^-  ,\underline{a}^+ }_{\textup{jump}}  \left(\mathsf{Low}_j(\ell,r)\cap \bigcap_{i=1}^j\mathsf{Flct}_i(\ell,r) \right)+\mathbb{P}^{1,j,[\ell,r],\underline{a}^-  ,\underline{a}^+ }_{\textup{jump}}  \left( \bigcup_{i=1}^j\mathsf{Flct}^{\textup{c}}_i(\ell,r) \right)\\
\leq &2^{-1}.
\end{align*}

Step 3: In this step, we show that

$$\mathbb{P}^{1,j,[\ell,r],a^-  ,a^+ }_{\textup{jump} }    ( \mathsf{S}_j(\ell,r)  \, |\,  \mathsf{Low}^{\textup{c}}_j(\ell,r))\geq D^{-1}e^{-DT^{5/2}}.$$

Consider the event $\mathsf{S}_j'(\ell,r)$ being defined by

* $  \mathcal{L}_i(x)\geq  \mathcal{L}_{i+1}(x)+(k+3/2-j)T^{1/2}\ \textup{for all}\ i\in [1,j-1]_{\mathbb{Z}}\ \textup{and}\ x\in P_{j+1}\cap (\ell,r)$,

* $ \mathcal{L}_j(x)\geq  \textup{Tent}_{j+1}(x)+8k(2k+5/2-2j)sT \ \textup{for all} \ x\in P_{j+1}\cap (\ell,r).$

Compared to $\mathsf{S}_j(\ell,r)$, $\mathsf{S}_j'(\ell,r)$ requires a larger gap between  $\mathcal{L}$ at points in $ P_{j+1}\cap (\ell,r)$.

In order to apply Theorem <ref>, we rewrite the following objects. Set

\begin{align*}
\breve{f}(x)\coloneqq\textup{Tent}_{j+1}(x)-\Delta_k,\ \breve{\mathcal{L}}_i(x):=\mathcal{L}_i(x),\ \breve{P}=P_{j+1}\cap (\ell,r),\ \breve{a}^\pm_i= {a}^\pm_i.
\end{align*}

It can easily verified that

\begin{align*}
\mathsf{Low}_j^{\textup{c}}=&\{\breve{\mathcal{L}}_i(p)>\breve{f}(p)  \ \textup{for}\ i\in [1,\mathbb{Z}]\ \textup{and}\ p\in  \breve{P}\},\\
\mathsf{S}'_{j }(\ell,r) =& \{\breve{\mathcal{L}}_i(p)\geq  \breve{\mathcal{L}}_{i+1}(p)+(k+3/2-j)T^{1/2}\ \textup{for all}\ i\in [1,j-1]_{\mathbb{Z}}\ \textup{and}\ p\in \breve{P} \}\\
&\cap \{ \breve{\mathcal{L}}_j(x)\geq  \breve{ f }(x)+8k(2k+5/2-2j)sT+\Delta_k \ \textup{for all} \ x\in \breve{P}_{j+1} .\}.
\end{align*}

From the conditions in $bS_j(\ell,r)$

\begin{align*}
& \breve{a}^\pm_i-\breve{a}^\pm_{i+1}\geq   (k+2-j)T^{1/2}\ \textup{  for all}\  i\in [1,j-1]_{\mathbb{Z}},\\
&\breve{a}^-_j-\breve{f}(\breve{\ell}),\breve{a}^+_j -\breve{f}(\breve{r}) \geq   8k(2k+3-2j)sT+\Delta_k
\end{align*}

\begin{align*}
\breve{a}^\pm_1\leq jT^2.
\end{align*}

Together with (<ref>) and (<ref>), the assumptions (<ref>) and (<ref>) hold.  By Theorem <ref>, we conclude that

\begin{align*}
\mathbb{P}^{1,j,[\ell,r],a^-  ,a^+ }_{\free }  &( \mathsf{S}'_j(\ell,r) \, |\,  \mathsf{Low}^{\textup{c}}_j(\ell,r))\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

Resampling the curves between pole points in $P_{j+1}\cap(\ell,r)$, we deduce

\begin{align*}
\mathbb{P}^{1,j,[\ell,r],a^-  ,a^+ }_{\free }  &( \mathsf{S}_j(\ell,r) \, |\,  \mathsf{Low}^{\textup{c}}_j(\ell,r))\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

When $\mathsf{S}_j(\ell,r)$ occurs, $\prod_{i=1}^j W_{\textup{jump}}^{i,[\ell,r]}\geq D^{-1}$. Therefore

\begin{align*}
&\mathbb{P}^{1,j,[\ell,r],a^-  ,a^+ }_{\textup{jump} }    ( \mathsf{S}_j(\ell,r)  \, |\,  \mathsf{Low}^{\textup{c}}_j(\ell,r)) \geq  D^{-1}e^{-DT^{5/2}}.
\end{align*}

In conclusion,

\begin{align*}
&\mathbb{P}^{1,j,[\ell,r],a^-,a^+}_{\textup{jump}} (\mathsf{S}_j(\ell,r)\cap\mathsf{Up}_j(\ell,r))\geq  \mathbb{P}^{1,j,[\ell,r],a^-  ,a^+ }_{\textup{jump}  }   ( \mathsf{S}_j(\ell,r) )-\mathbb{P}^{1,j,[\ell,r],a^-,a^+}_{\textup{jump}} ( \mathsf{Up}^{\textup{c}}_j(\ell,r))\\
\geq&  \mathbb{P}^{1,j,[\ell,r],a^-  ,a^+ }_{\textup{jump} }    ( \mathsf{S}_j(\ell,r)  \, |\,  \mathsf{Low}^{\textup{c}}_j(\ell,r))\times  \mathbb{P}^{1,j,[\ell,r],a^-  ,a^+ }_{\textup{jump}  }   (\mathsf{Low}^{\textup{c}}_j(\ell,r)  )-\mathbb{P}^{1,j,[\ell,r],a^-,a^+}_{\textup{jump}} ( \mathsf{Up}^{\textup{c}}_j(\ell,r))\\
\geq  &D^{-1}e^{-DT^{5/2}}-De^{-D^{-1}T^3}\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

The proof is finished.

§.§ Proof of Proposition <ref>
In this subsection, we combine Propositions <ref> and <ref> to prove Proposition <ref>. Proposition <ref> follows easily from Proposition <ref>.

For $j\in [1,k]_{\mathbb{Z}}$, Let $\mathsf{F}_j$ be the event

* $J_i(x)> \ug(x) \ \textup{for all}\ i\in [1,j]_\mathbb{Z}\ \textup{and}\ x\in [\ell_i+d',r_i-d']$,

* $ J_j(x)> \textup{Tent}_{j+1}(x)+16k(k+1-j)sT\ \textup{for all}\ x\in [\ell_{j}+1 ,r_{j}-1]$,

* $ J_{i}(x)> J_{i+1}(x)+(k+1-j)T^{1/2}\ \textup{for all}\ i\in [1,j-1]_\mathbb{Z}\ \textup{and}\ x\in [\ell_{i+1},r_{i+1}] $,

* $J_1(x)< (j+1)T^2\ \textup{for all}\ x\in [\ell_{1} ,r_{1}]$.

Roughly speaking, $\mathsf{F}_j$ is the event that $J_i> \ug $ in $[\ell_i+d',r_i-d']$ and $(J_1,J_2,\dots J_j)$ stays ordered.

There exists a constant $D=D(k,s)$ such that for $E$ large enough, the following statement holds. For all $j\in [1,k]_{\mathbb{Z}}$,

\begin{equation}
\mathbb{P}_{J}(\mathsf{F}_j)\geq D^{-1}e^{-DT^{5/2}}.
\end{equation}

It follows from $\mathsf{F}_k\subset\mathsf{NoTouch}$.

We use induction on $j$ and start with $j=1$. Comparing $\mathsf{A}_1$ and $\mathsf{F}_1$, the extra requirement in $\mathsf{F}_1$ is that $J_1(x)< 2T^2$. By a stopping domain argument  we can show that $\mathbb{P}_J(J_1(x)< 2T^2)\geq 1-De^{-D^{-1}T^3}$. See Step 1 in the proof of Proposition <ref> for a similar argument. We omit the detail here. For $T$ large enough, it holds that
$$\mathbb{P}(\mathsf{F}_1)\geq \mathbb{P}(\mathsf{A}_1)-De^{-D^{-1}T^3}\geq D^{-1}e^{-DT^{5/2}}.$$

Let $j\in [2,k]_{\mathbb{Z}}$ and assume that

\begin{align*}
\mathbb{P}_{J}(\mathsf{F}_{j-1})\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

Under the law of $\mathbb{P}_J$, $\mathsf{F}_{j-1}$ and $\mathsf{A}_j$ are independent. From Propositions <ref>, it holds that

\begin{align*}
\mathbb{P}_{J}(\mathsf{F}_{j-1}\cap\mathsf{A}_j)=\mathbb{P}_{J}(\mathsf{F}_{j-1} )\times \mathbb{P}_{J}( \mathsf{A}_j )\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

In other words, with probability $e^{-T^{5/2}}$, $J_1,\dots,J_{j+1}$ are well-separated and $J_j$  jumps over $\ug$.

Comparing $\mathsf{F}_{j-1}\cap\mathsf{A}_j$ with $\mathsf{F}_j$, the only requirement in $\mathsf{F}_j$ which is not guaranteed by $\mathsf{F}_{j-1}\cap\mathsf{A}_j$ is

\begin{align}\label{extra}
J_{j-1}(x)>J_{j }(x)+(k+1-j)T^{1/2}\ \textup{for all}\ x\in [\ell_j,r_j].
\end{align}

In other words, the only situation $\mathsf{F}_{j-1}\cap\mathsf{A}_j\cap \mathsf{F}^{\textup{c}}_{j}$ can occur is that $J_{j}$ raises to high and intersects with $J_{j-1}$. We use a stopping domain argument to resolve this issue. In particular, we will consider the stopping domain given by the leftmost and rightmost locations where (<ref>) fails.

Define a random set $I'_{j,-}\subset [\ell_1,r_1]$ which captures the requirements in $\mathsf{F}_{j-1}\cap\mathsf{A}_j$ from the left. Explicitly, $I'_{j,-}\subset [\ell_1,r_1]$ is the the collection of $x_0\in [\ell_1,r_1]$ such that

* $J_i(x)> \ug (x)\ \textup{for all}\ i\in [1,j]_\mathbb{Z}\ \textup{and}\ x\in [\ell_i+d',r_i-d']\cap (-\infty,x_0]$.

* $J_{i}(x)> J_{i+1}(x)+(k+2-j)T^{1/2}\ \textup{for all}\ i\in [1,j-2]_\mathbb{Z}\ \textup{and}\ x\in [\ell_{i+1},r_{i+1}] \cap (-\infty,x_0]$.

* $J_{j-1}(x)> \textup{Tent}_{j }(x)+16k(k+2-j)sT\ \textup{for all}\ x\in [\ell_{j-1}+1 ,r_{j-1}-1]\cap (-\infty,x_0]$.

* $J_{j }(x)> \textup{Tent}_{j+1 }(x)+16k(k+1-j)sT\ \textup{for all}\ x\in [\ell_{j}+1 ,r_{j}-1]\cap (-\infty,x_0]$.

* $ J_1(x)< jT^2\ \textup{for all}\ x\in [\ell_{1} ,r_{1}] \cap (-\infty,x_0]$.

Furthermore, we define $I''_{j,-}$ to be the subset of $I'_{j,-}$ with an extra requirement that

\begin{align*}
\ J_{j-1}(x)> J_{j}(x)+(k+2-j)T^{1/2}\ \textup{for all}\ x\in [\ell_{j},r_{j}] \cap (-\infty,x_0].
\end{align*}

Define $I'_{j,+},I''_{j,+}\subset [\ell_1,r_1]$ by replacing $(-\infty,x_0]$ with $[x_0,\infty)$.
\begin{align*}
\mathfrak{l}''_j\coloneqq\left\{ \begin{array}{cc}
\sup I''_{j,-},\ & \textup{if}\ I''_{j,-}\neq \phi\\
\ell_1, & \textup{if}\ I''_{j,-}= \phi.
\end{array}\right.\ \textup{and}\ \ \mathfrak{r}''_j \coloneq\left\{ \begin{array}{cc}
\inf I''_{j,+},  & \textup{if}\ I''_{j,+}\neq \phi\\
r_1, & \textup{if}\ I''_{j,+}= \phi.
\end{array}\right.
\end{align*}

As $\mathsf{F}_{j-1}\cap \mathsf{A}_j$ occurs, $ I'_{j,\pm} =[\ell_1,r_1]$. Moreover, $\ell_i\in I''_{j,-}$. It can be shown as the following. By the second requirement of $\mathsf{F}_{j-1}$, it holds that

\begin{align*}
J_{j-1}(\ell_j)\geq  &\textup{Tent}_{j}( \ell_j  )+16k(k+2-j)sT.
\end{align*}

By (<ref>), it holds that

\begin{align*}
\textup{Tent}_{j}( \ell_j  )+16k(k+2-j)sT \geq &\ug(\ell_j )-3ksT+16k(k+2-j)sT.
\end{align*}

Because $\ug(\ell_j )=\max\{f_j(\ell_j),f_{j+1}(\ell_j)\}\geq f_j(\ell_j)=J_j(\ell_j)$, we have

\begin{align*}
\ug(\ell_j )-3ksT+16k(k+2-j)sT  \geq  J_{j}(\ell_j )-3ksT+16k(k+2-j)sT  >  J_{j}(\ell_j )+(k+2-j)T^{1/2}.
\end{align*}

Similarly, $\mathsf{F}_{j-1}\cap \mathsf{A}_j$  implies $r_i\in I''_{j,+}$.

Next, we discuss the implications of $\mathsf{F}_{j-1}\cap \mathsf{A}_j\cap \mathsf{F}^{\textup{c}}_{j }$. $\mathsf{F}_{j-1}\cap \mathsf{A}_j\cap \mathsf{F}^{\textup{c}}_{j }$ implies (<ref>) fails at a point in $(\ell_j,r_j)$. In particular, $\ell_j<\mathfrak{l}''_j <\mathfrak{r}''_j<r_j$,

\begin{align*}
\end{align*}

\begin{align*}
\end{align*}

Moreover, all the conditions in $I'_{j,\pm}$ hold at $\mathfrak{l}''_j$ and $\mathfrak{r}''_j$.

Define the event $\mathsf{  K }_j$ by

\begin{align*}
\mathsf{  K }_j \coloneqq \{\ell_j<\mathfrak{l}''_j <\mathfrak{r}''_j<r_j\}\cap\{(J_1(\mathfrak{l}''_j),\dots,J_j(\mathfrak{l}''_j),J_1(\mathfrak{r}''_j),\dots,J_j(\mathfrak{r}''_j)) \in bS_j(\mathfrak{l}''_j,\mathfrak{r}''_j)\}.
\end{align*}

Here $bS_j(\ell,r)$  defined by (<ref>), (<ref>) and (<ref>). We claim that $\mathsf{F}_{j-1}\cap \mathsf{A}_j\cap \mathsf{F}^{\textup{c}}_{j } \subset \mathsf{ {K}}_j$. Comparing the condition in $bS_j(\mathfrak{l}''_j,\mathfrak{r}''_j)$ and  in $I'_{j,\pm}$, we only need to check

\begin{align*}
\mathsf{F}_{j-1}\cap \mathsf{A}_j\cap \mathsf{F}^{\textup{c}}_{j } \subset  &\{ J_j(\mathfrak{l}''_j)\geq \textup{Tent}_{j+1}(\mathfrak{l}''_j) +8k(k+3-2j)sT  \}\\
&\cap\{ J_j(\mathfrak{r}''_j)\geq \textup{Tent}_{j+1}(\mathfrak{r}''_j) +8k(k+3-2j)sT  \}.
\end{align*}

We give the proof for $\mathfrak{l}''_j$. The argument for $\mathfrak{r}''_j$ is similar. Suppose $\mathfrak{l}''_j\in [\ell_j+1,r_j-1]$, then the assertion holds by the fourth requirement of $I'_{j,-}$. Assume $\mathfrak{l}''_j\in [\ell_j,\ell_j+1)$. By the third requirement of $I_{j,-}'$,

\begin{align*}
J_{j-1}(\mathfrak{l}''_j)\geq &\textup{Tent}_j(\mathfrak{l}''_j)+16k(k+2-j)sT.
\end{align*}

Because of (<ref>),

\begin{align*}
\textup{Tent}_j(\mathfrak{l}''_j)+16k(k+2-j)sT \geq&   \textup{Tent}_j(\ell_j)-2ksT +16k(k+2-j)sT.
\end{align*}

From  (<ref>) and $ \ug(\ell_j)\geq f_{j+1}(\ell_j)=\textup{Tent}_{j+1}(\ell_j)$,

\begin{align*}
\textup{Tent}_j(\ell_j)-2ksT +16k(k+2-j)sT \geq   \textup{Tent}_{j+1}(\ell_j)-5ksT +16k(k+2-j)sT.
\end{align*}

Using (<ref>), again, we have

\begin{align*}
J_{j-1}(\mathfrak{l}''_j) \geq&  \textup{Tent}_{j+1}(\mathfrak{l}''_j )-7ksT +16k(k+2-j)sT.
\end{align*}

As a result, $\mathsf{F}_{j-1}\cap \mathsf{A}_j\cap \mathsf{F}^{\textup{c}}_{j }  $ implies

\begin{align*}
J_j(\mathfrak{l}''_j)= &J_{j-1}(\mathfrak{l}''_j)-(k+2-j)T^{1/2}\\
\geq&  \textup{Tent}_{j+1}(\mathfrak{l}''_j )-8ksT +16k(k+2-j)sT\\
\end{align*}

The argument for $\mathfrak{l}''_j\in (r_j-1,r_j]$ is the similar.

Next, we check that
$$\mathsf{S}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)\cap\mathsf{Up}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)\cap \mathsf{K}_j \subset \mathsf{F}_j\cap  {\mathsf{K}}_j.$$

The events $\mathsf{S}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)$ and  $\mathsf{Up}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)$ are defined in (<ref>) and (<ref>). The definition of $\mathsf{K}_j$ ensures the requirements in $\mathsf{F}_j$ holds outside the interval $(\mathfrak{l}''_j,\mathfrak{r}''_j)$. $\mathsf{S}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)\cap\mathsf{Up}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)$ then takes cares of the interval $[ \mathfrak{l}''_j,\mathfrak{r}''_j  ]$.

We are ready to give a lower bound for $\mathbb{P}_{J }(\mathsf{F}_j)$. Let $\delta\coloneqq  \mathbb{P}_{J}(\mathsf{F}_{j-1}\cap \mathsf{A}_j\cap \mathsf{F}^{\textup{c}}_{j } ).$ From the above discussion,

\begin{align*}
\mathbb{P}_{J }(\mathsf{F}_j)\geq  \mathbb{P}_{J }(\mathsf{F}_j\cap {\mathsf{K}}_j )\geq &\mathbb{P}_{J }(\mathsf{S}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)\cap\mathsf{Up}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)\cap  {\mathsf{K}}_j ).
\end{align*}

From the strong Markov property and Proposition <ref>, it holds that

\begin{align*}
\mathbb{P}_{J }(\mathsf{S}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)\cap\mathsf{Up}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)\cap  {\mathsf{K}}_j )= &\mathbb{P}_{J }(\mathsf{S}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)\cap\mathsf{Up}_j(\mathfrak{l}''_j,\mathfrak{r}''_j)\,|\,  {\mathsf{K}}_j ) \times \mathbb{P}_J( {\mathsf{K}}_j) \\
\geq &D^{-1}e^{-DT^{5/2}}\times \mathbb{P}_{J}(\mathsf{K}_j).
\end{align*}

Together with $ \mathsf{F}_{j-1}\cap \mathsf{A}_j\cap \mathsf{F}^{\textup{c}}_{j } \subset\mathsf{K}_j$, we have

\begin{align*}
\mathbb{P}_{J }(\mathsf{F}_j) \geq \delta\times D^{-1}e^{-DT^{5/2}} .
\end{align*}

On the other hand,

\begin{align*}
\mathbb{P}_{J}(\mathsf{F}_j)\geq \mathbb{P}_{J}(\mathsf{F}_{j-1}\cap \mathsf{A}_j\cap \mathsf{F}_{j } )= \mathbb{P}_{J}  (\mathsf{F}_{j-1}\cap \mathsf{A}_j )-\delta \geq D^{-1}e^{-D T^{5/2}}-\delta ,
\end{align*}

We conclude that

\begin{align*}
\mathbb{P}_{J}(\mathsf{F}_j)\geq \inf_{\delta \geq 0} \max\{ \delta \times D^{-1} e^{-DT^{5/2}},D^{-1}e^{-DT^{5/2}}-\delta\}\geq D^{-1}e^{-DT^{5/2}}.
\end{align*}

The induction argument is finished.

§ PROOF OF PROPOSITION <REF>

In this section we prove Proposition <ref>. The jump ensemble $J$ is defined through (<ref>). Recall that $A\subset {C}_{0,0}([0,s])$ is a Borel set and $\varepsilon=\mathbb{P}_{\free}(B^{[0,s]}\in A).$ The goal is to prove

\begin{align*}
\sup \{\mathbb{P}_{J}(  J_k^{[0,s]}\in A  )\, |\,  (\bar{\ell},\bar{r},f_{\textup{J}})\in \mathcal{G}  \}  \leq D\varepsilon \exp\left( D\left(\log\varepsilon^{-1}\right)^{5/6} \right).
\end{align*}

Starting from now, we fix a realization of the favorable event $(\bar{\ell},\bar{r},f_{\textup{J}})\in\mathcal{G}$ (see Definition <ref>). Suppose $P^*\cap (0,s)=\phi$, i.e. no pole in $(0,s)$. By the construction of the jump ensemble $J$, the law of $J_k^{[0,s]}$ is equivalent to that of a free Brownian bridge. Then $\mathbb{P}_{J}(J_k^{[0,s]}\in A)=\varepsilon$ and Proposition <ref> follows easily.

From now on we assume $P^*\cap (0,s)\neq \phi$. From the construction of the pole set $P^*$, see (<ref>) and (<ref>), $P^*\cap (0,s)$ consists of only one element, denoted by $p_0$. We enlarge the interval from $[0,s]$ to $[q_1,q_2]$ to prevent $p_0$ from being too close to the boundary. Let

\begin{align*}
q_1&\coloneqq p_0-\max\{p_0,\, 2^{-1}s\},\\
q_2&\coloneqq p_0+\max\{s-p_0,\, 2^{-1}s\}.
\end{align*}

Let $\pi:{C}_{0,0}([q_1,q_2])\to {C}_{0,0}([0,s])$ be a restriction map, defined as

\begin{align*}
\pi[h](x):=h(x)-s^{-1}xh(s)-s^{-1}(s-x)h(0).
\end{align*}

Note that $\pi\circ B^{[q_1,q_2]}  {=} B^{[0,s]}$ and $\pi\circ J_k^{[q_1,q_2]} {=} J_k^{[0,s]}$. For $A'=\pi^{-1}(A)$, it holds that

\begin{align*}
\mathbb{P}_{\free}(B^{[q_1,q_2]}\in A')=&\mathbb{P}_{\free}(B^{[0,s]}\in A)=\epsilon,\\
\mathbb{P}_{J}(J_k^{[q_1,q_2]}\in A')=&\mathbb{P}_{J}(J_k^{[0,s]}\in A).
\end{align*}

In the remaining of this section, we consider the event $\{J_k^{[q_1,q_2]}\in A'\}$ instead of $\{J_k^{[0,s]}\in A\}$.

Let $p_-=\max\{p\in P^*\,|\,p<p_0\}$ and $p_+=\min\{p\in P^*\,|\,p>p_0\}$ be the elements in $P^*$ next to $p_0$.
From (<ref>), it holds that

$$-2^{-1}T<p_-<q_1<p_0<q_2< p_+<2^{-1}T. $$

Recall $\ug(x)$ in (<ref>). Define the tent map

\begin{equation*}
\textup{Tent}(x)\coloneqq \left\{ \begin{array}{cc}
\ug(x) & x= p_0\ \textup{or}\ x=p_\pm,\\
\textup{linear inteperlation} & x\in (p_-,p_0 )\cup (p_0,p_+).
\end{array} \right.
\end{equation*}

We remark that $\textup{Tent}$ agrees with $\textup{Tent}_{j+1}$ in $[p_-,p_+]$ for all $j\in [1,k]_{\mathbb{Z}}$.

Recall that $\Delta_k=(k+2)\Delta+k\log k$. Define the quantities (see Figure <ref> for an illustration).

\begin{align*}
\underline{m}_0=&\textup{Tent}(p_0)-\Delta_k,\ \underline{m}_\pm=\textup{Tent}(p_\pm)- \Delta_k,\\
\underline{m}_1=& \textup{Tent}(q_1)- \Delta_k,\ \underline{m}_2=\textup{Tent}(q_2)- \Delta_k,\\
\sigma_1^2=&\frac{(p_0-q_1)(q_1- {p}_-)}{ {p_0}- {p}_-},\ \sigma_2^2=\frac{( {p}_+-q_2)(q_2- {p_0}) }{ {p}_+- {p_0}},\ \sigma_3^2=\frac{(q_2-p_0)(p_0-q_1)}{q_2-q_1}.
\end{align*}

Tent is constructed using values of $\ug$ at poles $p_{-}, p_0, p_{+}$ with $p_0$ being the single pole in $[0,s]$. We expect the values of $J_k$ stay around Tent, hence above Tent$-\Delta_k$ with high probability.

The following lemma shows that $J_k( {p_0} )\geq\underline{m}_0$ and $J_k( {p_\pm} )\geq\underline{m}_\pm$ occur with high probability. Define the event

\begin{align*}
\mathsf{Y}\coloneqq \left\{ J_k( {p_0} )\geq\underline{m}_0 ,\ J_k( {p}_\pm )\geq\underline{m}_\pm \right\}.
\end{align*}

There exists a constant $D=D(k)$ such that for $E$ large enough, we have

\begin{align*}
\mathbb{P}_J(\mathsf{Y}^{\textup{c}})\leq De^{-D^{-1}T^4}.
\end{align*}

The idea is that if $J_k( {p_0} )<\underline{m}_0$ or $J_k( {p_\pm} )<\underline{m}_\pm$ occurs, then with high probability $J_k$ will stay below $\ug-k\Delta-k\log k$ in an interval with length $d$. Because $d \, \mathbf{H}_t^{k,k+1}(k\Delta+k\log k)\approx T^4$, the Boltzmann weight $W_{\textup{jump}}^k$ is less than $e^{-T^4}$ which is unlikely. In below we fill in details of this argument.

From the definition of $\mathbb{P}_J$ ( see (<ref>) and (<ref>)), $J_k$ is independent of other curves and is distributed according to

\begin{align*}
\frac{{d} \mathbb{P}_{J_k}}{{d} \mathbb{P}^{k,k,(\ell_k,r_k),x_k,y_k}_{\free}}(\mathcal{L})=\frac{ W^{k }_{\textup{jump}}(\mathcal{L})}{ \mathbb{E}^{k,k,(\ell_k,r_k),x_k,y_k}_{\free}[ W^{k }_{\textup{jump}}]} .
\end{align*}

Here $x_k=f_k(\ell_k)$ and $y_k=f_k(r_k)$. We know that when $\mathsf{Y}^{\textup{c}}$ occurs, $W^{k }_{\textup{jump}}$ is, with high probability, less than $e^{-T^4}$. To conclude $\mathsf{Y}^{\textup{c}}$ is unlikely, we need to bound the normalizing constant. To do so, we turn off the interaction in $W^k_{\textup{jump}}$ except near $p_0$ and $p_\pm$ to get a new curve $J'_k$. The law of $J'_k$ satisfies the following Radon-Nikodym derivative relation,

\begin{align*}
\frac{{d} \mathbb{P}_{J'_k}}{{d} \mathbb{P}^{k,k,(\ell_k,r_k),x_k,y_k}_{\free}}(\mathcal{L})=\frac{ W'^{k }_{\textup{jump}}(\mathcal{L})}{ \mathbb{E}^{k,k,(\ell_k,r_k),x_k,y_k}_{\free}[ W'^{k }_{\textup{jump}}]} .
\end{align*}

Here the weight $W'^{k }_{\textup{jump}}$ is defined by

\begin{equation*}
\begin{split}
W'^{k}_{\textup{jump}}(\mathcal{L} )\coloneqq &  \exp\left( -\int_{[p_--d,p_-]\cup [p_0,p_0+d]\cup [p_+,p_++d] } \textbf{H}_t^{k,k+1}(\ug (x)-\mathcal{L}(x)) dx \right) .
\end{split}
\end{equation*}

By the stochastic monotonicity Lemma <ref>, it holds that

\begin{align*}
\mathbb{P}_{J_k}(\mathsf{Y}^{\textup{c}})\leq \mathbb{P}_{J'_k}(\mathsf{Y}^{\textup{c}}).
\end{align*}

From condition C2 and Remark <ref>, it is straightforward to check that

\begin{align*}
\mathbb{P}^{k,k,(\ell_k,r_k),x_k,y_k}_{\free}(\{B(x)\geq \ug(x)\ \textup{for}\ x\in [p_--d,p_-]\cup [p_0,p_0+d]\cup [p_+,p_++d]\})\geq C^{-1}e^{-CT^3}.
\end{align*}

Here $x_k=f_k(\ell_k)$ and $y_k=f_k(r_k)$. When this event occurs, $W'^k_{\textup{jump}}$ is bounded from below. As a result,

$$\mathbb{E}^{k,k,(\ell_k,r_k),x_k,y_k}_{\free}[ W'^{k }_{\textup{jump}}]\geq D^{-1}e^{-DT^3}.$$

Next, we bound the local fluctuation of $J'_k$. Recall the event $\mathsf{Flct}_k(\ell_k,r_k)$ is defined in (<ref>) as

\begin{equation*}
\mathsf{Flct}_k(\ell_k,r_k)\coloneqq \{|\mathcal{L}_{k}(x)-\mathcal{L}_k(y)|\leq \Delta\ \textup{for all}\ x,y\in [\ell_k,r_k]\ \textup{with}\ |x-y|\leq d \}.
\end{equation*}

Because $\Delta^2/d=T^2,$

$$\mathbb{P}^{k,k,(\ell_k,r_k),x_k,y_k}_{\free}(\mathsf{Flct}^{\textup{c}}_k(\ell_k,r_k))\leq D e^{-D^{-1}T^4}.$$

With the bound on the normalizing constant, it holds that

\begin{align*}
\mathbb{P}_{J'_k}(\mathsf{Flct}_k^{\textup{c}}(\ell_k,r_k))\leq De^{DT^3-D^{-1}T^4}\leq De^{-D^{-1}T^4} .
\end{align*}

When $\mathsf{Flct}_k(\ell_k,r_k)$ and $\mathsf{Y}^{\textup{c}}$ both occur, there exists an interval with length $d$ and on which $\ug(x)-J_k(x)\geq k\Delta+k\log k$. From

\begin{align*}
d\, \textbf{H}_t^{k,k+1}(k\Delta+k\log k)\geq d e^{\Delta }=64T^4(\log T)^2,
\end{align*}

we get

\begin{align*}
\mathbb{E}^{k,k,(\ell_k,r_k),x_k,y_k}_{\free}[ W'^{k }_{\textup{jump}}\cdot\mathbbm{1}_{\mathsf{Y}^{\textup{c}}}\cdot\mathbbm{1}_{\mathsf{Flct}_k(\ell_k,r_k) }]\leq De^{-D^{-1}T^4(\log T)^2}.
\end{align*}

Using the bound on the normalizing constant again, we conclude that

\begin{align*}
\mathbb{P}_{J'_k}(\mathsf{Y}^{\textup{c}})\leq &\mathbb{P}_{J'_k}(\mathsf{Y}^{\textup{c}}\cap \mathsf{Flct}_k(\ell_k,r_k))+\mathbb{P}_{J'}(\mathsf{Flct}_k^{\textup{c}}(\ell_k,r_k))\\
\leq &De^{-D^{-1}T^4(\log T)^2}+De^{-D^{-1}T^4 }.
\end{align*}

We now start to derive the expression (<ref>) about $\mathbb{P}_J(J_k^{[q_1,q_2]}\in A')$, which will be the starting point for the estimates. For any $(x_1,x_2)\in\mathbb{R}^2$, define

\begin{align*}
F(x_1,x_2)\coloneqq \mathbb{P}_J(J_k^{[q_1,q_2]}\in A'|J_k(q_1)=x_1,J_k(q_2)=x_2).
\end{align*}

Let $Q(x_1,x_2)$ be the joint p.d.f. of $(J_k(q_1), J_k(q_2))$. Then

\begin{align*}
\mathbb{P}_J(J_k^{[q_1,q_2]}\in A')=\int_{\mathbb{R}^2} F(x_1,x_2)Q(x_1,x_2)dx_1dx_2.
\end{align*}

From the definition (<ref>), $W^{k,[p_-,p_0]}_{\textup{jump}}(\mathcal{L}_k;f_\textup{J})=1$. Also,

\begin{align*}
&W^{k,[p_0,p_+]}_{\textup{jump}}(\mathcal{L}_k;f_\textup{J})=\exp\left( -\int_{[p_0,p_0+d]} \textbf{H}_t^{k,k+1}(\ug(x)-\mathcal{L}_k)\, dx \right).
\end{align*}

For any $\textbf{b}=(b_0,b_\pm)\in\mathbb{R}^3$, let

\begin{align*}
m_1(\mathbf{b})\coloneqq \frac{b_-  ({p}_0-q_1)+b_0 (q_1- {p}_-)}{ {p}_0- {p}_-},\  m_2(\mathbf{b})\coloneqq \frac{b_0 (  {p}_+-q_2) +b_+ (q_2- {p}_0) }{ {p}_+- {p}_0}.
\end{align*}

Conditioned on $J_k( {p}_0)=b_0$ and $J_k( {p}_\pm)=b_\pm$, $J_k(q_1)$ and $J_k(q_2)$ are independent. Let $\rho_{\sigma,m}(x)=(2\pi\sigma^2)^{-1/2}e^{-x^2/(2\sigma^2)}$ be the p.d.f. of a normal distribution with mean $m$ and variance $\sigma^2$. The p.d.f. of $J_k(q_1)$ is given by

\begin{align*}
\rho_{\sigma_1,m_1(\mathbf{b})}(x_1) =:  \rho_{\sigma_1,\underline{m}_1}(x_1) f_1(x_1;\mathbf{b}).
\end{align*}

The function $f_1(x_1;\mathbf{b})$ is defined through the last equality. Note that as $m_1(\mathbf{b})\geq \underline{m}_1$, it is straightforward to check that $f_1(x_1;\mathbf{b})$ is non-decreasing in $x_1$. Similarly, the p.d.f. of $J_k(q_2)$ is given by

\begin{align*}
&\rho_{\sigma_2,m_2(\mathbf{b})}(x_2)\left( \mathbb{E}_{\free}^{1,1,( {p}_0 ,q_2),b_0 ,x_2}[W^{k,[p_0,p_0+d]}_{\textup{jump}} ] \right)/\mathbb{E}_{\free}^{1,1,( {p}_0 , {p}_+),b_0,b_+}[W^{k,[p_0,p_0+d]}_{\textup{jump}}]  \\
=:  &\rho_{\sigma_2,\underline{m}_2}(x_2) f_2(x_2;\mathbf{b}).
\end{align*}

The function $f_2(x_2;\mathbf{b})$ is defined through the last equality and is non-decreasing provided $m_2(\mathbf{b})\geq \underline{m}_2$. Write $ \bar{Q}(\mathbf{b})$ for the joint p.d.f of $(J(p_0),J(p_\pm))$. Then

\begin{align*}
Q(x_1,x_2)=\int_{\mathbb{R}^3} \rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2)  f_1(x_1;\mathbf{b})f_2(x_2;\mathbf{b})\bar{Q}(\mathbf{b}) {d}\mathbf{b}.
\end{align*}

The following expression serves as a starting point for proving Proposition <ref>.

\begin{align}\label{eq:P_J(A)}
\mathbb{P}_J(J_k^{[q_1,q_2]}\in A')=\int_{\mathbb{R}^3} \int_{\mathbb{R}^2} F(x_1,x_2)\rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2)  f_1(x_1;\mathbf{b})f_2(x_2;\mathbf{b})\bar{Q}(\mathbf{b})\, {d}x_1  {d}x_2    {d}\mathbf{b}.
\end{align}

Let $\underline{ \mathbf{m}}\coloneqq (\underline{ {m}},\underline{ {m}}_\pm)$ and denote $\{\mathbf{b}\geq \underline{\mathbf{m}}\}\coloneqq \{b_0\geq \underline{m}_0,b_\pm\geq \underline{m}_\pm \}$. Because $F(x_1,x_2)\leq 1$, together with Lemma <ref>, we could bound $\mathbb{P}_J(J_k^{[q_1,q_2]}\in A' )$ by

\begin{align*}
&\int_{\mathbb{R}^3} \int_{\mathbb{R}^2} F(x_1,x_2)\rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2)  f_1(x_1;\mathbf{b})f_2(x_2;\mathbf{b})\mathbbm{1}_{\{\mathbf{b}\geq \underline{\mathbf{m}}\}}\bar{Q}(\mathbf{b}) {d}x_1 {d}x_2  {d}\mathbf{b}+\mathbb{P}_J(\mathsf{Y}^{\textup{c}})\\
\leq &\sup_{\mathbf{b}\geq \underline{\mathbf{m}}}\int_{\mathbb{R}^2} F(x_1,x_2)\rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2)  f_1(x_1;\mathbf{b})f_2(x_2;\mathbf{b}) {d}x_1 {d}x_2 +\mathbb{P}_J(\mathsf{Y}^{\textup{c}}).
\end{align*}

From now on we fix $\mathbf{b}\geq \underline{\mathbf{m}}$. The estimates we derive below are uniform in any such choice of $\mathbf{b}$. By the Gibbs property, $F(x_1,x_2)$ can be rewritten as

\begin{align*}
F(x_1,x_2)=&\mathbb{E}_{\free}^{k,k,(q_1,q_2),x_1,x_2}\left[\mathbbm{1}{\{B^{[q_1,q_2]}\in A\}} W^{k,[p_0,p_0+d]}_{\textup{jump}} \right]\left/\mathbb{E}_{\free}^{k,k,(q_1,q_2),x_1,x_2}\left[ W^{k,[p_0,p_0+d]}_{\textup{jump}} \right]\right.\\
\leq &\mathbb{P}^{k,k,(q_1,q_2),0,0}_{\free}(\{B^{[q_1,q_2]}\in A\})\bigg( \mathbb{E}_{\free}^{k,k,(q_1,q_2),x_1,x_2}[W^{k,[p_0,p_0+d]}_{\textup{jump}} ]\bigg)^{-1}.\\
=&\varepsilon\bigg( \mathbb{E}_{\free}^{1,1,(q_1,q_2),x_1,x_2}[ W^{k,[p_0,p_0+d]}_{\textup{jump}} ]\bigg)^{-1}.
\end{align*}

Here we have used $W^{k,[p_0,p_0+d]}_{\textup{jump}} \leq 1$.

$$\bar{F}(x_1,x_2)\coloneqq \min\left\{ 1,\varepsilon \bigg( \mathbb{E}_{\free}^{k,k,(q_1,q_2),x_1,x_2}[W^{k,[p_0,p_0+d]}_{\textup{jump}} ]\bigg)^{-1}\right\}.$$

Hence $F(x_1,x_2)\leq \bar{F}(x_1,x_2)$. Since $\bar{F}(x_1,x_2)$ is non-increasing in $x_1$ for fixed $x_2$ and and $f_1(x_1;\mathbf{b}) $ is non-decreasing, for any $x_2\in\mathbb{R}$,

\begin{align*}
&\int_{\mathbb{R} } \rho_{\sigma_1,\underline{m}_1}(x_1)  \bar{F}(x_1,x_2)f_1(x_1;\mathbf{b})  {d}x_1-\int_{\mathbb{R} } \rho_{\sigma_1,\underline{m}_1}(x_1)   \bar{F}(x_1,x_2)  {d}x_1\\
=&\left( \int_{\mathbb{R} } \rho_{\sigma_1,\underline{m}_1}(x_1)  \bar{F}(x_1,x_2)f_1(x_1;\mathbf{b})  {d}x_1\right)\left(\int_{\mathbb{R}} \rho_{\sigma_1,\underline{m}_1}(x_1')\,{d}x'_1 \right)\\
&-\int_{\mathbb{R} } \rho_{\sigma_1,\underline{m}_1}(x_1)   \bar{F}(x_1,x_2)  {d}x_1\left(\int_{\mathbb{R}} \rho_{\sigma_1,\underline{m}_1}(x_1')f_1(x'_1;\mathbf{b})\,{d}x'_1 \right)\\
=&   2^{-1}\int_{\mathbb{R}^2 } \rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_1,\underline{m}_1}(x'_1) (\bar{F}(x_1,x_2)-\bar{F}(x'_1,x_2) )(f_1(x_1;\mathbf{b}) -f_1(x'_1;\mathbf{b}) ) {d}x_1{d}x_1'\\
\leq &0.
\end{align*}

Since $\bar{F}(x_1,x_2)$ is non-increasing in $x_2$ for fixed $x_1$ and and $f_2(x_2;\mathbf{b}) $ is non-decreasing, we can repeat the same deduction to get

\begin{align*}
&\int_{\mathbb{R}^2} \bar{F}(x_1,x_2)\rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2)  f_1(x_1;\mathbf{b})f_2(x_2;\mathbf{b}) {d}x_1 {d}x_2\\
\leq &\int_{\mathbb{R}^2} \bar{F}(x_1,x_2)\rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2)   f_2(x_2;\mathbf{b}) {d}x_1 {d}x_2\\
\leq &\int_{\mathbb{R}^2} \bar{F}(x_1,x_2)\rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2) {d}x_1 {d}x_2\\
\leq & \int_{\underline{m}_2-\sigma_2T^{3/2}}^\infty\int_{\underline{m}_1-\sigma_1T^{3/2} }^\infty \bar{F}(x_1,x_2)\rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2) {d}x_1 {d}x_2+Ce^{-C^{-1}T^3}.
\end{align*}

We have used $\bar{F}(x_1,x_2)\leq 1$ and $\displaystyle\int_{-\infty}^{\underline{m}_i-\sigma_iT^{3/2}} \rho_{\sigma_i,\underline{m}_i}(x)dx\leq Ce^{-C^{-1}T^3}$ in the last inequality.

Now it suffices to estimate $\bigg( \mathbb{E}_{\free}^{k,k,(q_1,q_2),x_1,x_2}[ W^{k,[p_0,p_0+d]}_{\textup{jump}} ]\bigg)^{-1}$. Let $B$ be a Brownian bridge with $B(q_1)=x_1$ and $B(q_2)=x_2$. We start by considering an event which ensures $B(x)\geq \ug(x)$ for $x\in [p_0,p_0+d]$.

For $E$ large enough, the following statement holds. Let $B:[q_1,q_2]\to\mathbb{R}$ be a continuous curve with $B(q_2)=x_2$. Suppose that

* $x_2\geq \underline{m}_2-\sigma_2T^{3/2}$,

* $B({p}_0)\geq \ug( {p}_0)+3\Delta$,

* $\displaystyle\inf_{x\in [p_0,q_2]} B^{[p_0,q_2]}(x)  >-\Delta$.

Then $B(x)\geq \ug(x)$ in $[ {p}_0 ,{p}_0 +d]$.

It is sufficient to prove the assertion for the case $x_2=\underline{m}_2-\sigma_2T^{3/2}$ and $B(p_0)=g(p_0)+3\Delta$.

From Condition C3 and (<ref>), for all $x\in [p_0,p_0+d]$ we have

\begin{align*}
\ug(x)\leq   \ug(p_0)+ \Delta.
\end{align*}

Under the assumptions of this lemma, we bound the slope of the secant of $B$ in $[p_0,q_2]$ as

\begin{align*}
&\left| \frac{x_2-B(p_0)}{q_2-p_0} \right|=\left| \frac{(\underline{m}_2-\sigma_2T^{3/2})-(g(p_0)+3\Delta)}{q_2-p_0} \right|\\
=&\left| \frac{\textup{Tent}(q_2)-\textup{Tent}(p_0)-\sigma_2T^{3/2} - (k+5)\Delta-k\log k }{q_2-p_0} \right| .
\end{align*}

Using (<ref>), $\sigma_2\leq s^{1/2}$ and $q_2-p_0\geq 2^{-1}s$, the above is bounded by

\begin{align*}
2T+2T^{3/2}+2(k+5) \Delta+2(k\log k)T^2.
\end{align*}

For large enough $T$, it holds that

$$\left| \frac{x_2-B(p_0)}{q_2-p_0} \right|\leq 4T^{3/2}.$$

As a result, for $x\in [p_0,p_0+d]$, we have

\begin{align*}
B(x)=&B(p_0)+B^{[ {p}_0,q_2]}(x)+\frac{x_2-B(p_0)}{q_2-p_0} (x-p_0)\\
\geq & (\ug(p_0)+3\Delta)+(-\Delta)-4T^{3/2}d\\
\geq & \ug(x)+\left( \Delta -4T^{3/2}d \right)\\
=&\ug(x)+\left( 8\log T -256 T^{-5/2}(\log T)^2 \right).
\end{align*}

Here we have used $\Delta=8\log T$ and $d=64T^{-4}(\log T)^2$. For $T$ large enough, we conclude that $B(x)\geq \ug(x)$ for all $x\in [p_0,p_0+d]$.

When $B(x)\geq g(x)$ in $[p_0,p_0+d]$, we have
$$W^{k,[p_0,p_0+d]}_{\textup{jump}}(B)\geq e^{-d\textbf{H}_t^{k,k+1}(0)}\geq 2^{-1}.$$
Therefore, provided $x_2\geq \underline{m}_2-\sigma_2T^{3/2}$,

\begin{align*}
&\mathbb{E}_{\free}^{k,k,(q_1,q_2),x_1,x_2}[W^{k,[p_0,p_0+d]}_{\textup{jump}} ]\\
\geq &2^{-1}\mathbb{P}_{\free}^{k,k,(q_1,q_2),x_1,x_2}(B(p_0)\geq \ug(p_0)+3\Delta, \inf_{x\in [p_0,q_2]} B^{[p_0,q_2]}(x)  >-\Delta\ )\\
=&2^{-1}\mathbb{P}_{\free}^{k,k,(q_1,q_2),x_1,x_2}(B(p_0)\geq \ug(p_0)+3\Delta)\, \mathbb{P}_{\free} ( \inf_{x\in [p_0,q_2]} B^{[p_0,q_2]}(x)  >-\Delta\ )\\
\geq &4^{-1}\mathbb{P}_{\free}^{k,k,(q_1,q_2),x_1,x_2}(B(p_0)\geq \ug(p_0)+3\Delta).
\end{align*}

Under the law $\mathbb{P}_{\free}^{k,k,(q_1,q_2),x_1,x_2}$, $B(p_0)$ is a normal distribution with variance $\sigma_3^2$ and mean

$$m_3(x_1,x_2)\coloneqq \frac{x_1(q_2-p_0)+x_2(p_0-q_1) }{q_2-q_1} .$$

\begin{align*}
\mathbb{E}_{\free}^{k,k,(q_1,q_2),x_1,x_2}[W^{[p_0,p_0+d]}_{\textup{jump}} ] \geq 4^{-1}\int^{\infty}_{  \ug( {p}_0)+3\Delta-m_3(x_1,x_2)} \rho_{\sigma_3,0}(z_3)\,  {d}z_3\\
=:4^{-1}\nu_{\sigma_3,0}\left[  \ug( {p}_0)+3\Delta-m_3(x_1,x_2)\right].
\end{align*}

This implies

\begin{align*}
\bar{F}(x_1,x_2)\leq 4\varepsilon \nu_{\sigma_3,0}\left[  \ug( {p}_0)+3\Delta-m_3(x_1,x_2)\right]^{-1}.
\end{align*}

\begin{align*}
&\int_{\underline{m}_2-\sigma_2T^{3/2}}^\infty\int_{\underline{m}_1-\sigma_1T^{3/2} }^\infty \bar{F}(x_1,x_2)\rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2) {d}x_1 {d}x_2\\
\leq & 4\varepsilon \int_{\underline{m}_2-\sigma_2T^{3/2}}^\infty\int_{\underline{m}_1-\sigma_1T^{3/2} }^\infty \rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2)\nu_{\sigma_3,0}\left[  \ug( {p}_0)+3\Delta-m_3(x_1,x_2)\right]^{-1} {d}x_1 {d}x_2.
\end{align*}

To simplify the expression, we perform a change of variables $z_i=x_i-\underline{m}_i$.

\begin{align*}
m_3(x_1,x_2)=&\frac{z_1(q_2-p_0)+z_2 (p_0-q_1) }{q_2-q_1} +\frac{\text{Tent}(q_1)(q_2-p_0)+\text{Tent}(q_2) (p_0-q_1) }{q_2-q_1}    - \Delta_k.
\end{align*}

From (<ref>), the slope of $\text{Tent}$ is bounded by $\pm 2T$. Hence

\begin{align*}
&\frac{\text{Tent}(q_1)(q_2-p_0)+\text{Tent}(q_2) (p_0-q_1) }{q_2-q_1} \\
\geq &\frac{(\text{Tent}(p_0)-(p_0-q_1)2T)(q_2-p_0)+(\text{Tent}(p_0)-(q_2 -p_0)2T) (p_0-q_1) }{q_2-q_1}\\
=& \ug(p_0)-4\sigma_3^2 T.
\end{align*}

It follows that

\begin{align*}
m_3(x_1,x_2)\geq \frac{z_1(q_2-p_0)+z_2 (p_0-q_1) }{q_2-q_1} +\ug(p_0)-4\sigma_3^2 T- \Delta_k.
\end{align*}

Take $M= \sigma_3^{-1}(4\sigma_3^2 T+\Delta_k+3\Delta) $. In particular, $4\sigma_3^2 T+\Delta_k+3\Delta= \sigma_3 M .$ Then the upper bound becomes

\begin{align}\label{equ:long}
4\varepsilon \int_{ -\sigma_2T^{3/2}}^\infty\int_{ -\sigma_1T^{3/2} }^\infty  \rho_{\sigma_1,0}(z_1) \rho_{\sigma_2,0}(z_2)\cdot\bigg[ \nu_{\sigma_3,0}\left(-\frac{z_1(q_2-p_0)+z_2 (p_0-q_1) }{q_2-q_1} +\sigma_3 M \right)  \bigg]^{-1} dz_1dz_2.
\end{align}

Let $N_1$ and $N_2$ be two independent Gaussian random variables with mean zero and variances $\sigma_1^2$ and $\sigma_2^2$ respectively. Then
$$N_4\coloneqq \frac{ (q_2-p_0)N_1+ (p_0-q_1)N_2 }{q_2-q_1} $$

is a Gaussian random variable with mean zero and variance

$$\sigma^2_4\coloneqq \left(\frac{  q_2-p_0   }{q_2-q_1}\right)^2  \sigma_1^2+\left(\frac{  p_0-q_1   }{q_2-q_1}\right)^2 \sigma_2^2.$$
Moreover, as $N_1\geq -\sigma_1T^{3/2}$ and $N_2\geq -\sigma_2T^{3/2}$,

$$N_4\geq -\left(\left(\frac{  q_2-p_0   }{q_2-q_1}\right)\sigma_1+\left(\frac{  p_0-q_1   }{q_2-q_1}\right)\sigma_2\right)T^{3/2}.$$

Before continuing the estimate, we give a simple bound to the above.

* $4^{-1}\sigma_3^2\leq \sigma_4^2\leq \sigma_3^2$,

* $\left(\frac{  q_2-p_0   }{q_2-q_1}\right)\sigma_1+\left(\frac{  p_0-q_1   }{q_2-q_1}\right)\sigma_2 \leq 2\sigma_3  $

(1) From a direct computation,

\begin{align*}
\sigma_4^2=&\frac{(q_2-p_0)^2}{(q_2-q_1)^2}\times \frac{(p_0-q_1)(q_1-p_-)}{p_0-p_-}+\frac{(p_0-q_2)^2}{(q_2-q_1)^2}\times \frac{(p_+-q_2)(q_2-p_0)}{p_+-p_0}\\
=&\sigma_3^2\left( \frac{q_1-p_-}{p_0-p_-}\times\frac{q_2-p_0}{q_2-q_1}+\frac{p_+-q_2}{p_+-p_0}\times\frac{p_0-q_1}{q_2-q_1} \right) .
\end{align*}

Because $p_-<q_1<p_0<q_2<p_+ $, it holds that

\begin{align*}
\frac{q_1-p_-}{p_0-p_-}\leq 1,\  \frac{p_+-q_2}{p_+-p_0}\leq 1.
\end{align*}

\begin{align*}
\sigma_4^2\leq \sigma_3^2\left(  \frac{q_2-p_0}{q_2-q_1}+ \frac{p_0-q_1}{q_2-q_1} \right)=\sigma_3^2.
\end{align*}

Let $\alpha=\frac{q_2-p_0}{q_2-q_1}$ and $\beta=\frac{p_0-q_1}{q_2-q_1}$. Because $q_2-q_1\leq \frac{3}{2}s$ and $p_0-p_-\geq s$, it holds that

\begin{align*}
\frac{q_1-p_-}{p_0-p_-}=1-\frac{p_0-q_1 }{p_0-p_-}\geq 1-\frac{3}{2}\frac{p_0-q_1 }{q_2-q_1}=1-\frac{3}{2}\beta.
\end{align*}

\begin{align*}
\frac{p_+-q_2}{p_+-p_0}\geq 1-\frac{3}{2}\alpha.
\end{align*}

Together with $\alpha+\beta=1$, it holds that

\begin{align*}
&\frac{q_1-p_-}{p_0-p_-}\times\frac{q_2-p_0}{q_2-q_1}+\frac{p_+-q_2}{p_+-p_0}\times\frac{p_0-q_1}{q_2-q_1} \\
\geq  &\left(1-\frac{3}{2}\beta\right)\alpha+\left(1-\frac{3}{2}\alpha\right)\beta \\
=&3(\alpha-2^{-1})^2+4^{-1}\geq 4^{-1}.
\end{align*}

We conclude that $\sigma_4^2\geq 4^{-1}\sigma_3^2$.

(2) By a direct computation,

\begin{align*}
&\left(\frac{  q_2-p_0   }{q_2-q_1}\right)\sigma_1+\left(\frac{  p_0-q_1   }{q_2-q_1}\right)\sigma_2\\
=&\left(\frac{(q_2-p_0)(p_0-q_1)}{q_2-q_1} \right)^{1/2}\left( \left( \frac{(q_2-p_0)(q_1-p_-)}{(q_2-q_1)(p_0-p_-)} \right)^{1/2}+\left(\frac{(p_0-q_1)(p_+-q_2)}{(q_2-q_1)(p_+-p_0)} \right)^{1/2} \right)
\end{align*}

Using $\sigma_3^2=\frac{(q_2-p_0)(p_0-q_1)}{q_2-q_1}$, $\alpha=\frac{q_2-p_0}{q_2-q_1}$ and $\beta=\frac{p_0-q_1}{q_2-q_1}$, it becomes

\begin{align*}
\sigma_3\left(\alpha^{1/2}\left( \frac{  q_1-p_- }{  p_0-p_- } \right)^{1/2}+\beta^{1/2}\left(\frac{ p_+-q_2 }{ p_+-p_0 } \right)^{1/2} \right)
\end{align*}

Because $p_-<q_1<p_0<q_2<p_+ $, it holds that

\begin{align*}
\frac{q_1-p_-}{p_0-p_-}\leq 1,\  \frac{p_+-q_2}{p_+-p_0}\leq 1.
\end{align*}

\begin{align*}
\left(\frac{  q_2-p_0   }{q_2-q_1}\right)\sigma_1+\left(\frac{  p_0-q_1   }{q_2-q_1}\right)\sigma_2\leq \sigma_3(\alpha^{1/2}+\beta^{1/2})\leq 2\sigma_3.
\end{align*}

From Lemma <ref>, (<ref>) is bounded by

\begin{align*}
& 4\varepsilon\int_{-2\sigma_3T^{3/2} }^\infty\rho_{\sigma_4,0}(z_4)\cdot\bigg[ \nu_{\sigma_3,0}(-z_4+\sigma_3 M  )  \bigg]^{-1}\, dz_4\\
=& 4\varepsilon\int_{-\infty }^{ 2\sigma_3T^{3/2}}\rho_{\sigma_4,0}(z_4)\cdot\bigg[ \nu_{\sigma_3,0}( z_4+\sigma_3 M   )  \bigg]^{-1}\, dz_4.
\end{align*}

Because $\bigg[ \nu_{\sigma_3,0}( z_4+\sigma_3M   )  \bigg]^{-1}$ is non-decreasing in $z_4$,

\begin{align*}
&\int_{-\infty }^{0}\rho_{\sigma_4,0}(z_4)\cdot\bigg[ \nu_{\sigma_3,0}( z_4+\sigma_3M      )  \bigg]^{-1}\, dz_4\leq 2^{-1} \bigg[ \nu_{\sigma_3,0}( \sigma_3M  )  \bigg]^{-1}\\
\leq &2^{-1}Z^{-1}\int_{0 }^{ 2\sigma_3T^{3/2}}\rho_{\sigma_4,0}(z_4)\cdot\bigg[ \nu_{\sigma_3,0}( z_4+\sigma_3M   )  \bigg]^{-1}\, dz_4,
\end{align*}

\begin{align*}
Z=\int_{0 }^{ 2\sigma_3T^{3/2}}\rho_{\sigma_4,0}(z_4) \, dz_4.
\end{align*}

By Lemma <ref>, for large enough $T$, it holds that

\begin{align*}
Z\geq \int_{0 }^{ 2\sigma_4T^{3/2}}\rho_{\sigma_4,0}(z_4) \, dz_4\geq 4^{-1}.
\end{align*}

As a result, (<ref>) is bounded by

\begin{align*}
&12\varepsilon  \int_{0 }^{ 2\sigma_3T^{3/2}}\rho_{\sigma_4,0}(z_4)\cdot\bigg[ \nu_{\sigma_3,0}( z_4+\sigma_3M  )  \bigg]^{-1}\, dz_4.
\end{align*}

Next, we replace $\nu_{\sigma_3,0}( z_4+\sigma_3M  )$ with $\rho_{\sigma_3,0}(z_4+\sigma_3M)$. We first show that $M\geq 1$. Recall that $M= \sigma_3^{-1}(4\sigma_3^2 T+\Delta_k+3\Delta)$. Because $T\geq 10$,

\begin{align*}
M\geq 4\sigma_3 T+\sigma_3^{-1}\Delta\geq 4(T\Delta)^{1/2}\geq 1.
\end{align*}

For $z\geq 1$, [Wil91, Section 14.8] we have

\begin{align*}
\nu_{1,0}(z)\geq  \frac{z}{z^2+1}(2\pi)^{-1/2}e^{-z^2/2}\geq (2z)^{-1}(2\pi)^{-1/2}e^{-z^2/2}=(2z)^{-1}\rho_{1,0}(z).
\end{align*}

By the Brownian scaling, for $z\geq \sigma$ it holds that

$$\nu_{\sigma,0}(z)=\nu_{1,0}(\sigma^{-1}z)\geq \sigma (2z)^{-1}\rho_{1,0}(\sigma^{-1}z)=\sigma^2 (2z)^{-1}\rho_{\sigma,0}( z).$$

Because $M\geq 1$ and $z_4\geq 0$, we deduce

\begin{align*}
\nu_{\sigma_3,0}( z_4+\sigma_3 M   )\geq 2^{-1}\sigma_3^2 (z_4+\sigma_3 M)^{-1}\rho_{\sigma_3,0}(z_4+\sigma_3 M).
\end{align*}

Hence (<ref>)  is bounded by

\begin{align*}
&24\varepsilon  \sigma_3^{-2}\int_{0 }^{ 2\sigma_3T^{3/2}}\rho_{\sigma_4,0}(z_4)\rho_{\sigma_3,0}(z_4+\sigma_3 M )^{-1}   ( z_4+ \sigma_3 M )  \, dz_4\\
= &24\varepsilon\int_{0 }^{ 2\sigma_3T^{3/2}}\rho_{\sigma_4,0}(z_4)\rho_{\sigma_3,0}(z_4)^{-1}e^{ M\sigma_3^{-1}z_4+M^2/2 }  (\sigma_3^{-1} z_4+  M )\sigma_3^{-1}  \, dz_4 .
\end{align*}

Because $ 4^{-1}\sigma_3^2\leq \sigma_4^{2}\leq \sigma_3^{2}$,
$$\rho_{\sigma_4,0}(z_4)\rho_{\sigma_3,0}(z_4)^{-1}=\sigma_4^{-1}  \sigma_3 \exp( {-2^{-1}(\sigma_4^{-2}-\sigma_3^{-2})z_4^2}) \leq  2.$$

(<ref>) is bounded by

\begin{align*}
& 48\varepsilon\int_{0 }^{ 2\sigma_3T^{3/2}}e^{ M\sigma_3^{-1}z_4+M^2/2 }  (\sigma_3^{-1} z_4+  M )\sigma_3^{-1} \, dz_4\\
=& 48\varepsilon\times e^{M^2/2} \int_{0 }^{ 2 T^{3/2}} e^{Mw_4} (w_4+  M )  \, dw_4.\\
\leq & 48\varepsilon\times 2T^{3/2}(2T^{3/2}+M) e^{2MT^{3/2}+ M^2/2}.
\end{align*}

We proceed to bound $M$. Because $\frac{3 }{2}s\geq q_2-q_1 $ and $q_2-p_0,p_0-q_1\geq \frac{1}{2}s$, it holds that $\frac{3}{8}s\geq \sigma_3^2\geq \frac{1}{6}s$. Therefore $ M=\sigma_3^{-1}(4\sigma_3^2 T+\Delta_k+3\Delta)\leq DT $.  We arrive at

\begin{align*}
\eqref{equ:long} \leq\varepsilon\times De^{DT^{5/2}}.
\end{align*}

In conclusion,

\begin{align*}
\sup_{\mathbf{b}\geq \underline{\mathbf{m}}}\int_{\mathbb{R}^2} F(x_1,x_2)\rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2)  f_1(x_1;\mathbf{b})f_2(x_2;\mathbf{b})  {d}x_1  {d}x_2\leq \varepsilon\times   De^{DT^{5/2} }+Ce^{-C^{-1}T^3}.
\end{align*}

We are ready to complete the proof of Proposition <ref>.

From the above discussion,

\begin{align*}
\mathbb{P}_J(J_k^{[q_1,q_2]}\in A')\leq &\sup_{\mathbf{b}\geq \underline{\mathbf{m}}}\int_{\mathbb{R}^2} F(x_1,x_2)\rho_{\sigma_1,\underline{m}_1}(x_1)\rho_{\sigma_2,\underline{m}_2}(x_2)  f_1(x_1;\mathbf{b})f_2(x_2;\mathbf{b})  {d}x_1  {d}x_2+\mathbb{P}_J(\mathsf{Y}^{\textup{c}})\\
\leq & \varepsilon\times   De^{DT^{5/2}  }+Ce^{-C^{-1}T^3}+De^{-D^{-1}T^4}.
\end{align*}

Recall that $T=E\max\{\left(\log\varepsilon^{-1} \right)^{1/3},s\}.$ Therefore by taking $E$ large enough,

\begin{align*}
\mathbb{P}_J(J_k^{[0,s]}\in A)= \mathbb{P}_J(J_k^{[q_1,q_2]}\in A')\leq \varepsilon\times D\exp\left( D\left(\log\varepsilon^{-1}\right)^{5/6} \right).
\end{align*}

§ CONCAVE FUNCTIONS

In this section we record basic properties of concave functions. Throughout this section, $h:[\ell,r]\to\mathbb{R}$ is a continuous concave function.

For $x\in (\ell,r)$, the left/right derivative of $h$ at $x$ is defined by

\begin{align*}
h'_{\pm}(x)\coloneqq \lim_{y\to x^{\pm}}\frac{h(y)-h(x)}{y-x}.
\end{align*}

We may extend $h'_+(x)$to $x=\ell$ by allowing it to take value $\infty$. We may also extend $h'_-(r)$ to $x=r$ by allowing it to take value $-\infty$. $h'_{\pm}(x)$ are monotone non-increasing where they are defined. Moreover, for all $x \in (\ell,r)$, $h'_-(x )\geq h'_+(x )$.

$h'_+(x):[\ell,r)\to \mathbb{R}\cup\{\infty\}$ is right continuous and $h'_-(x):(\ell,r]\to \mathbb{R}\cup\{-\infty\}$ is left continuous.

Fix $x_0\in [\ell,r)$ and take $x_j \searrow x_0$. Given $y>x_0$, we have for $x_j<y$,

\begin{align*}
\frac{h(y)-h(x_j)}{y-x_j}\leq h'_+(x_j).
\end{align*}

By the continuity of $h(x)$ at $x_0$,

\begin{align*}
\frac{h(y)-h(x_0)}{y-x_0}\leq \liminf_{j\to\infty} h'_+(x_j).
\end{align*}

Hence $h'_+(x_0)\leq \displaystyle  \liminf_{j\to\infty} h'_+(x_j)$. By the monotonicity of $h'_+(x)$, $h'_+(x_0)\geq \displaystyle  \limsup_{j\to\infty} h'_+(x_j)$. This shows $h'_+(x)$ is right continuous. Left continuity of $h'_+(x)$ can be derived similarly.

Define the extreme points of $h(x)$ as

\begin{align*}
\text{xExt}(h)\coloneqq \{\ell,r \}\cup \left\{x\in (\ell,r)\ \big|\ h\big|_{[x-\delta,x+\delta]}\ \textup{is not linear for any}\ \delta>0  \right\}.
\end{align*}

For any $x_0\in [\ell,r]\setminus \textup{xExt}(h)$, by considering the largest interval containing $x_0$ in which $h(x)$ is linear, we get the following lemma.

For all $x_0\in [\ell,r]\setminus \textup{xExt}(h)$, there exist $x_1,x_2\in \textup{xExt}(h)$ such that $x_0\in (x_1,x_2)$ and $h\big|_{[x_1,x_2]}$ is linear.

As a direct corollary, we have

Let $\bar{h}:[\ell,r]\to \mathbb{R}$ be another concave function. Suppose $\bar{h}(x)\geq h(x)$ for all $x\in \textup{xExt}(h)$. Then $\bar{h}(x)\geq h(x)$ for all $x\in [\ell,r]$.

Let $P\subset \textup{xExt}(h)$ be a subset that satisfies the following properties

* $\ell ,r \in P$

* For any $y\in \textup{xExt}(h)\cap [\ell ,r ]$, there exists $x\in P$ such that $|x-y|<s$.

Suppose that $|h'_{\pm}|\leq m$ for some $m\geq 0$. Further assume that $\bar{h}:[\ell ,r ]\to\mathbb{R}$ is a concave function with $\bar{h}(x)\geq h(x)$ for all $x\in P\cap [\ell ,r ]$. Then $\bar{h}(x)+2 ms \geq h(x)$ for all $x\in [\ell ,r ]$.

It suffices to show that $\bar{h}(x)+2m \geq h(x)$ for all $x\in \textup{xExt}(h)$. Given $x\in \textup{xExt}(h)$. If $x\in P$, by the assumption $\bar{h}(x)\geq h(x)$. Suppose $x\in \textup{xExt}(h) \setminus P$. There exists $y\in P$ with $|x-y|< d$. We further assume that $x<y$. Because $\bar{h}(y)\geq h(y)$, $\bar{h}(\ell)\geq h(\ell)$ and $\bar{h}$ is concave, we have

\begin{align*}
\bar{h}(x )\geq&\frac{(y-x)\bar{h}(\ell)+(x-\ell)\bar{h}(y)}{y-\ell}
\geq  \frac{(y-x)  h (\ell)+(x-\ell) h (y)}{y-\ell}\\
=&h(y)+\frac{h(\ell)-h(y)}{y-\ell}(y-x)\geq h(y)-ms .
\end{align*}

Together with $h(x)\leq h(y)+ms  $, we obtain that $\bar{h}(x)\geq h(x)-2ms $. The argument for $y<x$ is similar. The proof is finished.

Let $m\in\mathbb{R}$. Assume that $\{x\in [\ell,r]\ |\ h'_+(x)\leq m\}$ is non-empty. Then

$$x_1\coloneqq\inf \{x\in [\ell,r]\ |\ h'_+(x)\leq m\}\in \textup{xExt}(h)$$

and $h'_+(x_1)\leq m$. Similarly, assume $\{x\in [\ell,r]\ |\ h'_-(x)\geq m\}$ is non-empty. Then
$$x_2\coloneqq\sup \{x\in [\ell,r]\ |\ h'_-(x)\geq m\}\in \textup{xExt}(h)$$

and $h'(x_2)\geq m$.

$h'_+(x_1)\leq m$ and $h'_-(x_2)\geq m$ follows from one-sided continuity of $h'_{\pm}.$ If $x_1\notin \textup{xExt}(h)$, then $h$ is linear around $x_1$ which contradicts the definition of $x_1$. The proof for $x_2$ is similar.

Fix $m\in\mathbb{R}$ and $[\ell',r']\subset (\ell,r)$. Define
$$\mathbf{l}(h)\coloneqq\left\{ \begin{array}{cc}
\inf\{x\in [\ell',r']\,|\, h'_+(x)\leq m \} & \{x\in [\ell',r']\,|\, h'_+(x)\leq m \}\neq \phi,\\
r' & \{x\in [\ell',r']\,|\, h'_+(x)\leq m \}=\phi.
\end{array}
\right..$$

$$\mathbf{r}(h)\coloneqq\left\{ \begin{array}{cc}
\sup\{x\in [\ell',r']\,|\, h'_-(x)\geq m \} & \{x\in [\ell',r']\,|\, h'_-(x)\geq m \} \neq \phi,\\
\ell' & \{x\in [\ell',r']\,|\, h'_-(x)\geq m \} =\phi.
\end{array}
\right..$$

The following lemma shows that $\textbf{l}$ and $\textbf{r}$ are semi-continuous with respect to the uniform topology.

Given a sequence of concave functions $h_i$ converging to $h_0$ uniformly on $[\ell,r]$, we have

\begin{align*}
\liminf_{i\to\infty} \mathbf{l}(h_i)\geq \mathbf{l}(h_0),\ \limsup_{i\to\infty} \mathbf{r}(h_i)\leq \mathbf{r}(h_0).
\end{align*}

We give the proof for $\mathbf{l}$. The argument for $\mathbf{r}$ is similar. We assume $\mathbf{l}(h_0)>\ell'$ otherwise the assertion is clearly true. For any $\ell'\leq x_0<\mathbf{l}(h_0))$, we have $(h_0)'_+(x_0)>m$. Hence there exists $\varepsilon>0$ such that

\begin{align*}
\varepsilon^{-1}( h_0(x_0+\varepsilon)-h_0(x_0))>m.
\end{align*}

Because $h_i$ converges to $h_0$ uniformly, for $i$ large enough,

\begin{align*}
\varepsilon^{-1}( h_i(x_0+\varepsilon)-h_i(x_0))>m.
\end{align*}

By the concavity of $h_i$, $(h_i)'_+(x_0)>m.$ For such $i$, $\mathbf{l}(h_i)\geq x_0$. Hence
$$\displaystyle\liminf_{i\to\infty} \mathbf{l}(h_i)\geq x_0.$$

Because the above holds for all $\ell'\leq x_0<\textbf{l}(h_0), $ we have

$$\displaystyle\liminf_{i\to\infty} \mathbf{l}(h_i)\geq \textbf{l}(h_0).$$

The proof is finished.

Let $g:[\ell ,r ]\to \mathbb{R}$ be an upper semi-continuous function. Let $h(x):[\ell,r]\to \mathbb{R}$ be the concave majorant of $g$ in $[\ell,r]$. In other words, for any $x_0\in [\ell,r]$

\begin{align*}
h(x_0)\coloneqq\inf\left\{ ax_0+b\ |\ ax+b\geq g(x)\ \textup{for all}\ x\in [\ell,r] \right\}.
\end{align*}

We note that $h(x)$ is continuous in $[\ell,r]$.

For all $x\in\textup{xExt}(h)\setminus\{\ell,r\}$, we have $h(x)=g(x)$.

Suppose that $x_0\in\textup{xExt}(h)\setminus\{\ell,r\}$ and that $h(x_0)-g(x_0)>0$. Let $L_{\pm}(x)=h'_\pm (x_0)(x-x_0)+h(x_0)$.  We have $L_{\pm}(x)\geq h(x)\geq g(x)$ for all $x\in [\ell,r]$. Suppose that $L_+(x)>g(x)$ for all $x\in (x_0,r]$. By the upper semi-continuity of $g(x)$,
$$\varepsilon=\displaystyle\inf_{[x_0,r]}(L_+(x)-g(x))>0. $$
Then for all $x\in [\ell,r]$,

$$ L_+(x)-\frac{\varepsilon}{r-x_0}(x-x_0)\geq g(x).$$

Hence $L_+(x)-\frac{\varepsilon}{r-x_0}(x-x_0)\geq h(x)$. This implies $h'_+(x_0)-\frac{\varepsilon}{r-x_0} \geq h'_+(x_0)$ which is a contradiction. Therefore there exists $x_+\in (x_0,r]$ such that $L_+(x_+)=g(x_+)$. Similarly, there exists $x_-\in [\ell,x_0)$ such that $L_-(x_-)=g(x_-)$. Because $x_0\in\textup{xExt}(h)$, we must have $h'_+(x_0)<h'_-(x_0)$. Take $L_0(x)=\frac{1}{2}(h'_+(x_0)+h'_-(x_0) )(x-x_0)+h(x_0)$. At $x=x_0$, $L_0(x_0)=h(x_0)>g(x_0)$. For $x\in (x_0,r]$, $L_0(x)> L_+(x)\geq g(x)$. For $x\in [\ell,x_0)$, $L_0(x)>L_-(x)\geq g(x)$. In particular,

\begin{align*}
\varepsilon_0= \inf_{x\in [\ell,r]} L_0(x)-g(x)>0.
\end{align*}

Then $h(x)\leq L_0(x)-\varepsilon_0$, which is a contradiction at $x=x_0$.

Next, we consider the situation that $h(x)$ or $g(x)$ is close to a parabola $-2^{-1}x^2$.

Let $\delta>0$ be a positive number. Let $h:[\ell,r]\to \mathbb{R}$ be a continuous concave function with $|h(x)+2^{-1}x^2|\leq 4^{-1}\delta^2$. Then for all $x\in [\ell+\delta ,r-\delta ]$, we have

\begin{equation}
-x-\delta \leq h'_+(x)\leq h'_-(x)\leq -x+\delta.
\end{equation}

Let $x\in [\ell+\delta,r-\delta].$ By the concavity of $h$, we have

\begin{align*}
h'_+(x)\geq \frac{h(x)-h(x-\delta )}{\delta}\geq \delta^{-1}\left( (2^{-1}x^2-4^{-1}\delta^2)-(2^{-1}(x-\delta)^2+4^{-1}\delta^2) \right)=-x-\delta.
\end{align*}

\begin{align*}
h'_-(x)\leq \frac{h(x+\delta)-h(x  )}{\delta}\leq \delta^{-1}\left( (2^{-1}(x+\delta)^2+4^{-1}\delta^2)-(2^{-1}x^2-4^{-1}\delta^2) \right)=-x+\delta.
\end{align*}

Let $g:[\ell,r]\to \mathbb{R}$ be a function with $|g(x)+2^{-1}x^2|\leq 4^{-1}\delta^2$. Take $x_0\in [\ell+3\delta,r-3\delta]$. Let $h_1$ and $h_2$ be the concave majorant of $g(x)$ in $[\ell,r]$ and $[x_0-3\delta,x_0+3\delta]$ respectively. Then $h_1(x_0)=h_2(x_0)$.

Directly from the definition, we have $h_1(x_0)\geq h_2(x_0)$. Consider the line

\begin{align*}
L(x)\coloneqq h_2(x_0)+(h'_{2})_+(x_0)(x-x_0).
\end{align*}

It suffices to show that $L(x)\geq g(x)$ for all $x\in [\ell,r]$. Because $|h_2(x)+2^{-1}x^2|\leq 4^{-1}\delta^2$ for all $x\in [x_0-\delta,x_0+\delta]$, from Lemma <ref> we have

\begin{align*}
|(h'_{2 })_+(x_0)+x_0|\leq \delta.
\end{align*}

As a result, for all $x\geq x_0$, we have

\begin{align*}
L(x)-g(x)&\geq \left[ -2^{-1}x_0^2-4^{-1}\delta^2+(-x_0+\delta)(x-x_0) \right]-[-2^{-1}x^2+4^{-1}\delta^2]\\
&=2^{-1}(x-x_0)^2-\delta (x-x_0)-2^{-1}\delta^2.
\end{align*}

In particular, for $x\geq x_0+3\delta$, we have $L(x)\geq g(x)$. Similar argument can show that $\ell(x)\geq g(x)$ for all $x\leq x_0-3\delta$. Together with $L(x)\geq h_2(x)\geq f(x)$ for all $x\in [x_0-3\delta,x_0+3\delta]$, we conclude that $L(x)\geq f(x)$ for all $x\in [\ell,r]$. Thus

\begin{align*}
h_2(x_0)=L(x_0)\geq h_1(x_0).
\end{align*}

The proof is finished.

Let $\ell<\ell' <r'<r$ be real numbers and $g:[\ell,r]\to \mathbb{R}$ be a function with $|g(x)+2^{-1}x^2|\leq 4^{-1}\delta^2$. Let $h_1$ and $h_2$ be the concave majorant of $g(x)$ in $[\ell,r]$ and $[\ell',r']$ respectively. Then $h_1(x)=h_2(x)$ for all $x\in [\ell'+3\delta,r'-3\delta]$.

§ MISCELLANEOUS

For any $k\in\mathbb{N}$ and $(y_1,y_2,\dots y_k, y_{k+1})\in\mathbb{R}^{k+1} $,

$$ -\sum_{j=1}^ke^{y_{j+1}-y_j}\leq -k e^{(y_{k+1}-y_1)/k} .$$

The assertion holds clearly when $k=1$. Assume $k\geq 2$ and the assertion holds for $k-1$. Then

\begin{align*}
-\sum_{j=1}^ke^{y_{j+1}-y_j}= -\sum_{j=1}^{k-1}e^{y_{j+1}-y_j}-e^{y_{k+1}-y_k}\leq -(k-1)e^{(y_k-y_1)/(k-1)}-e^{y_{k+1}-y_k}.
\end{align*}

It is straightforward to show that

$$-(k-1)e^{(y_k-y_1)/(k-1)}-e^{y_{k+1}-y_k}\leq -k e^{(y_{k+1}-y_1)/k}$$

and the inequality is achieved when $y_k=\frac{1}{k}y_1+\frac{k-1}{k}y_{k+1}$. The proof is finished.

Recall that $\textbf{H}_t(x)=e^{t^{1/3}x}$. For integers $1\leq j< k$, define

\begin{equation}\label{def:Hk}
\mathbf{H}_{t}^{j,k }(x)\coloneqq    (k-1)^{-1} e^{t^{1/3}x/{(k-j)}} .
\end{equation}

For any $k\in\mathbb{N}$, and $(y_1,y_2,\dots y_{k+1})\in\mathbb{R}^{k+1}$, we have

\begin{align*}
-\sum_{j=1}^k \mathbf{H}_t(y_{j+1}-y_j) \leq -\sum_{j=1}^k \mathbf{H}_{t}^{j,k+1}(y_{k+1}-y_j).
\end{align*}

When $k=1$, $\mathbf{H}_{t}^{1,2}(x)=\mathbf{H}_{t}(x)$ and the assertion clearly holds. Assume $k\geq 2$ and the assertion holds for $k-1$.

\begin{align*}
-\sum_{j=1}^k \mathbf{H}_t(y_{j+1}-y_j) &=-\frac{1}{k}\sum_{j=1}^k \mathbf{H}_t(y_{j+1}-y_j) -\frac{k-1}{k}\sum_{j=1}^k \mathbf{H}_t(y_{j+1}-y_j)\\
&\leq -\frac{1}{k}\sum_{j=1}^k \mathbf{H}_t(y_{j+1}-y_j) -\frac{k-1}{k}\sum_{j=2}^k \mathbf{H}_t(y_{j+1}-y_j).
\end{align*}

By Lemma <ref>,
$$-k^{-1}\displaystyle\sum_{j=1}^k \mathbf{H}_t(y_{j+1}-y_j)\leq -k\textbf{H}_t^{1,k+1}(y_{k+1}-y_1)\leq -\textbf{H}_t^{1,k+1}(y_{k+1}-y_1).$$

For $j\in [1,k-1]_\mathbb{Z}$, let $z_j=y_{j+1}$. Then by the induction hypothesis,

\begin{align*}
-\frac{k-1}{k}\sum_{j=2}^k \mathbf{H}_t(y_{j+1}-y_j)   =&-\frac{k-1}{k}\sum_{j=1}^{k-1} \mathbf{H}_t(z_{j+1}-z_j)\leq    -\frac{k-1}{k}\sum_{j=1}^{k-1} \mathbf{H}^{j,k}_t(z_{k}-z_j)\\
=&-\frac{k-1}{k}\sum_{j=2}^{k}  \mathbf{H}^{j-1,k}_t(y_{k+1}-y_j)=-\sum_{j=2}^{k}  \mathbf{H}^{j,k+1}_t(y_{k+1}-y_j).
\end{align*}

The proof is finished.

Fix $p_-<p<p_+$. Let $\mathring{L}  :[p_-,p_+]\to \mathbb{R}$ be a continuous function which is linear in $[p_-,p]$ and $[p ,p_+]$. Fix $q\in (p_-,p_+)$ and $b\in \mathbb{R}$. Let $L$ be a continuous function which is linear in $[p_-,q] $ and $[q  ,p_+]$ with $L(p_{\pm})=\mathring{L}(p_\pm)$ and $L(q)=b$. Define

\begin{align*}
\mathring{m}_\pm \coloneqq  \frac{\mathring{L}(p )-\mathring{L}(p_\pm)}{p -p_\pm},\ {m}_\pm\coloneqq  \frac{ {L}(q )-\mathring{L}(p_\pm)}{q -p_\pm}.
\end{align*}

Then for all $x\in [p_-,p_+]$,

\begin{equation}\label{equ:linear}
\mathring{L}  (x)-  L (x)\leq |L(p)-b|+|p-q|\max\left\{ |\mathring{m}_-|,|\mathring{m}_+|  \right\}
\end{equation}

Furthermore, suppose $|p-q|\leq 2^{-1}\min \{p-p_-,p_+-p\}$. Then

\begin{equation*}
\left| {m}_--\mathring{m}_- \right|\leq 2\left|\frac{ b-L (p) }{p-p_-}\right|+2\left|\frac{   p-q  }{ p-p_- }\right||\mathring{m}_-|.
\end{equation*}

\begin{equation*}
\left| {m}_+-\mathring{m}_+ \right|\leq     2\left|\frac{ b-L (p) }{p_+-p }\right|+2\left|\frac{   p-q  }{ p_+-p  }  \right||\mathring{m}_+|.
\end{equation*}

Without loss of generality, we assume that $q<p$. It suffice to prove (<ref>) at $x=p$ and $x=q$. For $x=q$,

\begin{align*}
\mathring{L}(q)- {L}(q)=[\mathring{L}(p)-\mathring{m}_-   (p-q)]-b\leq |\mathring{L} (p)-b|+|p-q||\mathring{m}_-|.
\end{align*}

For $x=p$,

\begin{align*}
\mathring{L}(p)- {L}(p)=&\mathring{L}(p)-\left[ \frac{(p_+-p)b+(p-q)\mathring{L}(p_+)}{p_+-q} \right]\\
=&\frac{p_+-p}{p_+-q}(\mathring{L}(p)-b)+(p-q) \frac{p_+-p}{p_+-q}\mathring{m}_+\\
\leq & |\mathring{L} (p)-b|+|p-q||\mathring{m}_+|.
\end{align*}

This shows (<ref>).

Under the assumption that $|p-q|\leq 2^{-1}\min \{p-p_-,p_+-p\}$,

\begin{align*}
\left| {m}_--\mathring{m}_- \right|= &\left| \frac{b-\mathring{L}(p_-)}{q-p_-}-\frac{\mathring{L}(p)-\mathring{L}(p_-)}{p-p_-}\right|\\
=&\left| \frac{b-\mathring{L}(p)}{q-p_-}+\frac{(p-q)(\mathring{L}(p)-\mathring{L}(p_-))}{(q-p_-)(p-p_-)} \right|\\
\leq &2\left|\frac{ b-L (p) }{p-p_-}\right|+2\left|\frac{   p-q  }{ p-p_- }\times\frac{\mathring{L}(p)-\mathring{L}(p_-)}{p-p_-} \right|\\
=&2\left|\frac{ b-L (p) }{p-p_-}\right|+2\left|\frac{   p-q  }{ p-p_- }\right||\mathring{m}_-|.
\end{align*}

The bound for $\left| {m}_+-\mathring{m}_+ \right|$ is similar.

Let $x_1<x_2\in\mathbb{R}$, $b_1,b_2\in\mathbb{R}$ and $m>0$. Assume that $|b_2-b_1|\leq m(x_2-x_1)$. Let

\begin{align*}
x_0=&2^{-1}(x_1+x_2)+(2m)^{-1}(b_2-b_1)\in [x_1,x_2],\\
\end{align*}

The point $(x_0,b_0)$ is determined by

\begin{align*}
\frac{b_0-b_1}{x_0-x_1}=m,\ \frac{b_2-b_0}{x_2-x_0}=-m.
\end{align*}

\begin{align*}
\mathbb{P}(B(x_0)\geq b_0\ |B(x_1)=b_1, B(x_2)=b_2)\geq  \mathbb{P}(N\geq  (x_2-x_1)^{1/2}m).
\end{align*}

Then mean and variance of $B(x_0)$ is given by

\begin{align*}
\sigma^2=&(4m^2)^{-1}((x_2-x_1) m^2-(x_2-x_1)^{-1}(b_2-b_1)^2).
\end{align*}

\begin{align*}
\sigma^{-1}(b_0-M)=((x_2-x_1)m^2-(x_2-x_1)^{-1}(b_2-b_1)^2)^{1/2}\leq (x_2-x_1)^{1/2}m .
\end{align*}

[1]
G. Barraquand, I. Corwin and E. Dimitrov. Spatial tightness at the edge of Gibbsian line ensembles. arxiv:2101.03045.

[2]
R. Basu, S. Ganguly and A. Hammond. Fractal geometry of $\textup{Airy}_2$ processes coupled via the Airy sheet. Ann. Probab., to appear.

[3]
I. Corwin and E. Dimitrov. Transversal fluctuations of the ASEP, stochastic six vertex model, and Hall-Littlewood Gibbsian line ensembles. Comm. Math. Phys., 363(2) (2018).

[4]
I. Corwin and P. Ghosal. Lower tail of the KPZ equation. Duke Math. J., 169(7) (2020).

[5]
I. Corwin and P. Ghosal. KPZ equation tails for general initial data. Electron.  J. Probab. 25 (2020).

[6]
I. Corwin and A. Hammond.
Brownian Gibbs property for Airy line ensembles. Invent. Math., 195  (2014).

[7]
I. Corwin and A. Hammond.
KPZ Line ensemble.
Probab. Theory Relat. Fields, 166 (2016).

[8]
J. Calvert, A. Hammond and M. Hegde. Brownian structure in the KPZ fixed point. arXiv:1912.00992.

[9]
I. Corwin. The Kardar-Parisi-Zhang equation and universality class. Random Matrices Theory Appl.,
1(1), 76 (2012).

[10]
E. Dimitrov and K. Matetski. Characterization of Brownian Gibbsian line ensembles. arxiv:2002.00684.

[11]
D. Dauvergne, J. Ortmann and B. Virág. The directed landscape. arxiv:1812.00309.

[12]
D. Dauvergne and B. Virág. Bulk properties of the Airy line ensemble. Ann. Probab. 49(4) (2021).

[13]
A. Hammond. Exponents governing the rarity of disjoint polymers in Brownian last passage percolations. Proc. London Math. Soc. 120(3) (2020).

[14]
A. Hammond. Modulus of continuity of polymer weight profiles in Brownian last passage percolation. Ann. Probab. 47(6) (2019).

[15]
A. Hammond. A patchwork quilt sewn from Brownian fabric: regularity of polymer weight profiles in Brownian last passage percolation. Forum Math. Pi, 7, e2 (2019).

[16]
A. Hammond. Brownian regularity for the Airy line ensemble, and multi-polymer watermelons in Brownian last passage percolation. Mem. Amer. Math. Soc., to appear.

[17] S. Janson, S.  Gaussian Hilbert spaces. Cambridge Texts in Mathematics 129. Cambridge Univ. Press, Cambridge.

[18]
K. Kardar, G. Parisi and Y.Z. Zhang. Dynamic scaling of growing interfaces. Phys. Rev. Lett., 56 (1986).

[19]
I. Karatzas and S. Shreve.
Brownian motion and stochastic calculus.
Volume 113 of Graduate Texts in Mathematics. Springer (1988).

[20]
C.H. Lun and J. Warren. Continuity and strict positivity of the multi-layer extension of the stochastic heat equation. Electron. J. Probab. Volume 25 (2020).

[21]
C. Mueller. On the support of solutions to the heat equation with noise. Stochastics Stochastics 37(4) (1991).

[22]
N. O’Connell and J. Warren. A Multi-Layer Extension of the Stochastic Heat Equation. Commun. Math. Phys. 341(1) (2016).

[23]
J. Quastel and H. Spohn. The one-dimensional KPZ equation and its universality class. J. Statist. Phys., 160(4) (2015).

[24]
J. Quastel and S. Sarkar. Convergence of exclusion processes and KPZ equation to the KPZ fixed point. arXiv:2008.06584.

[25]
B. Virág. The heat and the landscape I. arxiv.org/abs/2008.07241.

[26] Tightness of discrete Gibbsian line ensembles with exponential interaction Hamiltonians. arxiv:1909.00946.

[27] Tightness and local Fluctuation of the KPZ line ensemble. In progress.