In the case $u_{0}=w_{0}=\mathds{1}_{(-\infty,0]}$, using Markov’s inequality
we infer

$\texttt{P}_{x}^{\xi}\left(N^{\leq}(t,0)\geq
1\right)\leq\texttt{E}_{x}^{\xi}\left[N^{\leq}(t,0)\right]$

and thus $\overline{m}(t)\geq m(t)$ for all $t\geq 0$, which establishes the
first inequality in (1.13). The rest of this section will be dedicated to
deriving the second inequality in (1.13), i.e., that the front of the
randomized F-KPP equations lags behind the front of the solution of the
parabolic Anderson model at most logarithmically. We introduce some notation
and, recalling the notation $X^{\nu}$ introduced before (2.1), start with
considering certain “well-behaved” particles

$\begin{split}N_{s,u,t}^{\mathcal{L},a}:=\big{|}\big{\\{}\nu\in
N(s):X^{\nu}_{s}\leq 0,H^{\nu}_{k}\geq
u-T^{(a)}_{k}-5\chi_{1}(\overline{m}^{(a)}(t))\ \forall
k\in\\{1,\ldots,\lfloor&\overline{m}^{(a)}(t)\rfloor\\}\big{\\}}\big{|},\\\
&a>0,\ s,t,u\geq 0;\end{split}$ (4.1)

(4.1) here, $H^{\nu}_{k}:=\inf\\{t\geq 0:X^{\nu}_{t}=k\\}$, the random
variable $T^{(a)}_{k}$ has been defined in (3.43), and

$\displaystyle\chi_{b}(x)$
$\displaystyle:=\Cr{constprobapproxofTn}+b(1+\overline{K}+\Cr{constantlemmaT_{m}teqt})+\Cr{constapproxofTn}(\ln
x\vee 1),\quad x\in(0,\infty),\ b\in\mathbb{R},$ (4.2)

where $\Cr{constprobapproxofTn}$ and $\Cr{constapproxofTn}$ have been defined
in Lemma 3.14, $\overline{K}$ is taken from Corollary 3.17, the constant
$\Cr{constantlemmaT_{m}teqt}$ from Lemma 3.16. We abbreviate
$N_{t}^{\mathcal{L}}:=N_{t,t,t}^{\mathcal{L}}$ and call the particles
contributing to $N_{t}^{\mathcal{L}}$ _leading particles at time $t$_. Cauchy-
Schwarz immediately gives

$\texttt{P}_{x}^{\xi}\left(N^{\leq}(t,0)\geq
1\right)\geq\texttt{P}_{x}^{\xi}\left(N_{t}^{\mathcal{L}}\geq
1\right)\geq\frac{\texttt{E}_{x}^{\xi}\left[N_{t}^{\mathcal{L}}\right]^{2}}{\texttt{E}_{x}^{\xi}\big{[}(N_{t}^{\mathcal{L}})^{2}\big{]}}.$
(4.3)

The next two sections are dedicated to deriving an upper bound for the
denominator and a lower bound for the numerator of the right-hand side, both
for $x$ in a neighborhood of $\overline{m}(t)$.

###  4.1 First moment of leading particles

The biggest chunk of this section will consist of proving the following first
moment bound on the number of leading particles. Recall the notation
$\overline{m}^{(a)}(t)$ from (1.7).

######  Lemma 4.1.

For all $a>0$ there exists $\gamma_{1}=\gamma_{1}(a)\in(0,\infty)$ such that
$\mathbb{P}$-a.s., for all $t$ large enough

$\inf_{x\in[\overline{m}^{(a)}(t)-1,\overline{m}^{(a)}(t)+1]}\emph{{E}}_{x}^{\xi}\big{[}N_{t}^{\mathcal{L},a}\big{]}\geq
t^{-\gamma_{1}}.$

###### Proof.

Let $a>0$. To simplify notation, we will omit the index $a>0$ in the
quantities involved and write
$N_{s,u,t}^{\mathcal{L}}:=N_{s,u,t}^{\mathcal{L},a}$, $T^{(a)}_{x}:=T_{x}$,
$\overline{m}^{(a)}(t):=\overline{m}(t)$ from now on.

Let $A_{u,t}:=\\{H_{k}\geq u-T_{k}-5\chi_{1}(\overline{m}(t))\ \forall
k\in\\{1,\ldots,\lfloor m(t)\rfloor\\}\\}$, let $K$ be such that (3.16) holds
and set $\overline{t}:=T_{\lfloor\overline{m}(t)\rfloor}$. We obtain for all
$t$ large enough that

$\displaystyle\begin{split}&\inf_{x\in[\overline{m}(t)-1,\overline{m}(t)+1]}\texttt{E}_{x}^{\xi}\left[N_{t}^{\mathcal{L}}\right]\geq\frac{\inf_{x\in[\overline{m}(t)-1,\overline{m}(t)+1]}\texttt{E}_{x}^{\xi}\left[N_{t}^{\mathcal{L}}\right]}{2\texttt{E}_{\lfloor\overline{m}(t)\rfloor}^{\xi}\left[N^{\leq}(\overline{t},0)\right]}\geq\frac{c}{2}\frac{\texttt{E}_{\lfloor
m(t)\rfloor}^{\xi}\left[N_{t-1,t,t+1}^{\mathcal{L}}\right]}{\texttt{E}_{\lfloor
m(t)\rfloor}^{\xi}\left[N^{\leq}\left(\overline{t},0\right)\right]}\\\
&=\frac{c}{2}\frac{E_{\lfloor
m(t)\rfloor}\left[e^{\int_{0}^{t-1}\xi(B_{s})\mathop{}\\!\mathrm{d}s};B_{t-1}\leq
0,A_{t+1,t}\right]}{E_{\lfloor
m(t)\rfloor}\Big{[}e^{\int_{0}^{\overline{t}}\xi(B_{s})\mathop{}\\!\mathrm{d}s};B_{\overline{t}}\leq
0\Big{]}}\geq c_{1}\frac{E_{\lfloor
m(t)\rfloor}\left[e^{\int_{0}^{t-1}\zeta(B_{s})\mathop{}\\!\mathrm{d}s};B_{t-1}\leq
0,A_{t+1,t}\right]}{E_{\lfloor
m(t)\rfloor}\Big{[}e^{\int_{0}^{\overline{t}}\zeta(B_{s})\mathop{}\\!\mathrm{d}s};B_{\overline{t}}\leq
0\Big{]}};\end{split}$ (4.4)

here, the first inequality follows from the definition of
$T_{\lfloor\overline{m}(t)\rfloor}$, the second inequality is due to Lemma
A.7, the equality follows using Proposition 2.3 and the last inequality is due
to $\xi=\zeta+\texttt{es}$, as well as (3.51) which gives
$\overline{t}=T_{\lfloor\overline{m}(t)\rfloor}\leq
T_{\overline{m}(t)}+\overline{K}\leq t+\overline{K}$. The numerator can be
bounded from below by

$\displaystyle E_{\lfloor
m(t)\rfloor}\left[e^{\int_{0}^{t-1}\zeta(B_{s})\mathop{}\\!\mathrm{d}s};H_{0}\in[t-3\overline{K}-\Cr{constantlemmaT_{m}teqt},t-1],B_{t-1}\leq
0,A_{t+1,t}\right]$ $\displaystyle\geq
E_{\lfloor\overline{m}(t)\rfloor}\Big{[}e^{\int_{0}^{H_{0}}\zeta(B_{s})\mathop{}\\!\mathrm{d}s}\left.E_{0}\big{[}e^{\int_{0}^{r}\zeta(B_{s})\mathop{}\\!\mathrm{d}s};B_{t-1-r}\leq
0\big{]}\right|_{r=t-1-H_{0}};H_{0}\in[t-3\overline{K}-\Cr{constantlemmaT_{m}teqt},t-1],A_{t+1,t}\Big{]}$
$\displaystyle\geq
c_{2}E_{\lfloor\overline{m}(t)\rfloor}\Big{[}e^{\int_{0}^{H_{0}}\zeta(B_{s})\mathop{}\\!\mathrm{d}s};H_{0}\in[t-3\overline{K}-\Cr{constantlemmaT_{m}teqt},t-1],A_{t+1,t}\Big{]},$

where the second inequality is due to $\zeta\geq-(\texttt{es}-\texttt{ei})$
and $P_{0}(B_{s}\leq 0)\geq 1/2$ for all $s\geq 0$. Now using the inclusion
$\\{B_{\overline{t}}\leq 0\\}\subset\\{H_{0}\leq\overline{t}\\}$ in
combination with $\zeta\leq 0$, we infer
$E_{\lfloor\overline{m}(t)\rfloor}\big{[}e^{\int_{0}^{\overline{t}}\zeta(B_{s})\mathop{}\\!\mathrm{d}s};B_{\overline{t}}\leq
0\big{]}\leq
E_{\lfloor\overline{m}(t)\rfloor}\big{[}e^{\int_{0}^{H_{0}}\zeta(B_{s})\mathop{}\\!\mathrm{d}s};H_{0}\leq\overline{t}\big{]}$.
Thus, recalling $\overline{\eta}(v_{0})<0$ and (2.5), we can continue to lower
bound (4.4) via

$\displaystyle\inf_{x\in[\overline{m}(t)-1,\overline{m}(t)+1]}\texttt{E}_{x}^{\xi}\left[N_{t}^{\mathcal{L}}\right]\geq
c_{3}\frac{E_{\lfloor\overline{m}(t)\rfloor}^{\zeta,\overline{\eta}(v_{0})}\Big{[}e^{-\overline{\eta}(v_{0})H_{0}};H_{0}\in[t-3\overline{K}-\Cr{constantlemmaT_{m}teqt},t-1],A_{t+1,t}\Big{]}}{E_{\lfloor\overline{m}(t)\rfloor}^{\zeta,\overline{\eta}(v_{0})}\big{[}e^{-\overline{\eta}(v_{0})H_{0}};H_{0}\leq\overline{t}\big{]}}$
$\displaystyle\qquad\geq
c_{4}\frac{P_{\lfloor\overline{m}(t)\rfloor}^{\zeta,\overline{\eta}(v_{0})}\Big{(}H_{0}\in[t-3\overline{K}-\Cr{constantlemmaT_{m}teqt},t-1],\
H_{k}\geq t+1-T_{k}-5\chi_{1}(\overline{m}(t)),\,\forall 1\leq
k\leq\lfloor\overline{m}(t)\rfloor\Big{)}}{P_{\lfloor\overline{m}(t)\rfloor}^{\zeta,\overline{\eta}(v_{0})}\big{(}H_{0}\leq\overline{t}\big{)}}$
$\displaystyle\qquad\geq
c_{4}P_{\lfloor\overline{m}(t)\rfloor}^{\zeta,\overline{\eta}(v_{0})}\Big{(}H_{0}\in[\overline{t}-2\overline{K},\overline{t}-\overline{K}-1],\
H_{k}\geq\overline{t}-T_{k}-5\chi_{0}(\overline{m}(t)),\,\forall 1\leq
k\leq\lfloor\overline{m}(t)\rfloor\Big{)},$

where the last inequality is due to $t\geq
T_{\overline{m}(t)}\geq\overline{t}-\overline{K}$ and $\overline{t}\geq
T_{\overline{m}(t)}-\Cr{constantlemmaT_{m}teqt}-\overline{K}$ (by (3.51) and
(3.50)). Now as we recall that $\frac{\lfloor\overline{m}(t)\rfloor}{t}\to
v_{0}$, abbreviating $\eta=\overline{\eta}(v_{0})$,
$n:=\lfloor\overline{m}(t)\rfloor$ and thus $\overline{t}=T_{n}$, we see that
in order to finish the proof, it suffices to show that there exists
$\gamma\in(0,\infty)$ such that $\mathbb{P}$-a.s., for all $n\in\mathbb{N}$
large enough,

$P_{n}^{\zeta,\eta}\big{(}H_{0}\in[T_{n}-2\overline{K},T_{n}-\overline{K}-1],H_{k}\geq
T_{n}-T_{k}-5\chi_{0}(n)\;\forall k\in\left\\{1,\ldots,n\right\\}\big{)}\geq
n^{-\gamma}.$ (4.5)

Using the notation

$\displaystyle\widehat{H}^{(n)}_{k}$
$\displaystyle:=H_{k}-E_{n}^{\zeta,\eta}\left[H_{k}\right]\quad\text{ as well
as }\quad R_{k}^{(n)}:=T_{n}-T_{k}-E_{n}^{\zeta,\eta}\left[H_{k}\right],$
(4.6)

the probability in (4.5) can be rewritten as

$\displaystyle
P_{n}^{\zeta,\eta}\big{(}\widehat{H}^{(n)}_{0}\in[R_{0}^{(n)}-2\overline{K},R_{0}^{(n)}-\overline{K}-1],\widehat{H}^{(n)}_{k}\geq
R_{k}^{(n)}-5\chi_{0}(n)\ \forall k\in\left\\{1,\ldots,n\right\\}\big{)}.$
(4.7)

In order to facilitate computations, we approximate the sequence
$(R_{k}^{(n)})$ by a stationary one, setting

$\displaystyle\rho_{i}$
$\displaystyle:=\frac{L_{i}^{\zeta}(\eta)}{v_{0}L(\eta)}-(L_{i}^{\zeta})^{\prime}(\eta)=\frac{1}{v_{0}L(\eta)}\big{(}L_{i}^{\zeta}(\eta)-L(\eta)\big{)}-\big{(}E_{n}^{\zeta,\eta}[\tau_{i-1}]-\mathbb{E}[E_{n}^{\zeta,\eta}[\tau_{i-1}]]\big{)}\
\text{ and }\ $ (4.8) $\displaystyle{\widehat{R}_{k}^{(n)}}$
$\displaystyle:=\sum_{i=k+1}^{n}\rho_{i},\quad k<n,$ (4.9)

where $\tau_{i-1}=H_{i-1}-H_{i}$, and in the equality we used
$E_{n}^{\zeta,\eta}[\tau_{i-1}]=(L_{i}^{\zeta})^{\prime}(\eta)$ and
$\mathbb{E}[E_{n}^{\zeta,\eta}[H_{k}]]=\frac{n-k}{v_{0}}$. Applying inequality
(3.44) from Lemma 3.14 and using the identity
$\mathbb{E}[E_{n}^{\zeta,\eta}[H_{k}]]=\frac{n-k}{v_{0}},$ we get that
$\mathbb{P}$-a.s.,

$\begin{split}\big{|}R_{k}^{(n)}-\widehat{R}_{k}^{(n)}\big{|}&\leq
2\left(\Cr{constprobapproxofTn}+\Cr{constapproxofTn}\ln n\right)\quad\text{for
all $n\in\mathbb{N}$ and each $k\in\\{0,\ldots,n\\}.$}\end{split}$ (4.10)

From now on we will write $\chi:=\chi_{0}$. Then by (4.10), the probability in
(4.7) can be lower bounded by

$\displaystyle
P_{n}^{\zeta,\eta}\big{(}\widehat{H}^{(n)}_{0}\in[R_{0}^{(n)}-2\overline{K},R_{0}^{(n)}-\overline{K}-1];\widehat{H}^{(n)}_{k}\geq\widehat{R}_{k}^{(n)}-3\chi(n)\
\forall k\in\left\\{1,\ldots,n\right\\}\big{)}.$ (4.11)

Now, for every $n$, enlarging the underlying probability space if necessary,
we introduce two processes $(B_{t}^{(i,n)})_{t\geq 0}$, $i=1,2$, which are
independent from everything else and Brownian motions under $P_{n}$, starting
in $n$, and, without further formal definition, we tacitly assume in the
following that the tilting of the probability measure $P_{n}^{\zeta,\eta}$ of
our original Brownian motion also applies to $(B_{t}^{(i,n)})_{t\geq 0}$,
$i=1,2,$ in the obvious way. For $i=1,2$, let $H^{(i,n)}_{k}:=\inf\\{t\geq
0:B_{t}^{(i,n)}=k\\}$, $k\in\mathbb{Z}$, be the corresponding hitting times,
$\widehat{H}_{k}^{(i,n)}:={H}_{k}^{(i,n)}-E_{n}^{\zeta,\eta}[{H}_{k}^{(i,n)}]$
and let $\Sigma_{n}$ be a random variable which, under $P_{n}$, is uniformly
distributed on $\\{1,\ldots,n-1\\}$ and independent of everything else. We
define

$\displaystyle\beta_{k}^{(i,n)}$
$\displaystyle:=\widehat{H}^{(i,n)}_{k}-\widehat{R}_{k}^{(n)},\quad
k=n-1,n-2,\ldots,\quad i=1,2,$ $\displaystyle\beta_{k}^{(n)}$
$\displaystyle:=\begin{cases}\beta^{(1,n)}_{k},&\Sigma_{n}\leq k<n,\\\
\beta^{(1,n)}_{\Sigma_{n}}+\big{(}\beta^{(2,n)}_{k}-\beta^{(2,n)}_{\Sigma_{n}}\big{)},&k<\Sigma_{n}.\end{cases}$

The $\xi$-adaptedness of the process $(\widehat{R}_{k}^{(n)})_{k<n}$ implies
that the processes $(\beta^{(i,n)}_{k})_{k<n}$, $i=1,2$, are
$P_{n}^{\zeta,\eta}$-independent and have the same distribution as
$(\beta_{k}^{(n)})_{k<n}$. We can therefore rewrite (4.11) as

$\displaystyle P_{n}^{\zeta,\eta}\left(\beta_{k}^{(n)}\geq-3\chi(n)\ \forall
k\in\left\\{1,\ldots,n\right\\},\beta_{0}^{(n)}\in I_{n}\right),$ (4.12)

where
$I_{n}:=\big{[}R_{0}^{(n)}-\widehat{R}_{0}^{(n)}-2\overline{K},R_{0}^{(n)}-\widehat{R}_{0}^{(n)}-\overline{K}-1\big{]}.$
Due to (4.10) we have that $\mathbb{P}$-a.s. for all $n$ large enough,
$R_{0}^{(n)}-\widehat{R}_{0}^{(n)}-2\overline{K}\geq-3\chi(n),$ i.e.

$I_{n}\subset[-3\chi(n),\infty).$ (4.13)

For each $k\in\\{0,\ldots,n\\}$ we introduce

$\displaystyle\overline{\beta}^{(1,n)}_{k}$
$\displaystyle:=\beta^{(1,n)}_{n-1-k}-\beta^{(1,n)}_{n-1},\qquad\overline{\beta}^{(2,n)}_{k}:=\beta^{(2,n)}_{k}-\beta^{(2,n)}_{0},$

and note that

$\beta_{0}^{(n)}=\overline{\beta}_{n-1-\Sigma_{n}}^{(1,n)}-\overline{\beta}_{\Sigma_{n}}^{(2,n)}+\beta_{n-1}^{(1,n)}.$
(4.14)

An illustration of the various processes introduced above is given in Figure 2
below. Now the key to bound the probability in (4.12) is the following lemma.

######  Lemma 4.2.

  1. (a)

There exists $\gamma^{\prime}<\infty$ such that $\mathbb{P}$-a.s. for all $n$
large enough,

$\displaystyle P_{n}^{\zeta,\eta}\big{(}\overline{\beta}^{(1,n)}_{k}\geq 0\
\forall 0\leq k\leq n,\ \overline{\beta}^{(1,n)}_{n}\geq n^{1/4}\big{)}$
$\displaystyle\geq n^{-\gamma^{\prime}},\quad\text{ and }$ $\displaystyle
P_{n}^{\zeta,\eta}\big{(}\overline{\beta}^{(2,n)}_{k}\geq 0\ \forall 0\leq
k\leq n,\ \overline{\beta}^{(2,n)}_{n}\geq n^{1/4}\big{)}$ $\displaystyle\geq
n^{-\gamma^{\prime}}.$

  2. (b)

There exists $C(\gamma^{\prime})>0$ such that $\mathbb{P}$-a.s. for all $n$
large enough,

$\displaystyle P_{n}^{\zeta,\eta}\Big{(}\max_{1\leq k\leq
n,i\in\\{1,2\\}}\big{|}\beta_{k}^{(i,n)}-\beta_{k-1}^{(i,n)}\big{|}\leq
C(\gamma^{\prime})\ln n\Big{)}$ $\displaystyle\geq 1-n^{-3\gamma^{\prime}}.$
(4.15)

  3. (c)

Let $\delta\in(0,1)$. There exists $c>0$ such that for all $x\geq 1$ and all
$n\in\mathbb{Z},$

$\displaystyle
P_{n}^{\zeta,\eta}\big{(}\beta_{n-1}^{(1,n)}\in[x,x+\delta]\big{)}\geq c\delta
e^{-x/c}.$

Before proving Lemma 4.2 we will finish the current proof in order not to
interrupt the Reader. To this end let

$J_{n}:=\sup\big{\\{}k\in\\{1,\ldots,n-1\\}\,:\,I_{n}-\overline{\beta}_{n-k+1}^{(1,n)}+\overline{\beta}_{k}^{(2,n)}\subset[0,2C(\gamma^{\prime})\ln
n]\big{\\}},$

where as always $\sup\emptyset:=-\infty$. We have

$\displaystyle\left\\{\beta_{k}^{(n)}\geq-3\chi(n)\ \forall 0\leq k\leq n-1,\
\beta_{0}^{(n)}\in I_{n}\right\\}$
$\displaystyle\qquad\quad\supset\Big{(}\big{\\{}\overline{\beta}^{(1,n)}_{k}\geq
0\ \forall 0\leq k\leq n,\ \overline{\beta}^{(1,n)}_{n}\geq
n^{1/4}\big{\\}}\cap\big{\\{}\max_{1\leq k\leq
n}\big{|}\overline{\beta}_{k}^{(1,n)}-\overline{\beta}_{k-1}^{(1,n)}\big{|}\leq
C(\gamma^{\prime})\ln n\big{\\}}\Big{.}$
$\displaystyle\qquad\qquad\cap\big{\\{}\overline{\beta}^{(2,n)}_{k}\geq 0\
\forall 0\leq k\leq n,\ \overline{\beta}^{(2,n)}_{n}\geq
n^{1/4}\big{\\}}\cap\big{\\{}\max_{1\leq k\leq
n}\big{|}\overline{\beta}_{k}^{(2,n)}-\overline{\beta}_{k-1}^{(2,n)}\big{|}\leq
C(\gamma^{\prime})\ln n\big{\\}}$
$\displaystyle\qquad\qquad\Big{.}\cap\big{\\{}\beta_{n-1}^{(1,n)}\in
I_{n}-\overline{\beta}_{n-1-\Sigma_{n}}^{(1,n)}+\overline{\beta}_{\Sigma_{n}}^{(2,n)}\big{\\}}\cap\\{\Sigma_{n}=J_{n}\\}\Big{)}.$
(4.16)

Indeed, due to (4.14), the fifth event on the right-hand side of (4.16)
entails that $\beta_{0}^{(n)}\in I_{n}$ must hold. On the last two events on
the right-hand side of (4.16) we have $\beta^{(1,n)}_{n-1}\geq 0$ and thus the
first event on the right-hand side of (4.16) implies that $\beta^{(n)}_{k}$ is
non-negative for $k\geq\Sigma_{n}$. The third event then implies monotonicity
at times $k<\Sigma_{n}$. Since $I_{n}\subset[-3\chi(n),\infty)$ due to (4.13),
this gives the first condition on the left-hand side of (4.16).

Figure 2: Illustration of (4.16).

Now the first and third event on the right hand-side of (4.16) are independent
under $P_{n}^{\zeta,\eta}$ and their probabilities are bounded from below by
$n^{-\gamma^{\prime}}$ due to Lemma 4.2 a). Thus, as a consequence of Lemma
4.2 b), for $n$ large enough the probability of the first four events is
bounded from below by $n^{-2\gamma^{\prime}}-n^{-3\gamma^{\prime}}$.
Furthermore, the first four events imply that $J_{n}\in\\{1,\ldots,n-1\\}$.
Thus, due to Lemma 4.2, conditionally on the occurrence of the first four
events, the probability of the last two events on the right-hand side in
(4.16) can be bounded from below by $cn^{-1}e^{-C(\gamma^{\prime})\ln n/c}\geq
n^{-\gamma^{\prime\prime}}$ for $n$ large enough. The proof of (4.5) and thus
of Lemma 4.1 is completed by the choice
$\gamma_{1}>2\gamma^{\prime}+\gamma^{\prime\prime}$. ∎

###### Proof of Lemma 4.2.

$(b)$ Because $H_{k}^{(1,n)}-H_{k-1}^{(1,n)}\overset{d}{=}\tau_{k-1}$ under
$P_{n}^{\zeta,\eta}$, by recalling
$\mathbb{E}\big{[}E_{n}^{\zeta,\eta}[\tau_{k}]\big{]}=\frac{1}{n}$ and the
definition of $\widehat{R}_{k}^{(n)}$ from (4.8), we have
$\beta_{k}^{(1,n)}-\beta_{k-1}^{(1,n)}=\tau_{k-1}-\frac{L_{k}^{\zeta}(\eta)}{v_{0}L(\eta)}$.
Now $L_{k}^{\zeta}(\eta)$ is $\mathbb{P}$-a.s. bounded by Lemma A.1.
Furthermore, for all $\theta$ such that $|\theta|\leq|\eta^{*}|$ (where
$\triangle=[\eta_{*},\eta^{*}]$), we have

$0\leq
E_{n}^{\zeta,\eta}\big{[}e^{\theta\tau_{k-1}}\big{]}=E_{k}^{\zeta,\eta}\big{[}e^{\theta\tau_{k-1}}\big{]}\leq\left(E_{k}\big{[}e^{(-\texttt{es}+\texttt{ei}+\eta_{*})H_{k-1}}\big{]}\right)^{-1}=e^{\sqrt{2(\texttt{es}-\texttt{ei}+|\eta_{*}|)}}<\infty,$

where the last equation is due to [4, (2.0.1), p. 204]. I.e., $\tau_{k}$ has
uniform exponential moments under $P_{n}^{\zeta,\eta}$ and thus (4.15) follows
by a union bound in combination with the exponential Chebyshev inequality.

$(c)$ We have
$\beta_{n-1}^{(1,n)}=H_{n-1}^{(1,n)}-E_{n}^{\zeta,\eta}[H_{n-1}^{(1,n)}]-\widehat{R}_{n-1}^{(n)}=H_{n-1}^{(1,n)}-E_{n}^{\zeta,\eta}[\tau_{n-1}]-\rho_{n}$,
thus recalling definition (4.8), the event in (c) is equivalent to
$\\{H_{n-1}^{(1,n)}\in[x,x+\delta]+\frac{L_{n}^{\zeta}(\eta)}{v_{0}L(\eta)}\\}$.
Because $\frac{L_{n}^{\zeta}(\eta)}{v_{0}L(\eta)}$ is uniformly bounded and
non-negative, it suffices to check that for every $C>0,$ there exists $c>0$
such that
$\inf_{y\in[0,C]}P_{n}^{\zeta,\eta}(H_{n-1}^{(1,n)}\in[x+y,x+y+\delta])\geq
c\delta e^{-x/c}$ for all $x\geq 1$. Indeed, recalling (2.5), we can lower
bound

$\displaystyle P_{n}^{\zeta,\eta}(H_{n-1}^{(1,n)}\in[x+y,x+y+\delta])\geq
E_{n}\big{[}e^{-(\texttt{es}-\texttt{ei}-\eta_{*})H_{n-1}};H_{n-1}^{(1,n)}\in[x+y,x+y+\delta]\big{]}$
$\displaystyle\quad\geq
e^{-(\texttt{es}-\texttt{ei}-\eta_{*})(x+y+\delta)}P_{n}\big{(}H_{n-1}^{(1,n)}\in[x+y,x+y+\delta]\big{)}\geq\frac{\delta
e^{-(\texttt{es}-\texttt{ei}-\eta_{*})(x+y+\delta)}}{\sqrt{2\pi(x+y+\delta)^{3}}}e^{-\frac{1}{2(x+y)}},$

where the last inequality is due to [4, (2.0.2), p. 204]. Now since the latter
term can be lower bounded by $c\delta e^{-x/c}$, uniformly in $y\in[0,C],$ the
claim follows.

$(a)$ We will prove the second inequality, and explain the modifications that
are necessary to show the first one at the end of the proof. For later
reference it will serve our purposes to exclude some potentially bad behavior
of the process $(\widehat{R}_{k}^{(n)}-\widehat{R}_{k-1}^{(n)})_{k}$. To do
so, we take advantage of the next claim, the proof of which we provide after
concluding the proof of Lemma 4.2.

######  Claim 4.3.

For each $n\in\mathbb{Z}$, the sequence $(\rho_{i})_{i\in\mathbb{Z}}$ consists
of $\mathbb{P}$-centered and $\mathbb{P}$-stationary random variables, and the
family $(\rho_{i})_{i\in\mathbb{Z}}$ is bounded $\mathbb{P}$-a.s. In addition,
$\rho_{i}$ is $\mathcal{F}^{i-1}$-adapted and there exists
$\Cl{const_{p}ho}>0$ such that $\mathbb{P}$-a.s., for all $k,n\in\mathbb{Z}$,
$k<n$, we have

$\big{|}\mathbb{E}\big{[}\rho_{n-k}|\mathcal{F}^{n}\big{]}\big{|}\leq\Cr{const_{p}ho}\cdot\big{(}\psi(k/2)+e^{-k/\Cr{const_{p}ho}}\big{)}.$
(4.17)

Furthermore, there exists $\overline{\sigma}\in[0,\infty)$ such that
$n^{-1/2}\sum_{l=1}^{n}\rho_{l}$ and $n^{-1/2}\sum_{l=1}^{n}\rho_{-l}$
converge in $\mathbb{P}$-distribution to $\overline{\sigma}X$ as $n\to\infty$,
where $X\sim\mathcal{N}(0,1)$ is a standard Normal random variable.

Now due to (4.17), $(\rho_{i})_{i\in\mathbb{Z}}$ fulfills the conditions of
Corollary A.5. As a consequence we deduce that for $k\in\mathbb{N}$ and $x\geq
0$ we have $\mathbb{P}\big{(}\sum_{l=1}^{k}{\rho}_{l}\geq x\big{)}\leq
c_{1}e^{-\frac{x^{2}}{c_{1}k}},$ which, using stationarity, can be extended to
the maximal inequality (e.g. by [22, Theorem 1])

$\displaystyle\mathbb{P}\Big{(}\max_{0\leq k\leq
y}\big{(}\widehat{R}_{r+k}^{(n)}-\widehat{R}_{r}^{(n)}\big{)}\geq x\Big{)}$
$\displaystyle=\mathbb{P}\Big{(}\max_{0\leq k\leq
y}\sum_{l=0}^{k}{\rho}_{l}\geq x\Big{)}\leq
c_{2}e^{-\frac{x^{2}}{c_{2}y}}\quad\forall r,y\in\mathbb{Z},\ x\geq 0.$ (4.18)

Furthermore, recalling (4.6), (4.8) and (4.9), the increments of the process
$(\overline{\beta}^{(2,n)}_{k})_{k}$ can be written as

$\begin{split}\overline{\beta}^{(2,n)}_{k}-\overline{\beta}^{(2,n)}_{k-1}&=\big{(}H_{k}-H_{k-1}-E_{n}^{\zeta,\eta}[H_{k}-H_{k-1}]\big{)}-\Big{(}\sum_{i=k+1}^{n}\rho_{i}-\sum_{i=k}^{n}\rho_{i}\Big{)}\\\
&=\big{(}-\tau_{k-1}+(L_{k}^{\zeta})^{\prime}(\eta)\big{)}-\Big{(}-\frac{L_{k}^{\zeta}(\eta)}{v_{0}L(\eta)}+(L_{k}^{\zeta})^{\prime}(\eta)\Big{)}=\frac{L_{k}^{\zeta}(\eta)}{v_{0}L(\eta)}-\tau_{k-1}.\end{split}$

$\mathbb{P}$-a.s., by Lemma A.1 b), the last fraction in the previous display
is positive and uniformly bounded away from zero and infinity, whereas under
$P_{n}^{\zeta,\eta}$, $\tau_{k-1}$ is an absolutely continuous random variable
with positive density on $(0,\infty)$. Therefore, for the constant

$a:=\frac{1}{4}\sup_{k\in\mathbb{Z}}\operatorname*{ess\,inf}_{\xi}\big{(}\overline{\beta}_{k}^{(2,n)}-\overline{\beta}^{(2,n)}_{k-1}\big{)},$

we have $\operatorname*{ess\,inf}_{k,n\in\mathbb{Z}:\ k\leq n,\
\xi}P_{n}^{\zeta,\eta}(\overline{\beta}_{k}^{(n)}-\overline{\beta}_{k-1}^{(n)}\geq
2a)\geq\delta$ for some universal constant $\delta\in(0,1)$. We now split the
environment into $\overline{\xi}(j):=(\xi(l))_{l\geq j}$ and
$\underline{\xi}(j):=(\xi(l))_{l<j}$ and set $t_{0}=t_{-1}:=0$ as well as
$t_{i}:=2^{i}$ for $i\geq 1$. Furthermore, we introduce two constants:
$\overline{c}>0$, which is defined in (4.31) below, and $\overline{C}>0$,
which is independent of $\overline{c}$ and will be chosen large enough such
that the sums in (4.21) and (4.35) below are finite. For $i\geq 1$, we define
the random variables

$\displaystyle\begin{split}Z_{i}^{(n)}&:=\underset{\overline{\xi}(t_{i+1})}{\text{ess
inf}}\ \underset{x\geq
at_{i-1}^{1/2}}{\inf}P_{n}^{\zeta,\eta}\big{(}\overline{\beta}^{(2,n)}_{t_{i}}\geq
at_{i}^{1/2},\overline{\beta}_{k}^{(2,n)}\geq t_{i}^{1/4}\ \forall
k\in\\{t_{i-1},\ldots,t_{i}\\}\ \big{|}\
\overline{\beta}_{t_{i-1}}^{(2,n)}=x\big{)}\\\
&=\underset{\overline{\xi}(t_{i+1})}{\text{ess inf}}\
P_{n}^{\zeta,\eta}\big{(}\overline{\beta}^{(2,n)}_{t_{i}}\geq
at_{i}^{1/2},\overline{\beta}_{k}^{(2,n)}\geq t_{i}^{1/4}\ \forall
k\in\\{t_{i-1},\ldots,t_{i}\\}\ \big{|}\
\overline{\beta}_{t_{i-1}}^{(2,n)}=at_{i-1}^{1/2}\big{)},\end{split}$ (4.19)

where $\underset{\overline{\xi}(x)}{\text{ess inf}}$ means taking the
essential infimum with respect to $\overline{\xi}(x),$ and where the second
equality is due to the monotonicity of the first probability in (4.19) as a
function in $x.$ Thus, as a random variable, $Z_{i}^{(n)}$ is measurable with
respect to $\mathcal{F}_{t_{i+1}}$. Now since $\overline{\beta}_{k}^{(2,n)}$
is $\mathcal{F}^{k}$-measurable, we have that $Z_{i}^{(n)}$ is
$(\mathcal{F}^{t_{i-1}}\cap\mathcal{F}_{t_{i+1}})$-measurable. Setting
$i(n):=\ln_{2}\big{(}\lfloor\big{(}\overline{C}\ln
n\big{)}^{2}\rfloor\big{)}$, we further define

$\displaystyle Y^{(n)}$
$\displaystyle:=P_{n}^{\zeta,\eta}\Big{(}\overline{\beta}^{(2,n)}_{k}\geq 0\
\forall\lfloor\overline{C}\ln n\rfloor\leq k\leq t_{i(n)},\
\overline{\beta}^{(2,n)}_{t_{i(n)}}\geq a\lfloor\overline{C}\ln n\rfloor\
\Big{|}\ \overline{\beta}^{(2,n)}_{\lfloor\overline{C}\ln
n\rfloor}=2a\lfloor\overline{C}\ln n\rfloor\Big{)}.$

Writing $j(n):=\lceil\log_{2}(n)\rceil$, due to the Markov property of the
process $\overline{\beta}^{(2,n)}$ under $P_{n}^{\zeta,\eta}$, we have
$\mathbb{P}$-almost surely that for all $n$ large enough,

$\displaystyle P_{n}^{\zeta,\eta}\left(\overline{\beta}_{n}^{(2,n)}\geq
n^{1/4},\overline{\beta}_{k}^{(2,n)}\geq 0\ \forall k\leq n\right)$
$\displaystyle\geq\prod_{k=1}^{\lfloor\overline{C}\ln
n\rfloor}P_{n}^{\zeta,\eta}\big{(}\overline{\beta}^{(2,n)}_{k}-\overline{\beta}^{(2,n)}_{k-1}\geq
2a\big{)}\cdot Y^{(n)}\cdot\prod_{i=i(n)+1}^{j(n)}Z_{i}^{(n)}$
$\displaystyle\geq\delta^{\lfloor\overline{C}\ln n\rfloor}\cdot
Y^{(n)}\cdot\exp\Big{\\{}\sum_{i=i(n)+1}^{j(n)}\ln
Z_{i}^{(n)}\mathds{1}_{B_{i}^{(n)}}\Big{\\}},$ (4.20)

where the event

$\displaystyle B_{i}^{(n)}$
$\displaystyle:=\Big{\\{}\max_{{r\in[t_{i-1},t_{i}],\ 0\leq
k\leq\frac{5}{2}at_{i}^{1/2}/\overline{c}}}\big{(}\widehat{R}_{r+k}-\widehat{R}_{r}\big{)}<at_{i-1}^{1/2}/16\Big{\\}}$

occurs $\mathbb{P}$-almost surely for all $i\in[i(n),j(n)]$ and all $n$ large
enough. Indeed, by (4.18) we have

$\displaystyle\sum_{n}\sum_{i=i(n)}^{j(n)}\mathbb{P}\big{(}(B_{i}^{(n)})^{c}\big{)}$
$\displaystyle\leq\sum_{n}\sum_{i=i(n)}^{j(n)}t_{i-1}\mathbb{P}\Big{(}\max_{0\leq
k\leq\frac{5}{2}at_{i}^{1/2}/\overline{c}}\big{(}\widehat{R}_{k}-\widehat{R}_{0}\big{)}\geq
at_{i-1}^{1/2}/16\Big{)}$ $\displaystyle\leq
c_{3}\sum_{n}n\sum_{i=i(n)}^{j(n)}e^{-a^{2}\overline{c}t_{i-1}^{1/2}/c_{3}}\leq
c_{4}\sum_{n}n\log_{2}(n)e^{-a^{2}\overline{C}\ln n/c_{4}}<\infty,$ (4.21)

where the last inequality holds true for $\overline{C}$ large enough. Thus,
the Borel-Cantelli lemma implies that $\mathbb{P}$-a.s., for all $n$ large
enough the events $B_{i}^{(n)}$ occur for all $i\in[i(n),j(n)]$. Furthermore,
it is possible to show that $\mathbb{P}$-almost surely, for all $n$ large
enough we have $Y^{(n)}\geq n^{-\gamma^{\prime\prime}}$. We postpone a proof
of this fact, because it uses the same arguments as the following paragraph
and we will describe necessary adaptations afterwards, cf. page 4.1. Thus, for
the time being it remains to show that there exists $\widetilde{c}>0$ such
that $\mathbb{P}$-almost surely, for all $n$ large enough,

$\sum_{i=i(n)}^{j(n)}\ln(Z_{i}^{(n)})\mathds{1}_{B_{i}^{(n)}}\geq-\widetilde{c}\cdot
j(n).$ (4.22)

The second inequality in Lemma 4.2 (a) then follows from (4.20) with
$\gamma^{\prime}>\overline{C}\ln(1/\delta)+\gamma^{\prime\prime}+\widetilde{c}/\ln(2)$.

In order to establish (4.22), it is enough to show that there exist
$c^{\prime\prime},\theta>0$, independent of $\widetilde{c}$, such that for all
$i$ large enough,

$\sup_{n}\mathbb{E}\Big{[}e^{-\theta\ln(Z_{i}^{(n)})\mathds{1}_{B_{i}^{(n)}}}\Big{]}\leq
c^{\prime\prime}.$ (4.23)

Indeed, if (4.23) holds, setting
$\widetilde{Z}_{i}^{(n)}:=\ln(Z_{i}^{(n)})\mathds{1}_{B_{i}^{(n)}}$, by
Markov’s inequality we have

$\displaystyle\mathbb{P}$
$\displaystyle\Big{(}\sum_{i=i(n)}^{j(n)}\widetilde{Z}_{i}^{(n)}<-\widetilde{c}\cdot
j(n)\Big{)}\leq\mathbb{P}\Big{(}\sum_{k=0}^{3}\sum_{i=\lceil
i(n)/4\rceil}^{\lfloor\frac{j(n)}{4}\rfloor-1}\widetilde{Z}_{4i+k}^{(n)}<-\widetilde{c}\cdot
j(n)\Big{)}$
$\displaystyle\quad\leq\sum_{k=0}^{3}\mathbb{P}\Big{(}\sum_{i=\lceil
i(n)/4\rceil}^{\lfloor\frac{j(n)}{4}\rfloor-1}\widetilde{Z}_{4i+k}^{(n)}<-\widetilde{c}\cdot
j(n)/4\Big{)}\leq 4e^{-\theta\widetilde{c}\cdot
j(n)/4}\max_{k=1,\ldots,4}\mathbb{E}\Big{[}e^{-\theta\sum_{i=\lceil
i(n)/4\rceil}^{\lfloor\frac{j(n)}{4}\rfloor-1}\widetilde{Z}_{4i+k}^{(n)}}\Big{]}.$

We will only estimate the above expectation for the case $k=0;$ the cases
$k\in\\{1,2,3\\}$ can be estimated similarly. Now $\widetilde{Z}_{4i}^{(n)}$
is $\mathcal{F}^{t_{4i-1}}$-measurable, hence, also recalling
$t_{4i-1}-t_{4i-2}=2^{4i-2}$, by (• ‣ 1.4) we have

$\mathbb{E}\big{[}e^{-\theta\widetilde{Z}_{4i}^{(n)}}|\mathcal{F}_{t_{4i-2}}\big{]}\leq(1+\psi(2^{4i-2}))\mathbb{E}[e^{-\theta\widetilde{Z}_{4i}^{(n)}}].$

Since furthermore $\widetilde{Z}_{4(i-1)}^{(n)}$ is
$\mathcal{F}^{t_{4i-2}}$-measurable, we obtain via iterated conditioning that

$\displaystyle\mathbb{E}\Big{[}e^{-\theta\sum_{i=\lceil
i(n)/4\rceil}^{\lfloor\frac{j(n)}{4}\rfloor}\widetilde{Z}_{4i}^{(n)}}\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\mathbb{E}\big{[}\cdots\mathbb{E}\big{[}\mathbb{E}\big{[}e^{-\theta\sum_{i=\lceil
i(n)/4\rceil}^{\lfloor\frac{j(n)}{4}\rfloor}\widetilde{Z}_{4i}^{(n)}}\,|\,\mathcal{F}_{t_{4j(n)-2}}\big{]}\,|\,\mathcal{F}_{t_{4j(n)-6}}\big{]}\cdots\,|\,\mathcal{F}_{t_{2}}\big{]}\Big{]}$
$\displaystyle\leq\prod_{i=\lceil
i(n)/4\rceil}^{\lfloor\frac{j(n)}{4}\rfloor}(1+\psi(2^{4i-2}))\mathbb{E}[e^{-\theta\widetilde{Z}_{4i}^{(n)}}]\leq(c_{6}c^{\prime\prime})^{j(n)},$

for some $c_{6}>0$ and $n$ large enough. Choosing $\widetilde{c}$ large
enough, by a Borel-Cantelli argument similar to the proof of Lemma 2.5,
inequality (4.22) would follow.

Thus, in order to show (4.23), note that because

$Z_{i}^{(n)}=Z_{i}^{(n)}(\xi(\cdot))=Z_{i}^{(n-k)}(\xi(\cdot+k))\overset{d}{=}Z_{i}^{(n-k)}(\xi(\cdot))=Z_{i}^{(n-k)},$
(4.24)

we can drop the supremum in (4.23). In the following, we first choose $i$
large enough (and from then on fixed) such that several estimates in the
remaining part of the proof hold, and afterwards we adapt $n=n(i)$ to ensure
$0\leq i\leq\lceil\log_{2}(n)\rceil$. For simplicity, we write
$Z_{i}:=Z_{i}^{(n)}$, $\beta_{k}:=\overline{\beta}_{k}^{(2,n)}$,
$\widehat{H}_{k}:=H_{k}-E_{n}^{\zeta,\eta}[H_{k}]$,
$\widehat{R}_{k}:=\widehat{R}_{k}^{(n)}$ and define

$\displaystyle\overline{\rho}_{k}^{(j)}$
$\displaystyle:=\underset{\overline{\xi}(j)}{\text{ess sup}}\
\rho_{k},\quad\overline{R}_{k}^{(j)}:=\sum_{l=0}^{k}\overline{\rho}_{l}^{(j)},\quad
0\leq k\leq j.$

Furthermore, Thus, $\overline{\rho}_{k}^{(j)}$ is $\mathcal{F}_{j}$-measurable
and $\overline{R}_{k+l}^{(j)}-\overline{R}_{k}^{(j)}$ is
$(\mathcal{F}^{k}\cap\mathcal{F}_{j})$-measurable for all $l\geq 0$. Let
$M_{R}:=\operatorname*{ess\,sup}\rho_{0}$ and $L:=at_{i}^{1/2},$ and note that
the latter choice corresponds to diffusive scaling. Then we define

$\begin{split}r_{0}&:=t_{i-1},\quad m:=\frac{L}{16M_{R}},\quad
s_{0}:=\big{(}\inf\big{\\{}k\geq r_{0}:\
\overline{R}_{k}^{(k+m)}-\overline{R}_{r_{0}}^{(k+m)}\geq
L/8\big{\\}}-1\big{)}\wedge t_{i},\end{split}$ (4.25)

and for $j\geq 1$ let

$\displaystyle\begin{split}r_{j}&:=s_{j-1}+\Big{\lceil}\frac{L}{8M_{R}}\Big{\rceil},\\\
s_{j}&:=\big{(}\inf\big{\\{}k\geq
r_{j}:\overline{R}_{k}^{(k+m)}-\overline{R}_{r_{j}}^{(k+m)}\geq
L/8\big{\\}}-1\big{)}\wedge\left(r_{j}+(t_{i}-t_{i-1})\right).\end{split}$
(4.26)

Heuristically, $s_{j}$ is the first time after which the process
$\overline{R}$ (and thus $\widehat{R}$) increases at least by the amount $L/8$
after time $r_{j}$. Such large increments of $\widehat{R}$ are potentially
troublesome, since as a consequence, the process $\beta$ might decrease too
much and cause the event in the definition of $Z_{i}$ to have too small
probability. In order to cater for this inconvenience, we start noting that by
definition, $s_{j}-r_{j}$ is bounded by $t_{i}-t_{i-1}$ and
$\mathcal{F}_{s_{j}+m}$-measurable, and $r_{j+1}-(s_{j}+m)\geq m.$ Thus, by
condition (• ‣ 1.4), for every non-negative measurable function $f$ we notice
for later reference that

$\mathbb{E}[f(s_{j}-r_{j})\,|\,\mathcal{F}^{r_{j+1}}]\leq\big{(}1+\psi(m)\big{)}\mathbb{E}[f(s_{j}-r_{j})].$
(4.27)

Next, we define

$\displaystyle\mathcal{G}_{j}$ $\displaystyle:=\Big{\\{}\inf_{r_{j}\leq k\leq
s_{j}}(\widehat{H}_{k}-\widehat{H}_{r_{j}})\geq-L/8,\ \beta_{s_{j}}\geq
2L\Big{\\}},\quad\mathcal{G}_{j}^{\prime}:=\Big{\\{}\inf_{s_{j}\leq k\leq
r_{j+1}}(\widehat{H}_{k}-\widehat{H}_{s_{j}})\geq-L/8\Big{\\}},$
$\displaystyle J$
$\displaystyle:=\inf\big{\\{}j:s_{j}-r_{j}=t_{i}-t_{i-1}\big{\\}}\wedge\inf\\{j:s_{j}\geq
t_{i}\\},\quad\text{as well
as}\quad\mathcal{G}:=\bigcap_{j=0}^{J}\mathcal{G}_{j}\cap\bigcap_{j=0}^{J-1}\mathcal{G}_{j}^{\prime},$

and claim that

$Z_{i}\geq P_{n}^{\zeta,\eta}\big{(}\mathcal{G}\ |\
\beta_{t_{i-1}}=at_{i-1}^{1/2}\big{)}.$ (4.28)

Indeed, on $[r_{0},s_{0}]$, the process $\overline{R}$ (and thus also
$\widehat{R}$) increases by at most $L/8$, and the process $\widehat{H}$
decreases by at most $L/8$ on $\mathcal{G}_{0}$. Moreover, for $j\geq 1$, on
$[s_{j-1},r_{j}]$, the process $\widehat{R}$ increases by at most $L/8$, and
$\widehat{H}$ decreases by at most $L/8$ on $\mathcal{G}_{j}^{\prime}$.
Finally, on $[r_{j},s_{j}]$, the process $\overline{R}$ (and thus
$\widehat{R}$) increases by at most $L/8$, and on $\mathcal{G}_{j}$,
$\widehat{H}$ decreases by at most $L/8$ and $\beta_{s_{j}}\geq 2L$. All in
all, conditioning on $\beta_{t_{i-1}}=at_{i-1}^{1/2}=L/\sqrt{2}\geq L/2$, we
have $\beta_{k}\geq L/4\geq t_{i}^{1/4}$ for $k\in[r_{0},s_{0}]$ and
$\beta_{k}\geq L$ for all $k\in[s_{0},s_{J}]$. Since by definition, $s_{J}\geq
t_{i}$, we get $\beta_{t_{i}}\geq L=at_{i}^{1/2}$, implying (4.28).

Furthermore, we can continue to lower bound

$P_{n}^{\zeta,\eta}\big{(}\mathcal{G}\ |\
\beta_{t_{i-1}}=at_{i-1}^{1/2}\big{)}\geq
P_{n}^{\zeta,\eta}\left(\mathcal{G}_{0}\ |\
\beta_{r_{0}}=L/\sqrt{2}\right)\prod_{j=0}^{J-1}P_{n}^{\zeta,\eta}\left(\mathcal{G}_{j}^{\prime}\right)\prod_{j=1}^{J}P_{n}^{\zeta,\eta}\left(\mathcal{G}_{j}\
|\ \beta_{r_{j}}=2L\right).$ (4.29)

To see this, successively condition on $\beta_{r_{j}}\geq 2L,$ $j=1,\ldots,J,$
and use the Markov property of the process $\widehat{H}$ as well as the fact
that $x\mapsto
P_{n}^{\zeta,\eta}\left(\mathcal{G}_{j}\,|\,\beta_{r_{j}}=x\right)$ is
increasing. Then use the fact that under $P_{n}^{\zeta,\eta}$, the event
$\mathcal{G}_{j}^{\prime}$ is independent of $\beta_{r_{j}}$ by the
independence of the increments of $\widehat{H}$, $j=0,\ldots,J-1$.

In order to lower bound (4.29), observe that under $P_{n}^{\zeta,\eta}$, the
sequence $(\widehat{H}_{k+1}-\widehat{H}_{k})_{k\geq r_{j}}$ consists of
independent and centered random variables, whose $P_{n}^{\zeta,\eta}$-moment
generating function is finite in a neighborhood of zero. Thus, the central
limit theorem entails that for $i$ large enough we have
$P_{n}^{\zeta,\eta}(\mathcal{G}_{j}^{\prime})\geq 1/2$ for all relevant
choices of $j.$ Moreover,

$P_{n}^{\zeta,\eta}\left(\mathcal{G}_{j}\ |\ \beta_{r_{j}}=2L\right)\geq
P_{n}^{\zeta,\eta}\Big{(}\widehat{H}_{s_{j}}-\widehat{H}_{r_{j}}\geq
5L/2,\inf_{r_{j}\leq k\leq
s_{j}}(\widehat{H}_{k}-\widehat{H}_{r_{j}})\geq-L/8\Big{)}.$

We see that both events are nondecreasing in the (independent) increments of
$\widehat{H}$. By Harris’ inequality ([5, Theorem 2.15]) we get

$\displaystyle P_{n}^{\zeta,\eta}$
$\displaystyle\Big{(}\widehat{H}_{s_{j}}-\widehat{H}_{r_{j}}\geq
5L/2,\inf_{r_{j}\leq k\leq
s_{j}}(\widehat{H}_{k}-\widehat{H}_{r_{j}})\geq-L/8\Big{)}$ $\displaystyle\geq
P_{n}^{\zeta,\eta}\big{(}\widehat{H}_{s_{j}}-\widehat{H}_{r_{j}}\geq
5L/2\big{)}\cdot P_{n}^{\zeta,\eta}\Big{(}\inf_{r_{j}\leq k\leq
s_{j}}(\widehat{H}_{k}-\widehat{H}_{r_{j}})\geq-L/8\Big{)}.$

Recalling $s_{j}-r_{j}\leq t_{i}-t_{i-1}=\frac{L^{2}}{2a^{2}}$, a Gaussian
scaling yields $P_{n}^{\zeta,\eta}\Big{(}\inf_{r_{j}\leq k\leq
s_{j}}(\widehat{H}_{k}-\widehat{H}_{r_{j}})\geq-L/8\Big{)}\geq c_{7}>0$. To
bound the first factor, we recall that by (A.10) and (A.12), we have
$\mathbb{P}$-a.s.

$0\leq\overline{\rho}_{r_{j}+l}^{(r_{j}+k+m)}-\rho_{r_{j}+l}\leq
c_{8}(\psi(m/2)+e^{-m/c_{8}})\quad\text{for all }l\leq k\leq t_{i}/2.$

Because $m=\frac{L}{16M_{R}}$ and $t_{i}/2=\frac{L^{2}}{2a^{2}}$, due to (• ‣
1.4) we finally get for all $i$ (and thus $L$) large enough (due to
$\psi(x)\cdot x\to 0$ ($x\to\infty$), which itself is due to summability of
$\psi(k)$), that

$\begin{split}0&\leq\big{(}\overline{R}_{r_{j}+k}^{(r_{j}+k+m)}-\overline{R}_{r_{j}}^{(r_{j}+k+m)}\big{)}-\big{(}\widehat{R}_{r_{j}+k}-\widehat{R}_{r_{j}}\big{)}=\sum_{l=1}^{k}\big{(}\overline{\rho}_{r_{j}+l}^{(r_{j}+k+m)}-\rho_{r_{j}+l}\big{)}\\\
&\leq c_{8}L^{2}(\psi(L/16M_{R})+e^{-L/c_{8}})\leq L/16\quad\text{for all
}k\in\Big{\\{}0,\ldots,\frac{L^{2}}{2a^{2}}\Big{\\}}.\end{split}$ (4.30)

By $s_{j}-r_{j}\leq t_{i}-t_{i-1}=\frac{L^{2}}{2a^{2}}$ and (4.26), we see
that $s_{j}-r_{j}\geq L/16$ for all $i$ large enough. Recall that under
$P_{n}^{\zeta,\eta}$, the sequence $\widehat{H}_{s_{j}}-\widehat{H}_{r_{j}}$
is a sum of independent centered random variables, whose moment generating
function is uniformly bounded in a neighborhood of the origin. Then by [38,
(1)], we can apply [38, Theorem 4] in the following manner: Let $c^{\prime}>0$
be as in [38, Theorem 4] and in the notation of the latter theorem we choose
$k=s_{j}-r_{j}$,
$\alpha=\operatorname*{ess\,inf}_{\zeta,k<n}E_{n}^{\zeta,\eta}[(\widehat{H}_{k}-\widehat{H}_{k-1})^{2}]>0$,
$M=|\eta|/2$, $u_{1}=\ldots=u_{k}=1$ and

$\overline{c}:=c^{\prime}\cdot\frac{|\eta|}{2}\cdot\alpha$ (4.31)

Then a lower Bernstein-type inequality from [38, Theorem 4] gives that on
$B_{i}^{(n)}$ we have

$P_{n}^{\zeta,\eta}\big{(}\widehat{H}_{s_{j}}-\widehat{H}_{r_{j}}\geq
5L/2\big{)}\geq c_{9}e^{-\frac{L^{2}}{c_{9}(s_{j}-r_{j})}}.$ (4.32)

Note that the condition in $B_{i}^{(n)}$ makes [38, Theorem 4] applicable by
ensuring ’enough’ summands $s_{j}-r_{j}$ and is the main reason we have to
introduce the sets $B_{i}^{(n)}$. We will write $c=c_{7}\wedge c_{9}$ from now
on. Using (4.28) in combination with the lower bounds for the factors of
(4.29) just derived, the term in (4.23) can be bounded from above by

$\displaystyle\mathbb{E}$
$\displaystyle\big{[}e^{-\theta\ln(Z_{i})}\big{]}\leq
c_{10}\mathbb{E}\Big{[}\exp\Big{\\{}\theta
J\ln(2/c)+\theta\sum_{j=0}^{J}\frac{L^{2}}{c(s_{j}-r_{j})}\Big{\\}}\Big{]}$
$\displaystyle\quad\leq
c_{10}\sum_{k=0}^{\infty}\mathbb{E}\Big{[}\Big{(}\frac{2}{c}\Big{)}^{\theta
k}\exp\Big{\\{}\theta\frac{L^{2}}{c(s_{k}-r_{k})}\mathds{1}_{s_{k}-r_{k}=t_{i}-t_{i-1}}+\theta\sum_{j=0}^{k-1}\frac{L^{2}}{c(s_{j}-r_{j})}\mathds{1}_{s_{j}-r_{j}<t_{i}-t_{i-1}}\Big{\\}}\Big{]}$
$\displaystyle\quad\leq
c_{10}\sum_{k=0}^{\infty}\Big{(}\frac{2}{c}\Big{)}^{\theta
k}\big{(}1+\psi(m)\big{)}^{k}e^{\frac{2\theta
L^{2}}{ct_{i-1}}}\cdot\prod_{j=0}^{k-1}\mathbb{E}\Big{[}\exp\Big{\\{}\frac{\theta
L^{2}}{c(s_{j}-r_{j})}\Big{\\}}\mathds{1}_{s_{j}-r_{j}<t_{i}/2}\Big{]},$
(4.33)

where we recall $m$ from (4.26), and the last inequality is due to (4.27) in
combination with $t_{i}-t_{i-1}=t_{i}/2.$ For the latter expectation we have

$\displaystyle\mathbb{E}\left[\exp\Big{\\{}\frac{\theta
L^{2}}{c(s_{j}-r_{j})}\Big{\\}}\mathds{1}_{s_{j}-r_{j}<t_{i}/2}\right]=\int_{0}^{\infty}\mathbb{P}\Big{(}e^{\frac{\theta
L^{2}}{c(s_{j}-r_{j})}}\mathds{1}_{s_{j}-r_{j}<t_{i}/2}\geq
x\Big{)}\mathop{}\\!\mathrm{d}x$ $\displaystyle\leq e^{\frac{2\theta
L^{2}}{ct_{i}}}\mathbb{P}\Big{(}s_{j}-r_{j}<t_{i}/2\Big{)}+\int_{e^{\frac{2\theta
L^{2}}{ct_{i}}}}^{\infty}\mathbb{P}\Big{(}e^{\frac{\theta
L^{2}}{c(s_{j}-r_{j})}}\geq x\Big{)}\mathop{}\\!\mathrm{d}x.$ (4.34)

Substituting $x=e^{\frac{\theta L^{2}}{cy}}$, the second summand can be
written as

$\displaystyle\int_{0}^{t_{i}/2}\frac{\theta L^{2}}{cy^{2}}e^{\frac{\theta
L^{2}}{cy}}\mathbb{P}\left(s_{j}-r_{j}\leq y\right)\mathop{}\\!\mathrm{d}y.$

In order to obtain an upper bound, we start with the probability inside the
integral and get

$\displaystyle\mathbb{P}\left(s_{j}-r_{j}\leq y\right)$
$\displaystyle=\mathbb{P}\Big{(}\max_{1\leq k\leq
y}\sum_{l=1}^{k}\overline{\rho}_{r_{j}+l}^{(r_{j}+k+m)}\geq
L/8\Big{)}\leq\mathbb{P}\Big{(}\max_{1\leq k\leq
y}\sum_{l=1}^{k}{\rho}_{r_{j}+l}\geq L/16\Big{)}$ $\displaystyle\leq
c_{11}e^{-\frac{L^{2}}{c_{11}y}},\quad\forall\ y\in[0,t_{i}/2];$

here the first inequality is due to (4.30) and the last inequality due to
(4.18).

Putting these bounds together, the second summand in (4.34) can be bounded
from above by

$\displaystyle\int_{0}^{t_{i}/2}\frac{\theta L^{2}}{cy^{2}}e^{\frac{\theta
L^{2}}{cy}}c_{10}e^{-\frac{L^{2}}{yc_{10}}}\mathop{}\\!\mathrm{d}y$
$\displaystyle=c_{11}\int_{0}^{t_{i}/2}\frac{\theta
L^{2}}{cy^{2}}e^{\frac{L^{2}}{cy}(\theta-c_{12})}\mathop{}\\!\mathrm{d}y\leq
c_{13}\int_{0}^{1/2a^{2}}\frac{\theta}{z^{2}}e^{\frac{1}{cz}(\theta-
c_{12})}\mathop{}\\!\mathrm{d}z$ $\displaystyle\leq
c_{14}\theta\int_{0}^{\infty}e^{\frac{x(\theta-
c_{11})}{c}}\mathop{}\\!\mathrm{d}x.$

Now the latter term can be made arbitrarily close to zero by choosing
$\theta>0$ small enough. Furthermore, again choosing $\theta>0$ small enough,
the first term in (4.34) is strictly smaller than one by the central limit
theorem from Claim 4.3. Thus, $\theta>0$ can be chosen small enough such that
for all $i$ large enough, the sum on the right-hand side in (4.33) converges,
with a finite upper bound independent of $i.$

To show $Y^{(n)}\geq n^{-\gamma^{\prime\prime}}$, we adapt the strategy in the
proof of (4.22), i.e. set $L:=2a\overline{C}\ln n$,
$r_{0}=\lfloor\overline{C}\ln n\rfloor$ and
$J:=\inf\\{j:s_{j}\geq\lfloor(\overline{C}\ln n)^{2}\rfloor\\}$ and keep the
other definitions as in (4.25) and (4.26). Then by the same argument below
display (4.23), $Y^{(n)}\geq n^{-\gamma^{\prime\prime}}$ for some suitable
$\gamma^{\prime\prime}>0$ follows if $\mathbb{E}[e^{-\theta Y^{(n)}}]\leq c$
for some constant $c>0$, some small $\theta>0$ and all $n$ large enough. But
this follows (as in the argument leading to the definition of $B_{i}^{(n)}$),
if the process $(\widehat{R}_{k}-\widehat{R}_{k-1})_{k}$ does not decrease too
fast, see the Borel-Cantelli argument below display (4.32), which itself is a
consequence of

$\displaystyle\sum_{n}(\overline{C}\ln n)^{2}\cdot\mathbb{P}\big{(}\sup_{0\leq
k\leq
L/8\overline{c}}\big{(}\widehat{R}_{k}-\widehat{R}_{0}\big{)}\geq\frac{L}{8}\big{)}$
$\displaystyle\leq c_{15}\sum_{n}(\overline{C}\ln n)^{2}e^{-a\overline{C}\ln
n/c_{15}}<\infty.$ (4.35)

This completes the proof of the second inequality in Lemma 4.2 a).

We will now explain how to adapt the latter arguments for the proof of the
first inequality in a). We define
$\beta_{k}=\overline{\beta}^{(1,n)}_{k}-\overline{\beta}^{(1,n)}_{n-1}$. In
the definition of $Z_{i}^{(n)}$, we have to take the essential infimum over
$\overline{\xi}(n-t_{i-2})$ and have to replace the subscripts $k$ of
$\beta_{k}$ by $n-k$, i.e. “running backwards” from $n$. Thus, $Z_{i}^{(n)}$
is $(\mathcal{F}^{n-t_{i}}\cap\mathcal{F}_{n-t_{i-2}})$-measurable. It is then
enough to consider the case $n=0$ due to the argument in (4.24). Writing
$Z_{i}:=Z_{i}^{(0)}$, $\beta_{k}:=\beta_{k}^{(0)}$,
$\widehat{H}_{k}:=H_{k}-E_{0}^{\zeta,\eta}[H_{k}]$,
$\widehat{R}_{k}:=\widehat{R}_{k}^{(0)}$ and defining

$\displaystyle\overline{\rho}_{k}^{(j)}$
$\displaystyle:=\underset{\overline{\xi}(j)}{\text{ess sup}}\
\rho_{k},\qquad\overline{R}_{k}^{(j)}:=\sum_{l=k+1}^{0}\overline{\rho}_{l}^{(j)},\quad
k<0,\ k\in\mathbb{Z},$

we have to adapt the definitions of $r_{j}$ and $s_{j}$ by the expressions

$\displaystyle r_{0}$ $\displaystyle:=-t_{i-1},\qquad
s_{0}:=\big{(}\sup\big{\\{}k\leq r_{0}:\
\widehat{R}_{k}-\widehat{R}_{r_{0}}\geq L/8\big{\\}}+1\big{)}\vee(-t_{i}),$
$\displaystyle r_{j}$
$\displaystyle:=s_{j-1}-\Big{\lceil}\frac{L}{8M_{R}}\Big{\rceil},\quad
s_{j}^{\prime}:=s_{j-1}-\frac{L}{16M_{R}},\quad j\geq 1,$ $\displaystyle
s_{j}$ $\displaystyle:=\big{(}\sup\big{\\{}k\leq
r_{j}:\overline{R}_{k}^{(s^{\prime}_{j})}-\overline{R}_{r_{j}}^{(s_{j}^{\prime})}\geq
L/8\big{\\}}+1\big{)}\vee\left(r_{j}-(t_{i}-t_{i-1})\right),\quad j\geq 1,$
$\displaystyle J$
$\displaystyle:=\inf\\{j:s_{j}-r_{j}=-t_{i}\\}\vee\sup\\{j:s_{j}\leq-
t_{i}\\}.$

The remaining part of the proof essentially follows the same steps as for the
second inequality in a). ∎

###### Proof of Claim 4.3.

Boundedness, stationarity and adaptedness are direct consequences of the
corresponding properties of the sequences
$(L_{i}^{\zeta}(\eta))_{i\in\mathbb{Z}}$ and
$((L_{i}^{\zeta})^{\prime}(\eta))_{i\in\mathbb{Z}}$ and Lemma A.1. Display
(4.17) is due to Lemma A.2. To show that the central limit theorem, we note
that the sequence $(\rho_{i})_{i\in\mathbb{Z}}$ fulfills the same conditions
as the sequence $(\widetilde{L}_{i})_{i\in\mathbb{Z}}$ in the proof of Lemma
3.2 ∎

###  4.2 Second moment of leading particles

Recall the notation $N_{t}^{\mathcal{L}}$ from below (4.2) and that of
$\overline{m}(t)$ from (1.7). For the second moment of the leading particles
we now prove the following upper bound.

######  Lemma 4.4.

For every function $F$ fulfilling (PROB) and for every $a>0$, there exists
$\gamma_{2}=\gamma_{2}(F,a)<\infty$ such that $\mathbb{P}$-a.s., for all $t$
large enough,

$\sup_{x\in[\overline{m}^{(a)}(t)-1,\overline{m}^{(a)}(t)+1]}\emph{{E}}_{x}^{\xi}\big{[}\big{(}N_{t}^{\mathcal{L},a}\big{)}^{2}\big{]}\leq
t^{\gamma_{2}}.$ (4.36)

###### Proof.

We omit the superscript $a$ in the quantities involved and use the same
abbreviations as in the beginning of the proof of Lemma 4.1.

We want to show (4.36) with the help of the second-moment formula (FK-2). To
this end, define the function
$\varphi^{\xi}_{t}:[0,t]\to\mathbb{Z}\cup\\{-\infty\\}$,

$\displaystyle\varphi_{t}(s)$
$\displaystyle:=\varphi_{t}^{\xi}(s):=\lfloor\overline{m}(t)\rfloor\wedge\sup\big{\\{}k\in\mathbb{Z}:s\in\big{[}t-T_{k+1}-5\chi_{1}(\overline{m}(t)),t-T_{k}-5\chi_{1}(\overline{m}(t))\big{)}\big{\\}},$

where $\sup\emptyset:=-\infty$ and $\chi_{1}$ has been defined in (4.2). Due
to $T_{k}=0$ for all $k\leq 0$ (recall the notation $T_{k}$ from (3.42) and
(3.43)), we have $1\leq\varphi_{t}(0)\leq\lfloor\overline{m}(t)\rfloor.$
Furthermore, $\varphi_{t}(t)=-\infty$, because $T_{k}\geq 0$ and
$\chi_{1}^{\xi}(\overline{m}(t))\geq 0$. To apply (FK-2), the following upper
bound will prove useful.

######  Claim 4.5.

We have

$N_{t}^{\mathcal{L}}\leq\left|\left\\{\nu\in N(t):X_{t}^{\nu}\leq
0,X_{s}^{\nu}>\varphi_{t}(s)\ \forall s\in\left[0,t\right]\right\\}\right|$
(4.37)

and $\mathbb{P}$-a.s. for all $t$ large enough, the function $[0,t]\ni
s\mapsto\varphi_{t}(s)$ is a non-increasing, càdlàg step function.

In order not to hinder the flow of reading, we postpone the proof of the
latter claim to the end of the proof of Lemma 4.4.

Figure 3: Illustration of $\varphi_{t}$, which is the red line. We denote
$a_{t}(k):=t-T_{k}-5\chi^{\xi}(\overline{m}(t))$. Note that the sequence
$(a_{t}(k))_{k\in\mathbb{Z}}$ does not have to be monotone and thus the
interval $[a_{t}(k+1),a_{t}(k))$ might be empty. In this case the graph of
$\varphi_{t}$ jumps at least two steps at time $s=a_{t}(k)$.

By the Feynman-Kac formula (cf. Proposition 2.3) and (4.37), we have

$\displaystyle\begin{split}\texttt{E}_{x}^{\xi}\big{[}(N_{t}^{\mathcal{L}})^{2}\big{]}&\leq\texttt{E}_{x}^{\xi}\left[N_{t}^{\mathcal{L}}\right]+(m_{2}-2)\int_{0}^{t}E_{x}\left[e^{\int_{0}^{s}\xi(B_{r})\mathop{}\\!\mathrm{d}r}\xi(B_{s})\mathds{1}_{\left\\{B_{r}\geq\varphi_{t}(r)\
\forall r\in[0,s]\right\\}}\right.\\\
&\qquad\qquad\qquad\left.\times\left(E_{y}\left[e^{\int_{0}^{t-s}\xi(B_{r})\mathop{}\\!\mathrm{d}r}\mathds{1}_{\left\\{B_{r}\geq\varphi_{t}(r+s)\
\forall r\in[0,t-s]\right\\},B_{t-s}\leq
0}\right]\right)_{|_{y=B_{s}}}^{2}\right]\mathop{}\\!\mathrm{d}s.\end{split}$
(4.38)

For the first summand we have

$\displaystyle\sup_{x\in[\overline{m}(t)-1,\overline{m}(t)+1]}\texttt{E}_{x}^{\xi}\left[N_{t}^{\mathcal{L}}\right]$
$\displaystyle\leq\sup_{x\in[\overline{m}(t)-1,\overline{m}(t)+1]}\texttt{E}_{x}^{\xi}\left[N^{\leq}(t,0)\right]\leq
c_{1}\texttt{E}_{\overline{m}(t)+1}^{\xi}\left[N^{\leq}(t,0)\right]\leq\frac{c_{1}}{2},$
(4.39)

where the first inequality is due to the first inequality in (3.40) and the
last one due to the definition of $\overline{m}(t)$. Recall that the Markov
property provides us with

$E_{x}\left[e^{\int_{0}^{s}\xi(B_{r})\mathop{}\\!\mathrm{d}r}E_{y}\big{[}e^{\int_{0}^{t-s}\xi(B_{r})\mathop{}\\!\mathrm{d}r}\mathds{1}_{\left\\{B_{t-s}\leq
0\right\\}}\big{]}_{|_{y=B_{s}}}\right]=\texttt{E}_{x}^{\xi}[N^{\leq}(t,0)].$

Using $\xi\leq\texttt{es}$ and the two previous displays, the second summand
in (4.38) can thus be bounded from above by

$\displaystyle\begin{split}&\texttt{es}\cdot(m_{2}-2)\sup_{x\in[\overline{m}(t)-1,\overline{m}(t)+1]}\texttt{E}_{x}^{\xi}[N^{\leq}(t,0)]\cdot\int_{0}^{t}\sup_{y\geq\varphi_{t}(s)}E_{y}\big{[}e^{\int_{0}^{t-s}\xi(B_{r})\mathop{}\\!\mathrm{d}r};B_{t-s}\leq
0\big{]}\mathop{}\\!\mathrm{d}s\\\
&\quad\leq\frac{\texttt{es}(m_{2}-2)c_{1}}{2}\int_{0}^{t}\sup_{y\geq\varphi_{t}(s)}\texttt{E}_{y}^{\xi}[N^{\leq}(t-s,0)]\mathop{}\\!\mathrm{d}s.\end{split}$
(4.40)

It thus suffices to upper bound
$\sup_{y\geq\varphi_{t}(s)}\texttt{E}_{y}^{\xi}[N^{\leq}(t-s,0)]$ by a
polynomial in $t.$ We treat different areas for $s$ and $y$ separately and we
will need an additional claim, the proof of which will be provided after this
proof. It guarantees that the assumptions of the time perturbation Lemma 3.11
are satisfied in our setting.

Figure 4: Leading particles and the different areas in the proof of Lemma 4.4.

######  Claim 4.6.

There exists $\widetilde{C}_{1}\in(0,\infty)$ such that $\mathbb{P}$-a.s. for
all $t$ large enough and all $y\geq\widetilde{C}_{1}\chi_{1}(\overline{m}(t))$
we have
$\frac{y}{T_{y}-1},\frac{y}{T_{y}+\overline{K}+5\chi_{1}\left(\overline{m}(t)\right)}\in
V$, where $V$ is defined in (2.22). Furthermore, there exists
$\widetilde{C}_{2}=\widetilde{C}_{2}(\widetilde{C}_{1})\in(0,\infty)$ such
that $\mathbb{P}$-a.s. for all $t$ large enough and all
$s\in[0,t-\widetilde{C}_{2}\chi_{1}(\overline{m}(t)))$ we have
$\varphi_{t}(s)\geq\widetilde{C}_{1}\chi_{1}(\overline{m}(t))$.

We choose
$\gamma^{\prime}>5\Cr{const_{t}ime_{p}erturbation}\Cr{constapproxofTn}$. Then,
recalling the definition of $\chi_{1}$ from (4.2) and that $\mathbb{P}$-a.s.
$\frac{\overline{m}(t)}{t}\to v_{0}$ by Corollary 3.15, for $t$ large enough,
the statements from Claim 4.6 hold true, we have
$\Cr{const_{t}ime_{p}erturbation}e^{\Cr{const_{t}ime_{p}erturbation}(\overline{K}+1+5\chi_{1}(\overline{m}(t)))}\leq
t^{\gamma^{\prime}}$ and also $T_{\lfloor y\rfloor+1}\leq T_{y}+\overline{K}$
for all $y\geq\widetilde{C}_{1}\chi_{1}(\overline{m}(t))$ by Corollary 3.17.
$1)$ Let
$s\in\big{[}0,t-\widetilde{C}_{2}\chi_{1}\left(\overline{m}(t)\right)\big{)}$
and $y\geq\varphi_{t}(s)$. Then by Claim 4.6,
$y\geq\widetilde{C}_{1}\chi_{1}\left(\overline{m}(t)\right)$ and thus
$\frac{y}{T_{y}-1},\frac{y}{T_{y}+\overline{K}+5\chi_{1}\left(\overline{m}(t)\right)}\in
V$. By definition of $\varphi_{t}$ we have $s\geq t-T_{\lfloor
y\rfloor+1}-5\chi_{1}\left(\overline{m}(t)\right)$ and thus
$T_{y}+\overline{K}+5\chi_{1}\left(\overline{m}(t)\right)\geq T_{\lfloor
y\rfloor+1}+5\chi_{1}\left(\overline{m}(t)\right)\geq t-s$. Thus, by Lemma
A.6, we infer that $\texttt{E}_{y}^{\xi}\left[N^{\leq}(t-s,0)\right]\leq
2\texttt{E}_{y}^{\xi}\left[N^{\leq}\left(T_{y}+\overline{K}+5\chi_{1}\left(\overline{m}(t)\right),0\right)\right],$
and then the second inequality in (3.33) entails that for all $t$ large
enough,

$\displaystyle\sup_{\begin{subarray}{c}0\leq
s<t-\widetilde{C}_{2}\chi_{1}(\overline{m}(t))\\\
y\geq\varphi_{t}(s)\end{subarray}}\texttt{E}_{y}^{\xi}[N^{\leq}(t-s,0)]$
$\displaystyle\leq
2\Cr{const_{t}ime_{p}erturbation}e^{\Cr{const_{t}ime_{p}erturbation}(\overline{K}+1+5\chi_{1}(\overline{m}(t)))}\sup_{y\in\mathbb{R}}\texttt{E}_{y}^{\xi}\left[N^{\leq}((T_{y}-1)\vee
0,0)\right]\leq t^{\gamma^{\prime}}.$ (4.41)

$2)$ The remaining part of the domain above the graph of $\varphi_{t}$ not
controlled by $1)$ is a subset of

$\big{\\{}(s,y)\in[0,t]\times\mathbb{R}:t-\widetilde{C}_{2}\chi_{1}\left(\overline{m}(t)\right)\leq
s\leq t\big{\\}}.$

Recalling the definition of $\chi_{1}$ from (4.2) and that $\mathbb{P}$-a.s.,
$\frac{\overline{m}(t)}{t}\to v_{0}$ by Corollary 3.15, choosing
$\gamma^{\prime\prime}>\texttt{es}\widetilde{C}_{2}\Cr{constapproxofTn}$, on
the the above domain we get that $\mathbb{P}$-a.s., for all $t$ large enough,

$\displaystyle\texttt{E}_{y}^{\xi}\left[N^{\leq}\left(t-s,0\right)\right]\leq
2\texttt{E}_{y}^{\xi}\big{[}N^{\leq}\big{(}\widetilde{C}_{2}\chi_{1}\left(\overline{m}(t)\right),0\big{)}\big{]}\leq
2e^{\texttt{es}\widetilde{C}_{2}\chi_{1}\left(\overline{m}(t)\right)}\leq
t^{\gamma^{\prime\prime}}.$ (4.42)

To conclude the proof, defining
$\gamma_{2}:=1\vee\gamma^{\prime}\vee\gamma^{\prime\prime}+1$, inequalities
(4.39), (4.40) and the estimates (4.41) and (4.42) for the term
$E_{y}^{\xi}\left[N^{\leq}(t-s,0)\right]$ entail the statement of Lemma 4.4. ∎

###### Proof of Claim 4.5.

Let $a_{t}(k):=t-T_{k}-5\chi_{1}^{\xi}(\overline{m}(t))$ and recall the
definition of $N_{t}^{\mathcal{L}}$:

$N_{t}^{\mathcal{L}}=\left|\big{\\{}\nu\in N(t):\ X_{t}^{\nu}\leq 0,\
H_{k}^{\nu}\geq a_{t}(k)\ \forall\
k\in\\{1,\ldots,\lfloor\overline{m}(t)\rfloor\\}\big{\\}}\right|.$

To prove (4.37), note that $H_{k}^{\nu}\geq a_{t}(k)$ if and only if
$X_{s}^{\nu}>k$ for all $s<a_{t}(k)$. But the property $X_{s}^{\nu}>k$ for all
$s\in[0,a_{t}(k))$ and all $k\in\\{1,\ldots,\lfloor\overline{m}(t)\rfloor\\}$
implies that $X_{s}^{\nu}>\sup\\{k\in\mathbb{Z}:\
s\in[a_{t}(k+1),a_{t}(k))\\}\wedge\lfloor\overline{m}(t)\rfloor=\varphi_{t}(s)$
for all $s\in[0,t]$ and thus (4.37) is shown. The property of $\varphi_{t}$
being a càdlàg step-function is a direct consequence of the use of left-
closed, right-open intervals in the definition of $\varphi_{t}$. It remains to
show that $s\mapsto\varphi_{t}(s)$ is non-increasing. For this purpose, let us
first prove by induction in
$k=\lfloor\overline{m}(t)\rfloor,\lfloor\overline{m}(t)\rfloor-1,\ldots$ that
for all $t$ large enough and all $k\leq\overline{m}(t),$

$[0,a_{t}(k-1))\subset\bigcup_{l=k}^{\lfloor\overline{m}(t)\rfloor}[a_{t}(l),a_{t}(l-1))$
(4.43)

holds. By Corollary 3.17 and Lemma 3.16, there exist
$\overline{K},\Cr{constantlemmaT_{m}teqt}>0$ such that
$t-T_{\lfloor\overline{m}(t)\rfloor-1}\leq
t-T_{\overline{m}(t)}+\overline{K}\leq\Cr{constantlemmaT_{m}teqt}+\overline{K}$
and thus $a_{t}(\lfloor\overline{m}(t)\rfloor)\leq 0$ for all $t$ large
enough. Assume now that (4.43) holds for some $k\leq\overline{m}(t)$. Then

$\left[0,a_{t}(k-2)\right)\subset\left[0,a_{t}(k-1)\right)\cup\left[a_{t}(k-1),a_{t}(k-2)\right)\subset\bigcup_{l=k-1}^{\lfloor\overline{m}(t)\rfloor}[a_{t}(l),a_{t}(l-1)),$

where the last inclusion is due to induction hypothesis. Thus, we have shown
(4.43). Now let $0\leq s_{1}\leq s_{2}$. Assume there exists $k_{2}$ such that
$s_{2}\in\left[a_{t}(k_{2}),a_{t}(k_{2}-1)\right)$. Then by (4.43), there
exists $k_{1}\in\mathbb{Z}$ with $k_{2}\leq
k_{1}\leq\lfloor\overline{m}(t)\rfloor$, such that
$s_{1}\in\left[a_{t}(k_{1}),a_{t}(k_{1}-1)\right)$. By definition we get
$\varphi_{t}(s_{1})\geq\varphi_{t}(s_{2})$. If no such $k_{2}$ exists, then
$\varphi_{t}(s_{2})=-\infty\leq\varphi_{t}(s_{1})$. ∎

###### Proof of Claim 4.6.

We write $V=[v_{*},v^{*}]$. Since $\mathbb{P}$-a.s. we have
$\frac{y}{T_{y}}\to v_{0}\in\text{int}(V)$ by (3.45), it follows that
$\frac{y}{T_{y}-1},\frac{y}{T_{y}}\in V$ for all $y$ large enough. Among
others, there exists $\varepsilon=\varepsilon(v_{*},v^{*},v_{0})>0$ and
$\mathcal{N}^{\prime}(\xi)$ such that
$v_{*}(1+\varepsilon)\leq\frac{y}{T_{y}}\leq(1-\varepsilon)v^{*}$ for all
$y\geq\mathcal{N}^{\prime}$. Choosing
$\widetilde{C}_{1}>\frac{5v^{*}}{\varepsilon}$, this implies
$1\leq\frac{T_{y}+\overline{K}+5\chi_{1}\left(\overline{m}(t)\right)}{T_{y}}\leq
1+\varepsilon$ for all
$y\geq\widetilde{C}_{1}\cdot\chi_{1}\left(\overline{m}(t)\right)$ and all $t$
large enough. Thus, we get

$v_{*}\leq\frac{y}{T_{y}}\cdot\frac{T_{y}}{T_{y}+\overline{K}+5\chi_{1}\left(\overline{m}(t)\right)}\leq
v^{*}\quad\text{for all }y\geq\widetilde{C}_{1}\chi_{1}(\overline{m}(t))\text{
and all $t$ large enough}.$

This gives the first part of the Claim 4.6. For the second part, recall that
$T_{y}\leq\frac{1}{\underline{v}}y$ for all $y\geq\mathcal{N}^{\prime}$.
Furthermore, by the definition of $\varphi_{t}$ we have
$\varphi_{t}(s)\geq\lfloor y\rfloor+1$ for all $s\in\left[0,t-T_{\lfloor
y\rfloor}-5\chi_{1}\left(\overline{m}(t)\right)\right)$. Choosing
$y:=\lfloor\widetilde{C}_{1}5\chi_{1}\left(\overline{m}(t)\right)\rfloor$ and
$\widetilde{C}_{2}>\frac{\widetilde{C}_{1}}{\underline{v}}+1$, this implies
that for $t$ large enough we get
$\varphi_{t}(s)\geq\widetilde{C}_{1}5\chi_{1}\left(\overline{m}(t)\right)$ for
all $s\in\big{[}0,t-\widetilde{C}_{2}5\chi_{1}(\overline{m}(t))\big{)}$. ∎

###  4.3 Proof of Theorem 1.4

Recall $\overline{m}(t)=\sup\\{x\in\mathbb{R}:u(t,x)\geq 1/2\\}$ and
$m(t)=\sup\\{x\in\mathbb{R}:w(t,x)\geq 1/2\\}$, where $u$ and $v$ are the
solutions to (PAM) and $w$ to (F-KPP), respectively. We start with an
amplification result.

######  Lemma 4.7.

For every $(p_{k})_{k\in\mathbb{N}}$ fulfilling (PROB) there exists
$\Cl{const_{a}mpl}=\Cr{const_{a}mpl}((p_{k}))>1$ and $t_{0}>0$ such that
$\mathbb{P}$-a.s., for all $t\geq t_{0},$

$\sup_{x\in\mathbb{R}}\emph{{P}}_{x}^{\xi}\left(N(t,[x-1,x+1])\leq\Cr{const_{a}mpl}^{t}\right)\leq\Cr{const_{a}mpl}^{-t}.$

###### Proof.

For the proof it is enough to show the claim for binary branching with rate
$\xi(x)\equiv{\texttt{ei}^{\prime}}:=\texttt{ei}(1-p_{1})$ (which is the rate
of first branching into more than one particle) by a straightforward coupling
argument. Due to the spatial homogeneity of $\texttt{ei}^{\prime}$ it is
enough to show

$\texttt{P}_{0}^{\texttt{ei}^{\prime}}\left(N(t,[-1,1])\leq\Cr{const_{a}mpl}^{t}\right)\leq\Cr{const_{a}mpl}^{-t}$

for all $t\geq t_{0}$, where $\texttt{P}_{0}^{\texttt{ei}^{\prime}}$ is the
probability measure under which the branching Brownian motion starts with one
particle in $0$ and has constant branching rate ${\texttt{ei}^{\prime}}>0$.
Then for every $\varepsilon>0$ there exists $\delta=\delta(\varepsilon)>0$
such that

$\texttt{P}_{0}^{\texttt{ei}^{\prime}}\big{(}|N(t/3,[-\varepsilon,\varepsilon])||\geq\delta
t\big{)}\geq 1-e^{-\delta t}$ (4.44)

for all $t$ large enough. Indeed, the probability that the initial particle
does not leave the interval $[-\varepsilon t/2,\varepsilon t/2]$ before time
$t/3$ is at least $1-e^{-c_{1}\varepsilon^{2}t}$. If this happens, the
particle produces more than $t{\texttt{ei}^{\prime}}/4$ offsprings with
probability $1-e^{-c_{2}t}$ before time $t/3$, while each of these offsprings
does not leave the interval $[-\varepsilon t,\varepsilon t]$ before time $t/3$
with probability at least $1-e^{-c_{1}\varepsilon^{2}t/2}$. Combining these
observations and choosing $\delta(\varepsilon)>0$ small enough provides us
with (4.44). For a particle $\nu\in N(t/3)$ let $D^{\nu}(t/3+s)$ be the set of
offsprings of $\nu$ in the interval $[X_{t/3}^{\nu}-1,X_{t/3}^{\nu}+1]$ at
time $t/3+s$, $s\geq 0$. We will show the existence of some $p>0$ and $c>1$
such that

$\texttt{P}_{0}^{\texttt{ei}^{\prime}}\big{(}|D^{\nu}(t/3+s)|\geq
c^{s}\big{)}\geq p$ (4.45)

for all $s$ large enough. To obtain (4.45), let $r>0$ be such that

$\inf_{y\in[-1,+1]}\texttt{E}_{y}^{\texttt{ei}^{\prime}}\big{[}N(r,[-1,+1])\big{]}=:\mu>1,$

(the feasibility of such a choice of $r$ is a direct consequence of the
Feynman-Kac formula). For $\nu\in D_{\varepsilon}(t/3)$ consider the following
process under $\texttt{P}^{\texttt{ei}^{\prime}}_{0}$, conditionally on
$X_{t/3}^{\nu}$:

  * •

the process starts with one particle at position $X_{t/3}^{\nu}$;

  * •

between times $r(n-1)$ and $rn$, $n\in\mathbb{N}$, the process evolves as a
branching Brownian motion with branching rate ${\texttt{ei}^{\prime}}$;

  * •

at times $rn$, $n\in\mathbb{N}$, particles outside of the interval
$\big{[}X_{t/3}^{\nu}-1,X_{t/3}^{\nu}+1\big{]}$ are killed.

Using the Markov property, one readily observe that the number of particles of
the latter process stochastically dominates the number of particles of a
Galton-Watson process $(L_{n})_{n\in\mathbb{N}}$ which starts with one
particle and whose offspring distribution has expectation $\mu$. Then by [2,
Theorem 1, section I.5], the Galton-Watson process has positive probability to
survive, i.e. $P(L_{n}>0\ \forall n\in\mathbb{N})=:p_{1}>0$. Conditioned on
surviving, there exists $c>1$ such that

$P\big{(}L_{k}\geq c^{k}\,|\,L_{n}>0\ \forall
n\in\mathbb{N}\big{)}\geq\frac{1}{2}\quad\text{for all }k\in\mathbb{N}.$

One can see that for every $\nu\in N(t/3),$ inequality (4.45) holds true with
the choice $p:=p_{1}/2$ for all $s\in r\cdot\mathbb{N}$. By a straightforward
comparison argument, this extends to all $s\geq 0.$ Therefore, we can now
apply (4.44) and (4.45) in order to deduce

$\texttt{P}_{0}^{\texttt{ei}^{\prime}}\big{(}N(2t/3,[-\varepsilon
t,\varepsilon t])\geq c^{t}\big{)}\geq 1-e^{c^{\prime}(\varepsilon)t}.$ (4.46)

Furthermore, we have

$P_{\varepsilon t}\big{(}X_{t/3}\in[-1,1]\big{)}\geq
c_{3}t^{-1/2}e^{-3\varepsilon^{2}t/2}\geq\Big{(}\frac{1+c}{2}\Big{)}^{-t/3},$

for all $t$ large enough $\varepsilon>0$ small enough and $c>1$ suitable,
where for the last inequality we used that $\varepsilon$ does not depend on
$c$. The latter inequality and a large deviation statement then gives for all
$t\geq t_{4}\geq t_{3}$

$\texttt{P}_{0}^{\texttt{ei}^{\prime}}\Big{(}N(t,[-1,1])\geq\frac{1}{2}c^{t}\Big{(}\frac{1+c}{2}\Big{)}^{-t/3}\,\big{|}\,N(2t/3,[-\varepsilon
t,\varepsilon t])\geq c^{t}\Big{)}\geq 1-e^{-c_{4}t}.$ (4.47)

Thus, for $t\geq t_{0}$, where $t_{0}$ is chosen large enough, by (4.46) and
(4.47), in combination with $c>1,$ we infer the desired result. ∎

With the help of Lemmas 4.1, 4.4, and 4.7, it is now possible to state a
crucial result for the proof of Theorem 1.4.

######  Proposition 4.8.

For every $q>0$, $F$ satisfying (PROB) and $a>0$, there exist a constant
$\Cr{constanttheoremlogdist}=\Cr{constanttheoremlogdist}(q,F)\in(0,\infty)$, a
$\mathbb{P}$-a.s. finite $C=C(t)=C(t,q,F,\xi)>0$ and a $\mathbb{P}$-a.s.
finite random variable
$\Cl[index_{p}rob_{c}ont_{t}ime]{timetheoremlogdist}=\Cr{timetheoremlogdist}(a,q,F,\xi)$
such that for all $t\geq\Cr{timetheoremlogdist},$ we have
$C(t)\leq\Cr{constanttheoremlogdist}$ and

$\emph{{P}}^{\xi}_{\overline{m}^{a}(t)-C\ln
t}\left(N^{\leq}(t,0)\neq\emptyset\right)\geq 1-2t^{-q}.$

###### Proof.

For simplicity, we write $\overline{m}(t):=\overline{m}^{a}(t)$. Without loss
of generality, it is enough to show the claim for all
$q>2(\gamma_{1}+\gamma_{2})$, where $\gamma_{1}=\gamma_{1}(a)$ and
$\gamma_{2}=\gamma_{2}(a)$ are defined in Lemmas 4.1 and 4.4, respectively.
Let further $\Cr{const_{a}mpl}$ and $t_{0}$ be as in Lemma 4.7, and $c_{1}$ be
such that for $r:=c_{1}\ln t$ we have $\Cr{const_{a}mpl}^{-r}=t^{-q}$.

We claim that there exist $\Cr{constanttheoremlogdist},C(t)$ and
$\Cr{timetheoremlogdist}$ as above such that
$\overline{m}(t-r)=\overline{m}(t)-C(t)\ln t$,
$C(t)\leq\Cr{constanttheoremlogdist}$ and the conclusions of Lemmas 4.1 and
4.4 hold for all $t\geq\Cr{timetheoremlogdist}$. Indeed, writing
$u(t,x)=\texttt{E}_{x}^{\xi}[N^{\leq}(t,0)]$, by the time and space
perturbation Lemmas 3.11 and 3.12, defining
$c_{2}:=\Cr{const_{t}ime_{p}erturbation}\vee\Cr{const_{s}pace_{p}erturbation}$,
we deduce that

$\displaystyle u(t-r,\overline{m}(t)-\Cr{constanttheoremlogdist}\ln t)$
$\displaystyle\geq c_{2}^{-1}e^{\Cr{constanttheoremlogdist}\ln
t/c_{2}}u(t-r,\overline{m}(t))\geq c_{2}^{-2}e^{\Cr{constanttheoremlogdist}\ln
t/c_{2}-r/c_{2}}u(t,\overline{m}(t))\geq a,$ (4.48)

where for the last inequality we choose $\Cr{constanttheoremlogdist}$ and
$\Cr{timetheoremlogdist}=\Cr{timetheoremlogdist}(\varepsilon)$ large enough
such that the last inequality holds for all $t\geq\Cr{timetheoremlogdist}$. As
a consequence, we infer

$\overline{m}(t-r)\geq\overline{m}(t)-\Cr{constanttheoremlogdist}\ln t.$
(4.49)

By (3.33), we also get
$u(t,\overline{m}(t-r))\geq\Cr{const_{t}ime_{p}erturbation}^{-1}e^{r/\Cr{const_{t}ime_{p}erturbation}}u(t-r,\overline{m}(t-r))\geq
a$ for all $t$ large enough, i.e.

$\overline{m}(t-r)\leq\overline{m}(t).$

Thus, combining the two previous displays, we can find
$C(t)\leq\Cr{constanttheoremlogdist}$ such that
$\overline{m}(t-r)=\overline{m}(t)-C(t)\ln t$. Now let
$x:=\overline{m}(t)-C(t)\ln t=\overline{m}(t-r)$. Conditioning on whether
until time $r$ there are more or less than $\Cr{const_{a}mpl}^{r}$ particles
in $[x-1,x+1]$, we get for all $t\geq\Cr{timetheoremlogdist},$

$\displaystyle\texttt{P}_{x}^{\xi}\left(N^{\leq}(t,0)\geq 1\right)$
$\displaystyle\geq
1-\texttt{P}_{x}^{\xi}\big{(}N(r,[x-1,x+1])\leq\Cr{const_{a}mpl}^{r}\big{)}-\sup_{y\in[x-1,x+1]}\Big{(}\texttt{P}_{y}^{\xi}\left(N^{\leq}(t-r,0)=0\right)\Big{)}^{\Cr{const_{a}mpl}^{r}}$
$\displaystyle\geq
1-\Cr{const_{a}mpl}^{-r}-\sup_{y\in[x-1,x+1]}\Big{(}\texttt{P}^{\xi}_{y}\big{(}N^{\mathcal{L},a}_{t-r}=0\big{)}\Big{)}^{\Cr{const_{a}mpl}^{r}},$

using Lemma 4.7 in the last inequality. Now using Cauchy-Schwarz as in (4.3),
in combination with Lemmas 4.1 and 4.4, we infer

$\displaystyle\sup_{y\in[x-1,x+1]}\big{(}\texttt{P}^{\xi}_{y}(N^{\mathcal{L},a}_{t-r}=0)\big{)}^{\Cr{const_{a}mpl}^{r}}$
$\displaystyle\leq\left(1-t^{-2\gamma_{1}-\gamma_{2}}\right)^{t^{q}}\leq
t^{-q},$

where we adapt $\Cr{timetheoremlogdist}=\Cr{timetheoremlogdist}(a,q,\xi,q)$
such that the last inequality holds for all $t\geq\Cr{timetheoremlogdist}$. ∎

###### Proof of Theorem 1.4.

1) We first prove the result under the additional assumption that $F$ fulfills
(PROB). Let $w^{\xi,F,w_{0}}$ be the solution to (F-KPP) with initial
condition $w_{0}\in\mathcal{I}_{\text{F-KPP}},$ so in particular $0\leq
w_{0}\leq\mathds{1}_{(-\infty,0]}$. Because $F$ fulfills (PROB), by (McKean)
and the Markov property we infer

$\displaystyle w^{\xi,F,w_{0}}(t,x)$
$\displaystyle=\texttt{E}_{x}^{\xi}\Big{[}1-\prod_{\nu\in
N(t)}\big{(}1-w_{0}(X_{t}^{\nu})\big{)}\Big{]}\geq\texttt{E}_{x}^{\xi}\Big{[}1-\prod_{\nu\in
N(t)}\big{(}1-{w}_{0}(X_{t}^{\nu})\big{)};N^{\leq}(t-s,0)\geq 1\Big{]}$
$\displaystyle\geq\texttt{P}_{x}^{\xi}\big{(}N^{\leq}(t-s,0)\geq
1\big{)}\cdot\inf_{y\leq 0}w^{\texttt{ei},F,w_{0}}(s,y),$ (4.50)

where $w=w^{\texttt{ei},F,{w}_{0}}$ solves the homogeneous equation
$w_{t}=\frac{1}{2}w_{xx}+\texttt{ei}\cdot F(w)$ with initial condition
$w(0,\cdot)={w}_{0}$. Then we have
$w^{1,F,\widetilde{w}_{0}}=w^{\texttt{ei},F,w_{0}}(\frac{t}{\texttt{ei}},\frac{x}{\sqrt{\texttt{ei}}})$
with $\widetilde{w}_{0}(x):=w_{0}(x/\sqrt{\texttt{ei}})$. Because
$w^{\texttt{ei},F,w_{0}}(0,x)=0$ for $x>0$, conditions [8, (8.1) and (1.17)]
are fulfilled. Together with (KPP-INI) and [8, Theorem 3, p. 141],
$w^{1,F,\widetilde{w}_{0}}$ (and thus also $w^{\texttt{ei},F,w_{0}}$) is a
traveling wave solution, i.e., there exist
$m^{\texttt{ei}}(t)=\sqrt{2\texttt{ei}}t+o(t)$ and some $g$ fulfilling
$\lim_{x\to-\infty}g(x)=1$ and $\lim_{x\to\infty}g(x)=0$ such that

$\sup_{y}\big{|}w^{\texttt{ei},F,w_{0}}(t,y+m^{\texttt{ei}}(t))-g(y)\big{|}\to
0,\quad t\to\infty.$ (4.51)

Now let $\varepsilon\in(0,1)$ and choose $\delta>0$ such that
$\frac{\varepsilon}{1-\delta}\in(0,1)$. Then by (4.51) we get

$\inf_{y\leq
0}w^{\texttt{ei},F,w_{0}}(s,y)=\inf_{y\leq-m^{\texttt{ei}}(s)}w^{\texttt{ei},F,w_{0}}(s,y+m^{\texttt{ei}}(s))\geq
1-\delta\ \text{ for all }s\geq s_{0}(F,w_{0},\delta,\texttt{ei}),$

which, together with (4.50), gives

$m^{\xi,F,w_{0},\varepsilon}(t)\geq
m^{\xi,F,\mathds{1}_{(-\infty,0]},\frac{\varepsilon}{1-\delta}}(t-s_{0})\quad\text{for
all }t\geq s_{0}(F,w_{0},\delta,\texttt{ei}).$ (4.52)

The inequality

$m^{\xi,F,\mathds{1}_{(-\infty,0]},\frac{\varepsilon}{1-\delta}}(t-s_{0}){\geq}\
\overline{m}^{\xi,\mathds{1}_{(-\infty,0]},\frac{\varepsilon}{1-\delta}}(t-s_{0})-\Cr{constanttheoremlogdist}\ln(t),\quad\text{for
all }t\geq\mathcal{T}_{1}(\xi,F,\varepsilon,\delta),$

follows from Proposition 4.8. By Corollary 3.18, for $C^{\prime}>1$ from (PAM-
INI) we get

$\displaystyle\overline{m}^{\xi,\mathds{1}_{(-\infty,0]},\frac{\varepsilon}{1-\delta}}(t-s_{0})$
$\displaystyle\geq\overline{m}^{\xi,\mathds{1}_{(-\infty,0]},\frac{\varepsilon}{C^{\prime}}}(t-s_{0})-c_{1}(\varepsilon,\delta,C^{\prime})$
$\displaystyle\geq\overline{m}^{\xi,\mathds{1}_{(-\infty,0]},\frac{\varepsilon}{C}}(t)-c_{2}(\varepsilon,\delta,C^{\prime},s_{0})\quad\text{for
all }t\geq\mathcal{T}_{2}(\xi,\varepsilon,\delta),$

where the second inequality can be obtained similarly to the argument in
(4.48) and (4.49). Combining the above inequalities, we arrive at

$\displaystyle m^{\xi,F,w_{0},\varepsilon}(t)$
$\displaystyle\geq\overline{m}^{\xi,\mathds{1}_{(-\infty,0]},\frac{\varepsilon}{C^{\prime}}}(t)-C_{1}\ln(t)-c_{3}$
$\displaystyle=\overline{m}^{\xi,C^{\prime}\mathds{1}_{(-\infty,0]},\varepsilon}(t)-C_{1}\ln(t)-c_{3}\quad\text{for
all }t\geq\mathcal{T}_{3}(\xi).$

Now by (PAM-INI) every $u_{0}\in\mathcal{I}_{\text{PAM}}$ is upper bounded by
the function $C^{\prime}\mathds{1}_{(-\infty,0]},$ and since we have
$\overline{m}^{\xi,\mathds{1}_{(-\infty,0]},\varepsilon}(t)\geq
m^{\xi,F,w_{0},\varepsilon}(t)$ for all $\varepsilon\in(0,1)$ and
$w_{0}\in\mathcal{I}_{\text{F-KPP}}$, this finishes the proof for $F$
fulfilling (PROB).

2) Now let $F$ fulfill (SC) and $w_{0}$ be continuous. By a sandwich argument,
it suffices to show that there exists some function $G$ fulfilling (PROB),
such that $F(w)\geq G(w)$ for all $w\in[0,1]$. Indeed, by Corollary A.11 the
solutions $w^{F}$ and $w^{G}$ to

$w^{F}_{t}-\frac{1}{2}w^{F}_{xx}-\xi(x)F(w^{F})=0=w^{G}_{t}-\frac{1}{2}w^{G}_{xx}-\xi(x)G(w^{G})$

(which are classical by Proposition 2.1) fulfill $w^{F}\geq w^{G}$. As a
consequence, we infer that $m^{\xi,F,w_{0},\varepsilon}(t)\geq
m^{\xi,G,w_{0},\varepsilon}(t)\geq\overline{m}^{\xi,w_{0},\varepsilon}(t)-C_{1}\ln(t)$,
where the second inequality is due to step 1). The claim for arbitrary
$w_{0}\in\mathcal{I}_{\text{F-KPP}}$ is then true by an approximation argument
of $w_{0}$ by continuous functions, that is if $F\geq G$, by Remark 1.1 we
have $w^{w_{0},F}\geq w^{w_{0},G}$ and consequently
$m^{\xi,w_{0},F,\varepsilon}(t)\geq m^{\xi,w_{0},G,\varepsilon}(t)$ for all
$t\geq 0$ and we can conclude.

It remains to show that for every $F$ fulfilling (SC) there exists $G$
fulfilling (PROB) such that $F(x)\geq G(x)$ for all $x\in[0,1]$. To do so,
recall that (SC) implies that there exists $M\in\mathbb{N}$ such that

$1-F^{\prime}(x)\leq\frac{M}{2}x\quad\text{ and }\quad F(1-x)\geq
xM^{-1}\quad\text{ for all }x\in[0,M^{-1}].$ (4.53)

Define $G_{n}(x):=\frac{1-x}{n}\big{(}1-(1-x)^{n}\big{)}$, $x\in[0,1]$,
$n\in\mathbb{N}$. Then each $G_{n},$ $n\in\mathbb{N},$ satisfies (PROB) with
$p_{1}=1-n^{-1}$ and $p_{n+1}=n^{-1},$ and our goal is to show that $G_{n}\leq
F$ for all $x\in[0,1]$ and all $n$ large enough.

We start with noting that $G_{n+1}\leq G_{n}$ as functions on $x\in[0,1],$ for
all $n\in\mathbb{N},$ and that $G_{n}\downarrow 0$ uniformly as $n$ tends to
infinity. Thus, since $F$ is continuous and $F>0$ on $(0,1)$ due to (SC), we
only have to take care of the neighborhoods of $0$ and $1.$ From (4.53) we
immediately get $G_{M}(x)\leq(1-x)M^{-1}\leq F(x)$ for all $x\in[1-M^{-1},1]$.
To infer the desired inequality for $x\in[0,1-M^{-1}],$ Taylor expansion
yields

$\displaystyle(1-x)^{M}$ $\displaystyle\geq
1-Mx+\binom{M}{2}x^{2}-\binom{M}{3}x^{3}.$

Then for $M$ large enough and for all $x\in[0,M^{-1}]$ we get

$\displaystyle G_{M}(x)$
$\displaystyle\leq(1-x)\Big{(}x-\frac{M-1}{2}x^{2}+\frac{(M-1)(M-2)}{6}x^{3}\Big{)}\leq
x-\frac{M}{3}x^{2}$ $\displaystyle\leq
x-\frac{M}{4}x^{2}=\int_{0}^{x}(1-Mt/2)\mathop{}\\!\mathrm{d}t\leq\int_{0}^{x}F^{\prime}(t)\mathop{}\\!\mathrm{d}t=F(x),$

where the last inequality is due to (4.53) again. This finishes the proof. ∎

###  4.4 Remarks on $v_{c}$ and $v_{0}$

By the same argument as in [10, Lemma A.4], one can show that a rich class of
potentials $\xi$ satisfies (VEL), i.e. the inequality $v_{c}<v_{0}$. It is
natural to ask if $v_{c}<v_{0}$ is always fulfilled in our setting or not;
i.e., do there exist potentials which satisfy (Standing assumptions) but not
the inequality $v_{c}<v_{0}$?

In order to answer this question, we will take advantage of the following
result.

######  Claim 4.9.

$v_{0}=\inf_{\eta\leq 0}\frac{\eta-\text{\emph{{es}}}}{L(\eta)}.$ (4.54)

###### Proof.

As shown in [14, p. 514ff.], the function

$I:\,(0,\infty)\ni
y\mapsto\sup_{\eta\leq-\texttt{es}}\big{(}y\eta-L(\eta+\texttt{es})\big{)}$

is strictly decreasing, finite, convex, fulfills $\lim_{y\downarrow
0}I(y)=+\infty$ and $\lim_{y\uparrow\infty}I(y)=-\infty$, there exists a
unique $v^{*}>0$ such that $I(1/v^{*})=0$, and one has

$v^{*}=\inf_{\eta\leq 0}\frac{\eta-\texttt{es}}{L(\eta)}.$

We now show $v^{*}=v_{0}$. For this purpose, let $w$ and $u$ be solutions to
(F-KPP) and (PAM), respectively, both with initial condition
$\mathds{1}_{[-1,0]}$. Then $w\leq u$ and thus by [14, Lemma 7.6.3] and
Proposition A.3 we have for all $v>v^{*}$ that $\mathbb{P}$-a.s.,

$\displaystyle\Lambda(v)=\lim_{t\to\infty}\frac{1}{t}\ln u(t,vt)$
$\displaystyle\geq\liminf_{t\to\infty}\frac{1}{t}\ln w(t,vt)\geq-vI(1/v).$

Since $\Lambda$ and $I$ are continuous, passing to the limit $v\downarrow
v^{*}$ we deduce that $\Lambda(v^{*})\geq-v^{*}I(1/v^{*})=0$. Furthermore,
$\Lambda(v)<0$ for all $v>v_{0}$ and thus we infer $v_{0}\geq v^{*}.$

To get the converse inequality, we use that for all $v>0$ and all $\eta\leq 0$
we have

$\displaystyle u(t,vt)$
$\displaystyle=E_{vt}\big{[}e^{\int_{0}^{t}(\zeta(B_{s})+\texttt{es})\mathop{}\\!\mathrm{d}s};B_{t}\in[-1,0]\big{]}=E_{vt}\big{[}e^{\int_{0}^{t}(\zeta(B_{s})+\texttt{es})\mathop{}\\!\mathrm{d}s};B_{t}\in[-1,0],H_{0}\leq
t\big{]}$ $\displaystyle\leq
e^{(\texttt{es}-\eta)t}E_{vt}\big{[}e^{\int_{0}^{H_{0}}(\zeta(B_{s})+\eta)\mathop{}\\!\mathrm{d}s}\big{]}.$

In combination with (2.9) this yields that for all $v>0,$

$\displaystyle\Lambda(v)=\lim_{t\to\infty}\frac{1}{t}\ln u(t,vt)$
$\displaystyle\leq\inf_{\eta\leq
0}\big{(}(\texttt{es}-\eta)+vL(\eta)\big{)}=-v\sup_{\eta\leq-\texttt{es}}\big{(}\frac{\eta}{v}-L(\eta+\texttt{es})\big{)}=-vI(1/v).$

But then $\Lambda(v^{*})\leq-v^{*}I(1/v^{*})=0$ and thus we must have
$v^{*}\geq v_{0}$. ∎

We now formulate our main result on the relation between $v_{0}$ and $v_{c}.$
It implies that our results do not apply for all potentials fulfilling (BDD),
(STAT) and (• ‣ 1.4).

######  Proposition 4.10.

There exist potentials $\xi$ fulfilling (BDD), (STAT) and (• ‣ 1.4), and such
that $v_{c}>v_{0};$ i.e., condition (VEL) is violated.

###### Proof.

Recalling (4.54), the definition $v_{c}=\frac{1}{L^{\prime}(0-)}$ from Lemma
2.4 d), it is sufficient to show $L(0)+\texttt{es}\cdot L^{\prime}(0-)<0$,
which means

$\mathbb{E}\Big{[}\ln
E_{1}\big{[}e^{\int_{0}^{H_{0}}\zeta(B_{s})\mathop{}\\!\mathrm{d}s}\big{]}\Big{]}+\texttt{es}\cdot\mathbb{E}\Big{[}\frac{E_{1}\big{[}H_{0}e^{\int_{0}^{H_{0}}\zeta(B_{s})\mathop{}\\!\mathrm{d}s}\big{]}}{E_{1}\big{[}e^{\int_{0}^{H_{0}}\zeta(B_{s})\mathop{}\\!\mathrm{d}s}\big{]}}\Big{]}<0.$
(4.55)

To establish the latter, let $\widetilde{\omega}$ be a one-dimensional Poisson
point process with intensity one. In a slight abuse of notation,
$\widetilde{\omega}=(\widetilde{\omega}^{i})_{i}$ can be seen as a mapping
from $\Omega$ into the set of all locally finite point configurations; i.e.,
$\widetilde{\omega}=(\widetilde{\omega}^{i})_{i}$ can be interpreted as a
random set of countably many points in $\mathbb{R}$, satisfying
$|\\{i:\widetilde{\omega}^{i}\in B\\}|<\infty$ for every bounded Borel set and
$\widetilde{\omega}^{i}\neq\widetilde{\omega}^{j}$ for all $i\neq j$. See [32]
for further details. Now denote by $\omega=(\omega^{i})_{i}$ be the point
process that is obtained from $\widetilde{\omega}$ by deleting simultaneously
all points in $\widetilde{\omega}$ which have distance $1$ or less to their
nearest neighbor in $\widetilde{\omega}.$ (see [26, p. 47] for details). Let
$\varphi(x)$ be a mollifier with support $[-1/2,1/2]$, non-decreasing for
$x\leq 0$, non-increasing for $x\geq 0$ with $\varphi(0)=1$ and let
$\varphi^{(\varepsilon)}(x):=\varphi(x/\varepsilon)$, $\varepsilon>0$.
Finally, for $\varepsilon\in(0,1)$ and $a>0,$ define the potential
$\zeta(x)=\zeta^{(\varepsilon,a)}(x):=-a+a\sum_{i}\varphi^{(\varepsilon)}(x-\omega^{i})$.
One can easily check that $\zeta$ is the corresponding shifted potential as in
(2.3) of some $\xi$ fulfilling (BDD), (STAT) and (• ‣ 1.4), i.e.
$a=\texttt{es}-\texttt{ei}$. We will choose $a>\ln 2$,
$\texttt{ei}\in(0,\frac{a}{8})$ and $\varepsilon(a)>0$ suitably at the end of
the proof. Let us now consider both summands in (4.55) separately.

1) We observe that $\mathbb{P}$-a.s., $\zeta^{(\varepsilon,a)}\downarrow-a$ as
$\varepsilon\downarrow 0$ for all $x\in\mathbb{R},$ as well as by [4, (2.0.1),
p. 204]

$-\sqrt{2a}=\ln E_{1}\big{[}e^{-aH_{0}}\big{]}\leq\ln
E_{1}\big{[}e^{\int_{0}^{H_{0}}\zeta^{(\varepsilon,a)}(B_{s})\mathop{}\\!\mathrm{d}s}\big{]}\leq
0$

for all $\varepsilon\in(0,1).$ Thus, by dominated convergence, for all $a>0$
there exists $\varepsilon_{1}=\varepsilon_{1}(a)>0$, such that

$\mathbb{E}\Big{[}\ln
E_{1}\big{[}e^{\int_{0}^{H_{0}}\zeta^{(\varepsilon_{1},a)}(B_{s})\mathop{}\\!\mathrm{d}s}\big{]}\Big{]}\leq-\frac{3}{4}\sqrt{2a}.$
(4.56)

2) To bound the second summand in (4.55), we lower bound its denominator by

$E_{1}\big{[}e^{\int_{0}^{H_{0}}\zeta(B_{s})\mathop{}\\!\mathrm{d}s}\big{]}\geq
E_{1}\big{[}e^{-aH_{0}}\big{]}=e^{-\sqrt{2a}}.$ (4.57)

For the numerator, define $\mathbb{J}$ to be the set of possible point
configurations of the process $(\omega^{i})_{i}$. Let us first check that for
all $a>0$ there exists $\varepsilon=\varepsilon(a)>0$, such that

$\sup_{(\omega^{i})_{i}\in\mathbb{J}}E_{1}\big{[}H_{0}e^{\int_{0}^{H_{0}}(-a+a\sum_{i}\varphi^{(\varepsilon)}(B_{s}-\omega^{i}))\mathop{}\\!\mathrm{d}s}\big{]}<\infty.$
(4.58)

Indeed, letting
$g^{\varepsilon,(\omega^{i})_{i}}(x):=\sum_{i}\varphi^{(\varepsilon)}(x-\omega^{i})$,
we have

$\displaystyle\begin{split}E_{1}\big{[}H_{0}e^{\int_{0}^{H_{0}}(-a+a\cdot
g^{\varepsilon,(\omega^{i})_{i}}(B_{s}))\mathop{}\\!\mathrm{d}s}\big{]}&=\sum_{n=0}^{\infty}E_{1}\big{[}H_{0}e^{\int_{0}^{H_{0}}(-a+a\cdot
g^{\varepsilon,(\omega^{i})_{i}}(B_{s}))\mathop{}\\!\mathrm{d}s};H_{0}\in[n,n+1)\big{]}\\\
&\leq\sum_{n=0}^{\infty}(n+1)E_{1}\big{[}e^{\int_{0}^{n}(-a+a\cdot
g^{\varepsilon,(\omega^{i})_{i}}(B_{s}))\mathop{}\\!\mathrm{d}s}\big{]}.\end{split}$
(4.59)

Note that by the property of all point configurations in $\mathbb{J}$ to have
points with mutual distance at least one, we have

$\sup_{(\omega^{i})_{i}\in\mathbb{J}}\sup_{x\in\mathbb{R}}E_{x}\Big{[}a\int_{0}^{1}g^{\varepsilon,(\omega^{i})_{i}}(B_{s})\mathop{}\\!\mathrm{d}s\Big{]}\leq
a\int_{0}^{1}E_{0}\big{[}\mathds{1}_{A_{\varepsilon}}(B_{s})\big{]}\mathop{}\\!\mathrm{d}s\leq\frac{1}{2}$

for all $\varepsilon(a)>0$ small enough, where
$A_{\varepsilon}:=\bigcup_{i\in\mathbb{Z}}[-\varepsilon/2+i,\varepsilon/2+i]$.
Using Kasminskii’s lemma (cf. e.g. [34, Lemma 1.2.1]) we infer
$\sup_{x\in\mathbb{R}}E_{x}\big{[}e^{a\int_{0}^{1}g^{\varepsilon,(\omega^{i})_{i}}(B_{s})\mathop{}\\!\mathrm{d}s}\big{]}\leq
2$. An $(n-1)$-fold application of the Markov property at times $1,\ldots,n-1$
supplies us with
$\sup_{x\in\mathbb{R}}E_{x}\big{[}e^{a\int_{0}^{n}g^{\varepsilon,(\omega^{i})_{i}}(B_{s})\mathop{}\\!\mathrm{d}s}\big{]}\leq
2^{n}$ for all $n\in\mathbb{N}$ and all $(\omega^{i})_{i}\in\mathbb{J}$.
Plugging this into (4.59) we infer

$\displaystyle\sup_{(\omega^{i})_{i}\in\mathbb{J}}E_{1}\Big{[}H_{0}e^{\int_{0}^{H_{0}}(-a+a\cdot
g^{\varepsilon,(\omega^{i})_{i}}(B_{s}))\mathop{}\\!\mathrm{d}s}\Big{]}$
$\displaystyle\leq\sum_{n=0}^{\infty}(n+1)e^{-na}2^{n},$

so the right-hand side in (4.59) is finite, and (4.58) holds true for all
$a>\ln 2$ and $\varepsilon(a)$ small enough as well. Since
$g^{\varepsilon,(\omega^{i})_{i}}$ decreases $\mathbb{P}$-a.s. to $0$
monotonically as $\varepsilon\downarrow 0$, we infer

$\lim_{\varepsilon\downarrow
0}\mathbb{E}\Big{[}E_{1}\big{[}H_{0}e^{\int_{0}^{H_{0}}\zeta^{(\varepsilon,a)}(B_{s})\mathop{}\\!\mathrm{d}s}\big{]}\Big{]}=E_{1}\big{[}H_{0}e^{-aH_{0}}\big{]}=-\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}a}E_{1}\big{[}e^{-aH_{0}}\big{]}=-\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}a}(e^{-\sqrt{2a}})=\frac{1}{\sqrt{2a}}e^{-\sqrt{2a}},$
(4.60)

using [4, (2.0.1), p. 204] in the third equality. Thus, combining (4.57) and
(4.60) we infer that there exists $\varepsilon_{2}(a)>0$ such that the second
summand on the left-hand side of (4.55) is upper bounded by
$\texttt{es}\cdot\frac{4/3}{\sqrt{2a}}=(a+\texttt{ei})\cdot\frac{4/3}{\sqrt{2a}}.$
Using this in combination with (4.56), we infer that for all $a>\ln 2$ (which
is sufficient for (4.58)) and
$\varepsilon\in(0,\varepsilon_{1}(a)\wedge\varepsilon_{2}(a))$, choosing
$\texttt{ei}\in(0,\frac{a}{8}),$ we get that the left-hand side in (4.55) is
upper bounded by
$-\frac{3}{4}\sqrt{2a}+(a+\texttt{ei})\cdot\frac{4/3}{\sqrt{2a}}<0$. ∎

##  Appendix A Appendix

###  A.1 Properties of the logarithmic moment generating functions

######  Lemma A.1.

We recall that $P_{x}^{\zeta,\eta}$ has been defined in (2.6).

  1. (a)

$L,$ $L_{x}^{\zeta}$, and $\overline{L}_{x}^{\zeta},$ for $x\in\mathbb{R}$,
(see (2.7) to (2.8) for their definitions) are infinitely differentiable on
$(-\infty,0).$ Furthermore, for all $\eta<0$ we have

$\displaystyle\big{(}L_{x}^{\zeta}\big{)}^{\prime}(\eta)$
$\displaystyle=\frac{E_{x}\Big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}H_{\lceil
x\rceil-1}\Big{]}}{E_{x}\Big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}\Big{]}}=E_{x}^{\zeta,\eta}[\tau_{\lceil
x\rceil-1}],\quad x\in\mathbb{R},$ (A.1)
$\displaystyle\big{(}\overline{L}_{x}^{\zeta}\big{)}^{\prime}(\eta)$
$\displaystyle=\frac{1}{x}E_{x}^{\zeta,\eta}\big{[}H_{0}\big{]},\quad x>0,$
(A.2) $\displaystyle L^{\prime}(\eta)$
$\displaystyle=\mathbb{E}\Big{[}\frac{E_{1}\big{[}e^{\int_{0}^{H_{0}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}H_{0}\big{]}}{E_{1}\big{[}e^{\int_{0}^{H_{0}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}\big{]}}\Big{]}=\mathbb{E}\big{[}E_{1}^{\zeta,\eta}[H_{0}]\big{]},$
(A.3) and $\displaystyle\big{(}L_{x}^{\zeta}\big{)}^{\prime\prime}(\eta)$
$\displaystyle=\frac{E_{x}\big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}H_{\lceil
x\rceil-1}^{2}\big{]}}{E_{x}\big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}\big{]}}-\Bigg{(}\frac{E_{x}\big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}H_{\lceil
x\rceil-1}\big{]}}{E_{x}\big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}\big{]}}\Bigg{)}^{2}$
$\displaystyle=E_{x}^{\zeta,\eta}\big{[}\tau_{\lceil
x\rceil-1}^{2}\big{]}-\big{(}E_{x}^{\zeta,\eta}[\tau_{\lceil
x\rceil-1}]\big{)}^{2}=\text{\emph{Var}}_{x}^{\zeta,\eta}(\tau_{\lceil
x\rceil-1})>0,\quad x\in\mathbb{R},$ (A.4)
$\displaystyle\big{(}\overline{L}_{x}^{\zeta}\big{)}^{\prime\prime}(\eta)$
$\displaystyle=\frac{1}{x}\text{\emph{Var}}_{x}^{\zeta,\eta}(H_{0}),\quad
x>0,$ (A.5) $\displaystyle L^{\prime\prime}(\eta)$
$\displaystyle=\mathbb{E}\Bigg{[}\frac{E_{1}\big{[}e^{\int_{0}^{H_{0}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}H_{0}^{2}\big{]}}{E_{1}\big{[}e^{\int_{0}^{H_{0}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}\big{]}}-\Big{(}\frac{E_{1}\big{[}e^{\int_{0}^{H_{0}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}H_{0}\big{]}}{E_{1}\big{[}e^{\int_{0}^{H_{0}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}\big{]}}\Big{)}^{2}\Bigg{]}$
$\displaystyle=\mathbb{E}\Big{[}E_{1}^{\zeta,\eta}[H_{0}^{2}]-\big{(}E_{1}^{\zeta,\eta}[H_{0}]\big{)}^{2}\Big{]}=\mathbb{E}\big{[}\text{\emph{Var}}_{1}^{\zeta,\eta}(H_{0})\big{]}>0.$
(A.6)

  2. (b)

For each compact interval $\triangle\subset(-\infty,0)$ there exists a
constant $\Cl{constDerLMG}=\Cr{constDerLMG}(\triangle)>0$, such that the
following inequalities hold $\mathbb{P}$-a.s.:

$\displaystyle-\Cr{constDerLMG}\leq\inf_{\eta\in\triangle,x\geq
1}\big{\\{}L_{\lfloor
x\rfloor}^{\zeta}(\eta),\overline{L}_{x}^{\zeta}(\eta),L(\eta)\big{\\}}$
$\displaystyle\leq\sup_{\eta\in\triangle,x\geq 1}\big{\\{}L_{\lfloor
x\rfloor}^{\zeta}(\eta),\overline{L}_{x}^{\zeta}(\eta),L(\eta)\big{\\}}\leq-\Cr{constDerLMG}^{-1},$
$\displaystyle\Cr{constDerLMG}^{-1}\leq\inf_{\eta\in\triangle,x\geq
1}\big{\\{}(L_{\lfloor
x\rfloor}^{\zeta})^{\prime}(\eta),(\overline{L}_{x}^{\zeta})^{\prime}(\eta),L^{\prime}(\eta)\big{\\}}$
$\displaystyle\leq\sup_{\eta\in\triangle,x\geq 1}\big{\\{}(L_{\lfloor
x\rfloor}^{\zeta})^{\prime}(\eta),(\overline{L}_{x}^{\zeta})^{\prime}(\eta),L^{\prime}(\eta)\big{\\}}\leq\Cr{constDerLMG},$
$\displaystyle\Cr{constDerLMG}^{-1}\leq\inf_{\eta\in\triangle,x\geq
1}\big{\\{}(L_{\lfloor
x\rfloor}^{\zeta})^{\prime\prime}(\eta),(\overline{L}_{x}^{\zeta})^{\prime\prime}(\eta),L^{\prime\prime}(\eta)\big{\\}}$
$\displaystyle\leq\sup_{\eta\in\triangle,x\geq 1}\big{\\{}(L_{\lfloor
x\rfloor}^{\zeta})^{\prime\prime}(\eta),(\overline{L}_{x}^{\zeta})^{\prime\prime}(\eta),L^{\prime\prime}(\eta)\big{\\}}\leq\Cr{constDerLMG}.$

###### Proof.

$(a)$ Due to the convexity of the exponential function we have
$\big{|}\frac{e^{hx}-1}{h}\big{|}\leq xe^{hx}\vee 1$ for all $x\geq 0$ and
$h\in\mathbb{R}$. If we choose $h_{0}:=\frac{|\eta|}{2}$, then since
$\zeta\leq 0$, we have that for all $|h|\leq h_{0},$

$\Big{|}\frac{1}{h}e^{\int_{0}^{H_{y}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}\big{(}e^{hH_{y}}-1\big{)}\Big{|}\leq
e^{\eta H_{y}/2}(H_{y}\vee 1).$ (A.7)

Due to $H_{y}e^{\eta H_{y}}\leq\frac{1}{|\eta|}$ for all $H_{y}\geq 0$ and
$\eta<0,$ as well as $\lim_{h\to 0}\frac{e^{hH_{y}}-1}{h}=H_{y}e^{hH_{y}}$ for
all $H_{y}\geq 0$, dominated convergence yields for all $\eta<0$ that

$\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\eta}E_{x}\big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}\big{]}=E_{x}\big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}H_{\lceil
x\rceil-1}\big{]}.$ (A.8)

Then (A.1) is a consequence of the chain rule and the fact that the
expectation on the left-hand side in (A.8) is positive. Then (A.2) follows
from linearity of the derivative. To show (A.3), we have to apply dominated
convergence once more. This time, we additionally need that the expectation on
the left-hand side in (A.8) for $x=1$ is strictly bounded from below by the
constant
$E_{1}[e^{-(\texttt{es}-\texttt{ei}-\eta)H_{0}}]=e^{-\sqrt{2(\texttt{es}-\texttt{ei}-\eta)}}>0$
due to (BDD) and [4, (2.0.1), p. 204]. Using the mean value theorem and (A.7)
entail the existence of $c_{1}>0$ such that

$\sup_{|h|\leq|\eta|/2}\Big{|}\frac{1}{h}\big{(}L_{1}^{\zeta}(\eta+h)-L_{1}^{\zeta}(\eta)\big{)}\Big{|}\leq\sup_{|h|\leq|\eta|/2}(L_{1}^{\zeta})^{\prime}(\eta+h)\leq
c_{1}E_{1}\big{[}e^{\eta H_{0}/2}(H_{0}\vee 1)\big{]}\leq
c_{1}\big{(}\frac{2}{|\eta|}\vee 1\big{)}.$

Using (A.1) for $x=1$ and dominated convergence, we arrive at (A.3).

By induction and similar arguments as above, it follows that for all
$n\in\mathbb{N}$, the function $(-\infty,0)\ni\eta\mapsto
E_{x}\big{[}(H_{y})^{n}e^{\int_{0}^{H_{y}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}\big{]}$
is positive $\mathbb{P}$-a.s. and differentiable. Due to (BDD), one can
interchange expectation and differentiation in $\eta,$ yielding (A.4) and
$\eqref{eq:Lbar_x''}$. The first equality in (A.6) is then again a consequence
of dominated convergence. The strict inequalities in (A.4) and (A.6) are due
to the fact that under $P_{1},$ and thus also $\mathbb{P}$-a.s. under
$P_{1}^{\zeta,\eta}$, the random variable $H_{0}$ is non-degenerate.

To show (b), observe that the function $\zeta\mapsto
E_{x}\big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{s})+\eta)\mathop{}\\!\mathrm{d}s}\big{]}$ is
nondecreasing. Consequently, using the notation
$\triangle=[\eta_{*},\eta^{*}]$, we have

$\displaystyle-\infty$
$\displaystyle<e^{-\sqrt{2(\texttt{es}-\texttt{ei}+|\eta_{*}|})}\leq
E_{x}\big{[}e^{(\texttt{ei}-\texttt{es}+\eta_{*})H_{\lceil
x\rceil-1}}\big{]}\leq\inf_{\eta\in\triangle}\operatorname*{ess\,inf}_{\zeta}E_{x}\big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{s})+\eta)\mathop{}\\!\mathrm{d}s}\big{]}$
$\displaystyle\leq\sup_{\eta\in\triangle}\operatorname*{ess\,sup}_{\zeta}E_{x}\big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{s})+\eta)\mathop{}\\!\mathrm{d}s}\big{]}\leq
E_{x}\big{[}e^{\eta^{*}H_{\lceil x\rceil-1}}\big{]}=e^{(x-\lceil
x\rceil+1)\sqrt{2|\eta^{*}|}},$

where we used [4, (2.0.1), p. 204] for the second inequality and last
equality. Due to the inequality $e^{-xy}x\leq\frac{2}{y}e^{-xy/2}$ for all
$x\geq 0$ and $y>0$, these estimates can be used to derive similar bounds for
$E_{x}\big{[}e^{\int_{0}^{H_{\lceil
x\rceil-1}}(\zeta(B_{r})+\eta)\mathop{}\\!\mathrm{d}r}H_{\lceil
x\rceil-1}^{k}\big{]}$, $k=1,2$. Thus, estimating the numerator and the
denominator of the corresponding expressions in (A.1) to (A.6), we can
conclude. ∎

###  A.2 Exponential mixing

######  Lemma A.2.

Let $\mathcal{F}^{k}$ be as defined in (• ‣ 1.4), i.e.
$\mathcal{F}^{k}=\sigma(\xi(x):x\geq k)$, and let
$\triangle\subset(-\infty,0)$ be a compact interval. Then there exists a
constant $C_{\triangle}>0$ such that $\mathbb{P}$-a.s., for all
$i,j\in\mathbb{Z}$ with $i<j$, and all $\eta\in\triangle,$

$\displaystyle\big{|}\mathbb{E}\big{[}L_{i}^{\zeta}(\eta)|\mathcal{F}^{j}\big{]}-L(\eta)\big{|}$
$\displaystyle\leq
C_{\triangle}\cdot\Big{(}\psi\big{(}{\frac{j-i}{2}}\big{)}+e^{-(j-i)/C_{\triangle}}\Big{)},$
(A.9) $\displaystyle 0\leq\Big{(}\operatorname*{ess\,sup}_{\xi(k):k\geq
j}L_{i}^{\zeta}(\eta)\Big{)}-L_{i}^{\zeta}(\eta)$ $\displaystyle\leq
C_{\triangle}\cdot\Big{(}\psi\big{(}{\frac{j-i}{2}}\big{)}+e^{-(j-i)/C_{\triangle}}\Big{)},$
(A.10) as well as
$\displaystyle\Big{|}\mathbb{E}\big{[}(L_{i}^{\zeta})^{\prime}(\eta)|\mathcal{F}^{j}\big{]}-L^{\prime}(\eta)\Big{|}$
$\displaystyle\leq
C_{\triangle}\cdot\Big{(}\psi\big{(}{\frac{j-i}{2}}\big{)}+e^{-(j-i)/C_{\triangle}}\Big{)},$
(A.11) $\displaystyle 0\leq\Big{(}\operatorname*{ess\,sup}_{\xi(k):k\geq
j}\big{(}L_{i}^{\zeta}\big{)}^{\prime}(\eta)\Big{)}-\big{(}L_{i}^{\zeta}\big{)}^{\prime}(\eta)$
$\displaystyle\leq
C_{\triangle}\cdot\Big{(}\psi\big{(}{\frac{j-i}{2}}\big{)}+e^{-(j-i)/C_{\triangle}}\Big{)}.$
(A.12)

###### Proof.

By translation invariance, it is enough to prove (A.9) for $i=0$ and $j\geq 2$
(the case $j=1$ follows immediately from the uniform boundedness of
$L^{\zeta}_{0}$ and $L$ on $\triangle$ due to Lemma A.1). To show (A.9), let
$\eta\in\triangle$ and write $L_{0}^{\zeta}(\eta)=\ln(A+B)$ with

$\displaystyle A$
$\displaystyle=A(\eta):=E_{0}\Big{[}e^{\int_{0}^{H_{-1}}(\zeta(B_{s})+\eta)\mathop{}\\!\mathrm{d}s};\sup_{0\leq
s\leq H_{-1}}B_{s}<\lceil j/2\rceil\Big{]}$ and $\displaystyle B$
$\displaystyle=B(\eta):=E_{0}\Big{[}e^{\int_{0}^{H_{-1}}(\zeta(B_{s})+\eta)\mathop{}\\!\mathrm{d}s};\sup_{0\leq
s\leq H_{-1}}B_{s}\geq\lceil j/2\rceil\Big{]}.$

Then $A\leq E_{0}[e^{H_{-1}\eta}]\leq c_{\triangle,1}<1,$ and, using (2.3), at
the same time we have

$\displaystyle A$ $\displaystyle\geq
E_{0}\Big{[}e^{-(\texttt{es}-\texttt{ei}+|\eta|)H_{-1}};\sup_{0\leq s\leq
H_{-1}}B_{s}<\lceil j/2\rceil\Big{]}\geq c_{\triangle,2}>0\quad\text{for all
}j\geq 2.$

To bound $B$, we condition on $H_{-1}$ to happen before or after time $j$ and
use the reflection principle for Brownian motion and the tail estimate from
[7, Lemma 1.1] to infer for all $j\geq 2$

$\displaystyle 0\leq B$ $\displaystyle\leq P_{0}\Big{(}\sup_{0\leq s\leq
j}B_{s}\geq\lceil j/2\rceil\Big{)}+e^{\eta j}=2P_{0}(B_{j}\geq\lceil
j/2\rceil)+e^{\eta j}\leq 2e^{-j(1/8\wedge|\eta|)}.$

As $\ln(1+x)\leq x,$ the above implies that for all $j\geq 2$

$\ln(A)\leq
L_{0}^{\zeta}(\eta)=\ln(A)+\ln\Big{(}1+\frac{B}{A}\Big{)}\leq\ln(A)+c_{\triangle,3}e^{-j/c_{\triangle,3}}.$

Since $L_{0}^{\zeta}$ is continuous on $\triangle$, the latter display
$\mathbb{P}$-a.s. holds uniformly for all $\eta\in\triangle.$ Now $\ln(A)$ is
$\mathcal{F}_{\lceil j/2\rceil}$-measurable and bounded, so by (• ‣ 1.4)

$\displaystyle\sup_{\eta\in\triangle}\left|\mathbb{E}[L_{0}^{\zeta}(\eta)|\mathcal{F}^{j}]-L(\eta)\right|$
$\displaystyle\leq\sup_{\eta\in\triangle}\left|\mathbb{E}[\ln(A)-\mathbb{E}[\ln(A)]|\mathcal{F}^{j}]\right|+2c_{\triangle,3}e^{-j/c_{\triangle,3}}$
$\displaystyle\leq
C_{\triangle}\cdot\big{(}\psi(j/2)+e^{-j/C_{\triangle}}\big{)}.$ (A.13)

The proof of (A.11) is similar. Indeed, using the same notation we have
$(L_{0}^{\zeta})^{\prime}=\frac{A^{\prime}}{A+B}+\frac{B^{\prime}}{A+B}$. Then
by $A+B\geq c_{\triangle,2}$ and $e^{\eta
H_{-1}}H_{-1}\leq\frac{2}{|\eta|}e^{\eta H_{-1}/2}$, we can use above
calculation to conclude that $\frac{B^{\prime}}{A+B}$ decays exponentially to
$0$ as $j\to\infty$. Further,
$\frac{A^{\prime}}{A}-\frac{BA^{\prime}}{A(A+B)}=\frac{A^{\prime}}{A+B}\leq\frac{A^{\prime}}{A}$,
$F_{\lceil j/2\rceil}$-measurability of $\frac{A^{\prime}}{A}$ and above
estimates give a similar bound as in (A.13) for
$(L_{0}^{\zeta})^{\prime}(\eta)$ and $L^{\prime}(\eta)$ instead of
$(L_{0}^{\zeta})(\eta)$ and $L(\eta)$. Finally, (A.10) and (A.12) follow by
the same arguments as (A.9) and (A.11). ∎

###  A.3 Properties of the Lyapunov exponent

######  Proposition A.3.

Assume (BDD) and (STAT) and let $u^{(x,\delta)}$ be a solution to (PAM) with
initial condition $\mathds{1}_{[x-\delta,x+\delta]}$. Then the limit
$\Lambda(v):=\lim_{t\to\infty}\frac{1}{t}\ln u^{(x,\delta)}(t,vt)$,
$v\in\mathbb{R}$, exists $\mathbb{P}$-a.s., is non-random and independent of
$x\in\mathbb{R}$ and $\delta>0$. The function $[0,\infty)\ni
v\mapsto\Lambda(v)$ is concave, $\Lambda(0)=\text{\emph{$\texttt{es}$}}$ and
$\lim_{v\to\infty}\frac{\Lambda(v)}{v}=-\infty$. In particular, there exists a
unique $v_{0}>0$ such that $\Lambda(v_{0})=0$.

###### Proof.

Let $x=0$, $\lambda\in(0,1)$ and $v_{1},v_{2}\in\mathbb{R}$, and set
$v:=\lambda v_{1}+(1-\lambda)v_{2}$ and
$A:=[(1-\lambda)v_{2}t-\delta,(1-\lambda)v_{2}t+\delta]$. By (1.10), the
solution to (PAM), admits the Feynman-Kac representation
$u^{(x,\delta)}(t,vt)=E_{vt}\big{[}e^{\int_{0}^{t}\xi(B_{s})\mathop{}\\!\mathrm{d}s};B_{t}\in[x-\delta,x+\delta]\big{]}$.
Then by the Markov property, for all $y\in[vt-\delta,vt+\delta]$ we have

$\begin{split}\ln
E_{y}\big{[}e^{\int_{0}^{t}\xi(B_{s})\mathop{}\\!\mathrm{d}s};B_{t}\in[-\delta,\delta]\big{]}&\geq\ln
E_{y}\big{[}e^{\int_{0}^{\lambda
t}\xi(B_{s})\mathop{}\\!\mathrm{d}s};B_{\lambda t}\in A\big{]}\\\
&\quad+\inf_{z\in A}\ln
E_{z}\big{[}e^{\int_{0}^{(1-\lambda)t}\xi(B_{s})\mathop{}\\!\mathrm{d}s};B_{(1-\lambda)t}\in[-\delta,\delta]\big{]}.\end{split}$
(A.14)

Defining
$\overline{\mu}^{\delta}_{s,t}(v):=\inf_{y\in[vt-\delta,vt+\delta]}\ln
E_{y}\big{[}e^{\int_{0}^{t-s}\xi(B_{r})\mathop{}\\!\mathrm{d}r};B_{t-s}\in[vs-\delta,vs+\delta]\big{]}$,
$s<t$, by the same argument as in the last display one can see that
$\overline{\mu}^{\delta}_{s,t}(v)\geq\overline{\mu}^{\delta}_{s,u}(v)+\overline{\mu}^{\delta}_{u,t}(v)$
for all $s<u<t$. Furthermore for the $h$-lateral shift $\theta_{h}$ on $\xi$
(i.e. $\xi(\cdot)\circ\theta_{h}=\xi(\cdot+h)$) we have
$\overline{\mu}^{\delta}_{s,t}(v)\circ\theta_{h}=\overline{\mu}^{\delta}_{s+\frac{h}{v},t+\frac{h}{v}}(v)$
for every $h\in\mathbb{R}$, all $v\neq 0$ and $s<t$, and by (BDD) and a simple
Gaussian computation there exists $K_{\delta,v}\in(0,\infty)$ such that
$\overline{\mu}^{\delta}_{0,t}(v)\geq-K_{\delta,v}\cdot t$ for all $t\geq 1$.
By (BDD) we also have $\overline{\mu}_{0,t}^{\delta}(v)\leq\texttt{es}\cdot t$
and thus $\overline{\mu}_{0,t}^{\delta}(v)\in L^{1}$ for all $t>0$. Thus,
Kingman’s subadditive ergodic theorem [23] implies that the limit
$\Lambda^{\delta}(v):=\lim_{t\to\infty}\frac{1}{t}\overline{\mu}^{\delta}_{0,t}(v)$
exists $\mathbb{P}$-a.s. Furthermore, by (• ‣ 1.4), $\xi$ is mixing and thus
ergodic, so the $\Lambda^{\delta}(v)$ is non-random. By standard estimates,
the limit is independent of $\delta>0$ and $x\in\mathbb{R}$.

To show the concavity of $v\mapsto\Lambda(v)$, for $v_{1},v_{2}\geq 0$ and
dividing by $t$, the left-hand side of (A.14) converges $\mathbb{P}$-a.s. to
$\Lambda(\lambda v_{1}+(1-\lambda)v_{2})$, while the second term on the right-
hand side of (A.14) converges $\mathbb{P}$-a.s. to
$(1-\lambda)\Lambda(v_{2})$. Dividing by $t$ and using (STAT), the first term
on the right-hand side of (A.14) converges in distribution to the constant
$\lambda\Lambda(v_{1})$, proving the concavity of $v\mapsto\Lambda(v)$.

By [14, Theorem 7.5.2] we have $\lim_{t\to\infty}\frac{1}{t}\ln
u^{\mathds{1}_{(-\delta,\delta)}}(t,0)=\texttt{es}$, independent of
$\delta>0$, giving $\Lambda(0)=\texttt{es}$. Due to (BDD) and a standard
Gaussian computation we have $\lim_{v\to\infty}\frac{\Lambda(v)}{v}=-\infty$.
This, together with $\Lambda(0)=\texttt{es}>0$ and the concavity of
$v\mapsto\Lambda(v),$ implies the existence of a unique $v_{0}>0$, such that
$\Lambda(v_{0})=0$. ∎

###  A.4 A Hoeffding-type inequality for dependent random variables

######  Lemma A.4 ([33, Theorem 2.4]).

Let $(X_{i})_{i\in\mathbb{Z}}$ be a sequence of real-valued bounded random
variables, $\widetilde{\mathcal{F}}_{i}:=\sigma(X_{j}:j\leq i)$,
$S_{n}:=\sum_{i=1}^{n}X_{i},$ and let $(m_{1},\ldots,m_{n})$ be an $n$-tuple
of positive reals such that for all $i\in\\{1,\ldots,n\\},$

$\sup_{j\in\\{i,i+1,\ldots,n\\}}\Big{(}\|X_{i}^{2}\|_{\infty}+2\|X_{i}\sum_{k=i+1}^{j}\mathbb{E}[X_{k}|\widetilde{\mathcal{F}}_{i}]\|_{\infty}\Big{)}\leq
m_{i},$

with the convention
$\sum_{k=i+1}^{i}\mathbb{E}[X_{k}|\widetilde{\mathcal{F}}_{i}]=0$. Then for
every $x>0,$

$\mathbb{P}\left(|S_{n}|\geq
x\right)\leq\sqrt{e}\exp\left\\{-x^{2}/(2m_{1}+\cdots+2m_{n})\right\\}.$

Rearranging the quantities in the above result, we arrive at the following
corollary which we primarily pronounce explicitly since its formulation
tailor-made for our purposes.

######  Corollary A.5.

Let $(Y_{i})_{i\in\mathbb{Z}}$ be a sequence of real-valued bounded random
variables, $\widetilde{\mathcal{F}}^{k}:=\sigma(Y_{j}:j\geq k),$ and let
$(m_{1},\ldots,m_{n})$ be an $n$-tuple of positive real numbers such that for
all $i\in\\{1,\ldots,n\\},$

$\sup_{j\in\\{1,\ldots,i\\}}\Big{(}\|Y_{i}^{2}\|_{\infty}+2\Big{\|}Y_{i}\sum_{k=j}^{i-1}\mathbb{E}[Y_{k}|{\widetilde{\mathcal{F}}}^{i}]\Big{\|}_{\infty}\Big{)}\leq
m_{i},$ (A.15)

with the convention
$\sum_{k=i}^{i-1}\mathbb{E}[Y_{k}|\widetilde{\mathcal{F}}^{i}]=0$. Then for
every $x>0,$

$\mathbb{P}\Big{(}\Big{|}\sum_{i=1}^{n}Y_{i}\Big{|}\geq
x\Big{)}\leq\sqrt{e}\exp\left\\{-x^{2}/(2m_{1}+\cdots+2m_{n})\right\\}.$

Recall that $u^{u_{0}}$ denotes the solution to (PAM) with initial condition
$u_{0}$.

######  Lemma A.6.

For all $x\in\mathbb{R}$ and $0\leq s\leq t$ we have
$u^{\mathds{1}_{(-\infty,0]}}(s,x)\leq 2u^{\mathds{1}_{(-\infty,0]}}(t,x).$

###### Proof.

By the Feynman-Kac formula, $\xi\geq 0$ and the Markov property at time $s$ we
have

$\displaystyle u^{\mathds{1}_{(-\infty,0]}}(t,x)$
$\displaystyle=E_{x}\left[e^{\int_{0}^{t}\xi(B_{r})\mathop{}\\!\mathrm{d}r};B_{t}\leq
0\right]\geq
E_{x}\left[e^{\int_{0}^{s}\xi(B_{r})\mathop{}\\!\mathrm{d}r};B_{s}\leq
0,B_{t}\leq 0\right]$
$\displaystyle\geq\texttt{E}_{x}^{\xi}\left[N^{\leq}(s,0)\right]P_{0}(B_{t-s}\leq
0)=\frac{1}{2}u^{\mathds{1}_{(-\infty,0]}}(s,x).$

∎

######  Lemma A.7.

Let $N_{s,u,t}^{\mathcal{L}}$ be as in (4.1). Then there exists $c>0$ such
that $\mathbb{P}$-a.s.

$\emph{{E}}^{\xi}_{x}\big{[}N_{t,t,t}^{\mathcal{L}}\big{]}\geq
c\emph{{E}}^{\xi}_{\lfloor
x\rfloor}\big{[}N_{t-1,t+1,t}^{\mathcal{L}}\big{]}\quad\text{ for all }t\geq
1\text{ and }x\geq 1.$

###### Proof.

Write $A_{y,z,t}:=\big{\\{}H_{k}+y\geq z+t-T_{k}-5\chi_{1}(\overline{m}(t))\
\forall k\in\\{1,\ldots,\lfloor\overline{m}(t)\rfloor\\}\big{\\}}$ and
$p(r):=\inf_{y\leq 0}P_{y}(B_{r}\leq 0)=\frac{1}{2}$. Then by the Feynman-Kac
formula

$\displaystyle\texttt{E}^{\xi}_{x}\big{[}N_{t,t,t}^{\mathcal{L}}\big{]}$
$\displaystyle=E_{x}\big{[}e^{\int_{0}^{t}\xi(B_{s})\mathop{}\\!\mathrm{d}s};B_{t}\leq
0,A_{0,0,t}\big{]}$ $\displaystyle\geq E_{x}\Big{[}e^{\int_{H_{\lfloor
x\rfloor}}^{t-1+H_{\lfloor
x\rfloor}}\xi(B_{s})\mathop{}\\!\mathrm{d}s};B_{t}\leq 0,B_{t-1+H_{\lfloor
x\rfloor}}\leq 0,A_{H_{\lfloor x\rfloor},1,t},H_{\lfloor x\rfloor}\leq
1\Big{]}$ $\displaystyle\geq E_{x}\Big{[}E_{\lfloor
x\rfloor}\big{[}e^{\int_{0}^{t-1}\xi(B_{s})\mathop{}\\!\mathrm{d}s}\times\inf_{r\in[0,1]}p(r);B_{t-1}\leq
0,A_{0,1,t}\big{]};H_{\lfloor x\rfloor}\leq 1\Big{]}$
$\displaystyle\geq\frac{1}{2}P_{x}(H_{\lfloor x\rfloor}\leq 1)E_{\lfloor
x\rfloor}\big{[}e^{\int_{0}^{t-1}\xi(B_{s})\mathop{}\\!\mathrm{d}s};B_{t-1}\leq
0,A_{0,1,t}\big{]}=c\texttt{E}^{\xi}_{\lfloor
x\rfloor}\big{[}N_{t-1,t+1,t}^{\mathcal{L}}\big{]},$

with $c:=\frac{1}{2}P_{x}(H_{\lfloor x\rfloor}\leq 1)$. ∎

The next result is an inequality of Harnack-type flavor for the solution to
(PAM).

######  Lemma A.8.

There exists a constant $\Cl{Harnack}\in(0,\infty)$ such that
$\mathbb{P}$-a.s. for all $y\in\mathbb{R}$, $t\geq 1$ and all
$u_{0}\in\mathcal{I}_{\text{PAM}}$ we have

$u^{u_{0}}(t,y)\leq\Cr{Harnack}\inf_{x\in[y-1,y+1]}u^{u_{0}}(t+1,x).$ (A.16)

###### Proof.

For $x\in\mathbb{R}$ and $t>0$ let $f_{t,x}$ be the probability density of a
Brownian motion at time $t$, starting in $x$. Let us first show that for all
$y,z\in\mathbb{R}$ we have

$f_{1,y}(z)\leq\sqrt{2/e}\inf_{x\in[y-1,y+1]}f_{2,x}(z).$ (A.17)

Indeed, using $f_{t,y}(z)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{(z-y)^{2}}{2t}}$,
(A.17) follows from

$\displaystyle\inf_{x\in[y-1,y+1]}\left\\{2(z-y)^{2}-(z-x)^{2}\right\\}$
$\displaystyle=\begin{cases}2(z-y)^{2}-(z-y+1)^{2}=(z-y-1)^{2}-2,&\quad
y<z,\\\ 2(z-y)^{2}-(z-y-1)^{2}=(z-y+1)^{2}-2,&\quad y\geq z,\\\ \end{cases}$
$\displaystyle\geq-2.$

The Feynman-Kac formula in combination with the Markov property applied at
time $1$ now supplies us with

$\displaystyle u^{u_{0}}(t,y)$
$\displaystyle=E_{y}\big{[}e^{\int_{0}^{t}\xi(B_{r})\mathop{}\\!\mathrm{d}r}u_{0}(B_{t})\big{]}\leq
e^{\texttt{es}}\int_{\mathbb{R}}f_{1,y}(z)E_{z}\big{[}e^{\int_{0}^{t-1}\xi(B_{r})\mathop{}\\!\mathrm{d}r}u_{0}(B_{t-1})\big{]}\mathop{}\\!\mathrm{d}z$
$\displaystyle\leq
e^{\texttt{es}}\sqrt{2/e}\inf_{x\in[y-1,y+1]}\int_{\mathbb{R}}f_{2,x}(z)E_{z}\big{[}e^{\int_{0}^{t-1}\xi(B_{r})\mathop{}\\!\mathrm{d}r}u_{0}(B_{t-1})\big{]}\mathop{}\\!\mathrm{d}z$
$\displaystyle\leq
e^{\texttt{es}}\sqrt{2/e}\inf_{x\in[y-1,y+1]}u^{u_{0}}(t+1,x),$

where in the second inequality we used (A.17). Then (A.16) follows choosing
$\Cr{Harnack}=e^{\texttt{es}}\sqrt{2/e}$. ∎

###  A.5 Proof of the McKean representation

###### Proof of Proposition 2.1.

We start with considering $w_{0}$ continuous and $w_{0}(x)\in[0,1]$ for all
$x\in\mathbb{R}$. Define $u:=1-w$ and $u_{0}:=1-w_{0}$. Then by [21, Remark
4.4.4] (providing the existence of a $C^{1,2}$-solution to (PAM)) and [21,
Corollary 4.4.5] (Feynman-Kac representation), the function
$u^{(1)}(t,x)=E_{x}\big{[}e^{-\int_{0}^{t}\xi(B_{r})\mathop{}\\!\mathrm{d}r}u_{0}(B_{t})\big{]}$
is a classical solution to $u^{(1)}_{t}(t,x)=\frac{1}{2}\Delta
u^{(1)}(t,x)-\xi(x)u^{(1)}(t,x)$ and we have $u^{(1)}(0,x)=u_{0}(x)$.
Furthermore, by the same argument, for all $s<t$ and $k\in\mathbb{N}$ fixed,
the function

$\displaystyle u^{(2)}(t,x;s,k)$
$\displaystyle:=E_{x}\Big{[}\xi(B_{t-s})e^{-\int_{0}^{t-s}\xi(B_{r})\mathop{}\\!\mathrm{d}r}u(s,B_{t-s})^{k}\Big{]}$

is a classical solution to $u^{(2)}_{t}(t,x;s,k)=\frac{1}{2}\Delta
u^{(2)}(t,x;s,k)-\xi(x)u^{(2)}(t,x;s,k)$. By dominated convergence, taking
advantage of the uniform continuity of the first and second order derivatives,
for every fixed $t^{\prime}\leq t$, we can interchange limits to obtain the
identities

$\displaystyle\begin{split}\sum_{k}p_{k}\int_{0}^{t^{\prime}}\frac{\partial}{\partial
t}u^{(2)}(t,x;s,k)\mathop{}\\!\mathrm{d}s&=\frac{\partial}{\partial
t}\sum_{k}p_{k}\int_{0}^{t^{\prime}}u^{(2)}(t,x;s,k)\mathop{}\\!\mathrm{d}s,\\\
\sum_{k}p_{k}\int_{0}^{t^{\prime}}\big{(}\frac{1}{2}\Delta-\xi(x)\big{)}u^{(2)}(t,x;s,k)\mathop{}\\!\mathrm{d}s&=\big{(}\frac{1}{2}\Delta-\xi(x)\big{)}\sum_{k}p_{k}\int_{0}^{t^{\prime}}u^{(2)}(t,x;s,k)\mathop{}\\!\mathrm{d}s.\end{split}$
(A.18)

Conditioning on the first splitting time of the initial particle and on how
many children are born, we have

$\displaystyle u(t,x)$
$\displaystyle=E_{x}\big{[}e^{-\int_{0}^{t}\xi(B_{r})\mathop{}\\!\mathrm{d}r}u_{0}(B_{t})\big{]}+\sum_{k}p_{k}\int_{0}^{t}E_{x}\Big{[}\xi(B_{s})e^{-\int_{0}^{s}\xi(B_{s})}u(t-s,B_{s})^{k}\Big{]}\mathop{}\\!\mathrm{d}s$
$\displaystyle=u^{(1)}(t,x)+\sum_{k}p_{k}\int_{0}^{t}E_{x}\Big{[}\xi(B_{t-s})e^{-\int_{0}^{t-s}\xi(B_{r})\mathop{}\\!\mathrm{d}r}u(s,B_{t-s})^{k}\Big{]}\mathop{}\\!\mathrm{d}s,$

where we used the Markov property of the process $(B_{t})_{t\geq 0}$. We
deduce that for $h\neq 0,$

$\displaystyle\frac{1}{h}\big{(}u(t+h,x)-u(t,x)\big{)}$
$\displaystyle=\frac{1}{h}\big{(}u^{(1)}(t+h,x)-u^{(1)}(t,x)\big{)}$
$\displaystyle\quad+\sum_{k}p_{k}E_{x}\Big{[}\frac{1}{h}\int_{t}^{t+h}\xi(B_{t+h-s})e^{-\int_{0}^{t+h-s}\xi(B_{r})\mathop{}\\!\mathrm{d}r}u(s,B_{t+h-s})^{k}\mathop{}\\!\mathrm{d}s\Big{]}$
$\displaystyle\quad+\sum_{k}p_{k}\int_{0}^{t}\frac{1}{h}\big{(}u^{(2)}(t+h,x;s,k)-u^{(2)}(t,x;s,k)\big{)}\mathop{}\\!\mathrm{d}s.$

Invoking once again the arguments from the beginning of the proof, as $h$
tends to zero, the first summand converges to $\frac{1}{2}\Delta
u^{(1)}(t,x)-\xi(x)u^{(1)}(t,x)$ and by dominated convergence, the second
summand converges to $\xi(x)\sum_{k}p_{k}u(t,x)^{k}$. For the third term, we
mention that the integrand is uniformly bounded since $u^{(2)}_{t}$ is
continuous. Again, due to dominated convergence and (A.18), the latter term
converges to

$\displaystyle\big{(}\frac{1}{2}\Delta-\xi(x)\big{)}\sum_{k}p_{k}\int_{0}^{t}u^{(2)}(t,x;s,k)\mathop{}\\!\mathrm{d}s.$

All in all, combining these observations we arrive at

$\displaystyle u_{t}(t,x)$
$\displaystyle=u^{(1)}_{t}(t,x)+\xi(x)\sum_{k}p_{k}u(t,x)^{k}+\big{(}\frac{1}{2}\Delta-\xi(x)\big{)}\sum_{k}p_{k}\int_{0}^{t}u^{(2)}(t,x;s,k)\mathop{}\\!\mathrm{d}s$
$\displaystyle=\big{(}\frac{1}{2}\Delta-\xi(x)\big{)}u(t,x)+\xi(x)\sum_{k}p_{k}u(t,x)^{k},$

which is equivalent $w$ being a solution to (F-KPP). Thus we have shown the
claim for continuous $w_{0}$. By Remark 1.1, for
$w_{0}\in\mathcal{I}_{\text{F-KPP}}$ the result follows by approximation. ∎

###  A.6 Existence and monotonicity of solutions

We start with an existence result for generalized and also classical
solutions.

######  Proposition A.9 ([14, Theorem 7.4.1]).

Let $w_{0}$ be measurable, non-negative and bounded, and let $F$ fulfill (SC).
Then $\mathbb{P}$-a.s. there exists a unique, bounded generalized solution
$w\in C\big{(}(0,\infty)\times\mathbb{R}\big{)}$ to (F-KPP), which satisfies
$0\leq w(t,x)\leq 1\vee\sup_{x}w_{0}(x)$ for all
$(t,x)\in[0,\infty)\times\mathbb{R}$. If additionally $w_{0}$ is continuous
and $x\mapsto\xi(x,\omega)$ is Hölder continuous for $\mathbb{P}$-a.a.
$\omega\in\Omega$, then $\mathbb{P}$-a.s. the generalized solution $w$ is a
classical one, i.e. $w\in C^{1,2}\big{(}(0,\infty)\times\mathbb{R}\big{)}\cap
C\big{(}[0,\infty)\times\mathbb{R}\big{)}$ and $w$ solves (F-KPP).

For the next lemma, we introduce the differential operator

$(\mathcal{L}_{G}w)(t,x):=w_{t}(t,x)-\frac{1}{2}w_{xx}(t,x)-G(x,w(t,x)),$

where $G$ is uniformly Lipschitz-continuous in $w$, i.e., there exists a
constant $\alpha>0$, such that

$|G(x,u)-G(x,v)|\leq\alpha|u-v|\quad\forall\ x,u,v\in\mathbb{R}.$ (A.19)

The next lemma is in the spirit of [1, Proposition 2.1].

######  Lemma A.10.

Let $T>0$, $Q:=(0,T)\times\mathbb{R}$ and $G$ be such that (A.19) holds. Let
$w^{(1)}$ and $w^{(2)}$ be non-negative and bounded functions on
$\overline{Q}$, such that for $i\in\\{1,2\\}$, $w^{(i)}_{x}$ and
$w^{(i)}_{xx}$ are continuous on $Q,$ and such that $w^{(i)}_{t}$ exists in
$Q.$ If $\mathcal{L}_{G}w^{(1)}\leq\mathcal{L}_{G}w^{(2)}$ on $Q$ and $0\leq
w^{(1)}(0,x)\leq w^{(2)}(0,x)$ for all $x\in\mathbb{R}$, then also
$w^{(1)}\leq w^{(2)}$ on $Q.$

###### Proof.

From the assumptions we infer that

$\displaystyle w^{(2)}_{t}(t,x)-w^{(1)}_{t}(t,x)$
$\displaystyle\geq\frac{1}{2}(w^{(2)}_{xx}(t,x)-w^{(1)}_{xx}(t,x))+G(x,w^{(2)}(t,x))-G(x,w^{(1)}(t,x)).$

Then, recalling (A.19) and letting

$v(t,x):=e^{-2\alpha t}\big{(}w^{(2)}(t,x)-w^{(1)}(t,x)\big{)},$

we get

$\displaystyle v_{t}(t,x)-\frac{1}{2}v_{xx}(t,x)$ $\displaystyle\geq-2\alpha
v(t,x)+e^{-2\alpha t}\big{(}G(x,w^{(2)}(t,x))-G(x,w^{(1)}(t,x))\big{)}$
$\displaystyle\geq\big{(}-2-\text{sgn}(v(t,x))\big{)}\alpha\cdot v(t,x).$

Now the the first factor on the right-hand side is negative and bounded.
Applying the maximum principle [25, Theorem 8.1.4] to $-v$ then implies that
$v\geq 0$ on $Q$ and we can conclude. ∎

As an important application we get that the solution $w$ to
$\mathcal{L}_{G}w=0$ is monotone in $G$ and in the initial condition.

######  Corollary A.11.

Let $T>0$, $Q:=(0,T)\times\mathbb{R}$, and let $G_{1}$ and $G_{2}$ fulfill
${G_{1}}\leq{G_{2}}$ on $\mathbb{R}\times[0,\infty).$ Furthermore, assume that
$G_{2}$ satisfies (A.19). In addition, let $w^{(1)}$ and $w^{(2)}$ be non-
negative and bounded functions on $\overline{Q}$, such that for
$i\in\\{1,2\\}$, $w^{(i)}_{x}$ and $w^{(i)}_{xx}$ are continuous on $Q$ and
$w^{(i)}_{t}$ exist on $Q$. If
$\mathcal{L}_{G_{1}}w^{(1)}=\mathcal{L}_{G_{2}}w^{(2)}$ and
$w^{(1)}(0,\cdot)\leq w^{(2)}(0,\cdot)$ on $x\in\mathbb{R}$, then we have
$w^{(1)}\leq w^{(2)}$ on $Q.$

###### Proof.

Since the function $w^{(2)}$ is non-negative, we have
$\mathcal{L}_{G_{2}}w^{(2)}=\mathcal{L}_{G_{1}}w^{(1)}\geq\mathcal{L}_{G_{2}}w^{(1)}$.
Then by Lemma A.10, we have $w^{(1)}(t,x)\leq w^{(2)}(t,x)$ for all $(t,x)\in
Q$. ∎

######  Corollary A.12.

Let $G$ fulfill (A.19) and $G(x,0)=G(x,1)=0$ for all $x\in\mathbb{R}$. Let $w$
be a solution to $\mathcal{L}_{G}w=0$ with $0\leq w(0,x)\leq 1$. Then $0\leq
w(t,x)\leq 1$ for all $(t,x)\in[0,\infty)\times\mathbb{R}$.

###### Proof.

The functions $w^{(1)}(t,x)=0$ and $w^{(2)}(t,x)=1$ are solutions to
$\mathcal{L}w^{(1)}=\mathcal{L}w^{(2)}=0$ and $w^{(1)}(0,x)\leq w(0,x)\leq
w^{(2)}(0,x)$. The claim then follows from Lemma A.10. ∎

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