# Moderate deviations
for functionals over infinitely many
Rademacher random variables

Marius Butzek  Faculty of Mathematics, Ruhr University Bochum, Germany.
<EMAIL_ADDRESS>,  Peter Eichelsbacher  Faculty of Mathematics, Ruhr
University Bochum, Germany<EMAIL_ADDRESS>and  Benedikt Rednoß
Faculty of Mathematics, Ruhr University Bochum, Germany.
<EMAIL_ADDRESS>
###### Abstract.

In this paper, moderate deviations for normal approximation of functionals
over infinitely many Rademacher random variables are derived. They are based
on a bound for the Kolmogorov distance between a general Rademacher functional
and a Gaussian random variable, continued by an intensive study of the
behavior of operators from the Malliavin–Stein method along with the moment
generating function of the mentioned functional. As applications, subgraph
counting in the Erdős–Rényi random graph and infinite weighted 2-runs are
studied.

##  1\. Introduction and applications

The theory of moderate deviations goes back to H. Cramér in 1938: Knowing that
for an independently identically distributed (iid) sequence
$(X_{k})_{k\in\mathbb{N}}$ of random variables such that
$\mathbb{E}[X_{k}]=0,\mathbb{E}(X_{k}^{2})=1,W_{n}:=n^{-1/2}(X_{1}+...+X_{n})$
and $\Phi$ the distribution function of a standard normal random variable
$Z\sim\mathcal{N}(0,1)$, the statement

$\displaystyle\frac{\mathbb{P}(W_{n}>x)}{1-\Phi(x)}\longrightarrow
1\quad\text{for}\quad x=O(1)$ (1.1)

is valid, he was asking what happens if $x$ depends on $n\in\mathbb{N}$ such
that $x\rightarrow\infty$ for $n\rightarrow\infty$? Can you find an interval
such that (1.1) holds for $0\leq x\leq I(n),I(n)\rightarrow\infty$? The answer
was given by himself: Under the assumption
$\mathbb{E}[e^{t_{0}X_{1}}]<\infty\;\forall\;0\leq t\leq t_{0}$,

$\displaystyle\frac{\mathbb{P}(W_{n}>x)}{1-\Phi(x)}=1+O(1)n^{-1/2}(1+x^{3})\quad\text{for}\quad
0\leq x\leq n^{1/6},$ (1.2)

(see e.g. [6] and [16]). Reminiscent of (1.2) for a sequence
$(Y_{n})_{n\in\mathbb{N}}$ of random variables, such that
$Y_{n}\overset{d}{\rightarrow}Y$, the moderate deviation of Cramér-type is
given by

$\displaystyle\frac{\mathbb{P}(Y_{n}>x)}{\mathbb{P}(Y>x)}=1+\text{error
term}\rightarrow 1$

for $0\leq x\leq a_{n}$, where $a_{n}\rightarrow\infty$ for
$n\rightarrow\infty$.
In [23] Zhang was able to prove Cramér-type moderate deviations for unbounded
exchangeable pairs. He developed them by stopping the proof of the
corresponding Berry–Esseen-type inequalities, he had obtained before by
Stein’s method, at a certain point and continuing differently. Stein’s method
is a powerful tool by itself to derive upper bounds for differences of
probability distributions, originally for the normal distribution and later
extended to other distributions. Zhang rearranged the fragments of the so
called Stein-equation and the bound of its solution, a technique that was
already seen in [4], [10], [19] and [21].
Our ambition is to prove a similar result for functionals over infinitely many
independent Rademacher random variables taking values +1 and -1 only. For
these functionals a Kolmogorov bound in the context of normal approximation
was shown recently in [9, Theorem 3.1] such that the bounding terms can be
expressed in terms of operators of the so called Malliavin–Stein method.
Normal approximation of functionals over infinitely many Rademacher random
variables was derived already in [15], [12], [13] and [7]. Theorem 3.1 in [9]
will be our starting point.

###  1.1. Application to infinite weighted 2-runs

To begin with, we introduce some of the possible applications for our theorem.
In what follows, we use the usual big-O notation $O(.)$ with the meaning that
the implicit constant does not depend on the parameters in brackets. For a
sequence $a=(a_{i})_{i\in\mathbb{Z}}$ and $p>0$ we write
$\left\|a\right\|_{l^{p}(\mathbb{Z})}:=\left(\sum_{i\in\mathbb{Z}}\left|a_{i}\right|^{p}\right)^{1/p}$.

Let $X=(X_{i})_{i\in\mathbb{Z}}$ be a double sided sequence of i.i.d.
Rademacher random variables such that
$\mathbb{P}(X_{i}=1)=\mathbb{P}(X_{i}=-1)=1/2$ and let for each
$n\in\mathbb{N},(a^{(n)}_{i})_{i\in\mathbb{Z}}$ be a double-sided summable
111Usually 2-runs are definded with a square-summable sequence but this is not
enough here. sequence of real numbers.
The sequence $(F_{n})_{n\in\mathbb{N}}$ of standardized infinte weighted
2-runs is then defined as

$\displaystyle
F_{n}:=\frac{G_{n}-\mathbb{E}[G_{n}]}{\sqrt{\operatorname{Var}(G_{n})}},\quad
G_{n}:=\sum_{i\in\mathbb{Z}}a^{(n)}_{i}\xi_{i}\xi_{i+1},\quad n\in\mathbb{N},$

where $\xi_{i}:=(X_{i}+1)/2$ for $i\in\mathbb{Z}$.
As a motivation you can say that $G_{n}$ counts the weighted number of
subsequences of 1’s of length two in an infinite double-sided sequence of
Bernoulli trials.

######  Theorem 1.1.

Recall the definition of $F_{n}$ from above. Then

$\displaystyle\frac{\mathbb{P}(F_{n}>z)}{1-\Phi(z)}=1+O(1)(1+z^{2})\gamma_{n}(z),$
(1.3)

for $0\leq
z\leq\min\\{C_{n}^{-1/5},C_{n}^{-1/3},C_{n}^{-2/5},\operatorname{Var}(G_{n})^{1/2}\\}$
such that $(1+z^{2})\gamma_{n}(z)\leq 1$, where O(1) is bounded by a constant
only depending on the coefficient sequence $(a^{(n)}_{i})_{i\in\mathbb{Z}}$
and

$\displaystyle\gamma_{n}(z)$
$\displaystyle:=e^{O(1)z(\operatorname{Var}(G_{n}))^{-1/2}}\\!\left(z^{1/2}C_{n}^{1/2}+(1+z^{1/2}+z)C_{n}\right),\quad
C_{n}:=\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{2}}{(\operatorname{Var}(G_{n}))}.$

######  Remark 1.2.

Note that in general
$\operatorname{Var}(G_{n})=O\\!\left(\left\|a^{(n)}\right\|^{2}_{l^{2}(\mathbb{Z})}\right)$,
but to make (1.3) tend to 0 the condition $C_{n}\rightarrow 0$ for
$n\rightarrow\infty$ is sufficient. And in that case the upper bound of the
range is increasing in $n$.

######  Example 1.3.

We consider $a^{(n)}_{i}=\mathbbl{1}_{\left\\{1,...,n\right\\}}(i)\,\forall
i\in\mathbb{Z}$, which is obviously a summable sequence. Then
$\operatorname{Var}(G_{n})=O\\!\left(n\right)$ and $C_{n}=n^{-1/2}$. The
Moderate Deviation we get is

$\displaystyle\frac{\mathbb{P}(F_{n}>z)}{1-\Phi(z)}=1+O(1)(1+z^{2})\gamma_{n}(z),$
(1.4)

for $0\leq z\leq n^{1/10}$ such that $(1+z^{2})\gamma_{n}(z)\leq 1$, where

$\displaystyle\gamma_{n}(z)$
$\displaystyle:=e^{O(1)zn^{-1/2}}\\!\left(z^{1/2}n^{-1/4}+(1+z^{1/2}+z)n^{-1/2}\right).$

###  1.2. Application to subgraph counts in the Erdős–Rényi random graph

As a further application we derive a Cramér-type moderate deviation result for
the subgraph counting statistic in the Erdős–Rényi random graph $G(n,p)$.

Consider a random graph on $n\in\mathbb{N}$ vertices. Each possible edge
between two vertices is included with probability $p\in(0,1)$ independently of
all other edges, where $p$ may depend on $n$ even though we will not make this
visible in our notation.

Let ${G_{0}}$ be a fixed graph with at least one edge. A subgraph $H\subset
G(n,p)$ is called a _copy_ of ${G_{0}}$ in $G(n,p)$ if it is isomorphic to
${G_{0}}$. Note that we are calling two graphs $G_{1}=(V_{1},E_{1})$ and
$G_{2}=(V_{2},E_{2})$ _isomorphic_ , if there is an edge-preserving bijection
$f:V_{1}\rightarrow V_{2}$ between their sets of vertices, such that two
vertices $v,w\in V_{1}$ are joined by an edge $\\{v,w\\}\in E_{1}$ in $G_{1}$
if and only if the vertices $f(v),f(w)\in V_{2}$ are joined by an edge
$\\{f(v),f(w)\\}\in E_{2}$ in $G_{2}$.

We are interested in the standardized number $W$ of copies of ${G_{0}}$ in
$G(n,p)$.

It is well known under which necessary and sufficient assumption $W$ is
asymptotically normal. To state this condition and further results, we will
use the notation $q:=1-p$ as well as

$\displaystyle\Psi_{\min}:=\Psi_{\min}({G_{0}}):=\min_{\begin{subarray}{c}H\subset{G_{0}}\\\
e_{H}\geq 1\end{subarray}}\\{n^{v_{H}}\cdot p^{e_{H}}\\},$

with $H\subset{G_{0}}$ a subgraph of ${G_{0}}$, $v_{H}$ the number of vertices
of $H$, and $e_{H}$ the number of edges of $H$. Further, $a_{H}$ denotes the
number of automorphisms of $H$. Some very important yet easy to prove bounds
for $\Psi_{\min}$ are

$\displaystyle n^{2}p^{e_{{G_{0}}}}\leq\Psi_{\min}\leq n^{2}p.$

Theorem 2 in [20] states that $W$ is asymptotically normal if and only if

$\displaystyle q\cdot\Psi_{\min}\xlongrightarrow{n\to\infty}\infty$

The best known convergence rate regarding the bounded Wasserstein distance
$d_{\mathrm{bWass}}$ is presented in Theorem 2 in [3]: There is a constant
$C_{G_{0}}$ only depending on ${G_{0}}$ so that

$\displaystyle d_{\mathrm{bWass}}(W,N)$
$\displaystyle\leq\frac{C_{G_{0}}}{\sqrt{q\Psi_{\min}}},$

where $N\sim\mathcal{N}(0,1)$ is standard normal distributed. The same bound
holds true for the Wasserstein distance, which can be shown by only a slight
modification of the original proof in [3], and for the Kolmogorov distance,
which was a long standing problem that has been lately solved using different
approaches by [18], [22], [8] and [9].

A refined Cramér-type moderate deviation result for the subgraph counting
statistic in the Erdős–Rényi random graph is shown in Proposition 2.3 in [10]:
For any fixed constant $c_{0}>0$ there is a constant $C_{c_{0},{G_{0}}}$ only
depending on $c_{0}$ and ${G_{0}}$, so that

$\displaystyle\max\bigg{\\{}\bigg{|}\frac{\mathbb{P}(W<-t)}{\Phi(-t)\exp(-\frac{\gamma
t^{3}}{6})}-1\bigg{|},\bigg{|}\frac{\mathbb{P}(W>t)}{(1-\Phi(t))\exp(\frac{\gamma
t^{3}}{6})}-1\bigg{|}\bigg{\\}}\leq
C_{c_{0},{G_{0}}}\cdot\bigg{(}\frac{\Psi_{\min}}{q\cdot
p^{2e_{{G_{0}}}}}\bigg{)}^{\frac{5}{2}}\cdot\frac{1+t^{2}}{n^{6}}$

for

$\displaystyle 0\leq t\leq c_{0}\cdot\bigg{(}\frac{q\cdot
p^{2e_{{G_{0}}}}}{\Psi_{\min}}\bigg{)}^{\frac{5}{4}}\cdot n^{3},$

where $\Phi$ is the distribution function of the standard normal distribution,
and $\gamma=\mathbb{E}[W^{3}]$.

Another Cramér-type moderate deviation result is known from Theorem 3.1 in
[22]: There is a constant $C_{G_{0}}$ only depending on ${G_{0}}$, so that

$\displaystyle\bigg{|}\frac{\mathbb{P}(W>t)}{1-\Phi(t)}-1\bigg{|}\leq
C_{G_{0}}\cdot(1+t^{2})\cdot b_{n}(p,t)$ (1.5)

with

$\displaystyle b_{n}(p,t):={}$
$\displaystyle\begin{cases}\displaystyle\frac{1+t}{\sqrt{\Psi_{\min}}},&0<p<\frac{1}{2},\\\\[15.00002pt]
\displaystyle\frac{1+\frac{t}{\sqrt{q}}}{n\sqrt{p}},&\frac{1}{2}<p<1,\end{cases}$

for

$\displaystyle 0\leq t\leq\frac{q\cdot
p^{e_{{G_{0}}}}}{\sqrt{\Psi_{\min}}}\cdot n^{2}$

such that

$\displaystyle(1+t^{2})b_{n}(p,t)\leq 1.$

For any subgraph $H\subset{G_{0}}$ with at least one edge but more than two
vertices, there is $n^{v_{H}}p^{e_{H}}\geq
n^{2}p\cdot\frac{n^{v_{H}-2}}{2^{e_{H}-1}}\geq
n^{2}p\cdot\frac{n}{2^{e_{{G_{0}}}-1}}$ in case of $\frac{1}{2}<p<1$. Hence,
we know that $\Psi_{\min}=n^{2}p$ for $\frac{1}{2}<p<1$ and $n\geq
2^{e_{{G_{0}}}-1}$. We further know that
$\frac{1+t}{\sqrt{\Psi_{\min}}}\leq\frac{1+\frac{t}{\sqrt{q}}}{\sqrt{q\Psi_{\min}}}\leq
2\cdot\frac{1+t}{\sqrt{\Psi_{\min}}}$ in case of $0<p<\frac{1}{2}$. Therefore,
as long as $n\geq 2^{e_{{G_{0}}}-1}$, (1.5) can be rewritten without loss of
sharpness in the resulting rate:

$\displaystyle\bigg{|}\frac{\mathbb{P}(W>t)}{1-\Phi(t)}-1\bigg{|}$
$\displaystyle\leq 2\cdot
C_{G_{0}}\cdot\frac{(1+t^{2})\big{(}1+\frac{t}{\sqrt{q}}\big{)}}{\sqrt{q\Psi_{\min}}}.$
(1.6)

With our approach we are able to improve on the result of [22]. We will prove
the following result.

######  Theorem 1.4 (Subgraph counts in the Erdős–Rényi random graph — general
result).

Let ${G_{0}}$ be a graph with at least one edge. Let $W$ be the standardized
number of copies of ${G_{0}}$ in the Erdős–Rényi random graph $G(n,p)$. And
assume that $n\geq 4v_{{G_{0}}}^{2}$. Then for all $t\geq 0$ there is

$\displaystyle\bigg{|}\frac{\mathbb{P}(W>t)}{1-\Phi(t)}-1\bigg{|}\leq
50c_{G_{0}}\cdot\exp\Big{(}c_{G_{0}}\cdot t^{2}\cdot
s(t)\Big{)}\cdot(1+t^{2})\cdot s(t),$

with

$\displaystyle s(t)$
$\displaystyle=\bigg{(}1+\frac{t}{\min\\{\sqrt{\Psi_{\min}},1\\}}\bigg{)}\frac{1}{\sqrt{q\Psi_{\min}}}\cdot\exp\bigg{(}\frac{5Dt}{\sigma}\bigg{)}$
(1.7) and $\displaystyle c_{G_{0}}$
$\displaystyle=\frac{2^{\frac{9}{2}+\frac{15}{2}e_{{G_{0}}}}\cdot(v_{{G_{0}}}!)^{4}\cdot
e_{{G_{0}}}}{a_{{G_{0}}}^{\frac{3}{2}}}.$ (1.8)

The result holds for all $t\geq 0$. Similar to the approach by [22], the
result can be simplified by restricting $t$ to be smaller than a suitable
bound:

######  Corollary 1.5 (Subgraph counts in the Erdős–Rényi random graph —
bounded domain).

Let ${G_{0}}$ be a graph with at least one edge. Let $W$ be the standardized
number of copies of ${G_{0}}$ in the Erdős–Rényi random graph $G(n,p)$. Let
$c_{1},c_{2}>0$ be positive numbers. Assume that

$\displaystyle n$ $\displaystyle\geq 4v_{{G_{0}}}^{2},$ $\displaystyle 0\leq
t$ $\displaystyle\leq c_{1}\cdot\frac{n^{2}\cdot
p^{e_{{G_{0}}}}\cdot\sqrt{q}}{\sqrt{\Psi_{\min}}},$ $\displaystyle t^{2}\cdot
s(t)$ $\displaystyle\leq c_{2},$

with $s(t)$ as given in (1.7). Then

$\displaystyle\bigg{|}\frac{\mathbb{P}(W>t)}{1-\Phi(t)}-1\bigg{|}\leq
C_{c_{1},c_{2},{G_{0}}}\cdot\frac{1+t^{3}}{\sqrt{q\Psi_{\min}}},$

where

$\displaystyle C_{c_{1},c_{2},{G_{0}}}$ $\displaystyle=100(1+c_{1})\cdot
c_{G_{0}}\cdot e^{5c_{1}\hat{c}_{G_{0}}+c_{2}c_{G_{0}}},$

with $\hat{c}_{G_{0}}=\sqrt{2}\cdot\frac{\sqrt{v_{{G_{0}}}!}\cdot
v_{{G_{0}}}^{2}\cdot e_{{G_{0}}}}{\sqrt{a_{{G_{0}}}}}$ and $c_{G_{0}}$ as
given in (1.8).

In case of $\liminf_{n\rightarrow\infty}q>0$, our result is of the same order
as the result of [22] presented in (1.6). However, in case of
$\liminf_{n\rightarrow\infty}q=0$, our result yields the better rate.

We give an overview how the remaining parts of this paper are structured. In
Section 2 we list important operators used in Malliavin calculus and give a
short introduction to Stein’s method. The new moderate deviations for general
non-linear functionals of possibly infinite Rademacher random variables are
presented in Section 3. The proof of Theorem 1.1 is shown in Section 4 and the
proofs of Theorem 1.4 and Corollary 1.5 follow in Section 5.

##  2\. Preliminaries

In this section we list all the definitions and notions we deal with, in
particular the operators from Malliavin–Stein method. Since this is just a
summary, we refer to [14] for details and to [1] for further results related
to the topic. We start with $X=(X_{k})_{k\in\mathbb{N}}$, a sequence of
Rademacher random variables, e.g. $\forall k\in\mathbb{N}$:

$\displaystyle\mathbb{P}(X_{k}=1)$ $\displaystyle=p_{k}\in(0,1)$
$\displaystyle\mathbb{P}(X_{k}=-1)$ $\displaystyle=q_{k}=1-p_{k},$

and, if needed, the standardized random variable

$\displaystyle Y_{k}=\frac{X_{k}-p_{k}+q_{k}}{2\sqrt{q_{k}p_{k}}}.$

We are interested in random variables $F\in
L^{2}(\Omega,\sigma(X),\mathbb{P})\subset L^{1}(\Omega,\sigma(X),\mathbb{P})$.
Writing $F=f(X)=f(X_{1},X_{2},...)$ we define the Discrete Gradient $D_{k}F$
of $F$ at $k$th coordinate:

$\displaystyle D_{k}F$
$\displaystyle:=\sqrt{p_{k}q_{k}}(F_{k}^{+}-F_{k}^{-}),$ $\displaystyle DF$
$\displaystyle:=(D_{1}F,D_{2}F,...),$

where $F_{k}^{+}:=f(X_{1},...,X_{k-1},+1,X_{k+1},...)$ and
$F_{k}^{-}:=f(X_{1},...,X_{k-1},-1,X_{k+1},...),\,k\in\mathbb{N}$. For the
main result we will consider

$\displaystyle F\in\mathbb{D}^{1,2}:=\\{F\in
L^{2}(\Omega,\sigma(X),\mathbb{P})|\,\mathbb{E}[\left\|DF\right\|_{l^{2}(\mathbb{N})}^{2}]<\infty\\},$

where

$\displaystyle\mathbb{E}[\left\|DF\right\|_{l^{2}(\mathbb{N})}^{2}]:=\mathbb{E}\left[\sum_{k\in\mathbb{N}}(D_{k}F)^{2}\right].$

The adjoint operator $\delta$ of $D$ is characterized by

$\displaystyle\mathbb{E}[\langle DF,u\rangle]=\mathbb{E}[F\delta(u)],\quad
F\in\mathbb{D}^{1,2},u\in Dom(\delta).$

We will use that in our setting $F$ can be written as
$F=\mathbb{E}[F]+\sum_{n=1}^{\infty}J_{n}(f_{n})$, and

$\displaystyle
J_{n}(f_{n})=\sum_{(i_{1},...,i_{n})\in\mathbb{N}^{n}}f(i_{1},...,i_{n})Y_{i_{1}}Y_{i_{2}}...Y_{i_{n}}\mathbbl{1}_{\left\\{i_{j}\neq
i_{k}\,\text{for}\,j\neq k\right\\}}$

is the nth discrete multiple integral. Supposing $F$ has this representation
we define by

$\displaystyle LF$ $\displaystyle:=\sum_{n=1}^{\infty}-nJ_{n}(f_{n})$
$\displaystyle L^{-1}F$
$\displaystyle:=\sum_{n=1}^{\infty}-\frac{1}{n}J_{n}(f_{n}),$

the Ornstein–Uhlenbeck operator $L$ and the pseudo-inverse Ornstein–Uhlenbeck
operator $L^{-1}$. You can show that $F\in Dom(L)$ is equivalent to
$F\in\mathbb{D}^{1,2}$ and $DF\in Dom(\delta)$; in this case, it is $L=-\delta
D$. At last we recall the main ideas of Stein’s method for normal
approximation, starting with the important characterisation

$\displaystyle
Z\sim\mathcal{N}(0,1)\Leftrightarrow\mathbb{E}[f^{\prime}(Z)-Zf(Z)]=0$ (2.1)

for all continous differentiable $f$ such that the appearing expectations
exist. So if a random variable is in some sense close to $\mathcal{N}(0,1)$,
it is likely that the righthandside of (2.1) is close to 0. This motivates the
Stein-equation, written in the case of Kolmogorov distance, namely

$\displaystyle f_{z}^{\prime}(F)-Ff_{z}(F)=\mathbbl{1}_{\left\\{F\leq
z\right\\}}-\Phi(z),$

respectively

$\displaystyle\mathbb{E}[f_{z}^{\prime}(F)-Ff_{z}(F)]=\mathbb{P}(F\leq
z)-\Phi(z).$

The solution to this equation is given by [2]

$\displaystyle f_{z}(w)=\begin{cases}\frac{\Phi(w)(1-\Phi(z))}{p(w)}&w\leq
z,\\\ \frac{\Phi(z)(1-\Phi(w))}{p(w)}&\,w>z,\end{cases}$

where $p(w)=e^{-w^{2}/2}/\sqrt{2\pi}$ is the density of $\mathcal{N}(0,1)$.
Later on we will use the following bounds [5]:

$\displaystyle\frac{1-\Phi(w)}{p(w)}\leq\max\left\\{\frac{1}{w},\frac{\sqrt{2\pi}}{2}\right\\},\quad
w>0,$
$\displaystyle\left|f_{z}(w)\right|\leq\frac{\sqrt{2\pi}}{2}(1-\Phi(z)),\quad\left|wf_{z}(w)\right|\leq
1-\Phi(z),\quad w<0,$

and

$\displaystyle\left|f_{z}(w)\right|\leq\frac{\sqrt{2\pi}}{2}\Phi(z),\quad\left|wf_{z}(w)\right|\leq
1,\quad w>z.$

##  3\. Main Results

Now we present the main theorem of our paper.

######  Theorem 3.1 (Moderate deviations for Rademacher functionals).

Let $F\in\mathbb{D}^{1,2}$ with $\mathbb{E}[F]=0,\operatorname{Var}(F)=1$, and

$\displaystyle
Ff_{z}(F)+\mathbbl{1}_{\left\\{F>z\right\\}}\in\mathbb{D}^{1,2}\quad\forall
z\in\mathbb{R},$ $\displaystyle\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\in
Dom(\delta).$

Assume that there exists a constant $A>0$ and increasing fuctions
$\gamma_{1}(t),\gamma_{2}(t)$ such that $\mathbb{E}[e^{tF}]<\infty$ and

$\displaystyle\mathbb{E}\left[\left|1-\langle
DF,-DL^{-1}F\rangle\right|e^{tF}\right]$
$\displaystyle\leq\gamma_{1}(t)\mathbb{E}\left[e^{tF}\right],$ (A1)
$\displaystyle\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|e^{tF}\right]$
$\displaystyle\leq\gamma_{2}(t)\mathbb{E}\left[e^{tF}\right],$ (A2)

for all $0\leq t\leq A$. For $d_{0}\geq 0$, let

$\displaystyle A_{0}(d_{0}):=\max\left\\{0\leq t\leq
A:\frac{t^{2}}{2}\\!\left(\gamma_{1}(t)+\gamma_{2}(t)\right)\leq
d_{0}\right\\}.$

Then, for any $d_{0}\geq 0$,

$\displaystyle\left|\frac{\mathbb{P}(F>z)}{1-\Phi(z)}-1\right|\leq
25e^{d_{0}}(1+z^{2})\\!\left(\gamma_{1}(z)+\gamma_{2}(z)\right)$

provided that $0\leq z\leq A_{0}(d_{0})$.

In consequence, the following result is achieved.

######  Theorem 3.2.

Under the assumptions from Theorem 3.1, there is

$\displaystyle\left|\frac{\mathbb{P}(F>z)}{1-\Phi(z)}-1\right|$
$\displaystyle\leq
25e^{\frac{z^{2}}{2}\\!\left(\gamma_{1}(z)+\gamma_{2}(z)\right)}(1+z^{2})\\!\left(\gamma_{1}(z)+\gamma_{2}(z)\right)$

for all $0\leq z\leq A$.

###### Proof of Theorem 3.1.

We note at first that

$\displaystyle\left|\mathbb{P}(F>z)-(1-\Phi(z))\right|$
$\displaystyle=\left|1-\mathbb{P}(F\leq
z)-(1-\Phi(z))\right|=\left|\Phi(z)-\mathbb{P}(F\leq z)\right|.$

By Stein’s method and the proof of [9, Theorem 3.1] we have for
$z\in\mathbb{R}$

$\displaystyle\left|\mathbb{P}(F>z)-(1-\Phi(z))\right|$
$\displaystyle=\left|\mathbb{E}[f_{z}^{\prime}(F)-Ff_{z}(F)]\right|\leq
J_{1}+J_{2}$

with

$\displaystyle J_{1}$
$\displaystyle:=\mathbb{E}\left|f_{z}^{\prime}(F)\\!\left(1-\langle
DF,-DL^{-1}F\rangle\right)\right|,$ $\displaystyle J_{2}$
$\displaystyle:=\mathbb{E}\left[(Ff_{z}(F)+\mathbbl{1}_{\left\\{F>z\right\\}})\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right].$

For the upcoming estimation we can split $J_{2}$ into two terms, namely

$\displaystyle\left|J_{2}\right|$
$\displaystyle\leq\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|\left|Ff_{z}(F)+\mathbbl{1}_{\left\\{F>z\right\\}}\right|\right]\leq
J_{21}+J_{22}$

with

$\displaystyle J_{21}$
$\displaystyle:=\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|\left|Ff_{z}(F)\right|\right],$
$\displaystyle J_{22}$
$\displaystyle:=\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|\mathbbl{1}_{\left\\{F>z\right\\}}\right].$

Using the same arguments as in the proof of [23, Proposition 4.1] we have

$\displaystyle J_{21}\leq J_{23}+J_{24}+J_{25}$

with

$\displaystyle J_{23}$
$\displaystyle:=(1-\Phi(z))\cdot\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|\mathbbl{1}_{\left\\{F<0\right\\}}\right],$
$\displaystyle J_{24}$
$\displaystyle:=\sqrt{2\pi}\cdot(1-\Phi(z))\cdot\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|Fe^{F^{2}/2}\mathbbl{1}_{\left\\{0\leq
F\leq z\right\\}}\right],$ $\displaystyle J_{25}$
$\displaystyle:=\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|\mathbbl{1}_{\left\\{F>z\right\\}}\right]=J_{22}.$

Thus,

$\displaystyle\left|J_{2}\right|$ $\displaystyle\leq J_{23}+J_{24}+2\cdot
J_{25}$ (3.1)

For $J_{23}$, by condition (A2) with $t=0$ and noting that $\gamma_{2}$ is
increasing,

$\displaystyle\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|\mathbbl{1}_{\left\\{F<0\right\\}}\right]$
$\displaystyle\leq\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|\right]$
$\displaystyle\leq\gamma_{2}(0)\leq e^{d_{0}}\gamma_{2}(z).$ (3.2)

For $J_{24}$, by Lemma 3.4, we have

$\displaystyle\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|Fe^{F^{2}/2}\mathbbl{1}_{\left\\{0\leq
F\leq z\right\\}}\right]\leq 6e^{d_{0}}(1+z^{2})\gamma_{2}(z).$ (3.3)

For $J_{25}$, by condition (A2) and (3.6), for $0\leq z\leq A_{0}(d_{0})$,

$\displaystyle J_{25}$
$\displaystyle=\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|\mathbbl{1}_{\left\\{F>z\right\\}}e^{zF}e^{-zF}\right]$
$\displaystyle\leq\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|\mathbbl{1}_{\left\\{F>z\right\\}}e^{zF}\right]e^{-z^{2}}$
$\displaystyle\leq\gamma_{2}(z)\mathbb{E}\left[e^{zF}\right]e^{-z^{2}}$
$\displaystyle\leq e^{d_{0}}\gamma_{2}(z)e^{-z^{2}/2}.$

We recall that for $z>0$

$\displaystyle e^{-z^{2}/2}$
$\displaystyle\leq\sqrt{2\pi}\cdot(1+z)\cdot(1-\Phi(z))\leq\frac{3\sqrt{2\pi}}{2}\cdot(1+z^{2})\cdot(1-\Phi(z)).$

Then, for $0\leq z\leq A_{0}(d_{0})$,

$\displaystyle\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|\mathbbl{1}_{\left\\{F>z\right\\}}\right]$
$\displaystyle\leq\frac{3e^{d_{0}}\sqrt{2\pi}}{2}(1+z^{2})\gamma_{2}(z)(1-\Phi(z)).$
(3.4)

Therefore, combining (3.1) - (3.4), for $0\leq z\leq A_{0}(d_{0})$, we have

$\displaystyle\left|J_{2}\right|$
$\displaystyle\leq\\!\left(1+6\sqrt{2\pi}+3\sqrt{2\pi}\right)e^{d_{0}}(1+z^{2})\gamma_{2}(z)(1-\Phi(z))\leq
25e^{d_{0}}(1+z^{2})\gamma_{2}(z)(1-\Phi(z)).$

For the remaining term $J_{1}$ we have a similar approach after using again
Stein’s equation:

$\displaystyle\left|J_{1}\right|$
$\displaystyle=\mathbb{E}\left|f_{z}^{\prime}(F)\\!\left(1-\langle
DF,-DL^{-1}F\rangle\right)\right|$
$\displaystyle\leq\mathbb{E}\left[\left|f_{z}^{\prime}(F)\right|\left|1-\langle
DF,-DL^{-1}F\rangle\right|\right]$ $\displaystyle\leq J_{11}+J_{12}+J_{13}$

with

$\displaystyle J_{11}$
$\displaystyle:=\mathbb{E}\left[\left|Ff_{z}(F)\right|\left|1-\langle
DF,-DL^{-1}F\rangle\right|\right],$ $\displaystyle J_{12}$
$\displaystyle:=\mathbb{E}\left[(1-\Phi(z))\left|1-\langle
DF,-DL^{-1}F\rangle\right|\right],$ $\displaystyle J_{13}$
$\displaystyle:=\mathbb{E}\left[\mathbbl{1}_{\left\\{F>z\right\\}}\left|1-\langle
DF,-DL^{-1}F\rangle\right|\right].$

From here on we can identify any of these terms with a corresponding term from
the first part of the proof, namely $J_{21}-J_{25}$. Therefore, combining
these modified estimations, for $0\leq z\leq A_{0}(d_{0})$, we have

$\displaystyle\left|J_{1}\right|$
$\displaystyle\leq\\!\left(1+1+6\sqrt{2\pi}+3\sqrt{2\pi}\right)e^{d_{0}}(1+z^{2})\gamma_{1}(z)(1-\Phi(z))\leq
25e^{d_{0}}(1+z^{2})\gamma_{1}(z)(1-\Phi(z)).$

All in all, we have shown, for $0\leq z\leq A_{0}(d_{0})$,

$\displaystyle\left|\mathbb{P}(F>z)-(1-\Phi(z))\right|$ $\displaystyle\leq
25e^{d_{0}}(1+z^{2})(\gamma_{1}(z)+\gamma_{2}(z))(1-\Phi(z))$

or equivalent

$\displaystyle\left|\frac{\mathbb{P}(F>z)}{1-\Phi(z)}-1\right|$
$\displaystyle\leq 25e^{d_{0}}(1+z^{2})(\gamma_{1}(z)+\gamma_{2}(z)).\qed$

###### Proof of Theorem 3.2.

Let $0\leq z_{0}\leq A$ be fixed. Choose
$d_{0}=\frac{z_{0}^{2}}{2}\\!\left(\gamma_{1}(z_{0})+\gamma_{2}(z_{0})\right)$.
Per definition there is $0\leq z_{0}\leq A_{0}(d_{0})$. Hence, we may apply
Theorem 3.1, which then implies

$\displaystyle\left|\frac{\mathbb{P}(F>z_{0})}{1-\Phi(z_{0})}-1\right|$
$\displaystyle\leq
25e^{d_{0}}(1+z_{0}^{2})\\!\left(\gamma_{1}(z_{0})+\gamma_{2}(z_{0})\right)$
$\displaystyle=25e^{\frac{z_{0}^{2}}{2}\\!\left(\gamma_{1}(z_{0})+\gamma_{2}(z_{0})\right)}(1+z_{0}^{2})\\!\left(\gamma_{1}(z_{0})+\gamma_{2}(z_{0})\right).\qed$

######  Lemma 3.3 (Bound for the moment generating function).

Under the assumptions of Theorem 3.1, for $0\leq t\leq A$, we have

$\displaystyle\mathbb{E}\left[e^{tF}\right]\leq\exp\left\\{\frac{t^{2}}{2}(1+\gamma_{1}(t)+\gamma_{2}(t))\right\\}.$
(3.5)

Then, for $0\leq t\leq A_{0}(d_{0})$,

$\displaystyle\mathbb{E}\left[e^{tF}\right]\leq e^{d_{0}}e^{t^{2}/2}.$ (3.6)

###### Proof.

Let $h(t):=\mathbb{E}\left[e^{tF}\right]$. We recall that
$\mathbb{E}\left[e^{tF}\right]<\infty$ for $0\leq t\leq A$, and so, by the
continuity of the exponential funtion, we have
$h^{\prime}(t)=\mathbb{E}\left[Fe^{tF}\right]$. It follows that

$\displaystyle\mathbb{E}\left[Fe^{tF}\right]$
$\displaystyle=\mathbb{E}\left[(LL^{-1}F)e^{tF}\right]$
$\displaystyle=\mathbb{E}\left[(-\delta DL^{-1}F)e^{tF}\right]$
$\displaystyle=\mathbb{E}\left[\langle De^{tF},-DL^{-1}F\rangle\right]$
$\displaystyle=\mathbb{E}\left[\langle
tR,-DL^{-1}F\rangle\right]+\mathbb{E}\left[\langle
te^{tF}DF,-DL^{-1}F\rangle\right]$ $\displaystyle\leq
t\mathbb{E}\left[e^{tF}\right]+t\mathbb{E}\left|\langle
R,-DL^{-1}F\rangle\right|+t\mathbb{E}\left[\left|1-\langle
DF,-DL^{-1}F\rangle\right|e^{tF}\right]$ (3.7)

using for $k\in\mathbb{N}$

$\displaystyle D_{k}e^{tF}$
$\displaystyle=\sqrt{p_{k}q_{k}}\left[e^{tF_{k}^{+}}-e^{tF_{k}^{-}}\right]$
$\displaystyle=t\sqrt{p_{k}q_{k}}\int_{F_{k}^{-}}^{F_{k}^{+}}e^{tu}du$
$\displaystyle=t\sqrt{p_{k}q_{k}}\int_{F_{k}^{-}}^{F_{k}^{+}}\left[e^{tu}-e^{tF}\right]du+te^{tF}D_{k}F$
$\displaystyle=:tR_{k}+te^{tF}D_{k}F$

and $R=(R_{1},R_{2},\ldots)$. We can estimate $R_{k}$ as follows. Without loss
of generality $F_{k}^{-}\leq F\leq F_{k}^{+}$ (for the other case you just
have to change the sign).

$\displaystyle R_{k}$
$\displaystyle=\sqrt{p_{k}q_{k}}\int_{F_{k}^{-}}^{F_{k}^{+}}\left[e^{tu}-e^{tF}\right]du$
$\displaystyle\leq\sqrt{p_{k}q_{k}}\int_{F_{k}^{-}}^{F_{k}^{+}}\left[e^{tF_{k}^{+}}-e^{tF_{k}^{-}}\right]du$
$\displaystyle=\sqrt{p_{k}q_{k}}\left[e^{tF_{k}^{+}}-e^{tF_{k}^{-}}\right]\int_{F_{k}^{-}}^{F_{k}^{+}}du$
$\displaystyle=D_{k}e^{tF}\cdot\frac{1}{\sqrt{p_{k}q_{k}}}D_{k}F$

and by combining both cases

$\displaystyle\left|R_{k}\right|$
$\displaystyle\leq\frac{1}{\sqrt{p_{k}q_{k}}}D_{k}e^{tF}\cdot D_{k}F.$

By condition (A2) and our last inequality we get

$\displaystyle t\mathbb{E}\left|\langle R,-DL^{-1}F\rangle\right|$
$\displaystyle\leq
t\mathbb{E}\left[\langle\left|R\right|,\left|DL^{-1}F\right|\rangle\right]$
$\displaystyle\leq t\mathbb{E}\left[\langle
De^{tF},\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\rangle\right]$
$\displaystyle=t\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|e^{tF}\right]$
$\displaystyle\leq t\gamma_{2}(t)\mathbb{E}\left[e^{tF}\right].$ (3.8)

By condition (A1), for $0\leq t\leq A$,

$\displaystyle t\mathbb{E}\left[\left|1-\langle
DF,-DL^{-1}F\rangle\right|e^{tF}\right]\leq
t\gamma_{1}(t)\mathbb{E}\left[e^{tF}\right].$ (3.9)

Combining (3), (3) and (3.9), we have for $0\leq t\leq A$,

$\displaystyle h^{\prime}(t)$ $\displaystyle=\mathbb{E}\left[Fe^{tF}\right]$
$\displaystyle\leq th(t)+\\{t(\gamma_{1}(t)+\gamma_{2}(t))\\}h(t)$
$\displaystyle=\\{1+\gamma_{1}(t)+\gamma_{2}(t)\\}\,t\,h(t).$

Having in mind that $h(0)=1$, and $\gamma_{1}$ and $\gamma_{2}$ are
increasing, we complete the proof of (3.5) by solving the foregoing
differential inequality:

$\displaystyle\log(h(t))$
$\displaystyle=\int_{0}^{t}\frac{h^{\prime}(s)}{h(s)}ds$
$\displaystyle\leq\int_{0}^{t}(1+\gamma_{1}(s)+\gamma_{2}(s))sds$
$\displaystyle\leq\int_{0}^{t}(1+\gamma_{1}(t)+\gamma_{2}(t))sds$
$\displaystyle=\frac{t^{2}}{2}\\!\left(1+\gamma_{1}(t)+\gamma_{2}(t)\right),$

now we apply $\exp(.)$ on both sides. At last, (3.6) follows immediately from
(3.5) by definition of $A_{0}(d_{0})$. ∎

######  Lemma 3.4.

Under the assumptions of Theorem 3.1, we have for $0\leq z\leq A_{0}(d_{0})$,

$\displaystyle\mathbb{E}\left[\left|1-\langle
DF,-DL^{-1}F\rangle\right|Fe^{F^{2}/2}\mathbbl{1}_{\left\\{0\leq F\leq
z\right\\}}\right]\leq 6e^{d_{0}}(1+z^{2})\gamma_{1}(z)$ (3.10)

and

$\displaystyle\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|Fe^{F^{2}/2}\mathbbl{1}_{\left\\{0\leq
F\leq z\right\\}}\right]\leq 6e^{d_{0}}(1+z^{2})\gamma_{2}(z).$ (3.11)

###### Proof.

Same as [23] we apply the idea in [4, Lemma 5.2] for this proof. For
$a\in\mathbb{R}$, denote $[a]=\max\\{n\in\mathbb{N}:n\leq a\\}$. Next, we
define $H:=1-\langle DF,-DL^{-1}F\rangle$.

$\displaystyle\mathbb{E}\left[\left|H\right|Fe^{F^{2}/2}\mathbbl{1}_{\left\\{0\leq
F\leq z\right\\}}\right]$
$\displaystyle=\sum_{j=1}^{[z]}\mathbb{E}\left[\left|H\right|Fe^{F^{2}/2}\mathbbl{1}_{\left\\{j-1\leq
F\leq
j\right\\}}\right]+\mathbb{E}\left[\left|H\right|Fe^{F^{2}/2}\mathbbl{1}_{\left\\{[z]\leq
F\leq z\right\\}}\right].$

For the first term we get

$\displaystyle\sum_{j=1}^{[z]}\mathbb{E}\left[\left|H\right|Fe^{F^{2}/2}\mathbbl{1}_{\left\\{j-1\leq
F\leq j\right\\}}\right]$
$\displaystyle\leq\sum_{j=1}^{[z]}je^{(j-1)^{2}/2-j(j-1)}\mathbb{E}\left[\left|H\right|e^{jF}\mathbbl{1}_{\left\\{j-1\leq
F\leq j\right\\}}\right]$ $\displaystyle\leq
3\sum_{j=1}^{[z]}je^{-j^{2}/2}\mathbb{E}\left[\left|H\right|e^{jF}\mathbbl{1}_{\left\\{j-1\leq
F\leq j\right\\}}\right]$

and similarly, for the second

$\displaystyle\mathbb{E}\left[\left|H\right|Fe^{F^{2}/2}\mathbbl{1}_{\left\\{[z]\leq
F\leq z\right\\}}\right]$ $\displaystyle\leq
ze^{[z]^{2}/2-[z]z}\mathbb{E}\left[\left|H\right|e^{zF}\mathbbl{1}_{\left\\{[z]\leq
F\leq z\right\\}}\right]$ $\displaystyle\leq
3ze^{-z^{2}/2}\mathbb{E}\left[\left|H\right|e^{zF}\mathbbl{1}_{\left\\{[z]\leq
F\leq z\right\\}}\right].$

For both terms, we used similar manipulations, namely for $j-1\leq F\leq j$:

  * •

$e^{(j-1)^{2}/2-j(j-1)}=e^{j^{2}/2-j+1/2-j^{2}+j}=e^{-j^{2}/2}e^{1/2}\leq
3e^{-j^{2}/2}$.

  * •

$e^{(j-F)^{2}/2}\leq e^{1/2}\Leftrightarrow e^{F^{2}/2}\leq
e^{-j^{2}/2+jF+1/2}\Leftrightarrow e^{F^{2}/2}\leq
e^{(j-1)^{2}/2-j(j-1)}e^{jF}$.

And for $[z]\leq F\leq z$:

  * •

$e^{[z]^{2}/2-[z]z}=e^{[z]^{2}/2-[z]z+z^{2}/2-z^{2}/2}=e^{(z-[z])^{2}/2}e^{-z^{2}/2}\leq
3e^{-z^{2}/2}$.

  * •

$e^{(z-F)^{2}/2}\leq e^{(z-[z])^{2}/2}\Leftrightarrow e^{F^{2}/2}\leq
e^{(z-[z])^{2}/2+zF-z^{2}/2}\Leftrightarrow e^{F^{2}/2}\leq
e^{[z]^{2}/2-[z]z}e^{zF}$.

By condition (A1) and (3.6), and recalling that $\gamma_{1}$ is increasing,
for any $0\leq x\leq z\leq A_{0}(d_{0})$

$\displaystyle e^{-x^{2}/2}\mathbb{E}\left[\left|1-\langle
DF,-DL^{-1}F\rangle\right|e^{xF}\right]$
$\displaystyle\leq\gamma_{1}(x)\mathbb{E}[e^{xF-x^{2}/2}]$ $\displaystyle\leq
e^{d_{0}}\gamma_{1}(x)\leq e^{d_{0}}\gamma_{1}(z).$

By the foregoing inequalities,

$\displaystyle\mathbb{E}\left[\left|H\right|Fe^{F^{2}/2}\mathbbl{1}_{\left\\{0\leq
F\leq z\right\\}}\right]$ $\displaystyle\leq
3e^{d_{0}}\gamma_{1}(z)\\!\left(\sum_{j=1}^{[z]}j+z\right)\leq
6e^{d_{0}}(1+z^{2})\gamma_{1}(z).$

The other statement of the lemma is completely analogue. ∎

##  4\. Proofs I: Infinite weighted 2-runs

###### Proof of Theorem 1.1.

In what follows, we show that all the assumptions of Theorem 3.1 are verified.
Since the coefficient sequence $(a^{(n)}_{i})_{i\in\mathbb{Z}}$ is in
$l^{1}(\mathbb{Z})$ we have
$\left\|a^{(n)}\right\|_{l^{p}(\mathbb{Z})}<\infty\,\forall p\in\mathbb{N}$
and the random variable

$\displaystyle
F:=F_{n}=\frac{1}{\sqrt{\operatorname{Var}(G_{n})}}\sum_{i\in\mathbb{Z}}a^{(n)}_{i}\left[\xi_{i}\xi_{i+1}-\frac{1}{4}\right]=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\sum_{i\in\mathbb{Z}}a^{(n)}_{i}\left[X_{i}+X_{i}X_{i+1}+X_{i+1}\right]$

is bounded. In particular we will use
$\mathbb{E}\left|G_{n}\right|\leq\left\|a^{(n)}\right\|_{l^{1}(\mathbb{Z})}$
and

$\displaystyle\frac{1}{16}\left\|a^{(n)}\right\|_{l^{2}(\mathbb{Z})}^{2}\leq\operatorname{Var}(G_{n})=\frac{3}{16}\sum_{i\in\mathbb{Z}}(a^{(n)}_{i})^{2}+\frac{1}{8}\sum_{i\in\mathbb{Z}}a^{(n)}_{i}a^{(n)}_{i+1}\leq\frac{5}{16}\left\|a^{(n)}\right\|_{l^{2}(\mathbb{Z})}^{2}.$

Note that although here the Rademacher random variables are indexed by
$\mathbb{Z}$ instead of $\mathbb{N}$, Theorem 3.1 can be fully carried to this
setting. According to [9, Theorem 1.1] and its proof, the assumptions

  * •

$F\in L^{2}(\Omega,\sigma(X),\mathbb{P})$,

  * •

$Ff_{z}(F)+\mathbbl{1}_{\left\\{F>z\right\\}}\in\mathbb{D}^{1,2}\quad\forall
z\in\mathbb{R}$,

  * •

$\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\in Dom(\delta)$

are valid. Now we start to compute the terms appearing in (A1) and (A2). By
definition

$\displaystyle F_{k}^{+}$
$\displaystyle=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(\sum_{\begin{subarray}{c}i\in\mathbb{Z}\\\
i\neq
k-1,k\end{subarray}}a^{(n)}_{i}\left[X_{i}+X_{i}X_{i+1}+X_{i+1}\right]+a^{(n)}_{k-1}(2X_{k-1}+1)+a^{(n)}_{k}(2X_{k+1}+1)\right)$
$\displaystyle F_{k}^{-}$
$\displaystyle=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(\sum_{\begin{subarray}{c}i\in\mathbb{Z}\\\
i\neq
k-1,k\end{subarray}}a^{(n)}_{i}\left[X_{i}+X_{i}X_{i+1}+X_{i+1}\right]-a^{(n)}_{k-1}-a^{(n)}_{k}\right)$

and we get (for fixed $n\in\mathbb{N}$)

$\displaystyle D_{k}F$
$\displaystyle=\frac{1}{2}\\!\left(F_{k}^{+}-F_{k}^{-}\right)$
$\displaystyle=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(a^{(n)}_{k-1}(X_{k-1}+1)+a^{(n)}_{k}(X_{k+1}+1)\right).$

On the other hand it is

$\displaystyle-L^{-1}F=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\sum_{i\in\mathbb{Z}}a^{(n)}_{i}\left[X_{i}+\frac{1}{2}X_{i}X_{i+1}+X_{i+1}\right]$

and so

$\displaystyle-
D_{k}L^{-1}F=\frac{1}{8\sqrt{\operatorname{Var}(G_{n})}}\\!\left(a^{(n)}_{k-1}(X_{k-1}+2)+a^{(n)}_{k}(X_{k+1}+2)\right).$

With these expressions we can compute the scalar product

$\displaystyle\langle
DF,-DL^{-1}F\rangle=\frac{1}{32\operatorname{Var}(G_{n})}\sum_{k\in\mathbb{Z}}$
$\displaystyle(a^{(n)}_{k-1})^{2}\left[X_{k-1}^{2}+3X_{k-1}+2\right]$
$\displaystyle+(a^{(n)}_{k})^{2}\left[X_{k+1}^{2}+3X_{k+1}+2\right]$
$\displaystyle+a^{(n)}_{k-1}a^{(n)}_{k}\left[3X_{k-1}+2X_{k-1}X_{k+1}+3X_{k+1}+4\right].$

According to [12, (2.13)] it is

$\displaystyle 1=\mathbb{E}[\langle
DF,-DL^{-1}F\rangle]=\frac{1}{32\operatorname{Var}(G_{n})}\sum_{k\in\mathbb{Z}}3(a^{(n)}_{k-1})^{2}+4a^{(n)}_{k-1}a^{(n)}_{k}+3(a^{(n)}_{k})^{2}.$

We use Cauchy–Schwarz inequality for

$\displaystyle\mathbb{E}\left[\left|1-\langle
DF,-DL^{-1}F\rangle\right|e^{tF}\right]\leq\\!\left(\mathbb{E}\left[\\!\left(1-\langle
DF,-DL^{-1}F\rangle\right)^{2}e^{tF}\right]\right)^{\frac{1}{2}}\\!\left(\mathbb{E}\left[e^{tF}\right]\right)^{\frac{1}{2}}$

and deal with the double sum resulting from the square of

$\displaystyle\langle
DF,-DL^{-1}F\rangle-1=\frac{1}{32\operatorname{Var}(G_{n})}\sum_{k\in\mathbb{Z}}$
$\displaystyle(a^{(n)}_{k-1})^{2}3X_{k-1}+(a^{(n)}_{k})^{2}3X_{k+1}$
$\displaystyle+a^{(n)}_{k-1}a^{(n)}_{k}\left[3X_{k-1}+2X_{k-1}X_{k+1}+3X_{k+1}\right].$

Then we can write $\\!\left(1-\langle
DF,-DL^{-1}F\rangle\right)^{2}=B_{1}+\ldots+B_{9}$ with

$\displaystyle B_{1}$
$\displaystyle=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k-1})^{2}(a^{(n)}_{l-1})^{2}X_{k-1}X_{l-1},$
$\displaystyle B_{2}$
$\displaystyle=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}X_{k}X_{l},$
$\displaystyle B_{3}$
$\displaystyle=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k-1})^{2}(a^{(n)}_{l})^{2}X_{k-1}X_{l},$
$\displaystyle B_{4}$
$\displaystyle=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k})^{2}(a^{(n)}_{l-1})^{2}X_{k}X_{l-1},$
$\displaystyle B_{5}$
$\displaystyle=\frac{3}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k-1})^{2}(a^{(n)}_{l-1})(a^{(n)}_{l})X_{k-1}\left[3X_{l-1}+2X_{l-1}X_{l+1}+3X_{l+1},\right]$
$\displaystyle B_{6}$
$\displaystyle=\frac{3}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k})^{2}(a^{(n)}_{l-1})(a^{(n)}_{l})X_{k}\left[3X_{l-1}+2X_{l-1}X_{l+1}+3X_{l+1}\right],$
$\displaystyle B_{7}$
$\displaystyle=\frac{3}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k-1})(a^{(n)}_{k})(a^{(n)}_{l-1})^{2}X_{l-1}\left[3X_{k-1}+2X_{k-1}X_{k+1}+3X_{k+1},\right]$
$\displaystyle B_{8}$
$\displaystyle=\frac{3}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k-1})(a^{(n)}_{k})(a^{(n)}_{l})^{2}X_{l}\left[3X_{k-1}+2X_{k-1}X_{k+1}+3X_{k+1}\right],$

where $B_{1}=B_{2}=B_{3}=B_{4}$ and $B_{5}=B_{6}=B_{7}=B_{8}$ by symmetry and
change of variables. The last missing term is given by

$\displaystyle
B_{9}=\frac{1}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k-1})(a^{(n)}_{k})(a^{(n)}_{l-1})(a^{(n)}_{l})$
$\displaystyle\cdot\left[3X_{k-1}+2X_{k-1}X_{k+1}+3X_{k+1}\right]$
$\displaystyle\cdot\left[3X_{l-1}+2X_{l-1}X_{l+1}+3X_{l+1}\right].$

So basically we have to deal with three classes of subterms in total. Since
they will be multiplied with $e^{tF}$, we have to study

$\displaystyle\mathbb{E}[X_{i}e^{tF}],\quad\mathbb{E}[X_{i}X_{j}e^{tF}],\quad\mathbb{E}[X_{i}X_{j}X_{k}e^{tF}],\quad\mathbb{E}[X_{i}X_{j}X_{k}X_{l}e^{tF}]$

for $i\neq j\neq k\neq l$ — if two or more indices are equal, it is just one
of the terms from before or immediately $\mathbb{E}[e^{tF}]$. This is done in
Lemma 4.1 and we will refer to it, in particular the inequalities shown in its
proof. Now we are ready to deal with all three classes of subterms and choose
$B_{2},B_{6}$ and $B_{9}$ as representatives: First class of subterms,
$B_{1}-B_{4}$:

$\displaystyle\mathbb{E}[B_{2}e^{tF}]=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\Bigg{(}\sum_{k\in\mathbb{Z}}(a^{(n)}_{k})^{4}\mathbb{E}[e^{tF}]$
$\displaystyle+\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
\left|k-l\right|=1\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\mathbb{E}[X_{k}X_{l}e^{tF}]$
$\displaystyle+\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
\left|k-l\right|\geq
2\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\mathbb{E}[X_{k}X_{l}e^{tF}]\Bigg{)}.$
(4.1)

The first one is the easiest by

$\displaystyle
B_{21}:=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k\in\mathbb{Z}}(a^{(n)}_{k})^{4}\mathbb{E}[e^{tF}]=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}\mathbb{E}[e^{tF}].$

By using (4), (4.12) and (4.13) it remains to bound

$\displaystyle B_{22}$
$\displaystyle:=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
\left|k-l\right|=1\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\mathbb{E}[X_{k}X_{l}e^{tF}]$
$\displaystyle\leq\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\\!\left(B_{221}+B_{222}+B_{223}\right)\mathbb{E}\left[e^{tF}\right]e^{ct},$
$\displaystyle B_{23}$
$\displaystyle:=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
\left|k-l\right|\geq
2\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\mathbb{E}[X_{k}X_{l}e^{tF}]$
$\displaystyle\leq\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\\!\left(B_{231}+B_{232}\right)\mathbb{E}\left[e^{tF}\right]e^{ct},$

such that

$\displaystyle B_{221}$
$\displaystyle:=\frac{Ct}{4\sqrt{\operatorname{Var}(G_{n})}}\cdot\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
\left|k-l\right|=1\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\left|a^{(n)}_{\min(k,l)}\right|,$
$\displaystyle B_{222}$
$\displaystyle:=\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
\left|k-l\right|=1\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\cdot\sum_{\begin{subarray}{c}m_{1}\in\\{k-1,k,l-1,l\\}\\\
n_{1}\in\\{k-2,k+1,l-2,l+1\\}\end{subarray}}\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|,$
$\displaystyle B_{223}$
$\displaystyle:=\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
\left|k-l\right|=1\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\cdot\sum_{\begin{subarray}{c}m_{2},n_{2}\in\\{k-1,k,l-1,l\\}\end{subarray}}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|,$
$\displaystyle B_{231}$
$\displaystyle:=\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
\left|k-l\right|\geq
2\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\cdot\sum_{\begin{subarray}{c}m_{1}\in\\{k-1,k,l-1,l\\}\\\
n_{1}\in\\{k-2,k+1,l-2,l+1\\}\end{subarray}}\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|,$
$\displaystyle B_{232}$
$\displaystyle:=\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
\left|k-l\right|\geq
2\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\cdot\sum_{\begin{subarray}{c}m_{2},n_{2}\in\\{k-1,k,l-1,l\\}\end{subarray}}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|.$

Then by AGM inequality

$\displaystyle B_{221}$
$\displaystyle=\frac{Ct}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{k-1})^{2}\left|a^{(n)}_{k-1}\right|+\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{k+1})^{2}\left|a^{(n)}_{k}\right|\right)$
$\displaystyle=\frac{Ct}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}\sqrt[5]{\left|a^{(n)}_{k}\right|^{10}\left|a^{(n)}_{k-1}\right|^{10}\left|a^{(n)}_{k-1}\right|^{5}}+\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}\sqrt[5]{\left|a^{(n)}_{k}\right|^{10}\left|a^{(n)}_{k+1}\right|^{10}\left|a^{(n)}_{k}\right|^{5}}\right)$
$\displaystyle\leq\frac{Ct}{20\sqrt{\operatorname{Var}(G_{n})}}\\!\left(\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}2\left|a^{(n)}_{k}\right|^{5}+3\left|a^{(n)}_{k-1}\right|^{5}+\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}3\left|a^{(n)}_{k}\right|^{5}+2\left|a^{(n)}_{k+1}\right|^{5}\right)$
$\displaystyle\leq\frac{Ct}{\sqrt{\operatorname{Var}(G_{n})}}\left\|a^{(n)}\right\|_{l^{5}(\mathbb{Z})}^{5}.$

If we look at $B_{222}-B_{232}$, we can change the order of summation since
all summands are non-negative. And we become even bigger if we add the missing
indices:

$\displaystyle B_{223}$
$\displaystyle\leq\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\end{subarray}}\sum_{\begin{subarray}{c}m_{2}\in\\{k-1,k,l-1,l\\}\end{subarray}}\sum_{\begin{subarray}{c}n_{2}\in\\{k-1,k,l-1,l\\}\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|.$

From here on we treat different cases, but every time we can use AGM
inequality: Case 1: $m_{2}\in\\{k-1,k\\}$ and $n_{2}\in\\{k-1,k\\}$:

$\displaystyle\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}\sum_{\begin{subarray}{c}l\in\mathbb{Z}\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|$
$\displaystyle=\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}(a^{(n)}_{k})^{2}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|\sum_{\begin{subarray}{c}l\in\mathbb{Z}\end{subarray}}(a^{(n)}_{l})^{2}$
$\displaystyle\leq
C\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}\left\|a^{(n)}\right\|_{l^{2}(\mathbb{Z})}^{2}.$

Case 2: $m_{2}\in\\{l-1,l\\}$ and $n_{2}\in\\{l-1,l\\}$:

$\displaystyle\sum_{\begin{subarray}{c}l\in\mathbb{Z}\end{subarray}}\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|$
$\displaystyle=\sum_{\begin{subarray}{c}l\in\mathbb{Z}\end{subarray}}(a^{(n)}_{l})^{2}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}(a^{(n)}_{k})^{2}$
$\displaystyle\leq
C\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}\left\|a^{(n)}\right\|_{l^{2}(\mathbb{Z})}^{2}.$

Case 3: $m_{2}\in\\{k-1,k\\}$ and $n_{2}\in\\{l-1,l\\}$:

$\displaystyle\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}\sum_{\begin{subarray}{c}l\in\mathbb{Z}\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|$
$\displaystyle=\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}(a^{(n)}_{k})^{2}\left|a^{(n)}_{m_{2}}\right|\sum_{\begin{subarray}{c}l\in\mathbb{Z}\end{subarray}}(a^{(n)}_{l})^{2}\left|a^{(n)}_{n_{2}}\right|$
$\displaystyle\leq C\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{6}.$

Case 4: $m_{2}\in\\{l-1,l\\}$ and $n_{2}\in\\{k-1,k\\}$:

$\displaystyle\sum_{\begin{subarray}{c}l\in\mathbb{Z}\end{subarray}}\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}(a^{(n)}_{k})^{2}(a^{(n)}_{l})^{2}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|$
$\displaystyle=\sum_{\begin{subarray}{c}l\in\mathbb{Z}\end{subarray}}(a^{(n)}_{l})^{2}\left|a^{(n)}_{m_{2}}\right|\sum_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}(a^{(n)}_{k})^{2}\left|a^{(n)}_{n_{2}}\right|$
$\displaystyle\leq C\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{6}.$

According to Remark 1.2 in case 1 and 2 the norm
$\left\|a^{(n)}\right\|_{l^{2}(\mathbb{Z})}^{2}$ vanishes directly with the
variance in the prefactor. Summarizing for $B_{223}$:

$\displaystyle B_{223}$
$\displaystyle\leq\frac{Ct^{2}}{32}\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}+\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{6}.$

Analogue we get basically bounds of the same order for $B_{222},B_{231}$ and
$B_{232}$. Combining our bounds for the subterms of (4) gives us

$\displaystyle\mathbb{E}[B_{2}e^{tF}]\leq
Ce^{ct}\\!\left(t^{2}\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{6}}{(\operatorname{Var}(G_{n}))^{3}}+(1+t^{2})\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}}{(\operatorname{Var}(G_{n}))^{2}}+t\frac{\left\|a^{(n)}\right\|_{l^{5}(\mathbb{Z})}^{5}}{(\operatorname{Var}(G_{n}))^{5/2}}\right)\mathbb{E}[e^{tF}].$

Second class of subterms, $B_{5}-B_{8}$: We write
$\mathbb{E}[B_{6}e^{tF}]=B_{61}+B_{62}+B_{63}$ such that

$\displaystyle B_{61}$
$\displaystyle:=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k})^{2}(a^{(n)}_{l-1})(a^{(n)}_{l})\mathbb{E}[X_{k}X_{l-1}e^{tF}],$
$\displaystyle B_{62}$
$\displaystyle:=\frac{6}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k})^{2}(a^{(n)}_{l-1})(a^{(n)}_{l})\mathbb{E}[X_{k}X_{l-1}X_{l+1}e^{tF}],$
$\displaystyle B_{63}$
$\displaystyle:=\frac{9}{1024(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k})^{2}(a^{(n)}_{l-1})(a^{(n)}_{l})\mathbb{E}[X_{k}X_{l+1}e^{tF}].$

$B_{61}$ and $B_{63}$ are analogue to $B_{22}$ and $B_{23}$ since they have
the same structure: Two coefficients with k-index, two coefficients with
l-index, one $X$ with k-index and one $X$ with l-index. And with that the
arguments are the same. Looking at the remaining $B_{62}$ two indices of the
$X^{\prime}s$ are equal if and only if $k=l+1$ or $k=l-1$. In the latter case
$B_{62}$ reduces to

$\displaystyle\frac{C}{(\operatorname{Var}(G_{n}))^{2}}\sum_{k\in\mathbb{Z}}(a^{(n)}_{k})^{3}(a^{(n)}_{k+1})\mathbb{E}[X_{k+2}e^{tF}],$

so we can use (4) and (4.10) giving us upper bounds of order

$\displaystyle
O\\!\left(t\frac{\left\|a^{(n)}\right\|_{l^{5}(\mathbb{Z})}^{5}}{(\operatorname{Var}(G_{n}))^{5/2}}\right)\quad\text{and}\quad
O\\!\left(t^{2}\frac{\left\|a^{(n)}\right\|_{l^{6}(\mathbb{Z})}^{6}}{(\operatorname{Var}(G_{n}))^{3}}\right).$

And the same for $k=l+1$. At last, if neither $k=l-1$ nor $k=l+1$ we can use
(4.14) and (4.15) giving us upper bounds of order

$\displaystyle
O\\!\left(t^{2}\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{6}}{(\operatorname{Var}(G_{n}))^{3}}\right)\quad\text{and}\quad
O\\!\left(t^{2}\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}}{(\operatorname{Var}(G_{n}))^{2}}\right).$

Combining our bounds for $B_{61},B_{62}$ and $B_{63}$ we get

$\displaystyle\mathbb{E}[B_{6}e^{tF}]\leq
Ce^{ct}\\!\left(t^{2}\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{6}}{(\operatorname{Var}(G_{n}))^{3}}+t^{2}\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}}{(\operatorname{Var}(G_{n}))^{2}}+t\frac{\left\|a^{(n)}\right\|_{l^{5}(\mathbb{Z})}^{5}}{(\operatorname{Var}(G_{n}))^{5/2}}+t^{2}\frac{\left\|a^{(n)}\right\|_{l^{6}(\mathbb{Z})}^{6}}{(\operatorname{Var}(G_{n}))^{3}}\right)\mathbb{E}[e^{tF}].$

Third class of subterms: It consists only of $B_{9}$ and so we have to deal
with $\mathbb{E}[B_{9}e^{tF}]$. Multiplying all the $X^{\prime}s$ inside we
get products of lengths two,three and four. The first two cases are already
solved and a product of length four appears only one time, namely

$\displaystyle\frac{C}{(\operatorname{Var}(G_{n}))^{2}}\sum_{k,l\in\mathbb{Z}}(a^{(n)}_{k-1})(a^{(n)}_{k})(a^{(n)}_{l-1})(a^{(n)}_{l})\mathbb{E}[X_{k-1}X_{k+1}X_{l-1}X_{l+1}e^{tF}].$

We have two pairs of two equal indices of the $X^{\prime}s$ if and only if
$k=l$ and then we are in the situation of $B_{21}$. Note that it is impossible
that three or more indices are equal. If two indices are equal and two indices
are different, e.g. $k-1=l+1$ we are in the $2X$-case and can use (4), (4.12)
and (4.13). At last, if all four indices are different, most of the work is
done by (4.16) and (4.17) leading to upper bounds of order

$\displaystyle
O\\!\left(t^{2}\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{6}}{(\operatorname{Var}(G_{n}))^{3}}\right)\quad\text{and}\quad
O\\!\left(t^{2}\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}}{(\operatorname{Var}(G_{n}))^{2}}\right).$

Combining our bounds from all the different cases we get

$\displaystyle\mathbb{E}[B_{9}e^{tF}]\leq
Ce^{ct}\\!\left(t^{2}\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{6}}{(\operatorname{Var}(G_{n}))^{3}}+(1+t^{2})\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}}{(\operatorname{Var}(G_{n}))^{2}}+t\frac{\left\|a^{(n)}\right\|_{l^{5}(\mathbb{Z})}^{5}}{(\operatorname{Var}(G_{n}))^{5/2}}+t^{2}\frac{\left\|a^{(n)}\right\|_{l^{6}(\mathbb{Z})}^{6}}{(\operatorname{Var}(G_{n}))^{3}}\right)\mathbb{E}[e^{tF}].$

Summarizing everything we have done so far a bound as in condition (A1) is
obtained by

$\displaystyle\mathbb{E}\left[\left|1-\langle
DF,-DL^{-1}F\rangle\right|e^{tF}\right]$
$\displaystyle\leq\\!\left(\mathbb{E}\left[\\!\left(1-\langle
DF,-DL^{-1}F\rangle\right)^{2}e^{tF}\right]\right)^{\frac{1}{2}}\\!\left(\mathbb{E}\left[e^{tF}\right]\right)^{\frac{1}{2}}$
$\displaystyle=\\!\left(\mathbb{E}\left[\sum_{i=1}^{9}B_{i}e^{tF}\right]\right)^{\frac{1}{2}}\\!\left(\mathbb{E}\left[e^{tF}\right]\right)^{\frac{1}{2}}$
$\displaystyle\leq\sum_{i=1}^{9}\\!\left(\mathbb{E}\left[B_{i}e^{tF}\right]\right)^{\frac{1}{2}}\\!\left(\mathbb{E}\left[e^{tF}\right]\right)^{\frac{1}{2}}$
$\displaystyle\leq\widetilde{\gamma_{1}}(t)\mathbb{E}\left[e^{tF}\right]$

such that

$\displaystyle\widetilde{\gamma_{1}}(t)$
$\displaystyle=Ce^{ct}\\!\left(t\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{3}}{(\operatorname{Var}(G_{n}))^{3/2}}+(1+t)\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{2}}{(\operatorname{Var}(G_{n}))}+t^{1/2}\frac{\left\|a^{(n)}\right\|_{l^{5}(\mathbb{Z})}^{5/2}}{(\operatorname{Var}(G_{n}))^{5/4}}+t\frac{\left\|a^{(n)}\right\|_{l^{6}(\mathbb{Z})}^{3}}{(\operatorname{Var}(G_{n}))^{3/2}}\right)$
$\displaystyle=:Ce^{ct}\\!\left(tC_{n,1}+(1+t)C_{n,2}+t^{1/2}C_{n,3}+tC_{n,4}\right).$

We now move on and show that a bound as in condition (A2) exists: Again, by
Cauchy–Schwarz inequality

$\displaystyle\mathbb{E}\left[\left|\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right|e^{tF}\right]\leq\\!\left(\mathbb{E}\left[\\!\left(\delta\\!\left(\frac{1}{\sqrt{pq}}DF\left|DL^{-1}F\right|\right)\right)^{2}e^{tF}\right]\right)^{\frac{1}{2}}\\!\left(\mathbb{E}\left[e^{tF}\right]\right)^{\frac{1}{2}}.$

By [17, Corollary 9.9] it is $\delta(u)=\sum_{k=0}^{\infty}Y_{k}u_{k}$, where
$Y_{k}$ is the $kth$ centered and normalized Rademacher random variable. In
our case $Y_{k}=X_{k}$ and the corollary can be applied since
$u_{k}:=D_{k}F\left|D_{k}L^{-1}F\right|/\sqrt{p_{k}q_{k}}$ does not depend on
$X_{k}$. Then

$\displaystyle\\!\left(\mathbb{E}\left[\\!\left(\delta\\!\left(u\right)\right)^{2}e^{tF}\right]\right)^{\frac{1}{2}}$
$\displaystyle=\\!\left(\sum_{k,l\in\mathbb{Z}}\mathbb{E}\left[u_{k}u_{l}X_{k}X_{l}e^{tF}\right]\right)^{\frac{1}{2}}$
$\displaystyle\leq\\!\left(\sum_{k\in\mathbb{Z}}\mathbb{E}\left[u_{k}^{2}e^{tF}\right]\right)^{\frac{1}{2}}+\\!\left(\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
k\neq
l\end{subarray}}\mathbb{E}\left[u_{k}u_{l}X_{k}X_{l}e^{tF}\right]\right)^{\frac{1}{2}}.$
(4.2)

For the upcoming computations we recall

$\displaystyle D_{k}F$
$\displaystyle=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(a^{(n)}_{k-1}(X_{k-1}+1)+a^{(n)}_{k}(X_{k+1}+1)\right),$
$\displaystyle-D_{k}L^{-1}F$
$\displaystyle=\frac{1}{8\sqrt{\operatorname{Var}(G_{n})}}\\!\left(a^{(n)}_{k-1}(X_{k-1}+2)+a^{(n)}_{k}(X_{k+1}+2)\right).$

and so

$\displaystyle D_{k}F(-D_{k}L^{-1}F)=\frac{1}{32\operatorname{Var}(G_{n})}$
$\displaystyle(a^{(n)}_{k-1})^{2}\left[3X_{k-1}+3\right]+(a^{(n)}_{k})^{2}\left[3X_{k+1}+3\right]$
$\displaystyle+a^{(n)}_{k-1}a^{(n)}_{k}\left[3X_{k-1}+2X_{k-1}X_{k+1}+3X_{k+1}+4\right].$

The square of the righthandside is of a familiar form: Every summand consists
of a product of length four of coefficients with index $k\pm...$ multiplied
with something bounded, and so as before we get immediately or by AGM-
inequality

$\displaystyle\sum_{k\in\mathbb{Z}}\mathbb{E}\left[u_{k}^{2}e^{tF}\right]\leq
C\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}}{(\operatorname{Var}(G_{n}))^{2}}\mathbb{E}\left[e^{tF}\right].$

For the remaining term of (4) we adapt the strategy that is used in the proof
of Lemma 4.1 — see its beginning for a detailed explanation. Set

$\displaystyle A_{k}:=a^{(n)}_{k}\left[X_{k}+X_{k}X_{k+1}+X_{k+1}\right]$

so that

$\displaystyle
F_{a}=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(A_{k-2}+A_{k-1}+A_{k}+A_{k+1}+A_{l-2}+A_{l-1}+A_{l}+A_{l+1}\right)$

and by Taylor expansion

$\displaystyle\mathbb{E}[u_{k}u_{l}X_{k}X_{l}e^{tF}]$
$\displaystyle=\mathbb{E}[u_{k}u_{l}X_{k}X_{l}e^{tF_{a}}e^{tF_{u}}]$
$\displaystyle=\mathbb{E}[u_{k}u_{l}X_{k}X_{l}e^{tF_{u}}]+t\mathbb{E}[u_{k}u_{l}X_{k}X_{l}F_{a}e^{tF_{u}}]+\mathbb{E}[u_{k}u_{l}X_{k}X_{l}R_{2}(tF_{a})e^{tF_{u}}].$

0-order-term: By independence

$\displaystyle\mathbb{E}[u_{k}u_{l}X_{k}X_{l}e^{tF_{u}}]=\mathbb{E}[u_{k}u_{l}X_{k}X_{l}]\mathbb{E}[e^{tF_{u}}].$

Since by definition $X_{k}$ and $X_{l}$ respectively $u_{k}$ are independent,
we just have to check whether the same goes for $X_{k}$ and $u_{l}$, which is
leading to two cases. Case 1: $l\notin\\{k-1,k+1\\}$

$\displaystyle\mathbb{E}[u_{k}u_{l}X_{k}X_{l}]=\mathbb{E}[X_{k}]\mathbb{E}[u_{k}u_{l}X_{l}]=0.$

Case 2: $l\in\\{k-1,k+1\\}$

$\displaystyle\sum_{k\in\mathbb{Z}}\mathbb{E}[u_{k}u_{l}X_{k}X_{l}]\mathbb{E}[e^{tF_{u}}]\leq
C\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}}{(\operatorname{Var}(G_{n}))^{2}}\mathbb{E}\left[e^{tF}\right]e^{ct},$

following our usual argumentation. 1st-order-term: We split $F_{u}$ in the
same manner as before, $F_{u}=F_{u_{a}}+F_{u_{u}}$, such that
$F_{u_{a}}=\\!\left(A_{k-3}+A_{k+2}+A_{l-3}+A_{l+2}\right)/4\sqrt{\operatorname{Var}(G_{n})}$
and use another Taylor expansion of degree 1. Then

$\displaystyle t\mathbb{E}[u_{k}u_{l}X_{k}X_{l}F_{a}e^{tF_{u}}]$
$\displaystyle=t\mathbb{E}[u_{k}u_{l}X_{k}X_{l}F_{a}e^{tF_{u_{a}}}e^{tF_{u_{u}}}]$
$\displaystyle=t\mathbb{E}[u_{k}u_{l}X_{k}X_{l}F_{a}e^{tF_{u_{u}}}]+t\mathbb{E}[u_{k}u_{l}X_{k}X_{l}F_{a}R_{1}(tF_{u_{a}})e^{tF_{u_{u}}}].$
(4.3)

We can bound the first term of (4) by using

$\displaystyle
tu_{k}u_{l}X_{k}X_{l}F_{a}\leq\frac{Ct}{(\operatorname{Var}(G_{n}))^{5/2}}\quad\sum_{\mathclap{\begin{subarray}{c}i_{1},i_{2}\in\\{k-1,k\\}\\\
i_{3},i_{4}\in\\{l-1,l\\}\\\
i_{5}\in\\{k-2,k-1,k,k+1,l-2,l-1,l,l+1\\}\end{subarray}}}\quad
a_{i_{1}}^{(n)}a_{i_{2}}^{(n)}a_{i_{3}}^{(n)}a_{i_{4}}^{(n)}a_{i_{5}}^{(n)}$

and every summand consists of a product of length five of either three
coefficients with index $k\pm...$ and two coefficients with index $l\pm...$ or
the other way around. In both cases we get

$\displaystyle\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\ k\neq
l\end{subarray}}t\mathbb{E}[u_{k}u_{l}X_{k}X_{l}F_{a}e^{tF_{u_{u}}}]$
$\displaystyle\leq\frac{Ct}{(\operatorname{Var}(G_{n}))^{5/2}}\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{3}\left\|a^{(n)}\right\|_{l^{2}(\mathbb{Z})}^{2}\mathbb{E}\left[e^{tF}\right]e^{ct}$
$\displaystyle=Ct\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{3}}{(\operatorname{Var}(G_{n}))^{3/2}}\mathbb{E}\left[e^{tF}\right]e^{ct}.$

Having in mind that $\left|R_{1}(tF_{u_{a}})\right|\leq
t\left|F_{u_{a}}\right|e^{tF_{u_{a}}}$ we can bound the second term of (4) by
using

$\displaystyle
tu_{k}u_{l}X_{k}X_{l}F_{a}R_{1}(tF_{u_{a}})\leq\frac{Ct^{2}}{(\operatorname{Var}(G_{n}))^{3}}\quad\sum_{\mathclap{\begin{subarray}{c}i_{1},i_{2}\in\\{k-1,k\\}\\\
i_{3},i_{4}\in\\{l-1,l\\}\\\ i_{5}\in\\{k-2,k-1,k,k+1,l-2,l-1,l,l+1\\}\\\
i_{6}\in\\{k-3,k+2,l-3,l+2\\}\end{subarray}}}\quad
a_{i_{1}}^{(n)}a_{i_{2}}^{(n)}a_{i_{3}}^{(n)}a_{i_{4}}^{(n)}a_{i_{5}}^{(n)}a_{i_{6}}^{(n)}$

and every summand consists of a product of length six of either three
coefficients with index $k\pm...$ and three coefficients with index $l\pm...$,
or four coefficients with index $k\pm...$ and two coefficients with index
$l\pm...$or the other way around. Combining the cases we get

$\displaystyle\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\ k\neq
l\end{subarray}}t\mathbb{E}[u_{k}u_{l}X_{k}X_{l}F_{a}R_{1}(tF_{u_{a}})e^{tF_{u_{u}}}]$
$\displaystyle\leq
Ct^{2}\\!\left(\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}}{(\operatorname{Var}(G_{n}))^{2}}+\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{6}}{(\operatorname{Var}(G_{n}))^{3}}\right)\mathbb{E}\left[e^{tF}\right]e^{ct}.$

2nd-order-term: Finally, having in mind that $\left|R_{2}(tF_{a})\right|\leq
t^{2}F_{a}^{2}e^{tF_{a}}/2$ we can bound the last term of our original Taylor
expansion by using

$\displaystyle
u_{k}u_{l}X_{k}X_{l}F_{a}R_{2}(tF_{u_{a}})\leq\frac{Ct^{2}}{(\operatorname{Var}(G_{n}))^{3}}\quad\sum_{\mathclap{\begin{subarray}{c}i_{1},i_{2}\in\\{k-1,k\\}\\\
i_{3},i_{4}\in\\{l-1,l\\}\\\
i_{5},i_{6}\in\\{k-2,k-1,k,k+1,l-2,l-1,l,l+1\\}\end{subarray}}}\quad
a_{i_{1}}^{(n)}a_{i_{2}}^{(n)}a_{i_{3}}^{(n)}a_{i_{4}}^{(n)}a_{i_{5}}^{(n)}a_{i_{6}}^{(n)}$

and get the analogue bound

$\displaystyle\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\ k\neq
l\end{subarray}}\mathbb{E}[u_{k}u_{l}X_{k}X_{l}F_{a}R_{2}(tF_{a})e^{tF_{u}}]$
$\displaystyle\leq
Ct^{2}\\!\left(\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{4}}{(\operatorname{Var}(G_{n}))^{2}}+\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{6}}{(\operatorname{Var}(G_{n}))^{3}}\right)\mathbb{E}\left[e^{tF}\right]e^{ct}.$

Summarizing everything we have done so far a bound as in condition (A2) is
obtained by

$\displaystyle\mathbb{E}\left[\left|\delta\\!\left(u\right)\right|e^{tF}\right]$
$\displaystyle\leq\\!\left(\mathbb{E}\left[\\!\left(\delta\\!\left(u\right)\right)^{2}e^{tF}\right]\right)^{\frac{1}{2}}\\!\left(\mathbb{E}\left[e^{tF}\right]\right)^{\frac{1}{2}}$
$\displaystyle\leq\\!\left(\\!\left(\sum_{k\in\mathbb{Z}}\mathbb{E}\left[u_{k}^{2}e^{tF}\right]\right)^{\frac{1}{2}}+\\!\left(\sum_{\begin{subarray}{c}k,l\in\mathbb{Z}\\\
k\neq
l\end{subarray}}\mathbb{E}\left[u_{k}u_{l}X_{k}X_{l}e^{tF}\right]\right)^{\frac{1}{2}}\right)\\!\left(\mathbb{E}\left[e^{tF}\right]\right)^{\frac{1}{2}}$
$\displaystyle\leq\widetilde{\gamma_{2}}(t)\mathbb{E}\left[e^{tF}\right]$

such that

$\displaystyle\widetilde{\gamma_{2}}(t)$
$\displaystyle=Ce^{ct}\\!\left(t^{1/2}\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{3/2}}{(\operatorname{Var}(G_{n}))^{3/4}}+t\frac{\left\|a^{(n)}\right\|_{l^{3}(\mathbb{Z})}^{3}}{(\operatorname{Var}(G_{n}))^{3/2}}+(1+t)\frac{\left\|a^{(n)}\right\|_{l^{4}(\mathbb{Z})}^{2}}{(\operatorname{Var}(G_{n}))}\right)$
$\displaystyle=:Ce^{ct}\\!\left(t^{1/2}C_{n,1}^{1/2}+tC_{n,1}+(1+t)C_{n,2}\right).$

In a final step we want to simplify our bounds by comparing the constants
$C_{n,i}$ with each other. To do so, we will use, for $m\geq 2$:

$\displaystyle\left\|a^{(n)}\right\|_{l^{m}(\mathbb{Z})}^{m}$
$\displaystyle=\sum_{k\in\mathbb{Z}}\\!\left(\left|a_{k}\right|^{m-1}\left|a_{k}\right|\right)$
$\displaystyle\leq\sqrt{\sum_{k\in\mathbb{Z}}\\!\left(\left|a_{k}\right|^{m-1}\right)^{2}\sum_{k\in\mathbb{Z}}\left|a_{k}\right|^{2}}$
$\displaystyle=\sqrt{\sum_{k\in\mathbb{Z}}\\!\left(\left|a_{k}\right|^{m-1}\right)^{2}}\cdot
C\cdot(\operatorname{Var}(G_{n}))^{1/2}$
$\displaystyle\leq\sum_{k\in\mathbb{Z}}\left|a_{k}\right|^{m-1}\cdot
C\cdot(\operatorname{Var}(G_{n}))^{1/2}$
$\displaystyle=\left\|a^{(n)}\right\|_{l^{m-1}(\mathbb{Z})}^{m-1}\cdot
C\cdot(\operatorname{Var}(G_{n}))^{1/2}$

by Cauchy–Schwarz inequality. This estimation implies

$\displaystyle C_{n,1}$
$\displaystyle=\sum_{k\in\mathbb{Z}}\left|a_{k}\right|^{3}(\operatorname{Var}(G_{n}))^{-3/2}\leq\\!\left(\sum_{k\in\mathbb{Z}}\left|a_{k}\right|^{4}\right)^{1/2}(\operatorname{Var}(G_{n}))^{-1}=C_{n,2},$
$\displaystyle C_{n,3}$
$\displaystyle=\\!\left(\sum_{k\in\mathbb{Z}}\left|a_{k}\right|^{5}\right)^{1/2}(\operatorname{Var}(G_{n}))^{-5/4}\leq\\!\left(\sum_{k\in\mathbb{Z}}\left|a_{k}\right|^{4}(\operatorname{Var}(G_{n}))^{1/2}\right)^{1/2}(\operatorname{Var}(G_{n}))^{-5/4}=C_{n,2},$
$\displaystyle C_{n,4}$
$\displaystyle=\\!\left(\sum_{k\in\mathbb{Z}}\left|a_{k}\right|^{6}\right)^{1/2}(\operatorname{Var}(G_{n}))^{-3/2}\leq\\!\left(\sum_{k\in\mathbb{Z}}\left|a_{k}\right|^{5}(\operatorname{Var}(G_{n}))^{1/2}\right)^{1/2}(\operatorname{Var}(G_{n}))^{-3/2}\leq
C_{n,2}.$

So, we choose $\gamma_{1}(t):=Ce^{ct}\\!\left((1+t^{1/2}+t)C_{n}\right)$ and
$\gamma_{2}(t):=Ce^{ct}\\!\left(t^{1/2}C_{n}^{1/2}+(1+t)C_{n}\right)$ for (A1)
and (A2), respectively, and $C_{n}:=C_{n,2}$. ∎

######  Lemma 4.1.

In the setting of Theorem 1.1 we have for $i\neq j\neq k\neq l\in\mathbb{Z}$

$\displaystyle\left|\mathbb{E}[X_{i}e^{tF}]\right|\leq$
$\displaystyle\;\;\;\frac{Ct}{4\sqrt{\operatorname{Var}(G_{n})}}\cdot\\!\left(\left|a^{(n)}_{i-1}\right|+\left|a^{(n)}_{i}\right|\right)\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}$
$\displaystyle+\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{1}\in\\{i-1,i\\}\\\
n_{1}\in\\{i-2,i+1\\}\end{subarray}}\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}$
$\displaystyle+\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{2},n_{2}\in\\{i-1,i\\}\end{subarray}}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.4) $\displaystyle\left|\mathbb{E}[X_{i}X_{j}e^{tF}]\right|\leq$
$\displaystyle\;\;\;\frac{Ct}{4\sqrt{\operatorname{Var}(G_{n})}}\cdot\left|a^{(n)}_{\min(i,j)}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}\mathbbl{1}_{\left\\{\left|i-j\right|=1\right\\}}$
$\displaystyle+\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{1}\in\\{i-1,i,j-1\\}\\\
n_{1}\in\\{i-2,i+1,j-2,j+1\\}\end{subarray}}\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}$
$\displaystyle+\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{2},n_{2}\in\\{i-1,i,j-1,j\\}\end{subarray}}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.5) $\displaystyle\left|\mathbb{E}[X_{i}X_{j}X_{k}e^{tF}]\right|\leq$
$\displaystyle\;\;\;\;\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{1}\in\\{i-1,i,j-1,j,k-1,k\\}\\\
n_{1}\in\\{i-2,i+1,j-2,j+1,k-2,k+1\\}\end{subarray}}\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}$
$\displaystyle+\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{2},n_{2}\in\\{i-1,i,j-1,j,k-1,k\\}\end{subarray}}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.6) $\displaystyle\left|\mathbb{E}[X_{i}X_{j}X_{k}X_{l}e^{tF}]\right|\leq$
$\displaystyle\;\;\;\;\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\qquad\qquad\sum_{\mathclap{\begin{subarray}{c}m_{1}\in\\{i-1,i,j-1,j,k-1,k,l-1,l\\}\\\
n_{1}\in\\{i-2,i+1,j-2,j+1,k-2,k+1,l-2,l+1\\}\end{subarray}}}\qquad\qquad\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}$
$\displaystyle+\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\cdot\qquad\qquad\sum_{\mathclap{\begin{subarray}{c}m_{2},n_{2}\in\\{i-1,i,j-1,j,k-1,k,l-1,l\\}\end{subarray}}}\qquad\quad\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.7)

###### Proof.

The first key element of our strategy is to split $F$ into $F_{a}$, the
summands that depend on the $X^{\prime}s$ multiplied with $e^{tF}$, and
$F_{u}$, the summands that are independent. You should have in mind that
$F_{a}$ and $F_{u}$ are not necessarily independent from each other. To get
this dependency structure under control we will make use of several Taylor
expansions of the exponential. Note that there are remainder functions
$r_{1},r_{2}:\mathbb{R}\rightarrow\mathbb{R}$ so that $e^{x}=1+x\cdot
r_{1}(x)=1+x+\frac{x^{2}}{2}\cdot r_{2}(x)$ with $|r_{1}(x)|,|r_{2}(x)|\leq
e^{\max\\{0,x\\}}\leq e^{|x|}$ for all $x\in\mathbb{R}$. So, the second key
element is an iterated Taylor expansion on $e^{tF}$ according to the following
scheme: For a finite index set $I$, let there be real numbers $(x_{j})_{j\in
I}$, $(y_{j})_{j\in I}$ and $z$. Then by iterated Taylor expansion there is

$\displaystyle e^{z+\sum_{j\in I}x_{j}}$ $\displaystyle=1\cdot
e^{z}+\sum_{j\in I}x_{j}\cdot e^{z}+\frac{1}{2}\Big{(}\sum_{j\in
I}x_{j}\Big{)}^{2}r_{2}\Big{(}\sum_{j\in I}x_{j}\Big{)}\cdot e^{z}$
$\displaystyle=1\cdot e^{z}+\sum_{j\in I}x_{j}\cdot 1\cdot
e^{z-y_{j}}+\sum_{j\in I}x_{j}\cdot y_{j}r_{1}(y_{j})\cdot
e^{z-y_{j}}+\frac{1}{2}\Big{(}\sum_{j\in
I}x_{j}\Big{)}^{2}r_{2}\Big{(}\sum_{j\in I}x_{j}\Big{)}\cdot e^{z}.$ (4.8)

We remind on the short notation
$A_{k}:=a^{(n)}_{k}\left[X_{k}+X_{k}X_{k+1}+X_{k+1}\right]$ for the upcoming
computations. In the case of $\mathbb{E}[X_{i}e^{tF}]$:

$\displaystyle
F_{a}=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(A_{i-1}+A_{i}\right)$

and by independence

$\displaystyle\mathbb{E}[X_{i}e^{tF}]=\mathbb{E}[X_{i}e^{tF_{a}}e^{tF_{u}}]$
$\displaystyle=\mathbb{E}[X_{i}e^{tF_{u}}]+t\mathbb{E}[X_{i}F_{a}e^{tF_{u}}]+t^{2}\mathbb{E}[X_{i}F_{a}^{2}r_{2}(tF_{a})e^{tF_{u}}/2]$
$\displaystyle=t\mathbb{E}[X_{i}F_{a}e^{tF_{u}}]+t^{2}\mathbb{E}[X_{i}F_{a}^{2}r_{2}(tF_{a})e^{tF_{u}}/2],$

where we chose $z=F_{u}$ and $\sum_{j\in I}x_{j}=F_{a}$ — note that $I$ will
increase with every case since the number of multiplied $X^{\prime}s$
increases. For the first order term we split $F_{u}$ in the same manner as
before, $F_{u}=F_{u_{a}}+F_{u_{u}}$, such that
$F_{u_{a}}=\\!\left(A_{i-2}+A_{i+1}\right)/4\sqrt{\operatorname{Var}(G_{n})}$
and use the iteration from 4. Then

$\displaystyle
X_{i}F_{a}=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(a^{(n)}_{i-1}(X_{i-1}X_{i}+X_{i-1}+1)+a^{(n)}_{i}(1+X_{i+1}+X_{i}X_{i+1})\right)$

and

$\displaystyle t\mathbb{E}[X_{i}F_{a}e^{tF_{u}}]$
$\displaystyle=t\mathbb{E}\left[X_{i}F_{a}e^{tF_{u_{a}}}e^{tF_{u_{u}}}\right]$
$\displaystyle=t\mathbb{E}\left[X_{i}F_{a}e^{tF_{u_{u}}}\right]+t^{2}\mathbb{E}\left[X_{i}F_{a}F_{u_{a}}r_{1}(tF_{u_{a}})e^{tF_{u_{u}}}\right],$

so $y_{j}=F_{u_{a}}$. From here on we get $e^{tF}$ back by bounding the
difference of the independent part and $F$, e.g.

$\displaystyle\left|F_{u_{u}}-F\right|=\left|F_{u_{a}}+F_{a}\right|\leq
c=O\\!\left(\frac{1}{\sqrt{\operatorname{Var}(G_{n})}}\right)\rightarrow
0\;\text{as}\;n\rightarrow\infty.$

Since the exact constant is not important, we always write just $c$ if we use
that type of estimation, and in the same way $C$ for prefactors. Thus

$\displaystyle t\left|\mathbb{E}[X_{i}F_{a}e^{tF_{u}}]\right|\leq$
$\displaystyle\frac{Ct}{4\sqrt{\operatorname{Var}(G_{n})}}\cdot\\!\left(\left|a^{(n)}_{i-1}\right|+\left|a^{(n)}_{i}\right|\right)\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}$
$\displaystyle+\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{1}\in\\{i-1,i\\}\\\
n_{1}\in\\{i-2,i+1\\}\end{subarray}}\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.9)

For the second order term we just bound

$\displaystyle
t^{2}\left|\mathbb{E}[X_{i}F_{a}^{2}r_{2}(tF_{a})e^{tF_{u}}/2]\right|$
$\displaystyle\leq\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{2},n_{2}\in\\{i-1,i\\}\end{subarray}}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.10)

In the case of $\mathbb{E}[X_{i}X_{j}e^{tF}]$:

$\displaystyle
F_{a}=\begin{cases}\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(A_{i}+A_{j-1}+A_{j}\right),&i=j+1,\\\
\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(A_{i-1}+A_{i}+A_{j-1}+A_{j}\right),&\left|i-j\right|\geq
2,\\\
\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(A_{i-1}+A_{i}+A_{j}\right),&j=i+1,\end{cases}$

and

$\displaystyle\mathbb{E}[X_{i}X_{j}e^{tF}]$
$\displaystyle=t\mathbb{E}[X_{i}X_{j}F_{a}e^{tF_{u}}]+t^{2}\mathbb{E}[X_{i}X_{j}F_{a}^{2}r_{2}(tF_{a})e^{tF_{u}}/2].$

For the first order term we compute as a preparation

$\displaystyle
X_{i}X_{j}F_{a}=a^{(n)}_{i-1}(X_{i-1}X_{i}X_{j}+X_{i-1}X_{j}+X_{j})$
$\displaystyle+a^{(n)}_{i}(X_{j}+X_{i+1}X_{j}+X_{i}X_{i+1}X_{j})$
$\displaystyle+a^{(n)}_{j-1}(X_{i}X_{j-1}X_{j}+X_{i}X_{j-1}+X_{i})$
$\displaystyle+a^{(n)}_{j}(X_{i}+X_{i}X_{j+1}+X_{i}X_{j}X_{j+1}).$

In particular we have to consider the special case $\left|i-j\right|=1$ and
assume $i=j+1$. If not, we just have to swap $i$ and $j$. Under our assumption
the last equation reduces to

$\displaystyle X_{i}X_{j}F_{a}=a^{(n)}_{i-1}(X_{i}+1+X_{j})$
$\displaystyle+a^{(n)}_{i}(X_{j}+X_{i+1}X_{j}+X_{i}X_{i+1}X_{j})$
$\displaystyle+a^{(n)}_{j-1}(X_{i}X_{j-1}X_{j}+X_{i}X_{j-1}+X_{i}).$

From here on we assume that the indices apperaring in upcoming $F_{a}$’s and
$F_{u_{a}}$’s are all different. If not, there is only an effect on the number
of coefficients and so the constants, but not on the order of our bound.
Having that in mind we split $F_{u}$ in the same manner as before,
$F_{u}=F_{u_{a}}+F_{u_{u}}$, such that
$F_{u_{a}}=\\!\left(A_{i-2}+A_{i+1}+A_{j-2}+A_{j+1}\right)/4\sqrt{\operatorname{Var}(G_{n})}$.
Then

$\displaystyle t\mathbb{E}[X_{i}X_{j}F_{a}e^{tF_{u}}]$
$\displaystyle=t\mathbb{E}[X_{i}X_{j}F_{a}e^{tF_{u_{u}}}]+t^{2}\mathbb{E}[X_{i}X_{j}F_{a}F_{u_{a}}r_{1}(tF_{u_{a}})e^{tF_{u_{u}}}]$

and thus for $i=j+1$

$\displaystyle t\left|\mathbb{E}[X_{i}X_{j}F_{a}e^{tF_{u}}]\right|\leq$
$\displaystyle\frac{Ct}{4\sqrt{\operatorname{Var}(G_{n})}}\cdot\left|a^{(n)}_{i-1}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}$
$\displaystyle+\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{1}\in\\{i-1,i,j-1\\}\\\
n_{1}\in\\{i-2,i+1,j-2,j+1\\}\end{subarray}}\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.11)

In the case $\left|i-j\right|\geq 2$ the first term of the last inequality
does not appear since all the indices in $X_{i}X_{j}F_{a}$ are different:

$\displaystyle
t\left|\mathbb{E}[X_{i}X_{j}F_{a}e^{tF_{u}}]\right|\leq\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{1}\in\\{i-1,i,j-1,j\\}\\\
n_{1}\in\\{i-2,i+1,j-2,j+1\\}\end{subarray}}\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.12)

For the second order term we just bound

$\displaystyle
t^{2}\left|\mathbb{E}[X_{i}X_{j}F_{a}^{2}r_{2}(tF_{a})e^{tF_{u}}/2]\right|$
$\displaystyle\leq\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{2},n_{2}\in\\{i-1,i,j-1,j\\}\end{subarray}}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.13)

In the case of $\mathbb{E}[X_{i}X_{j}X_{k}e^{tF}]$:

$\displaystyle
F_{a}=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(A_{i-1}+A_{i}+A_{j-1}+A_{j}+A_{k-1}+A_{k}\right)$

and

$\displaystyle\mathbb{E}[X_{i}X_{j}X_{k}e^{tF}]$
$\displaystyle=t\mathbb{E}[X_{i}X_{j}X_{k}F_{a}e^{tF_{u}}]+t^{2}\mathbb{E}[X_{i}X_{j}X_{k}F_{a}^{2}r_{2}(tF_{a})e^{tF_{u}}/2].$

For the first order term we compute as a preparation

$\displaystyle X_{i}X_{j}X_{k}F_{a}=$ $\displaystyle
a^{(n)}_{i-1}(X_{i-1}X_{i}X_{j}X_{k}+X_{i-1}X_{j}X_{k}+X_{j}X_{k})$
$\displaystyle+a^{(n)}_{i}(X_{j}X_{k}+X_{i+1}X_{j}X_{k}+X_{i}X_{i+1}X_{j}X_{k})$
$\displaystyle+a^{(n)}_{j-1}(X_{i}X_{j-1}X_{j}X_{k}+X_{i}X_{j-1}X_{k}+X_{i}X_{k})$
$\displaystyle+a^{(n)}_{j}(X_{i}X_{k}+X_{i}X_{j+1}X_{k}+X_{i}X_{j}X_{j+1}X_{k})$
$\displaystyle+a^{(n)}_{k-1}(X_{i}X_{j}X_{k-1}X_{k}+X_{i}X_{j}X_{k-1}+X_{i}X_{j})$
$\displaystyle+a^{(n)}_{k}(X_{i}X_{j}+X_{i}X_{j}X_{k+1}+X_{i}X_{j}X_{k}X_{k+1}).$

And by our assumption $i\neq j\neq k$ in every summand at least one $X$ will
remain. We split $F_{u}=F_{u_{a}}+F_{u_{u}}$, such that
$F_{u_{a}}=\\!\left(A_{i-2}+A_{i+1}+A_{j-2}+A_{j+1}+A_{k-2}+A_{k+1}\right)/4\sqrt{\operatorname{Var}(G_{n})}$.
Then

$\displaystyle t\mathbb{E}[X_{i}X_{j}X_{k}F_{a}e^{tF_{u}}]$
$\displaystyle=t\mathbb{E}[X_{i}X_{j}X_{k}F_{a}e^{tF_{u_{a}}}e^{tF_{u_{u}}}]$
$\displaystyle=t\mathbb{E}[X_{i}X_{j}X_{k}F_{a}e^{tF_{u_{u}}}]+t^{2}\mathbb{E}[X_{i}X_{j}X_{k}F_{a}F_{u_{a}}r_{1}(tF_{u_{a}})e^{tF_{u_{u}}}]$
$\displaystyle=t^{2}\mathbb{E}[X_{i}X_{j}X_{k}F_{a}F_{u_{a}}r_{1}(tF_{u_{a}})e^{tF_{u_{u}}}]$

by independence and thus

$\displaystyle
t\left|\mathbb{E}[X_{i}X_{j}X_{k}F_{a}e^{tF_{u}}]\right|\leq\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{1}\in\\{i-1,i,j-1,j,k-1,k\\}\\\
n_{1}\in\\{i-2,i+1,j-2,j+1,k-2,k+1\\}\end{subarray}}\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.14)

For the second order term we just bound

$\displaystyle
t^{2}\left|\mathbb{E}[X_{i}X_{j}X_{k}F_{a}^{2}r_{2}(tF_{a})e^{tF_{u}}/2]\right|$
$\displaystyle\leq\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\cdot\sum_{\begin{subarray}{c}m_{2},n_{2}\in\\{i-1,i,j-1,j,k-1,k\\}\end{subarray}}\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.15)

In the case of $\mathbb{E}[X_{i}X_{j}X_{k}X_{l}e^{tF}]$:

$\displaystyle
F_{a}=\frac{1}{4\sqrt{\operatorname{Var}(G_{n})}}\\!\left(A_{i-1}+A_{i}+A_{j-1}+A_{j}+A_{k-1}+A_{k}+A_{l-1}+A_{l}\right)$

and

$\displaystyle\mathbb{E}[X_{i}X_{j}X_{k}X_{l}e^{tF}]$
$\displaystyle=t\mathbb{E}[X_{i}X_{j}X_{k}X_{l}F_{a}e^{tF_{u}}]+t^{2}\mathbb{E}[X_{i}X_{j}X_{k}X_{l}F_{a}^{2}r_{2}(tF_{a})e^{tF_{u}}/2].$

For the first order term we compute as a preparation

$\displaystyle X_{i}X_{j}X_{k}X_{l}F_{a}=$ $\displaystyle
a^{(n)}_{i-1}(X_{i-1}X_{i}X_{j}X_{k}X_{l}+X_{i-1}X_{j}X_{k}X_{l}+X_{j}X_{k}X_{l})$
$\displaystyle+a^{(n)}_{i}(X_{j}X_{k}X_{l}+X_{i+1}X_{j}X_{k}X_{l}+X_{i}X_{i+1}X_{j}X_{k}X_{l})$
$\displaystyle+a^{(n)}_{j-1}(X_{i}X_{j-1}X_{j}X_{k}X_{l}+X_{i}X_{j-1}X_{k}X_{l}+X_{i}X_{k}X_{l})$
$\displaystyle+a^{(n)}_{j}(X_{i}X_{k}X_{l}+X_{i}X_{j+1}X_{k}X_{l}+X_{i}X_{j}X_{j+1}X_{k}X_{l})$
$\displaystyle+a^{(n)}_{k-1}(X_{i}X_{j}X_{k-1}X_{k}X_{l}+X_{i}X_{j}X_{k-1}X_{l}+X_{i}X_{j}X_{l})$
$\displaystyle+a^{(n)}_{k}(X_{i}X_{j}X_{l}+X_{i}X_{j}X_{k+1}X_{l}+X_{i}X_{j}X_{k}X_{k+1}X_{l})$
$\displaystyle+a^{(n)}_{l-1}(X_{i}X_{j}X_{k}X_{l-1}X_{l}+X_{i}X_{j}X_{k}X_{l-1}+X_{i}X_{j}X_{k})$
$\displaystyle+a^{(n)}_{k}(X_{i}X_{j}X_{k}+X_{i}X_{j}X_{k}X_{l+1}+X_{i}X_{j}X_{k}X_{l}X_{l+1}).$

And by our assumption $i\neq j\neq k\neq l$ in every summand at least one $X$
will remain. $F_{u_{a}}$ is given by
$F_{u_{a}}=\\!\left(A_{i-2}+A_{i+1}+A_{j-2}+A_{j+1}+A_{k-2}+A_{k+1}+A_{l-2}+A_{l+1}\right)/4\sqrt{\operatorname{Var}(G_{n})}$
this time. Then

$\displaystyle t\mathbb{E}[X_{i}X_{j}X_{k}X_{l}F_{a}e^{tF_{u}}]$
$\displaystyle=t^{2}\mathbb{E}[X_{i}X_{j}X_{k}X_{l}F_{a}F_{u_{a}}r_{1}(tF_{u_{a}})e^{tF_{u_{u}}}]$

by independence and thus

$\displaystyle
t\left|\mathbb{E}[X_{i}X_{j}X_{k}X_{l}F_{a}e^{tF_{u}}]\right|\leq\frac{Ct^{2}}{16\operatorname{Var}(G_{n})}\cdot\qquad\qquad\sum_{\mathclap{\begin{subarray}{c}m_{1}\in\\{i-1,i,j-1,j,k-1,k,l-1,l\\}\\\
n_{1}\in\\{i-2,i+1,j-2,j+1,k-2,k+1,l-2,l+1\\}\end{subarray}}}\qquad\qquad\left|a^{(n)}_{m_{1}}\right|\left|a^{(n)}_{n_{1}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.16)

For the second order term we just bound

$\displaystyle
t^{2}\left|\mathbb{E}[X_{i}X_{j}X_{k}X_{l}F_{a}^{2}r_{2}(tF_{a})e^{tF_{u}}/2]\right|$
$\displaystyle\leq\frac{Ct^{2}}{32\operatorname{Var}(G_{n})}\cdot\qquad\quad\sum_{\mathclap{\begin{subarray}{c}m_{2},n_{2}\in\\{i-1,i,j-1,j,k-1,k,l-1,l\\}\end{subarray}}}\qquad\quad\left|a^{(n)}_{m_{2}}\right|\left|a^{(n)}_{n_{2}}\right|\cdot\mathbb{E}\left[e^{tF}\right]e^{ct}.$
(4.17)

∎

##  5\. Proofs II: Subgraph counts in the Erdős–Rényi random graph

There are $\binom{n}{2}$ possible edges in the Erdős–Rényi random graph.
Hence, we can describe it by using as many Rademacher random variables: Let
$E$ be the set of all possible edges of $G(n,p)$ and let $(X_{k})_{k\in E}$ be
a set of independent Rademacher random variables, in which $X_{k}=1$ indicates
the presence of edge $k$ in $G(n,p)$. Thus, $\mathbb{P}(X_{k}=1)=p$ for all
$k\in E$. In our calculations, we will make use of the following rescaled
versions of $(X_{k})_{k\in E}$: For $k\in E$ we define
$B_{k}=\frac{1}{2}(X_{k}+1)$, which is Bernoulli($p$)-distributed, and
$Y_{k}=(pq)^{-\frac{1}{2}}(B_{k}-p)$, which is standardized. Further, for any
subset $A\subset E$, we shorten the notation for the product of $(B_{k})_{k\in
A}$ or $(Y_{k})_{k\in A}$ by defining $B_{A}:=\prod_{k\in A}B_{k}$ and
$Y_{A}:=\prod_{k\in A}Y_{k}$, where $B_{\emptyset}=Y_{\emptyset}=1$ by
convention. For any integrable random variable $Z$, let
$Z^{c}:=Z-\mathbb{E}[Z]$ denote the centered version of $Z$.

Let ${G_{0}}$ be a graph with at least one edge. Neither the standardized
number of copies of ${G_{0}}$ in $G(n,p)$ nor $\Psi_{\min}$ do depend on the
number of isolated vertices of ${G_{0}}$, see Lemma 4.3 in [8]. Hence, without
loss of generality, we may assume that ${G_{0}}$ does not have isolated
vertices. This way, every copy of ${G_{0}}$ in $G(n,p)$ can be identified by
its set of edges. Therefore, it will be useful to simplify our notation:

For any set of edges $\Gamma\subset E$ let $\operatorname{graph}(\Gamma)$
denote the graph consisting of all edges given by $\Gamma$ and all necessary
vertices. Let $v_{\Gamma}:=v_{\operatorname{graph}(\Gamma)}$,
$e_{\Gamma}:=e_{\operatorname{graph}(\Gamma)}$, and
$a_{\Gamma}:=a_{\operatorname{graph}(\Gamma)}$. We will call $\Gamma$ to be a
_copy_ of ${G_{0}}$ if $\operatorname{graph}(\Gamma)$ is a copy of ${G_{0}}$.
Let $\mathcal{M}:=\\{\Gamma\subset E\,|\,\Gamma$ is a copy of ${G_{0}}\\}$ be
the set of all possible copies of ${G_{0}}$ in $G(n,p)$, and let
$\mathcal{M}_{k}:=\\{\Gamma\in\mathcal{M}\,|\,k\in\Gamma\\}$ be the set of all
possible copies that contain a specific edge $k\in E$. Further, for any non-
empty $A\subset E$ let $\mathcal{M}_{A}:=\bigcup_{k\in A}\mathcal{M}_{k}$ be
the set of all possible copies that contain at least one of the edges given by
$A$. This last definition will be mainly used in case of
$A=\Gamma\in\mathcal{M}$ being a copy of ${G_{0}}$. In this case,
$\mathcal{M}_{\Gamma}$ contains all possible copies of ${G_{0}}$ that have at
least one edge in common with $\Gamma$. We will call $\mathcal{M}_{\Gamma}$ to
be the _neighborhood_ of $\Gamma$, and the elements of $\mathcal{M}_{\Gamma}$
to be the _neighbors_ of $\Gamma$. Due to symmetry, the cardinality of the
neighborhood $\mathcal{M}_{\Gamma}$ does not depend on the choice of
$\Gamma\in\mathcal{M}$. We will denote this cardinality by $D$.

The number of copies of ${G_{0}}$ in $G(n,p)$ is given by
$\sum_{\Gamma\in\mathcal{M}}B_{\Gamma}$. Its standardization is

$\displaystyle W$
$\displaystyle:=\frac{1}{\sigma}\sum_{\Gamma\in\mathcal{M}}(B_{\Gamma}-\mathbb{E}[B_{\Gamma}])=\frac{1}{\sigma}\sum_{\Gamma\in\mathcal{M}}B_{\Gamma}^{c},$

where $\sigma^{2}:=\operatorname{Var}(\sum_{\Gamma\in\mathcal{M}}B_{\Gamma})$.
From Lemma 4.2 in [8] we know that

$\displaystyle\sigma^{2}$ $\displaystyle\geq\frac{q}{2\cdot v_{{G_{0}}}!\cdot
a_{{G_{0}}}}\cdot\frac{n^{2v_{{G_{0}}}}p^{2e_{{G_{0}}}}}{\Psi_{\min}}$ (5.1)

for $n\geq 4v_{{G_{0}}}^{2}$.

The main goal of this section is to prove Theorem 1.4 and Corollary 1.5, which
have been presented in our introductory section. These proofs are postponed to
the end of this section. Before, we present several lemmas that deal with some
basic yet very important properties of our random variables. These results
will be useful repeatedly in the proof of Theorem 1.4.

We start by exploring the behavior of our most important operators, the
discrete gradient $D_{k}$ with $k\in E$ and the pseudo-inverse Ornstein-
Uhlenbeck operator $L^{-1}$.

######  Lemma 5.1.

For any subset $A\subset E$ and any edge $k\in E$ there is

$\displaystyle(i)$ $\displaystyle D_{k}B_{A}^{c}$
$\displaystyle=\sqrt{pq}\cdot\mathbbl{1}_{\left\\{k\in A\right\\}}\cdot
B_{A\backslash\\{k\\}},$ $\displaystyle(ii)$ $\displaystyle-
D_{k}L^{-1}B_{A}^{c}$ $\displaystyle=\sqrt{pq}\cdot\mathbbl{1}_{\left\\{k\in
A\right\\}}\cdot\sum_{\\{k\\}\subset\alpha\subset
A}\frac{p^{|A|-|\alpha|}}{|A|\cdot\binom{|A|-1}{|\alpha|-1}}B_{\alpha\backslash\\{k\\}}.$

In particular, all expressions above are non-negative.

######  Remark 5.2.

In the application of Lemma 5.1 $(ii)$ it will be useful to keep in mind that

$\displaystyle\sum_{\\{k\\}\subset\alpha\subset
A}\frac{1}{|A|\cdot\binom{|A|-1}{|\alpha|-1}}$
$\displaystyle=\sum_{\alpha\subset
A\backslash\\{k\\}}\frac{1}{|A|\cdot\binom{|A|-1}{|\alpha|}}=\sum_{i=0}^{|A|-1}\bigg{(}\frac{1}{|A|}\sum_{\begin{subarray}{c}\alpha\subset
A\backslash\\{k\\}\\\
|\alpha|=i\end{subarray}}\frac{1}{\binom{|A|-1}{i}}\bigg{)}=\sum_{i=0}^{|A|-1}\frac{1}{|A|}=1$

for any non-empty subset $A\subset E$ and any edge $k\in A$.

###### Proof of Lemma 5.1.

First, we note that
$D_{k}B_{A}^{c}=D_{k}(B_{A}-\mathbb{E}[B_{A}])=D_{k}B_{A}$. If $k\not\in A$,
$B_{A}$ is independent of $X_{k}$ so that $D_{k}B_{A}=0$. If $k\in A$, there
is $B_{A}=B_{A\backslash\\{k\\}}\cdot
B_{k}=B_{A\backslash\\{k\\}}\cdot\frac{1}{2}(X_{k}+1)$ so that
$D_{k}B_{A}=\sqrt{pq}\cdot(B_{A\backslash\\{k\\}}\cdot
1-B_{A\backslash\\{k\\}}\cdot 0)$. This proves $(i)$.

To prove the second part of our statement, we need to represent $B_{A}$ with
regard to the standardized random variables
$Y_{\ell}=(pq)^{-\frac{1}{2}}(B_{\ell}-p)$, $\ell\in E$. By expansion of the
product, we see that

$\displaystyle B_{A}^{c}$
$\displaystyle=B_{A}-\mathbb{E}[B_{A}]=\prod_{\ell\in
A}\big{(}(pq)^{\frac{1}{2}}Y_{\ell}+p\big{)}-p^{|A|}=\sum_{\emptyset\neq\alpha\subset
A}p^{|A|-\frac{1}{2}|\alpha|}q^{\frac{1}{2}|\alpha|}Y_{\alpha}.$

Now we are able to describe how the pseudo-inverse Ornstein–Uhlenbeck operator
works on this expression. We get

$\displaystyle-L^{-1}B_{A}^{c}$
$\displaystyle=\sum_{\emptyset\neq\alpha\subset
A}\frac{p^{|A|-\frac{1}{2}|\alpha|}q^{\frac{1}{2}|\alpha|}}{|\alpha|}Y_{\alpha}.$

Having applied the operator, we now want to transform the result back to a
representation using the Bernoulli variables $(B_{\ell})_{\ell\in E}$. Thus,

$\displaystyle-L^{-1}B_{A}^{c}$
$\displaystyle=\sum_{\emptyset\neq\alpha\subset
A}\bigg{(}\frac{p^{|A|-\frac{1}{2}|\alpha|}q^{\frac{1}{2}|\alpha|}}{|\alpha|}\cdot\prod_{\ell\in\alpha}\Big{(}(pq)^{-\frac{1}{2}}(B_{\ell}-p)\Big{)}\bigg{)}=\sum_{\emptyset\neq\alpha\subset
A}\sum_{\hat{\alpha}\subset\alpha}\frac{p^{|A|-|\hat{\alpha}|}}{|\alpha|}(-1)^{|\alpha|-|\hat{\alpha}|}B_{\hat{\alpha}}.$

The double sum can be rearranged so that we can first choose $\hat{\alpha}$ as
an arbitrary subset of $A$, and then fill up the rest
$\tilde{\alpha}:=\alpha\backslash\hat{\alpha}\subset A\backslash\hat{\alpha}$.
However, using this rearrangement, we have to explicitly exclude the case in
which $\hat{\alpha}=\tilde{\alpha}=\emptyset$. This leads to

$\displaystyle-L^{-1}B_{A}^{c}$ $\displaystyle=\sum_{\hat{\alpha}\subset
A}\sum_{\begin{subarray}{c}\tilde{\alpha}\subset A\backslash\hat{\alpha}\\\
\hat{\alpha}\cup\tilde{\alpha}\neq\emptyset\end{subarray}}\frac{p^{|A|-|\hat{\alpha}|}}{|\hat{\alpha}|+|\tilde{\alpha}|}(-1)^{|\tilde{\alpha}|}B_{\hat{\alpha}}$
$\displaystyle=\sum_{\emptyset\neq\tilde{\alpha}\subset
A}\frac{p^{|A|}}{|\tilde{\alpha}|}(-1)^{|\tilde{\alpha}|}B_{\emptyset}+\sum_{\emptyset\neq\hat{\alpha}\subset
A}\sum_{\tilde{\alpha}\subset
A\backslash\hat{\alpha}}\frac{p^{|A|-|\hat{\alpha}|}}{|\hat{\alpha}|+|\tilde{\alpha}|}(-1)^{|\tilde{\alpha}|}B_{\hat{\alpha}}$
$\displaystyle=p^{|A|}\sum_{i=1}^{|A|}\binom{|A|}{i}\frac{(-1)^{i}}{i}+\sum_{\emptyset\neq\hat{\alpha}\subset
A}\Bigg{(}p^{|A|-|\hat{\alpha}|}B_{\hat{\alpha}}\sum_{i=0}^{|A|-|\hat{\alpha}|}\binom{|A|-|\hat{\alpha}|}{i}\frac{(-1)^{i}}{|\hat{\alpha}|+i}\Bigg{)}.$

The first term above is not random but deterministic and hence not of
interest, as it will disappear after the application of the discrete gradient.
To handle the second term, we use the Beta-function, which is usually defined
via $\operatorname{Beta}(x,y)=\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$ for all
$x,y\in\mathbb{C}$ that have a positive real part. It is well known that in
case of $x,y\in\mathbb{N}$ there is
$\sum_{i=0}^{x-1}\binom{x-1}{i}\frac{(-1)^{i}}{y+i}=\operatorname{Beta}(x,y)=(x+y-1)^{-1}\cdot\binom{x+y-2}{y-1}^{-1}$.
Application of these results on $x=|A|-|\hat{\alpha}|+1$ and
$y=|\hat{\alpha}|$ yields

$\displaystyle\sum_{i=0}^{|A|-|\hat{\alpha}|}\binom{|A|-|\hat{\alpha}|}{i}\frac{(-1)^{i}}{|\hat{\alpha}|+i}$
$\displaystyle=\frac{1}{|A|\cdot\binom{|A|-1}{|\hat{\alpha}|-1}}$

so that

$\displaystyle-L^{-1}B_{A}^{c}$
$\displaystyle=p^{|A|}\sum_{i=1}^{|A|}\binom{|A|}{i}\frac{(-1)^{i}}{i}+\sum_{\emptyset\neq\hat{\alpha}\subset
A}\frac{p^{|A|-|\hat{\alpha}|}}{|A|\cdot\binom{|A|-1}{|\hat{\alpha}|-1}}B_{\hat{\alpha}}.$

When applying the discrete gradient, the deterministic term cancels itself
out, while the second term can be handled the same way as shown in the first
part of this proof. We arrive at

$\displaystyle-D_{k}L^{-1}B_{A}^{c}$
$\displaystyle=\sqrt{pq}\cdot\sum_{\emptyset\neq\hat{\alpha}\subset
A}\frac{p^{|A|-|\hat{\alpha}|}}{|A|\cdot\binom{|A|-1}{|\hat{\alpha}|-1}}B_{\hat{\alpha}\backslash\\{k\\}}\mathbbl{1}_{\left\\{k\in\hat{\alpha}\right\\}}$
$\displaystyle=\sqrt{pq}\cdot\mathbbl{1}_{\left\\{k\in
A\right\\}}\cdot\sum_{\\{k\\}\subset\alpha\subset
A}\frac{p^{|A|-|\alpha|}}{|A|\cdot\binom{|A|-1}{|\alpha|-1}}B_{\alpha\backslash\\{k\\}}.$

This proves the second part of our statement. ∎

Next, we want to improve our understanding of the correlation of our Bernoulli
random variables $\\{B_{A}\\}_{A\subset E}$.

######  Lemma 5.3.

For every $i\in\mathbb{N}$ let $A_{i}\subset E$ be a set of edges. And let
$I\subset\mathbb{N}$ be a finite index set. Then the following statements
hold:

$\displaystyle 0$
$\displaystyle\;\leq\;\mathbb{E}[B_{A_{1}}]\mathbb{E}[B_{A_{2}}]$
$\displaystyle\;\leq\;\mathbb{E}[B_{A_{1}\cup A_{2}}],$ ($i$) $\displaystyle
0$ $\displaystyle\;\leq\;\mathbb{E}[B_{A_{1}}^{c}B_{A_{2}}^{c}]$
$\displaystyle\;\leq\;\mathbb{E}[B_{A_{1}\cup A_{2}}],$ ($ii$) $\displaystyle
0$ $\displaystyle\;\leq\;\mathbb{E}[B_{A_{1}}^{c}B_{A_{2}}^{c}B_{A_{3}}^{c}]$
$\displaystyle\;\leq\;\mathbb{E}[B_{A_{1}\cup A_{2}\cup A_{3}}],$ ($iii$)
$\displaystyle 0$ $\displaystyle\;\leq\;\mathbb{E}\Big{[}\Big{|}\prod_{i\in
I}B_{A_{i}}^{c}\Big{|}\Big{]}$
$\displaystyle\;\leq\;2^{|I|}\mathbb{E}[B_{\bigcup_{i\in I}A_{i}}].$ ($iv$)

###### Proof.

Obviously, $\mathbb{E}[B_{A_{i}}]=p^{|A_{i}|}$ is non-negative for all
$i\in\mathbb{N}$. Further,

$\displaystyle\mathbb{E}[B_{A_{1}}]\mathbb{E}[B_{A_{2}}]$
$\displaystyle=p^{|A_{1}|+|A_{2}|}\leq p^{|A_{1}\cup
A_{2}|}=\mathbb{E}[B_{A_{1}\cup A_{2}}].$

This proves $(i)$. To varify $(ii)$, see

$\displaystyle\mathbb{E}[B_{A_{1}}^{c}B_{A_{2}}^{c}]$
$\displaystyle=\mathbb{E}\big{[}\big{(}B_{A_{1}}-\mathbb{E}[B_{A_{1}}]\big{)}\cdot\big{(}B_{A_{2}}-\mathbb{E}[B_{A_{2}}]\big{)}\big{]}=\mathbb{E}[B_{A_{1}\cup
A_{2}}]-\mathbb{E}[B_{A_{1}}]\mathbb{E}[B_{A_{1}}].$

Hence, $(ii)$ follows from $(i)$. Analogously, due to $(i)$ there is

$\displaystyle\mathbb{E}[B_{A_{1}}^{c}B_{A_{2}}^{c}B_{A_{3}}^{c}]={}$
$\displaystyle\mathbb{E}\big{[}\big{(}B_{A_{1}}-\mathbb{E}[B_{A_{1}}]\big{)}\cdot\big{(}B_{A_{2}}-\mathbb{E}[B_{A_{2}}]\big{)}\cdot\big{(}B_{A_{3}}-\mathbb{E}[B_{A_{3}}]\big{)}\big{]}$
$\displaystyle={}$ $\displaystyle\mathbb{E}[B_{A_{1}\cup A_{2}\cup
A_{3}}]-\mathbb{E}[B_{A_{1}}]\big{(}\mathbb{E}[B_{A_{2}\cup
A_{3}}]-\mathbb{E}[B_{A_{2}}]\mathbb{E}[B_{A_{3}}]\big{)}$
$\displaystyle\phantom{\mathbb{E}[B_{A_{1}\cup A_{2}\cup
A_{3}}]}-\mathbb{E}[B_{A_{2}}]\big{(}\mathbb{E}[B_{A_{1}\cup
A_{3}}]-\mathbb{E}[B_{A_{1}}]\mathbb{E}[B_{A_{3}}]\big{)}$
$\displaystyle\phantom{\mathbb{E}[B_{A_{1}\cup A_{2}\cup
A_{3}}]}-\mathbb{E}[B_{A_{3}}]\big{(}\mathbb{E}[B_{A_{1}\cup
A_{2}}]-\mathbb{E}[B_{A_{1}}]\mathbb{E}[B_{A_{2}}]\big{)}$
$\displaystyle\phantom{\mathbb{E}[B_{A_{1}\cup A_{2}\cup
A_{3}}]}-\mathbb{E}[B_{A_{1}}]\mathbb{E}[B_{A_{2}}]\mathbb{E}[B_{A_{3}}]$
$\displaystyle\leq{}$ $\displaystyle\mathbb{E}[B_{A_{1}\cup A_{2}\cup
A_{3}}],$

which proves $(iii)$. Finally, by expansion of the product, we see

$\displaystyle\mathbb{E}\Big{[}\Big{|}\prod_{i\in
I}B_{A_{i}}^{c}\Big{|}\Big{]}$ $\displaystyle\leq\sum_{\hat{I}\subset
I}\bigg{(}\mathbb{E}\Big{[}\prod_{i\in\hat{I}}B_{A_{i}}\Big{]}\cdot\prod_{i\in
I\backslash\hat{I}}\mathbb{E}[B_{A_{i}}]\bigg{)}\leq\sum_{\hat{I}\subset
I}\mathbb{E}[B_{\bigcup_{i\in I}A_{i}}]=2^{|I|}\mathbb{E}[B_{\bigcup_{i\in
I}A_{i}}],$

where we again used our result from $(i)$. This finishes the proof. ∎

We will later have to deal with sums of powers of $p$, in which the structure
of the sum is determined via the neighborhood structure of the copies of
${G_{0}}$. These sums together with the variance $\sigma^{2}$ are the main
reason for the appearance of $\Psi_{\min}$ in our results, and will now be
examined in detail.

######  Definition 5.4.

Let $m\in\mathbb{N}$ be a natural number, and let
$M=(\Gamma_{i})_{i=1}^{m}\subset\mathcal{M}$ be an ordered set of copies, in
which each copy is a neighbor of at least one of its predecessors, i.e.
$\Gamma_{1}\in\mathcal{M}$ and
$\Gamma_{i}\in\bigcup_{j=1}^{i-1}\mathcal{M}_{\Gamma_{j}}$ for all $2\leq
i\leq m$. We then call $M$ to be a _set of connected copies_ of size $m$.

######  Lemma 5.5.

Let $m$ and $\hat{m}$ be natural numbers, and let $\bar{m}=m+\hat{m}$. Then

$\displaystyle(i)$
$\displaystyle\sum_{\mathclap{M:|M|=m\text{\,con.\,cop.}}}p^{|\bigcup_{\Gamma\in
M}\Gamma|}$ $\displaystyle\leq c_{m}\cdot\frac{n^{m\cdot v_{{G_{0}}}}\cdot
p^{m\cdot e_{{G_{0}}}}}{\Psi_{\min}^{m-1}},$ $\displaystyle(ii)$
$\displaystyle\sum_{\mathclap{\begin{subarray}{c}M:|M|=m\text{\,con.\,cop.}\\\
\hat{M}:|\hat{M}|=\hat{m}\text{\,con.\,cop.}\end{subarray}}}p^{|\bigcup_{\Gamma\in
M\cup\hat{M}}\Gamma|}$ $\displaystyle\leq
c_{\bar{m}}\cdot\frac{n^{\bar{m}\cdot v_{{G_{0}}}}\cdot p^{\bar{m}\cdot
e_{{G_{0}}}}}{\Psi_{\min}^{\bar{m}-2}\cdot\min\\{\Psi_{\min},1\\}},$

where the sums run over all sets of connected copies of size $m$ or $\hat{m}$,
respectively, and where the constants are given by
$c_{k}:=(v_{{G_{0}}}!)^{k-1}\cdot 2^{\frac{1}{2}k(k-1)\cdot e_{{G_{0}}}}\cdot
a_{{G_{0}}}^{-k}$ for $k\in\mathbb{N}$.

This lemma summarizes a counting technique that is commonly used in the field
of random subgraph counting, see e.g. (3) in [20], (3.10) in [3], and (3.10)
in [11]. However, in contrast to these references, we formulate the result of
this counting strategy as an independent lemma for an arbitrary number of
connected copies of ${G_{0}}$.

###### Proof of Lemma 5.5.

To prove $(i)$, let us first assume that $m=1$. In this case, every set of
connected copies $M$ consists of only one copy $\Gamma_{1}$ that can be
arbitrarily chosen among all copies in $\mathcal{M}$. Therefore, we are
interested in the size of $\mathcal{M}$: There are $\binom{n}{v_{{G_{0}}}}$
possibilities to choose $v_{{G_{0}}}$ vertices. On every set of $v_{{G_{0}}}$
vertices, we can find $\frac{v_{{G_{0}}}!}{a_{{G_{0}}}}$ copies of ${G_{0}}$.
So there are
$\frac{v_{{G_{0}}}!}{a_{{G_{0}}}}\cdot\binom{n}{v_{{G_{0}}}}\leq\frac{n^{v_{{G_{0}}}}}{a_{{G_{0}}}}$
possible choices for $\Gamma_{1}\in\mathcal{M}$. Hence,

$\displaystyle\sum_{\mathclap{\begin{subarray}{c}M:|M|=1\\\
\text{\,con.\,cop.}\end{subarray}}}p^{|\bigcup_{\Gamma\in M}\Gamma|}$
$\displaystyle=\sum_{\mathclap{\Gamma\in\mathcal{M}}}p^{|\Gamma|}\leq\frac{n^{v_{{G_{0}}}}}{a_{{G_{0}}}}\cdot
p^{e_{G_{0}}}=c_{1}\cdot\frac{n^{1\cdot v_{{G_{0}}}}\cdot p^{1\cdot
e_{{G_{0}}}}}{\Psi_{\min}^{0}}.$

Next, we assume that $(i)$ holds true for some fixed $m\in\mathbb{N}$ and we
want to prove that $(i)$ still holds if $m$ is increased by $1$. If $\hat{M}$
is a set of connected copies of size $m+1$, then the first $m$ elements of
$\hat{M}$ form a set of connected copies of size $m$. Hence,

$\displaystyle\sum_{\mathclap{\begin{subarray}{c}\hat{M}:|\hat{M}|=m+1\\\
\text{\,con.\,cop.}\end{subarray}}}p^{|\bigcup_{\Gamma\in\hat{M}}\Gamma|}$
$\displaystyle=\;\;\;\sum_{\mathclap{\begin{subarray}{c}M:|M|=m\\\
\text{\,con.\,cop.}\end{subarray}}}p^{|\bigcup_{\Gamma\in
M}\Gamma|}\;\sum_{\mathclap{\Gamma_{m+1}\in\bigcup_{\Gamma\in
M}\mathcal{M}_{\Gamma}}}p^{|\Gamma_{m+1}\backslash\bigcup_{\Gamma\in
M}\Gamma|}.$

Given $M$ a set of connected copies of size $m$, $\Gamma_{m+1}$ has to be a
neighbor of at least one $\Gamma\in M$. In particular, the intersection
$\Gamma_{m+1}\cap\bigcup_{\Gamma\in M}\Gamma$ may not be empty. To understand
the resulting structure, we fix a non-empty subset $H\subset\bigcup_{\Gamma\in
M}\Gamma$ and we count the possible choices for $\Gamma_{m+1}$, whose
intersection with $\bigcup_{\Gamma\in M}\Gamma$ equals $H$. Assuming that $H$
is small enough that it can be completed to a copy of ${G_{0}}$, there are
$\binom{n-v_{H}}{v_{{G_{0}}}-v_{H}}$ possibilities to complete the $v_{H}$
vertices of $H$ to a set of $v_{{G_{0}}}$ vertices. On such a set there are
$\frac{v_{{G_{0}}}!}{a_{{G_{0}}}}$ copies of ${G_{0}}$. Hence, we can find at
most
$\frac{v_{{G_{0}}}!}{a_{{G_{0}}}}\cdot\binom{n-v_{H}}{v_{{G_{0}}}-v_{H}}\leq\frac{v_{{G_{0}}}!}{a_{{G_{0}}}}\cdot
n^{v_{{G_{0}}}-v_{H}}$ possible copies $\Gamma_{m+1}\in\mathcal{M}$ with
$H=\Gamma_{m+1}\cap\bigcup_{\Gamma\in M}\Gamma$. Further note that
$|\Gamma_{m+1}\backslash\bigcup_{\Gamma\in M}\Gamma|=|\Gamma_{m+1}\backslash
H|=e_{{G_{0}}}-e_{H}$. Thus,

$\displaystyle\sum_{\mathclap{\begin{subarray}{c}M:|M|=m\\\
\text{\,con.\,cop.}\end{subarray}}}p^{|\bigcup_{\Gamma\in
M}\Gamma|}\;\sum_{\mathclap{\Gamma_{m+1}\in\bigcup_{\Gamma\in
M}\mathcal{M}_{\Gamma}}}p^{|\Gamma_{m+1}\backslash\bigcup_{\Gamma\in
M}\Gamma|}$
$\displaystyle\leq\;\;\;\sum_{\mathclap{\begin{subarray}{c}M:|M|=m\\\
\text{\,con.\,cop.}\end{subarray}}}p^{|\bigcup_{\Gamma\in
M}\Gamma|}\;\sum_{\mathclap{H\underset{{G_{0}}}{\subset}\bigcup_{\Gamma\in
M}\mathcal{M}_{\Gamma}}}v_{{G_{0}}}!\cdot a_{{G_{0}}}^{-1}\cdot
n^{v_{{G_{0}}}-v_{H}}\cdot p^{e_{{G_{0}}}-e_{H}},$

where $H\underset{{G_{0}}}{\subset}\bigcup_{\Gamma\in M}\mathcal{M}_{\Gamma}$
denotes that $H$ is a non-empty subset of $\bigcup_{\Gamma\in
M}\mathcal{M}_{\Gamma}$ while additionally being isomorphic to a subset of
${G_{0}}$. There are less than $2^{m\cdot e_{{G_{0}}}}$ sets with these
properties. For each of them there is $n^{v_{H}}p^{e_{H}}\geq\Psi_{\min}$, so
that

$\displaystyle\sum_{\mathclap{\begin{subarray}{c}M:|M|=m\\\
\text{\,con.\,cop.}\end{subarray}}}p^{|\bigcup_{\Gamma\in
M}\Gamma|}\;\sum_{\mathclap{H\underset{{G_{0}}}{\subset}\bigcup_{\Gamma\in
M}\mathcal{M}_{\Gamma}}}v_{{G_{0}}}!\cdot a_{{G_{0}}}^{-1}\cdot
n^{v_{{G_{0}}}-v_{H}}\cdot p^{e_{{G_{0}}}-e_{H}}$
$\displaystyle\leq\;\;\;\sum_{\mathclap{\begin{subarray}{c}M:|M|=m\\\
\text{\,con.\,cop.}\end{subarray}}}p^{|\bigcup_{\Gamma\in M}\Gamma|}\cdot
2^{m\cdot
e_{{G_{0}}}}\cdot\frac{v_{{G_{0}}}!}{a_{{G_{0}}}}\cdot\frac{n^{v_{{G_{0}}}}p^{e_{{G_{0}}}}}{\Psi_{\min}}.$

As we assumed that $(i)$ holds true for $m$, we arrive at

$\displaystyle\sum_{\mathclap{\begin{subarray}{c}M:|M|=m\\\
\text{\,con.\,cop.}\end{subarray}}}p^{|\bigcup_{\Gamma\in M}\Gamma|}\cdot
2^{m\cdot
e_{{G_{0}}}}\cdot\frac{v_{{G_{0}}}!}{a_{{G_{0}}}}\cdot\frac{n^{v_{{G_{0}}}}p^{e_{{G_{0}}}}}{\Psi_{\min}}$
$\displaystyle\leq c_{m}\cdot\frac{n^{m\cdot v_{{G_{0}}}}\cdot p^{m\cdot
e_{{G_{0}}}}}{\Psi_{\min}^{m-1}}\cdot 2^{m\cdot
e_{{G_{0}}}}\cdot\frac{v_{{G_{0}}}!}{a_{{G_{0}}}}\cdot\frac{n^{v_{{G_{0}}}}p^{e_{{G_{0}}}}}{\Psi_{\min}}$
$\displaystyle=c_{m+1}\cdot\frac{n^{(m+1)\cdot v_{{G_{0}}}}\cdot p^{(m+1)\cdot
e_{{G_{0}}}}}{\Psi_{\min}^{(m+1)-1}}.$

This finishes the proof of $(i)$. To prove the second part, we want to adapt
our strategy from above. Let $M=(\Gamma_{i})_{i=1}^{m}\subset\mathcal{M}$ be a
set of connected copies of size $m$, and
$\hat{M}=(\hat{\Gamma}_{i})_{i=1}^{\hat{m}}\subset\mathcal{M}$ be a set of
connected copies of size $\hat{m}$. Further, let
$\bar{M}:=(\Gamma_{1},\dots,\Gamma_{m},\hat{\Gamma}_{1},\dots,\hat{\Gamma}_{\hat{m}})$
be the set created by joining $M$ and $\hat{M}$. If we could be sure that
$\bar{M}$ is a set of connected copies, we could directly apply $(i)$.
However, we can not generally assume that $\bar{M}$ is a set of connected
copies. In fact, $\bar{M}$ is a set of connected copies of size $m+\hat{m}$ if
and only if $\hat{\Gamma}_{1}\in\bigcup_{i=1}^{m}\mathcal{M}_{\Gamma_{i}}$.
Hence, we have to modify our strategy from $(i)$ only with respect to
$\hat{H}=\hat{\Gamma}_{1}\cap\bigcup_{i=1}^{m}\mathcal{M}_{\Gamma_{i}}$, which
can be empty in this case. Therefore, the lower bound $n^{v_{\hat{H}}}\cdot
p^{e_{\hat{H}}}\geq\Psi_{\min}$ does not necessarily hold. We have to take the
case $\hat{H}=\emptyset$ into account, which results in $n^{v_{\hat{H}}}\cdot
p^{e_{\hat{H}}}\geq\min\\{\Psi_{\min},1\\}$. This proves the second part of
the statement. ∎

Before we turn to the main proofs, we want to exemplify two bounding
strategies in connection with the moment generating function. We will
repeatedly use these strategies, later.

######  Lemma 5.6.

Let $A_{1},A_{2}\subset\mathcal{M}$ be sets of copies of ${G_{0}}$, and let
$F$ be a non-negative functional of $\\{B_{\Gamma}\\}_{\Gamma\in A_{2}}$, so
that $F$ is a random variable independent from
$\\{B_{\Gamma}\\}_{\Gamma\in\mathcal{M}\backslash\hat{A}_{2}}$, where
$\hat{A}_{2}:=\bigcup_{\Gamma\in A_{2}}\mathcal{M}_{\Gamma}$ is the set of all
copies of ${G_{0}}$ that depend on at least one element from $A_{2}$, in
particular $A_{2}\subset\hat{A}_{2}$. Then for all $t\geq 0$

$\displaystyle\mathbb{E}\Big{[}F\cdot
e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash
A_{1}}B_{\Gamma}^{c}}\Big{]}$ $\displaystyle\leq\mathbb{E}[F]\cdot
e^{\frac{t}{\sigma}|A_{1}\cup\hat{A}_{2}|}\cdot\mathbb{E}[e^{tW}].$ In
particular, there is
$\displaystyle\mathbb{E}\Big{[}e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash
A_{1}}B_{\Gamma}^{c}}\Big{]}$ $\displaystyle\leq
e^{\frac{t}{\sigma}|A_{1}|}\cdot\mathbb{E}[e^{tW}].$

###### Proof.

There is

$\displaystyle\mathbb{E}\Big{[}F\cdot
e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash
A_{1}}B_{\Gamma}^{c}}\Big{]}$ $\displaystyle=\mathbb{E}\Big{[}F\cdot
e^{\frac{t}{\sigma}\sum_{\Gamma\in\hat{A}_{2}\backslash
A_{1}}B_{\Gamma}^{c}}\cdot
e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash(A_{1}\cup\hat{A}_{2})}B_{\Gamma}^{c}}\Big{]}$
$\displaystyle\leq\mathbb{E}\Big{[}F\cdot
e^{\frac{t}{\sigma}\sum_{\Gamma\in\hat{A}_{2}\backslash
A_{1}}(1-p^{e_{{G_{0}}}})}\cdot
e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash(A_{1}\cup\hat{A}_{2})}B_{\Gamma}^{c}}\Big{]}$
$\displaystyle=\mathbb{E}[F]\cdot e^{\frac{t}{\sigma}|\hat{A}_{2}\backslash
A_{1}|(1-p^{e_{{G_{0}}}})}\cdot\mathbb{E}\Big{[}e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash(A_{1}\cup\hat{A}_{2})}B_{\Gamma}^{c}}\Big{]}.$

Further,

$\displaystyle\mathbb{E}\Big{[}e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash(A_{1}\cup\hat{A}_{2})}B_{\Gamma}^{c}}\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}e^{-\frac{t}{\sigma}\sum_{\Gamma\in
A_{1}\cup\hat{A}_{2}}B_{\Gamma}^{c}}\cdot e^{tW}\Big{]}\leq
e^{\frac{t}{\sigma}|A_{1}\cup\hat{A}_{2}|p^{e_{{G_{0}}}}}\cdot\mathbb{E}[e^{tW}].$

This proves the first inequality. The second inequality follows from the
special case of $F=1$ with $A_{2}=\emptyset$. ∎

We are now prepared to prove Theorem 1.4 and Corollary 1.5.

###### Proof of Theorem 1.4.

This proof ist based on our theoretical results presented in Theorem 3.1.
Since $W$ depends on only finitely many Rademacher random variables we may
apply this theorem as long as we find suitable functions $\gamma_{1}$ and
$\gamma_{2}$. In the application of this theorem, the quantity
$U=(U_{k})_{k\in E}$ with

$\displaystyle U_{k}:={}$ $\displaystyle\frac{1}{\sqrt{pq}}\cdot
D_{k}W\cdot|D_{k}L^{-1}W|$ $\displaystyle={}$
$\displaystyle\frac{1}{\sqrt{pq}}\cdot D_{k}W\cdot(-D_{k}L^{-1}W),$ (5.2)

$k\in E$, is of special importance, where $|D_{k}L^{-1}W|=-D_{k}L^{-1}W$ holds
true due to the results from Lemma 5.1.

Due to the application of the discrete gradient, $U_{k}$ does not depend on
$Y_{k}$. For this special case it is known by Corollary 9.9 in [17] that
$\delta(U)=\sum_{k\in E}Y_{k}U_{k}$. We further note that
$\mathbb{E}[\sqrt{pq}\sum_{k\in E}U_{k}]=1$, which is implied by (2.13) in
[12] in case of $f$ being the identity.

These considerations allow us to rephrase Theorem 3.1 with the notation
introduced in (5.2): To apply this theorem, we need to construct functions
$\gamma_{1}$ and $\gamma_{2}$ so that

$\displaystyle\mathbb{E}\Big{[}\Big{|}\sqrt{pq}\sum_{k\in
E}U_{k}^{c}\Big{|}\cdot e^{tW}\Big{]}$
$\displaystyle\leq\gamma_{1}(t)\cdot\mathbb{E}[e^{tW}],$ (A1’)
$\displaystyle\mathbb{E}\Big{[}\Big{|}\sum_{k\in E}Y_{k}U_{k}\Big{|}\cdot
e^{tW}\Big{]}$ $\displaystyle\leq\gamma_{2}(t)\cdot\mathbb{E}[e^{tW}],$ (A2’)

for $0\leq t\leq A$, where $A\geq 0$.

For the construction of suitable $\gamma_{1}$ and $\gamma_{2}$, a further
decomposition of $(U_{k})_{k\in E}$ will be usefull, which can be achieved by
application of Lemma 5.1: We decompose

$\displaystyle U_{k}$ $\displaystyle=\frac{1}{\sqrt{pq}}\cdot
D_{k}W\cdot(-D_{k}L^{-1}W)$
$\displaystyle=\frac{1}{\sqrt{pq}\cdot\sigma^{2}}\cdot\sum_{\Gamma_{1},\Gamma_{2}\in\mathcal{M}}D_{k}B_{\Gamma_{1}}^{c}\cdot\big{(}-D_{k}L^{-1}B_{\Gamma_{2}}^{c}\big{)}$
$\displaystyle=\frac{\sqrt{pq}}{\sigma^{2}}\cdot\sum_{\Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}}\sum_{\\{k\\}\subset\alpha_{2}\subset\Gamma_{2}}\frac{p^{|\Gamma_{2}|-|\alpha_{2}|}}{|\Gamma_{2}|\cdot\binom{|\Gamma_{2}|-1}{|\alpha_{2}|-1}}B_{(\Gamma_{1}\cup\alpha_{2})\backslash\\{k\\}}$
$\displaystyle=\frac{\sqrt{pq}}{\sigma^{2}}\cdot\sum_{\Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}}V_{k,\Gamma_{1},\Gamma_{2}}$
(5.3) with $\displaystyle V_{k,\Gamma_{1},\Gamma_{2}}$
$\displaystyle:=\sum_{\\{k\\}\subset\alpha_{2}\subset\Gamma_{2}}\frac{p^{|\Gamma_{2}|-|\alpha_{2}|}}{|\Gamma_{2}|\cdot\binom{|\Gamma_{2}|-1}{|\alpha_{2}|-1}}B_{(\Gamma_{1}\cup\alpha_{2})\backslash\\{k\\}}\geq
0.$ (5.4)

Construction of $\gamma_{1}$: To verify that assumption (A1’) can be fulfilled
with some function $\gamma_{1}$—that is yet to be constructed—we start by
applying the Cauchy–Schwarz inequality and using the decomposition of $U_{k}$
from (5.3). This results in

$\displaystyle\mathbb{E}\Big{[}\Big{|}\sqrt{pq}\sum_{k\in
E}U_{k}^{c}\Big{|}\cdot e^{tW}\Big{]}$
$\displaystyle\leq\mathbb{E}\Big{[}\Big{(}\sqrt{pq}\sum_{k\in
E}U_{k}^{c}\Big{)}^{2}e^{tW}\Big{]}^{\frac{1}{2}}\cdot\mathbb{E}[e^{tW}]^{\frac{1}{2}}$
$\displaystyle=\Big{(}pq\sum_{k,\ell\in
E}\mathbb{E}\Big{[}U_{k}^{c}U_{\ell}^{c}e^{tW}\Big{]}\Big{)}^{\frac{1}{2}}\cdot\mathbb{E}[e^{tW}]^{\frac{1}{2}}$
$\displaystyle=\bigg{(}\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\sum_{k,\ell\in
E}\sum_{\Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}}\sum_{\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}}\mathbb{E}\Big{[}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}e^{tW}\Big{]}\bigg{)}^{\frac{1}{2}}\cdot\mathbb{E}[e^{tW}]^{\frac{1}{2}}$

for all $t\geq 0$. Next, we apply the iterated Taylor expansion that was
introduced in (4) to $e^{tW}$ by choosing
$x_{\Gamma}=\frac{t}{\sigma}B_{\Gamma}^{c}$ and
$y_{\Gamma}=\frac{t}{\sigma}\sum_{\Gamma_{5}\in\mathcal{M}_{\Gamma}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}}B_{\Gamma_{5}}^{c}$
for $\Gamma\in
I=\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}$, and
$z=\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}}B_{\Gamma}^{c}$.
Due to the underlying dependency structure this expansion results in

$\displaystyle\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\sum_{k,\ell\in
E}\sum_{\Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}}\sum_{\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}}\mathbb{E}\Big{[}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}e^{tW}\Big{]}$
$\displaystyle=S_{1,1}(t)+S_{1,2}(t)+S_{1,3}(t)+S_{1,4}(t)$

with

$\displaystyle S_{1,1}(t)$
$\displaystyle:=\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\sum_{\mathclap{\begin{subarray}{c}k,\ell\in
E\\\ \Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}\\\
\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}\end{subarray}}}\mathbb{E}\Big{[}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\Big{]}\cdot\mathbb{E}\Big{[}e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}}B_{\Gamma}^{c}}\Big{]},$
$\displaystyle S_{1,2}(t)$
$\displaystyle:=\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\frac{t}{\sigma}\cdot\sum_{\mathclap{\begin{subarray}{c}k,\ell\in
E\\\ \Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}\\\
\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}\\\
\Gamma_{5}\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}\mathbb{E}\Big{[}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\cdot
B_{\Gamma_{5}}^{c}\Big{]}\cdot\mathbb{E}\Big{[}e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}}}B_{\Gamma}^{c}}\Big{]},$
$\displaystyle S_{1,3}(t)$
$\displaystyle:=\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\frac{t^{2}}{\sigma^{2}}\cdot\sum_{\mathclap{\begin{subarray}{c}k,\ell\in
E\\\ \Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}\\\
\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}\\\
\Gamma_{5}\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}\\\
\Gamma_{6}\in\mathcal{M}_{\Gamma_{5}}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}\mathbb{E}\Bigg{[}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\cdot
B_{\Gamma_{5}}^{c}B_{\Gamma_{6}}^{c}\cdot
r_{1}\Big{(}\frac{t}{\sigma}\;\;\;\;\;\;\;\;\;\;\;\;\sum_{\mathclap{\Gamma\in\mathcal{M}_{\Gamma_{5}}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}}}B_{\Gamma}^{c}\;\;\;\;\;\;\;\Big{)}$
$\displaystyle\hskip 256.0748pt\cdot
e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}}}B_{\Gamma}^{c}}\Bigg{]},$
$\displaystyle S_{1,4}(t)$
$\displaystyle:=\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\frac{t^{2}}{2\sigma^{2}}\cdot\sum_{\mathclap{\begin{subarray}{c}k,\ell\in
E\\\ \Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}\\\
\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}\\\
\Gamma_{5},\Gamma_{6}\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}\mathbb{E}\Bigg{[}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\cdot
B_{\Gamma_{5}}^{c}B_{\Gamma_{6}}^{c}\cdot
r_{2}\Big{(}\frac{t}{\sigma}\;\;\;\;\;\;\;\;\sum_{\mathclap{\Gamma\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}}}B_{\Gamma}^{c}\;\;\;\Big{)}$
$\displaystyle\hskip 256.0748pt\cdot
e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}}B_{\Gamma}^{c}}\Bigg{]},$

so that

$\displaystyle\mathbb{E}\Big{[}\Big{|}\sqrt{pq}\sum_{k\in
E}U_{k}^{c}\Big{|}\cdot e^{tW}\Big{]}$
$\displaystyle\leq\big{(}S_{1,1}(t)+S_{1,2}(t)+S_{1,3}(t)+S_{1,4}(t)\big{)}^{\frac{1}{2}}\cdot\mathbb{E}[e^{tW}]^{\frac{1}{2}}.$

The advantage of this expansion based on [22] is that $S_{1,1}$ and $S_{1,2}$
each possess a comfortable dependency structure that allowed us to split the
expectations, while $S_{1,3}$ and $S_{1,4}$ are of second order and therefore
allow more rough estimates without a loss in the resulting rate, as we will
see later in this proof.

We now have to derive bounds for $S_{1,1}$, $S_{1,2}$, $S_{1,3}$, and
$S_{1,4}$.

To derive a bound for $S_{1,1}$, we first fix $k,\ell\in E$,
$\Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}$, and
$\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}$. If $\Gamma_{1}\cup\Gamma_{2}$ is
disjoint from $\Gamma_{3}\cup\Gamma_{4}$, then $V_{k,\Gamma_{1},\Gamma_{2}}$
and $V_{\ell,\Gamma_{3},\Gamma_{4}}$ are independent, so that
$\mathbb{E}\Big{[}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\Big{]}=0$.
Otherwise, using our knowledge from Lemma 5.3 about the correlation behavior
of the random variables $\\{B_{A}\\}_{A\subset E}$ we see that
$\mathbb{E}[V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}]\geq
0$ and

$\displaystyle\mathbb{E}\Big{[}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\Big{]}$
$\displaystyle=\sum_{\begin{subarray}{c}\\{k\\}\subset\alpha_{2}\subset\Gamma_{2}\\\
\\{\ell\\}\subset\alpha_{4}\subset\Gamma_{4}\end{subarray}}\frac{\mathbb{E}[B_{\Gamma_{2}\backslash\alpha_{2}}]}{|\Gamma_{2}|\cdot\binom{|\Gamma_{2}|-1}{|\alpha_{2}|-1}}\frac{\mathbb{E}[B_{\Gamma_{4}\backslash\alpha_{4}}]}{|\Gamma_{4}|\cdot\binom{|\Gamma_{4}|-1}{|\alpha_{4}|-1}}\mathbb{E}\Big{[}B_{(\Gamma_{1}\cup\alpha_{2})\backslash\\{k\\}}^{c}B_{(\Gamma_{3}\cup\alpha_{4})\backslash\\{\ell\\}}^{c}\Big{]}$
$\displaystyle\leq\sum_{\begin{subarray}{c}\\{k\\}\subset\alpha_{2}\subset\Gamma_{2}\\\
\\{\ell\\}\subset\alpha_{4}\subset\Gamma_{4}\end{subarray}}\frac{1}{|\Gamma_{2}|\cdot\binom{|\Gamma_{2}|-1}{|\alpha_{2}|-1}}\frac{1}{|\Gamma_{4}|\cdot\binom{|\Gamma_{4}|-1}{|\alpha_{4}|-1}}\mathbb{E}\Big{[}B_{\big{(}(\Gamma_{1}\cup\alpha_{2})\backslash\\{k\\}\big{)}\cup(\Gamma_{2}\backslash\alpha_{2})\cup\big{(}(\Gamma_{3}\cup\alpha_{4})\backslash\\{\ell\\}\big{)}\cup(\Gamma_{4}\backslash\alpha_{4})}\Big{]}$
$\displaystyle\leq\sum_{\begin{subarray}{c}\\{k\\}\subset\alpha_{2}\subset\Gamma_{2}\\\
\\{\ell\\}\subset\alpha_{4}\subset\Gamma_{4}\end{subarray}}\frac{1}{|\Gamma_{2}|\cdot\binom{|\Gamma_{2}|-1}{|\alpha_{2}|-1}}\frac{1}{|\Gamma_{4}|\cdot\binom{|\Gamma_{4}|-1}{|\alpha_{4}|-1}}p^{|\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}|-2}\;=\;p^{|\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}|-2},$

where in the last step we used what we already noted in Remark 5.2. The second
inequality from Lemma 5.6 implies

$\displaystyle\mathbb{E}\Big{[}e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}}B_{\Gamma}^{c}}\Big{]}$
$\displaystyle\leq
e^{\frac{t}{\sigma}|\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}|}\cdot\mathbb{E}[e^{tW}]\leq
e^{\frac{4Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}].$

Putting these partial results together yields $S_{1,1}(t)\geq 0$ and

$\displaystyle S_{1,1}(t)$
$\displaystyle\leq\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\sum_{\mathclap{\begin{subarray}{c}k,\ell\in
E\\\ \Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}\\\
\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}\end{subarray}}}\mathbbl{1}_{\left\\{(\Gamma_{1}\cup\Gamma_{2})\cap(\Gamma_{3}\cup\Gamma_{4})\neq\emptyset\right\\}}p^{|\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}|-2}\cdot
e^{\frac{4Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$
$\displaystyle\leq\frac{q^{2}}{\sigma^{4}}\cdot
2\sum_{\mathclap{\begin{subarray}{c}\Gamma_{1}\in\mathcal{M}\\\
\Gamma_{2}\in\mathcal{M}_{\Gamma_{1}}\\\
\Gamma_{3}\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}}\\\
\Gamma_{4}\in\mathcal{M}_{\Gamma_{3}}\end{subarray}}}\;\;\;\;\;\;\sum_{\mathclap{\begin{subarray}{c}k\in\Gamma_{1}\cap\Gamma_{2}\\\
\ell\in\Gamma_{3}\cap\Gamma_{4}\end{subarray}}}p^{|\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}|}\cdot
e^{\frac{4Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}],$

where we used that due to symmetry the sum over
$\mathbbl{1}_{\left\\{(\Gamma_{1}\cup\Gamma_{2})\cap(\Gamma_{3}\cup\Gamma_{4})\neq\emptyset\right\\}}$
is smaller than or equal to $2$ times the sum over
$\mathbbl{1}_{\left\\{(\Gamma_{1}\cup\Gamma_{2})\cap\Gamma_{3}\neq\emptyset\right\\}}=\mathbbl{1}_{\left\\{\Gamma_{3}\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}}\right\\}}$.
We further reordered the indices to point out that the first sum runs over a
subset of all sets of connected copies of size $4$ in the sense of Definition
5.4, while the second sum runs over not more than $e_{{G_{0}}}^{2}$ summands.
Lemma 5.5 therefore yields

$\displaystyle S_{1,1}(t)$ $\displaystyle\leq\frac{q^{2}}{\sigma^{4}}\cdot
2\cdot\frac{(v_{{G_{0}}}!)^{3}\cdot
2^{6e_{{G_{0}}}}}{a_{{G_{0}}}^{4}}\cdot\frac{n^{4v_{{G_{0}}}}\cdot
p^{4e_{{G_{0}}}}}{\Psi_{\min}^{3}}\cdot e_{{G_{0}}}^{2}\cdot
e^{\frac{4Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$ $\displaystyle\leq
c_{1,1}\cdot\frac{1}{\Psi_{\min}}\cdot
e^{\frac{4Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$

with $\displaystyle
c_{1,1}:=\frac{2^{3+6e_{{G_{0}}}}\cdot(v_{{G_{0}}}!)^{5}\cdot
e_{{G_{0}}}^{2}}{a_{{G_{0}}}^{2}}$. For the last step we used the lower bound
for $\sigma$ presented in (5.1), which we may use since we assume that $n\geq
4v_{{G_{0}}}^{2}$.

The calculations for $S_{1,2}$ are largely analogous. We find $S_{1,2}(t)\geq
0$ and

$\displaystyle S_{1,2}(t)$
$\displaystyle\leq\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\frac{t}{\sigma}\cdot
2\sum_{\mathclap{\begin{subarray}{c}k,\ell\in E\\\
\Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}\\\
\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}\\\
\Gamma_{5}\in\mathcal{M}_{\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}\mathbb{E}\Big{[}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\cdot
B_{\Gamma_{5}}^{c}\Big{]}\cdot\mathbb{E}\Big{[}e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}}}B_{\Gamma}^{c}}\Big{]}$
$\displaystyle\leq\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\frac{t}{\sigma}\cdot
2\sum_{\mathclap{\begin{subarray}{c}k,\ell\in E\\\
\Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}\\\
\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}\\\
\Gamma_{5}\in\mathcal{M}_{\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}\mathbbl{1}_{\left\\{(\Gamma_{1}\cup\Gamma_{2})\cap(\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5})\neq\emptyset\right\\}}p^{|\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}|-2}\cdot
e^{\frac{t}{\sigma}|\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}}|}\cdot\mathbb{E}[e^{tW}]$
$\displaystyle\leq\frac{q^{2}}{\sigma^{4}}\cdot\frac{t}{\sigma}\cdot
2\bigg{(}\;2\sum_{\mathclap{\begin{subarray}{c}\Gamma_{1}\in\mathcal{M}\\\
\Gamma_{2}\in\mathcal{M}_{\Gamma_{1}}\\\
\Gamma_{3}\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}}\\\
\Gamma_{4}\in\mathcal{M}_{\Gamma_{3}}\\\
\Gamma_{5}\in\mathcal{M}_{\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}\;\;\;\;\;\;\sum_{\mathclap{\begin{subarray}{c}k\in\Gamma_{1}\cap\Gamma_{2}\\\
\ell\in\Gamma_{3}\cap\Gamma_{4}\end{subarray}}}p^{|\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}|}+2\sum_{\mathclap{\begin{subarray}{c}\Gamma_{1}\in\mathcal{M}\\\
\Gamma_{2}\in\mathcal{M}_{\Gamma_{1}}\\\
\Gamma_{5}\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}}\\\
\Gamma_{3}\in\mathcal{M}_{\Gamma_{5}}\\\
\Gamma_{4}\in\mathcal{M}_{\Gamma_{3}}\end{subarray}}}\;\;\;\;\;\;\sum_{\mathclap{\begin{subarray}{c}k\in\Gamma_{1}\cap\Gamma_{2}\\\
\ell\in\Gamma_{3}\cap\Gamma_{4}\end{subarray}}}p^{|\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}|}\bigg{)}\cdot
e^{\frac{5Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$
$\displaystyle\leq\frac{q^{2}}{\sigma^{4}}\cdot\frac{t}{\sigma}\cdot
8\cdot\frac{(v_{{G_{0}}}!)^{4}\cdot
2^{10e_{{G_{0}}}}}{a_{{G_{0}}}^{5}}\cdot\frac{n^{5v_{{G_{0}}}}\cdot
p^{5e_{{G_{0}}}}}{\Psi_{\min}^{4}}\cdot e_{{G_{0}}}^{2}\cdot
e^{\frac{5Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$ $\displaystyle\leq
c_{1,2}\cdot\frac{t}{\sqrt{q\Psi_{\min}}}\cdot\frac{1}{\Psi_{\min}}\cdot
e^{\frac{5Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$

with $\displaystyle
c_{1,2}:=\frac{2^{\frac{11}{2}+10e_{{G_{0}}}}\cdot(v_{{G_{0}}}!)^{\frac{13}{2}}\cdot
e_{{G_{0}}}^{2}}{a_{{G_{0}}}^{\frac{5}{2}}}$.

To construct a bound for $S_{1,3}$ we have to slightly modify our approach.
Due to symmetry, we can rename the variables if necessary to make sure that
$\Gamma_{5}\in\mathcal{M}_{\Gamma_{3}\cup\Gamma_{4}}$. However, due to the
influence of the remainder function $r_{1}$ the relevant expectation can be
non-zero even if $\Gamma_{1}\cup\Gamma_{2}$ and
$\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}\cup\Gamma_{6}$ are disjoint. This
disadvantage can later be compensated for by using the additional
$\sigma^{-2}$. We start by taking the absolute to get rid of the remainder
$r_{1}$. Recall that $t\geq 0$ and $r_{1}(x)\leq e^{|x|}$ for all
$x\in\mathbb{R}$, so that

$\displaystyle\Big{|}r_{1}\Big{(}\frac{t}{\sigma}\;\;\;\;\;\;\;\;\;\;\;\;\;\sum_{\mathclap{\Gamma\in\mathcal{M}_{\Gamma_{5}}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}}}B_{\Gamma}^{c}\;\;\;\;\;\;\;\Big{)}\Big{|}$
$\displaystyle\leq e^{\frac{Dt}{\sigma}}.$

Hence, by application of Lemma 5.6 and Lemma 5.3, there is

$\displaystyle|S_{1,3}(t)|$
$\displaystyle\leq\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\frac{t^{2}}{\sigma^{2}}\cdot
2\sum_{\mathclap{\begin{subarray}{c}k,\ell\in E\\\
\Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}\\\
\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}\\\
\Gamma_{5}\in\mathcal{M}_{\Gamma_{3}\cup\Gamma_{4}}\\\
\Gamma_{6}\in\mathcal{M}_{\Gamma_{5}}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}\mathbb{E}\bigg{[}\big{|}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\cdot
B_{\Gamma_{5}}^{c}B_{\Gamma_{6}}^{c}\big{|}\cdot e^{\frac{Dt}{\sigma}}\cdot
e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}}}B_{\Gamma}^{c}}\bigg{]}$
$\displaystyle\leq\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\frac{t^{2}}{\sigma^{2}}\cdot
2\sum_{\mathclap{\begin{subarray}{c}\Gamma_{1}\in\mathcal{M}\\\
\Gamma_{2}\in\mathcal{M}_{\Gamma_{1}}\\\ \Gamma_{3}\in\mathcal{M}\\\
\Gamma_{4}\in\mathcal{M}_{\Gamma_{3}}\\\
\Gamma_{5}\in\mathcal{M}_{\Gamma_{3}\cup\Gamma_{4}}\\\
\Gamma_{6}\in\mathcal{M}_{\Gamma_{5}}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}\;\;\;\;\;\;\sum_{\mathclap{\begin{subarray}{c}k\in\Gamma_{1}\cap\Gamma_{2}\\\
\ell\in\Gamma_{3}\cap\Gamma_{4}\end{subarray}}}\mathbb{E}\Big{[}\big{|}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\cdot
B_{\Gamma_{5}}^{c}B_{\Gamma_{6}}^{c}\big{|}\Big{]}\cdot
e^{\frac{Dt}{\sigma}}\cdot e^{\frac{6Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$
$\displaystyle\leq\frac{q^{2}}{\sigma^{4}}\cdot\frac{t^{2}}{\sigma^{2}}\cdot
2\sum_{\mathclap{\begin{subarray}{c}\Gamma_{1}\in\mathcal{M}\\\
\Gamma_{2}\in\mathcal{M}_{\Gamma_{1}}\\\ \Gamma_{3}\in\mathcal{M}\\\
\Gamma_{4}\in\mathcal{M}_{\Gamma_{3}}\\\
\Gamma_{5}\in\mathcal{M}_{\Gamma_{3}\cup\Gamma_{4}}\\\
\Gamma_{6}\in\mathcal{M}_{\Gamma_{5}}\end{subarray}}}e_{{G_{0}}}^{2}2^{4}p^{|\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}\cup\Gamma_{6}|}\cdot
e^{\frac{7Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}].$

As already mentioned, we can not be sure that $\Gamma_{1}\cup\Gamma_{2}$ and
$\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}\cup\Gamma_{6}$ are not disjoint.
Therefore, in the sum above $\Gamma_{3}$ runs over the the complete set
$\mathcal{M}$, so that $(\Gamma_{1},\Gamma_{2})$ and
$(\Gamma_{3},\Gamma_{4},\Gamma_{5},\Gamma_{6})$ are sets of connected copies
while in general their union
$(\Gamma_{1},\Gamma_{2},\Gamma_{3},\Gamma_{4},\Gamma_{5},\Gamma_{6})$ is not.
This case is covered by the second inequality from Lemma 5.5, so that

$\displaystyle|S_{1,3}(t)|$
$\displaystyle\leq\frac{q^{2}}{\sigma^{4}}\cdot\frac{t^{2}}{\sigma^{2}}\cdot
2^{5}\cdot\frac{(v_{{G_{0}}}!)^{5}\cdot
2^{15e_{{G_{0}}}}}{a_{{G_{0}}}^{6}}\cdot\frac{n^{6v_{{G_{0}}}}\cdot
p^{6e_{{G_{0}}}}}{\Psi_{\min}^{4}\cdot\min\\{\Psi_{\min},1\\}}\cdot
e_{{G_{0}}}^{2}\cdot e^{\frac{7Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$
$\displaystyle\leq
c_{1,3}\cdot\frac{t^{2}}{q\min\\{\Psi_{\min},1\\}}\cdot\frac{1}{\Psi_{\min}}\cdot
e^{\frac{7Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$

with $\displaystyle
c_{1,3}:=\frac{2^{8+15e_{{G_{0}}}}\cdot(v_{{G_{0}}}!)^{8}\cdot
e_{{G_{0}}}^{2}}{a_{{G_{0}}}^{3}}$.

The calculations for $S_{1,4}$ are largely analogous to $S_{1,3}$ again. We
get

$\displaystyle|S_{1,4}(t)|$
$\displaystyle\leq\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\frac{t^{2}}{2\sigma^{2}}\cdot\sum_{\mathclap{\begin{subarray}{c}k,\ell\in
E\\\ \Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}\\\
\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}\\\
\Gamma_{5},\Gamma_{6}\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}\mathbb{E}\bigg{[}\big{|}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\cdot
B_{\Gamma_{5}}^{c}B_{\Gamma_{6}}^{c}\big{|}\cdot e^{\frac{4Dt}{\sigma}}\cdot
e^{\frac{t}{\sigma}\sum_{\Gamma\in\mathcal{M}\backslash\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}}B_{\Gamma}^{c}}\bigg{]}$
$\displaystyle\leq\frac{p^{2}q^{2}}{\sigma^{4}}\cdot\frac{t^{2}}{2\sigma^{2}}\cdot\sum_{\mathclap{\begin{subarray}{c}\Gamma_{1}\in\mathcal{M}\\\
\Gamma_{2}\in\mathcal{M}_{\Gamma_{1}}\\\ \Gamma_{3}\in\mathcal{M}\\\
\Gamma_{4}\in\mathcal{M}_{\Gamma_{3}}\\\
\Gamma_{5},\Gamma_{6}\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}\;\;\;\;\;\;\sum_{\mathclap{\begin{subarray}{c}k\in\Gamma_{1}\cap\Gamma_{2}\\\
\ell\in\Gamma_{3}\cap\Gamma_{4}\end{subarray}}}\mathbb{E}\Big{[}\big{|}V_{k,\Gamma_{1},\Gamma_{2}}^{c}V_{\ell,\Gamma_{3},\Gamma_{4}}^{c}\cdot
B_{\Gamma_{5}}^{c}B_{\Gamma_{6}}^{c}\big{|}\Big{]}\cdot
e^{\frac{4Dt}{\sigma}}\cdot e^{\frac{6Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$
$\displaystyle\leq\frac{q^{2}}{\sigma^{4}}\cdot\frac{t^{2}}{2\sigma^{2}}\bigg{(}2\sum_{\mathclap{\begin{subarray}{c}\Gamma_{1}\in\mathcal{M}\\\
\Gamma_{2}\in\mathcal{M}_{\Gamma_{1}}\\\ \Gamma_{3}\in\mathcal{M}\\\
\Gamma_{4}\in\mathcal{M}_{\Gamma_{3}}\\\
\Gamma_{5},\Gamma_{6}\in\mathcal{M}_{\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}e_{{G_{0}}}^{2}2^{4}p^{|\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}\cup\Gamma_{6}|}+2\sum_{\mathclap{\begin{subarray}{c}\Gamma_{1}\in\mathcal{M}\\\
\Gamma_{2}\in\mathcal{M}_{\Gamma_{1}}\\\ \Gamma_{3}\in\mathcal{M}\\\
\Gamma_{4}\in\mathcal{M}_{\Gamma_{3}}\\\
\Gamma_{5}\in\mathcal{M}_{\Gamma_{1}\cup\Gamma_{2}}\\\
\Gamma_{6}\in\mathcal{M}_{\Gamma_{3}\cup\Gamma_{4}}\end{subarray}}}e_{{G_{0}}}^{2}2^{4}p^{|\Gamma_{1}\cup\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4}\cup\Gamma_{5}\cup\Gamma_{6}|}\bigg{)}e^{\frac{10Dt}{\sigma}}\mathbb{E}[e^{tW}]$
$\displaystyle\leq\frac{q^{2}}{\sigma^{4}}\cdot\frac{t^{2}}{\sigma^{2}}\cdot
2^{5}\cdot\frac{(v_{{G_{0}}}!)^{5}\cdot
2^{15e_{{G_{0}}}}}{a_{{G_{0}}}^{6}}\cdot\frac{n^{6v_{{G_{0}}}}\cdot
p^{6e_{{G_{0}}}}}{\Psi_{\min}^{4}\cdot\min\\{\Psi_{\min},1\\}}\cdot
e_{{G_{0}}}^{2}\cdot e^{\frac{10Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$
$\displaystyle\leq
c_{1,4}\cdot\frac{t^{2}}{q\min\\{\Psi_{\min},1\\}}\cdot\frac{1}{\Psi_{\min}}\cdot
e^{\frac{10Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$

with $\displaystyle
c_{1,4}:=c_{1,3}=\frac{2^{8+15e_{{G_{0}}}}\cdot(v_{{G_{0}}}!)^{8}\cdot
e_{{G_{0}}}^{2}}{a_{{G_{0}}}^{3}}$.

Having bounded $S_{1,1}$, $S_{1,2}$, $S_{1,3}$, and $S_{1,4}$, we note that
$c_{1,3}=\max\\{c_{1,1},c_{1,2},c_{1,3},c_{1,4}\\}$ because of
$a_{{G_{0}}}\leq v_{{G_{0}}}!$. In conclusion, for all $t\geq 0$ it holds that

$\displaystyle\mathbb{E}\Big{[}\Big{|}\sqrt{pq}\sum_{k\in
E}U_{k}^{c}\Big{|}\cdot e^{tW}\Big{]}$
$\displaystyle\leq\big{(}S_{1,1}(t)+S_{1,2}(t)+S_{1,3}(t)+S_{1,4}(t)\big{)}^{\frac{1}{2}}\cdot\mathbb{E}[e^{tW}]^{\frac{1}{2}}$
$\displaystyle\leq\bigg{(}c_{1,1}\cdot\frac{1}{\Psi_{\min}}\cdot
e^{\frac{4Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$ $\displaystyle\hskip
10.00002pt+c_{1,2}\cdot\frac{t}{\sqrt{q\Psi_{\min}}}\cdot\frac{1}{\Psi_{\min}}\cdot
e^{\frac{5Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$ $\displaystyle\hskip
10.00002pt+c_{1,3}\cdot\frac{t^{2}}{q\min\\{\Psi_{\min},1\\}}\cdot\frac{1}{\Psi_{\min}}\cdot
e^{\frac{7Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$ $\displaystyle\hskip
10.00002pt+c_{1,4}\cdot\frac{t^{2}}{q\min\\{\Psi_{\min},1\\}}\cdot\frac{1}{\Psi_{\min}}\cdot
e^{\frac{10Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]\bigg{)}^{\frac{1}{2}}\cdot\mathbb{E}[e^{tW}]^{\frac{1}{2}}$
$\displaystyle\leq\sqrt{c_{1,3}}\cdot\bigg{(}1+\frac{t}{\sqrt{\Psi_{\min}}}+2\frac{t^{2}}{\min\\{\Psi_{\min},1\\}}\bigg{)}^{\frac{1}{2}}\frac{1}{\sqrt{q\Psi_{\min}}}\cdot
e^{\frac{5Dt}{\sigma}}\cdot\mathbb{E}[e^{tW}]$
$\displaystyle\leq\gamma_{1}(t)\cdot\mathbb{E}[e^{tW}]$

with

$\displaystyle\gamma_{1}(t):={}$ $\displaystyle
c_{A1}\cdot\bigg{(}1+\frac{t}{\min\\{\sqrt{\Psi_{\min}},1\\}}\bigg{)}\frac{1}{\sqrt{q\Psi_{\min}}}\cdot
e^{\frac{5Dt}{\sigma}},$ (5.5)

where $\displaystyle
c_{A1}:=\sqrt{2c_{1,3}}=\frac{2^{\frac{9}{2}+\frac{15}{2}e_{{G_{0}}}}\cdot(v_{{G_{0}}}!)^{4}\cdot
e_{{G_{0}}}}{a_{{G_{0}}}^{\frac{3}{2}}}$.

Construction of $\gamma_{2}$: To verify that assumption (A2’) can be fulfilled
with some function $\gamma_{2}$ we again start by applying the Cauchy–Schwarz
inequality and using the decomposition of $U_{k}$ from (5.3), which results in

$\displaystyle\mathbb{E}\Big{[}\Big{|}\sum_{k\in E}Y_{k}U_{k}\Big{|}\cdot
e^{tW}\Big{]}$ $\displaystyle\leq\mathbb{E}\Big{[}\Big{(}\sum_{k\in
E}Y_{k}U_{k}\Big{)}^{2}e^{tW}\Big{]}^{\frac{1}{2}}\cdot\mathbb{E}[e^{tW}]^{\frac{1}{2}}$
$\displaystyle=\Big{(}\sum_{k,\ell\in
E}\mathbb{E}\Big{[}Y_{k}Y_{\ell}U_{k}U_{\ell}e^{tW}\Big{]}\Big{)}^{\frac{1}{2}}\cdot\mathbb{E}[e^{tW}]^{\frac{1}{2}}$
$\displaystyle=\bigg{(}\frac{1}{\sigma^{4}}\cdot\sum_{k,\ell\in
E}\sum_{\begin{subarray}{c}\Gamma_{1},\Gamma_{2}\in\mathcal{M}_{k}\\\
\Gamma_{3},\Gamma_{4}\in\mathcal{M}_{\ell}\end{subarray}}\mathbb{E}\Big{[}B_{k}^{c}B_{\ell}^{c}\cdot
V_{k,\Gamma_{1},\Gamma_{2}}V_{\ell,\Gamma_{3},\Gamma_{4}}\cdot
e^{tW}\Big{]}\bigg{)}^{\frac{1}{2}}\cdot\mathbb{E}[e^{tW}]^{\frac{1}{2}}$

for all $t\geq 0$. We split the sum over $k,\ell\in E$ into two parts. The