# Strong limit of processes constructed from a renewal process

Xavier Bardina and Carles Rovira111 X. Bardina and C. Rovira are supported by
the grant PID2021-123733NB-I00 from SEIDI, Ministerio de Economia y
Competividad.

###### Abstract

We construct a family of processes, from a renewal process, that have
realizations that converge almost surely to the Brownian motion, uniformly on
the unit time interval. Finally we compute the rate of convergence in a
particular case.

MSC(2010): 60F17, 60G15

Keywords: strong convergence, renewal process, Brownian motion

##  1 Introduction

In this paper we study realizations of processes that converge almost surely,
uniformly on the unit time interval, to the standard Brownian motion. In the
mathematical literature we can find papers studying the strong convergence of
random walks or the process usually called as uniform transport processes. Our
aim is to deal with extensions of the uniform transport process.

The uniform transport process, introduced by Kac in [10], can be written as

$y_{n}(t)=\frac{1}{n}(-1)^{A}\int_{0}^{n^{2}t}(-1)^{N(u)}du,$

where $N=\\{N(t),\,t\geq 0\\}$ is a standard Poisson process and
$A\sim\textrm{Bernoulli}\left(\frac{1}{2}\right)$ independent of the Poisson
process $N$.

Griego, Heath and Ruiz-Moncayo [9] showed that these processes converge
strongly and uniformly on bounded time intervals to Brownian motion. Gorostiza
and Griego [8] and Csörgő and Horváth [2] obtained a rate of convergence. More
precisely, in [8] it is proved that there exist versions of the transport
processes $\tilde{y}_{n}$ on the same probability space as a given Brownian
motion $(y(t))_{t\geq 0}$ such that, for each $q>0$,

$P\left(\sup_{a\leq t\leq b}|y_{n}(t)-y(t)|>{C}n^{\frac{1}{2}}\left(\log
n\right)^{\frac{5}{2}}\right)=o\left({{n}^{-q}}\right),$

as $n\to\infty$ and where $C$ is a positive constant depending on $a$, $b$ and
$q$. These bounds are improved in [11] using an explicit computation in the
Skorohod embedding problem. Furthermore, we can find several papers (see for
instance [7], [1], [3], [4], [5], [6]) where the authors defined a sequence of
processes, obtained as modifications of the uniform transport process, that
converges strongly to some Gaussian processes uniformly on bounded intervals.

Nevertheless, all these papers are based on processes built from a Poisson
process. Let us recall that Poisson process has jump times with exponential
laws and so, we are able to use all the particular properties of this
distribution.

Our aim is to deal with jump times that don’t have exponential law. We
consider extensions of the uniform transport using a reward renewal process
instead of a Poisson process. As far as the authors know, these type of
processes has not been studied.

We consider

$x_{n}(t)=h(n)\int_{0}^{g(n)t}(-1)^{T(u)}du,$

where $T=\\{T(t),\,t\geq 0\\}$ is a renewal reward process and $h$ and $g$ are
nonnegative functions defined on ${\mathbb{N}}$.

We will show that for a wide class of renewal reward processes we have, when
$n$ goes to $\infty$, the strong convergence of these processes to a standard
Brownian motion. We also deal with the rate of convergence. Unfortunately we
are not able to get a general result since the proofs heavily rely on the
specific distribution of the jump times. We will compute the rate of
convergence when the jump times have uniform distribution, showing that the
method used in [8] can be adapted for non exponential times. All these results
give us new ways to simulate the behaviour of the standard Brownian motion.

The paper is organized in the following way. Section 2 is devoted to define
the processes and to give the main results. In Section 3 we prove the strong
convergence theorem. The study of the rate of convergence is given in Section
4.

##  2 Definitions and main result

Consider $(U_{m})_{m\geq 1}$ be a sequence of independent random variables
which take on only nonnegative values. We also assume that they are
identically distributed with $P(U_{1}=0)<1$ and $E((U_{1})^{4})<\infty.$

For each $k\geq 1$ consider the renewal sequence $S_{k}=U_{1}+\cdots+U_{k}$
and the counting renewal function

$L(t)=\sum_{k=1}^{\infty}{\rm l}\hskip-5.97527pt1_{\,[0,t]}(S_{k}),$

that is the counting function of the number of renewals in $[0,t].$

Set $\\{\eta_{m}\\}_{m\geq 0}$ a sequence of independent identically
distributed random variables with law Bernoulli($\frac{1}{2}$), independent of
$\\{U_{m}\\}_{m\geq 1}$. Then, we will deal with the renewal reward process
defined as

$T(t)=\eta_{0}+\sum_{k=1}^{\infty}\eta_{k}\,\,{\rm
l}\hskip-5.97527pt1_{\,[0,t]}(S_{k})=\sum_{l=0}^{L(t)}\eta_{l}.$

Then, given a strictly positive function $\beta$ we define

$T_{n}(t)=T_{\beta(n)}(t)=T\big{(}\frac{t}{\beta(n)}\big{)}=\eta_{0}+\sum_{k=1}^{\infty}\eta_{k}\,\,{\rm
l}\hskip-5.97527pt1_{\,[0,\frac{t}{\beta(n)}]}(S_{k})=\eta_{0}+\sum_{k=1}^{\infty}\eta_{k}\,\,{\rm
l}\hskip-5.97527pt1_{\,[0,t]}(\beta(n)S_{k}).$ (1)

Notice that putting $U_{m}^{n}=\beta(n)\times U_{m}$ for all $m\geq 1$, we
have that

$\beta(n)\times S_{k}=U_{1}^{n}+\cdots+U_{k}^{n}.$

Our aim is to study the convergence of the processes

$x_{n}(t)=\Big{(}\beta(n)\frac{{\mathbf{E}}((U_{1})^{2})}{{\mathbf{E}}(U_{1})}\Big{)}^{-\frac{1}{2}}\int_{0}^{t}(-1)^{T_{\beta(n)}(u)}du=\frac{1}{G(n)}\int_{0}^{t}(-1)^{T_{\beta(n)}(u)}du,$
(2)

where

$G(n)=\Big{(}\beta(n)\frac{{\mathbf{E}}((U_{1})^{2})}{{\mathbf{E}}(U_{1})}\Big{)}^{\frac{1}{2}},$

with

$\sum_{n\geq 1}\beta(n)<\infty.$

Obviously, we can write

$\displaystyle
x_{n}(t)=\frac{1}{G(n)}\int_{0}^{t}(-1)^{T(\frac{u}{\beta(n)})}du=\frac{1}{G(n)}\beta(n)\int_{0}^{\frac{t}{\beta(n)}}(-1)^{T(v)}dv$
$\displaystyle\qquad=\Big{(}\frac{{\mathbf{E}}(U_{1})}{{\mathbf{E}}((U_{1})^{2})}\Big{)}^{\frac{1}{2}}\beta(n)^{\frac{1}{2}}\int_{0}^{\frac{t}{\beta(n)}}(-1)^{T(v)}dv.$

Our next result gives the strong convergence of realizations of our processes
$\\{x_{n}(t);\,t\in[0,1]\\}$ and states as follows:

######  Theorem 2.1.

There exists realizations of the process $x_{n}$ on the same probability space
as a standard Brownian motion $\\{x(t),t\geq 0\\}$ such that

$\lim_{n\rightarrow\infty}\max_{0\leq t\leq 1}|x_{n}(t)-x(t)|=0\quad a.s.$

###### Proof.

See Section 3. ∎

Observe that we are assuming that the jumps occurs with times that follows a
family of nonnegative independent identically distributed random variables
$\\{U_{m}^{n}\\}_{m\geq 1}$.

##  3 Proof of strong convergence

In this section, we will prove the strong convergence when $n$ tends to
$\infty$ of the processes $\\{x_{n}(t);\,t\in[0,1]\\}$ defined in Section 2.

Proof of Theorem 2.1. We will follow the methodology used in [9].

Let $(\Omega,\cal{F},\cal{P})$ be the probability space for a standard
Brownian motion $\\{x_{t},t\geq 0\\}$ with $x(0)=0$ and let us define:

  1. 1.

for each $n>0$, $\\{\xi_{m}^{n}\\}_{m\geq 1}$ a sequence of nonnegative
independent identically distributed random variables, independent of the
Brownian motion $x$, such that

$G(n)\times\xi_{m}^{n}\sim U_{m}^{n}.$ (3)

  2. 2.

$\\{k_{m}\\}_{m\geq 1}$ a sequence of independent identically distributed
random variables such that $P(k_{1}=1)=P(k_{1}=-1)=\frac{1}{2}$ , independent
of $x$ and $\\{\xi_{m}^{n}\\}_{m\geq 1}$ for all $n$.

Notice that

$\xi_{m}^{n}\sim\frac{\beta(n)}{G(n)}\times
U_{m}=\beta(n)^{\frac{1}{2}}\frac{{\mathbf{E}}(U_{1})^{\frac{1}{2}}}{E((U_{1})^{2})^{\frac{1}{2}}}\times
U_{m}.$

So

${\mathbf{E}}(\xi_{m}^{n})=\beta(n)^{\frac{1}{2}}\frac{{\mathbf{E}}(U_{1})^{\frac{3}{2}}}{E((U_{1})^{2})^{\frac{1}{2}}},\qquad{\mathbf{E}}((\xi_{m}^{n})^{2})=\beta(n){\mathbf{E}}(U_{1})$

and

${\mathbf{E}}((\xi_{m}^{n})^{4})=\beta(n)^{2}\frac{{\mathbf{E}}((U_{1})^{4}){\mathbf{E}}(U_{1})^{2}}{{\mathbf{E}}((U_{1})^{2})^{2}}$

By Skorokhod’s theorem ([12] page 163) for each $n\geq 1$ there exists a
sequence $\sigma_{1}^{n},\sigma_{2}^{n},...$ of nonnegative independent random
variables on $(\Omega,\cal{F},\cal{P})$ so that the sequence
$x(\sigma_{1}^{n}),x(\sigma_{1}^{n}+\sigma_{2}^{n}),...,$ has the same
distribution as $k_{1}\xi_{1}^{n},k_{1}\xi_{1}^{n}+k_{2}\xi_{2}^{n},...,$ and,
for each $m$,

  1. 1.

${\mathbf{E}}(\sigma_{m}^{n})=Var(k_{m}\xi_{m}^{n})={\mathbf{E}}((\xi_{m}^{n})^{2})=\beta(n){\mathbf{E}}(U_{1}),$

  2. 2.

There exists $L_{2}$ such that

$Var(\sigma_{m}^{n})\leq{\mathbf{E}}((\sigma_{m}^{n})^{2})\leq
L_{2}{\mathbf{E}}((\xi_{m}^{n})^{4})=L_{2}\beta(n)^{2}\frac{{\mathbf{E}}((U_{1})^{4}){\mathbf{E}}(U_{1})^{2}}{{\mathbf{E}}((U_{1})^{2})^{2}}.$

For each $n$ we define $\gamma_{0}^{n}\equiv 0$ and for each $m$

$\gamma_{m}^{n}=G(n)\left|x\left(\sum_{j=0}^{m}\sigma_{j}^{n}\right)-x\left(\sum_{j=0}^{m-1}\sigma_{j}^{n}\right)\right|,$

where $\sigma_{0}^{n}\equiv 0$.

Then, from (3) it follows that the random variables
$\gamma_{1}^{n},\gamma_{2}^{n},...,$ are independent with the same
distribution that $U_{1}^{n},U_{2}^{n},\ldots,$ and

$E(\gamma_{m}^{n})=E(U_{m}^{n})=\beta(n)E(U_{1})$

and

$Var(\gamma_{m}^{n})=\beta(n)^{2}Var(U_{1}).$

Now, we define $x_{n}(t),t\geq 0$ to be piecewise linear satisfying

$x_{n}\left(\sum_{j=1}^{m}\gamma_{j}^{n}\right)=x\left(\sum_{j=1}^{m}\sigma_{j}^{n}\right),\qquad
m\geq 1$ (4)

and $x_{n}(0)\equiv 0$. Observe that the process $x_{n}$ has slope
$\pm|G(n)|^{-1}$ in the interval
$[\sum_{j=1}^{m-1}\gamma_{j}^{n},\sum_{j=1}^{m}\gamma_{j}^{n}]$.

On the other hand, let $\Gamma_{m}^{n}=\sum_{j=1}^{m}\gamma_{j}^{n}$. We get
that the increments $\Gamma_{m}^{n}-\Gamma_{m-1}^{n}$, for each $m$, with
$\Gamma_{0}^{n}\equiv 0$, are independent and have law
$G(n)\times\xi_{1}^{m}\sim U_{1}^{n}$. Moreover the probability that
$x\left(\sum_{j=0}^{m}\sigma_{j}^{n}\right)-x\left(\sum_{j=0}^{m-1}\sigma_{j}^{n}\right)$
is positive is $\frac{1}{2}$, independent of the past up to time
$\sum_{j=0}^{m-1}\sigma_{j}^{n}.$ Thus $x_{n}$ is a realization of the process
(2).

Set $H(n):=\beta(n)E(U_{1})$. Recalling that
$\gamma_{0}^{n}\equiv\sigma_{0}^{n}\equiv 0$, by (4) and the uniform
continuity of Brownian motion on $[0,1]$, we have almost surely

$\displaystyle\lim_{n\rightarrow\infty}\,\,\max_{0\leq t\leq
1}\left|x_{n}(t)-x(t)\right|$ $\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\,\,\max_{0\leq
m\leq\frac{1}{H(n)}}\left|x_{n}\left(\sum_{j=0}^{m}\gamma_{j}^{n}\right)-x\left(\sum_{j=0}^{m}\gamma_{j}^{n}\right)\right|$
$\displaystyle=$ $\displaystyle\lim_{n\rightarrow\infty}\,\,\max_{0\leq
m\leq\frac{1}{H(n)}}\left|x\left(\sum_{j=0}^{m}\sigma_{j}^{n}\right)-x\left(\sum_{j=0}^{m}\gamma_{j}^{n}\right)\right|,$

and it reduces the proof to check that,

$\lim_{n\rightarrow\infty}\,\max_{1\leq
m\leq\frac{1}{H(n)}}\left|\gamma_{1}^{n}+\dots+\gamma_{m}^{n}-mH(n)\right|=0\quad
a.s.,$

and that

$\lim_{n\rightarrow\infty}\,\max_{1\leq
m\leq\frac{1}{H(n)}}\left|\sigma_{1}^{n}+\dots+\sigma_{m}^{n}-mH(n)\right|=0\quad
a.s.,$

The first limit can be obtained easily by Borel-Cantelli lemma since by
Kolmogorov’s inequality, for each $\alpha>0$, we have

$\displaystyle P\left(\max_{1\leq
m\leq\frac{1}{H(n)}}\left|\gamma_{1}^{n}+\dots+\gamma_{m}^{n}-mH(n)\right|\geq\alpha\right)\leq\frac{1}{\alpha^{2}}\sum_{m=1}^{[\frac{1}{H(n)}]}Var(\gamma_{k}^{n})$
$\displaystyle\qquad\leq\frac{1}{\alpha^{2}}\sum_{m=1}^{\infty}\beta(n)^{2}Var(U_{1})<\infty.$

We can study the second limit repeting the same arguments as before. Using the
bouns obtained from Skorohod’s theorem, for each $\alpha>0$, we have

$\displaystyle P\left(\max_{1\leq
m\leq\frac{1}{H(n)}}\left|\sigma_{1}^{n}+\dots+\sigma_{m}^{n}-mH(n)\right|\geq\alpha\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{\alpha^{2}}\sum_{m=1}^{[\frac{1}{H(n)}]+1}Var(\sigma_{m}^{n})<\infty$

$\square$

##  4 Rate of convergence

In this section we will prove the rate of convergence of the processes
$x_{n}(t)$ in a particular case. We consider $U_{m}\sim U(0,1)$ for all $m\geq
1$ and $\beta(n)=n^{-k}$ with $k>1$. Then

$\displaystyle
G(n)=\frac{2^{\frac{1}{2}}}{3^{\frac{1}{2}}n^{\frac{k}{2}}},\qquad
H(n)=\frac{1}{2n^{k}},$

and

$U_{m}^{n}\sim U(0,n^{-k}),\qquad\gamma_{m}^{n}\sim
U(0,n^{-k}),\qquad\xi_{m}^{n}\sim
U(0,\frac{3^{\frac{1}{2}}}{2^{\frac{1}{2}}}n^{-\frac{k}{2}}).$

######  Theorem 4.1.

Assume $U_{m}\sim U(0,1)$ for all $m\geq 1$ and $\beta(n)=n^{-k}$ with $k>1$.
Then, for all $q>0$,

$P\left(\max_{0\leq t\leq
1}|x_{n}(t)-x(t)|>\alpha\,n^{-\frac{k}{4}}\left(\log{n}\right)^{\frac{3}{2}}\right)=o(n^{-q})\qquad\mbox{as}\quad
n\rightarrow\infty$

where $\alpha$ is a positive constant depending on $q$.

Since the proof follows the structure of part b) of Theorem 1 in [8], we give
a sketch of the proof.

Proof of Theorem 4.1. Recall that $\gamma_{0}^{n}\equiv\sigma_{0}^{n}\equiv 0$
and define

$\Gamma_{m}^{n}=\sum_{j=0}^{m}\gamma_{j}^{n}\qquad\mbox{and}\qquad\Lambda_{m}^{n}=\sum_{j=0}^{m}\sigma_{j}^{n}.$

Notice that $x_{n}(\Gamma_{m}^{n})=x(\Lambda_{m}^{n})$. Set

$J^{n}\equiv\max_{0\leq m\leq\frac{1}{H(n)}}\,\max_{0\leq
r\leq\gamma_{m+1}^{n}}\big{|}x_{n}(\Gamma_{m}^{n}+r)-x(\Gamma_{m}^{n}+r)\big{|}.$

Since $x_{n}$ is piecewise linear and using the definition of
$\gamma_{m}^{n}$, notice that

$\displaystyle x_{n}(\Gamma_{m}^{n}+r)$ $\displaystyle=$ $\displaystyle
x(\Lambda_{m}^{n})+\frac{x(\Lambda_{m+1}^{n})-x(\Lambda_{m}^{n})}{\gamma_{m+1}^{n}}\,r$
$\displaystyle=$ $\displaystyle
x(\Lambda_{m}^{n})+\frac{1}{G(n)}\times\operatorname{sgn}\Big{(}x(\Lambda_{m+1}^{n})-x(\Lambda_{m}^{n})\Big{)}r.$

Thus,

$\displaystyle J^{n}$ $\displaystyle\leq$ $\displaystyle\max_{0\leq
m\leq\frac{1}{H(n)}}\Big{|}x(\Lambda_{m}^{n})-x\Big{(}mH(n)\Big{)}\Big{|}+\max_{0\leq
m\leq\frac{1}{H(n)}}\Big{|}x(\Gamma_{m}^{n})-x\Big{(}mH(n)\Big{)}\Big{|}$
$\displaystyle+\max_{0\leq m\leq\frac{1}{H(n)}}\,\max_{0\leq
r\leq\gamma_{m+1}^{n}}\big{|}x(\Gamma_{m}^{n})-x(\Gamma_{m}^{n}+r)\big{|}+\max_{1\leq
m\leq\frac{1}{H(n)}+1}\frac{1}{G(n)}\gamma_{m}^{n}$ $\displaystyle:=$
$\displaystyle J_{1}^{n}+J_{2}^{n}+J_{3}^{n}+J_{4}^{n},$

and for any $a_{n}>0$,

$P(J^{n}>a_{n})\leq\sum_{j=1}^{4}P\Big{(}J_{j}^{n}>\frac{a_{n}}{4}\Big{)}:=I_{1}^{n}+I_{2}^{n}+I_{3}^{n}+I_{4}^{n}.$

We will study the four terms separately.

1\. Study of the term $I_{4}^{n}.$ Since $\gamma_{m}^{n}$’s are independent
variables with law $\sim U(0,n^{-k})$,

$\displaystyle I_{4}^{n}$ $\displaystyle\leq$ $\displaystyle
P\left(\max_{1\leq
m\leq\frac{1}{H(n)}+1}\gamma_{m}^{n}>\frac{a_{n}}{2^{\frac{3}{2}}3^{\frac{1}{2}}n^{\frac{k}{2}}}\right)=1-P\left(\gamma_{m}^{n}\leq\frac{a_{n}}{2^{\frac{3}{2}}3^{\frac{1}{2}}n^{\frac{k}{2}}}\right)^{[\frac{1}{H(n)}]+1}=0,$

when $n$ is big enough for $a_{n}$ of the type
$\alpha\,n^{-\frac{k}{4}}\left(\log n\right)^{\beta}$, with $\alpha$ and
$\beta$ positive arbitrary fixed constants.

2\. Study of the term $I_{1}^{n}.$ Let $\delta_{n}>0$.We can write

$\displaystyle I_{1}^{n}$ $\displaystyle\leq$ $\displaystyle
P\left(\max_{0\leq
m\leq\frac{1}{H(n)}}\,\max_{|s|\leq\delta_{n}}\Big{|}x\Big{(}mH(n)+s\Big{)}-x\Big{(}mH(n)\Big{)}\Big{|}>\frac{a_{n}}{4}\right)$
$\displaystyle+P\left(\max_{1\leq
m\leq\frac{1}{H(n)}}\Big{|}\Lambda_{m}^{n}-mH(n)\Big{|}>\delta_{n}\right)$
$\displaystyle=$ $\displaystyle I_{11}^{n}+I_{12}^{n}.$

2.1. Study of the term $I_{12}^{n}.$ Notice that

$\displaystyle I_{12}^{n}$ $\displaystyle=$ $\displaystyle P\left(\max_{1\leq
m\leq\frac{1}{H(n)}}\bigg{|}\sum_{j=1}^{m}\Big{(}\frac{1}{H(n)}\sigma_{j}^{n}-1\Big{)}\bigg{|}>\frac{\delta_{n}}{H(n)}\right)$
(5) $\displaystyle\leq$
$\displaystyle\left(\frac{H(n)}{\delta_{n}}\right)^{2p}\,{\mathbf{E}}\left[\left(\sum_{m=1}^{[\frac{1}{H(n)}]}\bigg{(}\frac{1}{H(n)}\sigma_{m}^{n}-1\bigg{)}\right)^{2p}\right],$

for any $p\geq 1$, by Doob’s martingale inequality.

Set $Y_{m}:=\frac{1}{H(n)}\sigma_{m}^{n}-1$. Using Hölder’s inequality, we
obtain

$\displaystyle{\mathbf{E}}\left[\Bigg{(}\sum_{m=1}^{[\frac{1}{H(n)}]}Y_{m}\Bigg{)}^{2p}\right]$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}|u|=2p\\\ u_{m}\neq
1\,\forall m\end{subarray}}{2p\choose u}{\mathbf{E}}\Big{(}Y_{1}^{u_{1}}\cdots
Y_{[\frac{1}{H(n)}]}^{u_{[\frac{1}{H(n)}]}}\Big{)}$ $\displaystyle\leq$
$\displaystyle\sum_{\begin{subarray}{c}|u|=2p\\\ u_{m}\neq 1\,\forall
m\end{subarray}}{2p\choose
u}\big{[}{\mathbf{E}}\big{(}Y_{1}^{2p}\big{)}\big{]}^{u_{1}/2p}\cdots\big{[}{\mathbf{E}}\big{(}Y_{[\frac{1}{H(n)}]}^{2p}\big{)}\big{]}^{u_{[\frac{1}{H(n)}]}/2p}.$

where $u=(u_{1},\dots,u_{[\frac{1}{H(n)}]})$ with
$|u|=u_{1}+\cdots+u_{[}\frac{1}{H(n)]}$ and

${2p\choose u}=\frac{(2p)!}{u_{1}!\cdots u_{[\frac{1}{H(n)}]}!}.$

Notice that in the first equality we have used that if $u_{m}=1$ for any $m$,
then ${\mathbf{E}}\big{(}Y_{1}^{u_{1}}\cdots
Y_{[\frac{1}{H(n)}]}^{u_{[\frac{1}{H(n)}]}}\big{)}=0$. On the other hand, by
the estimates given by Skorohod’s theorem (see [8]), we have

$\displaystyle{\mathbf{E}}[(\sigma_{m}^{n})^{2p}]\leq
2(2p)!{\mathbf{E}}\Big{[}\,(k_{i}\xi_{m}^{n})^{4p}\Big{]}\leq
2(2p)!\,\frac{1}{(4p+1)}3^{2p}\left(\frac{1}{2n^{k}}\right)^{2p}.$

So, using the inequality $|a+b|^{2p}\leq 2^{2p}(|a|^{2p}+|b|^{2p})$, we obtain

${\mathbf{E}}\Big{(}Y_{m}^{2p}\Big{)}\leq(2p)!\,6^{2p}.$ (7)

Finally from a lemma in page 298 in [8] (see also Lemma 5-1 in [1]) we obtain
that, for

$p\leq 1+\frac{\log 2}{\log\big{[}1+(2n^{-k}-n^{-2k})^{\frac{1}{2}}\big{]}},$
(8)

we get that

$\displaystyle\sum_{\begin{subarray}{c}|u|=2p\\\ u_{i}\neq 1\,\forall
i\end{subarray}}{2p\choose u}\leq 2^{2p}(2p)!\left(2n^{k}\right)^{p}.$ (9)

Therefore, for $p$ as above, putting together (5), (4), (7) and (9) and
applying Stirling formula,
$k!=\sqrt{2\pi}\,k^{k+\frac{1}{2}}e^{-k}e^{\frac{a}{12k}}$, with $0<a<1$, we
obtain

$\displaystyle I_{12}^{n}$ $\displaystyle\leq$
$\displaystyle\,(\delta_{n})^{-2p}\,n^{-kp}\,6^{2p}\,2^{p}\Big{[}\sqrt{2\pi}(2p)^{2p+\frac{1}{2}}e^{-2p}e^{\frac{a}{24p}}\Big{]}^{2}$
$\displaystyle\leq$ $\displaystyle
K_{1}^{p}\,(\delta_{n})^{-2p}\,n^{-kp}\,p^{4p+1}$

where $K_{1}$ is a constant.

Let us impose now $K_{1}^{p}\,(\delta_{n})^{-2p}\,n^{-kp}\,p^{4p+1}=n^{-2q}$
and $p=\big{[}\log n\big{]}$. Observe that this $p$ fulfills condition on $p$
of inequality (8). We get

$\displaystyle\delta_{n}=K_{2}\,n^{q/[\log{n}]-\frac{k}{2}}\,\left[\log{n}\right]^{2+1/(2[\log{n}])},$
(10)

where $K_{2}=\sqrt{K_{1}}$ is a constant. Clearly, with this $\delta_{n}$, it
follows that $I_{12}^{n}=o(n^{-q})$.

2.4. Study of the term $I_{11}^{n}.$ As in Theorem 1 in [8], for big $n$ and
using a Doob’s martingale inequality for Brownian motion we get

$\displaystyle I_{11}^{n}$ $\displaystyle\leq$
$\displaystyle\frac{1}{H(n)}P\left(\max_{|s|\leq\delta_{n}}\big{|}x(s)\big{|}>\frac{a_{n}}{4}\right)\leq
8n^{k}P\left(\max_{0\leq s\leq\delta_{n}}x(s)>\frac{a_{n}}{4}\right)$
$\displaystyle\leq$ $\displaystyle
8n^{k}\exp\left(-\Big{(}\frac{a_{n}}{4}\Big{)}^{2}\frac{1}{2\delta_{n}}\right).$

Condition $8n^{k}\exp\big{(}-(a_{n})^{2}/(32\delta_{n})\big{)}=8n^{-2q}$
yields that

$a_{n}=K_{3}\,n^{-k/4}\,n^{q/2[\log{n}]}\,\big{(}\log{n}\big{)}^{1+1/4[\log{n}]}\,\big{(}\log{n}\big{)}^{1/2},$

where $K_{3}$ is a constant depending on $q$. Notice that

$a_{n}=\alpha\,n^{-k/4}\,\big{(}\log{n}\big{)}^{3/2},$

for big $n$, where $\alpha$ is a constant that depends on $q$, satisfies such
a condition. Thus, with $\delta_{n}$ as in (10), it follows that
$I_{11}^{n}=o(n^{-q})$.

3\. Study of the term $I_{2}^{n}$. For our $\delta_{n}>0$, we have

$\displaystyle I_{2}^{n}$ $\displaystyle\leq$ $\displaystyle
P\left(\max_{0\leq
m\leq\frac{1}{H(n)}}\,\max_{|s|\leq\delta_{n}}\Big{|}x\Big{(}mH(n)+s\Big{)}-x\Big{(}mH(n)\Big{)}\Big{|}>\frac{a_{n}}{4}\right)$
$\displaystyle+P\left(\max_{0\leq
m\leq\frac{1}{H(n)}}\Big{|}\Gamma_{m}^{n}-mH(n)\Big{|}>\delta_{n}\right)=I_{21}^{n}+I_{22}^{n}.$

On one hand, observe that $I_{21}^{n}=I_{11}^{n}$, thus
$I_{21}^{n}=o(n^{-q})$.

On the other hand, applying Doob’s martingale inequality

$\displaystyle I_{22}^{n}$ $\displaystyle=$ $\displaystyle P\left(\max_{0\leq
m\leq\frac{1}{H(n)}}\Bigg{|}\sum_{j=0}^{m}\bigg{(}\frac{1}{H(n)}\gamma_{j}^{n}-1\bigg{)}\Bigg{|}>\frac{\delta_{n}}{H(n)}\right)$
$\displaystyle\leq$
$\displaystyle\left(\frac{H(n)^{2}}{\delta_{n}}\right)^{2p}\,{\mathbf{E}}\left[\left(\sum_{m=1}^{[\frac{1}{H(n)}]}\bigg{(}\frac{1}{H(n)}\gamma_{m}^{n}-1\bigg{)}\right)^{2p}\right].$

Set $V_{m}:=\frac{1}{H(n)}\gamma_{m}^{n}-1$. Notice that $V_{m}$’s are
independent and centered random variables with

$\displaystyle{\mathbf{E}}\Big{(}V_{m}^{2p}\Big{)}$ $\displaystyle\leq$
$\displaystyle
2^{2p}\left(\bigg{(}2n^{k}\bigg{)}^{2p}{\mathbf{E}}\big{[}(\gamma_{m}^{n})^{2p}\big{]}+1\right)\leq
2\cdot 4^{2p}\,\frac{1}{(2p+1)}.$

Then using an inequality of the type of (4) and following the same arguments
that in the study of $I_{12}^{n}$, we get that $I_{22}^{n}=o(n^{-q})$.

4\. Study of the term $I_{3}^{n}$. For $\delta_{n}>0$ defined in (10) and
$a_{n}$ of the type $\alpha\,n^{-\frac{k}{4}}\left(\log
n\right)^{\frac{3}{2}}$

$\displaystyle I_{3}^{n}$ $\displaystyle\leq$ $\displaystyle
P\left(\max_{0\leq
m\leq\frac{1}{H(n)}}\,\max_{|r|\leq\delta_{n}}\big{|}x(\Gamma_{m}^{n})-x(\Gamma_{m}^{n}+r)\big{|}>\frac{a_{n}}{4}\right)$
$\displaystyle+P\left(\max_{1\leq
m\leq\frac{1}{H(n)}+1}\gamma_{m}^{n}>\delta_{n}\right):=I_{31}^{n}+I_{32}^{n}.$

On one hand, $I_{31}^{n}=o(n^{-q})$ is proved in the same way as $I_{11}^{n}$.
On the other hand, for $n$ big enough, $I_{32}^{n}=0$, similarly as we have
proved for $I_{4}^{n}$.

We have checked now that all the terms in our decomposition are of order
$n^{-q}$. The proof of Theorem 4.1 can be completed following the same
computations that in [8] (see also Theorem 3.2 in [1])

$\square$

## References

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Xavier Bardina

Departament de Matemàtiques, Facultat de Ciències,

Edifici C, Universitat Autònoma de Barcelona, 08193 Bellaterra.

<EMAIL_ADDRESS>

Carles Rovira

Facultat de Matemàtiques, Universitat de Barcelona,

Gran Via 585, 08007 Barcelona.

<EMAIL_ADDRESS>