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\begin{document}

\begin{center}
{\bf \Large On probabilities of small deviations
for compound sum processes}
\end{center}


\begin{center}
Andrei N. Frolov and
Alexander I. Martikainen
\end{center}

\begin{center}
Department of Mathematics and Mechanics,\\
St.~Petersburg State University
\end{center}


\bigskip
\bigskip
This talk is based mainly on the results obtained in our joint work with
Prof. J. Steinebach (K\"{o}ln, Germany).


%\bigskip
%{\it Keywords:} Probabilities of small deviations; Compound sum
%process; Compound Poisson process; Renewal counting process;
%Chung's law of the iterated logarithm

\bigskip
\bigskip
{\bf Abstract.}
The logarithmic asymptotics of small
deviation probabilities for Poisson and compound Poisson processes
are well known because they have independent increments.
We study such asymptotics for compound sum processes of much more general
structure. Our results are also new for renewal processes.
In our theorems, the renewal process and summands,
generating the compound sum process, may be dependent.


\newpage

\bigskip

{\bf 1. Introduction}

\bigskip

Let $(X,Y), (X_1,Y_1), (X_2,Y_2), \ldots$ be a sequence of
i.i.d. random vectors with
\begin{eqnarray*} %\label{ex}
&& \mathE X = \mu \in (0,\infty),\\
&& %\label{ey}
\mathP(Y>0)=1 \; \mbox{and} \;
\mathE Y = \lambda \in (0,\infty).
\end{eqnarray*}
Note that $X$ and $Y$ may be dependent.

Put $S_0=R_0=0,$
$$S_n=X_1+\cdots+X_n,\; R_n=Y_1+\cdots+Y_n,\; n \geqslant 1.$$


Define the compound sum process as
\begin{equation} \nonumber %\label{St}
  S(t) = S_{N(t)}, \;\; \hbox{where} \;\;
   N(t)=\max\{n:R_n\leqslant t\},\;\; t \geq 0.
\end{equation}
$N(t)$ is the renewal process generated by Y's.%the sequence $\{Y_i\}$.

%that is
%\begin{equation}\nonumber %\label{Nt}
%  N(t)=\max\{n:R_n\leqslant t\},\qquad t \geq 0.
%\end{equation}

%\medskip
If $\mathP (Y_1 > x) = e^{-x/\lambda},$ $x>0,$ then
$N(t)$ is a Poisson process.
If %, in addition,
$N(t)$ and $\{X_i\}$ are independent, then $S(t)$ is called a
compound Poisson process.

%\medskip

%Compound sum processes play an important role in actuarial and
%financial mathematics (cf., e.g., Embrechts and Kl\"uppelberg (1993)).
%They describe, for example, the processes of total claims
%of an insurance company in a collective risk model. In the latter
%case, $X_i$ may denote the amount of the $i$-th claim,
%$N(t)$ the number of claims up to time $t,$ and
%the $Y_i$'s are the random times between successive claims.
%
%\medskip


%In what follows,
We will describe the logarithmic asymptotics of
%probabilities of small deviations of the process $\{S(t)\},$ i.e.,
%of
\begin{equation}\nonumber %\label{PT}
P_T = \mathP \Big(
\sup_{0\leqslant t\leqslant T} \Big|S(t)-\frac{\mu}{\lambda}\,t\Big|
\leqslant x_T \Big),
\end{equation}
where $\{x_T\}$ is a positive real function such that
$x_T \rightarrow \infty$ and $x_T^2 = o(T),$ as $T\to\infty.$
%
%(%
\newline \indent Note that
if X and Y are independent, $\mathE S(t) \sim t\mu/\lambda$
as $t \to \infty$ and
%It is not difficult to check that
$\mathE S(t) = t\mu/\lambda$ for compound Poisson process.%)

\newpage
$P_T$ is the probability that a trajectory of the compound sum process,
centered at $t\mu/\lambda$, lies within a symmetric strip (around zero)
of width $2 x_T.$ The same results may be stated for a more
general class of sets instead of strips.
%(see, e.g., Mogul'skii~(1974) for further details and methods).

\medskip
%\newpage
If $\mathP(Y_1=1)=1,$ then $S(t)=S_n,$ for $t \in [n,n+1),$
%in which case
the problem has been investigated by Mogul'skii (1974).

\newpage
\medskip
\noin
{\bf Theorem A (Mogul'skii).} {\it Let $\{ \xi_i \}$ be a sequence
of \,i.i.d. zero mean random variables from the domain of
attraction of a strictly stable law $F_{\alpha}$ with index
$\alpha\in (1,2]$,
%and $F_{\alpha}(0) \in (0,1),$
i.e.
$(\xi_1+\cdots+\xi_n)/B_n$ converges weakly to $F_{\alpha},$
where $\{ B_n \}$ is a sequence of positive constants.

\noin
Then, for every sequence $\{x_n\}$ of positive numbers such that
$x_n \rightarrow \infty$ and $x_n = o(B_n),$
\begin{equation*}\label{Mogul}
 \log \mathP \Big(
\max_{1\leqslant k\leqslant n} \left|\xi_1+\cdots+\xi_k\right|
\leqslant x_n \Big) \sim
- \, C \, \frac{n}{x_n^{\alpha}} \, L(x_n),% \big(1+o(1)\big),
\end{equation*}
where the function $L(x)= x^{\alpha-2}\,\mathE \xi_1^2 I\{ |\xi_1|
< x\}$ is slowly varying $($at infinity$\,),$ and $C$ is an
absolute positive constant, depending only on the
distribution~$F_{\alpha}.$ If $\alpha=2$, then $C=\pi^2/8.$}

\medskip\noin
Here and in the sequel, \newline
$I\{ B \}$ denotes the indicator function of an event~$B$, \newline
%the symbols $\limsup$, $\liminf$,
O, o, $\sim$
$\rightarrow$ etc. are used for $n\rightarrow\infty$ or $T\rightarrow\infty$.
%unless otherwise stated.

Such results for the Poisson process and compound Poisson
processes follow from results for processes with independent
increments in Mogul'skii (1974). (See, also
Borovkov and Mogul'ski (1989, 1991).)

For compound sum processes our results are new,
even in the case $\mathP(X_1=1)=1,$ that is for renewal processes.

\newpage
\bigskip
{\bf 2. Results}
\bigskip


%In what follows, the symbols $\limsup$, $\liminf$, O, o,
%$\rightarrow$ etc. are used for $T\rightarrow\infty,$ unless
%otherwise stated.

\medskip
\begin{theorem}\label{th1}
{\it Assume that either $\mathE e^{h Y}<\infty,$ for some
$h>0,$ or that $X$ is positive and $\mathE e^{h X}<\infty,$ for some $h>0.$
Moreover, let the random variables $\xi_i = X_i/\mu-
Y_i/\lambda,$ $i\geqslant 1,$ satisfy the conditions of Theorem A.

\noin
Then, for every positive function $\{x_T\}$ such that
$x_T \rightarrow\infty$ and $x_T = o(B_{[T]}),$ %there is a
%positive constant $C$ such that
\begin{equation}\nonumber %\label{upbnd}
\hspace*{-5mm}
 \log \mathP \Big(
\sup_{0\leqslant t\leqslant T} \Big|S(t)-\frac{\mu}{\lambda} \, t \Big| \leqslant x_T
\Big) \leqslant - C \frac{\mu^{\alpha}}{\lambda} \frac{T}{x_T^{\alpha}} L(x_T)
\big(1+o(1)\big),
\end{equation}
where $C$ is the constant from Theorem A.
}
\end{theorem}

\medskip
We cannot omit the one-sided Cram\'{e}r condition in Theorem \ref{th1}.
Nevertheless one can obtain a weaker upper bound.

\medskip
\begin{remark}\label{r1}
{\it Assume that the random variables $\xi_i = X_i/\mu-
Y_i/\lambda$ satisfy the conditions of Theorem A.

\noin
Then, for every positive function $\{x_T\}$ such that
$x_T \rightarrow\infty$, $x_T = o\big(B_{[T]}\big)$ and $x_T o\big(T^{1/(\alpha+1)}\big),$ %there is a $C>0$ such that
\begin{equation*}
\hspace*{-5mm}
 \log \mathP \Big(
\sup_{0\leqslant t\leqslant T} \Big|S(t)-\frac{\mu}{\lambda}\,t\Big|
\leqslant x_T \Big) \leqslant
- C \frac{\mu^{\alpha}}{2 \lambda}
\frac{T}{x_T^{\alpha+1}} L(x_T) \big(1+o(1)\big),
\end{equation*}
where $C$ is the constant from Theorem A.
}
\end{remark}

\medskip
Let us now turn to lower bounds for $P_T.$ We first consider the
case of finite variances.

\medskip
\begin{theorem}\label{th2}
{\it Assume that %$\mathE X^2<\infty$, $\mathE Y^2<\infty$ and
one of the following two conditions holds:

\noindent
{\ups (i)} at least one of the random variables $X$ and $Y$ is bounded,
and another one has a finite variance;

\noindent
{\ups (ii)} $\mathE X^3<\infty$, $\mathE Y^3<\infty$ and
\begin{eqnarray}
\label{m1}
%\hspace*{-\parindent}
&& \mathE X I\{ Y>u\}
\leqslant \frac{\mu}{\lambda}\Big(1-\frac{A}{u}\Big) \,\mathE Y I\{ Y>u\},
\\[2mm]
\label{m2}
%\hspace*{-\parindent}
&& \mathE Y I\{ X>w\}
\leqslant  \frac{\lambda}{\mu} \Big(1-\frac{A}{w}\Big) \,\mathE X I\{ X>w\},
\end{eqnarray}
for some $A>0$, all large $u$ and all $w$ close to $\esup X$.

Then, for every positive function $\{x_T\}$ such that
$x_T \to \infty$ and
$x_T = o\big(\sqrt{T}\big),$
\begin{equation*}%\label{lwbnd1}
 \log \mathP \Big(
\sup_{0\leqslant t\leqslant T} \Big|S(t)-\frac{\mu}{\lambda}\,t \Big|
\leqslant x_T \Big) \geqslant -\, \frac{\pi^2}{8} \,\frac{\mu^2 v^2}{\lambda}
\,\frac{T}{x_T^2} \big(1+o(1)\big),
\end{equation*}
if $v^2=\mathbf{Var}(X/\mu-Y/\lambda)>0.$ }
\end{theorem}

\medskip
\noin
Conditions (\ref{m1}) and (\ref{m2}) are satisfied, if either

\noindent
(i') $X$ and $Y$ are independent and $X$ is nondegenerate;
%
\newline\noindent
(ii')  $Y$ is bounded.


\medskip
\noin
Theorems \ref{th1} and \ref{th2} yield the following result.

\medskip
\begin{corollary}\label{c1}
{\it Assume that at least one of the following three conditions holds:

\begin{itemize}
\item[\ups (i'')\,\,] $X$ is bounded and $\mathE e^{h Y}<\infty,$ for some
$h>0,$
\item[\ups (ii'')\,] $Y$ is bounded and $\mathE X^2<\infty,$
\item[\ups (iii'')]%$\mathrm{ess sup} X = \mathrm{ess sup} Y =\infty$,
%$\mathE X^3<\infty,$
$\mathE e^{h Y}<\infty,$ for some $h>0,$
and {\ups (ii)} is satisfied.
\end{itemize}
Then, for every positive function $\{x_T\}$ such that $x_T \to \infty$ and
$x_T = o\big(\sqrt{T}\big),$
\begin{equation*}
 \log \mathP \Big(
\sup_{0\leqslant t\leqslant T} \Big|S(t)-\frac{\mu}{\lambda}\,t \Big|
\leqslant x_T \Big) ~\sim~ - \,\frac{\pi^{2}}{8}
\,\frac{\mu^2 v^2}{\lambda}\,\frac{T}{x_T^{2}},
\end{equation*}
if $v^2=\mathbf{Var}(X/\mu-Y/\lambda)>0.$ }
\end{corollary}

The next result follows immediately from Corollary \ref{c1} and
duality principle in Theorem \ref{th5}.

\medskip
\begin{corollary}\label{c2}
{\it Assume that $\mathE X^2<\infty$, $\mathE Y^2<\infty$ and
$v^2>0.$ Moreover, let $X$ and $Y$ be positive and either
$X$ or $Y$ be bounded.

\noin
Then, for every positive function $x_T$ such that $x_T \to \infty$ and
$x_T = o\big(\sqrt{T}\big),$
\begin{equation*}
 \log \mathP \Big(
\sup_{0\leqslant t\leqslant T} \Big|S(t)-\frac{\mu}{\lambda}\,t \Big|
\leqslant x_T \Big) ~\sim~ - \,\frac{\pi^{2}}{8}
\,\frac{\mu^2 v^2}{\lambda} \,\frac{T}{x_T^{2}}.
\end{equation*}
}
\end{corollary}

\medskip
For the special case of $\mathP(X=1)=1,$ Corollary~\ref{c2}
implies the following result for renewal counting processes.

\medskip
\begin{corollary}\label{c3}
{\it Assume that $0<\mathbf{Var}(Y)=\sigma^2_Y<\infty$.

\noin
Then, for every positive function $\{x_T\}$ such that $x_T \to \infty$ and
$x_T = o\big(\sqrt{T}\big),$
\begin{equation*}
 \log \mathP \Big(
\sup_{0\leqslant t\leqslant T} \Big|N(t)-\frac{t}{\lambda} \Big|
\leqslant x_T \Big) ~\sim~
- \,\frac{\pi^{2}}{8}\,\frac{\sigma^2_Y}{\lambda^3}\,\frac{T}{x_T^{2}}.
\end{equation*}
}
\end{corollary}

\newpage
\medskip
We now turn to the case, when $\mathE X^2=\infty$ or
$\mathE Y^2=\infty.$


\medskip
\begin{theorem}\label{th3}
{\it Assume that $X$ and $Y$ are independent. If the
random variables $X$ and $Y$ belong to the domains of attraction
of stable laws, i.e.
\begin{equation*}%\label{fal}
  \mathP \left( S_n - n \mu < x B_n \right) \Rightarrow
F_{\alpha}(x), \;\; %as \quad n\rightarrow \infty,
%\mbox{for some} ~0<\alpha \leqslant 2
\alpha\in(1,2]
~\mbox{and}~ B_n>0,
\end{equation*}
and
\begin{equation*}%\label{fbe}
  \mathP \left( R_n - n \lambda < x B'_n \right) \Rightarrow
F_{\beta}(x), \;\; %as \quad n\rightarrow \infty,
%\mbox{for some} ~0<\beta \leqslant 2
\beta\in(1,2]~\mbox{and}~ B'_n>0,
\end{equation*}
then, for every positive function $\{x_T\}$ such that
$x_T \to \infty$ and
$x_T = o\big(\max\{B_{[T]}, B'_{[T]}\}\big),$
\begin{eqnarray*}%\label{tri}
&&\hs{-7mm}\log \mathP \Big(
\sup_{0\leqslant t\leqslant T} \Big|S(t)-\frac{\mu}{\lambda}\,t \Big|
\leqslant x_T \Big)  \hspace*{7\parindent}\nonumber
\\
&&\hspace*{1mm}\geqslant - \,C \,\frac{\mu^{2}}{\lambda}
\,\frac{T}{x_T^{2}}
\int\limits_{-x_T/\mu} ^{x_T/\mu} u^2 \,d\big(F_{X/\mu}(u)+F_{Y/\lambda}(u)\big)
\big(1+o(1)\big),
\end{eqnarray*}
where $C$ is an absolute positive constant, depending only on the
distributions $F_{\alpha}$ and $F_{\beta}$.

Moreover, if $\alpha \neq \beta,$ then %{\ups (\ref{tri})}
the latter bound holds
with the same absolute constant $C$ as in Theorem~A, depending on
$F_{\min\{\alpha,\beta\}}$. %and $C=\pi^2/8,$
%when \newline $\max\{\alpha, \beta\}=2.$
}
\end{theorem}

\medskip
If $X$ is positive, then
the proof of Theorem~\ref{th1} is based on the duality
between the compound sum processes $S(t)$ and
\begin{equation*}
  R(t) = R_{M(t)+1},\;\;
\mbox{\ups where} \;\;
  M(t)=\max\{n: S_n\leqslant t\},\; t \geq 0.
\end{equation*}
This duality is of independent interest.

\medskip
\begin{theorem}\label{th5}
{\it Assume that $\mathP(X>0)=1$. Then,
\begin{eqnarray*}
&&
\hs{-20mm}\mathP \Big(
\sup_{0\leqslant t\leqslant T} \Big|S(t)-\frac{\mu}{\lambda} t\Big|
< x_T \Big)
%\hspace*{3\parindent}
\leqslant
\mathP \Big(
\sup_{0\leqslant t\leqslant \frac{\mu}{\lambda} T- x_T}
\Big|R(t)-\frac{\lambda}{\mu} t \Big|
\leqslant \frac{\lambda}{\mu} x_T \Big),
%\label{slr}
\\[2mm]
&&
\hs{-20mm}\mathP \Big(
\sup_{0\leqslant t\leqslant T} \Big|S(t)-\frac{\mu}{\lambda} t \Big|
\leqslant x_T \Big)
\geqslant
\mathP \Big(
\sup_{0\leqslant t\leqslant \frac{\mu}{\lambda} T+ x_T}
\Big|R(t)-\frac{\lambda}{\mu} t \Big|
< \frac{\lambda}{\mu} x_T \Big).
%\label{rls}
\end{eqnarray*}
}
\end{theorem}

\end{document}

\vs{10mm}
\stepcounter{section}\noin
{\bf \arabic{section}. Proofs}

\vs{5mm}
\setcounter{equation}{0}
Without loss of generality we may assume in the sequel that
$\mu=\lambda=1.$ Otherwise, if either $\lambda\neq 1$ or $\mu\neq 1,$ put
$X'_i=X_i/\mu,$ $Y'_i=Y_i/\lambda,$ $S'_k=\sum_{i=1}^k X'_i,$
$R'_k=\sum_{i=1}^k Y'_i,$ $N'(t)=\max\{n:R'_n\leqslant t\}=N(\lambda
t),$ $S'(t)=S'_{N'(t)}.$ Then,
\begin{equation*}
  P_T ~=~\mathP \Big(
\sup_{0\leqslant t\leqslant T/\lambda} |S'(t)- t|
\leqslant \frac{x_T}{\mu} \Big).
\end{equation*}
Now, suppose that $\mu=\lambda=1$ and set
\begin{equation*}
  V_T ~=~ \sup_{0\leqslant t\leqslant T} |S(t)- t |.
\end{equation*}
Since the trajectories of $\{S(t)\}$ are step functions with jumps in
the random points $R_1, R_2, \ldots,$ we have
\begin{equation}\label{vt1}
  V_T ~=~ \max\{ M_{N(T)}, M'_{N(T)}, |S_{N(T)}-T|\},
\end{equation}
where $M_t=\max_{1\leqslant k\leqslant t} |S_{k}-R_k|$,
$M'_t= \max_{1\leqslant k\leqslant t} |S_{k-1}-R_k|$, $t\geqslant 1$.
Relation (\ref{vt1}) implies that
\begin{equation}\label{vt2}
  V_T ~\leqslant~ \max\{ M_{N(T)}, M'_{N(T)+1}  \}.
\end{equation}

\bigskip
\noin
{\bf Proof of Theorem~\ref{th1}.} We first assume that $\mathE
e^{h Y}< \infty,$ for some $h>0.$

\noin
Take $c \in (0,1)$.
By (\ref{vt1}), we have
\begin{eqnarray*}
P_T &\leqslant &\mathP \big(  M_{N(T)}
\leqslant x_T\big)
\\[1mm]
&\leqslant & \mathP \big(  M_{N(T)}
\leqslant x_T, N(T)\geqslant [c T] \big)+
\mathP \big(N(T) < [c T] \big)
\\[1mm]
&\leqslant & \mathP \big(  M_{[c T]}
\leqslant x_T \big)+
\mathP \big( R_{[c T] } > T \big).
\end{eqnarray*}
By Theorem A, the right-hand side of (\ref{upbnd}), with $T$
replaced by $[c T],$ provides the logarithmic asymptotic of the
first term. The second one is negligible since, in view of the
Chebyshev inequality, it is exponentially small. On letting
$c\searrow 1,$ assertion (\ref{upbnd}) is proved.

If $X$ is positive and $\mathE e^{h Y}< \infty,$ for some
$h>0,$ the desired conclusion follows from the duality
principle of Theorem~\ref{th5}. \hfill $\Box$





\bigskip
\noin
{\bf Proof of Remark~\ref{r1}.} We have
\begin{eqnarray*}
P_T &=& \mathP \Big( V_T \leqslant x_T,
\max_{1\leqslant k\leqslant N(T)} |X_{k}| \leqslant 2 x_T,
\max_{1\leqslant k\leqslant N(T)} Y_k \leqslant 2 x_T
\Big)
\\
&=& \mathP \Big( V_T \leqslant x_T,
\max_{1\leqslant k\leqslant N(T)} |X_{k}| \leqslant 2 x_T,
\max_{1\leqslant k\leqslant N(T)} Y_k \leqslant 2 x_T,
N(T) \geqslant \frac{T}{2 x_T}\Big)
\\
&\leqslant & \mathP \Big( M_{N(T)}\leqslant x_T,
N(T) \geqslant \frac{T}{2 x_T}\Big)
\leqslant \mathP \Big( M_{\frac{T}{2 x_T}}\leqslant x_T
\Big).
\end{eqnarray*}
Hence the result follows from Theorem~A. \hfill $\Box$

\bigskip\noin
{\bf Proof of Theorem~\ref{th2}.} Take $c >1$ and
$0<\varepsilon<1.$ By (\ref{vt2}), we have
\begin{eqnarray}
\hspace*{-\parindent}
\nonumber
P_T &\geqslant &\mathP\big( M_{N(T)} \leqslant x_T, M'_{N(T)+1}
\leqslant x_T\big)
\\[1mm]
%\hspace*{-\parindent}
\nonumber
&\geqslant &\mathP\Big(
M_{N(T)} \leqslant x_T, M'_{N(T)+1} \leqslant x_T,
\max_{1\leqslant k\leqslant N(T)+1} Y_k \leqslant \varepsilon x_T
\Big)
\\
%\hspace*{-\parindent}
\nonumber
&\geqslant &\mathP\Big( M_{N(T)} \leqslant (1-\varepsilon) x_T,
\max_{1\leqslant k\leqslant N(T)+1} Y_k \leqslant \varepsilon x_T
\Big)
\\
%\hspace*{-\parindent}
\nonumber
&\geqslant &\mathP\Big(
M_{N(T)} \leqslant (1-\varepsilon) x_T,
\max_{1\leqslant k\leqslant N(T)+1} Y_k \leqslant \varepsilon x_T,
N(T) < c T \Big)
\end{eqnarray}
\begin{eqnarray}
%\hspace*{-\parindent}
 \nonumber
&\geqslant &\mathP\Big( M_{ c T} \leqslant (1-\varepsilon) x_T,
\max_{1\leqslant k\leqslant c T +1} Y_k \leqslant \varepsilon x_T,
N(T) < c T \Big)
\\
%\hspace*{-\parindent}
\nonumber
&\geqslant &\mathP\Big(M_{ c T} \leqslant (1-\varepsilon) x_T,
\max_{1\leqslant k\leqslant c T +1} Y_k \leqslant \varepsilon x_T \Big) -
\mathP\big(N(T) \geqslant [c T] \big)
\\
%\hspace*{-\parindent}
\nonumber
&=&
\mathP\Big(M_{ c T} \leqslant (1-\varepsilon) x_T,
\max_{1\leqslant k\leqslant c T} Y_k \leqslant \varepsilon x_T \Big)
\mathP\big(Y \leqslant \varepsilon x_T \big)
\\[-1mm]
&&
\hs{75mm}-~\mathP\big( R_{[cT]} \leqslant T \big)
\label{pt0}
\\[2mm]
%\hspace*{-\parindent}
\label{pt}
&\geqslant & P'_T \big(1+o(1)\big) - e^{-\delta T},
\end{eqnarray}
for some $\delta>0,$ where
$P'_T = \mathP\big(M_{ c T} \leqslant (1-\varepsilon) x_T,
\max_{1\leqslant k\leqslant c T} Y_k \leqslant \varepsilon x_T
\big).$
Here we also used the estimate
\begin{equation*}
  \mathP\left( R_{[cT]} \leqslant T \right) ~\leqslant~ e^{-\delta T}
\end{equation*}
which follows, for nonnegative $Y$ with $\mathE Y = 1 < c,$ from
Markov's inequality.

Assume that condition (i) holds. If $Y$ is bounded, then we
obtain from (\ref{pt0}),
\begin{eqnarray}\label{pt1}
  P_T &\geqslant& \mathP\big(M_{ c T} \leqslant (1-\varepsilon) x_T \big)
- e^{-\delta T},
\end{eqnarray}
for large $T,$ and the desired conclusion follows from Theorem A.
When $X$ is bounded, a similar reasoning yields (\ref{pt0}), if
$\max_{1\leqslant k\leqslant c T} Y_k$ is replaced by
$\max_{1\leqslant k\leqslant c T} |X_k|,$ and $Y$ by $X,$
respectively.
The remaining arguments are the same as before.

\medskip
Let now condition (ii) be satisfied.
Without loss of generality we assume that $\esup Y =\infty$.

We present the detailed argument for the case $\esup X =\infty$.
If $\esup X <\infty$, the proof follows the same pattern and is omitted.

We first assume that both marginal distributions (of $X$ and
$Y$) are continuous. Put
\begin{eqnarray*}
  &&\hs{-7mm}H_1(u)=\mathE X I\{-\varepsilon x_T \leqslant X \leqslant u,
  Y \leqslant \varepsilon x_T\},
\\[1mm]
 &&\hs{-7mm}H_2(u)=\mathE Y I\{-\varepsilon x_T \leqslant X \leqslant u,
  Y \leqslant \varepsilon x_T\},
\end{eqnarray*}
and suppose that
\begin{eqnarray}\label{h1h2}
&&\hs{-7mm} H_1(\varepsilon x_T) >  H_2(\varepsilon x_T).
\end{eqnarray}

By (\ref{m2}), we have $\mathE Y I\{X>u\} < \mathE X I\{X>u\},$
for $u \geqslant u_0,$ which, in view of $\mathE Y=\mathE X=1,$
gives
$\mathE Y I\{X \leqslant u\} > \mathE X I\{X \leqslant
u\},$ $u \geqslant u_0.$ Taking into account that
$H_1(u) \rightarrow \mathE X I\{X \leqslant u\}$ and
$H_2(u) \rightarrow \mathE Y I\{X \leqslant u\},$ as
$T\rightarrow\infty,$ we conclude that $H_1(u)<H_2(u),$ for all $u \geqslant u_0$ and
$T\geqslant T_u.$

\noin
Note that $H_1(u)$ and $H_2(u)$ are continuous. Indeed, e.g.,
\begin{equation*}
  H_2(u+)-H_2(u-) = \mathE Y I\{ X=u, X \geqslant -\,\varepsilon x_T,
  Y \leqslant \varepsilon x_T\} =0.
\end{equation*}
Assume $T$ to be such that $T \geqslant T_{u_0},$ and
(\ref{h1h2}). Then, there exist $\{u_T\}$ satisfying $u_T
\rightarrow\infty,$
$u_T \leqslant \varepsilon x_T$ and
$H_1(u_T)=H_2(u_T)$.
\\
We have (\ref{pt}) and
\begin{eqnarray}
\nonumber
P'_T &\geqslant& \mathP\big(M_{ c T} \leqslant (1-\varepsilon)
x_T, Y_k \leqslant \varepsilon x_T,
- \,\varepsilon x_T \leqslant X_k \leqslant u_T, 1\leqslant k\leqslant c T
 \big)
\\
\nonumber
&\geqslant& \mathP\Big(
\max_{1\leqslant k\leqslant c T} |\bar{S}_{k}-\bar{R}_{k}|
\leqslant (1-\varepsilon) x_T \Big)
\big(\mathP( Y \leqslant \varepsilon x_T,
- \,\varepsilon x_T \leqslant X \leqslant u_T)\big)^{cT}
\\
&=& \label{pt2}
P''_T \,Q_T^{c T},
\end{eqnarray}
where
$\bar{S}_{k}=\sum_{i=1} ^{k} \bar{X}_{i},$
$\bar{R}_{k}=\sum_{i=1} ^{k} \bar{Y}_{i},$
and $(\bar{X}, \bar{Y}),$ $\{ (\bar{X}_{i}, \bar{Y}_{i})\}$ are
i.i.d. random vectors having (conditional) distribution
\begin{eqnarray*}
\bar{F}(x,y)= \frac{1}{Q_T}\,
\mathP (X<x, - \,\varepsilon x_T \leqslant X \leqslant u_T,
Y<y, Y \leqslant \varepsilon x_T).
\end{eqnarray*}
By the definition of $H_1(u)$, $H_2(u)$ and $u_T,$ we obtain
\begin{eqnarray*}
&&\hspace*{-2\parindent}\mathE X I\{ X\geqslant u_T\}
\\[1mm]
&\leqslant &
\mathE X I\{ X\geqslant u_T \vee X< -\,\varepsilon x_T \vee Y>\varepsilon x_T\}
\\
&&\hspace*{6\parindent}
+ ~\mathE |X| I\{ X< -\,\varepsilon x_T \}+
\mathE |X| I\{Y>\varepsilon x_T\}
\\[1mm]
&=& E_0+E_1+E_2 ~=~ 1-H_1(u_T)+E_1+E_2~=~1-H_2(u_T)+E_1+E_2
\\[1mm]
&=& \mathE Y I\{ X\geqslant u_T \vee X< -\,\varepsilon x_T \vee Y>\varepsilon x_T\}
+E_1+E_2
\\[1mm]
&\leqslant &
\mathE Y I\{ X\geqslant u_T \}+\mathE Y I\{ X< -\,\varepsilon x_T \}+
\mathE Y I\{Y>\varepsilon x_T\}+E_1+E_2
\\[1mm]
&=& \mathE Y I\{ X\geqslant u_T \}+E'_1+E'_2+E_1+E_2.
\end{eqnarray*}
Since $\mathE X^3 <\infty$ and $\mathE Y^3 <\infty,$ we get
$E_1=o\big(x_T^{-2}\big)$ and $E'_2=o\big(x_T^{-2}\big)$ from
Chebyshev's inequality. By a combination of the Cauchy-Bunyakovski
and Chebyshev inequalities,
\begin{eqnarray*}
E_2 &\leqslant &\big(\mathE |X|^3 \big)^{1/3}
\big( \mathP (Y > \varepsilon x_T)\big)^{2/3}\\
&\leqslant &
\big(\mathE |X|^3 \big)^{1/3}
\Big( \frac{1}{\varepsilon^3 x_T^3}
\mathE Y^3 I\{Y > \varepsilon x_T\} \Big)^{2/3}
=~o\Big(\frac{1}{x_T^2} \Big).
\end{eqnarray*}
Similarly, we conclude that $E'_1=o\big(x_T^{-2}\big).$ Hence
\begin{equation*}
  \mathE X I\{ X\geqslant u_T\} ~\leqslant~
\mathE Y I\{ X\geqslant u_T \}+o\Big(\frac{1}{x_T^2} \Big).
\end{equation*}
By (\ref{m2}), $\mathE X I\{ X\geqslant u_T\} =o\big(u_T
x_T^{-2}\big),$ which yields $\mathP ( X\geqslant u_T)
=o\big(x_T^{-2}\big).$

\noin
It follows
\begin{eqnarray*}
Q_T &\geqslant& 1-\mathP(Y > \varepsilon x_T) - \mathP(X > u_T)-
\mathP(X<- \,\varepsilon x_T)=1+o\Big(\frac{1}{x_T^2} \Big),
\end{eqnarray*}
and therefore,
\begin{eqnarray}
\label{qt}
Q_T^{c T} &=& \exp\{o(T x_T^{-2})\}.
\end{eqnarray}
Note that $ \mathE ( \bar{X}-\bar{Y}) = (H_1(u_T)-
H_2(u_T))/Q_T=0$.
\\
By the Skorohod embedding,
\begin{eqnarray*}
\{ \bar{S}_{k}-\bar{R}_{k}, 1\leqslant k\leqslant c T \}
\stackrel{\mathcal{D}}{=}
%\frac{\underline{\mathcal{D}}}{ }
\{ W(\tau_k), 1\leqslant k\leqslant c T \},
\end{eqnarray*}
where $\{W(t)\}$ is a standard Wiener process,
$\tau_k=\nu_1+\cdots+\nu_k,$ and $\nu$, $\{\nu_j\}$ are i.i.d.
random variables such that $ \mathE \nu =\mathE \bar{Z}^2$,
$ \mathE \nu^k \leqslant 4 k! \,\mathE \bar{Z}^{2k}$,
$k=2,3,\ldots$, with $\bar{Z}=\bar{X}-\bar{Y}$.
Hence
\begin{eqnarray}
&&\hs{-15mm}P_T''~=~\mathP\Big(
\max_{1\leqslant k\leqslant c T} |W(\tau_k)| \leqslant (1-\varepsilon) x_T
\Big)
\nonumber
\\
%\hspace*{-2\parindent}
\nonumber
&&\hs{-7mm}\geqslant ~\mathP\Big(
\sup_{0\leqslant t\leqslant \mathE \tau_{[c T]} + \beta c T} |W(t)|
\leqslant (1-\varepsilon) x_T \Big)-
\mathP\big( \tau_{[c T]}>\mathE \tau_{[c T]} + \beta c T\big)
\\
%\hspace*{-2\parindent}
\label{pt3}
&&\hs{-7mm}\geqslant ~\mathP\Big(
\sup_{0\leqslant t\leqslant c T (v^2 + 2 \beta)} |W(t)|
\leqslant (1-\varepsilon) x_T
\Big)- \mathP\big( \tau_{[c T]}>\mathE \tau_{[c T]} + \beta c T\big).
\end{eqnarray}
Here we used the estimate $\mathE \tau_{[c T]} = [c T]
\,\mathbf{Var}\bar{Z} \leqslant c T (v^2 +\beta),$ for all large
$T.$
\\
It is well-known that
\begin{eqnarray}\label{wt}
&&\hs{-10mm}\mathP\Big(
\sup_{0\leqslant t\leqslant c T (v^2 + 2 \beta)} |W(t)|
\leqslant (1-\varepsilon) x_T
\Big) = e^{-\frac{\pi^2}{8} \frac{c T (v^2 + 2 \beta)}{(1-\varepsilon)^2 x_T^2}
 \left(1+o(1)\right)}.
\end{eqnarray}
Since $ \mathE |\nu-\mathE\nu|^k
\leqslant 8 k! \,\mathE \bar{Z}^{2k}
\leqslant 8 k! (2 \varepsilon x_T)^{2k-4}\mathE \bar{Z}^4$,
one has, for $h(2 \varepsilon x_T)^2=\frac 12,$
\begin{eqnarray*}
\mathE e^{h (\nu-\mathE\nu)} &=& 1+\sum_{k=2}^{\infty}
\frac{h^k \mathE (\nu-\mathE\nu)^k}{k!}
%\\ &&
~\leqslant~ 1+ 8 h^2 \mathE \bar{Z}^4
\sum_{k=2}^{\infty} \big(h (2 \varepsilon x_T)^2\big)^{k-2}
\\
&=& 1+16 h^2 \mathE \bar{Z}^4 \leqslant
e^{16 h^2 \mathE \bar{Z}^4 }.
\end{eqnarray*}
Hence
\begin{eqnarray*}
&&
\hs{-10mm}
\mathP\big( \tau_{[c T]}>\mathE \tau_{[c T]} + \beta c T\big)
~\leqslant~ e^{(16 h^2 \mathE \bar{Z}^4 -h \beta) c T }
%\\ &&
~=~e^{ - \frac{c T \beta}{2 (2 \varepsilon x_T)^2}
\left(1+o(1)\right)},
\end{eqnarray*}
where we took into account the definition of $h,$
$\mathE \bar{Z}^4 \leqslant 2 \varepsilon x_T
\mathE|\bar{Z}|^3$ and
$\mathE|\bar{Z}|^3\rightarrow\mathE|X-Y|^3,$ as
$T\to\infty.$
It follows that
\begin{eqnarray}\label{tct}
&&\hs{-\parindent}\mathP\big( \tau_{[c T]}>\mathE \tau_{[c T]} + \beta c T
\big)~=~
o\Big(e^{-\frac{T}{x_T^2}}\Big),
\end{eqnarray}
for $\beta>8 \varepsilon^2/c$.

A combination of (\ref{pt}) and (\ref{pt2})---(\ref{tct}) results
in
%(\ref{qt}), (\ref{pt2}), (\ref{pt3}),
%(\ref{wt}) and (\ref{tct}),
\begin{eqnarray*}
P_T &\geqslant& P''_T \,Q_T^{c T} (1+o(1)) - e^{-\delta T}
%\\ &&
~\geqslant~ e^{-\frac{\pi^2}{8} \frac{c T (v^2 + 2
\beta)}{(1-\varepsilon)^2 x_T^2}
 \left(1+o(1)\right)}
\end{eqnarray*}
for all large $T$ such that (\ref{h1h2}) holds.

If $T$ satisfies the opposite inequality to~(\ref{h1h2}), we
obtain the latter bound in a similar way using (\ref{m1})
instead of (\ref{m2}).

The conclusion of Theorem~\ref{th2} now follows from letting
$\varepsilon\searrow 0,$ $c\searrow 1,$ and then $\beta\searrow 0.$

Next, we discuss the case when the distributions, either of $X$
or $Y$ or both, have atoms.

Set $X'_k=X_k+\eta_k$, $Y'_k=Y_k+\eta_k,$ where $\eta$ and $\{
\eta_k \}$ are \,i.i.d. random variables, uniformly
distributed on $[-\kappa,0],$ and independent of
$\{ (X_k,Y_k)\}.$ Then,
\begin{eqnarray}\label{pt4}
P'_T &\geqslant& \mathP\Big( \max_{1\leqslant k\leqslant c T }
|S'_k-R'_k|
 \leqslant (1-\varepsilon) x_T,
\max_{1\leqslant k\leqslant c T } Y'_k \leqslant
\frac{\varepsilon}{2} x_T \Big),
\end{eqnarray}
for large $T$. Here $S'_k-R'_k=(X'_1-Y'_1)+\cdots+(X'_k-Y'_k)=S_k-R_k$.

Note that the distributions of $X'$ and $Y'$ are continuous.

We want to apply the above argument for estimating the right-hand side
of (\ref{pt4}). To this end, we have to check that conditions (\ref{m1}) and
(\ref{m2}) hold with $(X,Y)$ being replaced by $(X',Y')$.
Since $\eta$ is negative, we have
\begin{eqnarray*}
&&
\hs{-10mm}\mathE Y' I\{X'>u \}~=~ \mathE Y I\{X>u-\eta \}
+\mathE \eta I\{X'>u \} \\ %&&
&=&\mathE \mathE_{\eta} Y I\{X>u-\eta \}+\mathE \eta I\{X'>u \}
\\
&\leqslant& \mathE \Big(1-\frac{A}{u-\eta}\Big) \mathE_{\eta} X I\{X>u-\eta \}+\mathE \eta I\{X'>u \}
\\ %&&
&\leqslant&  \Big(1-\frac{A'}{u}\Big) \mathE X I\{X'>u \}+\mathE \eta I\{X'>u \}
%\\
~\leqslant~ \Big(1-\frac{A'}{u}\Big) \mathE X' I\{X'>u \}.
\end{eqnarray*}
This gives (\ref{m2}) for $(X',Y')$.
Assumption (\ref{m1}) for $(X',Y')$ can be obtained in the same way.

So, an application of the argument for the continuous case, and
then letting $\kappa\searrow 0,$ completes the proof.
\hfill $\Box$

\bigskip
\noin
{\bf Proof of Theorem \ref{th3}.} Take $c>1.$ Assume that (i)
holds. As in the proof of Theorem \ref{th2}, we get (\ref{pt}) and
\begin{eqnarray*}
&&\hspace*{-\parindent} P_T ~\geqslant
\\
&&\hspace*{-\parindent} \mathP \Big(
  \max_{1\leqslant k\leqslant c T} |S_{k}-k| \leqslant (1-\varepsilon)^2 x_T,
\max_{1\leqslant k\leqslant c T} |R_k-k|
\leqslant (1-\varepsilon) \varepsilon x_T,
N(T)< c T \Big).
\end{eqnarray*}
The inequality $\max_{1\leqslant k\leqslant c T} |R_k-k|
\leqslant (1-\varepsilon) \varepsilon x_T$ implies that
$R_{cT} > c T +o(T),$ which, in turn, yields $N(T)< c T$ for large $T.$
Hence,
\begin{eqnarray*}
&&\hspace*{-\parindent}  P_T ~\geqslant~ \mathP \Big(
  \max_{1\leqslant k\leqslant c T} |S_{k}-k| \leqslant (1-\varepsilon)^2 x_T
\Big) \mathP \Big(
\max_{1\leqslant k\leqslant c T} |R_k-k|
\leqslant (1-\varepsilon) \varepsilon x_T\Big).
\end{eqnarray*}
The result now follows from Mogul'skii (1974).

If (ii) holds, the proof is similar to the first part. We only
have to exchange
$X$ and $Y$ formally. Details are omitted. \hfill $\Box$


\bigskip
\noin
{\bf Proof of Theorem~\ref{th5}.} We shall make repeated use of
the fact that, for any $k=0,1,\ldots$ and $t>0,$
\begin{eqnarray*}
&&\hspace*{-\parindent} N(t)\geqslant k~~\Longleftrightarrow ~~R_k\leqslant t \quad
\wedge \quad N(t)< k~~\Longleftrightarrow ~~R_k >t,
\\[1mm]
&& \hspace*{-\parindent}M(t)\geqslant k~~\Longleftrightarrow ~~S_k\leqslant t \quad
\wedge \quad M(t)< k~~\Longleftrightarrow ~~S_k >t,
\end{eqnarray*}
Note that
\begin{eqnarray*}
&&\hspace*{-\parindent} t-\frac{\lambda}{\mu}\big(\mltm\big)^+ ~\leqslant~
\frac{\lambda}{\mu} \,x_T.
\end{eqnarray*}
We have
\begin{eqnarray*}
&&\hs{-12mm}
\sup_{0\leqslant t\leqslant T} \big|S(t)-\frac{\mu}{\lambda}\,t\big| < x_T
\\
&\Longrightarrow& ~ S_{N(t)} < \mltp ~~\wedge~~ S_{N(t)} > \big(\mltm\big)^+
\quad\forall~t\in[0,T]
\\
&\Longrightarrow& ~ M\big(\mltp\big) \geqslant N(t) ~~\wedge~~  M\big((\mltm)^+\big) < N(t)
\quad\forall~t\in[0,T]
\\
&\Longrightarrow& ~ M\big(\mltp\big)+1 > N(t) ~~\wedge~~
\\
&& \hspace*{5\parindent}
M\big((\mltm)^+\big)+1 \leqslant N(t)
\quad\forall~t\in[0,T]
\\
&\Longrightarrow& ~ R_{M\left(\mltp\right)+1} > t ~~\wedge~~  R_{M\left((\mltm)^+\right)+1} \leqslant t
\quad\forall~t\in[0,T]
\\
&\Longrightarrow& ~ R_{M\left(\mltp\right)+1} - \,\frac{\lambda}{\mu}\big(\mltp\big)
> -\,\frac{\lambda}{\mu}\,x_T
 ~~\wedge~~
\\
&& \hspace*{3.5\parindent}
 R_{M\left((\mltm)^+\right)+1} - \frac{\lambda}{\mu}\big(\mltm\big)^+
 \leqslant \frac{\lambda}{\mu}\,x_T
\quad\forall~t\in[0,T].
\end{eqnarray*}
The last inequalities combined with
\begin{eqnarray*}
&&\hs{-\parindent} R_{M(u)+1} ~\geqslant~
-\,\frac{\lambda}{\mu}\,x_T,
\end{eqnarray*}
for $u \in [0,x_T],$ imply that
\begin{eqnarray*}
&&\hs{-\parindent}\sup_{0\leqslant t\leqslant \frac{\mu}{\lambda}\,T - x_T}
\big|R(t)-\frac{\lambda}{\mu}\,t \big|
~\leqslant~ \frac{\lambda}{\mu}\,x_T,
\end{eqnarray*}
which proves (\ref{slr}).
\\[1mm]
For the final step of the proof, note that $t-(t-x_T)^+\leqslant
x_T.$ Then,
\begin{eqnarray*}
&&\hspace*{-2.5\parindent}
\sup_{0\leqslant t\leqslant \frac{\mu}{\lambda} \,T + x_T}
\big|R(t)-\frac{\lambda}{\mu}\,t \big|
< \frac{\lambda}{\mu}\,x_T
\\
&\Longrightarrow& ~
R_{M(t)+1} < \frac{\lambda}{\mu} \big(t+x_T\big) ~~\wedge~~
\\
&&\hspace*{4\parindent}
R_{M(t)+1} >
\frac{\lambda}{\mu} \big(t-x_T\big)^+
\quad\forall~t\in\big[0,\frac{\mu}{\lambda}\,T + x_T\big]
\\
&\Longrightarrow& ~
N\big(\frac{\lambda}{\mu} (x_T+t)\big)\geqslant M(t)+1 ~~\wedge~~
\\
&&\hspace*{4\parindent}
N\big(\frac{\lambda}{\mu} (t-x_T)^+\big) < M(t)+1
\quad\forall~t\in\big[0,\frac{\mu}{\lambda}\,T+ x_T\big]
\\
&\Longrightarrow& ~
N\big(\frac{\lambda}{\mu} (x_T+t)\big)\geqslant M(t)+1 ~~\wedge~~
\\
&&\hspace*{4\parindent}
N\big(\frac{\lambda}{\mu} (t-x_T)^+\big) \leqslant M(t)
\quad\forall~t\in\big[0,\frac{\mu}{\lambda}\,T + x_T\big]
\\
&\Longrightarrow& ~
S_{N\left(\frac{\lambda}{\mu} (x_T+t)\right)}> t ~~\wedge~~
S_{N\left(\frac{\lambda}{\mu} (t-x_T)^+\right)} \leqslant t
\quad\forall~t\in\big[0,\frac{\mu}{\lambda}\,T + x_T\big]
\\
&\Longrightarrow& ~
S_{N\left(\frac{\lambda}{\mu} (x_T+t)\right)}-(x_T+t)> - \,x_T
~~\wedge~~
\\
&& \hspace*{4\parindent}
S_{N\left(\frac{\lambda}{\mu} (t-x_T)^+\right)}- (t-x_T)^+
\leqslant x_T
\quad\forall~t\in\big[0,\frac{\mu}{\lambda}\,T + x_T\big].
\end{eqnarray*}
The last inequalities together with
\begin{eqnarray*}
&&\hs{-10mm}S_{N(v)} -\frac{\mu}{\lambda}\, v ~\geqslant~ - \,x_T
\end{eqnarray*}
for $v \in [0,\frac{\lambda}{\mu}\,x_T]$, yield (\ref{rls}).
\hfill $\Box$

\bigskip
\bigskip
\newpage

\hs{-4mm}
{\bf References}
\footnotesize

%\begin{thebibliography}{12}

\parindent0mm
{\begin{itemize}
\item[{[1]}]
A.A. Borovkov, A.A. Mogulskii, On probabilities of small
deviations for stochastic processes, Trudy Inst. Math. SOAN SSSR,
13\,(1989), 147--168. (In Russian)
\item[{[2]}]
A.A. Borovkov, A.A.
Mogulskii, On probabilities of small deviations for stochastic
processes, Sib. Adv. Math., 1\,(1991), 39--63.
\item[{[3]}]
P. Embrechts, C. Kl\"uppelberg, Some aspects of
insurance mathematics, Theory Probab. Appl., 38\,(1993),
374--416. (In Russian)
\item[{[4]}]
M. Ledoux, Isoperimetry and Gaussian analysis,
Lectures on Probability Theory and Statistics, Lecture Notes in
Mathematics, Vol. 1648, Springer, Berlin, 1996, pp. 165--294.
\item[{[5]}]
W.V. Li, Q.-M. Shao, Gaussian processes: inequalities, small ball
probabilities and applications, in: C.R.Rao, D. Shanbhag (Eds.),
Stochastic Processes: Theory and Methods, Handbook of Statistics,
Vol. 19, North-Holland, Amsterdam, 2001, pp. 533--597.
\item[{[6]}]
M.A. Lifshits, Asymptotic behavior of small ball probabilities,
in: B. Grigelionis (Ed.), Proceedings of the Seventh Vilnius
Conference on Probability Theory and Mathematical Statistics,
VSP/TEV. Vilnius, 1999, pp. 453--468.
\item[{[7]}]
A.A. Mogul'skii, Small deviations in a space of trajectories,
Theory Probab. Appl., 19\,(1974), 726--736.

\end{itemize}
}

\end{document}
