\magnification=1200
\input amstex
\hsize=16truecm
\parskip=3pt plus 1pt
\TagsOnRight
\define\e{\varepsilon}
 
\centerline{\bf ALMOST SURE LIMIT THEOREMS AND THE}
\centerline{\bf INVARIANCE PRINCIPLE}
\centerline{P\'eter Major}
\bigskip\noindent
{\bf Almost sure limit theorems}\medskip\noindent
The following ``almost sure central limit theorem" is a popular
subject in recent research. It can be formulated in the following way:
Let $X_1$, $X_2$,\dots be iid.\ random variables, $EX_1=0$, $EX_1^2=1$,
$S_n=\sum\limits_{k=1}^n X_k$. Then
$$
\lim_{n\to\infty}\frac1{\log n}\sum_{n=1}^nI\left(\frac{S_n}{\sqrt
n}<x\right)\to \Phi(x)
$$
for all numbers $x$, where $I(\cdot)$ denotes the indicator function,
and $\Phi(x)$ is the standard normal distribution function.
 
In this talk we discuss a generalization (also known in the literature)
of this result and try to explain why the result is natural. We
formulate and prove a functional limit theorem version of this result.
We prove it by first proving it for the Wiener process, and show that
this special case together wih the invariance principle implies this
result for sums of independent random variables. In the case of Wiener
process we rewrite the statement in an equivalent form to the
Ornstein--Uhlenbeck process, and this version of the result follows
from the ergodic theorem. We shall also show that great part of these
statements can be proved for stable processes. We discuss these cases
too. Here some open problems appear about the approximation of sums of
independent random variables in the domain of attraction of a stable
law by a stable process. We discuss this problem too. In particular, we
discuss a probabilistic interpretation of the L\'evy--Hin\v cin formula.
We start our discussion by the functional limit theorem version of the
``almost sure invariance principle" for the Wiener process. But we
shall formulate it in a general form valid for all stable processes.
 
Let $X(t)$, $t>0$, be a stable process with parameter $\alpha$,
$0<\alpha\le2$. This means that this process has the following property:
$$
X(u)\overset \Delta \to =\frac {X(Tu)}{T^{1/\alpha}},\quad u>0, \tag$*$
$$
for all $T>0$, where $\overset\Delta\to =$ means that the processes at
at the two sides have the same distribution. We return to this
formula later. Define the processes $X_t=X_t(u)$, $t>0$, $0\le u\le1$,
$$
X_t(u)=\frac{X(ut)}{t^{1/\alpha}}.
$$
\medskip\noindent{\bf Claim.}
{\it For all bounded and continuous functionals $\Cal F$ over the space
$\text{\rm D}([0,1])$, ($\text{\rm C}([0,1])$ for $\alpha=2$ when the
Wiener process is considered)
$$
\frac1{\log T}\int_1^T \frac1t \Cal F(X_t)\,dt\to E\Cal F(X_1)\quad
\text{if }T\to\infty.
$$
In general, for all $\omega$ define the random probability measures
$\mu_{T,\omega}$ in the following way: Consider the probability space
$([1,T], \text{\rm Borel } \sigma\text{\rm-algebra}, \rho_T)$,
$\rho_T(dt)=\dfrac{dt}{t\log
T}$, the $\text{\rm D}([0,1])$ valued random variable
$U_t=U_{t,\omega}=X_t(\cdots,\omega)$, $1\le t\le T$, and let
$\mu_{T,\omega}$ be the distribution of
the random variable $U_{t,\omega}$. Then $\mu_{T,\omega}$ converges
in the weak topology to the distribution of the stable process $X_t$,
$0\le t\le 1$.}
\medskip\noindent {\it Proof:}\/
Introduce the ``generalized Ornstein--Uhlenbeck" process
$$
Z(t)=\frac{X(e^t)}{e^{t/\alpha}},\quad -\infty<t<\infty,
$$
and observe that it is a stationary process. Indeed, for a fixed number
$T>0$
$$
Z(t+T)=\frac{X(e^{t+T})}{e^{(t+T)/\alpha}}\overset \Delta\to=
\frac{X(e^{t})}{e^{(t+T)/\alpha}e^{-T/\alpha}}=
\frac{X(e^{t})}{e^{t/\alpha}}=Z(t).
$$
Moreover, the distribution of the process $Z(t)$ is ergodic with the
usual shift. This follows for instance from the fact that the
zero--one law for sums of iid.\ random variables implies that the
tail-$\sigma$ algebra of the process $Z(t)$ is trivial. For fixed
$s$ define the process $Z_s$,
$$
Z_s(v)=Z(s+v),\quad -\infty<v<\infty,
$$
and given a process $Z=Z(v)$, $-\infty<v<\infty$, put
$$
Z^0=Z^0(v)=\cases Z(v) &\text{if }v\le 0\\
0 &\text{if }v> 0
\endcases.
$$
We rewrite the statement of the claim with the process $Z$ instead of
the process $X$. To do this introduce the functional $\Cal G$, ($\Cal
G=\Cal G(\Cal F)$) over the space of functions on the real line as
$\Cal G(z)=\Cal F(x_z)$ with the function $x_z=x_z(u)$, $0\le u\le1$,
$x_z(u)=u^{1/\alpha}z^0(\log u)$. Then we have with the change of
variables $s=\log t$
$$
\frac1{\log T}\int_1^T \frac1t \Cal F(X_t)\,dt=
\frac1{\log T}\int_0^{\log T}\Cal F(X_{e^s})\,ds=
\frac1{\log T}\int_0^{\log T}\Cal G(Z_s)\,ds
$$
since $\Cal G(Z_s)=\Cal F(X_{e^s})$. Indeed, for $0\le u\le1$
$$
x_{Z_s}(u)=u^{1/\alpha}Z^0_s(\log u)=u^{1/\alpha}Z(s+\log
u)=u^{1/\alpha} \frac{X(e^{s+\log u})}{e^{(s+\log
u)/\alpha}}=\frac{X(ue^s)}{e^{s/\alpha}},\quad 0\le u\le1,
$$
hence $x_{Z_s}=X_{e^s}$.
 
Finally, the ergodic theorem implies that the last expression has a
limit which equals $E\Cal G(Z_0)=E\Cal F(X_1)$.
 
In the proof of the second part of the claim we discuss only the case
$\alpha=2$, i.e.\ the case when the stable process is a Wiener process.
Moreover, we give only a sketch of the proof. The problem is the
following: The exceptional set of the bad $\omega$-s, where the first
part of the statement does not hold  has zero probability, but it may
depend on the choice of the functional $\Cal F$. We have to show that
it can be chosen independently of this functional. Some compactness type
argument is needed which states that there is a countable ``dense"
subset of the bounded continuous functionals $\Cal F$ by which all such
functionals can be well approximated, so if the first part of the claim
holds for the $\Cal F$-s in this ``dense subset" of the functionals,
then it holds for all of them. The problem with these approach is that
the space $C([0,1])$ (or $D([0,1])$ in the general case) is very large.
But this argument can be rescued by showing that for almost all
$\omega$-s the measures $\mu_{T,\omega}$ have the following property:
For all $\e>0$ there is a compact subset $K=K(\e,\omega)$ of $C([0,1])$
such that $\mu_{T,\omega}(K)>1-\e$. (Then the continuous functionals
over $K$ are uniformly continuous, and the compactness argument can be
worked out even if it is not quite trivial.)
 
It can be proved that for almost all $\omega$ and all $\eta>0$ there is
some $L=L(\eta,\omega)$ and $\delta=\delta(\eta,\omega)$ such that
$$
\mu_{T,\omega}\left(\sup_{|u-v|<\delta}|x(u)-x(v)|<\eta,\;\sup_{u\in K}
|x(u)|<L\right)>1-\eta
$$
for all $T>1$. Then the existence of a compact set $K$ in $C([0,1])$
with the required properties follows from the last relation. This can be
proved by applying the first part of the claim with the choice of $\Cal
F(x)=\sup\limits_{u\in [0,1]} |x(u)|$ and $\Cal F(x)=\Cal
F_{\delta}(x)=\sup\limits_{|u-v|<\delta}|x(u)-x(v)|$ with sufficiently
small $\delta$. Then the relation $\lim\limits_{T\to\infty}\Cal
F(\mu_{T,\omega})=\Cal F(\mu_w)$, where $\mu_w$ denotes the Wiener
measure in $C([0,1])$, implies that $\int \Cal F(x)\mu_{T,\omega}
(\,dx)<L'$ for the supremum functional and $\int\Cal F_\delta(x)
\mu_{T,\omega}(\,dx)<\eta^2/4$ with an appropriate $\delta>0$ if
$T>T(\omega)$. This implies the last relation for $T>T(\omega)$, and
this is the hard part of the statement. Then some extra work shows that
by a possible choice of the parameters  $L$ and $\delta$ the relation
holds for all $T>1$.
 
Let $X_1$, $X_2$, \dots be a sequence of iid. random variables,
$EX_1=0$, $EX_1^2=1$, put $S_n=\sum\limits_{k=1}^n X_k$, and define the
random poligon $S(t)$ in the following way: $S(t)=S_n$ if $t=n$ is
integer, and $S(t)=(t-n)S_n+(n+1-t)S_{n+1}$ for $n<t<n+1$. For all $t>1$
define the random continuous function $S_t=S_t(u)$ in the interval
$0\le u\le1$ by the formula
$$
S_t(u)=\frac{1}{\sqrt t}S(tu).
$$
Let us denote the stable process $X(t)$ with parameter $\alpha=2$, which
is actually the Wiener process, by $W(t)$ and the processes $X_t$
defined through $X(t)$ by $W_t$. It is proved that if
$EX_1^{2+\e}<\infty$ with some $\e>0$, then
$$
\sup_{0\le u\le 1}(S_t(u)-W_t(u))=o(t^{-\e/2(2+\e)})\quad\text{with
probability 1 if }t\to\infty.
$$
(This statement is equivalent to the relation $S_n-W_n=o(n^{1/(2+\e)})$
with probability 1.) Hence if $\Cal F$ is a bounded continuous
functional on $C([0,1])$, then $\lim\limits_{t\to\infty}(\Cal
F(S_t)-\Cal F(W_t))\to 0$, as $t\to\infty$. This statement clearly
implies that
$$
\lim_{T\to\infty}\left[
\frac1{\log T}\int_1^T\frac 1t\Cal F(S_t)\,dt-
\frac1{\log T}\int_1^T\frac 1t\Cal F(W_t)\,dt\right]\to 0\quad\text{as }
T\to\infty
$$
with probability 1, and the ``almost sure invariance
principle" for the Wiener process implies this law for the appropriate
process defined through the iid.\ random variables~$X_i$.
 
If we assume no more about $X_1$ than the existence of two moments, then
the above argument must be refined. In this case an approximation
$$
S_n-W(A_n)=o(n^{1/2})\quad\text{with probability 1, as }n\to \infty,
$$
is possible, where $A_n=\sum\limits_{k=1}^nD^2(X_kI(|X_k|<\sqrt n))$,
i.e.\ the
approximating Wiener process is scaled by the sum of the variance of the
appropriately truncated random variables $X_k$.  Here the scaling $A_n$
cannot be replaced by $n$ without violating the strength approximation.
The sequence $S_n$ is as well approximable by $W(n)$ as the sequence
$W(A_n)$. Define the function $A(t)$, $t>0$, so that it equals $A_n$ if
$t=n$, i.e.\ $t$ is an integer and it is the linear approximation
between $A_n$ and $A_{n+1}$ if $n<t<n+1$. The relations $A(t)/t\to1$
and $\dfrac d{dt} A(t)\to1$ hold as $t\to\infty$. But these relations
cannot be strenghtened without additional conditions.
 
We shall explain how the ``almost sure invariance principle" can be
proved in this case. We do not work out all details.
 
The function  $S_t$ can be replaced by $W_{A(t)}$, and it is enough to
prove the following statement:
$$
\lim_{T\to\infty}\left[
\frac1{\log T}\int_1^T\frac 1t\Cal F(W_{A(t)})\,dt-
\frac1{\log T}\int_1^T\frac 1t\Cal F(W_t)\,dt\right]\to 0\quad\text{as }
T\to\infty.
$$
It is not difficult to prove this statement with the help of change of
variables $s=A(t)$, because the properties of $A(t)$ imply that
$\frac{ds}s=\frac{dt}t(1+o(1))$, $s(T)=T(1+o(1))$, and the functional
$\Cal F$ is bounded. We also sketch another proof.
 
Given a fixed $\e>0$ define the set $A=A_\e$ of ``not well approximable
points" $t$ at level $\e$
$$
A=A_\e(W(\omega)=\{t\: t>1, \sup_{u\in [0,1]}|W_t(u)-W_{A(t)}(u)|>\e\}.
$$
it is enough to prove that this set $A$ is thin. More explicitly, it is
enough to show that
$$
\lim_{T\to\infty}\int_{A\cap[1,T]}\frac1t\,dt=0.
$$
To prove this statement it is enough to show the following: Fix some
$\eta>0$, the intervals $B_k=B_k(\eta)=[(1+\eta)^k,(1+\eta)^{k+1}]$.
Then the relation
$$
\lim_{n\to\infty}\frac1n\#\{k\: B_k\cap A\neq\emptyset\}=0.
$$
Since $\lim\limits_{t\to\infty}\dfrac{t-A(t)}t=0$, for arbitrary large
$L>0$
there is a threshold $k_0=k_0(L)$ such that a number $k>k_0$ belongs
to the bad set, i.e.\ $B_k\cap A\neq\emptyset$ if
$$
\sup\Sb 0\le u<v\le
(1+\eta)^{k+1}\\|u-v|<\frac{(1+\eta)^{k+1}}L\endSb
|W(u)-W(v)|>C\e(1+\eta)^{(k+1)/2}<\e
$$
 
Hence it is enough to bound the random variable
$$
S_n=\frac1n\sum_{k=1}^nI\left(\sup\Sb 0\le u<v<(1+\eta)^k\\
|u-v|<\frac1L(1+\eta)^k\endSb |W(v)-W(u)|>\e(1+\eta)^{k/2}\right).
$$
and to show that
$$
\lim_{n\to\infty}S_n\le \text{const.}\, \e
$$
if $\e\gg L^{-1}$. To bound this expression observe that
$$
\align
&I\left(\sup\Sb 0\le u<v<(1+\eta)^k\\
|u-v|<\frac1L(1+\eta)^k\endSb |W(v)-W(u)|>\e(1+\eta)^{k/2}\right)\\
&\qquad<\sum_{j=1}^kI\left(\sup\Sb (1+\eta)^{j-1}\le u<v<(1+\eta)^j\\
|u-v|<\frac1L(1+\eta)^k\endSb |W(v)-W(u)|>\e(1+\eta)^{k/2}\right)\\
&\qquad\qquad+I\left(\sup\Sb 0\le u<v<1\\
|u-v|<\frac1L(1+\eta)^k\endSb |W(v)-W(u)|>\e(1+\eta)^{k/2}\right),
\endalign
$$
hence
$$
S_n\le\zeta(n)=\frac1n\left(\zeta_0+\sum_{j=1}^n\zeta_j\right)
$$
with
$$
\zeta_j=\sum_{k=j}^nI\left(\sup\Sb (1+\eta)^{j-1}\le u<v<(1+\eta)^j\\
|u-v|<\frac1L(1+\eta)^k\endSb |W(v)-W(u)|>\e(1+\eta)^{k/2}\right)
$$
for $k=1,\dots,n$, and
$$
\zeta_0=\sum_{k=1}^n I\left(\sup\Sb 0\le u<v<1\\
|u-v|<\frac1L(1+\eta)^k\endSb |W(v)-W(u)|>\e(1+\eta)^{k/2}\right).
$$
Some calculation shows that
$$
\align
&I\left(\sup\Sb (1+\eta)^{j-1}\le u<v<(1+\eta)^j\\
|u-v|<\frac1L(1+\eta)^k\endSb |W(v)-W(u)|>\e(1+\eta)^{k/2}\right)\\
&\qquad<\min\left(\exp\left\{-\frac{\e^2(1+\eta)^{k-j}}{5}\right\},
\exp\left\{-\frac{L\e^2}{5}\right\}\right)\\
&\qquad<\exp\left\{-\frac{\e^2\sqrt
L(1+\eta)^{(k-j)/2}}5\right\} <e^{-(1+\eta)^{(k-j)/2}/\e^2}
\endalign
$$
if $L>\frac5{\e^4}$. The minimum appeared in this estimate, because we
separated the cases when $(1+\eta)^{k-j}>L$ and $(1+\eta)^{k-j}<L$.
These estimates enable us to estimate the moments of the variables
$\zeta_j$. They are sums of variables taking values 0 or 1. It can be
proved that because of the convexity of the functions $x^p$ the maximum
is taken when the indicator functions of these summands are embedded
into each other. It is not difficult to prove that
$$
E\zeta_j ^p<\text{const.}\,\e\quad\text{for all }p=1,2,\dots,
$$
where the const.\ depends only on $p$ but not on $j$. Because of the
independence of the variables $\zeta_j$ these relations imply that
$$
E\zeta(n)<\e\text{const.}\,n^{-1},\quad\text{and
}E(\zeta(n)-E\zeta(n))^{2p}<\text{const.}\,n^{-p}\text{ for }p\ge1.
$$
Hence $\lim\limits_{n\to0}(\zeta(n)-E\zeta(n))=0$, and
$$
\limsup_{n\to 0}S_n\le \limsup_{n\to0}B_n\le \text{const.}\,\e
$$
By working out the details one can prove the ``almost sure invariance
principle" in this case, too.
 
Our goal with this example was to show that the result can be proved in
the general case through an invariance principle, but one has to find
the right formulation of it which is adapted to the problem. So we turn
to the second part of this talk.
 
How can we prove, using the approach of invariance principles, to prove
the corresponding result for partial sums of iid.\ random variables in
the domain of attraction of a stable law? In particular, how  can one
get a good approximation of this partial sums by a stable process? To
answer the last question first we discuss the L\'evy--Hin\v cin formula,
in particular its probabilistic content.
\bigskip\noindent
{\bf On the L\'evy--Hin\v cin formula:}\medskip\noindent
In the investigation of possible limit theorems the solution of the
following problem plays an important role: Describe class of distribution
functions $F(x)$ whith the following property: For all positive
integers $n$ there are iid.\ random variables $X_1, X_2,\dots,X_n$
such that the distribution of $X_1+\dots+X_n-B_n$ with an appropriate
constant $B_n$ has distribution function $F(x)$. These distribution
functions are called infinitely divisible, and they are described by
the L\'evy--Hin\v cin formula. Let us explain its content.
 
It is clear that the normal distribution function is infinitely
divisible. The Poisson distribution also has this property, since the
sum of two independent Poisson distributed random variables with
parameters $\mu$ and $\lambda$ is again Poissonian distributed with
parameter $\mu+\lambda$. it is also clear that if $X_1$, $X_2$,\dots,
$X_k$ are independent infinitely distributed random variables, then the
linear combination $c_1X_1+\cdots+c_kX_k-B_k$ is again infinitely
divisible. The limit of infinitely divisible distribution is again
infinitely divisible. The content of the L\'evy--Hin\v cin formula is that
taking the linear combination of independent Poisson distributed random
variables plus an independent normally distributed random variable, and
then taking the closure of the distributions obtained in this way we get
all infinitely distributions. We shall give a description of this
result. First we restrict our attention to infinitely divisible
distributions without a normal component. We give a probabilistic
description. Given an infinitely distributed distibution function
$F(x)$ we construct a process $X(t)$, $t>0$, with independent
stationary increment  in such a way that $X(0)=0$, and $X(1)$ has
distribution $F(x)$. (Then the sum $X(1)=\sum\limits_{i=1}^n[X(\frac
in)-X(\frac{i-1}n)]$ gives the decomposition of the distribution in the
requested way.)
 
Let us first consider the special case when $F(x)$ is the distribution
of the linear combination $\sum\limits_{k=1}^n c_k\eta_k$ where
$\eta_1,\dots,\eta_n$ are independent random variables, and $\eta_k$ has
Poisson distribution with parameter $\lambda_k$, $k=1,\dots,n$.
 
Let us consider the positive half-plane $R^2_+=(-\infty,\infty)\times
[0,\infty)$ and introduce the following (product) measure on it: Let
$\nu$ be the measure (concentrated in finitely many points) on
$[-\infty,\infty]$ such that $\mu_k(c_k)=\lambda_k$, $k=1,\dots,n$, and
the remaining set has zero measure. Let $\nu=\mu\times$the Lebesgue
measure on $[0,\infty]$. Let $P$ be a Poisson process on $R^2_+$ with
measure $\nu$. For an elementary event $\omega$ let the sample
$\{\xi_j(\omega)=(\xi_j^{(1)}(\omega),\xi_j^{(2)}(\omega))\}$,
$j=1,2,\dots$ be the value of this Poisson process at $\omega$. Then we
define
$$
X(T)=\sum_{\xi_j^{(2)}\le T} \xi_j^{(1)},\quad T>0.\tag$**$
$$
Then $X(1)$ is the linear combination of independent Poisson
distributed random variables with the required weights and parameters,
and $X(T)$ is a stochastic process with independent stationary process.
 
Then the natural way to construct the closure of these processes is the
following: Let $\mu$ be a (not necessarily finite) measure $\mu$, which
is called the spectral measure, on $(-\infty,\infty)$, and $P(\omega)=
(\xi_j(\omega)=(\xi_j^{(1)}(\omega),\xi_j^{(2)}(\omega))$ be a Poisson
process with counting measure $\nu=\mu\times$the Lebesgue measure on
$[0,\infty)$. Define the process $X(T)$ by the formula $(**)$.
 
This formula is almost correct, but it must be refined because the sum
in $(**)$ may be divergent, hence meaningless. The three series theorem
about the necessary and sufficient condition for the convergence of sum
of independent random variables can help to clarify when this sum  is
convergent. (Actually we have certain freedom. We can add some constant
to the approximating sums, and this may help to  satisfy the
convergence.)
 
Let the spectral measure $\mu$ be such that
$$
\lim_{x\to\infty}\mu([x,\infty)\cup(-\infty,-x])=0\quad\text{and}\quad
\int_{-\e}^ex^2\mu(dx)\to0
$$
as $\e\to\infty$. For $0<a<b<\infty$ or $-\infty<a<b<0$ define
$$
X^{(a,b)}(T)=\sum\Sb{\xi_j^{(2)}\le
T}\\a<\xi_j^{(1)}<b\endSb \xi_j^{(1)},\quad T>0.
$$
then define the process
$$
X(T)=\lim_{\e\to0}[X^{(\e,1/\e)}(T)-EX^{(\e,1/\e)}(T)+
X^{(-1/\e,-\e)}(T)-EX^{(-1/\e,-\e)}(T)],\quad T>0
$$
is meaningful, and it is a process with independent stationary
increments. The L\'evy--Hin\v cin formula in the present context
can be formulated in the following way: All infinitely divisibe
distributions without a normal component can be represented as the
distribution of $X(1)+\,$const. of the above constructed process with
some spectral measure $\mu$. The class of all infinitely distributed
random distributions are the distributions obtained in such a way
convolved with a normal distribution. This representation shows that
the logarithm of the characteristic function of an infinitely divisible
distribution is the mixture of the logarithm of the caracteristic
function of Poisson distributed random variables with different
intensities multiplied by different constants minus constant times
$x^2$ (the logarithm of the Fourier transform of the normal
distribution) plus some constant. The mixture is taken by the spectral
measure $\mu$. This is the classical form of the L\'evy--Hin\v cin
formula.
 
An important subclass of the infinitely distributions are the stable
distributions with the following property: For all $n=1,2,\dots$ a
random variable $X$ with stable distribution has the following property:
$$
X\overset\Delta\to=\frac1{A_n}(X_1+\dots+X_n-B_n)
$$
with some appropriate $A_n$ and $B_n$, where $X_1,\dots,X_n$ are
independent random variables with the same distribution as $X$. It can
be proved that the above relation may hold only with $A_n=n^{-1/\alpha}$,
$0<\alpha\le2$. For $\alpha=2$ only the normal distribution function
satisfies this property. For $0<\alpha<2$ this relation holds if $X$ has
the representation given for general infinitely distributed random
variable with a spectral density $\mu$ of very special form. Namely, it
must be homogeneous with parameter $\alpha$. More explicitly, $\mu$ has
a density function $C_1x^{-\alpha-1}$, for $x>0$, and
$C_2|x|^{-\alpha-1}$, for $x<0$, $C_1\ge0$, $C_2\ge0$, $C_1+C_2>0$.
 
To understand the behaviour of the stable process whose spectral density
function is $x^{-\alpha-1}$ define the map $S_u$, $u>0$, of
$R^2_+=(-\infty,\infty)\times[0,\infty)$,
$$
S_u(x,t)=(u^{1/\alpha}x,ut),
$$
and observe that $S_u$ is a measure preserving transformation of
$(R^2_+,\mu(dx)dt)$. This implies that the distribution of the Poisson
process $(\xi_n^{(1)},\xi_n^{(2)})$ with counting measure $\mu(dx)dt$
agrees in distribution with the distribution of the process
$(u^{1/\alpha}\xi_n^{(1)},u\xi_n^{(2)})$. It can be proved by means of
this property why the infinitely divisible process constructed with the
above spectral measure satisfies the condition demanded for a stable
process. But to make a complete proof one has to clarify certain
convergence problem. In this investigation the cases $0<x<1$, $x=1$ and
$1<x<2$ has to be handled differently. In the case $1<\alpha<2$ the
process
$$
X(T)=\lim_{\e\to0}\sum\Sb\xi_j^{(2)}\le T\\ \e<|\xi_j^{(1)}|<1/\e
\endSb \xi_j^{(1)},\quad T>0,
$$
i.e.\ the sum is meaningful without any centralization, and the formula
($*$) holds in this case. If $0<\alpha<1$, then $X(T)$ can be defined
with the following centralization:
$$
X(T)=\lim_{\e\to0}\left[\sum\Sb\xi_j^{(2)}\le T\\ \e<|\xi_j^{(1)}|<1/\e
\endSb \xi_j^{(1)}
-E\sum\Sb\xi_j^{(2)}\le T\\ \e<|\xi_j^{(1)}|<1/\e
\endSb \xi_j^{(1)}\right],\quad T>0.
$$
Then $EX(T)=0$, and $(*)$ holds in this case too. The situation
$\alpha=1$ is different. The reason for the special behaviour is related
to the fact that the integral $\int x^{-1}\,dx$ is divergent both in
zero and infinity. In this case $X(T)$ can be defined in the following
way:
$$
X(T)=\lim_{\e\to0}\left[\sum\Sb\xi_j^{(2)}\le T\\ \e<|\xi_j^{(1)}|<1/\e
\endSb \xi_j^{(1)}
-E\sum\Sb\xi_j^{(2)}\le T\\ \e<|\xi_j^{(1)}|<A
\endSb \xi_j^{(1)}\right],\quad T>0.
$$
Then some calculation shows that the process $X(T)$ satisfies the
following version of~($*$):
$$
X(tT)\overset\Delta\to=TX(t)+\gamma tT\log T
$$
with $\gamma=C_1-C_2$. In this case introduce the process
$$
\bar X(t)=X(t)-\gamma t\log t.
$$
Then
$$
\align
\frac{\bar X(tT)}T=\frac{X(tT)-\gamma tT\log (tT)}T&\overset\Delta\to
=\frac {TX(t)+\gamma tT\log T-\gamma tT\log(tT)}T\\
&=X(t)-\gamma t\log t=\bar X(t).
\endalign
$$
Hence the same argument can be applied in this case too if the process
$X(t)$ is replaced by $\bar X(t)$.
 
We also want to prove an invariance principle which implies that a
sequence of of partial sums of independent random variables in the
domain of attraction of a stable law satisfies the same ``almost sure
limit theorem" as the stable process. Let us first recall the following
theorem.
\medskip\noindent
{\bf Approximation of partial sum of iid.\ random variables in the
domain of attraction of a stable law}
\medskip\noindent
A distribution function $F(x)$ belongs to the stable law with
spectral density with spectral density function $C_1\alpha
x^{-\alpha-1}$ for $x>0$, and $C_2\alpha |x|^{-\alpha-1}$ for $x<0$, if
$F(x)=C_1L(x)x^{-\alpha}(1+o(1))$ for $x\to\infty$, and
$1-F(x)=C_2L(|x|)|x|^{-\alpha}(1+o(1))$ for $x\to-\infty$, where
the normalization has to be chosen in the following way. If $\xi_n$,
$n=1,2,\dots$, are iid.\ random variables with distribution function
$F(x)$, then $n^{-1/\alpha}\tilde L(n^{1/\alpha})^{-1/\alpha}
\sum\limits_{k=1}^n\xi_k-B_n$ with an appropriate centralizing constant
$B_n$, where the function $\tilde L(n)$ defined by the equation
$\tilde L(n^{1/\alpha})=L(n^{1/\alpha}\tilde L(n^{1/\alpha})
^{1/\alpha})$, converges in distribution to the stable law with the
prescribed spectral density. We briefly explain that if the distribution
of the sequence $\bar\xi_n$, $n=1,2,\dots$, of iid.\ random variables
has a distribution function $F(x)$ in the domain of a stable law with
the above spectral density, then for all $n$ a sequence $\eta_k$,
$k=1,2,\dots,n$, of iid.\ random variables can be constructed whose
distribution is the stable law together with a sequence $\xi_k$,
$k=1,2,\dots,n$, of iid.\ random  variables with the same distribution
as the sequence $\bar\xi_n$ in such a way that
$$
\sup_{1\le k\le n}\left[n^{-1/\alpha}\sum_{j=1}^k\eta_j
-n^{-1/\alpha}\tilde L^{-1/\alpha}(n^{1/\alpha})
\sum_{j=1}^k\xi_j-B_k\right]\Rightarrow 0, \tag1
$$
with appropriate centralizing constants $C_k$ as $n\to\infty$, where
$\Rightarrow$ denotes convergence in probability. Then we shall
consider a version of this result which enables us to prove that the
``almost sure limit theorem" for stable laws also implies this result
for partial sums of iid.\ random variables in the domain of attraction
of the underlying stable law.
\medskip\noindent {\it Proof:}\/ Consider a Poisson process $P(x,t)$
on $R_2^+=(-\infty,\infty)\times[0,\infty)$ with counting measure
$C_1\alpha x^{-\alpha-1}dx\,dt$ for $x>0$, and
$C_2\alpha|x|^{-\alpha-1}dx\,dt$ for $x<0$. Let
$\zeta_j=(\zeta_j^{(1)},\zeta_j^{(2)})$, $j=1,2,\dots$, denote
a sample of this Poisson process. Fix some small $\e>0$, (actually,
$\e=\e_n\to 0$ slowly as $n\to\infty$) and define the following
quantities:
 
For all $k=1,\dots,n$ define the random variable $\hat \eta_k$ in the
following way:
$$
\hat\eta_k=\cases 0&\text{if the sample of the Poisson process }P(x,t)
\text{ in the domain} \\ &\qquad
((-\infty,-\e n^{1/\alpha}]\cup[\,\e
n^{1/\alpha},\infty))\times[k-1,k)\text{ is empty}\\
\zeta_j^{(1)}&\text{with }\zeta_j\in
((-\infty,-\e n^{1/\alpha}]\cup[\,\e
n^{1/\alpha},\infty))\times[n-1,n))\text{ and }\\ &\qquad\zeta_j^{(2)}=
\min\limits_{\zeta_k\in((-\infty,-\e]\cup[\e,\infty))\times[k-1,k))}
\zeta_k^{(2)}\text{ otherwise}
\endcases   .
$$
Then the random variables $\hat\eta_k$, $k=1,\dots,n$, are independent
$P(\hat\eta^k=0)=1-(C_1+C_2)\e^{-\alpha}n^{-1}$,
$P(\hat\eta_k>x)=C_1 x^{-\alpha}$ if $x>\e n^{1/\alpha}$, and
$P(\hat\eta_k<x)=C_2|x|^{-\alpha}dx$ if $x<-\e n^{1/\alpha}$.
 
We define the iid.\ random variables $\xi_1,\dots,\xi_n$ with
distribution $F(x)$ in the following way. Let
$\xi_1^{(\e)},\dots,\xi_n^{(\e)}$ be a sequence of iid.\ random
variables, independent of the Poisson process $P(x,t)$ with the
distribution $P(\xi_1^{(\e)}\in A)=\dfrac{\rho(A\cap[L_-^\e,L_+^\e])}
{\rho([L_-^\e,L_+^\e])}$, where $\rho$ is the measure induced by the
distribution function $F(x)$, and the constants $L^\e_-$ and $L_+^\e$
are defined by the equation $1-F(L_+^\e)=C_1\e^{-\alpha}n^{-1} $ and
$F(L_-^\e)=C_2\e^{-\alpha}n^{-1}$. Define $\bar\xi_k^{(\e)}$ in the
following way: $\bar\xi_k^{(\e)}=\bar F^{-1}(1-C_1(\tilde
L(n^{1/\alpha}))^{1/\alpha}\hat\eta_k^{-\alpha})$ for $\hat\eta_k>0$,
and $\bar\xi_k=F^{-1}(C_2(\tilde L(n^{1/\alpha}))^{1/\alpha}
\hat\eta_k^{-\alpha})$ for $\hat\eta_k<0$, where $\bar F(x)=1-F(x)$.
The random variable $\xi_k$ is the quantile transform of $\hat\eta_k$,
and $\bar\xi_k^{(\e)}=H((\tilde L(n^{1/\alpha}))^{1/\alpha}\hat\eta_k)$
with  a function $H(x)$ such that $H(x)=x(1+o(1))$ for
$\e n^{1/\alpha}<|x|<\e^{-1}n^{1/\alpha}$. This relation holds, because
$$
\align
1-F(x)=C_1x^{-\alpha}L(x)(1+o(1))&=C_1x^{-\alpha}L(n^{1/\alpha}\tilde
L(n^{1/\alpha})^{1/\alpha})(1+o(1))\\
&=C_1x^{-\alpha}\tilde L(n^{1/\alpha})(1+o(1)),
\endalign
$$
for $\e(n\tilde L(n^{1/\alpha}))^{1/\alpha}<x<\e^{-1}(n\tilde
L(n^{1/\alpha}))^{1/\alpha}$, and a similar statement holds for $x<0$ too.
 
Then the sequence $\xi_k=I(\hat\eta_k=0)\xi_k^{(\e)}+\bar\xi_k^{(\e)}$
is a sequence of iid.\ random variables with distribution $F(x)$.
Indeed, because of the independence of $I(\hat\eta_k=0)$ and
$\xi_k^{(\e)}$ both $\xi_k\cap [L_-^\e,L_+^\e]$ and $\xi_k\cap(R\setminus
[L_-^\e,L_+^\e]$ have the prescribed distribution. We claim that this
sequence $\xi_k$ and $\eta_k$ constructed in the way explained in the
construction of the infinitely divisible processes, i.e.
$$
\eta_k=\sum_{k-1\le \zeta^{(2)}_j<k} \zeta_j^{(1)}-\text{possible
centralization}
$$
satisfy (1).
 
To prove this statement let us make the following observations. Define
$$
\hat\eta_k=\sum_{k-1\le \zeta^{(2)}_j<k,\;|\zeta^{(1)}_j|>\e
n^{1/\alpha}} \zeta_j^{(1)}.
$$
Then
$$
P\left(\sum_{k=1}^p(\hat\eta_k-\bar\eta_k)=0\text{ for
all }1\le p\le n\right)\to 1.
$$
Indeed,
$$
\align
&\sum_{k=1}^p(\hat\eta_k-\bar\eta_k)\neq0\text{ for some }1\le p\le n\\
&\qquad\subset \bigcup_{k=1}^n \{P(x,t)
\text{ contains at least two points in }\\
&\qquad\qquad((-\infty,-\e n^{1/\alpha})\cup (\e
n^{1/\alpha},\infty))\times[k-1,k)\}
\endalign
$$
and
$$
\align
&P\biggl(\bigcup_{k=1}^n \{P(x,t)\text{ contains at least two points in
}\\
&\qquad((-\infty,-\e n^{1/\alpha})\cup (-\e
n^{1/\alpha},\infty))\times[k-1,k)\biggr)\\
&\qquad <nP(P(x,t)\text{ contains at least two points in }\\
&\qquad\qquad((-\infty,-\e n^{1/\alpha})\cup (\e
n^{1/\alpha},\infty))\times[k-1,k))<(C_1+C_2)^2\e^{-2\alpha}n^{-1}\to 0
\endalign
$$
as $n\to0$.
 
We also show that
$$
n^{-1/\alpha}\sup_{1\le k\le n}\sum_{j=1}^k(\hat\eta_j
-\tilde L(n^{1/\alpha})^{-1/\alpha}\bar\xi_j^{(\e)})\Rightarrow 0
$$
if $\e\to 0$ and $\e=\e_n$ tends to 0 sufficiently slowly as
$n\to\infty$. Indeed, let us consider the contribution of those terms
where $|\hat\eta_j|>\e^{-1}n^{1/\alpha}$ and
$\e n^{1/\alpha}\le |\hat\eta_j|\le \e^{-1}n^{1/\alpha}$. (The function
$\xi_j^{(\e)}$ is a deterministic function of $\hat\eta_j$.
$$
\left\{\omega\: \sup_{1\le k\le n}\sum_{j=1}^k(\hat\eta_j
-\tilde L(n^{1/\alpha})^{-1/\alpha}\bar\xi_j^{(\e)}\neq0\right\}
\subset\bigcup_{j=1}^n\{|\hat\eta_j|>\e^{-1}n^{1/\alpha}\},
$$
and
$$
P(\subset\bigcup_{j=1}^n\{|\hat\eta_j|>\e^{-1}n^{1/\alpha}\})\le
\sum_{j=1}^nP(|\hat\eta_j|>\e^{-1}n^{1/\alpha})<\e_n^{\alpha}\to0
$$
as $n\to\infty$. On the other hand, for each term
$$
n^{-1/\alpha}(\hat\eta_j
-\tilde L(n^{1/\alpha})^{-1/\alpha}\bar\xi_j^{(\e)}
I(\e n^{1/\alpha}\le |\hat\eta_j|\le \e^{-1}n^{1/\alpha})\Rightarrow 0,
$$
and the distribution of the last expression depends only on $n$, but not
on $j$. (We need certain uniformness of the convergence.) The last
relation holds, since $n^{-1/\alpha}(\hat\eta_j-\tilde L(n^{1/\alpha})
^{-1/\alpha}\bar\xi_j^{(\e)})= n^{-1/\alpha}(\hat\eta_j
-\tilde L(n^{1/\alpha})^{-1/\alpha}H(\hat\eta_j^{(\e)}))=o(1)$ on this
set. Since the number of indices $k$ for which $\hat\eta_k\neq0$ is
stochastically bounded, also the expression
$$
n^{-1/\alpha}\sup_{1\le k\le n}\sum_{j=1}^k(\hat\eta_j-\tilde
L(n^{1/\alpha})^{-1/\alpha}\bar\xi_j^{(\e)})\Rightarrow0
$$
as $n\to0$. To complete the proof it is enough to show that
$$
\sup_{1\le k\le n}n^{-1/\alpha}\tilde L(n^{1/\alpha})^{-1/\alpha}
\sum_{j=1}^k I(\hat\eta_j=0)\bar\xi_j^{(\e)}-
E I(\hat\eta_j=0)\bar\xi_j^{(\e)}\Rightarrow 0
$$
and
$$
\sup_{1\le k\le n}n^{-1/\alpha}\sum_{l=1}^k\tilde\eta_l-
E\tilde\eta_l\Rightarrow 0,
$$
where
$$
\tilde\eta_l=X(l)-\bar\eta_l=\sum\Sb l\le \zeta^{(2)}_j<l+1\\
|\zeta^{(1)}_j|<\e n^{1/\alpha}\endSb \zeta^{(1)}_j-\text{appropriate
centralization}.
$$
These relations follow from the Kolmogorov inequality and the following
bound on the variance of the summands in the appropriate expressions.
$$
\align
E(I(\hat\eta_j=0)\bar\xi_j^{(\e)})^2&\le\text{const.}\,\int_{-\e
n^{1/\alpha} \tilde L(n^{1/\alpha})^{-1}}^{\e n^{1/\alpha} \tilde
L(n^{1/\alpha})^{-1}}\min(1,x^{1-\alpha}\tilde L(x))\,dx\\
&<\text{const.}\,\e^{2-\alpha}n^{-1+2/\alpha}
\tilde L(n^{1/\alpha})^{2/\alpha},
\endalign
$$
hence
$$
\align
P&\left(\sup_{1\le k\le n}n^{-1/\alpha}\tilde L(n^{1/\alpha})^{-1/\alpha}
\sum_{j=1}^k I(\hat\eta_j=0)\bar\xi_j^{(\e)}-
E I(\hat\eta_j=0)\bar\xi_j^{(\e)}>u\right)\\
&\qquad<\text{const.}\,\e^{2-\alpha} u^{-2}
\endalign
$$
which tends to zero as $n\to\infty$ if $\e=\e_n\to0$. Similarly,
$$
E(\tilde\eta_l-E\tilde\eta_l)^2\int_{-\e n^{1/\alpha}}^{\e n^{1/\alpha}}
x^{1-\alpha}\,dx<\text{const.}\,\e^{2-\alpha}n^{-1+2/\alpha},
$$
and the second expression can be estimated in the same way.
 
Let us briefly discuss the centralization needed in this result. If
$0<\alpha<1$, then no centralization is needed. Also no centralization
is needed for $1<\alpha<2$ if the random variables $\eta_n$ with the
stable law and the random variables $\xi_n$ with a distribuion in its
domain of attraction are centralized so that $E\eta_n=E\xi_n=0$. If
$\alpha=1$, then this centralization problem cannot be answered in such
a simple way. It can be given in the following way: For all $A>0$ define
teh  the random variables $\eta_n=\eta_n(A)$ by the formula
$$
\eta_n(A)=\lim_{\e\to0}\left[\sum\Sb n<\zeta^{(2)}_j<n+1\\
\e<|\zeta^{(1)}_j|<e^{-1}\endSb \zeta^{(1)}_j-
E\sum\Sb n<\zeta^{(2)}_j<n+1\\
\e<|\zeta^{(1)}_j|<A\endSb \zeta^{(1)}_j\right],
$$
and put $E\xi_n(A)=E\xi_nI(|\xi_n|<A)$. Then the centralization constant
$B_n$ equals $n\lim\limits_{A\to\infty}[E\xi_n(A)-E\eta_n(A)]$. We omit
the proof of this statement.
\bigskip\noindent
We formulate the following result which enables one to get an invariance
principle for the ``almost sure limit theorem" for stable laws together
with a sketch of proof.\medskip\noindent
{\bf Theorem.} {\it If a sequence of iid. random variables with
distribution function $F(x)$ is in the domain of attraction (with
norming $n^\alpha \tilde L(n)$) of a stable law with index $\alpha$,
$0<\alpha<2$, then a sequence of iid.\ random variables $\xi_n$,
$n=1,2,\dots$, with distribution function $F(x)$ and a stable process
$X(t)$, $t\ge0$, can be constructed in such a way that the partial sums
$S_n=\sum\limits_{k=1}^n\xi_k$ and the process $X(t)$ satisfy the
relation
$$
\lim_{N\to\infty}\frac1{\log N}\sum_{n=1}^N\frac1n I\left(
\frac1{(n\tilde L(n^{1/\alpha}))^{1/\alpha}}\sup_{1\le k\le n} \left|
S_k-X(k(\tilde L(k^{1/\alpha}))-B_k\right|>\delta\right)=0\quad\text
{\rm a. s.}
$$
for all $\delta>0$ with appropriate centralizing constants $B_k$.
This statement also holds if $\delta=\delta_n$ tends to zero sufficiently
slowly as $n\to\infty$.}
\medskip\noindent{\it Sketch of proof:}\/   The construction is a
natural modification of the previous one. The main difference is that
the set $((-\infty,-\e n^{1/\alpha}]\cup[\,\e n^{1/\alpha},\infty))
\times[k-1,k)$ in the definition of $\hat\eta_k$ is replaced by
$$
((-\infty,-\e (k\tilde L(k^{1/\alpha}))^{1/\alpha}]\cup[\,\e(k\tilde
L(k^{1/\alpha}))^{1/\alpha},\infty))\times[(k-1)^{1/\alpha}\tilde
L((k-1)^{1/\alpha}),k^{1/\alpha}\tilde L(k^{1/\alpha})).
$$
Then we define the random variables $\tilde\eta_k$ similarly to the
definition in the previous proof. It equals the first coordinate of
the sample points if the previously defined set is non-empty, and
equals zero, if it is empty. Then simple calculation shows that the
probability of the event that this set contains at least two points is
less than const.$\times k^{-2}$, hence by the Borel--Cantelli lemma this
event happens only for finitely many indices $k$ with probability one.
This implies that only a negligible error is committed if in the sum of
the first coordinates of the sample points $\zeta_j$ appearing in the
definition of $X(t)$ the sum of the first coordinates of the sample
points appearing in the set of the definition of $\tilde\eta_k$ is
replaced by $\tilde\eta_k$. Let us also remark that
$P(\tilde\eta_k>x)\sim C_1\tilde L(k^{1/\alpha})x^{-\alpha}$ for
$x>\e(k\tilde L(k^{1/\alpha}))^{1/\alpha}$,
$P(\tilde\eta_k<x)\sim C_2\tilde L(k^{1/\alpha})|x|^{-\alpha}$ for $x<-\e
(k\tilde L(k^{1/\alpha}))^{1/\alpha}$, and $\tilde\eta_k=0$ if
$-\e(k\tilde L(k^{1/\alpha}))^{1/\alpha}<\eta_k<\e
(k\tilde L(k^{1/\alpha}))^{1/\alpha}$.
 
We define $\bar\xi_k^{(\e)}$ by the same quantile transform of
$(k^{1/\alpha}\tilde L(k^{1/\alpha})-(k-1)^{1/\alpha}\tilde
L((k-1)^{1/\alpha}))\hat\eta_k$ as in the previous case. Here again
$\e=\e_k$ tends to zero sufficiently slowly as $k\to\infty$. Because of
the properties of the distribution functions in the domain of
attraction of a stable law
$$
\sup_{\e(k\tilde L(k^{1/\alpha}))^{1/\alpha}<|\tilde\eta_k|<\e^{-1}
(k\tilde L(k^{1/\alpha}))^{1/\alpha}}\frac{|\bar\xi_k-\tilde\eta_k|}
{(n\tilde L(n^{1/\alpha}))^{1/\alpha}}\to0\quad
\text{a. s. \ as }k\to\infty,
$$
because
$$
\align
1-F(x)=C_1x^{-\alpha}L(x)(1+o(1))&=C_1x^{-\alpha}L(k^{1/\alpha}\tilde
L(k^{1/\alpha})^{1/\alpha})(1+o(1))\\
&=C_1x^{-\alpha}\tilde L(k^{1/\alpha})(1+o(1)),
\endalign
$$
for $\e(k\tilde L(k^{1/\alpha}))^{1/\alpha}<x<\e^{-1}(k\tilde
L(k^{1/\alpha}))^{1/\alpha}$, and a similar statement holds for $x<0$ too.
 
We want to prove by means of this estimate that
$$
\frac1{\log N}\sum_{n=1}^N\frac 1nI\left(\sum_{k=1}^n
\frac{|\bar\xi_k-\tilde\eta_k|
I(\e(k\tilde L(k^{1/\alpha}))^{1/\alpha}<|\tilde\eta_k|<
\e^{-1}(k\tilde L(k^{1/\alpha}))^{1/\alpha})
}{(n\tilde L(n^{1/\alpha}))^{1/\alpha}}>\delta
\right)\to0
$$
almost surely for all $\delta>0$ as $N\to\infty$.
 
Put $\bar Y_n=|\bar\xi_n-\tilde\eta_n| I(\e(n\tilde L(n^{1/\alpha}))
^{1/\alpha}<|\tilde\eta_n|<\e^{-1}(n\tilde L(n^{1/\alpha}))^{1/\alpha})$,
$Y_n=\bar Y_n-E\bar Y_n$, $U_n=\sum\limits_{l=1}^n Y_k$ and
$T_k=\dfrac{\sup\limits_{2^k\le m<2^{k+1}}(S_m-S_{2^k})}{2^{k/\alpha}
\tilde L(2^{k/\alpha})^{1/\alpha}}$. Then $E\bar Y_n<\text{const.}\,\e
n^{-1-1/\alpha}\tilde L(n^{1/\alpha})^{-1/\alpha}$ and $EY_n^2<\text
{const.}\,\e n^{-1-2/\alpha}\tilde L(n^{1/\alpha}))^{2/\alpha}$.
Hence it follows from the Kolmogorov inequality that
$$
P(T_k>x)<\frac{\text{const.}\,\e}{x^2}.
$$
The required inequality follows from the following lemma applied for
the sequence $\pm Y_n$, whose conditions are satisfied with $\beta=2$
and the observation that because of the bound on $U_k$ the estimate
remains valid if $\pm Y_k$ is replaced by $\pm \bar Y_k$ and $\delta$ by
$2\delta$.
\medskip\noindent
{\bf Lemma.}  {\it Let $Y_n$, $n=1,2,\dots$, be a sequence of independent
random variables. Define $U_n=\sum\limits_{k=1}^n Y_k$,  $n=1,2,\dots$,
and $T_k=\dfrac{\sup\limits_{2^{k}\le
m<2^{k+1}}U_m-U_{2^k}}{2^{k/\alpha}\tilde L(2^{k/\alpha})^{1/\alpha}}$,
$k=0,1,\dots$. Assume that the inequality
$$
P(T_k>x)<\text{\rm const.}{\e}{x^{-\beta}}
$$
holds with some $\beta>0$ and $\e=\e_k\to0$ as $k\to\infty$ for all
$x>0$.
Then for all $\delta>0$
$$
\lim_{N\to\infty}\frac1{\log N} \sum_{n=1}^ N\frac1n I\left(\frac1
{n^{1/\alpha}\tilde L(n^{1/\alpha})^{1/\alpha}}\sup_{1\le k\le n}
U_k>\delta\right)=0\quad\text{\rm a. s.}
$$
for all $\delta>0$. (This relation also holds if $\delta=\delta_n$ tends
to zero sufficiently slowly as $n\to\infty$.)}\medskip\noindent
{\it Proof of the Lemma:}\/ Let $2^{m-1}<n\le 2^m$. Then
$$
\frac1{n^{1/\alpha}\tilde L(n^{1/\alpha})^{1/\alpha}}\sup_{1\le k\le n}
U_k\le \text{const.}\,\sum_{k=0}^m \frac {T_k}{2^{(m-k)/\alpha}}
\left(\frac{\tilde L(2^{k\alpha})}{\tilde L(2^{m\alpha})}\right)
^{1/\alpha}.
$$
Hence it is enough to prove that
$$
\lim_{m\to\infty}\frac 1m\sum_{p=0}^m I\left(\sum_{k=0}^p
2^{-(p-k)/2\alpha} T_k>\delta\right)=0 \quad\text{a. s.}.
$$
We have
$$
I\left(\sum_{k=0}^p2^{-(p-k)/2\alpha} T_k>\delta\right)\le
\sum_{k=0}^pI(2^{-(p-k)/4\alpha}T_k>c)
$$
with $c=(1-2^{-\alpha/4})\delta$. Hence
$$
\align
&\sum_{p=0}^mI\left(\sum_{k=0}^p2^{-(p-k)/2\alpha} T_k>\delta\right)\le
\sum_{p=0}^m\sum_{k=0}^pI(2^{-(p-k)/4\alpha}T_k>c)\\
&\qquad=\sum_{k=0}^m\sum_{p=k}^mI(2^{-(p-k)/4\alpha}T_k>c)
\le\sum_{k=0}^m\sum_{p=0}^\infty
I(2^{-p/4\alpha}T_k>c)=\sum_{k=0}^m Z_k
\endalign
$$
with $Z_k=\sum\limits_{p=0}^\infty I(2^{-p/4\alpha}T_k>c)$. Then the
random variables $Z_k$ are independent, and
$EZ_k\le\text{const.}\,\e\sum\limits_{p=0}^\infty2^{-\beta p/4\alpha}<
\text{const.}\,\e$, and similarly $EZ_k^2<\text{const.}\,\e$. Hence by
the law of large numbers
$$
\lim_{m\to\infty}\frac1m\sum_{k=1}^m (Z_k-EZ_k)=0\quad\text{a. s.},
$$
and since $EZ_k\to 0$, also
$$
\lim_{m\to\infty}\frac1m\sum_{k=1}^m Z_k=0\quad\text{a. s.}
$$
These relations imply the lemma. \medskip
 
Let us define the random variables $\tilde \xi_k =\xi_k
I(u_k<\xi_k<v_k)$  with $u_k=F^{-1}(-\e n^{k/\alpha})$ and
$v_k=F^{-1}(-\e n^{k/\alpha})$. Let us also define
$$
\eta'_k=\sum\Sb |\zeta_j^{(1)}|<\e k^{1/\alpha}\tilde L(k^{1/\alpha})
^{1/\alpha}\\
(k-1)\tilde L((k-1)^{1/\alpha})<\zeta_j^{(2)}<k\tilde L(k^{1/\alpha})
\endSb \zeta_j^{(1)},
$$
where $\zeta_j$ is the Poisson process which appears in the definition
of the stable process $x(t)$. We want to prove by means of the Lemma
that
$$
\frac1{\log N}\sum_{n=1}^N\frac 1nI\left(\left|\sum_{k=1}^n\frac{\tilde
\xi_k-E\tilde\xi_k}{(n\tilde L(n^{1/\alpha}))^{1/\alpha}}\right|
>\delta \right)\to0\quad\text{a. s.}
$$
and
$$
\frac1{\log N}\sum_{n=1}^N\frac 1nI\left(\left|\sum_{k=1}^n
\frac{\eta'_k-E\eta'_k}{(n\tilde L(n^{1/\alpha}))^{1/\alpha}}\right|>
\delta \right)\to0\quad\text{a. s.}
$$
for all $\delta>0$ as $N\to\infty$. Let us observe that
for $Y_k=\eta'_k-E\eta'_k$ or $Y_k=\tilde\xi_k-E\tilde\xi_k$
$EY_k^2<\text{const.}\,\e^{2-\alpha}n^{2/\alpha-1}\tilde
L(n^{1/\alpha})^{2/\alpha}$. These bounds are the same as in the case
when the analogous estimate was proved. Hence the same proof works in
this case too.
 
Finally we want to prove that the large values of the summands in the
Theorem is negligible. More explicitly we want to prove, by applying
again the Lemma, the following statements. Define the sequence of
independent random variables $Y_k$, $k=1,2,\dots$, in one of the
following four forms:
$Y_k=\tilde\eta_kI(\tilde\eta_k>\e^{-1}(k\tilde L(k^{1/\alpha})
^{1/\alpha})$,
$Y_k=-\tilde\eta_kI(\tilde\eta_k<-\e^{-1}(k\tilde L(k^{1/\alpha})
^{1/\alpha})$,
$Y_k=\xi_kI(\xi_k>\bar F^{-1}(\e^{\alpha}k^{-1})$,
$Y_k=-\xi_kI(\xi_k< F^{-1}(\e^{\alpha}k^{-1})$, where $\bar
F(x)=1-F(x)$. Put $U_n=\sum\limits_{k=1}^n Y_k$. Then we want to show
that
$$
\lim_{N\to\infty}\frac1{\log N} \sum_{n=1}^ N\frac1n I\left(\frac1
{n^{1/\alpha}\tilde L(n^{1/\alpha})^{1/\alpha}}\sup_{1\le k\le n}
U_k>\delta\right)=0\quad\text{\rm a. s.}
$$
for all $\delta>0$.
 
Let us apply the lemma. Since the random variables $Y_k$ are
non-negative,
$$
T_k=\dfrac{\sum\limits_{m=2^k}^{2^{k+1}-1}Y_m}
{2^{k/\alpha}\tilde L(2^{k/\alpha})^{1/\alpha}},
$$
and
$P(T_k>x)<2^kP(Y_{2^k}>{2^{k/\alpha}\tilde L(2^{k/\alpha})
^{1/\alpha}}x)$. On the other hand
$$
P(Y_{2^k}>x)<\text{const.}\,2^{-k}x^{-\alpha}
$$
if $Y_{2^k}$ is defined through $\pm \tilde\eta_{2^k}$ and
$$
P(Y_{2^k}>x)<\text{const.}\,2^{-k}
x^{-\alpha}\frac{L(2^{k/\alpha}x\tilde L(2^{k/\alpha})^{1/\alpha})}
{\tilde L(2^{k/\alpha})}<\text{const.}\,2^{-k}x^{-3x/4}
$$
if $Y_{2^k}$ is defined through $\pm \xi_{2^k}$, since
$$
L(2^{k/\alpha}x\tilde
L(2^{k/\alpha})^{1/\alpha})<\text{const.}\,x^{\alpha/4}L(2^{k/\alpha}
\tilde L(2^{k/\alpha})^{1/\alpha})=\text{const.}\,
x^{\alpha/4}\tilde L(2^{k/\alpha}).
$$
The inequality $P(Y_{2^k}>x)<\text{const.}\,2^{-k}\e^{\alpha}$ also
holds for $x>0$, because by the definition of $Y_k$ it can take
positive values only if $\pm\xi_k$ or $\pm\tilde\eta_k$ is bigger than
a number described in the definition of $Y_k$. These estimates imply
that
$$
P(T_k>x)\le2^kP(Y_{2^k}>x)\le\text{const.}\,\min(x^{-3\alpha/4},
\e^{\alpha})\le\text{const.}\,\e^{\alpha/3}x^{-\alpha/2}.
$$
Hence the Lemma can be applied to get the needed estimates with
$\beta=\alpha/2$ and $\e^{\alpha/3}$ instead of $\e$. These estimates
together imply the Theorem.
 
\bye
