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% tr2latex was written by Kamal Al-Yahya at Stanford University
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\documentclass{article}
\include{macros}

% Extra Macros
\newcommand{\C}{\ensuremath{\mathbb{C}}} % complex numbers
\newcommand{\law}{\ensuremath{\mathbf{L}}}
\newcommand{\normal}{\ensuremath{\mathcal{N}}}
\newcommand{\toinf}{\ensuremath{\rightarrow \infty}}
\newcommand{\tozero}{\ensuremath{\rightarrow 0}}
\newcommand{\eps}{\ensuremath{\epsilon}}
\newcommand{\ind}[1]{\1 \left(#1\right)}
% Fairly local macros
\newcommand{\threecond}{the Triangular Array Conditions}
\newcommand{\bs}{\ensuremath{\mathbf{s}}}

\begin{document}
\lecture{10}{Setup for the Central Limit Theorem}{David S. Rosenberg}{drosen@stat.berkeley.edu}

  See Durrett 2nd ed pages 116-118 for an equivalent formulation and a
  proof using characteristic functions. That proof leans on the
  continuity theorem for characteristic functions, (3.4) on page 99,
  which in turn relies on the Helly selection theorem (2.5) on page
  88.  The present approach, due to Lindeberg, is more elementary in
  that it does not require these tools. But note that the basic idea
  in both arguments is to estimate the expected value of a smooth
  function of a sum of independent variables using a Taylor expansion
  with error bound.

  \section{Triangular Arrays}
  
  Roughly speaking, a sum of many small independent 
  random variables will be nearly normally distributed.
  To formulate a limit theorem of this kind, we must consider 
  sums of more and more smaller and smaller random variables.
  Therefore, throughout this section we shall study the sequence of sums
  $$
  S_i\ =\ \sum_j X_{ij} ,
  $$

obtained by summing the rows of a \emph{triangular array} of random variables
$$
\begin{array}{l}
  X_{11},X_{12},\ldots,X_{1n_1} \\
  X_{21},X_{22},\ldots \ldots,X_{2n_2} \\
  X_{31},X_{32},\ldots \ldots \ldots,X_{3n_3} \\
  \vdots \hspace{.5in} \vdots \hspace{.5in} \vdots  \hspace{.5in} \vdots 
\end{array}
$$

It will be assumed throughout that triangular arrays satisfy 3
\emph{Triangular Array Conditions\footnote{This is not standard
    terminology, but is used here as a simple referent for these
    conditions.}}:
\begin{enumerate}
\item \label{indeprows} for each $i$, the $n_i$ random variables
  $X_{i1},X_{i2},\ldots,X_{in_{i}}$ in the $i$th row are mutually
  independent,n
\item \label{zeromean}$\E(X_{ij})=0$ for all $i,j$, and
\item \label{varsumone}$\sum_{j} \E X_{ij}^2=1$ for all $i$.
\end{enumerate}

%\begin{eqnarray}
% & &  \forall i, \text{ the } n_i \text{ r.v.s }
%   X_{i1},X_{i2},\ldots,X_{in_{i}} \text{in the $i$th row are mutually
%  independent,} \\
% & & \E(X_{ij})=0 \text{ for all } i,j \text{, and}\\
% & & \sum_{j} \E X_{ij}^2=1 \text{ for all $i$.}
%\end{eqnarray}

Here the row index $i$ should always be taken to range over
$1,2,3,\ldots$, while the column index $j$ ranges from $1$ to $n_i$.
It is \emph{not} assumed that the r.vs in each row are identically
distributed. And it is \emph{not} assumed that different rows are independent.
(Different rows could even be defined on different probability spaces.)
For motivation, see section **********  below for how such a triangular array is
set up in the most important application to partial sums
$X_1+X_2+\cdots+X_n$ obtained from a sequence of independent r.v.s 
$X_1,X_2,\ldots$

It will usually be the case that $n_1<n_2<\cdots$, whence the term
triangular.  It is not necessary to assume this however.

\section{The Lindeberg Condition and Some Consequences}
We will write $\law(X)$ to denote the \emph{law} or \emph{distribution} of a
random variable $X$. $\normal(0,\sigma^2)$ is the normal distribution
with mean $0$ and variance $\sigma^2$.

\begin{theorem}[Lindebergs Theorem] Suppose that in addition to
  \threecond, the triangular array satisfies \emph{Lindebergs
  Condition}:
\begin{equation}
\label{lindeberg}
\forall \eps>0,\, \lim_{i\toinf} 
\sum_{j=1}^{n_i} \E [X_{ij}^2
  \ind{|X_{ij}|>\eps}]=0
\end{equation}
Then, as $i\toinf$, $\law(S_i)\rightarrow \normal(0,1)$.
\end{theorem}
This theorem will be proved in Section ********* below.  For an
alternative proof using characteristic functions, see Billingsley
Sec. 27. 

The Lindeberg condition makes precise the sense in which the r.v.s
must be small.  It says that for arbitrarily small $\eps>0$, the
contribution to the total row variance from the terms with absolute
value greater than $\eps$ becomes negligible as you go down the rows.
The Lindeberg condition implies that the maximum contribution to the
variance from any of the individual terms in a row becomes negligible
as you go down the rows.  We see this as follows:
\begin{eqnarray}
  X_{ij}^2 &\leq& \eps^2 + X_{ij}^2 \ind{|X_{ij}|>\eps} \\
  \E X_{ij}^2 &\leq& \eps^2 + \E [X_{ij}^2 \ind{|X_{ij}|>\eps}] \\
  \E X_{ij}^2 &\leq& \eps^2 + \sum_{j} \E [X_{ij}^2
              \ind{|X_{ij}|>\eps}] \text{, which is independent of $j$, so...} \\
  \max_j \E X_{ij}^2 &\leq& \eps^2 + \sum_{j} \E [X_{ij}^2 \ind{|X_{ij}|>\eps}] \label{eq:maxVarIneq}
\end{eqnarray}
The Lindeberg condition says that, as we go down the rows
(i.e. $i\toinf$), the summation on the RHS tends to zero. Since 
inequality (\ref{eq:maxVarIneq}) holds for all $\eps >0$, we get
\begin{equation}
\label{eq:maxVarToZero}
\lim_{i\toinf} \max_j \E X_{ij}^2 = 0,
\end{equation}

A consequence of (\ref{eq:maxVarToZero}) and condition \ref{varsumone}
($\sum_{j} \E X_{ij}^2=1$ for all $i$) is that $n_i \toinf$ as
$i\toinf$.  Another consequence follows from the application of
(\ref{eq:maxVarToZero}) to Chebychevs inequality. We have for all $\eps >0$,
$$
\P (|X_{ij}| >\eps ) \leq \frac{\E (X_{ij}^2)}{\eps^2}
$$ 
Taking the maximum over $j$ and $i\toinf$, we get
\begin{equation}
\label{eq:unifAsympNeglig}
 \forall \eps >0,\,\lim_{i\toinf} \max_j \P(|X_{ij}|>\eps)=0.
\end{equation}

An array with property (\ref{eq:unifAsympNeglig}) is said to be \emph{uniformly
  asymptotically negligible (UAN)}, and there is a striking converse
to Lindebergs theorem:

\begin{theorem}[Fellers Theorem]
If a triangular array satisfies the \threecond\, and is UAN,
then $\law(S_i)\rightarrow \normal(0,1)$ [if and] only if Lindebergs condition
(\ref{lindeberg}) holds.
\end{theorem}
\begin{proof}
  See Billingsley, Theorem 27.4.
\end{proof}

For UAN arrays there is a more elaborate CLT with infinitely divisible
laws as limits - well return to this in later lectures.

Just note for now that
\begin{enumerate}
\item it is possible to get normal limits from UAN triangular arrays
  with infinite variances, and that 
\item it is possible to get a $\normal(0,\sigma^2)$ limit with
  $\sigma^2<1$ for an array satisfying \threecond.
\end{enumerate}

\section{The Lyapounov Condition}
A condition stronger than Lindebergs that is often easier
to check is the \emph{Lyapounov condition}:
\begin{equation}
  \label{eq:lyapounovCondition}
  \exists \delta>0 \text{ such that } \lim_{i\toinf} \sum_{j} \E
  |X_{ij}|^{2+\delta} = 0
\end{equation}

\begin{lemma}\label{lyapounovImpliesLindeberg} Lyapounovs condition implies Lindebergs condition.
\end{lemma}
\begin{proof}
  Fix any
  $\eps,\delta>0$.  For any r.v. $|X|>\eps$, we have
\begin{equation}
  X^2 = \frac{|X|^{2+\delta}}{|X|^\delta} \leq \frac{|X|^{2+\delta}}{\eps^\delta}
\end{equation}
Thus for any r.v. $X$ we have
$$
\E [X^2 \ind{|X|>\eps}] \leq \frac{\E {|X|}^{2+\delta }}{\eps ^\delta}.
$$
Take $X=X_{ij}$ to be the elements of our triangular array, and take
$\delta$ to be the value from Lyapounovs condition. Then we can sum
over $j$ on the RHS and take the limit as $i\toinf$ on both sides to
get Lindebergs condition.
\end{proof}

\begin{theorem}[Lyapounovs Theorem] If a triangular array satisfies
  \threecond\ and the Lyapounov condition
  (\ref{eq:lyapounovCondition}), then $\law(S_i)\rightarrow
  \normal(0,1)$.
\end{theorem}

\section{Preliminaries to the proof of Lindebergs Theorem}
The key property of the normal distribution is
\begin{theorem} \label{thm:sumNormalIsNormal}
  If $X$ and $Y$ are independent with $\law(X)=\normal(0,\sigma^2)$
  and $\law(Y)=\normal(0,\tau^2)$, then
  $\law(X+Y)=\normal(0,\sigma^2+\tau^2)$.
\end{theorem}
\begin{proofsketch}
Either
\begin{enumerate}
\item use the formula for the convolution of densities, or
\item use characteristic or moment generating functions, or
\item use the radial symmetry of the joint density function of i.i.d.
  $\normal(0, \sigma^2+\tau^2)$ r.v.s $U$ and $V$ to argue that
  $\law(U \sin \theta + V \cos \theta)=\law(U)$. Take
  $\sin(\theta)=\frac{\sigma^2}{\sigma^2+\tau^2}$
\end{enumerate}
\end{proofsketch}

The key characterization of convergence in distribution is
\begin{theorem}
  \label{thm:bndThirdMomConvergDetermclass}
  $\law(S_i)\rightarrow \law(Z)$ if and only if
  $\lim_{i\toinf} \E f(S_i) = \E f(Z)$ 
  for all $f\in \C ^3(-\infty,\infty)$, 
  the set of functions from \emph{reals} to \emph{reals} with three
  bounded continuous derivatives.
\end{theorem}
\begin{proof}
Mimic the proof of Theorem 6 on page 4.2 of notes on the convergence of probability laws.
????????????????????????????????????????????????????????????
\end{proof}

\section{\label{sec:proof-lyap-theor-delt-one}Proof of Lyapounovs Theorem  for $\delta=1$}
This illustrates the general idea and avoids a few tricky details.
With $n$ fixed, let $X_1,X_2,\ldots,X_n$ be independent random
variables, not necessarily identically distributed.  Suppose $\E
X_j=0$ and let $\sigma_j^2=\E (X_j^2)<\infty$. Then for
$S=X_1+\cdots+X_n$ we have $\var S = \sum_{j=1}^n \sigma_j^2$. Let
$\sigma^2=\var S$.
Note:
\begin{enumerate}
\item If $\law(X_j)$ is $\normal(0,\sigma_j^2)$, then $\law(S)$ is
  $\normal(0,\sigma^2)$ by Theorem \ref{thm:sumNormalIsNormal}.
\item Given independent r.v.s $X_1,\ldots,X_n$ with arbitrary
  distributions, we can always construct a new sequence
  $Z_1,\ldots,Z_n$ of \emph{normal} r.v.s with matching means and
  variances so that $$Z_1,Z_2,\ldots,Z_n,X_1,X_2,\ldots,X_n$$ are
  mutually independent. This may involve changing the basic
  probability space, but that doesnt matter because the distribution
  of $S$ is determined by the joint distribution of $(X_1,\ldots,X_n)$,
  which remains the same.
\end{enumerate}

Let
\begin{eqnarray*}
  S := S_0 &:=& X_1+X_2+X_3+\cdots+X_n,\\
  S_1 &:=& Z_1 + X_2+X_3+\cdots+X_n,\\
  S_2&:=& Z_1 + Z_2 + X_3 + \cdots + X_n,\\
  \vdots &\vdots&\vdots\\
  T:=S_n&:=&Z_1+Z_2+Z_3+\cdots+Z_n.
\end{eqnarray*}

We want to show that $\law(S)$ is ``close to $\law(T)$, which is
$\normal(0,\sigma^2)$, i.e., that $\E f(S)$ is ``close to $\E f(T)$
for all $f\in \C ^3(-\infty,\infty)$ with uniform bound $K$ on
$|f^{(i)}|,\;i=0,1,2,3$.

Clearly,
\begin{equation}
\label{eq:stTriangleIneq}  |\E f(S) - \E f(T)| \leq \sum_{j=1}^n |\E f(S_j) - \E f(S_{j-1})|.
\end{equation}
Let $R_j$ be the sum of the common terms in $S_{j-1}$ and $S_j$. Then
$S_{j-1}=R_j+X_j$ and $S_j=R_j+Z_j$. Note that by construction
\begin{equation}
\label{eq:rxrzIndependent}
  R_j \text{ and } X_j \text{ are independent, as are } R_j \text{ and
  }Z_j
\end{equation}

We need to compare $\E f(R_j + X_j)$ and $\E f(R_j+Z_j)$. By the
Taylor series expansion up to the third term,
$$
f(R_j+X_j)=f(R_j)+X_j
f^{(1)}(R_j)+\frac{X_j^2}{2!}f^{(2)}(R_j)+\frac{X_j^3}{3!}f^{(3)}(\alpha),
$$
where $\alpha \in (R_j,R_j+X_j)$. And the same is true with $Z_j$
instead of $X_j$. So, assuming that the $X$s have third moments, we
can take expectations in each of these identites and subtract the
resulting equations. We get the following:
\begin{enumerate}
  \item Since $\E X_j=0=\E X_j$, and by the independence of $X_j,R_j,$
  and $Z_j$ (\ref{eq:rxrzIndependent}), we have $\E (X_j
  f^{(1)}(R_j))=0=\E(Z_j f^{(1)}(R_j))$.
\item Since $\var X_j = \var Z_j$, and by (\ref{eq:rxrzIndependent}),
  we have $\E(X_j^2 f^{(2)}(R_j))=\E(Z_j^2 f^{(2)}(R_j))$.
\end{enumerate}
Thus the first and second order terms cancel, so we are left with the
last inequality below (the first two equalities summarize the previous
paragraphs):
\begin{eqnarray}
  |\E f(S_j) - \E f(S_{j-1})| &=&  |\E f(R_j+X_j) - \E f(R_j+Z_j)| \\
  \label{eq:thirdMomentEquality} &=&
  \left|\E \frac{X_j^3}{3!}f^{(3)}(\alpha)
    - \E \frac{Z_j^3}{3!}f^{(3)}(\alpha)\right|\\
  \label{eq:thirdMomentIneq}
  &\leq& \frac{K}{6}(\E |X_j|^3 +
  \E|Z_j|^3)
\end{eqnarray}
where $K$ is the bound on the derivatives of $f$. Now
\begin{eqnarray}
  \E |Z_j|^3  &=& 2 \int_0^\infty z^3 \frac{1}{\sqrt{2 \pi}\sigma_j}
  \exp\{-z^2/(2\sigma_j^2)\}\,dz \\
  &=&  2 \int_0^\infty \sigma_j^3 x^3 \frac{1}{\sqrt{2 \pi}}
  \exp\{-x^2/2\}\,dx\\
\label{eq:ZthirdMomentIscSigmaCubed}  &=& c \sigma_j^3
\end{eqnarray}
where
$$
c =   2 \int_0^\infty x^3 \frac{1}{\sqrt{2 \pi}}
\exp\{-x^2/2\}\,dx  =2\cdot\frac{2}{\sqrt{2\pi}} <\infty
$$
and since $(\E |X|^2)^\frac{1}{2} \leq (\E |X|^3)^\frac{1}{3}$ for
any random variable $X$, we have $\sigma_j^3 \leq \E |X_j|^3$ for
each $j$. Thus $\E|Z_j|^3 = c \sigma_j^3\leq c \E |X_j|^3$, for each
$j$. Applying this to (\ref{eq:thirdMomentIneq}), we get
$$
\frac{K}{6}(\E |X_j|^3 +
\E|Z_j|^3) \leq \frac{K}{6}\E |X_j|^3 (1+c).
$$
Now, from (\ref{eq:stTriangleIneq}), we get
\begin{equation}
\label{eq:newThirdMomentIneq}
  |\E f(S) - \E f(T)| \leq \frac{(c+1)K}{6} \sum_{j=1}^n \E |X_j|^3.
\end{equation}

To summarize, we have proved:
\begin{lemma}
  Let $X_1,\ldots,X_n$ be independent random variables with $\E X_j=0$ and
  $\E|X_j|^3<\infty$. Let $S=X_1+\cdots+X_n$ and let $T$ be
  $\normal(0,\sigma^2)$, where $\sigma^2$ is the variance of $S$. Then 
  (\ref{eq:newThirdMomentIneq}) holds for every function $f$ with
  three continuous derivatives bounded by $\pm K$.
\end{lemma}

Now check using \ref{thm:bndThirdMomConvergDetermclass} that
Lyapounovs theorem for $\delta=1$ is obtained by applying the Lemma
to the rows of a triangular array satisfying the \threecond and the
Lyapounov condition (\ref{eq:lyapounovCondition}) for $\delta=1$.

\section{Proof of Lindebergs Central Limit Theorem}
For Lyapounovs version of the CLT, we looked at a triangular array
$\{X_{ij}\}$ with $\E X_{ij}=0$, $\E X_{ij}^2=\sigma_{ij}^2$,
$\sum_{j=1}^{n_i}\sigma_{ij}^2=1$. Taking $S_i =
X_{i1}+X_{i2}+\cdots+X_{in}$, we saw that we could prove
$\law(S_i)\rightarrow \normal(0,1)$ assuming that $\lim_{i\toinf}
\sum_{k=1}^{n_i} \E |X_{ij}|^3=0$

This is a condition on third moments - we would like to see if a
weaker condition will suffice.  We used third moments in a Taylor
series expansion as follows:
\begin{equation}
\label{eq:taylorThreeTerms}
f(R+X)=f(R)+X f^{(1)}(R)+\frac{X^2}{2!}f^{(2)}(R)+\frac{X^3}{3!}f^{(3)}(\alpha),
\end{equation}
where $\alpha \in (R,R+X)$.

Roughly, without the third moments assumption, the above expression is
``bad when $X$ is large -- although the first two moments exist, we
might have $\E |X|^3 = \infty$.  The idea now is to use the form in
equation (\ref{eq:taylorThreeTerms}) when $X$ is small and to make use
of
\begin{equation}
  \label{eq:taylorTwoTerms}
  f(R+X)=f(R)+X f^{(1)}(R)+\frac{X^2}{2!}f^{(2)}(\beta)
\end{equation}
where $\beta \in (R,R+X)$, when $X$ is large.
  
Equating these expansions (\ref{eq:taylorThreeTerms}) and
(\ref{eq:taylorTwoTerms}) for $f(R+X)$, we get an alternative form for
the remainder in (\ref{eq:taylorThreeTerms}):
\begin{eqnarray}
  \frac{X^3}{6}f^{(3)}(\alpha) &=&
  \frac{X^2}{2}f^{(2)}(\beta)-\frac{X^2}{2}f^{(2)}(R)\\
  &=&   \frac{X^2}{2}[f^{(2)}(\beta)-f^{(2)}(R)]\ind{|X|>\eps}
  + \frac{X^3}{6}f^{(3)}(\alpha)\ind{|X|\leq \eps}
\end{eqnarray}
for $\eps>0$. Thus, for $f$ with $|f^{(i)}|\leq K$ for $i=2,3$, we get
\begin{eqnarray}
  \left| \frac{X^3}{6}f^{(3)}(\alpha) \right| &\leq&
  K X^2 \ind{|X|>\eps}   + \frac{K}{6}|X|^3 \ind{|X|\leq \eps}\\
\label{eq:newThirdTermTaylorBound}  &\leq&  K X^2 \ind{|X|>\eps}   + \frac{K}{6}\eps X^2,
 \end{eqnarray}
an alternative to the upper bound $\frac{K}{6} |X|^3$, which we used
in (\ref{eq:thirdMomentIneq}).

Now we return to the setup of section \ref{sec:proof-lyap-theor-delt-one}
and use our new result to get more refined bounds. From
(\ref{eq:stTriangleIneq}) and (\ref{eq:thirdMomentEquality}), we had
$$
|\E f(S) - \E f(T)| \leq
\sum_{j=1}^{n_j} \left|\E \frac{X_j^3}{6}f^{(3)}(\alpha)
  - \E \frac{Z_j^3}{6}f^{(3)}(\alpha)\right|
$$
Using the triangle inequality, the new bound for $X_j^3$
(\ref{eq:newThirdTermTaylorBound}), the assumption that $|f^{(3)}|<K$,
and $\E |Z_j|^3 = c \sigma_j^3$ (\ref{eq:ZthirdMomentIscSigmaCubed}), we
get
\begin{eqnarray}
|\E f(S) - \E f(T)| &\leq&
\sum_{j=1}^{n} \left[ K \E X_j^2 \ind{|X_j|>\eps}  +\frac{K}{6}\eps
  \E X_j^2\right]  + \sum_{j=1}^{n} \frac{K}{6}c \sigma_j^3 \\
&=& K \sum_{j=1}^{n} \E X_j^2 \ind{|X_j|>\eps}  +\frac{K}{6}\eps
  \sigma^2  +  \frac{cK}{6} \sum_{j=1}^{n} \sigma_j^3
\end{eqnarray}
As $i\toinf$ (i.e. we go down the rows of the triangular array), the
first term goes to zero by the Lindeberg condition. The last term
goes to zero since
$$
\sum_{j=1}^{n(i)} \sigma_{ij}^3 \leq \left(\max_{1\leq j\leq n(i)}
  \sigma_{ij}\right)\sum_{j=1}^{n(i)}
\sigma_{ij}^2=\sigma^2 \max_{1\leq j\leq n(i)} \sigma_{ij},
$$
which tends to zero by (\ref{eq:maxVarToZero}). Only
$\frac{K}{6}\eps \sigma^2$ remains, and letting $\eps\tozero$
finishes the argument.

\section{Applications}
Let $S_n = X_1+X_2+\cdots+X_n$ where $X_1,X_2,\ldots$ is a sequence of
independent, possibly non-identically distributed r.v.s, each with
mean $0$.  Let $\var X = \sigma_j^2$ and $\bs_n^2=\sigma_{j=1}^n
\sigma_j^2$. We want to know when
$\law(S_i/\bs_i)\rightarrow\normal(0,1)$. Check Lindebergs condition
for the triangular array $X_{ij}=X_j/\bs_i,\;j=1,2,\ldots,i$. Then
$S_i$ in the Lindeberg CLT is replaced by $S_i/\bs_i$, and the
Lindeberg condition becomes
\begin{eqnarray}
  \lim_{n \toinf} \sum_{j=1}^n \E \left[ \frac{X_j^2}{\bs_n^2}
  \ind{\left|\frac{X_j}{\bs_n} \right|>\eps}\right]
&=& 0,\text{ for all }\eps>0, \\
\text{i.e.}\;\;  \lim_{n \toinf} \frac{1}{\bs_n^2} \sum_{j=1}^n \E \left[
  X_j^2  \ind{|X_j|>\eps \bs_n} \right]
&=& 0,\text{ for all }\eps>0,
\end{eqnarray}  

Examples where the Lindeberg condition holds:
\begin{enumerate}
\item The i.i.d. case where $\bs_n^2 = n \sigma^2$:
  $$
  \frac{1}{n\sigma^2} \sum_{j=1}^n \E [X_j^2 \ind{|X_j|>\eps \sigma
    \sqrt{n}}]
  = \frac{1}{\sigma^2} \E [X_1^2 \ind{|X_1|>\eps \sigma \sqrt{n}}],
  $$
  and since $\E X_1^2 <\infty$, we can use the dominated convergence
  theorem to conclude that the Lindeberg condition holds.
\item Lyapounovs condition
  $$
  \lim_{n\toinf} \frac{1}{\bs_n^{2+\delta}} \sum_{j=1}^n \E
  |X_j|^{2+\delta}=0 \text{ for some } \delta>0
  $$
  implies Lindebergs condition. The proof of this is given
  (essentially) in Lemma \ref{lyapounovImpliesLindeberg}.
\item If $X_1,X_2,\ldots$ are uniformly bounded: $|X_j|\leq M$ for all
  $j$, and $\bs_n\uparrow \infty$. Fix $\eps >0$. For $n$ so large
  that $\bs_n \geq M/\eps $, we have
  $$
  \ind{|X_j|>\eps \bs_n} = \ind{|X_j|>M}=0\text{ for all }j.
  $$
  Hence the Lindeberg condition is satisfied.
\end{enumerate}
\end{document}
