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 \begin{document}


\

\bigskip

\centerline{\huge\kaishu 北~~京~~师~~范~~大~~学}

\bigskip\bigskip\bigskip

\centerline{\Huge\bf 本科生毕业论文(设计)}

\thispagestyle{empty}

\bigskip\bigskip\bigskip\bigskip

{\Large\heiti 毕业论文(设计)题目: \qquad\qquad\qquad\qquad}

\bigskip\bigskip

\centerline{\huge\heiti\bf \underline{\qquad
概率分布和特征函数\qquad}}


\vspace{1cm}

{\large
 \begin{eqnarray*}
 \mbox{\heiti 院~~\,~(系):} && \mbox{数学科学学院} \\
 \mbox{\heiti 专~~~~~~业:} && \mbox{概率论与数理统计} \\
 \mbox{\heiti 学~~~~~~号:} && \mbox{\qquad} \\
 \mbox{\heiti 学生姓名:} && \mbox{段丽琼} \\
 \mbox{\heiti 指导教师:} && \mbox{李增沪} \\
 \mbox{\heiti 职~~~~~~称:} && \mbox{教~~~授} \\
 \mbox{\heiti 所在单位:} && \mbox{数学科学学院}
 \end{eqnarray*}
}


\vspace{1cm}

\centerline{2005 年 5 月 8 日}


\newpage

\tableofcontents


\pagenumbering{roman}


\newpage


\pagenumbering{arabic}


\newpage\addcontentsline{toc}{section}{中英文摘要}

\section*{中英文摘要}

\songti

\bigskip\bigskip\bigskip\bigskip

{\narrower{\narrower

\centerline{\large\heiti\bf 概率分布和特征函数}

\bigskip

\centerline{段丽琼}

\bigskip

\noindent{\heiti 摘要:} 在这篇文章中,
我们对于概率分布的特征函数的基本理论给出了自封的处理.
主要的结论包括: 逆转公式, 唯一性定理及连续性 定理.
对于这些重要结论的应用, 我们也进行了简单的介绍.

\bigskip

\noindent{\heiti 关键词:} 概率空间, 随机变量, 分布函数, 特征函数,
逆转公式, 唯一性定理, 连续性定理.


\bigskip\bigskip\bigskip\bigskip


\centerline{\large\bf Distributions and Characteristic Functions}

\bigskip

\centerline{DUAN Liqiong}


\bigskip

\noindent{\bf Abstract.} In this work, we give a self-contained
treatment of the basic theory of characteristic functions of
probability distributions. The main results include: the inversion
theorem, the uniqueness theorem and the continuity theorem. Some
applications of those important results are also briefly discussed.

\bigskip

\noindent{\bf Keyword.} Probability space, random variable,
distribution function, characteristic function, inversion theorem,
uniqueness theorem, continuity theorem.

\par}\par}


\newpage




\centerline{\LARGE\bf Distributions and Characteristic Functions}

\bigskip\bigskip

\centerline{DUAN Liqiong}

\bigskip\bigskip


\section{Introduction}

A distribution function determines uniquely a probability measure on
the Euclidean space, which is certainly an important concept in
probability theory. The characteristic function of a probability
distribution is simply the Fourier transformation of the later.
Since the characteristic function has much better analytical
properties, it plays a particularly important role in the study of
probability distributions.


In this work, we give a self-contained treatment of the basic
theory of characteristic functions of probability distributions.
We recall the concepts of distributions and characteristic
functions and some of their elementary properties in section 2.
Then we prove the inversion and the uniqueness theorems in section
3, which show the importance of characteristic functions. The
complete convergence of distribution functions are discussed in
section 4. A necessary and sufficient condition for the complete
convergence, namely continuity theorem, is provided in section 5. The
materials are mainly based on refs.\ [1,2]. However, we provide
many more details in the proofs which were omitted in those
references.



\section{Definitions and basic properties}

We first recall some basic concepts in probability theory.

\bigskip

\noindent{\bf Definition 2.1 (Probability Space).}~ Let $\Omega$
be a nonempty set, $\mathcal F$ a $\sigma$-field of subsets of
$\Omega$, and $P$ a measure defined on $\mathcal F$ satisfying
$P(\Omega)=1$. Then the triple $(\Omega, {\mathcal F}, P)$ is
called a {\it probability space}. The elements of $\mathcal F$ are
called {\it events}. In particular, $\Omega$ is the {\it sure
event}. A singleton $\{\omega\}$, if it belongs $\mathcal F$, is
called an {\it elementary event}. The symbol $\emptyset$ denotes
the empty set and is known as the {\it null} or {\it impossible
event}.

\bigskip

\noindent{\bf Example 2.1.}\quad Let $\Omega=\{\omega_{j}: j \ge
1\}$ and let $\mathcal F$ be the $\sigma$-filed consisting of all
subsets of $\Omega$. Let $\{p_{j}: j \ge 1\}$ be any sequence of
nonnegative real numbers satisfying $\sum_{j=1}^{\infty} p_
 {j}=1$. Define $P$ on $\mathcal F$ by
 $$
P(E)=\sum_ {\omega_{j}\in E}p_ {j} \qquad E\in \mathcal F.
 $$
Then $P$ is a probability measure on $(\Omega, \mathcal F)$, and
$(\Omega, \mathcal F, P)$ is a probability space.

\bigskip

\noindent{\bf Definition 2.2 (Random Variable).}~ Let ($\Omega$,
$\mathcal F$, $P$) be a probability space. A real-valued function
$X$ defined on $\Omega$ is called a {\it random variable} if
 $$
X ^ {-1}(E)= \{ \omega \in \Omega: X(\omega) \in E \}\in \mathcal
F
 $$
for all $E \in \mathcal B$, where $\mathcal B$ is the
$\sigma$-field of Borel sets in $\Re = (-\infty, \infty)$. That
is, a random variable $X$ is a measurable transformation of
($\Omega$, $\mathcal F$, $P$) into ($\Re$, $\mathcal B$).


\bigskip

\noindent{\bf Definition 2.3 (Probability Distribution).}~ A random
variable $X$ defined on ($\Omega$, $\mathcal F$, $P$) induces a
probability measure $P_ X$ on $\mathcal B$ by
 $$
P_{ X} (E)= P\{X^{-1}(E)\} \qquad (E \in \mathcal B ),
 $$
which is called the {\it probability distribution} or, simply, the
{\it distribution} of $X$.


\bigskip

\noindent{\bf Definition 2.4 (Distribution Function).}~ For every
$x\in \Re$ set
 $$
F_{ X} (x)=P_X(- \infty,x \rbrack = P\{ \omega \in \Omega:
X(\omega)\le x \}.
 $$
We call $F_ X$ the distribution function of the random variable $X$.

\bigskip

In the following, we write $\{X \le x \}$ for the event $ \{
\omega \in \Omega:X(\omega)\le x \}$. We first prove the following
elementary property of distribution functions.


\bigskip

\noindent{\bf Theorem 2.1.}~ The distribution function $F$ of a
random variable $X$ is a nondecreasing, right-continuous function
on $\Re$ which satisfies
 $$
F(-\infty)=\lim_{x\to-\infty} F(x)=0
 $$
and
 $$
F(+\infty)=\lim_{x\to+\infty}F(x)=1.
 $$

\bigskip

\noindent{\bf Proof.}~ For every $x\in\Re$ and $h>0$,
 $$
F(x+h)-F(x)=P\{x<X \le x+h\}\ge 0,
 $$
so that $F$ is nondecreasing.

Next, let $\{h_{n}\}$ be a sequence of real numbers such that $0<h_
{n}\downarrow{0}$, as $n\to\infty$. Then, for every
 $$
F(x+h_ {n})-F(x)=P\{x<X\le x+h_ {n}\}.
 $$
It follows from the continuity property of $P$ that
 $$
 \lim_{n\to\infty}[F(x+h_{n})-F(x)]=0,
 $$
and hence that $F$ is right-continuous.

Finally, for every $N \ge 1$ we have
 $$
F(N)-F(-N)=P\{-N<X\le N\}.
 $$
Taking the limit of both sides as $N\to \infty$ and using the
continuity property once again, we conclude that
 $$
F(+\infty)-F(-\infty)=1.
 $$
But since $0\le F(x)\le 1$ for every $x\in \Re$, it follows that
$F(-\infty)=0$ and $F(+\infty)=1$.


\bigskip

\noindent{\bf Definition 2.5 (Characteristic Function).}~ Any
random variable $X$ on a probability space ($\Omega$, $\mathcal
F$, $P$) has an associated characteristic function $\varphi_
 {X}(t)$ defined for all real $t$ by
 $$
\varphi(t)=\varphi_{ X}(t)=\int^{\infty}_{-\infty}e^{itx}\
dF(x),\qquad t\in\Re.
 $$
Note that the module of integrand on the right-hand side is no
more than 1. Then for any probability distribution, its
corresponding characteristic function is determined uniquely.


\bigskip

\noindent{\bf Example 2.2.}~ For the discrete distribution, its
corresponding characteristic function is
 $$
\varphi(t)=\sum_{k}e^{itx_ {k}}p_{k},\qquad {\rm where}\ p
_{k}=P\{X=x_{k}\}.
 $$
We can easily get the characteristic function corresponding to
Bernoulli distribution
 $$
\varphi(t)=q+pe^{it};
 $$
the characteristic function corresponding to Poisson distribution
 $$
\varphi(t)=\sum_ {k=0}^{\infty}e^{itk}\frac{\lambda
^{k}}{k!}e^{-\lambda}=e^{\lambda (e^{it}-1)};
 $$
the characteristic function corresponding to geometric
distribution
 $$
\varphi(t)=\sum_
 {k=1}^{\infty}e^{itk}q^{k-1}p=\frac{pe^{it}}{1-qe^{it}}.
 $$


\bigskip

\noindent{\bf Definition 2.6 (Probability Density Function).}~
Suppose $X$ is a random variable of the probability space. If
there exists a function $f(x)$ satisfying

(1) \qquad $f(x)\ge 0$;

(2) \qquad $\int_{-\infty}^{\infty}f(x)\ dx=1$, \\

\noindent and $F(x)=\int_ {-\infty}^{x}f(t)\ dt$, then we call
$f(x)$ is the corresponding probability density function, and $X$ is
continuous.

\bigskip

\noindent{\bf Proposition 2.1.}~ The characteristic function
$\varphi$ defined on $\Re$ has the following properties:

(1)\quad $\vert \varphi(t)\vert\le\varphi(0)=1$;

(2)\quad $\varphi(-t)= \overline {\varphi}(t)$, where
$\overline{\varphi}$ is the complex conjugate of $\varphi$;

(3)\quad $\varphi_{aX+b}=\varphi_{X}(at)e^{ibt}$, for every t$\in$
$\Re$, a,b $\in$ $\Re$;

(4)\quad $\varphi$ is uniformly continuous on $\Re$, that is,
 $$
\vert \varphi(t+h)-\varphi(t) \vert \le \int^{\infty}
_{-\infty}\vert e^{ixh}-1 \vert\ dF(x) \to 0 , \quad {\rm as}\ h
\rightarrow 0.
 $$

\noindent{\bf Proof.}~ Clearly, (1), (2) and (3) follow from the
definition. We show that $\varphi$ is uniformly continuous on
$\Re$. Let $t\in$ $\Re$ and $h \in$ $\Re$. Then
 \begin{eqnarray*}
\vert \varphi(t+h)-\varphi(t)\vert
 &=&
\Big\vert \int^{\infty}_{-\infty}e^{i(t+h)x}-e^{itx}\
dF(x)\Big\vert\\
 &\le&
 \int_{-\infty}^{\infty}\vert e^{ihx}-1\vert\ dF(x)\\
 &=&
\int^{\infty}_ {-\infty}\vert(\cos \ hx-1)+i\sin \ hx\vert\ dF(x)\\
 &=&
\int_{-\infty}^{\infty}\sqrt{2-2\cos\ hx}\ dF(x)\\
 &=&
2\int^{\infty}_{-\infty}\Big\vert \sin \frac{hx}{2}\Big\vert\ dF(x),
 \end{eqnarray*}
where the right side is independent of $t$. Using the
Lebesgue-dominated convergence theorem, we see that
 $$
\sup_{t\in \Re} \vert \varphi(t+h)-\varphi(t)\vert \to 0\qquad {\rm
as}\ h \to 0.
 $$


\section{Inversion Theorem}


Now we prove two results which show the importance of
characteristic function in probability theorem. The first of these
is the inversion theorem, which enables us to compute the
distribution function from its characteristic function. The second
result is the uniqueness theorem, which allows us to work with
characteristic functions instead of distribution functions in the
study of limit theorems.

\bigskip

\noindent{\bf Theorem 3.1 (Inversion Theorem). } Let $F$ be a
distribution function, and let $\varphi$ be the associated
characteristic function. Then the relation
 $$
F(a+h)-F(a-h)=\lim_{T\to \infty} \frac{1}{\pi}\int_{-T}^{T}
\frac{\sin\ ht}{t}e^{-ita}\varphi(t)\ dt \eqno(1)
 $$
holds for $a\in\Re$ and $h > 0$ whenever the points $a \mp h \in
C_
 {F}$.

\bigskip

\noindent{\bf Proof. } For the proof we first note that the integral
$\int^{x}_{0}(\sin\ t/t)\ dt$ exists, is bounded for all $x> 0$, and
tends to the limit $\pi$/2 as $x\to $+ $\infty$. We set
 $$
\theta(h,T)=\frac{2}{\pi}\int_{0}^{T}\frac{\sin\ ht}{t} \ dt.
 $$
Clearly $\theta(h,T)$ is uniformly bounded for all $h\in \Re$ and
all $T>0$ such that
 $$
\theta(-h,T)=-\theta(h,T).
 $$
Moreover, the relation
 $$
\lim_{T\to \infty}\theta(h,T)=\left \{
 \begin{array}{cc}
 1 &\quad \hbox {if $h>0$,} \\
 0 &\quad \hbox{if $h=0$,} \\
 -1 &\quad \hbox{if $h<0$,}
 \end{array}
 \right.
 $$
holds. For $T>0$ we set
 \begin{eqnarray*}
I_{T}&=&\frac{1}{\pi}\int_{-T}^{T}\frac{\sin\
ht}{t}e^{-ita}\varphi(t)\ dt\\
&=&\frac 1{\pi}\int^{T}_{-T}\frac{\sin\ ht}{t}e^{-ita} \ \Big
[\int^{\infty}_ {-\infty}e^{itx}\ dF(x)\Big]\ dt,
 \end{eqnarray*}
and, noting that for $h>0$
$$
 \Big\vert\frac{\sin\ ht}{t}e^{it(x-a)}\Big\vert \le h,
 $$
we conclude that $I_{T}$ exists and is finite. Using Fubini's
theorem to change the order of integration, we obtain
 \begin{eqnarray*}
I_{T}&=&\frac{1}{\pi}\int^{\infty}_ {-\infty}\ dF(x)\int^{T}_
{-T}\frac{\sin\ ht}{t}e^{it(x-a)}\ dt\\
&=&\frac{2}\pi\int^{\infty}_{-\infty}\ dF(x)\int^{T}_ {0}\frac{\sin\
ht}t\cos[(x-a)t]\ dt\\
&=&\int^{\infty}_ {-\infty}g(x,T) \ dF(x),
 \end{eqnarray*}
where
 \begin{eqnarray*}
 g(x,T)&=&\frac{2}\pi\int^{T}_ {0}\frac{\sin\ ht}t\cos\ [(x-a)t]\ dt\\
 &=&\frac{1}\pi\int^{T}_{0}\frac{\sin\ [(x-a+h)t]}t\
 dt-\frac{1}\pi\int^{T}_{0}\frac{\sin\ [(x-a-h)t]}t\ dt\\
 &=&\frac{1}2\theta(x-a+h,T)-\frac{1}2\theta(x-a-h,T).
 \end{eqnarray*}
Clearly $g(x,T)$ is uniformly bounded for all $x\in\Re$ and all $T>
0$. Moreover,
 $$
\lim_{T\to \infty}g(x,T)=\left\{
\begin{array}{cc}
0 &\quad \hbox{if $x<a-h$,}\\
\frac1{2}&\quad \hbox{if $x=a-h$,}\\
1 &\quad\quad\quad\qquad \hbox{if $a-h<x<a+h$,}\\
\frac1{2}&\quad \hbox{if $x=a+h$,}\\
0 &\quad \hbox{if $x>a+h$.}
\end{array}
\right.
 $$
Using the Lebesgue dominated convergence theorem, we can take the
limit as $T \to \infty$ under the integral sign and thus obtain
\begin{eqnarray*}
\lim_{T\to \infty}I_{T} &=&\int^{\infty}_{-\infty}\Big[\lim
_{T\to\infty}g(x,T)\Big]\ dF(x)\\
&=&\int^{a+h}_{a-h}\ dF(x)=F(a+h)-F(a-h),
 \end{eqnarray*}
 since $a\mp h$
$\in $$C_{F}$. This completes the proof.
\bigskip

\noindent{\bf Corollary 3.1.}~ Let $a, b$ $\in$ $C_{F}$ be such that
$a<b$. Then
 $$
F(b)-F(a)=\lim_{T\to\infty}\frac 1{2 \pi}\int^{T}
{-T}\frac{e^{-ita}-e^{-itb}}{it}\varphi(t)\ dt.
 \eqno(2)
 $$

\bigskip

\noindent{\bf Proof.}\quad Letting $a=(b'+a')/2$, $h=(b'-a')/2$ in
(1), we obtain
 $$
F(b')-F(a')=\lim_{T\to\infty}\frac{1}{\pi}\int_{-T}^{T}\frac{\sin\
\Big(\frac{b'-a'}2\Big)t}{t}\
e^{-it\Big(\frac{b'+a'}2\Big)}\varphi(t)\ dt.
 $$
Using Euler Formula, we can get
 \begin{eqnarray*}
 & & \sin\ \Big(\frac{b'-a'}2\Big)t\ e^{-it\Big(\frac{b'+a'}2\Big)}\\
 &=&\sin\ \Big(\frac{b'-a'}2\Big)t\ \cos\ \Big(\frac{b'+a'}2\Big)t -
 i\sin\ \Big(\frac{b'-a'}2\Big)t\ \sin\ \Big(\frac{b'+a'}2\Big)t\\
 &=&\frac1{2}\Big(\sin\ b't - \sin\ a't +i\cos\ b't - i\cos\ a't\Big)\\
 &=&\frac{e^{-ita'}-e^{-itb'}}{2i}.
 \end{eqnarray*}
This completes the proof.

 \bigskip

\noindent{\bf Corollary 3.2 (Uniqueness Theorem).}\quad Let $F_{1}$
and $F_{2}$ be two distribution functions with characteristic
functions $\varphi_{1}$ and $\varphi_{2}$, respectively. Suppose
that $\varphi_{1}(t)=\varphi_{2}(t)$ for every $t\in\Re$. Then
$F_{1} \equiv F_{2}$.

\bigskip

\noindent{\bf Proof.}\quad Let $a, b$ $\in$ $C_ {F_ {1}} \cap C_ {F_
{2}}$ be such that $a<b$. Then (2) yields
 $$
F_{1}(b)-F_{1}(a)=F_{2}(b)-F_{2}(a).
 $$
Letting $a\to\-\infty$, we obtain $F_{1}(b)=F_{2}(b)$, so that
$F_{1}=F_{2}$ on $C_{F_{1}} \cap C_{F_{2}}$. Hence $F_{1}\equiv F_
{2}$.

\bigskip

\noindent{\bf Corollary 3.3 (Convolution Theorem).}~ Let $F$,$F
_{1}$,$F_{2}$ be three distribution functions with characteristic
functions $\varphi$,$\varphi_{1}$,$\varphi_{2}$, respectively. Then
 $$
F=F_{1} * F_{2}(x) \Longleftrightarrow \varphi= \varphi_ {1}\varphi
_{2},
 $$
where
 $$F_{1}*F_{2}= \int_{-\infty}^{\infty}F_{1}(x-y)\ dF_
{2}(y)\qquad x\in
 \Re,
 $$
is said to be the convolution of $F_{1}$ and $F_{2}$.


\bigskip

\noindent{\bf Proof.}\quad Suppose that $F=F_{1}*F_{2}$. Let $[a,b]$
be an arbitrary closed bounded interval in $\Re$, and let
 $$
a=x_{0,N}<x_{1,N}<...<x_{\nu N,N}=b
 $$
be a sequence of subdivisions of $[a,b]$ such that
 $$
\triangle_{N}=\max_{1\le \nu \le \nu N}(x_{\nu,N}-x_{\nu-1,N})\to
0\qquad {\rm as} \ N\to \infty.
 $$
Using the definition of the Riemann-Stieltjes integral, we obtain
for $t\in \Re$
 \begin{eqnarray*}
 &&\int_{a}^{b}e^{itx}\ dF(x)=\lim_{N\to\infty}\sum_{\nu=1}^{\nu
 N}e^{itx_{\nu,N}}[F(x_{\nu,N})-F(x_{\nu-1,N})]\\
 &=&\lim_{N\to\infty}\int_{-\infty}^{\infty}\sum_{\nu=1}^{\nu
 N}e^{it(x_{\nu,N}-y)}[F_{1}(x_{\nu,N}-y)-F_{1}(x_{\nu-1,N}-y)]e^{ity}\
 dF_{2}(y).
 \end{eqnarray*}
We note that
 \begin{eqnarray*}
 &&\bigg\vert\sum_{\nu=1}^{\nu N}e^{it(x  {\nu
 ,N}-y)}[F_{1}(x_{\nu,N}-y)-F_{1}(x_{\nu-1,N}-y)]\bigg\vert\\
 &\le& \sum_{\nu=1}^{\nu N}[F_{1}(x_{\nu,N}-y)-F_{1}(x_{\nu-1,N}-y)]\\
 &=&F_{1}(b)-F_{1}(a)\le 1,
 \end{eqnarray*}
so that the integrand on the right-hand side of ($\star$) is
uniformly bounded. Using the Lebesgue dominated convergence theorem,
we see that for $t\in\Re$
 $$
\int_{a}^{b}e^{itx}\ dF(x)=\int_{-\infty}^{\infty}\bigg [\int_
{a-y}^{b-y}e^{itx} \ dF_{1}(x)\bigg] e^{ity}\ dF_{2}(y).
 $$
Next we take the limits on both sides as $a\to-\infty$ and
$b\to+\infty$ to obtain $\varphi=\varphi_{1}\varphi_{2}$.


Conversely, suppose that $\varphi=\varphi_{1}\varphi_{2}$. Let $G=F
_{1}*F_{2}$, and let $\theta$ be the characteristic function of $G$.
Then from what has been shown above it follows that $\theta
=\varphi_{1}\varphi_{2}=\varphi$. It follows at once from the
uniqueness theorem that $G=F$.


\bigskip

\noindent{\bf Theorem 3.2 (Fourier Inversion Theorem). } Suppose
that $\varphi$ is absolutely integrable on $\Re$. Then the
corresponding distribution function $F$ is absolutely continuous.
Moreover, the probability density function $f=F'$ exists, is
bounded and uniformly continuous on $\Re$, and is given by
 $$
 f(x)=\frac1{2\pi}\int ^{\infty}_{- \infty} e^{-itx} \varphi(t)\ dt,\qquad
 x \in \Re. \eqno(3)
 $$

\noindent{\bf Proof. } Since $\int^{\infty}_{-\infty} \vert
\varphi(t) \vert\ dt <\infty$, the integrand on the right-hand side
of (1) is dominated by an absolutely integrable function.
Consequently we can write
 $$
F(x+h)-F(x-h)=\frac{1}\pi\int^{\infty}_{-\infty}\frac{\sin\ ht}t
e^{-itx}\varphi(t)\ dt \eqno(4)
 $$
whenever $ x\mp h \in C_{F}$. Taking the limits of both sides as $ h
\downarrow 0$, we obtain
 $$
F(x)-F(x-0)=0.
 $$
It follows that $F$ is continuous on $\Re$. Using a similar argument
on
 $$
\frac{F(x+h)-F(x-h)}{2h}=\frac{1}{2\pi}\int_{-\infty}^{\infty}
\frac{\sin\ ht}{ht}e^{-itx}\varphi(t)\ dt,
 $$
we conclude that $F$ is differentiable everywhere on $\Re$ and for
$x\in\Re$ we have
 \begin{eqnarray*}
F'(x)&=&\frac 1{2\pi}\int^{\infty}_{-\infty}\lim_{h \downarrow
0}\frac{\sin\ ht}{ht}e^{-itx}\varphi(t)\ dt\\
 &=&
\frac 1{2\pi}\int^{\infty}_{-\infty}e^{-itx}\varphi(t)\ dt.
 \end{eqnarray*}
Hence $F$ is absolutely continuous, and $f=F'$ exists and is bounded
on $\Re$. Finally we note that
 $$
\vert f(x+h)-f(x)\vert\le\frac1{\pi}\int^{\infty}_{-\infty}
\Big\vert\sin\ \frac{ht}2\Big\vert\vert\varphi(t)\vert\ dt,
 $$
so that the uniform continuity of $f$ follows from the Lebesgue
dominated convergence theorem.



\section{Convergence of Distributions}

First we prove the weak compactness criterion due to Helly.

\bigskip

\noindent{\bf Theorem 4.1 (Helly's Theorem).}\ quad Let $\{F_{n}\}$
be a sequence of uniformly bounded, nondecreasing, right-continuous
functions. Then $\{F_{n}\}$ contains a subsequence $\{F_{nk}\}$
which converges weakly to a bounded, nondecreasing, right-continuous
function.

\bigskip

For the proof of Theorem 4.1 we need the following two lemmas.

\bigskip

\noindent{\bf Lemma 4.1. }\quad Let $\{F_{n}\}$ be as above.
Suppose, in addition, that $\{F_{n}\}$ converges to a function $G$
pointwise on a set $D$ which is everywhere dense in $\Re$. Then
$\{F_{n}\}$ converges to $G$ at all continuity points of $G$.

\bigskip

\noindent{\bf Proof.}\quad Let $x,x',x''\in\Re$ be such that
$x'<x<x''$. Then for every $n\ge 1$
 $$
F_{n}(x')\le F_{n}(x) \le F_{n}(x'').
 $$
Let $x',x''\in D$. Letting $n\to \infty$, we obtain
 $$
G(x')\le \liminf_ {n\to\infty}F_{n}(x)\le
\limsup_{n\to\infty}F_{n}(x)\le G(x'').
 $$
Now letting $x'\uparrow x$ and $x''\downarrow x$, we obtain the
required result.

\bigskip

\noindent{\bf Lemma 4.2.}\quad Let $\{F_{n}\}$ be as above, and let
$D=\{x_{\nu}:\nu \ge 1\}$ be a countable subset of $\Re$. Then
$\{F_{n}\}$ contains a subsequence $\{F_{nk}\}$ which converges at
every point of $D$.

\bigskip

\noindent{\bf Proof.}\quad Consider the set of real numbers $\{F_
{n}(x_{1}),n\ge 1 \}$. Clearly this is a bounded set, and hence,
accoring to the Bolzano-Weierstrass theorem, $\{F_{n}\}$ contains a
subsequence $\{F^{(1)}_{n}\}$ such that $\{F^{(1)}_{n}(x_{1})\}$
converges to some real number, say $a_{1}$. Next consider the
sequence of real numbers $\{F_{n}^{(1)}(x_{2}),n \ge 1 \}$. If we
proceed as above, it follows that there exists a subsequence $\{F_
{n}^{(2)}\}\subset \{F_{n}^{(1)}\} $ such that $\{F_{n}^{(2)}(x_
{2})\}$ converges to a real number, say $a_{2}$. Clearly $F
_{n}^{(2)}(x_{1})\to a_{1}$ as $n\to \infty$. We continue this
process indefinitely and thus obtain a countable set of convergent
subsequences $\{F_{n}^{(k)}\}$ such that
 $$
\{F_{n}\}\supset \{F_{n}^{(1)}\}\supset \{F_{n}^{(2)}\} \supset
...\supset \{F_{n}^{(k)}\} \supset ...\ ,
 $$
and, moreover, the relation
 $$
\lim_{n\to \infty}F_{n}^{(k)}(x_{k})=a_{k}
 $$
holds for all $k \ge 1$. Now consider the diagonal sequence $\{F_
{n}^{(n)},n\ge 1\}$. We show that, for every $x\in D$, $\{F_
 {n}^{(n)}(x),n\ge 1\}$ converges. Let $k\ge 1$ be fixed, and
consider the point $x_{k}\in D$. Then $\{F_{n}^{(n)}(x_{k}),n\ge
k\}$ clearly is a subsequence of $\{F_{n}^{(k)}(x_{k}),n\ge 1\}$, so
that $F_{n}^{(n)}(x_{k})\to a_{k}$ as $n\to \infty$. This completes
the proof of Lemma 4.2.


\bigskip

\noindent{\bf Proof of Theorem 4.1.}\quad Let $D=\{x_{\nu}:\nu \ge
1\}$ be a countable set which is everywhere dense in $\Re$. Then
from Lemma 4.2 $\{F_{n}\}\supset\{F_{nk}\}$ such that $\{F_{nk}\}$
converges at every point of $D$. Define $G$ on $\Re$ by setting
 $$
G(x_{\nu})=\lim_{k\to\infty}F_{nk}(x_{\nu})\qquad {\rm for}\
x_{\nu}\in D
 $$
and
 $$
G(x)=\sup_{{x {\nu}\le x}\atop {x_{\nu}\in D}}G(x_{\nu})\qquad {\rm
for}\ x\in\Re-D
 $$
Clearly $G$ is bounded. We show that $G$ is nondecreasing.

Case 1.\quad Let $x\in D$ and $y\in D$ such that $x<y$. Clearly
$G(x)\le G(y)$ from the definition of $G$.

Case 2.\quad Let $x\notin D$ and $y\in D$ be such that $x<y$. Let
$x_{\nu}\in D$ be such that $x_{\nu}<x$. Then $x_{\nu}<y$ so that
$G(x_{\nu})\le G(y)$ from Case 1. Hence
 $$
G(x)=\sup_{{x_{\nu}<x} \atop {x_{\nu}\in D}}G(x_{\nu})\le G(y).
 $$

Case 3.\quad Let $x\in D$ and $y\notin D$ such that $x<y$. Clearly
 $$
G(y)=\sup_{{x_{\nu}<y}\atop {x_{\nu}\in D}}G(x_{\nu})\ge G(x).
 $$

Case 4.\quad Let $x\notin D$ and $y\notin D$ with $x<y$. Then
 $$
\{x_{\nu} \in D:x_{\nu}<x\}\subset \{x_{\nu}\in D:x_{\nu}<y\}
 $$
so that $G(x)\le G(y)$.

We note that $\{F_{nk}\}$ converges to $G$ on $D$. In view of Lemma
4.1, $\{F_{nk}\}$ converges to $G$ at all continuity points of $G$.


Next we define $F$ by setting
 \begin{eqnarray*}
F(x)=G(x) &&\qquad {\rm if} \ x \in C  {G},\\
F(x)=G(x+0)&&\qquad {\rm if} \ x \notin C  {G}.
 \end{eqnarray*}
Clearly $F$ is a bounded, nondecreasing, right-continuous function,
and, moreover, $F_{nk}\stackrel {\omega} {\to} F$. This completes
the proof of Theorem 4.1.


\bigskip

\noindent{\bf Theorem 4.2 (Helly-Bray Theorem).}\quad Let $g$ be a
real-valued, continuous function defined on a closed interval
$[a,b]$, and let $\{F_{n}\}$ be a sequence of uniformly bounded,
nondecreasing, right-continuous functions which converges weakly to
some function $F$ on $[a,b]$, where $a,b$ $\in$ $C_{F}$ . Then
 $$
\lim_{n\to\infty}\int_{a}^{b}g\ dF_{n}=\int_{a}^{b}g\ dF.
 $$


\bigskip

\noindent{\bf Proof.}\quad It follows from Theorem 4.1 that $F$ is
bounded and nondecreasing. Let $\mu_{n}$ and $\mu$ denotes the
finite measures corresponding to $F_{n}$ and $F$, respectively. Let
$a=x_{0,N}<x_{1,N}<...<x_{\nu N,N}=b$ be a sequence of subdivisions
of $[a,b]$ such that $$\triangle_{N}= \max_{1\le \nu\le \nu
N}(x_{\nu,N}-x_{\nu-1,N})\to 0 \quad {\rm as} \ N\to \infty.$$ Set
 $$
g_{N}(x)=\sum_{\nu=1}^{\nu N}g(x_{\nu,N})\chi  {(x_{\nu-1,N},x
_{\nu,N}]}(x),\qquad x\in[a,b].
 $$
Then $g$ can be uniformly approximated on $[a,b]$ by the sequence
$\{g_{N}\}$, that is,
 $$
\sup_{a\le x\le b}\vert g_{N}(x)-g(x)\vert \to 0\qquad {\rm as} \ N
\to \infty.
 $$
Since $\{g_{N}\}$ is uniformly bounded, we have from the Lebesgue
dominated convergence theorem
 $$
\int_{a}^{b}g\ dF_{n}=\int_{(a,b]}g\
d\mu_{n}=\lim_{N\to\infty}\int_{(a,b]}g_{N}\ d\mu_{n}
 $$
for every $n\ge 1$. Similarly
 $$
\int_{a}^{b}g\ dF=\int_{(a,b]}g\ d\mu =
\lim_{N\to\infty}\int_{(a,b]}g_{N}\ d\mu.
 $$
We first show that for every $N \ge 1$,
 $$
\lim_{n\to \infty}\int_{(a,b]}g_{N}\ d\mu_{n}=\int_{(a,b]}g_{N}\
d\mu.
 $$
For this purpose select the points of subdivision
$x_{\nu,N}(\nu=1,2,...,\nu {N};N=1,2,...)$ to be the continuity
points of $F$. Since $F_{n} \stackrel {\omega}{\to}F$, we have
 \begin{eqnarray*}
\lim_{n\to\infty}\mu_{n}(x_{\nu-1,N},x_{\nu,N}]&=&
\lim_{n\to\infty}[F_{n}(x_{\nu,N})-F  {n}(x_{\nu-1,N})]\\
 &=&
F(x_{\nu,N})-F(x_{\nu-1,N})\\
 &=&
\mu(x_{\nu-1,N},x_{\nu,N}).
 \end{eqnarray*}
Hence
 \begin{eqnarray*}
\lim_{n\to \infty}\int_{(a,b]}g_{N}\ d\mu_{n}&=& \lim
_{n\to\infty}\sum_{\nu=1}^{\nu
 N}g(x_{\nu,N})\mu_{n}(x_{\nu-1,N},x_{\nu,N}]\\
 &=&
\sum_{\nu=1}^{\nu N}g(x_{\nu,N})\mu(x_{\nu-1,N},x_{\nu,N}]\\
 &=&
\int_{(a,b]}g_{N} \ d \mu.
 \end{eqnarray*}
Writing $M_{N}=\sup_{a\le x\le b}\vert g_{N}(x)-g(x)\vert$, we see
that
 \begin{eqnarray*}
 & &\bigg \vert\int_{a}^{b}g\ dF_{n}-\int_{a}^{b}g\ dF \bigg\vert
 \\
 &\le&
\int_{(a,b]}\vert g-g_{n}\vert \
d\mu_{n}+\bigg\vert\int_{(a,b]}g_{N}\ d\mu_{n}-\int_{(a,b]}g_{N}\
d\mu \bigg\vert+\int_{(a,b]}\vert g-g_{N}\vert\ d\mu\\
 &\le&
M_{N}\mu_{n}(a,b]+\bigg\vert\int_{a}^{b}g_{N}\
d\mu_{n}-\int_{(a,b]}g_{N}\ d\mu\bigg \vert+M_{N}\mu(a,b].
 \end{eqnarray*}
Since $M_{N}\to 0$ as $N\to\infty$, by letting $n\to\infty$ and then
$N\to\infty$ we obtain the result.

\bigskip

\noindent{\bf Theorem 4.3 (Extended Helly-Bray Theorem).}\quad Let
$g$ be a bounded, real-valued, continuous function on $\Re$. Let
$\{F_{n}\}$ be a sequence of uniformly bounded, nondecreasing,
right-continuous functions which converges completely to some
function $F$ on $\Re$. Then
 $$
\lim_{n\to\infty}\int_{-\infty}^{\infty}g \
dF_{n}=\int_{-\infty}^{\infty}g \ dF.
 $$

\bigskip

\noindent{\bf Proof.}\quad Clearly $F$ is a bounded, nondecreasing,
and right-continuous. Let $a,b \in C_{F}$ such that $a<b$, and let
 $$
M=\sup_{x\in\Re}\vert g(x)\vert.
 $$
Then
 \begin{eqnarray*}
 & &\Big \vert \int_{-\infty}^{\infty}g \ dF_{n}-
 \int_{-\infty}^{\infty}g\ dF \Big \vert \\
 &\le&
\Big \vert \int_{-\infty}^{\infty}g\ dF_{n}- \int^{b}_{a}g
 \ dF_{n} \Big \vert \\
 &+&
\Big \vert \int^{b}_{a} g \ dF_{n}-\int_{a}^{b}g\ dF \Big \vert
+\Big \vert \int_{a}^{b}g\ dF -\int_{-\infty}^{\infty}g\
dF\Big \vert \\
 &\le&
M[F_{n}(a)-F_{n}(-\infty)+F_{n}(+\infty)-F_{n}(b)]+\Big \vert
\int_{a}^{b}g\ dF_{n}-\int_{a}^{b}g\ dF\Big \vert \\
 &+&
M[F(a)-F({-\infty})+F(+\infty)-F(b)].
 \end{eqnarray*}

Since $\lim_{n\to\infty}F_{n}(\pm \infty)=F(\pm \infty)$,
$\lim_{n\to\infty} F_{n}(a)=F(a)$, and $\lim_{n\to\infty}
F_{n}(b)=F(b)$, the first and the third term on the right-hand side
of the last inequality $\to\infty$ as $n\to\infty$, $a\to -\infty$,
and $b\to +\infty$, while the second term $\to 0$ as $n\to \infty$
in view of Theorem 4.2. Hence we have the proof.


\bigskip

\section{Continuity Theorem}

The following continuity theorem gives a necessary and sufficient
condition for the complete convergence of a sequence of
distribution functions to a distribution function.

\bigskip

\noindent{\bf Theorem 5.1 (Continuity Theorem). } Let \{$F_{n}$\} be
a sequence of distribution functions, and let \{$\varphi_{n}$\} be
the sequence of corresponding characteristic functions. Then $\{F_
{n}\}$ converges completely to a distribution function $F$ if and
only if $\varphi_{n}\to\varphi$ as $n\to\infty$ on $\Re$, where
$\varphi$ is continuous at $t=0$. In this case the limit function
$\varphi$ is the characteristic function of the limit distribution
function $F$.

\bigskip

\noindent{\bf Proof. } First suppose that $F_{n} \stackrel {c}{\to}
F$, where $F$ is a distribution function. Then it follows from
Theorem 4.2 that for every $t\in\Re$
 $$
 \int_{\Re}\cos \ tx \ dF_{n}(x) \to \int_{\Re} \cos \ tx\ dF(x)
 $$
and
 $$
 \int_{\Re}\sin \ tx\ dF_{n}(x) \to \int_{\Re}\sin\ tx\ dF(x),
 $$
so that
 $$
\varphi_{n}(t)\to\varphi(t)\qquad {\rm for \ every}\ t\in\Re\quad
{\rm as}\ n\to\infty,
 $$
where $\varphi$ is the characteristic function of $F$.

Conversely, suppose that $\varphi_{n}\to\varphi$ on $\Re$, where
$\varphi$ is continuous at $t=0$. We first show that, if $F$ is a
distribution function with characteristic function $\varphi$, then
for every $h>0$
 $$
\int_{0}^{h}F(y)\ dy-\int_{-h}^{0}F(y)\ dy = \frac{1}\pi
\int^{\infty}_{-\infty} \frac{1-\cos\ ht}{t^2}\varphi(t)\ dt.
\eqno(5)
 $$
Let $a>0$, and let $G$ be the uniform distribution on $[-a,a]$ with
characteristic function $\theta(t)=(\sin \ at) /at$. Consider the
distribution function $H=F*G$. Clearly $H$ is continuous on $\Re$
and is given by
 $$
H(x)=\frac1{2a}\int^{x+a}_{x-a}F(y)\ dy. \eqno(6)
 $$
Let $\psi$ be the characteristic function of $H$. Then it follows
from the Corollary 3.3
 $$
\psi(t)=\varphi(t)\theta(t)=\varphi(t)\frac{\sin\ at}{at}.
 $$
Applying the inversion theorem to $H$ and $\psi$, we obtain for all
$x\in\Re$
 \begin{eqnarray*}
H(x+a)-H(x-a)&=&\lim_{T\to\infty}\frac1{\pi}\int^{T}_{-T}\frac{\sin^2\
at}{at^2}e^{-itx}\varphi(t)\ dt\\
 &=&
\frac1{2 \pi a}\int^{\infty}_{-\infty}\frac{1-\cos\
2at}{t^2}e^{-itx}\varphi(t)\ dt.
 \end{eqnarray*}
In particular, for $x=0$ we have
 $$
H(a)-H(-a)=\frac1{2 \pi a} \int^{\infty}_{-\infty}\frac{1-\cos\
2at}{t^2}\varphi(t)\ dt. \eqno(7)
 $$
The proof of (5) then follows at once on substituting (6) in (7) and
writing $h=2a$.

We now return to the proof of the theorem. Since $\{F_{n}\}$ is a
sequence of distribution functions, Helly's theorem implies that
there exists a subsequence $\{F_{nk}\}$ of $\{F_{n}\}$ which
converges weakly to a bounded, nondecreasing, and right-continuous
function $F$. We show that $F$ is a distribution function. From (5)
we obtain
 $$
\int^{h}_{0}F_{nk}(y)\ dy - \int^{0}_{-h}F_{nk}(y)\ dy = \frac1{\pi}
\int_{-\infty}^{\infty}\frac{1-\cos\ ht}{t^2}\varphi_{nk}(t)\ dt
 $$
for every $h>0$. We now take the limit of both sides as
$k\to\infty$. First note that the integrals on the left-hand side
are taken over finite intervals, and, moreover, the $\{F_{nk}\}$ are
uniformly bounded and converge to $F$ a.e. on $\Re$. In view of the
Lebesgue dominated convergence theorem we can take the limit under
the integral sign as $k\to\infty$. On the other hand, the integrand
on the right-hand side is dominated by the absolutely integrable
function $(1-\cos\ ht)/t^2$, so that we can again take the limit
under the integral sign. Thus we obtain
 $$
\int_{0}^{h}F(y)\ dy-\int_{-h}^{0}F(y)\
dy=\frac1{\pi}\int_{-\infty}^{\infty}\frac{1-\cos\
ht}{t^2}\varphi(t)\ dt.
 $$
Dividing both sides by $h$, we have
 $$
\frac1{h}\Big[\int_{0}^{h}F(y)\ dy-\int_{-h}^{0}F(y)\
dy\Big]=\frac1{\pi}\int_{-\infty}^{\infty}\frac{1-\cos\
t}{t^2}\varphi\Big(\frac t{h}\Big)\ dt. \eqno(8)
 $$
Since $\varphi$ is continuous at $t=0$, we have
 $$
\lim_{h\to\infty}\varphi\Big(\frac
t{h}\Big)=\varphi(0)=\lim_{n\to\infty}\varphi_{n}(0)=1.
 $$
Letting $h\to\infty$ in (8), we obtain
 $$
F(+\infty)-F(-\infty)=\frac 1{\pi}\int_{-\infty}^{\infty}
\frac{1-\cos\ t}{t^2}\ dt=1.
 $$
Since $0\le F \le 1$, it follows that $F(-\infty)=0$ and
$F(+\infty)=1$, so that $F$ is a distribution function. We then
conclude from the first part that $\varphi$ is the characteristic
function corresponding to $F$.


Now suppose that $\{F_{n}\}$ contains another subsequence which
converges to a limit, say $F*$. Then, proceeding as above, we see
that $F^*$ is a distribution function and $\varphi$ is the
corresponding characteristic function. By the uniqueness theorem it
follows that $F=F^*$, which implies that every weakly convergent
subsequence of $\{F_{n} \}$ has the same limit $F$. This shows that
$F_{n} \stackrel {c}{\to} F$ and that $\varphi$ is the
characteristic function of $F$.

 


\addcontentsline{toc}{section}{References}

\section*{References}

\bigskip

\begin{enumerate}

\bibitem{}
Chow, Y.S. and Teicher, H.: {\rm Probability Theory: Independence,
Interchangeability, Martingales}. 2nd Edition. Spring-Verlag, Berlin
(1979).

\bibitem{}
Laha, R.G. and Rohatgi, V.K.: {\rm Probability Theory}. John Wiley
\& Sons, New York (1988).

\end{enumerate}

\end{document}
