% scribe: Cristine Pinto
% lastupdate: 26 October 2005
% lecture: 16
% references: Durrett, Section 2.3
% title: Continuity Theorem for Characteristic Functions
% keywords: characteristic functions, inversion formula, inversion formula of characteristic function, Helly's selection theorem, tightness, determining class, uniqueness theorem, Continuity Theorem 
% end

\documentclass[12pt, letterpaper]{article}

\include{macros}
\def\vcv{\stackrel{\scriptscriptstyle v}{\longrightarrow}}     % vague convergnece

\begin{document}

\lecture{16}{Continuity Theorem for Characteristic Functions}
{Cristine Pinto}{cristine@econ.berkeley.edu}

References: \cite{durrett}, section 2.3.

\section{Review of the Inversion Formula}
% Keywords: characteristic functions, inversion formula, inversion formula of characteristic function
% end
Recall that if $X$ is a random variable, it's characteristic function is 
\[\varphi_X(t) =\E \left[ e^{itX} \right].\]
In the last lecture, we proved the Uniqueness Theorem for characteristic 
functions.  We also learned the \emph{inversion formula}:
if $\int |\varphi_{X}(t)| < \infty$, then X has a bounded continuous density
%
\[f_X(x)=\frac{1}{2\pi}\int  e^{-itx}\varphi_{X}(t)dt.\]
%

\section{Continuity Theorem for Characteristic Functions}
Suppose we have a sequence of distributions on the line $(\P_n)$ with 
characteristic functions
%
\[\varphi_n(t)=\int e^{itx} \cdot\P_n(dx)=\E \left[ e^{itX_n} \right]\] 
%
for $X_n \sim \P_n$.
We want to be able to tell that $\P_n$ converges in distribution to 
some limit on $\R$ by looking at $\varphi_X(t)$.
Recall that if $\P_n \dcv \P$ then 
%
\[\int f d\P_n\longrightarrow \int f d\P 
\mbox{ for every bounded continuous function } f.\]
%
So in particular for $f(x)=e^{itx}$, we get
$\varphi_n(t)\longrightarrow\varphi(t)$ as  $\ n\longrightarrow\infty$ 
where $\varphi(t)=\int e^{itx} \P(dx)$, the characteristic function of 
$\P$.

Consider the converse.  Suppose we have a sequence $\P_n$ with $\varphi_n(t)$ 
and $\varphi_n(t)\longrightarrow\varphi(t)$ as $n\longrightarrow\infty$ 
for some function $\varphi(t)$.
Without imposing further assumptions, we cannot conclude that 
$\P_n \dcv \P$ where $\P$ has a continuous characteristic function 
$\varphi(t)$.

\begin{example}
Let $\P_n=N(0,n)$, so that $\varphi_n(t)=e^{-\frac{1}{2}n{t^2}}$, $t\in\R$.
Notice that $\varphi_n(t)\longrightarrow \1(t=0)$, but
\[\P_n\vcv\frac{1}{2}\delta_{-\infty}+\frac{1}{2}\delta_{+\infty}.\]
\end{example}

Recall the idea of $\emph{tightness}$: $(\P_n)$ is tight if:
\[\lim_{x \rightarrow \infty}\sup_n\P_n(-x,x)^c=0.\]
We present the first (easy) version of the continuity theorem for 
characteristic functions.

\begin{theorem}
\label{easyctythm}
Let $\P_n$ be probability measures on $\R$ with c.f.\ $\varphi_n$. If:
\begin{enumerate}
\item $\lim_{n \rightarrow \infty}\varphi_n(t)=\varphi(t)$ exists for every 
$t\in\R$; and
\item $\P_n$ is tight,
\end{enumerate}
then $\P_n\longrightarrow \P$ where $\P$ is  a probability measure on $\R$ 
with c.f.\ $\varphi$.
\end{theorem}

\begin{proof} By Helly's selection theorem, to prove that there exists a $\P$ 
such that $\P_n\longrightarrow \P$, it is enough to show that there exists a 
$\P$ such that every subsequence of $(\P_n)$ has a further subsequence which 
converges to $\P$.

To find a suitable $\P$, recall the general theorem: Let $\mathcal C$ be a 
collection of bounded continuous functions which is determining. If:
\begin{enumerate}
\item $\lim_{n \rightarrow \infty}\int f d\P_n$ exists for all 
$f\in\mathcal C$; and
\item $(\P_n)$ is tight,
\end{enumerate}
then:
\[\P_n\dcv \P \mbox{ where } \int f d\P=\lim_{n \rightarrow \infty}\int f d\P_n,\, \forall f\in\mathcal C\] 
Apply this general theorem to
\[ \mathcal C = \left\{ f \mbox{ of the form } f(x) = \sin(tx), \,t \in \R  
\mbox{ or } f(x)=\cos(tx), \, t \in \R \right\}. \]
$\mathcal C$ is determining by the uniqueness theorem for c.f.'s.
\end{proof}


This form of the continuity theorem is adequate for most applications, 
such as the CLT. Usually in the CLT we try to show:
\[Z_n=\frac{S_n}{\sqrt{\E(S_n^2)}}\dcv N(0,1)\]
In this case, $\P_n$ is the distribution of $Z_n$ with $\E(Z_n)=0$. Clearly, 
$\P_n$ is tight:
\[\P_n(-x,x)^c=\P(|Z_n|>x)\leq\frac {\E(Z_n^2)}{x^2}=\frac{1}{x^2},\]
which decreases to $0$ as $x \uparrow \infty$.


So, if we continue to assume condition 1 of theorem \ref{easyctythm}, 
condition 2 
implies that there exists some $\P$ with c.f.\ $\varphi$ such that:
\begin{enumerate}
\item[(2a)] $\varphi$ is the characteristic function of \emph{some} 
distribution.  This also implies
\item[(2b)] $\varphi$ is continuous as a function of $t$.  This, in 
turn, implies
\item[(2c)] $t\longrightarrow\varphi(t)$ is continuous at $t=0$.
\end{enumerate}

Paul L\'evy found that with condition 1, (2b) is \emph{equivalent} to 
condition 2 of theorem \ref{easyctythm}.

\begin{theorem}(L\'evy Continuity Theorem for c.f.'s): 
Given $\P_n$ with c.f. $\varphi_n$, if:
\begin{enumerate}
\item $lim_{n \rightarrow \infty} \varphi_n(t)=\varphi(t)$ exists for all $
t\in\R$; and 
\item $t\longrightarrow\varphi(t)$ is continuous at $t=0$,
\end{enumerate}
then
\[\P_n\dcv\P \mbox{ with } \int e^{itx} \,\P(dx)= \varphi(t).\]
\end{theorem}

\begin{proof}
In the proof, we just need to show that continuity of $\varphi$ at $0$ implies
$(\P_n)$ is tight. For that, it's most convincing to use a genuine bound 
(\cite{durrett}, p.\ 98):
%
\[\frac{1}{u}\cdot\int^{-u}_{u}[1-\varphi_n(t)]dt \geq\P_n\left(\frac{-2}{u},\frac{2}{u}\right)^c\]
%

Recall that $\varphi_n(t)\longrightarrow\varphi(t)$ as 
$n\longrightarrow\infty$ and $\varphi_n(t)$ is continuous at $t=0$, 
$\varphi(0)=1$.  As $n\longrightarrow\infty$, then,
\[\frac{1}{u}\int^{u}_{-u}[1-\varphi_n(t)]dt \longrightarrow\frac{1}{u}\int^{u}_{-u}[1-\varphi(t)] dt,\]
and as $u\longrightarrow 0$,
\[\frac{1}{u}\int^{u}_{-u}[1-\varphi(t)] dt\longrightarrow 0.\]

Fix $\epsilon>0$ and choose $u$ small enough so that 
$\frac{1}{u}\int^{u}_{-u}[1-\varphi(t)]dt< \epsilon$.
Choose $N$ large enough that:
\[\frac{1}{u}\int^{u}_{-u}|1-\varphi_n(t)|dt<2\epsilon \mbox{ for } n\geq N.\]

Now we have
\[\P_n\left(-\frac{2}{u},\frac{2}{u}\right)^c\leq2\epsilon \mbox{ for all }
n\geq N,\]
and hence $\lim_{x\rightarrow \infty}\sup_{n} \P_n(-x,x)^c=0$ as desired.
\end{proof}

\section{Exercises}

\begin{exercise} [Extra credit problem] 
\label{extracreditprob}
Suppose a sequence $(\P_n)$ of probability measure on $\R$ such that
$\lim_{n \rightarrow \infty}\int f d\P_n$ exists and $\in\R$ for every 
bounded continuous $f$.
Then (you check): there exists a unique probability measure $\P$ such that 
$\P_n\dcv\P$. Consequently:
\[\int f d\P=\lim_{n \rightarrow \infty}\int f d\P_n.\]
\end{exercise}

\begin{exercise}
What happens if we replace $\R$ in exercise \ref{extracreditprob} by $\R^n$ or 
a generic metric space?
\end{exercise}

\begin{example}[Related to the exercises above]
Consider:
\[\mathcal C = \left\{f:f \mbox{ is bounded, continuous, and has a compact 
support}\right\}\]
(Note that if $f$ has compact support, then $f(x)=0$ for $|x|>B$ for some 
$B\geq 0$.)  Check that $\mathcal C$ is a determining class.


Consider 
\[\mathcal C_0 =\left\{f:f(0)=0 \mbox{ and $f$ is continuous with compact 
support}\right\};\]
$\mathcal C_0$ is also a determining class.

Let
\begin{equation*}
\P_n = \left\{
\begin{array}{ll}
\delta_n & \mbox{ if $n$ is even}\\
\delta_0 & \mbox{ if $n$ is odd}
\end{array}\right.
\end{equation*}
In this case,
$\int f d\P_n \longrightarrow \int f d\delta_0$ for all $f\in\mathcal C$, but
$\P_n \nrightarrow \P_0$.

The moral of this example is that to prove that $\P_n\dcv\P$ for some 
probability $\P$ on $\R$, 
it is not enough to just show that $\int f d\P_n\longrightarrow\int f d\P$ 
for all $f$ in a determining class.
\end{example}

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