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\begin{document}
\title{\textbf{Report in Probability}}
\author{Wang Chung-Chen}
\date{2013.06}
\maketitle

Recall that the \textbf{characteristic function} of a given random variable $X$ is given by 
\begin{align}
\varphi(t) := Ee^{itX} = E\cos tX + iE\sin tX.\notag
\end{align}
with several properties about its values. An important theorem called \textbf{Inversion Formula} says that:

\begin{prop}[Theorem 3.3.4.]
Let $\varphi(t) = \int e^{itx} \mu(dx)$ where $\mu$ is a probability measure. If $a<b$, then 
\begin{align}
\lim_{T\to\infty}(2\pi)^{-1}\int_{-T}^T\frac{e^{-ita}-e^{-itb}}{it}\varphi(t)dt = \mu(a,b) + \frac{1}{2}\mu(\{a,b\}).\notag
\end{align}
\end{prop}
We shall use this formula to show that 



\begin{thm}
Let $X$ be a random variable. If $\int |\varphi(t)|dt <\infty$, then 
\begin{itemize}
\item[(1)] $\mu$  has a density. (in fact it is bounded continuous and is expressed by $f(y)=\frac{1}{2\pi}\int e^{-ity}\varphi(t)dt$.) 
\item[(2)] Then $\varphi(t)\to 0$ as $t\to \infty$, 
\item[(3)] and then $\mu$ has no point masses. i.e. there is no $x$ such that $\mu(\{x\})\ne 0$. 
\end{itemize}
Moreover, none of the inverse of $(1),(2),(3)$ is necessarily true. 
\end{thm}

\noindent\textsc{Proof of (1)}. This follows from textbook, theorem 3.3.5. $\int_a^b e^{-ity}dy = \frac{e^{-ita}-e^{-itb}}{it}$ and absolute property imply that 
\begin{align}
\left| \frac{e^{-ita}-e^{-itb}}{it}\right| = \left|\int_a^b e^{-ity} dy\right|\leq |b-a|.\notag
\end{align}
Since, in fact,  $\int_{-T}^T \frac{e^{-ita} - e^{-itb}}{it}\varphi(t)dt$ and $b-a$ are real, 
\begin{align}
\mu(a,b) + \frac{1}{2}\mu(\{a,b\}) = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{-ita} - e^{-itb}}{it}\varphi(t)dt\leq \frac{b-a}{2\pi}\int_{-\infty}^\infty |\varphi(t)|dt.\notag
\end{align}
I claim that $\mu$ has no point masses, for if $a$ is a point mass, then 
\begin{align}
0< \frac{1}{2}\mu(\{a\})\leq \frac{b-a}{2\pi}\int_{-\infty}^\infty|\varphi(t)|dt. \notag
\end{align}
Let $b\to a$ , then $a$ is not a point mass, a contradiction. Thus by inversion formula and Fubini's theorem,
\begin{align}
\mu(x,x+h)&= \frac{1}{2\pi}\int \frac{e^{-itx}-e^{-it(x+h)}}{it}\varphi(t)dt\notag\\
&=\frac{1}{2\pi}\int\left( \int_x^{x+h} e^{-ity}dy\right)\varphi(t)dt 
= \int_x^{x+h}\left(\frac{1}{2\pi}\int e^{-ity}\varphi(t)dt\right)dy. \notag
\end{align}
So $f(y) = \frac{1}{2\pi}\int e^{-ity}\varphi(t)dt$ is the distribution. On the other hand, it converse fails. If we consider the \textbf{Exponential distribution} (example 3.3.6), who has density $e^{-x}$ and ch.f. $\frac{1}{1-it}$, but the integral 
\begin{align}
\int_{-\infty}^\infty \frac{1}{\sqrt{1+t^2}} dt = \sinh^{-1} t \big| _{-\infty}^\infty=\infty\notag
\end{align}
is infinity. $\,\,\,\,\,\square$

\text{ }

\noindent\textsc{Proof of (2)}. We recall Riemann-Lebesgue Lemma that 
\begin{lem}
If $g$ is integrable then 
\begin{align}
\lim_{\lambda\to\infty}\int g(x)\cos \lambda x dx &= 0, \notag\\
\lim_{\lambda\to\infty}\int g(x)\sin \lambda x dx &= 0. \notag
\end{align}
\end{lem}
\noindent\textsc{Proof of Lemma}. Since every function $g(x)$ can be written as the limit of a simple function sequence $g_n(x)$, where the integration of their difference can be less than an arbitrarily given $\varepsilon > 0$. That is, $\int |g(x) - g_N(x)| dx < \varepsilon$ (Zygmund pp.54) . 
\begin{align}
\left|\int g(x)\cos nx dx\right| &\leq \left| \int g_N(x)\cos nx dx\right| + \int |g(x)-g_N(x)|dx\notag\\
&\leq\frac{2K}{n} + \varepsilon,\notag
\end{align}
so if $n>>0$ we can conclude that $\int g(x)\cos nx dx=0$. $\,\,\,\,\,\,\square$

\text{ }

Then consider 
\begin{align}
\varphi(t) = \int\left(\cos (tX) + i\sin (tX)\right)d\mu.\notag
\end{align}
By above lemma, as $t\to\infty$, $\varphi(t)\to 0$. Its converse also fails. Let $\Omega = [0,1]$ with Lebesgue measure. Let 
\begin{align}
X(\omega)=
\begin{cases}
 \omega &\text{as $0\leq\omega\leq 0.5$}\\
2\omega - 0.5 &\text{as $0.5\leq \omega\leq 1$}.
\end{cases}
\notag
\end{align}
Then the distribution function $F(x)$ has a non-differentiable point at $\frac{1}{2}$ and so cannot be expressed by $\int_{-\infty}^x f(y) dy$ for some $f$. Meanwhile, 
\begin{align}
\varphi(t) = Ee^{-itX} &= \int_0^\frac{1}{2} e^{-it\omega} d\omega+\int_\frac{1}{2}^1 e^{-it(2\omega - 0.5)}d\omega\notag\\
&=\left( \frac{e^{-it\omega}}{-it}\right)\big|_0^\frac{1}{2} + \left(\frac{e^{-it(2\omega -0.5)}}{-2it}\right)\big|_\frac{1}{2}^1\to 0\notag
\end{align}
as $t\to \infty$. $\,\,\,\,\,\,\square$

\text{ }

\noindent\textsc{Proof of (3)}. We need the following properties.
\begin{lem}
\begin{itemize}
\item[(i)] $\mu(\{a\}) = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T e^{-ita}\varphi(t) dt$. 
\item[(ii)] If $P(X\in hZ) = 1$ where $h > 0$, then its ch.f. has $\varphi(\frac{2\pi}{h} + t) = \varphi(t)$ , so for $x\in hZ$, 
\begin{align}
P(X=x)=\frac{h}{2\pi}\int_\frac{-\pi}{h}^\frac{\pi}{h} e^{-itx} \varphi(t)dt\notag
\end{align}
\item[(iii)] Write $X=Y+b$, then $Ee^{itX} = e^{itb}Ee^{itY}$. So if $P(X\in b+hZ)=1$, the inversion formula in (ii) is alid for $x\in b+hZ$. 
\end{itemize}
\end{lem}


\begin{prop}
If $X$ and $Y$ are independent and have ch.f. $\varphi$ and distribution $\mu$. Then 
\begin{align}
\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T |\varphi(t)|^2 dt = P(X-Y=0)=\sum_x \mu(\{x\})^2.\notag
\end{align}
\end{prop}

To prove this proposition, we need the following.
\begin{prop}
(i) If $X$ and $Y$ are independent with distributions $\mu$ and $\nu$ then $P(X+Y=0)=\sum_y \mu(\{-y\})\nu(\{y\})$. (ii) If $X$ has continuous distribution then $P(X=Y)=0$. 
\end{prop}



\noindent\textsc{Proof of original proposition}. 
Then $X-Y$ has ch.f $\varphi\cdot \bar{\varphi} = |\varphi|^2$. Let $a=0$ in (i) of the last theorem, then 
\begin{align}
P(X-Y = 0 ) =\lim_{T\to\infty} \frac{1}{2T}\int_{-T}^T|\varphi(t)|^2 dt. \notag
\end{align}
By exercise 2.1.8, $P(X-Y=0) = \sum_x \mu(\{x\})^2$. $\,\,\,\,\,\square$

\text{ } 

Back to its proof. We show that $\frac{1}{T}\int_0^T|\varphi(t)|^2dt\to 0$ as $T\to\infty$. Given any $\varepsilon>0$, by condition there is an $M>0$ such that $|\varphi(t)|<\sqrt\varepsilon$ for any $T>M$. At this time, 
\begin{align}
0\leq \frac{1}{T}\int_{0}^T|\varphi(t)|^2dt& =\frac{1}{T}\int_0^M|\varphi(t)|^2dt+\int_M^T|\varphi(t)|^2dt\notag\\
&\leq \frac{1}{T}\int_0^M|\varphi(t)|^2dt+\int_M^T\varepsilon dt\leq\frac{1}{T}\int_0^M|\varphi(t)|^2dt+\varepsilon \notag.
\end{align}
Then $0\leq\liminf_{T\to\infty}\frac{1}{T}\int_0^M|\varphi(t)|^2dt+\varepsilon\leq\limsup_{T\to\infty}\frac{1}{T}\int_0^M|\varphi(t)|^2dt+\varepsilon=\varepsilon$. Since $\varepsilon$ is arbitrary, $\lim_{T\to\infty}\frac{1}{T}\int_0^T|\varphi(t)|^2dt=0$. Hence by last theorem, there is no point masses. The converse is false. Give a random variable $P(X=0)=P(X=0.5)=0.5$. Then we claim that  $X$ has ch.f.
\begin{align}
\varphi(t) = \prod_{j=1}^\infty \frac{1+e^{-it2\cdot 3^{-j}}}{2}.\notag
\end{align}
Since $e^{2\pi j} = 1 $ for all $j\in\mathbb{N}$, 
\begin{align}
\varphi(3^k \pi) = \prod_{j=1}^\infty\frac{1+e^{i2\pi\cdot 3^{k-j}}}{2} = \prod_{r=1}^\infty \frac{1+e^{i2\pi\cdot 3^{-r}}}{2} = \varphi{\pi}.\notag
\end{align}
Since $\varphi(\pi)\ne 0$, $\lim_{t\to\infty}\varphi(t)\ne 0$. $\,\,\,\,\,\,\square$

\text{ }

\noindent\textsc{Proof of Claim}.  It follows from the lemma which states that if $X_1, X_2,\cdots$ are independent and $S_n = X_1 + \cdots + X_n$, and $\varphi_j$ is the ch.f. of $X_j$ and $S_n\to S_\infty$ a.s. then $S_\infty$ has ch.f. $\prod_{j=1}^\infty \varphi_j(t)$. 


The proof is like this: Firstly we show that if $X_n\to X$ in probability and then $X_n\Rightarrow X$. Supposely that $X_n\to X$ in probability, then choose a bounded continuous $g$. Then $Eg(X_n)\to Eg(X)$ by BCT. Since $g$ is arbitrary, by theorem $X_n\Rightarrow X$. 

By basic properties of ch.f. $S_n$ has ch.f. $\prod_{j=1}^n\varphi_j(t)$. By condition plus the last paragraph $S_n\Rightarrow S_\infty$. By continuity theorem that $\prod_{j=1}^\infty \varphi_(t)\to \varphi$, which is the ch.f. of $S_\infty$. $\,\,\,\,\,\square$

\begin{thebibliography}{100} 
\bibitem{ }  Rick Durrett, ``Probability : theory and examples",  Cambridge University Press, 2010.
\end{thebibliography}

\end{document}