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\title[Stat 609 Lecture 9]{Stat 609: Mathematical Statistics \\ Lecture 9}
\author[UW-Madison]{Jun Shao}
\institute[Statistics]{%
%  Department of Statistics\\
%  University of Wisconsin\\
%  Madison, WI 53706, USA\\
%~~~\\
  }
\date[2014]{}

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\AtBeginSection[]{\frame{\frametitle{Outline}\tableofcontents[current]}}
\def\back{\!\!\!\!\!}

\begin{document}

%\frame{\titlepage}

\frame {\vspace{-1mm}
\frametitle{Lecture 9: Characteristic functions}
\vspace{-2mm}
\begin{alertblock}{}
The mgf's are useful, but they sometimes are not finite. 
\vspace{1mm}

Another function that can characterise a distribution is the so-called characteristic function, which is always well defined but it is a complex function. 
\end{alertblock}
\vspace{-2mm}
\begin{block}{Definition (characteristic functions).}
The \cblue{characteristic function} (chf) of the distribution of a random variable $X$  is
defined as\vspace{-2mm}
\[ \phi_X (t) = E (e^{ \dot{\imath}  t  X })
= E[ \cos ( t X)] + \dot{\imath}  E [\sin (tX)] , \quad\quad  t
\in {\cal R} \vspace{-2mm} \]
where $\dot{\imath}  = \sqrt{-1} $ and $e^{ \dot{\imath}  s } = \cos (s) + \dot{\imath} \sin (s)$. 
\end{block}
\vspace{-2mm}
\begin{alertblock}{}<2->
\begin{itemize}
\item
If the mgf $M_X(t)$ of $X$ is finite in a neighborhood of
$0$, then the chf of $X$ is \vspace{-2.5mm} 
\[\phi_X (t)  = M_X(\dot{\imath} t), \quad\quad t \in {\cal R} \vspace{-2.5mm} \]
\item
If $Y = aX+b$ for constants $a$ and $b$, then the chf of $Y$ in terms of the chf $\phi_X(t)$ of $X$ is 
\vspace{-3mm} 
\[ \phi_Y(t) = e^{\dot{\imath}bt} \phi_X(at), \quad\quad t \in {\cal R} \]
\end{itemize}  
\end{alertblock}
}
\frame{\vspace{-2mm} 
\begin{block}{Theorem C1.}
If a random variable $X$ has finite $E|X|^r$ for a positive integer $r$, then \vspace{-2mm} 
\[ \frac{d^r \phi_X(t)}{dt^r}\bigg|_{t=0} = \dot{\imath}^r E(X^r)  \]
\end{block}
\vspace{-2mm}
\begin{block}{Proof.}
We give a proof for the case where $X$ has pdf $f_X(x)$.
\vspace{1mm} 
 
If we can exchange the differentiation and integration, then \vspace{-2mm} 
\[ \frac{d^r \phi_X(t)}{dt^r} = \frac{d^r }{dt^r} \int_{-\infty}^\infty e^{\dot{\imath}tx} f_X(x) dx 
= \int_{-\infty}^\infty  \frac{d^r }{dt^r} e^{\dot{\imath}tx} f_X(x) dx
= \int_{-\infty}^\infty  \dot{\imath}^r x^r e^{\dot{\imath}tx} f_X(x) dx \vspace{-2mm} \]
and the result follows by setting $t=0$ at the beginning and end of the above expression. 
\vspace{1mm}

The exchange is justified by the dominated convergence theorem, since \vspace{-1mm}
\[ \bigg| \frac{d^r }{dt^r} e^{\dot{\imath}tx} f_X(x)\bigg| 
= \bigg|  \dot{\imath}^r x^r e^{\dot{\imath}tx}  \bigg| f_X(x) \leq |x|^r f_X(x) \vspace{-2mm} \]
and \vspace{-1mm} 
\[ E|X|^r = \int_{-\infty}^\infty |x|^r f_X(x) dx < \infty  \]
\end{block}}
\frame{
\begin{alertblock}{}
Thus, like the mgf, the chf can be used to calculate the moments, but we have to first know the existence of moments.  
\end{alertblock}\vspace{-2mm}
\begin{block}{Example.}<2->
Let $X$ be a random variable with pdf \vspace{-2mm} 
\[ f_X(x) = \frac{1}{\pi (1+x^2)} , \quad \quad x \in {\cal R} \vspace{-2mm} \]
From the Fourier transformation, \vspace{-2mm} 
\[ \int_0^\infty \frac{\cos (tx) }{1+x^2} dx = \frac{\pi}{2} e^{-t} , \quad \quad t>0 \vspace{-2mm} \]
we obtain that, when $t>0$, \vspace{-2mm} 
\beqrs
\phi_X(t) \back & = & \back E[\cos (tX) ] + \dot{\imath} E[ \sin(tX)] \\
\back & = & \back \int_{-\infty}^\infty \frac{\cos (tx)}{\pi(1+x^2)} dx + \dot{\imath} 
\int_{-\infty}^\infty \frac{\sin (tx)}{\pi(1+x^2)} dx \\
\back & = & \back \frac{2}{\pi} \int_{-\infty}^\infty \frac{\cos (tx)}{1+x^2} dx = e^{-t}
\eeqrs
$~~$ \vspace{-6mm} 

since $\sin (s)$ is an odd function and $\cos (s)$ is an even function. 
\end{block}
}
\frame{
\begin{block}{}
For $t < 0$, \vspace{-2mm} 
\[ \phi_X(t)  =  E[\cos (tX) ] + \dot{\imath} E[ \sin (tX)] 
= E[\cos (-tX) ] = e^t = e^{-|t|} \vspace{-2mm} \]
When $t=0$, $\phi_X(0)= 1$ for any chf, and hence we obtain that 
\vspace{-2mm} 
\[ \phi_X(t) = e^{-|t|} \quad\quad t \in {\cal R} \vspace{-2mm} \]  
Note that this chf is not differentiable at $t=0$ and previously we showed that the expectation of $X$ does not exist. 
\end{block}
\vspace{-2mm}
\begin{alertblock}{Uniqueness and inversion}<2->
Similar to the mgf, the chf characterizes the distribution in the sense that there is a one-to-one correspondence between chf and cdf. 
\vspace{1mm} 

We first establish an inversion formula. 
\end{alertblock}
\vspace{-2mm}
\begin{alertblock}{Theorem C2}<2->
Let $X$ be a random variable with cdf $F$  and chf $\phi$. 

For any real numbers $y<u$, \vspace{-2mm} 
\[  P(y<X<u) + \frac{P(X=y)+P(X=u)}{2} 
= \lim_{T \rightarrow \infty} \int_{-T}^T \frac{e^{-\dot{\imath} ty} -e^{-\dot{\imath} tu} }{2\pi \dot{\imath} t} \phi(t) dt \] 
\end{alertblock}
}
\frame{\vspace{-2mm} 
\begin{alertblock}{Proof.}
By definition, \vspace{-2mm} 
\beqrs
\int_{-T}^T \frac{e^{-\dot{\imath} ty} -e^{-\dot{\imath} tu} }{2\pi \dot{\imath} t} \phi(t) dt
\back & = & \back \int_{-T}^T \frac{e^{-\dot{\imath} ty} -e^{-\dot{\imath} tu} }{2\pi \dot{\imath} t} E(e^{\dot{\imath}tX}) dt \\
\back & = & \back E \int_{-T}^T \frac{e^{-\dot{\imath} ty} -e^{-\dot{\imath} tu} }{2\pi \dot{\imath} t} e^{\dot{\imath}tX} dt \\
\back & = & \back E [ g_T(u,y,X) ]
\eeqrs
$~~$ \vspace{-7mm} 

where 
the switch of $E$ and $\int_{-T}^T$ is justified by the fact that \vspace{-2mm}
\[ \bigg| \frac{e^{-\dot{\imath} ty} -e^{-\dot{\imath} tu} }{2\pi \dot{\imath} t} e^{\dot{\imath}tX} \bigg| \leq 1 \vspace{-2mm} \]
and
\vspace{-2mm}
\beqrs
 g_T( u, y, x) \back & = & \back
\int_{-T}^T \frac{e^{\dot{\imath} t(x-y)} -e^{\dot{\imath} t(x-u)}}{2\pi \dot{\imath} t} dt \\
 \back & = & \back
 \frac{1}{\pi} \int_0^T \frac{\sin (t(x-y)) - \sin (t(x-u))}{t} dt
 \eeqrs
$~~$ \vspace{-4mm} 

because $\frac{\sin (t(x-y)) - \sin (t(x-u))}{t} $ is even and 
$\frac{\cos (t(x-y)) - \cos (t(x-u))}{t} $ is odd. 
\end{alertblock}}
\frame{
\begin{alertblock}{}
By Fourier transformation, \vspace{-2mm} 
\[ \int_0^\infty \frac{\sin (st )}{t} dt = \frac{\pi }{2} {\rm sign} (s) \vspace{-2mm} \]
Hence \vspace{-3mm} 
\[ \lim_{T \rightarrow \infty} g_T(u, y, x) = \left\{ 
\begin{array}{ll} 
- \frac{1}{2} - (-\frac{1}{2} ) = 0 & \quad x < y \vspace{1mm} \\
0 - (-\frac{1}{2} ) = \frac{1}{2}  & \quad x = y \vspace{1mm} \\
 \frac{1}{2} - (-\frac{1}{2} ) = 1 & \quad  y < x < u \vspace{1mm} \\
 \frac{1}{2} -  0 = \frac{1}{2}  & \quad x = u \vspace{1mm} \\
 \frac{1}{2} - \frac{1}{2}  = 0 & \quad x >u  \end{array} \right.\vspace{-2mm} \]
Also,  
\[ 0 \leq {\rm sign}(s) \int_0^T \frac{\sin (s t)}{t} dt \leq \int_0^\pi \frac{\sin t}{t} dt \quad \quad T>0 \vspace{-1mm} \]
which implies that \vspace{-2mm} 
\[ | g_T( u, y, x)| \leq \frac{2}{\pi} \int_0^\pi \frac{\sin t}{t} dt \quad \quad T>0 \vspace{-2mm} \]
By the dominated convergence theorem, 
\end{alertblock}}
\frame{\vspace{-2mm} 
\begin{alertblock}{}
\vspace{-5mm} 
\beqrs
 \lim_{T \rightarrow \infty} E\left[ g_T(u,y,X)\right] \back & = & \back  
E \left[ \lim_{T \rightarrow \infty}  g_T(u,y,X)\right] \\
\back & = & \back E \left[ I(y<X<u) + \frac{1}{2} I(X=y) + \frac{1}{2}I(X=u) \right] \\
\back & = & \back P(y<X<u) + \frac{P(X=y) +P(X=u)}{2}  
\eeqrs
$~~$ \vspace{-6mm} 

This proves the theorem. 
\end{alertblock}\vspace{-2mm}
\begin{block}{}<2->
Some consequences of Theorem C2 are given in the following result. 
\end{block}
\vspace{-2mm}
\begin{block}{Theorem 3C.}<2->
Let $F$ be a cdf with chf $\phi$. 
\begin{itemize}
\item[(i)]
If $F$ is continuous at $y$ and $u$ with $y <u$, then \vspace{-2mm} 
\[ F(x) -F(y) = \lim_{T \rightarrow \infty} 
\int_{-T}^T \frac{e^{-\dot{\imath} ty} -e^{-\dot{\imath} tu} }{2\pi \dot{\imath} t} \phi(t) dt \vspace{-2mm} \]
\item[(ii)] 
If $F$ is continuous at $x$, then \vspace{-2mm} 
\[ F(x)  = \lim_{y \rightarrow -\infty} \lim_{T \rightarrow \infty}
\int_{-T}^T \frac{e^{-\dot{\imath} ty} -e^{-\dot{\imath} tu} }{2\pi \dot{\imath} t} \phi(t) dt  \]
\end{itemize}
\end{block}
}
\frame{
\begin{block}{}
\begin{itemize}
\item[(iii)]
There is a one-to-one correspondence between cdf and chf. 
\item[(iv)]
If $F$ is continuous at $x$, then \vspace{-2mm}
\[ F(x)  = \lim_{y \rightarrow -\infty} \lim_{T \rightarrow \infty}
\int_{-T}^T \frac{e^{-\dot{\imath} ty} -e^{-\dot{\imath} tu} }{2\pi \dot{\imath} t} \phi(t) dt  \]
\end{itemize}

\end{block}}
\frame{\vspace{-2mm}
\begin{block}{}
In fact, the condition that $M_X(t)$ exists at $\pm t$ ensures that $M_X(s)$ has the power series expansion \vspace{-3mm}
\[ M_X(s) = \sum_{k=0}^\infty \frac{E(X^k) s^k }{k!} \quad \quad -t <s < t \vspace{-3mm}\]
If the distribution of $X$ is symmetric (about 0), i.e., $X$ and $-X$ have the same distribution, then \vspace{-3mm}
\[ M_X(t) = E(e^{tX}) = E(e^{t(-X)}) = E(e^{-tX}) = M_X(-t) \vspace{-3mm} \]
i.e., $M_X(t)$ is an even function and $M_X(t)$ exists at $\pm t$ is the same as
$M_X(t)$ exists at a $t>0$.
\end{block}
\vspace{-2mm}
\begin{alertblock}{Example 2.3.8 (Gamma mgf)}<2->
Let $X$ have the gamma pdf\vspace{-3mm}
\[ f_X(x) = \frac{1}{\Gamma (\alpha )\beta^\alpha} x^{\alpha -1} e^{-x/\beta} , \quad x>0 , \vspace{-3mm}\]
where $\alpha >0$ and $\beta >0$ are two constants and
\vspace{-3mm}
\[ \Gamma (\alpha ) = \int_0^\infty x^{\alpha -1} e^{-x} dx  \quad \quad \alpha > 0 \vspace{-3mm}\]
is the so-called gamma function.
\end{alertblock}
}
\frame{
\begin{alertblock}{}
If $t < 1/\beta$,  \vspace{-3mm}
\beqrs
M_X(t) &=& E(e^{tX}) = \frac{1}{\Gamma(\alpha )\beta^\alpha} \int_0^\infty e^{tx} x^{\alpha -1} e^{-x/\beta} dx \\
& = & \frac{1}{\Gamma(\alpha )\beta^\alpha} \int_0^\infty  x^{\alpha -1} e^{-x/(\frac{\beta}{1-\beta t})} dx \\
& = &  \frac{(\frac{\beta}{1-\beta t})^\alpha}{\Gamma(\alpha )\beta^\alpha} \int_0^\infty  s^{\alpha -1} e^{-s} ds =  \frac{1}{(1-\beta t)^\alpha}
\eeqrs
If $t \geq 1/\beta$, then $E(e^{tX}) = \infty$.
\vspace{1.5mm}

We can obtain\vspace{-2mm}
\[ E(X) = \frac{d}{dt}M_X(t)\bigg|_{t=0} = \frac{\alpha \beta }{(1-\beta t)^{\alpha +1}}\bigg|_{t=0} = \alpha \beta  \vspace{-2mm}\]
For any integer $n>1$, \vspace{-2mm}
\beqrs
E(X^n) &=& \frac{d^n}{dt^n}M_X(t)\bigg|_{t=0} = \frac{\alpha (\alpha+1)\cdots (\alpha+n-1) \beta^n }{(1-\beta t)^{\alpha +n}}\bigg|_{t=0} \\
& = &  \alpha
(\alpha+1)\cdots (\alpha +n-1) \beta^n
\eeqrs
\end{alertblock}}
\end{document}
