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\title[Stat 609 Lecture 20]{Stat 609: Mathematical Statistics \\ Lecture 20}
\author[UW-Madison]{Jun Shao}
\institute[Statistics]{%
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%  Madison, WI 53706, USA\\
%~~~\\
  }
\date[2014]{}

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\AtBeginSection[]{\frame{\frametitle{Outline}\tableofcontents[current]}}

\begin{document}

%\frame{\titlepage}

\frame {\vspace{-1mm}
\frametitle{Lecture 20: Multivariate mgf's and chf's}\vspace{-2mm}
\begin{alertblock}{Multivariate mgf and chf}
For an $n$-dimensional random vector $X$, its mgf is defined as
\vspace{-3mm}
\[ M_X(t) = E(e^{t'X}), \quad \quad t \in {\cal R}^n \vspace{-3mm}  \]
and its chf is defined as
\vspace{-3mm}
\[ \phi_X(t) = E(e^{\dot{\imath} t'X}), \quad \quad t \in {\cal R}^n  \]
\end{alertblock}
\vspace{-2mm}
\begin{block}{Simple properties of mgf's and chf's}<2->
\begin{itemize}
\item
$M_X(t)$ is either finite or $\infty$, and $M_X(0)=1$.
\item
The chf $\phi_X(t)$ is a continuous function of $t \in {\cal R}^n$ and $\phi_X(0)=1$.
\item
 For any fixed $k \times n$ matrix $A$ and vector $b \in {\cal R}^k $, the mgf and chf of the random vector $AX+b$ are, respectively, \vspace{-2mm}
\[ M_{AX+b}(t)= e^{b't}M_X(A't)  \quad \mbox{and} \quad
 \phi_{AX+b}(t)= e^{\dot{\imath} b't}\phi_X(A't),  \quad t \in {\cal R}^k \vspace{-2mm} \]
In particular, if $Y$ is the first $m$ components of $X$, $m < n$, then the mgf and chf of $Y$ are, respectively, \vspace{-3mm}
\[ M_Y(s) = M_X((s,0))  \quad \mbox{and} \quad \phi_Y(s) = \phi_X((s,0)),  \quad s\in {\cal R}^m , \  0 \in {\cal R}^{n-m}\]
\end{itemize}
\end{block}}
\frame{
\begin{block}{Calculation of moments}
If $X=(X_1,...,X_n)$ and  $M_X(t)$ is finite in a neighborhood of $0$, then
$ E(X_1^{r_1}\cdots X_n^{r_n})$ is finite
for any nonnegative integers $r_1,...,r_n$, and
\vspace{-2mm}
\[ E(X_1^{r_1}\cdots X_n^{r_n}) =
\frac{\partial^{r_1+\cdots +r_n}
 M_X(t)}{\partial t_1^{r_1} \cdots \partial t_n^{r_n}}
\bigg|_{t=0}  \vspace{-2mm} \]
In particular,\vspace{-1mm}
\[ \frac{\partial M_X (t)}{
 \partial t } \bigg|_{t=0} = E(X) = \left( \br{c}
E(X_1) \\
\vdots \\
E(X_n) \er \right)  \vspace{-2mm}\]
and
\vspace{-2mm}
\[ \frac{\partial^2 M_X (t)}{ \partial t\partial t' }
 \bigg|_{t=0} = E(X X' )  = \left( \br{cccc}
 E(X_1^2) & E(X_1X_2) & \cdots & E(X_1X_n) \\
 E(X_2X_1) & E(X_2^2) & \cdots & E(X_2X_n) \\
 \cdots & \cdots & \cdots & \cdots \\
 E(X_nX_1) & E(X_nX_2) & \cdots & E(X_n^2) \er \right) \]
When the mgf is not finite in any neighborhood of 0, then we can use the chf
to calculate moments.
\end{block}}
\frame{\vspace{-2mm}
\begin{block}{}
 Suppose that $E|X_1^{r_1}\cdots X_k^{r_n}|< \infty$
for some nonnegative integers $r_1,...,r_n$ with  $r=r_1+\cdots + r_n$.
\vspace{1mm}

Since\vspace{-2mm}
\[\bigg| \frac{\partial^r e^{\dot{\imath} t' X}}{\partial t_1^{r_1}
\cdots \partial t_k^{r_n} }\bigg| = \bigg| \dot{\imath}^{r} X_1^{r_1}\cdots X_k^{r_n}
e^{\dot{\imath}t' X} \bigg| \leq | X_1^{r_1}\cdots X_k^{r_n} | \vspace{-2mm} \]
 which is integrable, we can switch integration and differentiation to get \vspace{-2mm}
\[
\frac{\partial^r \phi_X(t)}{\partial t_1^{r_1}
\cdots \partial t_k^{r_n} } = \dot{\imath}^{r}
E \left( X_1^{r_1}\cdots X_k^{r_n}
e^{\dot{\imath} t' X}\right) \vspace{-1mm} \]
and\vspace{-1mm}
\[\frac{\partial^r \phi_X(t)}{\partial t_1^{r_1}
\cdots \partial t_k^{r_n} } \bigg|_{t=0}
 = \dot{\imath}^{r} E ( X_1^{r_1}\cdots X_k^{r_n} ). \vspace{-1mm}\]
In particular,\vspace{-1mm}
\[\frac{\partial \phi_X (t)}{ \partial t }\bigg|_{t=0} = \dot{\imath} EX ,
\quad  \quad
\frac{\partial^2 \phi_X (t)}{ \partial t\partial t^\tau } \bigg|_{t=0}
= -E(XX' )  \]
\end{block}\vspace{-2.5mm}
\begin{alertblock}{Theorem M1.}<2->
If the mgf $M_X(t)$ of an $n$-dimensional random vector $X$ is finite in a neighborhood of 0, then the chf of $X$ is 
$\phi_X(t) = M_X(\dot{\imath} t)$, $t \in {\cal R}^n$.
\end{alertblock}}
\frame{\vspace{-2mm}
\begin{alertblock}{Proof.}
We first prove this for a univariate $X$.
\vspace{1mm}

If $M_X(t)$ is finite in a neighborhood of 0, $(-\delta , \delta )$, $0<\delta < 1$, then $M_X(t)$ is differentiable of all order
in $(-\delta , \delta )$, $E|X|^r < \infty$ for all $r=1,2...$, and 
\vspace{-3mm}
\[ M_X ( t) = \sum_{k=0}^\infty
\frac{E(X^k) t^k }{k!},  \quad \quad t \in (-\delta , \delta ) \vspace{-3mm}\]
Using the inequality \vspace{-3mm}
\[ \bigg| e^{\dot{\imath} sx} \left[ e^{\dot{\imath}  tx}-
\sum_{k=0}^n \frac{(\dot{\imath} tx)^k}{k!} \right] \bigg|
\leq \frac{|tx|^{n+1}}{(n+1)!} \quad \quad t , s \in {\cal R}  \vspace{-2mm} \]
we obtain that \vspace{-2mm}
\[ \bigg| \phi_X (s+t) -
\sum_{k=0}^n \frac{\dot{\imath}^k t^k}{k!}  E(X^ke^{\dot{\imath}sX})  \bigg| \leq
\frac{|t|^{n+1}E|X|^{n+1}}{(n+1)!} ,  \quad \quad |t| < \delta, \ s \in {\cal R}  \vspace{-2mm}\]
which implies that \vspace{-4mm}
\[ \phi_X(s+ t) = \sum_{k=0}^\infty \frac{\phi_X^{(k)}(s)  t^k}{k!} \quad
\quad |t| < \delta , \  s \in {\cal R} \vspace{-3mm} \]
where $g^{(k)}(s) = d^k g(s) / ds^k$.
\end{alertblock}}
\frame{
\begin{alertblock}{}
Setting $s=0$, we obtain that \vspace{-2mm}
\[ \phi_X(t) = \sum_{k=0}^\infty \frac{\phi_X^{(k)}(0)  t^k}{k!}
= \sum_{k=0}^\infty \frac{\dot{\imath}^k E(X^k) t^k}{k!}
= \sum_{k=0}^\infty \frac{E(X^k) (\dot{\imath}t)^k}{k!}
= M_X(\dot{\imath} t) \quad |t| < \delta \vspace{-2mm} \]
For $|t| < \delta $ and $ |s| < \delta $,  \vspace{-2mm}
\beqrs
 \phi_X(s+ t) \back & = & \back \sum_{k=0}^\infty \frac{\phi_X^{(k)}(s)  t^k}{k!}
= \sum_{k=0}^\infty \frac{ t^k}{k!} \sum_{j=k}^\infty \frac{\dot{\imath}^j E(X^j)s^{j-k} }{(j-k)!} \\
\back & = & \back
\sum_{j=0}^\infty \sum_{k=0}^j \dot{\imath}^j E(X^j)  \frac{t^k s^{j-k} }{k! (j-k)!}
= \sum_{j=0}^\infty \sum_{k=0}^j \frac{\dot{\imath}^j E(X^j)(s+t)^j}{j!}
\eeqrs
$~$ \vspace{-5mm}

This proves that \vspace{-1mm}
\[ \phi_X(s) = M_X(\dot{\imath} s), \quad\quad |s| < 2\delta \vspace{-2mm} \]
Continuing this process, we can show that $\phi_X(s) = M_X(\dot{\imath} s)$ for $|s| < 3 \delta$, $|s| < 4 \delta$,..., and 
hence, $\phi_X(t) = M_X( \dot{\imath} t)$ for $t \in {\cal R}$.
\vspace{1mm}

Consider now an $n$-dimensional $X$.
\vspace{1mm}

From the proof result for the univariate case, we have \vspace{-2mm} 
\[ \phi_X(t) = E(e^{\dot{\imath}t'X}) = \phi_{t'X}(1) = 
M_{t'X}(\dot{\imath}) = M_X(\dot{\imath} t), \quad \quad t \in {\cal R}^n. \]
\end{alertblock}}
\frame{
\begin{block}{Theorem M2 (uniqueness)}
\begin{itemize}
\item[(i)]
Random vectors  $X$ and $Y$ have the same distribution iff the chf's $\phi_X(t)=\phi_Y(t)$ for $t \in {\cal R}^n$.
\item[(ii)]
If the mgf's of random vectors $X$ and $Y$ satisfy
$M_X(t)=M_Y(t)< \infty $ for $t$ in a neighborhood of 0, then $X$ and $Y$ have the same distribution.
\end{itemize}
\end{block}\vspace{-2mm}
\begin{block}{Proof.} 
Part (i) follows from the following multivariate inversion formula:
for any $A = (a_1,b_1) \times \cdots \times (a_n,b_n)$ such that the distribution of 
an $n$-dimensional random vector is continuous at all points in the boundary of $A$, \vspace{-2mm}
\[ P\big(X \in A \big) = \lim_{c \rightarrow \infty}
 \int_{-c}^c\cdots \int_{-c}^c \frac{\phi_X (t_1,...,t_n)}{(-1)^{n/2}
(2\pi )^n}\prod_{i=1}^n
\frac{e^{-\dot{\imath} t_ia_i}-e^{-\dot{\imath} t_ib_i}}{t_i} dt_i .\vspace{-1mm}  \]
By making use of the proved result for the case of $n=1$, however, we can have the following simpler proof. 

If $X $ and $Y$ do not have the same distribution, then by
part (i) there exists $t \in {\cal R}^k$ such that
$\phi_X(t) \neq \phi_Y(t)$. Then $\phi_{t^\tau X} (1) \neq
\phi_{t^\tau Y}(1)$, which implies that $P_{t^\tau X} \neq P_{t^\tau Y}$.
But $\psi_X=\psi_Y< \infty$ in a neighborhood of $0 \in {\cal R}^k$
implies that $\psi_{t^\tau X} = \psi_{t^\tau Y}< \infty$ in a
neighborhood of $0 \in {\cal R}$ and, by
the proved result for $k=1$,  $P_{t^\tau X} = P_{t^\tau Y}$.
This contradiction shows that $P_X=P_Y$. 
\end{block}}
\frame{
\begin{block}{}

Suppose that $X_1,X_2,...$ is a sequence of random vectors with mgf's $M_{X_n}(t)$ and \vspace{-2mm}
\[ \lim_{n\rightarrow \infty} M_{X_n}(t) = M_X(t) < \infty \quad \mbox{for all $t$ in a neighborhood of 0} \vspace{-2mm}\]
where  $M_X(t) $ is the mgf of a random vector $X$. Then, \vspace{-2mm}
\[ \lim_{n\rightarrow \infty} F_{X_n}(x) = F_X(x)\vspace{-2mm} \]
for all $x$ where
$F_X(x)$ is continuous.

Suppose that $X_1,X_2,...$ is a sequence of random vectors with mgf's $M_{X_n}(t)$ and \vspace{-2mm}
\[ \lim_{n\rightarrow \infty} M_{X_n}(t) = M_X(t) < \infty \quad \mbox{for all $t$ in a neighborhood of 0} \vspace{-2mm}\]
where  $M_X(t) $ is the mgf of a random vector $X$. Then, \vspace{-2mm}
\[ \lim_{n\rightarrow \infty} F_{X_n}(x) = F_X(x)\vspace{-2mm} \]
for all $x$ where
$F_X(x)$ is continuous.
\end{block}
}
\frame{
\begin{alertblock}{}<2->
We now consider some applications of mgf's in a family of multivariate distributions that is an extension of the univariate normal distribution family.
\end{alertblock}
\vspace{-2mm}
\begin{alertblock}{$n$-dimensional multivariate normal distribution}<2->
Let $\mu \in {\cal R}^n$,
$\Sigma$ be a positive definite $n\times n$ matrix, and
$|\Sigma |$ be the determinant of $\Sigma$.
\vspace{1mm}

The following pdf on ${\cal R}^n$,
 \vspace{-2mm}
\[  f(x) = \frac{1}{(2\pi)^{n/2} |\Sigma |^{1/2}}
\exp\left( - \frac{( x-\mu)' \Sigma^{-1} (x-\mu )}{2}\right) , \quad \quad x \in {\cal R}^n , \vspace{-2mm}\]
is called the $n$-dimensional normal pdf and the corresponding distribution
is called the $n$-dimensional normal distribution and denoted by $N(\mu, \Sigma )$.
\end{alertblock}}
\frame{
\begin{alertblock}{}
If $X \sim N(0,  \Sigma )$, then, for any $t \in {\cal R}^n$, \vspace{-2mm}
\beqrs
 M_X(t) \back &=& \back \frac{1}{(2\pi)^{n/2} |\Sigma |^{1/2}} \int_{{\cal R}^n} e^{t'x}
\exp\left( - \frac{ x' \Sigma^{-1} x}{2}\right) dx \\
\back &= & \back \frac{e^{t'\Sigma t/2}}{(2\pi)^{n/2} |\Sigma |^{1/2}}
\int_{{\cal R}^n} \exp\left( - \frac{( x- \Sigma t)' \Sigma^{-1} (x- \Sigma t  )}{2}\right) dx \\
\back & = &  \back e^{t'\Sigma t/2}
\eeqrs
$~$ \vspace{-7mm}

If $X \sim N(\mu,  \Sigma )$,  an application of the property of mgf and the previous result leads to
\[ M_X(t)=  e^{\mu't + t' \Sigma t/2} \quad \quad \quad t \in {\cal R}^n  \]
Also, from the property of mgf, we can differentiate $M_X(t)$ to get \vspace{-2mm}
\beqrs
 E(X) \back & = & \back \frac{\partial M_X (t)}{
 \partial t } \bigg|_{t=0} =  \left( \mu + \Sigma t \right)  e^{\mu't  + t' \Sigma t/2}\bigg|_{t=0}  = \mu \\
 & & \\
    E(X X' ) \back & = & \back  \frac{\partial^2 M_X (t)}{ \partial t\partial t' } \bigg|_{t=0}
 =  \frac{\partial}{\partial t} \left( \mu + \Sigma t \right)  e^{\mu't  + t' \Sigma t/2} \bigg|_{t=0}
     \eeqrs
 \end{alertblock}}
 \frame{
 \begin{alertblock}{} \vspace{-4mm}
\beqrs
 \back & = & \back    \left[ \Sigma   e^{\mu't  + t' \Sigma t/2} +
 ( \mu + \Sigma t ) ( \mu + \Sigma t )'  e^{\mu't  + t' \Sigma t/2}\right] \bigg|_{t=0} \\
 \back & = & \back \Sigma + \mu\mu'
 \eeqrs
$~$ \vspace{-7mm}

 Therefore, \vspace{-1mm}
 \[  {\rm Var} (X)  =  E(XX') - E(X)E(X') = \Sigma \]
\end{alertblock}
\vspace{-2mm}
\begin{block}{}
\begin{itemize}
\item
Like the univariate case, $\mu$ and $\Sigma$ in $N(\mu, \Sigma)$ are respectively
the mean vector and covariance matrix of the distribution.
\item
Another important consequence from the previous result is that the components of $X \sim N(\mu, \Sigma )$
are independent iff $\Sigma$ is a diagonal matrix, i.e., components of $X$ are uncorrelated.
\end{itemize}
\end{block}


}
\frame{
\begin{block}{Proof of the theorem}<2>
An $n$-dimensional random vector $X$ is  $N(\mu, \Sigma )$ iff for any $n$-dimensional constant vector $c$,
$c' X \sim N(c'\mu , c'\Sigma c)$.
\end{block}
}
\frame{\vspace{-2mm}
\begin{block}{}
\begin{itemize}
\item
If $X \sim N(\mu,\Sigma )$, then $M_X(t) = e^{\mu't + t' \Sigma t/2}$.

For any $c \in {\cal R}^n$, by the properties of mgf, the mgf of $c'X$ is \vspace{-2mm}
\[ M_{c'X} (t) = M_X(ct) = e^{\mu'(ct) + (ct)' \Sigma (ct)/2} =
e^{(c'\mu) t + (c'\Sigma c) t^2/2} \quad \quad t \in {\cal R} \vspace{-2mm} \]
which is the mgf of $N(c'\mu, c'\Sigma c)$.

By uniqueness, $c'X \sim N(c'\mu, c'\Sigma c)$.
\item
If $c'X \sim N(c'\mu, c'\Sigma c)$ for any $c \in {\cal R}^n$, then
$t'X \sim N(t'\mu, t'\Sigma t)$ for any $t \in {\cal R}^n$ and\vspace{-1mm}
\[ M_{t'X}(s) = e^{(t'\mu )s + (t'\Sigma t)s^2/2} \quad \quad s \in {\cal R} \vspace{-1mm} \]
Letting $s= 1$, we obtain \vspace{-1mm}
\[ M_{t'X}(1) = e^{(t'\mu ) + (t'\Sigma t)/2} = E(e^{t'X}) = M_X(t) \quad \quad t \in {\cal R}^n  \vspace{-1mm} \]
By uniqueness, $X \sim N(\mu,\Sigma )$.
\end{itemize}
\end{block}\vspace{-2mm}


}
\end{document}


