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%
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%
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\title[Stat 609 Lecture 15]{Stat 609: Mathematical Statistics \\ Lecture 15}
\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]}}

\begin{document}

%\frame{\titlepage}

\frame {
\frametitle{Lecture 15: Joint distributions, mgf's, and chf's}\vspace{-2mm}
\begin{block}{Bivariate normal distribution}
The bivariate normal distribution has the following pdf on ${\cal R}^2$: \vspace{-2mm}
\[ f(x,y)=  \frac{\exp \left( - \frac{(x - \mu_1)^2}{2 \sigma_1^2(1-\rho^2)}
+ \frac{\rho (x-\mu_1)(y-\mu_2)}{\sigma_1\sigma_2(1-\rho^2)}
- \frac{(y - \mu_2 )^2}{2 \sigma_2^2(1-\rho^2)} \right)}{2\pi \sigma_1\sigma_2 \sqrt{1-\rho^2}}\quad \quad (x,y) \in {\cal R}^2 \vspace{-2mm}\]
where $\mu_1 \in {\cal R}$, $\mu_2\in {\cal R}$, $\sigma_1>0$, $\sigma_2>0$, and
$-1 < \rho < 1$ are constants.
\vspace{-4mm}

\begin{figure}[h]
\begin{center}
%\caption{Graph of $y=\sin^2(x)$ in Example 2.1.2.}
\includegraphics[height=2.9in, width=4.5in]{figure400.eps}
\end{center}
\end{figure}
\end{block}}
\frame{
\begin{block}{}
What is the marginal pdf of $X$? \vspace{-2mm}
\beqrs
f_X(x) &=& \int_{-\infty}^\infty f(x,y) dy  =
 \frac{\exp \left( - \frac{(x - \mu_1)^2}{2 \sigma_1^2(1-\rho^2)}\right)}{2\pi \sigma_1\sigma_2 \sqrt{1-\rho^2}} \\
 & & \times \int_{-\infty}^\infty
\exp \left( \frac{\rho (x-\mu_1)(y-\mu_2)}{\sigma_1\sigma_2(1-\rho^2)}
- \frac{(y - \mu_2 )^2}{2 \sigma_2^2(1-\rho^2)} \right) dy \\
& = & \frac{\exp \left( - \frac{(x - \mu_1)^2}{2 \sigma_1^2}\right)}{2\pi \sigma_1\sigma_2 \sqrt{1-\rho^2}}
\int_{-\infty}^\infty
\exp \left( - \frac{[(y - \mu_2 )-\frac{\rho\sigma_2}{\sigma_1} (x-\mu_1)]^2}{2 \sigma_2^2(1-\rho^2)} \right) dy \\
& = & \frac{1}{\sqrt{2\pi} \sigma_1} \exp \left( - \frac{(x - \mu_1)^2}{2 \sigma_1^2}\right)
\eeqrs
which is the pdf of $N(\mu_1,\sigma_1^2)$ (as expected).
\vspace{1mm}

Similarly, the marginal pdf of $Y$ is that of $N(\mu_2,\sigma_2^2)$.
\vspace{1mm}

Is the converse true? That is, if both $X$ and $Y$ are normal,
should the joint pdf of $(X,Y)$ be always normal?
\end{block}
}
\frame{\vspace{-2mm}
\begin{alertblock}{A counter-example}
Consider  \vspace{-2mm}
\[ f(x,y) = \frac{1}{2\pi} e^{-(x^2+y^2)/2} [1+ g(x,y)] , \quad \quad (x,y) \in {\cal R}^2 \vspace{-2mm} \]
where
\[ g(x,y) = \left\{ \br{ll}
xy & \quad -1<x<1 , \ -1<y<1 \\
0 & \quad \mbox{otherwise} \er \right. \vspace{-1mm} \]
Then $f(x,y) \geq 0$. (Is it a pdf?) \vspace{-2mm}
\beqrs
  \int_{-\infty}^\infty f(x,y)dy  &=&
\frac{1}{2\pi} \left[ \int_{-\infty}^\infty e^{-(x^2+y^2)/2} dy +
\int_{-\infty}^\infty   e^{-(x^2+y^2)/2} g(x,y) dy \right]  \\
& = & \frac{1}{2\pi}e^{-x^2/2} \left[ \int_{-\infty}^\infty e^{-y^2/2} dy
  + x I(|x|<1)\int_{-1}^1   e^{-y^2/2} y dy \right]  \\
  & = & \frac{1}{\sqrt{2\pi}}e^{-x^2/2}
 \eeqrs
 i.e., the marginal of $X$ is $N(0,1)$ (which also shows that $f(x,y)$ is a pdf).

Similarly, the pdf of $Y$ is $N(0,1)$.
\vspace{1mm}

However, this $f(x,y)$ is certainly not a bivariate normal pdf.
\end{alertblock}
}
\frame{\vspace{-2mm}
\begin{block}{}
Other joint distributions may not have the property that the marginal distributions are of the same type as the joint distribution.
\end{block}\vspace{-2mm}
\begin{alertblock}{Uniform distributions}
A general $n$-dimensional uniform distribution on a set $A \subset {\cal R}^n$ has pdf \vspace{-2mm}
\[ f(x) = \left\{ \br{ll}
1/C & \quad x \in A \\
0 & \quad x \not \in A \er \right. \quad\quad\quad \quad  C = \int_A dx \vspace{-2mm} \]
For $n=1$ and $A =$ an interval, $C$ is the length of the interval.

For $n=2$, $C$ is the area of the set $A$.

Consider for example $n=2$ and $A$ is the rectangle \vspace{-2mm}
\[ A = [a, b] \times [c,d] = \{ (x,y) \in {\cal R}^2: \ a \leq x \leq b , \ c \leq y \leq d \} \vspace{-2mm} \]
where $a$, $b$, $c$ and $d$ are constants.

In this case, the two marginal distributions are uniform distributions on intervals $[a,b]$ and $[c,d]$, since, when $x \in [a,b]$, \vspace{-2mm}
\[ \int_{-\infty}^\infty f(x,y) dy = \frac{1}{C} \int_c^d dy = \frac{1}{(b-a)(d-c)} \int_c^d dy = \frac{1}{b-a}  \vspace{-2mm} \]
and     $\int_{-\infty}^\infty f(x,y) dy =0$ when $x \not \in [a,b]$.
\end{alertblock}
}
\frame{\vspace{-2mm}
\begin{alertblock}{}
However, if $A$ is a disk \vspace{-2mm}
\[ A = \{ (x,y) \in {\cal R}^2: \ x^2+y^2 \leq 1 \} \vspace{-2mm} \]
then $C = \pi$ and, when $x \in [-1,1]$, \vspace{-2mm}
\[ \int_{-\infty}^\infty f(x,y) dy = \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\pi} dy = \frac{2\sqrt{1-x^2}}{\pi} \vspace{-2mm} \]
and $\int_{-\infty}^\infty f(x,y) dy = 0$ when $x \not \in [-1,1]$.

This shows that the marginal distributions are not uniform.
\vspace{-5mm}
\begin{figure}[h]
\begin{center}
%\caption{Graph of $y=\sin^2(x)$ in Example 2.1.2.}
\includegraphics[height=3.5in, width=4.5in]{figure404.eps}
\end{center}
\end{figure}
\end{alertblock}
}
\frame{
\begin{block}{$n$-dimensional multivariate normal distribution}
The $n$-dimensional normal distribution, denoted by $N(\mu, \Sigma )$,
 has the following pdf on ${\cal R}^n$: \vspace{-2mm}
\[  f(x) = \frac{1}{(2\pi)^{n/2} |\Sigma |^{1/2}}
\exp\left( - \frac{( x-\mu)' \Sigma^{-1} (x-\mu )}{2}\right) , \quad \quad x \in {\cal R}^n ,\]
where
$\mu \in {\cal R}^n$,
$\Sigma$ is a positive definite $n\times n$ matrix,
$|\Sigma |$ is the determinant of $\Sigma$
and $c'$ denotes the transpose of a column vector $c$.
\end{block}\vspace{-2mm}
\begin{alertblock}{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{alertblock}\vspace{-2mm}
\begin{alertblock}{}<3->
To prove this theorem, we need the tool of mgf.
\end{alertblock}\vspace{-2mm}
\begin{block}{Multivariate mgf}<4->
For an $n$-dimensional random vector, its mgf is defined as (if the integral exists)
\vspace{-1mm}
\[ M_X(t) = E(e^{t'X}), \quad \quad t \in {\cal R}^n \]
\end{block}
}
\frame{\vspace{-3mm}
\begin{block}{Properties of mgf's}
\begin{itemize}
\item
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{-3mm}
\[ \frac{\partial M_X (t)}{
 \partial t } \bigg|_{t=0} = E(X) = \left( \br{c}
E(X_1) \\
\vdots \\
E(X_n) \er \right)  \vspace{-2mm}\]
\[ \frac{\partial^2 M_X (t)}{ \partial t\partial t' }
 \bigg|_{t=0} = E(X X' )  = \left( \br{cccc}
 E(X_1^2) & E(X_1X_2) & \cdots & E(X_1X_n) \\
 E(X_2X_1) & E(X_2^2) & \cdots & E(X_2X_n) \\
 \cdots & \cdots & \cdots & \cdots \\
 E(X_nX_1) & E(X_nX_2) & \cdots & E(X_n^2) \er \right) \vspace{-1mm} \]
 \item
 For any fixed $k \times n$ matrix $A$ and vector $b \in {\cal R}^k $, the mgf of the random vector $A'X+b$ is \vspace{-1mm}
\[ M_{A'X+b}(t)= e^{b't}M_X(At) \quad\quad t \in {\cal R}^k \]
\end{itemize}
\end{block}}
\frame{\vspace{-2mm}
\begin{block}{}
\begin{itemize}
\item
For random vectors $X$ and $Y$, if $M_X(t)=M_Y(t)< \infty $ for $t$ in a neighborhood of 0, then $X$ and $Y$ have the same distribution.
\item
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{itemize}
\end{block}\vspace{-2mm}
\begin{alertblock}{$n$-dimensional multivariate normal distribution}<2->
If $X \sim N(0,  \Sigma )$, then \vspace{-2mm}
\beqrs
 M_X(t) &=& \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 \\
&= & \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 \\
& = &  e^{t'\Sigma t/2}
\eeqrs
\end{alertblock}
}
\frame{
\begin{alertblock}{}
If $X \sim N(\mu,  \Sigma )$,  an application of the property of mgf leads to \vspace{-2mm}
\beqrs
M_X(t) &=&  e^{\mu't + t' \Sigma t/2} \quad \quad \quad t \in {\cal R}^n  \\
 E(X) &=& \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' ) &=& \frac{\partial^2 M_X (t)}{ \partial t\partial t' }  \bigg|_{t=0} \\
  &=&  \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} \\
 & = & \Sigma + \mu\mu' \\
 {\rm Var} (X) & = & E(XX') - E(X)E(X') = \Sigma
 \eeqrs
\end{alertblock}\vspace{-2mm}
\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}
\begin{alertblock}{}<2->
The condition {\bf any} $c \in {\cal R}^n$ is important.
\vspace{1mm}

In fact, we can show that if $X \sim N(\mu, \Sigma )$ and $A$ is a fixed matrix and $b$ is a fixed vector, then
$A'X+b \sim N( A' \mu + b , A' \Sigma A)$.
\end{alertblock}}
\frame{
\begin{block}{The uniform distribution on $ [a, b] \times [c,d]$}
We have shown that the two marginal distributions are uniform distributions on intervals $[a,b]$ and $[c,d]$.

For non-zero constants $\xi $ and $\zeta$, is the distribution of $\xi X + \zeta Y$ a uniform distribution on some interval?

If $(e^{bt}-e^{at})/t$ is defined to be $b-a$ when $t=0$ for any constants $a < b$, then
\vspace{-2mm}
\beqrs
M_{X,Y}(t,s) &=& \int_a^b \int_c^d e^{tx+sy} \frac{1}{(b-a)(d-c)} dx dy \\
&=& \frac{(e^{bt}-e^{at})(e^{ds}-e^{cs})}{(b-a)(d-c)ts} \quad \quad s, t \in {\cal R} \vspace{-2mm}
\eeqrs
and\vspace{-2mm}
\[ M_{\xi X+ \zeta Y}(t) = E(e^{t(\xi X + \zeta Y)}) = \frac{(e^{b\xi t}-e^{a\xi t})(e^{d \zeta t}-e^{c \zeta t})}{(b-a)(d-c)\xi \zeta t^2} \quad \quad t \in {\cal R}
\vspace{-2mm} \]
This is not a mgf of a uniform distribution on an interval $[r,h]$, which is of the form
$(e^{ht}-e^{rt})/[t(h-r)]$ for $t \in {\cal R}$.
\end{block}
}
\end{document}


