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\title[Stat 609 Lecture 21]{Stat 609: Mathematical Statistics \\ Lecture 21}
\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 {\vspace{-2mm}
\frametitle{Lecture 21: Multivariate normal distributions}
\begin{alertblock}{}\vspace{-2mm}
We now  study properties of the multivariate normal distribution defined in the last lecture.
\end{alertblock}\vspace{-2mm}
\begin{alertblock}{Bivariate normal distributions}
First, we want to show that the definition for the general multivariate normal distribution with $n=2$ is consistent with the early definition of the bivariate normal distribution.
\vspace{1mm}

Let $\mu_1 \in {\cal R}$, $\mu_2\in {\cal R}$, $\sigma_1>0$, $\sigma_2>0$, and
$-1 < \rho < 1$ be constants, and
\vspace{-2mm}
\[ \mu = \left( \br{l}
\mu_1 \\
\mu_2 \er \right) \quad \quad \Sigma = \left( \br{cc}
\sigma_1^2 & \rho \sigma_1\sigma_2 \\
\rho \sigma_1\sigma_2 & \sigma_2^2 \er \right)
\quad\quad x = \left( \br{l}
x_1 \\
x_2 \er \right) \vspace{-3mm} \]
Then\vspace{-1mm}
\[ | \Sigma |  = \sigma_1^2 \sigma_2^2 - \rho^2 \sigma_1^2\sigma_2^2 = \sigma_1^2\sigma_2^2 (1-\rho^2)
\]
\[ \Sigma^{-1}  = \frac{1}{ \sigma_1^2\sigma_2^2 (1-\rho^2)}  \left( \br{cc}
\sigma_2^2 & - \rho  \sigma_1\sigma_2 \\
- \rho \sigma_1\sigma_2 & \sigma_1^2 \er \right)  \]
\[  ( x-\mu)' \Sigma^{-1} (x-\mu ) = \frac{1}{|\Sigma|} \!\!  \left( \br{c}
x_1 - \mu_1 \\
x_2 - \mu_2 \er \right)' \!\!\!
 \left( \br{cc}
\sigma_2^2 & - \rho \sigma_1\sigma_2 \\
- \rho \sigma_1\sigma_2 & \sigma_1^2 \er \right) \!\!\!
\left( \br{c}
x_1 - \mu_1 \\
x_2 - \mu_2 \er \right)
\]
\end{alertblock}}
\frame{\vspace{-2mm}
\begin{alertblock}{}\vspace{-5mm}
\beqrs
\back & = & \back
\frac{1}{|\Sigma|} \!\!  \left( \br{c}
x_1 - \mu_1 \\
x_2 - \mu_2 \er \right)' \!\!\!
 \left( \br{c}
\sigma_2^2(x_1-\mu_1) - \rho  \sigma_1\sigma_2 (x_2-\mu_2) \\
- \rho \sigma_1\sigma_2 (x_1-\mu_1) + \sigma_1^2 (x_2-\mu_2) \er \right) \\
\back & = & \back \frac{\sigma_2^2 (x_1-\mu_1)^2 - 2\rho \sigma_1\sigma_2(x_1-\mu_1)(x_2-\mu_2)
+ \sigma_1^2 (x_2-\mu_2)^2}{ \sigma_1^2\sigma_2^2 (1-\rho^2)} \\
\back & = & \back \frac{(x_1 - \mu_1)^2}{ \sigma_1^2(1-\rho^2)}
- \frac{2 \rho (x_1-\mu_1)(x_2-\mu_2)}{\sigma_1\sigma_2(1-\rho^2)}
+ \frac{(x_2 - \mu_2 )^2}{ \sigma_2^2(1-\rho^2)}
\eeqrs
$~$ \vspace{-4.5mm}

Thus, when $n=2$, \vspace{-3mm}
\beqrs
  f(x)\back &  = & \back \frac{1}{(2\pi)^{n/2} |\Sigma |^{1/2}}
\exp\left( - \frac{( x-\mu)' \Sigma^{-1} (x-\mu )}{2}\right) \\
\back & =  &\back
 \frac{\exp \left( - \frac{(x_1 - \mu_1)^2}{2 \sigma_1^2(1-\rho^2)}
+ \frac{\rho (x_1-\mu_1)(x_2-\mu_2)}{\sigma_1\sigma_2(1-\rho^2)}
- \frac{(x_2 - \mu_2 )^2}{2 \sigma_2^2(1-\rho^2)} \right)}{2\pi \sigma_1\sigma_2 \sqrt{1-\rho^2}}
\eeqrs
$~$ \vspace{-5.5mm}

which is the same as the bivariate normal pdf defined earlier.
\end{alertblock}
\vspace{-2.5mm}
\begin{block}{Transformation and marginals}<2->
If $X$ is an $n$-dimensional random vector $ \sim N(\mu, \Sigma)$ and $Y = AX + b$, where $A$ is a fixed $k \times n$ matrix
and $b$ is a fixed $k$-dimensional vector, then $Y \sim N( A\mu , A\Sigma A')$.
\end{block}
}
\frame{
\begin{block}{}
This can be proved by using multivariate transformation, or the mgf. 
\vspace{1mm}

In the last lecture we showed that the mgf of $X \sim N(\mu, \Sigma )$ is \vspace{-2mm} 
\[ M_X(t)=  e^{\mu't + t' \Sigma t/2} \quad \quad \quad t \in {\cal R}^n  \vspace{-2mm} \]
From the properties of mgf, $Y = AX + b$, 


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}
\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}{}
An $n$-dimensional random vector $X$ is  $N(\mu, \Sigma )$ iff for any $n$-dimensional constant vector $c$,
$c' X \sim N(c'\mu , c'\Sigma c)$.
\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}
}
\frame{
\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 $t \in {\cal R}$.

In general, we can show that,
if $(X,Y)$ is a random vector  $\sim N(\mu, \Sigma )$, \vspace{-2mm}
\[ \mu = \left( \br{cc}
\mu_1 \\
\mu_2 \er \right), \quad\quad \quad \Sigma = \left( \br{cc}
\Sigma_{11} & \Sigma_{12} \\
\Sigma_{21} & \Sigma_{22} \er \right) ,\vspace{-2mm} \]
then \vspace{-2mm}
\[ Y|X \sim N\left( \mu_2 +  \Sigma_{21} \Sigma_{11}^{-1} (X-\mu_1)  ,
\Sigma_{22}-\Sigma_{21}\Sigma^{-1}_{11}\Sigma_{12}\right) \]
It follows from the properties of normal distributions that \vspace{-2mm}
\[ E(Y|X) =  \mu_2 +  \Sigma_{21} \Sigma_{11}^{-1} (X-\mu_1) \quad \quad  {\rm Var}(Y|X) = \Sigma_{22}-\Sigma_{21}\Sigma^{-1}_{11}\Sigma_{12} \vspace{-2mm}
 \]
While the conditional mean depends on $X$, the conditional variance does not.
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