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%
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\title[Stat 609 Lecture 19]{Stat 609: Mathematical Statistics \\ Lecture 19}
\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 19: Multivariate expectation and covariance}
\begin{alertblock}{Multivariate expectation}
The expectation of a random vector $X=(X_1,...,X_n)$ is defined 
as $E(X) = ( E(X_1),...,E(X_n))$, provided that $E(X_i)$ exists for any $i$. 
\vspace{1mm}

When $M$ is a matrix whose $(i,j)$th element if a random variable $X_{ij}$, 
$E(M)$ is defined as the matrix whose $(i,j)$th element is $E(X_{ij})$,
provided that $E(X_{ij})$ exists for any $(i,j)$. 
\end{alertblock}\vspace{-2mm}
\begin{block}{Definition 4.5.1.}<2->
The covariance of random variables $X$ and $Y$ is defined as\vspace{-2mm}
\[ {\rm Cov}(X,Y) = E[(X-E(X)][Y-E(Y)] \vspace{-2mm}\]
provided that the expectation exists. 
\end{block}\vspace{-2mm}
\begin{block}{Definition 4.5.2.}<2->
The correlation (coefficient) of random variables $X$ and $Y$ is defined as\vspace{-2mm}
\[ \rho_{X,Y} = \frac{{\rm Cov}(X,Y)}{\sqrt{{\rm Var}(X){\rm Var}(Y)}} \]
\end{block}
}
\frame{\vspace{-2mm}
\begin{block}{}
\begin{itemize}
\item
By Cauchy-Schwartz's inequality, $[{\rm Cov}(X,Y)]^2 \leq {\rm Var}(X) {\rm Var}(Y)$ and, hence, 
$| \rho_{X,Y}| \leq 1$. 
\item
If large values of $X$ tend to be observed with large values of $Y$ and small values of $X$ with small values of $Y$, then 
Cov$(X,Y)>0$. 
\item
If large values of $X$ tend to be observed with small values of $Y$ and small values of $X$ with large values of $Y$, then
Cov$(X,Y)<0$.
\item
If Cov$(X,Y)=0$, then we say that $X$ and $Y$ are uncorrelated. 
\item
The correlation is a standardized value of the covariance.
\end{itemize}
\end{block}\vspace{-2mm}
\begin{alertblock}{Theorem 4.5.3.}<2->
For any random variables $X$ and $Y$, \vspace{-3mm}
\[ {\rm Cov}(X,Y) = E(XY) -E(X)E(Y) .\]
\end{alertblock}\vspace{-3mm}
\begin{alertblock}{}<2->
A direct calculation shows \vspace{-3mm}
\beqrs
 {\rm Cov}(X,Y) &=&  E[X-E(X)][Y-E(Y)] \\
 &=& E(XY) - E[XE(Y)]-E[YE(X)] + E(X)E(Y) \\ 
& =&   E(XY) - E(X)E(Y)-E(Y)E(X) + E(X)E(Y)\\
&=& E(XY) -  E(X)E(Y)
  \eeqrs
\end{alertblock}
}
\frame{
\begin{block}{Example 4.5.4.}
The joint pdf of $(X,Y)$ is \vspace{-2mm}
\[ f(x,y) = \left\{ \br{ll} 
1 & \quad 0<x< 1, \ x<y<x+1 \\
0 & \quad \mbox{otherwise} \er \right. \vspace{-2mm} \]
The marginal of $X$ is $uniform(0,1)$, since for $0<x<1$, \vspace{-2mm} 
\[ f_X(x) = \int_{-\infty}^\infty f(x,y) dy = \int_x^{x+1} dy = x+1-x = 1 \vspace{-2mm} \]
For $0<y<1$, \vspace{-3mm} 
\[  f_Y(y) = \int_{-\infty}^\infty f(x,y) dx = \int_0^{y} dx = y - 0 = y \vspace{-3mm} \]
and for $1 \leq y < 2$, \vspace{-3mm} 
\[  f_Y(y) = \int_{-\infty}^\infty f(x,y) dx = \int_{y-1}^1 dx = 1- (y-1)=2 - y \vspace{-2mm} \]
i.e., \vspace{-2mm} 
\[ f_Y(y) = \left\{ \br{ll} 
2-y & \quad 1 \leq  y < 2 \\
y & \quad 0< y< 1 \\
0 & \quad \mbox{otherwise} \er \right.  \]
\end{block}
}
\frame{\vspace{-2mm}
\begin{block}{}
Thus, $E(X)=1/2$ and Var$(X) = 1/12$, and \vspace{-2mm}
\[ E(Y) = \int_0^1 y^2 dy + \int_1^2 y(2-y) dy = \frac{1}{3} + 4 -1 - \frac{1}{3} (8-1) =1 \vspace{-2mm} \]
\[ {\rm Var} (Y) = E(Y^2) - 1 \!=\! \int_0^1 y^3 dy + \! \int_1^2 y^2(2-y) dy - 1 
 = \frac{1}{4} + \frac{14}{3} - \frac{15}{4} -1 = \frac{1}{6}  \vspace{-2mm} \]
 Also, \vspace{-2mm} 
 \[ E(XY) = \int_0^1 \int_x^{x+1}xy dy dx = \int_0^1 \left( x^2+ \frac{x}{2} \right) dx = \frac{1}{3} + \frac{1}{4} =\frac{7}{12}\vspace{-2mm} \]
 Hence, \vspace{-2mm} 
 \[ {\rm Cov}(X,Y) = E(XY) - E(X)E(Y) = \frac{7}{12} - \frac{1}{2} = \frac{1}{12} \vspace{-2mm} \]
 \[ \rho_{X,Y} = \frac{{\rm Cov}(X,Y) }{\sqrt{{\rm Var}(X){\rm Var}(Y)}} 
 = \frac{1/12}{\sqrt{1/12 \times 1/6}}\frac{1}{\sqrt{2}} \]
\end{block}\vspace{-2mm}
\begin{alertblock}{Theorem 4.5.6.}
If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then \vspace{-2mm} 
\[ {\rm Var}(aX+bY) = a^2 {\rm Var}(X) + b^2 {\rm Var}(Y) \]
\end{alertblock}
}
\frame{
\begin{alertblock}{}
Theorem 4.5.6 with $a=b=1$ implies that, if $X$ and $Y$ are positively correlated, then the variation in $X+Y$ is greater than the sum of the variations in $X$ and $Y$; but if they are negatively correlated, then the variation in $X+Y$ is less than the sum. 
\vspace{1mm}

This result is useful in statistical applications.
\end{alertblock}\vspace{-2mm}
\begin{block}{Theorem 4.5.7.}<2->
For any random variables $X$ and $Y$, $|\rho_{X,Y}|=1$ iff there exist constants $a \neq 0$ and $b$ such that 
$P(Y= aX+b)=1$, where $a>0$ if $\rho_{X,Y} =1$ and $a<0$ if $\rho_{X,Y} = -1$. 
\end{block}
\vspace{-2mm}
\begin{block}{}<2->
\begin{itemize}
\item
The proof of this theorem is actually discussed when we study Cauchy-Schwartz's inequality (when the equality holds).
\item
If there is a line, $y=ax+b$ with $a \neq 0$, such that the values of the 2-dimensional random vector $(X,Y)$ ahve a high probability of being near this line, then 
the correlation between $X$ and $Y$  
\end{itemize}
\end{block}
}
\frame{
\begin{block}{}
\begin{itemize}
\item
The condition expectation $E[g(X_{k+1},...,X_n)|X_1=x_1,...,X_k=x_k]$ is a function of $x_1,...,x_k$.
\item
It is an expectation of the conditional distribution. 
\item
Let $h(x_1,...,x_k) = E[g(X_{k+1},...,X_n)|X_1=x_1,...,X_k=x_k]$. \\
Then 
$h(X_1,...,X_k)$ is a random variable and is denoted by 
$E[g(X_{k+1},...,X_n)|X_1,...,X_k]$. 
\end{itemize}
\end{block} \vspace{-2mm}
\begin{alertblock}{Example 4.2.4.}<2->
Since $Y|X \sim exponential(X,1)$, the conditional expectation of $Y$ given $X$ is $E(Y|X) = 1+X$, and the conditional expectation of $[Y-E(Y|X)]^2$ given $X$ is \vspace{-3mm}
\beqrs
 E\{ [Y-E(Y|X)]^2|X\} &=& \int_X^\infty [y-(1+X)]^2 e^{-y}dy \\
 &=&  \int_X^\infty y^2 e^{-y}dy - (1+X)^2 \\
 &=& 1+ (1+X)^2 - (1+X)^2 = 1 
 \eeqrs
Note that the function $g$ may depend on $X$ (treated as a constant).   
\end{alertblock}}
\frame{
\begin{block}{Conditional mean and variance}
In general, if $X$ and $Y$ are random vectors, then the conditional mean and variance of $Y$ given $X$ are simply the mean and variance of the conditional distribution of $Y$ given $X$.

\vspace{1mm}
It can be shown that \vspace{-2mm}
\[ {\rm Var}(Y|X) = E(YY' |X) - [E(Y|X)][E(Y|X)]' \]
\end{block}
}
\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}


