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\title[Stat 609 Lecture 1]{Stat 609: Mathematical Statistics \\ Lecture 2}
\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 2: Measurable space, measure and probability}

\begin{alertblock}{Random experiment}
An experiment (in a general sense) that results in one of more than one outcomes
with uncertainty in which outcome will be the realization
\end{alertblock}
\begin{block}{Examples of random experiments}<2->
\begin{itemize}
\item
Tomorrow's weather: Sunny, Cloud, Rain
\item
Tomorrow's temperature: $(-M , M)$
\item
Number of car accidents in a city: 0, 1, 2, 3, ...
\item
The life time of a television set: $(0, \infty )$
\item
Toss a coin: Heads or Tails?
\end{itemize}
\end{block}
\begin{alertblock}{Statistical inference}<3->
Study
    \end{alertblock}
}
\frame{
\begin{alertblock}{Probability theory}
A basic tool for statistics, which explains a random experiment in terms of frequencies and their characteristics
\vspace{2mm}

For example, some past observations (data) may tell us that tomorrow's weather is sunny with 80\%, cloud with 15\%, and raining with 5\%.
\vspace{2mm}

To study statistical inference, we must have a very good knowledge in probability theory.
\end{alertblock}
\vspace{-2mm}
\begin{block}{Sample space $S$ or outcome space $\Omega$}<2->
A set containing all possible outcomes
    \end{block}
\vspace{-2mm}
\begin{alertblock}{Example}<2->
$S = \{Heads, Tails\}$, $= \{0, 1\}$, $= \{0, 1, 2, 3, ... \}$,
$ = (-M,M)$, $= {\cal R}$ (all real numbers), $= (0 , \infty )$
\vspace{2mm}

Discrete sample space: $S$ contains integers (finitely or infinitely many)
\vspace{2mm}

Continuous sample space: e.g., $S = [0,1]$, $ {\cal R}$, or ${\cal R}^2$.
\end{alertblock}
}
\frame{
\begin{block}{Subsets or sets}
In probability and statistics, we are interested in subsets of a given sample space $S$
(sometime called sets for simplicity).
\vspace{2mm}

$A \subset S$: $A$ is a subset of $S$.
\vspace{2mm}

Example: the life time of a television set is between 0 and 100 is a subset $A = (0, 100]$ of the sample space $S = (0, \infty )$.
\vspace{2mm}

Two very special subsets:
\begin{itemize}
\item
Empty set: $\emptyset $ is the subset of $S$ containing no element.
\item
The sample space $S$ itself is a subset of $S$ (the notation $A \subset S$ allows the special case of $A = S$).
\end{itemize}
\end{block}
\begin{alertblock}{Expressions of subsets}<2->
A subset is  expressed in words or math formulas inside of $\{ \ \ \}$.
\vspace{2mm}

Example:\\
$A = \{ x: \mbox{$x$ satisfies some condition}\}$\\
$A= \{ x: x \in {\cal R} \mbox{ and } x \geq 0 \} = [0, \infty )$
\end{alertblock}
}
\frame{ \vspace{-2mm}
\begin{alertblock}{Relationship}
\begin{itemize}
\item
$A\subset B$: $  x \in A \Rightarrow x \in B$
\item
$A=B $: $A \subset B$ and $B \subset A$
\item
Disjoint: $A$ and $B$ have no common elements
\end{itemize}
\end{alertblock}\vspace{-2mm}
\begin{block}{Example}<2->
$A = \{ \mbox{all even integers} \}$\\
$B = [2, \infty )$
\vspace{2mm}

Obviously, for any $x \in A$, $x \geq 2$ and thus $x \in B$. Hence $A \subset B$.
\end{block}\vspace{-2mm}
\begin{alertblock}{A more complicated example}<3->
Let \vspace{-2mm}
\[ A = \{ x \in {\cal R}^k: x^T x \leq c \} \vspace{-2mm}\]
where $c>0$ is a given constant, $x^T$ is the transpose of $x$, and $a^Tb$ is the usual vector product,
and\vspace{-2mm}
\[ B= \left\{ x \in {\cal R}^k:  (y^T x)^2 \leq c y^Ty \ \ \mbox{for all $y \in {\cal R}^k$}\right\}\vspace{-2mm}\]
We want to show $A=B$.
\end{alertblock}
}
\frame{
\begin{alertblock}{A more complicated example}
First, we show $A \subset B$.\\ For any $x \in A$, $x^Tx \leq c$ and by the Cauchy-Schwartz inequality,
\[  (y^T x)^2 \leq x^T x y^Ty \leq c (y^Ty) \]
for all $y \in {\cal R}^k$. \\
This means $x \in B$ and thus $A \subset B$.
\vspace{2mm}

Next, we show $B \subset A$. \\
It is obvious $0 \in B$ and $0 \in A$.\\
If $x \in B$ and $x \neq 0$,
\[ (y^T x)^2 \leq c y^Ty \ \ \mbox{for all $y \in {\cal R}^k$} \]
In particular, $y=x$ should satisfy this inequality, which means
\[ (x^T x)^2 \leq c x^Tx \quad \Rightarrow \quad x^Tx \leq c  \]
This means $x \in A$ and thus $B \subset A$.
\end{alertblock}
}
\frame{\vspace{-2mm}
\begin{block}{Set operation}
\begin{itemize}
\item
The union of $A$ and $B$: $A \cup B = \{ x\!\! : x \in A \mbox{ or } x \in B\}$
\item
The intersection of $A$ and $B$: $ A  \cap B = \{ x\!\! :  x  \in A
\mbox{ and } x \in B\}$

Disjoint $A$ and $B$: $A\cap B = \emptyset $ (the empty set)
\item
The complement of $A$: $A^c = \{ x: x \not \in A\}$, $S^c = \emptyset$
\item
Difference of $A$ and $B$: $A-B = A\cap B^c$
\end{itemize}
\end{block}\vspace{-2mm}
\begin{alertblock}{Venn Diagram}
\begin{figure}[htbp]
\begin{center}
\includegraphics[height=2.0in, width=3.0in]{vinn.eps}
%\centerline{\psfig{figure=vinn1.eps,height=2.1in,width=2.1in}}
\end{center}
\end{figure}
\end{alertblock}
}
\frame{
    \begin{block}{Definition 1.1}
     Let ${\cal F}$ be a collection of subsets of a
sample space $\Omega$.

${\cal F}$ is called a $\sigma$-field (or
$\sigma$-algebra) iff it has the following properties.\\
(i) The empty set $\emptyset \in {\cal F}$.\\
(ii) If $A \in {\cal F}$, then the complement $A^c \in {\cal F}$.\\
(iii) If $A_i \in {\cal F}$, $i=1,2,...$, then their union $\cup A_i
\in {\cal F}$.

$(\Omega , {\cal F})$ is a measurable space if ${\cal F}$ is a $\sigma$-field on $\Omega$
    \end{block}\vspace{-3mm}
\begin{alertblock}{Discussion}<4->
${\cal F}$ is a collection (set) of sets

Two trivial examples: ${\cal F}$ contains $\emptyset$ and $\Omega$
only; \\
${\cal F}$ contains all subsets of $\Omega$ (power set)
\end{alertblock}
}

\frame {
    \begin{alertblock}{Why do we need to consider other $\sigma$-field?}
We may be interested in a particular collection of sets

e.g., ${\cal F} =  \{ \emptyset , A, A^c , \Omega \}$, where $A \subset
\Omega$

${\cal C} =$ a collection of sets of interest

${\cal C}$ may not be a $\sigma$-field

$\sigma({\cal C})$: the smallest $\sigma$-field
containing ${\cal C}$ (the $\sigma$-field generated by ${\cal
C}$)

$\sigma({\cal C})= {\cal C}$ if ${\cal C}$ itself is a $\sigma$-field

$\Gamma = \{ {\cal F}: {\cal F} $ is a $\sigma$-field on
$\Omega$ and ${\cal C} \subset {\cal F} \}$

$\sigma({\cal C}) = \cap_{ {\cal F} \in \Gamma} {\cal F}$

 $\sigma( \{ A \})= \sigma( \{ A, A^c\}) = \sigma( \{ A, \Omega \})
= \sigma( \{ A, \emptyset \}) = \{ \emptyset , A, A^c , \Omega \}$
\end{alertblock}
\begin{overprint}
     \begin{block}{Borel $\sigma$-field}<2->
 ${\cal R}^k$: the $k$-dimensional Euclidean space
(${\cal R}^1 = {\cal R}$ is the real line)

${\cal O}$: the collection of all open sets

${\cal B}^k = \sigma ( {\cal O})$: the Borel $\sigma$-field on ${\cal R}^k$

${\cal B}^k = \sigma ( {\cal C})$, ${\cal C} $ is the collection of all closed sets

$C \in {\cal B}^k$, ${\cal B}_C= \{ C \cap B: B \in {\cal B}^k\}$
is the Borel $\sigma$-field on $C$
     \end{block}
\end{overprint}

}

\frame {
\begin{alertblock}{Measure}
Length, area, volume, ...
\end{alertblock}
\begin{overprint}
     \begin{block}{Definition 1.2.}<2->
Let $(\Omega, {\cal F})$ be a measurable space.

A set function $\nu$
defined on ${\cal F}$ is called a {\em
measure} iff it has the following properties.\\
(i) $0 \leq \nu(A) \leq \infty$ for any $A\in {\cal F}$.\\
(ii) $\nu ( \emptyset ) =0$.\\
(iii) If $A_i \in {\cal F}$, $i=1,2,...$, and $A_i$'s are disjoint,
i.e., $A_i \cap A_j = \emptyset$ for any $i \neq j$, then
\[ \nu\left( \bigcup_{i=1}^\infty A_i \right) = \sum_{i=1}^\infty \nu(A_i)
.  \]
$(\Omega , {\cal F}, \nu )$ is a measure space if $\nu$ is a measure on
a $\sigma$-field on $\Omega$.
\end{block}
\end{overprint}

}

\frame {
\begin{alertblock}{Conventions}
\begin{itemize}
  \item For any $x \in {\cal R}$, $\infty + x = \infty$, $x \, \infty =
\infty$ if $x>0$, $x \, \infty = - \infty$ if $x<0$, and $0 \,
\infty = 0$;
  \item $\infty + \infty = \infty$;
  \item $\infty^a = \infty$ for any $a>0$;
  \item $\infty - \infty$ or $\infty / \infty$ is not defined
\end{itemize}
\end{alertblock}
\begin{overprint}
\begin{block}{Probability measure}<2->
If $\nu ( \Omega )=1$, then $\nu$ is a probability measure.\\
We usually use notation $P$ instead of $\nu$.\\
$(\Omega , {\cal F}, P)$ is a probability space if $P$ is a probability measure on
a $\sigma$-field on $\Omega$.
\end{block}
\begin{alertblock}{Important examples of measures}<3->
\begin{itemize}
\item
Measures take $\infty$ as its value:
\[ \nu(A) =
\left\{ \begin{array}{ll}
    \infty & \quad A \in {\cal F}, A \neq \emptyset \\
    0 & \quad A=\emptyset.
  \end{array}
\right.
\]
\end{itemize}
\end{alertblock}
\end{overprint}
}

\frame {
\begin{alertblock}{Important examples of measures}
\begin{itemize}
\item
Point mass:

Let $x \in \Omega$ be a fixed point.
\[ \delta_x (A) =
\left\{ \begin{array}{ll}
    c & \quad x \in A \\
    0 & \quad x \not \in A.
  \end{array}
\right.
\]
\item
Counting measure:

Let $\Omega$ be a sample space, ${\cal F}$ the collection of all
subsets, and $\nu(A)$
 the number of elements in $A \in {\cal F}$ ($\nu(A)=\infty$ if $A$
contains infinitely many elements).

Then $\nu$ is a measure on
${\cal F}$ and is called the {\em counting measure}.
\item
Lebesgue measure:

There is a unique measure $m$ on $({\cal R}, {\cal B})$ that
satisfies $ m([a,b])=b-a$ for every finite interval $[a,b]$,
$-\infty < a \leq b < \infty$.

This is called the {\em Lebesgue
measure}.

If we restrict $m$ to the measurable space $([0,1],{\cal
B}_{[0,1]})$, then $m$ is a probability measure.
\end{itemize}
     \end{alertblock}
}

\frame {
\begin{block}{Proposition 1.1 (Properties of measures)}
 Let $(\Omega, {\cal F}, \nu)$ be a measure space.
\begin{enumerate}
  \item (Monotonicity). If $A \subset B$, then $\nu (A) \leq \nu(B)$.
  \item (Subadditivity). For any sequence $A_1,A_2,...$,
\[ \nu\left( \bigcup_{i=1}^\infty A_i \right) \leq
 \sum_{i=1}^\infty \nu(A_i).\]
  \item (Continuity). If $A_1 \subset A_2 \subset A_3 \subset \cdots$
(or $A_1 \supset A_2 \supset A_3 \supset \cdots$ and $\nu(A_1)<
\infty$), then
\[ \nu \left( \lim_{n\rightarrow \infty } A_n \right)
=\lim_{n\rightarrow \infty } \nu \left(  A_n \right), \] where
\[\lim_{n\rightarrow \infty }A_n = \bigcup_{i=1}^\infty A_i
\quad \left( \hbox{ or } = \bigcap_{i=1}^\infty A_i \ \right) . \]
\end{enumerate}
\end{block}
}

\frame {

\begin{alertblock}{Cumulative distribution function)}
Let $P$ be a probability measure on $({\cal R}, {\cal B})$.

The {\em cumulative distribution
function} (c.d.f.) of $P$ is defined to be
\[  F(x)= P\left( (-\infty, x]\right), \quad x \in {\cal R} \]
\end{alertblock}
\begin{overprint}
     \begin{block}{Proposition 1.2 (Properties of c.d.f.'s)}<2->
     \begin{enumerate}
       \item[(i)] Let $F$ be a c.d.f.\
on ${\cal R}$.
         \begin{enumerate}
           \item[(a)] $F(-\infty) = \lim_{x \rightarrow - \infty} F(x) =0$;
           \item[(b)] $F(\infty) = \lim_{x \rightarrow \infty} F(x) =1$;
           \item[(c)] $F$ is nondecreasing, i.e., $F(x)\leq F(y)$ if $x \leq y$;
           \item[(d)] $F$ is right continuous, i.e., $\lim_{y \rightarrow x,
y > x} F(y) = F(x)$.
         \end{enumerate}
       \item[(ii)] Suppose a real-valued function $F$ on ${\cal R}$ satisfies
(a)-(d) in part (i).

Then $F$ is the c.d.f.\ of a unique probability
measure on $({\cal R}, {\cal B})$.
     \end{enumerate}
     \end{block}
\end{overprint}

}
\frame { %\frametitle{Lecture 2: Product measure, measurable function and distribution}

\begin{block}{Product space}
${\cal I}= \{ 1,...,k\}$, $k$ is finite or $\infty$

$\Gamma_i$, $i \in {\cal I}$, are some sets

$ \prod_{i \in {\cal I}}\Gamma_i
= \Gamma_1 \times \cdots \times \Gamma_k = \{ (a_1,...,a_k): a_i \in
\Gamma_i, i\in {\cal I}\}$

${\cal R} \times {\cal R} = {\cal R}^2$,
${\cal R} \times {\cal R}\times {\cal R} = {\cal R}^3$
\end{block}
\begin{overprint}
    \begin{alertblock}{Product $\sigma$-field}<2->
Let $(\Omega_i , {\cal F}_i  )$, $i \in {\cal I}$, be measurable
spaces

$\prod_{i \in {\cal I}} {\cal F}_i
$ is not necessarily a $\sigma$-field

$\sigma\left(\prod_{i \in {\cal I}} {\cal F}_i \right)$ is called the {\em
product} $\sigma$-{\em field} on the {\em product space}
$\prod_{i\in {\cal I}} \Omega_i $

$\left(\prod_{i \in {\cal I}} \Omega_i , \sigma \left(\prod_{i \in {\cal I}}
{\cal F}_i\right)\right) $ is denoted by $\prod_{i \in {\cal I}} (
\Omega_i, {\cal F}_i)$

Example: $\prod_{i=1,...,k} ( {\cal R}, {\cal B}) = ({\cal R}^k, {\cal B}^k)$
    \end{alertblock}
\end{overprint}

}

\frame {
\begin{block}{Product measure}
Consider a rectangle $[a_1,b_1]\times [a_2,b_2] \subset {\cal R}^2$.

The
usual area of $[a_1,b_1]\times [a_2,b_2]$ is
\[  (b_1-a_1)(b_2-a_2) = m([a_1,b_1])m([a_2,b_2])
\]
Is $ m([a_1,b_1])m([a_2,b_2])$ the same as the value of a measure
defined on the product $\sigma$-field?
\end{block}
\begin{overprint}
\begin{alertblock}{$\sigma$-finite}<2->
A measure $\nu$ on $(\Omega, {\cal F})$ is
said to be $\sigma$-{\em finite} iff there exists a
sequence
 $\{ A_1,A_2,... \}$ such that
$\cup A_i = \Omega$ and $\nu (A_i) < \infty$ for all $i$

Any finite measure (such as a probability measure) is clearly
$\sigma$-finite

The Lebesgue measure on ${\cal R}$ is $\sigma$-finite,
since ${\cal R}= \cup A_n$ with $A_n=(-n,n)$, $n=1,2,...$

The counting measure in is $\sigma$-finite if and only if
$\Omega$ is countable

The measure $\nu (A) = \infty$ unless $A = \emptyset$ is not $\sigma$-finite
    \end{alertblock}
\end{overprint}
}

\frame {

\begin{block}{Proposition 1.3 (Product measure theorem)}
Let $(\Omega_i, {\cal F}_i, \nu_i )$, $i=1,...,k$, be measure spaces
with $\sigma$-finite measures, where $k \geq 2$ is an integer.

Then
there exists a unique $\sigma$-finite measure on the product
$\sigma$-field $\sigma({\cal F}_1 \times \cdots \times {\cal F}_k)$,
called the {\em product measure} and denoted by $\nu_1 \times \cdots
\times \nu_k$, such that \vspace{-1mm}
\[ \nu_1\times \cdots \times \nu_k (A_1\times \cdots \times A_k)
= \nu_1(A_1)\cdots \nu_k(A_k) \vspace{-1mm}\] for all $A_i \in {\cal
F}_i $, $i=1,...,k$.
\end{block}
\begin{overprint}
\begin{alertblock}{The {\em joint} c.d.f.}<2->
Let $P$ be a probability measure on $({\cal R}^k, {\cal B}^k)$.

The
{\em joint} c.d.f. of $P$ is defined by
\[
F(x_1,...,x_k) = P\left( (-\infty , x_1]\times \cdots \times (
-\infty, x_k] \right) , \quad x_i \in {\cal R}
\]
There is a one-to-one correspondence between probability measures
and joint c.d.f.'s on ${\cal R}^k$
    \end{alertblock}
\end{overprint}
}
\frame{
    \begin{block}{{\em Marginal} c.d.f.}
 If $F(x_1,...,x_k)$ is a joint c.d.f.,
then
\[ F_{i}(x) = \lim_{x_j\rightarrow \infty ,
j=1,...,i-1,i+1,...,k} F(x_1,...,x_{i-1},x,x_{i+1},...,x_k)  \] is a
c.d.f.\ and is called the $i$th {\em marginal} c.d.f.
    \end{block}
\begin{overprint}
    \begin{alertblock}{The c.d.f.\ and product measure}<2->
   \begin{itemize}
     \item Marginal c.d.f.'s are determined by their joint c.d.f.
     \item But a joint c.d.f.\ cannot be determined by $k$ marginal c.d.f.'s.
     \item If
\[
F(x_1,...,x_k) = F_{1}(x_1)\cdots F_{k}(x_k), \quad (x_1,...,x_k)
\in {\cal R}^k ,
\]
then the probability measure corresponding to $F$ is the product
measure $P_1 \times \cdots \times P_k$ with $P_i$ being the
probability measure corresponding to $F_i$
   \end{itemize}
     \end{alertblock}
\end{overprint}
}

\end{document}
