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%%%%%%%%%%%%%%% MY DEFINITIONS %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Put your own definitions
%
\def\VaR{{\rm VaR}}
\def\E{{\rm E}}
\def\beqrs{\begin{eqnarray*}}           %\beqrs
\def\eeqrs{\end{eqnarray*}}             %\eeqrs

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\title[Stat 609 Lecture 4]{Stat 609: Mathematical Statistics \\ Lecture 4}
\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}
\begin{alertblock}{Theorem 2.1.10}
Let $X$ have continuous cdf $F_X$ and $Y= F_X(X)$.
Then $Y$ is uniformly distributed on $(0,1)$, that is,
$P(Y \leq y ) = y$, $0<y<1$.
\end{alertblock}\vspace{-2mm}
\begin{alertblock}{}
This result is useful for generating random variables.

If we want to generate a random value $X$ from a cdf $F_X$, we need only generate
a uniform random number $Y$ (from the uniform distribution on $(0,1)$) and
solve for $X$ in the equation $F_X(X)= Y$.
\end{alertblock}\vspace{-2mm}
\begin{block}{The inverse of a cdf}
If $F_X$ is strictly increasing, then its inverse $F_X^{-1}$ is well defined by \vspace{-2mm}
\[ F_X^{-1} (y) =x \quad \quad \mbox{iff} \quad \quad F_X(x) = y \vspace{-2mm} \]
If $F_X$ is flat on an interval $(x_1,x_2]$ (see the figure), then we  define \vspace{-2mm}
\[ F_X^{-1} (y) = \inf \{ x: F_X(x) \geq y \} , \quad \quad 0 \leq y \leq 1 \vspace{-6mm} \]
\begin{itemize}\vspace{-1mm}
\item
$F_X^{-1}$ is increasing in $y$.\vspace{-1mm}
\item
 $F_X(F_X^{-1}(y)) =y $ when $F_X$ is continuous at $F_X^{-1}(y)$. \vspace{-1mm}
\item
$F_X^{-1}(y) \leq x $ iff $y \leq F_X(x)$.
\end{itemize}
\end{block}
}
\frame{\vspace{-2mm}
\begin{block}{}\vspace{-2mm}
\begin{figure}[h]
\begin{center}
%\caption{Graph of $y=\sin^2(x)$ in Example 2.1.2.}
\includegraphics[height=1.8in, width=3.5in]{img5.pdf}
\end{center}
\end{figure}
\vspace{-4mm}

In part b of the figure, $F_X^{-1}(y) = x_1$.
\end{block}\vspace{-2mm}
\begin{alertblock}{Proof of Theorem 2.1.10}
For $0<y<1$, \vspace{-2mm}
\[ \begin{array}{rll}
P(Y \leq y ) & = P(F_X(X) \leq y ) & \quad  \\
&= P(F_X^{-1}(F_X(X)) \leq F_X^{-1}(y)) & \quad  \mbox{($F_X^{-1}$ is increasing)} \\
& =P(X \leq F_X^{-1}(y)) & \quad  \mbox{($P(F_X^{-1}(F_X(X)) <X)=0$)} \\
&= F_X(F_X^{-1}(y)) & \quad  \\
&=  y & \quad  \mbox{(continuity of $F_X$)} \end{array}
\]
\end{alertblock}
}
\frame{\vspace{-1mm} 
\begin{block}{Probability integral transform}
If $U $ is uniformly distributed on $(0,1)$ and $F$ is a cdf on ${\cal R}$, then the random variable $X = F^{-1}(U)$ has cdf $F$.
\vspace{2mm}

We use the property $F^{-1}(y) \leq x $ iff $y \leq F(x)$.
\begin{itemize}
\item
If $y \leq F(x)$, then $x \geq \inf \{ t: F(t) \geq y \} = F^{-1}(y)$.
\item
If $F^{-1}(y) = \inf \{ t: F(t) \geq y \} \leq x$, then $F(F^{-1}(y)) \leq F(x)$. But\vspace{-2mm}
\[ F(F^{-1}(y)) \begin{array}{ll}
= y & \quad \mbox{if $F$ is continuous at $y$} \vspace{1mm} \\
> y & \quad \mbox{if $F$ has a jump at $y$} \end{array} \vspace{-2mm}\]
Thus, we must have $y \leq F^{-1}(y) \leq x $.
\end{itemize}
Then, for any $x \in {\cal R}$, \vspace{-2mm}
\[ P(X \leq x ) = P(F^{-1}(U) \leq x ) = P(U \leq F(x)) = F(x) \]
\end{block}
\vspace{-2mm}
}
\end{document}
