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\title[Stat 609 Lecture 9]{Stat 609: Mathematical Statistics \\ Lecture 9}
\author[UW-Madison]{Jun Shao}
\institute[Statistics]{%
%  Department of Statistics\\
%  University of Wisconsin\\
%  Madison, WI 53706, USA\\
%~~~\\
  }
\date[2015]{}

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\AtBeginSection[]{\frame{\frametitle{Outline}\tableofcontents[current]}}

\begin{document}

%\frame{\titlepage}

\frame {
\frametitle{Lecture 9: Exponential and location-scale families}
\vspace{-3mm}
\begin{alertblock}{Families of Distributions}
In statistics we are interested in some families of distributions, i.e., some collections of distributions.
\vspace{1mm}

For example, the family of binomial distributions
with $p \in (0,1)$ and a fixed $n$; the family of normal distributions with
$\mu \in {\cal R}$ and $\sigma >0$.
\end{alertblock}\vspace{-3mm}
\begin{block}{Exponential families}
A family of pdfs or pmfs indexed by $\theta$ is called an exponential family iff it can be expressed as\vspace{-2mm}
\[ f_\theta (x) = h(x)c(\theta ) \exp \left( \sum_{i=1}^k w_i(\theta ) t_i(x)\right) ,
\quad \theta \in \Theta , \vspace{-2mm}\]
where $\exp (x) = e^x$, $\Theta$ is the set of all values of $\theta$ (parameter space),
$h(x) \geq 0$ and $t_1(x),...,t_k(x)$ are functions of $x$ (not depending on $\theta$), and $c(\theta ) > 0$
and $w_1(\theta ),...,w_k(\theta )$ are functions of the possibly vector-valued $\theta $ (not depending on $x$).
\vspace{1mm}

Note that the expression for $f$  may not be unique.
\end{block}
}
\frame{
\begin{alertblock}{Example 3.4.1.}
To show that a family of pdf's or pmf's is an exponential family, we must identify the functions
$h(x)$, $t_i(x)$, $c(\theta )$, and $w_i(\theta )$ and show that the pdf or pmf has the given form.
\vspace{1mm}

The $binomial(n,p)$ distribution with $p \in (0,1)$ and a fixed $n$ has pmf \vspace{-2mm}
\[ {n \choose x}  p^x (1-p)^{n-x}
= {n \choose x}  (1-p)^{n}\exp \left( \log \left(\frac{p}{1-p}\right)x \right), \quad x=0,1,...,n.\vspace{-2mm}\]
Let $\theta =p$, $c(\theta ) = (1-p)^n$, $w_1(\theta )
 = \log (\frac{p}{1-p})$, $t_1(x) = x$, and
$h(x) = {n \choose x} $ for $ x=0,1,...,n$ and $=0$ otherwise.
\vspace{1mm}

Then, the binomial family with $p \in (0,1)$ and a fixed $n$ is an exponential family ($k=1$).\\
(Note that $p=0$ and $p=1$ are not included in the family.)
\vspace{1mm}

Other examples: Poisson, negative binomial, normal, gamma, beta,...
\end{alertblock}\vspace{-2mm}
\begin{block}{}
Exponential families have many nice properties.

The following result is useful since we can replace integration or summation by differentiation.
\end{block}
}
\frame{ \vspace{-1mm}
\begin{block}{Theorem 3.4.2.}
If $X$ has a pdf or pmf from an exponential family and $w_i(\theta )$'s are differentiable functions, then\vspace{-3mm}
\[ E\left( \sum_{i=1}^k \frac{\partial w_i(\theta )}{\partial \theta_j} t_i(X)\right) = - \frac{\partial \log c(\theta )}{\partial \theta_j} \vspace{-3mm}\]
where $\theta_j$ is the $j$th component of $\theta$, and\vspace{-3mm}
\[  {\rm Var}\left( \sum_{i=1}^k \frac{\partial w_i(\theta )}{\partial \theta_j} t_i(X)\right) = - \frac{\partial^2 \log c(\theta )}{\partial \theta_j^2}
- E\left( \sum_{i=1}^k \frac{\partial^2 w_i(\theta )}{\partial \theta_j^2} t_i(X)\right)  \]
\end{block}\vspace{-3mm}
\begin{block}{Proof.}
From the exponential family expression for $f_\theta (x)$, \vspace{-2mm}
\[ \log f_\theta (X) = \log h(X) + \log c(\theta ) + \sum_{i=1}^k w_i(\theta ) t_i(X) \vspace{-2mm} \]
Differentiating this expression leads to \vspace{-2mm}
\[ \frac{\partial \log f_\theta (X)}{\partial \theta_j}  = \frac{\partial  \log c(\theta )}{\partial \theta_j } + \sum_{i=1}^k \frac{\partial w_i(\theta )}{\partial \theta_j}  t_i(X)  \]
\end{block}}
\frame{
\begin{block}{}
Taking expectation, we obtain \vspace{-2mm}
\[ E\left(  \frac{\partial \log f_\theta (X)}{\partial \theta_j}\right) = \frac{\partial \log c(\theta )}{\partial \theta_j}
+ E \left(\sum_{i=1}^k \frac{\partial w_i(\theta )}{\partial \theta_j}  t_i(X)\right)  \vspace{-2mm}
 \]
If $f_\theta (x)$ is a pdf (the proof for pmf is similar), then the left side of the previous expression is
 \vspace{-1mm}
 \[ \int_{-\infty}^\infty \frac{\partial \log f_\theta (x) }{ \partial \theta_j} f_\theta (x) dx
 = \int_{-\infty}^\infty \frac{\partial f_\theta (x) }{\partial \theta_j} dx
 = \frac{\partial }{\partial \theta_j} \int_{-\infty}^\infty f_\theta (x)dx
 = \frac{\partial 1}{\partial \theta_j} = 0 \vspace{-1mm}\]
 We interchanged the differentiation and integration, which is justified under the exponential family assumption.
 
 This proves the first result. 
 \vspace{1mm}
 
Note that  \vspace{-4mm} 
\[ \frac{\partial^2 \log f_\theta (X)}{\partial \theta_j^2}  = \frac{\partial }{\partial \theta_j} 
\left[ \frac{\frac{\partial f_\theta (X)}{\partial \theta_j}}{f_\theta (X)}\right] 
=  \frac{\frac{\partial^2 f_\theta (X)}{\partial \theta_j^2}}{f_\theta (X)}
- \left[  \frac{\frac{\partial f_\theta (X)}{\partial \theta_j}}{f_\theta (X)}\right]^2  \vspace{-2mm}\]
Then
\end{block}}
\frame{
\begin{block}{}
\vspace{-2mm} 
\beqrs
E \left( \frac{\partial^2 \log f_\theta (X)}{\partial \theta_j^2}  \right) \back & = & \back 
\int_{-\infty}^\infty \left\{ \frac{\frac{\partial^2 f_\theta (X)}{\partial \theta_j^2}}{f_\theta (X)}
- \left[  \frac{\frac{\partial f_\theta (X)}{\partial \theta_j}}{f_\theta (X)}\right]^2 f_\theta (x)\right\} dx \\
\back & = & \back 
\int_{-\infty}^\infty \frac{\partial^2 f_\theta (X)}{\partial \theta_j^2}dx - \int_{-\infty}^\infty 
\left[ \frac{\partial \log f_\theta (X)}{\partial \theta_j}\right]^2  f_\theta (x) dx \\
\back & = & \back
- \int_{-\infty}^\infty
\left[ \frac{\partial  \log c(\theta )}{\partial \theta_j } + \sum_{i=1}^k \frac{\partial w_i(\theta )}{\partial \theta_j}  t_i(X)\right]^2  f_\theta (x) dx \\
\back & = & \back - {\rm Var}\left( \sum_{i=1}^k \frac{\partial w_i(\theta )}{\partial \theta_j}  t_i(X)\right) 
\eeqrs
which follows from the first result. 

Then the second result follows from \vspace{-2mm} 
\[ \frac{\partial^2 \log f_\theta (X)}{\partial \theta_j^2} = \frac{\partial^2  \log c(\theta )}{\partial \theta_j^2 } + \sum_{i=1}^k \frac{\partial^2 w_i(\theta )}{\partial \theta_j^2}  t_i(X)  \]
\end{block}}
\frame{
\begin{alertblock}{Example 3.4.4.}
If $X \sim N(\mu, \sigma^2)$, then $\theta = (\mu, \sigma )$ and\vspace{-3mm}
\[ f_\theta (x) = \frac{1}{\sqrt{2\pi} \sigma} \exp \left( -\frac{(x-\mu)^2}{2\sigma^2}\right)
= \frac{1}{\sqrt{2\pi} \sigma} \exp \left( - \frac{\mu^2}{2\sigma^2} \right) \exp\left(
  \frac{\mu x}{\sigma^2} - \frac{x^2}{2\sigma^2} \right) \vspace{-3mm} \]
  Let $h(x) =1$, $c(\theta ) = \frac{1}{\sqrt{2\pi} \sigma}\exp ( -\frac{\mu^2}{2\sigma^2})$,
  $w_1(\theta ) = 1/\sigma^2$, $w_2(\theta ) = \mu/\sigma^2$, $t_1(x) = -x^2/2$, and $t_2(x) =x$.

  Then this normal family is an exponential family with $k=2$.

  Applying Theorem 3.4.2, we obtain $E(X) = \mu$ from equation \vspace{-3mm}
  \[ - \frac{\partial \log c(\theta )}{\partial \mu} = \frac{\mu}{\sigma^2} =
  E\left( \sum_{i=1}^2 \frac{\partial w_i(\theta )}{\partial \mu} t_i(X)\right)
  = E\left( \frac{X}{\sigma^2} \right) \]
Also, \vspace{-3mm}
  \[ - \frac{\partial \log c(\theta )}{\partial \sigma } = \frac{\mu^2}{\sigma^3} + \frac{1}{\sigma}  =
  E\left( \sum_{i=1}^2 \frac{\partial w_i(\theta )}{\partial \sigma} t_i(X)\right)
  = E\left( \frac{X^2}{\sigma^3} -\frac{2\mu X}{\sigma^3} \right) \vspace{-3mm}\]
  Using $E(X) = \mu$, we obtain from this equation that Var$(X) = \sigma^2$.
\end{alertblock}}
\frame{\vspace{-2mm}
	\begin{block}{Beta distribution $beta(\alpha , \beta )$}
		For constants $\alpha >0$ and $\beta >0$, the $beta(\alpha , \beta )$ distribution has pdf \vspace{-2mm}
		\[ f(x) = \left\{ \begin{array}{ll}
		\frac{\Gamma (\alpha + \beta )}{\Gamma (\alpha ) \Gamma (\beta )}
		x^{\alpha -1}(1-x)^{ \beta-1} & \quad 0<x<1 \vspace{2mm} \\
		0 & \quad \mbox{otherwise} \end{array} \right. \vspace{-2mm} \]
		This is a pdf because  \vspace{-2mm}
		\[  \int_0^1 x^{\alpha -1} (1-x)^{\beta -1} dx = \frac{\Gamma (\alpha ) \Gamma (\beta )}{\Gamma (\alpha + \beta )} \]
		\vspace{-7cm}
		\begin{figure}[h]
			\begin{center}
				%\caption{Graph of $y=\sin^2(x)$ in Example 2.1.2.}
				\includegraphics[height=5in, width=4.5in]{figure335.pdf}
			\end{center}
		\end{figure}
	\end{block}}
	\frame{\vspace{-2mm}
		\begin{block}{}
			\begin{itemize}
				\item
				Since $\Gamma (2) = \Gamma (1) = 1$, $beta (1,1)$ is the same as $uniform(0,1)$.
				\item
				If $X \sim beta (\alpha , \beta )$, then $1-X \sim  beta (\beta, \alpha )$.
				\item
				The pdf of $beta (\alpha , \beta )$ can be increasing ($\alpha >1 , \beta =1$), decreasing ($\alpha = 1, \beta >1$),
				U-shaped ($\alpha < 1, \beta <1$), or unimodal ($\alpha >1, \beta >1$).
				\item
				If $\alpha = \beta$, then the pdf of $beta (\alpha , \beta )$ is symmetric about $\frac{1}{2}$.
				\item
				For any $r >0$, if $X \sim beta (\alpha , \beta )$, then \vspace{-2mm}
				\[ E(X^r) =  \frac{\Gamma (\alpha + \beta )}{\Gamma (\alpha ) \Gamma (\beta )}
				\int_0^1 x^{r+\alpha -1} (1-x)^{\beta -1} dx =
				\frac{\Gamma (\alpha + \beta )\Gamma (r+\alpha )}{\Gamma (\alpha )\Gamma (r+\alpha + \beta ) } \vspace{-2mm}\]
				In particular, \vspace{-2mm}
				\[ E(X) = \frac{\alpha }{\alpha + \beta } , \quad \quad E(X^2) = \frac{\alpha (\alpha +1)}{ (\alpha + \beta )(\alpha + \beta + 1 )}  \]
				Then
				\[ {\rm Var}(X) =  \frac{\alpha (\alpha +1)}{ (\alpha + \beta )(\alpha + \beta + 1 )}
				- \frac{\alpha^2 }{(\alpha + \beta )^2 } = \frac{\alpha \beta }{(\alpha + \beta )^2(\alpha + \beta + 1)} \]
			\end{itemize}
		\end{block}\vspace{-2mm}
		\begin{alertblock}{}
			The family of $beta(\alpha , \beta)$ distributions is an exponential family. 
				\end{alertblock}
	}
\frame{
\begin{block}{Natural exponential family}
If $\eta_i = w_i(\theta )$, $i=1,...,k$, and $\eta = (\eta_1,...,\eta_k)$, the form of $f_\theta$ in the exponential family becomes \vspace{-3mm}
\[ f^*_\eta (x) = h(x)c^*(\eta ) \exp \left( \sum_{i=1}^k \eta_i t_i(x) \right) \vspace{-3mm} \]
$\eta$ is called the natural parameter.

The set of $\eta$'s for which $f^*_\eta (x)$ is a well-defined pdf is called the natural parameter space.
\end{block}\vspace{-2mm}
\begin{alertblock}{Full or curved exponential families}
In an exponential family, if the dimension of $\theta$ is $k$ (there is an open set $\subset \Theta$), then
the family is a full exponential family. Otherwise the family is a curved exponential family.
\end{alertblock} \vspace{-2mm}
\begin{block}{}
An example of a full exponential family is $N(\mu , \sigma^2)$, $\mu \in {\cal R}$, $\sigma >0$.

An example of a curved exponential family is $N(\mu, \mu^2)$, $\mu \in {\cal R}$.
\end{block}}
\frame{
\begin{block}{How to show a family is not an exponential family}
It may be difficult to show that a family is not an exponential family.
\vspace{1mm}

We cannot say ``we are not able to express $f_\theta(x)$ in the form of an exponential family". 
\vspace{1mm}


If $f_\theta$, $\theta \in \Theta$ is an exponential family, then \vspace{-2mm}
\[ \{ x: f_\theta (x) > 0 \} = \{ x: h(x) > 0 \} \vspace{-2mm} \]
which does not depend on $\theta$ values.
\vspace{1mm}

This fact can be used to show a family is non-exponential, i.e.,
if $\{ x: f_\theta (x)>0\}$ depends on $\theta$, then $f_\theta $, $\theta \in \Theta$, is not an exponential family.
\vspace{1mm}

Consider the family of two parameters exponential distributions with pdf's\vspace{-2mm}
\[ f_\theta (x) = \left\{ \begin{array}{ll}
\lambda^{-1} e^{-(x-\mu)/\lambda} & \quad x > \mu \vspace{1mm} \\
0 & \quad x \leq \mu \end{array} \right. \quad \mu \in {\cal R}, \  \lambda > 0 \vspace{-2mm} \]
It is not an exponential family because \vspace{-2mm}
\[ \{ x: f_\theta (x) > 0 \} = \{ x: x > \mu \}  \]
\end{block}
}
\frame{\vspace{-2mm}
\begin{block}{Definition 3.5.2 (location family)}
Let $f(x)$ be a given pdf. The family of pdf's, $f(x-\mu)$, $\mu \in {\cal R}$, is called a location family with location parameter $\mu$.
\end{block}\vspace{-2mm}
\begin{alertblock}{}
\begin{itemize}
\item
Examples of location families are normal and 
Cauchy with location parameter $\mu \in {\cal R}$ and the other parameter $\sigma$ fixed.

Other examples are given later. 
\item
The pdf $f(x-\mu)$ is obtained by shifting the entire curve $f(x)$ by an amount $\mu$ (see the figure) without changing the structure of $f(x)$.
\item
It can be shown that $X \sim f(x- \mu)$ iff $X=Z+ \mu $ with $Z \sim f(x)$.
\end{itemize}
\end{alertblock}\vspace{-2mm}
\begin{alertblock}{}
\vspace{-8.8cm}
\begin{figure}[h]
\begin{center}
%\caption{Graph of $y=\sin^2(x)$ in Example 2.1.2.}
\includegraphics[height=7in, width=4.8in]{figure351.pdf}
\end{center}
\end{figure}
\end{alertblock}

}
\frame{\vspace{-2mm}
\begin{alertblock}{Definition 3.5.4 (scale family)}
Let $f(x)$ be a given pdf. The family of pdf's, $\sigma^{-1}f(x/\sigma )$, $\sigma >0$, is called a scale family with scale parameter $\sigma$.
\end{alertblock}\vspace{-3mm}
\begin{alertblock}{}
\begin{itemize}
\item
Examples of scale families are normal and Cauchy with scale parameter $\sigma >0$ and $\mu$ fixed,
$gamma (\alpha , \beta )$ with $\beta >0$ and $\alpha$ fixed.

Other examples are given later. 
\item
The pdf $\sigma^{-1} f(x/\sigma )$ is obtained by stretching ($\sigma >1$) or contracting ($\sigma <1$) the curve $f(x)$ while still maintaining the same shape.
\item
It can be shown that $X \sim \sigma^{-1} f(x/\sigma )$ iff $X=\sigma Z $ with $Z \sim f(x)$.
\end{itemize}
\end{alertblock}\vspace{-2mm}
\begin{alertblock}{}
\vspace{-6.4cm}
\begin{figure}[h]
\begin{center}
%\caption{Graph of $y=\sin^2(x)$ in Example 2.1.2.}
\includegraphics[height=4.9in, width=4.5in]{figure353.pdf}
\end{center}
\end{figure}
\end{alertblock}
}
\frame{
\begin{block}{Definition 3.5.5 (location-scale family)}
Let $f(x)$ be a given pdf. The family of pdf's, $\sigma^{-1}f((x-\mu)/\sigma )$, $\mu \in {\cal R}$, $\sigma >0$, is called a location-scale family with location parameter $\mu$ and scale parameter $\sigma$.
\end{block}\vspace{-2mm}
\begin{alertblock}{}
\begin{itemize}
\item
A location-scale family is a combination of a location family and a scale family:
it contains a sub-family that is a location family with any fixed $\sigma$, and a sub-family that is a scale
family with any fixed $\mu$.
\end{itemize}
\end{alertblock}\vspace{-2mm}
\begin{alertblock}{}
\vspace{-6.3cm}
\begin{figure}[h]
\begin{center}
%\caption{Graph of $y=\sin^2(x)$ in Example 2.1.2.}
\includegraphics[height=5in, width=4.5in]{figure354.pdf}
\end{center}
\end{figure}
\end{alertblock}}
\frame{
\begin{alertblock}{}
\begin{itemize}
\item
Examples of location-scale families are normal,
double exponential, Cauchy, logistic, and  two-parameter exponential  distributions with location parameter $\mu \in {\cal R}$ and
scale parameter $\sigma >0$. Except for the two-parameter exponential distribution, all others are symmetric about $\mu$.
\item
If $f(x)$ is symmetric about 0, then $\sigma^{-1}  f((x-\mu)/\sigma)$ is symmetric about $\mu$ and $\mu$ is the median of
$X \sim \sigma^{-1}  f((x-\mu)/\sigma)$; furthermore, if the expectation of $f(x)$ exists, then $\mu$ is the expectation of
$\sigma^{-1}  f((x-\mu)/\sigma)$.
\item
It can be shown that $X \sim \sigma^{-1} f((x- \mu)/\sigma) $ iff $X=\sigma Z + \mu$ with $Z \sim f(x)$; furthermore, if $E(Z^2)< \infty$, then
$E(X) = \sigma E(Z) + \mu$ and Var$(X ) = \sigma^2$Var$(Z)$.
\item
The pdf $f(x)$ in a location-scale family is standard iff the expectation $\int_{-\infty}^\infty xf(x)dx =0$ and the variance
$\int_{-\infty}^\infty x^2f(x) dx = 1$.
\item
Typically, we choose a standard $f(x)$ to generate a location-scale family, in which case $\mu$ and $\sigma^2$ are the expectation and variance
of  $\sigma^{-1}  f((x-\mu)/\sigma)$, respectively.
\end{itemize}
\end{alertblock}
}
\frame{ \vspace{-2mm}
	\begin{block}{Two parameter exponential distribution $exponential ( \mu, \beta )$}
		If $X \sim exponential (\beta )$ and $\mu \in {\cal R}$ is a constant, then the distribution of $Y=X+\mu$
		is called the two parameter exponential distribution and denoted by $exponential (\mu, \beta )$.
		
		Its pdf and cdf are (by transformation) \vspace{-2mm}
		\[ f(x) = \left\{ \begin{array}{ll}
		\frac{1}{\beta} e^{-(x-\mu)/\beta } & \quad x \geq \mu \vspace{2mm} \\
		0 & \quad x < \mu \end{array} \right. \quad \quad F(x) = \left\{ \begin{array}{ll}
		1-e^{-(x-\mu)/\beta } & \quad x \geq \mu \vspace{1mm} \\
		0 & \quad x < \mu \end{array} \right. \vspace{-2mm} \]
		and, if $Y \sim exponential(\mu, \beta )$, \vspace{-2mm}
		\[ E(Y)  \! = \! \mu+ \beta , \ \ {\rm Var}(Y) \! = \! \beta^2 , \ \ M_Y(t) \! =\! \frac{e^{\mu t}}{1-\beta t } , \ t < \frac{1}{\beta},
		\  \phi_Y(t) \!=\! \frac{e^{\dot{\imath} \mu t}}{1- \dot{\imath} \beta t } , \ t\in {\cal R} \]
	\end{block}\vspace{-2mm}
	\begin{alertblock}{Double exponential distribution $double$-$exponential (\mu , \sigma )$}
		By reflecting the pdf of $exponential (\mu, \sigma )$ around $\mu$, we obtain the $double$-$exponential(\mu, \sigma)$ pdf that is symmetric about $\mu$:\vspace{-2mm}
		\[ f(x) = \left\{ \begin{array}{ll}
		\frac{1}{2\sigma} e^{-(x-\mu)/\sigma } & \quad x \geq \mu \vspace{2mm} \\
		\frac{1}{2\sigma} e^{(x-\mu)/\sigma }  & \quad x < \mu \end{array} \right.
		= \frac{1}{2\sigma} e^{-|x-\mu|/\sigma} , \quad x \in {\cal R}   \]
	\end{alertblock}
}
\frame{
	\begin{alertblock}{}
		\begin{itemize}
			\item
			This pdf is not bell-shaped; in fact, it has a peak (a non-differentiable point) at $x= \mu$.
			\item
			Its cdf is \vspace{-1mm}
			\[ F(x) = \left\{ \begin{array}{ll}
			1-\frac{1}{2} e^{-(x-\mu)/\sigma } & \quad x \geq \mu \vspace{2mm} \\
			\frac{1}{2} e^{(x-\mu ) / \sigma}  & \quad x < \mu \end{array} \right.  \]
			\item
			If $X \sim double$-$exponential ( \mu, \sigma)$, then $Z = (X-\mu)/\sigma  \sim double$-$exponential (0,1)$.
			\item
			If $Z = (X-\mu)/\sigma  \sim double$-$exponential (0,1)$, then \vspace{-2mm}
			\[ E(Z) = \frac{1}{2}\int_{-\infty}^\infty x e^{-|x|}dx = 0 \vspace{-2mm}\]
			because $xe^{-|x|}$ is an odd function, and \vspace{-2mm}
			\[ {\rm Var}(Z)  = E(Z^2) = \frac{1}{2}\int_{-\infty}^\infty x^2 e^{-|x|}dx =
			\int_0^\infty x^2 e^{-x} dx = \Gamma (3) = 2 \vspace{-2mm}\]
			\item
			If $X \sim double$-$exponential ( \mu, \sigma)$, then $X = \sigma Z + \mu $,
			$Z = (X-\mu)/\sigma  \sim double$-$exponential (0,1)$, and
			\vspace{-2mm}
			\[ E(X) = E(\sigma Z + \mu ) = \mu , \quad \quad {\rm Var}(X) =
			{\rm Var}( \sigma Z+\mu) = \sigma^2 {\rm Var}(Z) = 2 \sigma^2 \]
		\end{itemize}
	\end{alertblock}
}
\frame{\vspace{-2mm}
	\begin{alertblock}{Logistic distribution $logistic(\mu, \sigma )$}
		For constants  $\mu \in {\cal R}$ and $\sigma >0$, the $logistic(\mu , \sigma )$ distribution has pdf \vspace{-2mm}
		\[ f(x) = \frac{e^{-(x- \mu )/\sigma} }{\sigma  [ 1+ e^{-(x-\mu)/\sigma } ]^2} , \quad x \in {\cal R} \vspace{-4mm} \]
		\begin{itemize}
			\item
			This pdf is again bell-shaped and symmetric about $\mu$.
			\item
			The cdf of $logistic(\mu, \sigma )$ has a close form: \vspace{-2mm}
			\[ F(x) = \int_{-\infty}^x f(t)dt = \frac{1}{1+e^{-(x-\mu)/\sigma} } , \quad x \in {\cal R} \vspace{-2mm} \]
			\item
			By symmetry, $E(X) = \mu$ if $X \sim logistic(\mu, \sigma )$.
			\item
			The variance of $X\sim logistic(\mu, \sigma )$ is not easy to obtain, but we give the result here:
			Var$(X) = \sigma^2\pi^2/3$.
		\end{itemize}
	\end{alertblock}
	\vspace{-2mm}
	\begin{block}{Pareto distribution $pareto(\alpha, \beta )$}
		For constants  $\alpha >0$ and $\beta >0$, the $pareto (\alpha , \sigma )$ distribution has pdf \vspace{-2mm}
		\[ f(x) = \left\{ \begin{array}{ll}
		\alpha \beta^\alpha x^{-(\alpha +1)} & \quad x > \beta \vspace{2mm} \\
		0 & \quad x \leq \beta \end{array} \right. \]
	\end{block}}
	\frame{\vspace{-2mm}
		\begin{block}{}
			\begin{itemize}
				\item
				First,  $f$ is indeed a pdf, because
				\vspace{-3mm}
				\[ \int_{-\infty}^\infty f(x) dx = \alpha \beta^\alpha \int_\beta^\infty x^{-(\alpha +1)} dx =
				\beta^\alpha x^{-\alpha} \bigg|_\infty^\beta = \beta^\alpha \beta^{-\alpha} = 1 \vspace{-2mm}\]
				\item
				Using a similar argument, we can obtain the cdf of $pareto(\alpha, \beta )$ as \vspace{-3mm}
				\[ F(x) = \left\{ \begin{array}{ll}
				1- \left( \frac{\beta}{x} \right)^\alpha & \quad x > \beta \vspace{2mm} \\
				0 & \quad x \leq \beta \end{array} \right.  \vspace{-2mm} \]
				\item
				Since the integral $\int_\beta^\infty x^{-t } dx$ is finite iff $t>1$,
				$E(X) = \infty$ if $\alpha \leq 1$ when $X \sim pareto(\alpha, \beta )$; if $\alpha >1$, then \vspace{-3mm}
				\[ E(X) = \alpha \beta^\alpha \int_\beta^\infty x^{-\alpha} dx
				= \frac{\alpha \beta^\alpha}{\alpha -1} x^{-(\alpha -1)} \bigg|_\infty^\beta =
				\frac{\alpha \beta^\alpha}{\alpha -1} \beta^{-(\alpha -1)} =
				\frac{\alpha \beta}{\alpha -1} \vspace{-2mm} \]
				\item
				Similarly, Var$(X) = \infty$ if $\alpha \leq 2$; and if $\alpha >2$, \vspace{-2mm}
				\[    E(X^2 ) = \alpha \beta^\alpha \int_\beta^\infty x^{-\alpha+1} dx
				= \frac{\alpha \beta^\alpha}{\alpha -2} x^{-\alpha +2} \bigg|_\infty^\beta =
				\frac{\alpha \beta^\alpha}{\alpha -2} \beta^{-\alpha +2} =
				\frac{\alpha \beta^2}{\alpha -2} \vspace{-2mm}\]
				\[ {\rm Var}(X) = E(X^2) - [E(X)]^2 =
				\frac{\alpha\beta^2}{\alpha -2} - \frac{\alpha^2\beta^2}{(\alpha -1)^2}
				= \frac{\alpha\beta^2}{(\alpha -1)^2 (\alpha -2)} \]
			\end{itemize}
		\end{block}
	}
	\frame{
		\begin{alertblock}{Weibull distribution $Weibull (\gamma , \beta )$}
			For constants $\gamma >0$ and $\beta >0$, if $X \sim exponential (\beta )$, then
			$Y = X^{1/\gamma } \sim Weibull (\gamma, \beta )$ with pdf \vspace{-2mm}
			\[ f(x) = \left\{ \begin{array}{ll}
			\frac{\gamma }{\beta} x^{\gamma -1} e^{-x^\gamma /\beta } & \quad x > 0 \vspace{2mm} \\
			0 & \quad x \leq 0 \end{array} \right. \vspace{-3mm} \]
			\begin{itemize}
				\item
				An example of $Y \sim Weibull (\gamma , \beta )$ is lifetime or failure time.
				\item
				If $Y \sim  Weibull (\gamma , \beta )$, then $X = Y^\gamma \sim exponential (\beta )$ and \vspace{-2mm}
				\beqrs
				E(Y) &=& E(X^{1/\gamma} ) = \frac{1}{\beta } \int_0^\infty x^{1/\gamma} e^{-x/\beta } dx \\
				& = & \beta^{1/\gamma } \int_0^\infty u^{1/\gamma} e^{u} du =
				\beta^{1/\gamma } \Gamma \left(\frac{1}{\gamma} + 1\right)
				\eeqrs
				Similarly, we can obtain that \vspace{-2mm}
				\[ {\rm Var}(Y) = \beta^{2/\gamma} \left\{ \Gamma \left( \frac{2}{\gamma} +1\right) -\left[ \Gamma \left( \frac{1}{\gamma} + 1 \right)\right]^2 \right\} \]
			\end{itemize}
		\end{alertblock}
	}

\end{document}
