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\title[Stat 609 Lecture 23]{Stat 609: Mathematical Statistics \\ Lecture 23}
\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 {\vspace{-2mm}
\frametitle{
Lecture 23: Minimal sufficiency}
\vspace{-1mm}
\begin{alertblock}{Maximal reduction without loss of information}
\begin{itemize}
  \item There are many sufficient statistics for a given problem. \vspace{-0.5mm}
  \item In fact, $X$ (the whole data set) is sufficient. \vspace{-0.5mm}
  \item If $T$ is a sufficient statistic and $T=\psi(S)$, where $\psi$ is a function and $S$ is another statistic, then $S$ is sufficient.

  For instance, if $X_1,\dots,X_n$ are iid with $P(X_i=1)=\theta$ and $P(X_i=0)=1-\theta$, then $(\sum_{i=1}^mX_i,\ \sum_{i=m+1}^nX_i)$ is sufficient for $\theta$, where $m$ is any fixed integer between 1 and $n$. \vspace{-0.5mm}
  \item If $T$ is sufficient and $T=\psi(S)$ with a measurable function $\psi$ that is
not one-to-one, then $T$ is more useful than $S$, since $T$ provides a further reduction of the data without loss of information. \vspace{-0.5mm}
  \item Is there a sufficient statistics that provides ``maximal" reduction of the data?
\end{itemize}
\end{alertblock}\vspace{-2mm}
    \begin{block}{Definition 6.2.11 (minimal sufficiency)}
  A sufficient statistic $T(X )$ is a minimal sufficient
statistic if, for any other sufficient statistic $U(X )$,
$T(x )$ is a function of $U(x )$.
\end{block}}
\frame{\vspace{-2mm}
\begin{block}{}
\vspace{-1mm}
\begin{itemize}
\item
$T(x )$ is a function of $U(x )$ iff  $U(x ) =
U(y )$ implies that $T(x ) = T(y )$ for any pair $(x,y)$.
\vspace{-0.5mm}
\item
In terms of the partitions of the range of the data set,
if $T(x )$ is a function of $U(x )$, then
\vspace{-3mm}
\[ \{ x: U(x) = t \} \subset  \{ x : T(x ) = t \}  \vspace{-3mm} \]
Thus, a minimal sufficient statistic
 achieves the greatest possible data reduction as a sufficient statistic.\vspace{-0.5mm}
\item
If both $T$ and $S$ are minimal sufficient statistics,
then by definition there is a one-to-one function $\psi$ such that $T=\psi(S)$; hence, the minimal sufficient statistic is unique in the sense that two statistics that are  functions of each other can be treated as one statistic.
\vspace{-0.5mm}
\item
 For example, if $T$ is minimal sufficient, then so is $(T,e^T)$,
 but no one is going to use  $(T,e^T)$.
\vspace{-0.5mm}
 \item
 If the range of $X$ is ${\cal R}^k$, then there exists a minimal sufficient statistic. 
\end{itemize}
     \end{block}\vspace{-2.5mm}
\begin{alertblock}{Example 6.2.15.}
 Let $X_1,...,X_n$  be iid form  the $uniform(\theta,\theta+1)$ distribution, where $\theta\in {\cal R}$ is an unknown parameter.
 \end{alertblock}}
\frame{
\begin{alertblock}{}
The joint pdf of $(X_1,...,X_n)$ is \vspace{-2mm}
 $$\prod_{i=1}^nI(\{ \theta < x_i < \theta+1\} )=I(\{ x_{(n)}-1< \theta < x_{(1)}\} ),
\ \ \ (x_1,...,x_n)\in {\cal R}^n,\vspace{-2mm} $$
 where $x_{(i)}$ denotes the $i$th ordered  value of $x_1,...,x_n$.
\vspace{1mm}

By the factorization theorem, $T=(X_{(1)},X_{(n)})$ is sufficient for $\theta$.
\vspace{1mm}

Note that\vspace{-2mm}
 $$x_{(1)}=\sup\{\theta:\ f_{\theta}(x)>0\}\ \hbox{ and } \ x_{(n)}=1+\inf\{\theta:\ f_{\theta}(x)>0\}. \vspace{-2mm}$$
If $S(X)$ is a statistic sufficient for $\theta$, then by the factorization theorem, there are functions $h$ and $g_{\theta}$ such that $f_{\theta}(x)=g_{\theta}(S(x))h(x)$.
\vspace{1mm}

For $x$ with $h(x)>0$,\vspace{-2mm}
 $$x_{(1)}=\sup\{\theta:\ g_{\theta}(S(x))>0\}\ \hbox{ and }x_{(n)}=1+\inf\{\theta:\ g_{\theta}(S(x))>0\}. \vspace{-2mm}$$
Hence, there is a function $\psi$ such that $T(x)=\psi(S(x))$ when $h(x)>0$.
\vspace{1mm}

Since $P( h(X) >0 )=1 $,  we conclude that $T$ is minimal sufficient.
\vspace{1mm}

In this example, the dimension of a minimal sufficient statistic is larger than the dimension of $\theta$.
\end{alertblock}}
\frame{
\begin{alertblock}{}
Finding a minimal sufficient statistic by definition is not convenient.
\vspace{1mm}

The next theorem is a useful tool.
     \end{alertblock}
\vspace{-2mm}
\begin{block}{Theorem 6.2.13.} Let $f_\theta (x )$ be the pmf or pdf of $X$. Suppose that
$T(x )$ is sufficient for $\theta$ and that,
for every pair $x$ and $y$ with at least one of
$f_\theta (x )$ and $ f_\theta (y )$ is not 0,
 $f_\theta (x )/f_\theta (y )$ does not
depend on $\theta$ implies $T(x )=T(y )$.
(Here, we define $a/0= \infty$ for any $a>0$.)
Then $T(X)$ is minimal sufficient for $\theta$.
\end{block}\vspace{-2mm}
\begin{block}{Proof.}
Let $U(X)$ be another sufficient statistic.

By the factorization theorem, there are functions $h$ and $g_\theta$ such that
$f_\theta (x)= g_\theta (U(x))h(x)$ for all $x$ and $\theta$.

For $x$ and $y$ such that
at least one of
$f_\theta (x )$ and $ f_\theta (y )$ is not 0, i.e., at least one of $h(x)$ and $h(y)$ is not 0,
if  $U(x)=U(y)$, then
\vspace{-2mm}
\[ \frac{f_\theta (x)}{f_\theta (y)} =\frac{g_\theta (U(x))h(x)}{g_\theta (U(y))h(y)} = \frac{h(x)}{h(y)} \vspace{-2mm}\]
which does not depend on $\theta$.
\end{block}}
\frame{
\begin{block}{}
By the assumption of the theorem, $T(x)=T(y)$.
\vspace{1mm}

This shows that there is a function $\psi$ such that $T(x)=\psi(S(x))$
\vspace{1mm}

Hence, $T$ is minimal sufficient for $\theta$
\end{block}\vspace{-2mm}
\begin{alertblock}{Example 6.2.15.}
We re-visit this example and show how to apply Theorem 6.2.13.

If $X_1,...,X_n$  are iid form  the $uniform(\theta,\theta+1)$ distribution, then the joint pdf is \vspace{-2mm}
 $$ f_\theta (x) = I(\{ x_{(n)}-1< \theta < x_{(1)}\} ),
\ \ \ x= (x_1,...,x_n)\in {\cal R}^n,\vspace{-2mm} $$
Let $x$ and $y $ be two data points.

Each of $f_\theta (x ) $ and $ f_\theta (y )$ only takes
two possible values, 0 and 1.

Let $A_x = (x_{(n)}-1 , x_{(1)})$ and $A_y =  (y_{(n)}-1 , y_{(1)})$.

Then\vspace{-2mm}
\[
 \frac{f_\theta(x ) }{f_\theta (y )} = \left\{ \br{ll}
0 & \quad \theta \not \in A_x, \theta \in A_y \\
1 & \quad  \theta  \in A_x, \theta \in A_y \\
\infty & \quad  \theta  \in A_x, \theta \not \in A_y  \er \right. \vspace{-2mm}\]
This depends on $\theta$ unless $A_x=A_y$.
\end{alertblock}}
\frame{\vspace{-2mm}
\begin{alertblock}{}
Thus, if the ratio does not
depend on $\theta$, we must have $x_{(1)}=y_{(1)}$ and
$x_{(n)}=y_{(n)}$.
\vspace{1mm}

Therefore, by Theorem 6.2.13, $(X_{(1)}, X_{(n)})$ is minimal sufficient for $\theta$.
\end{alertblock}\vspace{-2mm}
\begin{block}{Example 6.2.14.}
Let $X_1,...,X_n$ be iid from $N(\mu, \sigma^2)$ with unknown $\mu \in {\cal R}$ and $\sigma >0$.
\vspace{1mm}

Earlier,  we showed that $T=(\bar{X}, S^2 )$ is sufficient for $\theta = (\mu, \sigma^2)$.
\vspace{1mm}

Is $T$ minimal sufficient?
\vspace{1mm}

Let $f_\theta (x)$ be the joint pdf of the sample, $x$ and $y$ be two sample points.

Then
\vspace{-1mm}
\beqrs
 \frac{f_\theta (x)}{f_\theta (y)} &=&
\frac{(2\pi\sigma^2)^{-n/2} \exp ( - [(n(\bar{x} - \mu)^2+ (n-1)s^2_x]/(2\sigma^2))}{
(2\pi\sigma^2)^{-n/2} \exp ( - [(n(\bar{y} - \mu)^2+ (n-1)s^2_y]/(2\sigma^2))} \\
&=& \exp ( [-n(\bar{x}^2 -\bar{y}^2)+ 2n \mu (\bar{x}-\bar{y}) - (n-1)(s_x^2 -s_y^2)] /(2\sigma^2))
 \eeqrs
 where $\bar{x}$ and $s^2_x$ are the sample mean and variance based on the sample point $x$, and
 $\bar{y}$ and $s^2_y$ are the sample mean and variance based on the sample point $y$.
\end{block}
}
\frame{
\begin{block}{}
This ratio does not depend on $\theta = (\mu, \sigma^2)$ iff $\bar{x}=\bar{y}$ and $s_x^2=s_y^2$.
\vspace{1mm}

By Theorem 6.2.13, $T=(\bar{X},S^2)$ is minimal sufficient for $\theta = (\mu,\sigma^2)$.
\end{block}\vspace{-2mm}
\begin{alertblock}{}
Like the concept of sufficiency, minimal sufficiency also depends on the parameters we are interested in.
\vspace{1mm}

In Example 6.2.14, let us consider two sub-families.
\begin{itemize}
\item
 Suppose that it is known that $\sigma =1$.

 Then $T=(\bar{X},S^2)$ is still sufficient but not minimal sufficient for $\mu$.

 In fact, using Theorem 6.2.13 we can show that $\bar{X}$ is minimal sufficient for $\mu$ when $\sigma =1$
 and, $T= (\bar{X},S^2)$ is not a function of $\bar{X}$ (why?).
\item
Suppose that it is know that $\mu^2  =  \sigma^2$ so that we only have one unknown parameter $\mu \in {\cal R}$, $\mu \neq 0$.
(This is called a curved exponential family.)

Is $T=(\bar{X},S^2)$ minimal sufficient for $\theta = \mu$?

Note that $T$ is certainly sufficient for $\theta$.

The answer can be found as a special case of the following result.
\end{itemize}
\end{alertblock}
}
\frame { \frametitle{ }
\begin{block}{Minimal sufficient statistics in exponential families}
Let $X_1,...,X_n$ be iid from an exponential family with pdf
\vspace{-2mm}
$$f_{\theta}(x)=h(x)c(\theta ) \exp \left( \eta(\theta )' T(x) \right), \quad \quad \theta \in \Theta \vspace{-2mm} $$
where $\eta ( \theta )' = (w_1(\theta ),...,w_k(\theta ))$ and
$T(x)' = (t_1(x),...,t_k(x))$.

Suppose that there exists $\{\theta_0,\theta_1,\dots,\theta_k\}\subset\Theta$
such that the vectors $\eta_i=\eta(\theta_i)-\eta(\theta_0),$ $i=1,...,k$,
are linearly independent in $\cal{R}^k$.

(This is true if the family is of full rank).

We have shown that $T(X)$ is sufficient for $\theta$.

We now show that $T$ is in fact minimal sufficient for $\theta$.

Note that we can focus on the pints $x$ at which $h(x) >0$.

For any $\theta$,\vspace{-2mm}
\[ \frac{f_\theta (x)}{f_\theta (y)} = \exp \left(\eta(\theta )'[T(x)-T(y)]\right)  \frac{h(x)}{h(y)} \vspace{-2mm}\]
If this ratio equals $\phi (x,y)$ not depending on $\theta$, then  \vspace{-2mm}
\[ \eta(\theta )' [T(x)-T(y)] = \log (\phi(x,y))+\log (h(y)/h(x)) \vspace{-2mm} \]
which implies \vspace{-2mm}
\[ [\eta (\theta_i )-\eta (\theta_0)]' [T(x)-T(y)] = 0 \quad \quad i=1,...,k.  \]
\end{block}}
\frame{
\begin{block}{}
Since $\eta(\theta_i)-\eta(\theta_0)$, $i=1,...,k$, are linearly independent, we must have $T(x)=T(y)$.
\vspace{1mm}

By Theorem 6.2.13, $T$ is minimal sufficient for $\theta$.
\end{block}\vspace{-2mm}
    \begin{alertblock}{Application to curved normal family}
We can now show that $T=(\bar{X},S^2)$ minimal sufficient for $\theta = \mu$
when the sample is a random sample from $N(\mu, \mu^2)$, $\mu \in {\cal R}$, $\mu \neq 0$.

It can be shown that \vspace{-2mm}
\[ \eta (\theta ) = \left( \frac{n\mu}{\sigma^2} , - \frac{n-1}{2\sigma^2} \right)'
= \left( \frac{n}{ \mu } , - \frac{n-1}{2\mu^2} \right)' \vspace{-2mm} \]
Points $\theta_0 = (1,1)$, $\theta_1 = (-1,1)$, and $\theta_2 = (1/2,1/2)$ are in the parameter space, and \vspace{-2mm}
\[ \eta (\theta_1) - \eta (\theta_0 ) =
\left( \br{c}
-n \\
-(n-1)/2 \er \right)
-  \left( \br{c}
n \\
-(n-1)/2 \er \right)
= \left( \br{c}
-2n \\
0 \er \right) \]
\[ \eta (\theta_2) - \eta (\theta_0 ) =
\left( \br{c}
n/2 \\
-2(n-1) \er \right)
-  \left( \br{c}
n \\
-(n-1)/2 \er \right)
= \left( \br{c}
-n/2 \\
-3(n-1)/2 \er \right) \]
     \end{alertblock}
}
\frame{\vspace{-2mm} 
     \begin{alertblock}{}
Are these two vectors linearly independent?

If there are $c$ and $d$ such that \vspace{-2mm}
\[ c[\eta (\theta_1) - \eta (\theta_0 )] + d[\eta (\theta_2) - \eta (\theta_0 )] = 0 \vspace{-2mm} \]
i.e., \vspace{-2mm}
\[ c \left( \br{c}
-2n \\
0 \er \right) + d  \left( \br{c}
-n/2 \\
-3(n-1)/2 \er \right) = 0 \]
Then, we must have $d=0$ and then $c=0$.

That means vectors $\eta (\theta_1) - \eta (\theta_0 )$ and $\eta (\theta_2) - \eta (\theta_0 )$ are linearly independent
and the condition for the result of exponential family is satisfied.

Therefore,  $T=(\bar{X},S^2)$ is minimal sufficient for $\theta = \mu$.
     \end{alertblock}
     \vspace{-2mm} 
     \begin{block}{Example.}
     Let $(X_1,...,X_n)$ be a random sample from 
$Cauchy(\mu, \sigma) $, where
$\mu \in {\cal R}$ and $\sigma>0$ are unknown parameters.

The joint pdf of $(X_1, ..., X_n) $ is\vspace{-2mm} 
\[ f_{\mu, \sigma}(x ) = \frac{\sigma^n }{ \pi^n}
\prod_{i=1}^n \frac{1}{\sigma^2 + ( x_i-\mu) ^2}, \quad \quad  x=(x_1,...,x_n)\in {\cal R}^n .\]
\end{block}}
\frame{\vspace{-2mm} 
\begin{block}{}
For any $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$, if \vspace{-2mm} 
\[
f_{\mu, \sigma}(x )= \psi (x, y)f_{\mu, \sigma}(y ) \vspace{-2mm} \]
holds for any $\mu$ and $\sigma$, where $\psi$ does not depend on $(\mu,\sigma)$, then   \vspace{-2mm} 
\[ \prod_{i=1}^n \left [1+ \left ( y_i-\mu \right ) ^2 \right ]=
\psi (x, y) \prod_{i=1}^n \left [1+ \left (x_i-\mu \right ) ^2 \right ] \quad \quad \mu \in {\cal R} \vspace{-2mm}  \]
Both sides of the above identity are polynomials of  degree
$2n$ in $\mu$.

Comparison of the coefficients to  the highest terms gives
$ \psi( x, y) =1$ and hence \vspace{-3mm} 
\[
\prod_{i=1}^n \left [1+ \left ( y_i-\mu \right ) ^2 \right ]=
\prod_{i=1}^n \left [1+ \left (x_i-\mu \right ) ^2 \right ] \quad \quad \mu \in {\cal R} \vspace{-3mm} \]
As a polynomial of $\mu$, the left-hand side of the above identity
has $2n$ complex roots $x_i\pm \dot{\imath}$, $i=1, ..., n$, while the
right-hand side of the above identity has $2n$ complex roots
$y_i\pm \dot{\imath}$, $i=1, ... , n$. 

Since the two
sets of roots must agree, 
the ordered
values of $x_i$'s are the same as the ordered values of
$y_i$'s. 

By Theorem 6.2.13, the order statistics of $X_1,...,X_n$
is minimal sufficient for $(\mu,\sigma )$.
     \end{block}
}
\frame{\vspace{-2mm} 
	\begin{alertblock}{}
		The following result may be useful in establishing minimal sufficiency outside of the exponential family.
\end{alertblock}\vspace{-2mm}
	\begin{alertblock}{Theorem 6.6.5.}
		Let $f_0,f_1,f_2,...$ be a sequence of pdf's or pmf's
		on the range of $X$, all have the same support. 
		\begin{itemize}
			\item[a.]
			The statistic 
			\[ T (X) = \left(\frac{f_1(X)}{f_0(X)},\frac{f_2(X)}{f_0(X)},...\right) \vspace{-2mm}\]
			is minimal sufficient for the family 
			$\{ f_0, f_1,f_2,... \}$ of populations for $X$. 
			\item[b.]
			If ${\cal P}$ is a family of pdf's or pmf's for $X$ with common support, and 
		\\
		(i) $f_i \in {\cal P}$, $i=0,1,2,...$\\
		(ii) $T(X)$ is sufficient for ${\cal P}$, \\
		then $T$ is minimal sufficient for ${\cal P}$. 
							\end{itemize}
	\end{alertblock}\vspace{-2mm}
	\begin{block}{Proof of part a.}
			Let $g_i(T)=T_i$, the $i$th component of $T$, $ i=0,1,2,...$, and let $g_0(T)=1$.\vspace{1mm}  
			
			Then $f_i(x)=g_i(T(x))f_{0}(x)$, $i=0,1,2,...$ 
			\vspace{1mm}
			
			By the factorization theorem, $T$ is sufficient for ${\cal P}_0 = \{ f_0,f_1,f_2,...\}$. 
		\end{block}}
		\frame{\begin{block}{}
			Suppose that $S(X)$ is another sufficient statistic. 
			\vspace{1mm}
			
			By the factorization theorem, there are Borel functions $h$ and $\tilde g_i$ such that\vspace{-1mm} 
			\[ f_i(x)=\tilde g_i(S(x))h(x), \quad \ i=0,1,2,... \vspace{-2mm} \]
			Then \vspace{-1mm} 
			\[ T_i(x)=\frac{f_i(x)}{f_0(x)} = \frac{\tilde g_i(S(x))}{ \tilde g_0(S(x))},  \quad \ i=0,1,2,... \vspace{-1mm}\]
			 	Thus, $T$ is a function of $S$. 
			\vspace{1mm}
			 	
			 	By definition, $T$ is minimal sufficient for ${\cal P}_0$. 
	\end{block}\vspace{-2mm}
	\begin{block}{Proof of part b.}
		The two additional conditions are ${\cal P}_0 \subset {\cal P}$ and $T$ is sufficient for ${\cal P}$. 
		\vspace{1mm}
		
		Let $S$ be any sufficient statistic for ${\cal P}$. 
		\vspace{1mm}
		
		Then $S$ is also sufficient for ${\cal P}_0$.
		\vspace{1mm}
		
		From part a, $T$ is minimal sufficient for ${\cal P}_0$ so there is a function $\psi $ such that $T=\psi (S)$. 
		\vspace{1mm}
		
		By definition,  $T$ is minimal sufficient for ${\cal P}$ since $T$ is sufficient for ${\cal P}$. 
			\end{block}}
			\frame{\vspace{-2mm}
		\begin{alertblock}{}
			\begin{itemize}
				\item
		In fact, the result in Theorem 6.6.5 still holds if the common support condition is replaced by that the support of $f_0$ contains the support of any pdf or pmf in ${\cal P}$.
		\item
		In most applications, we can choose ${\cal P}_0$ containing finitely many elements. 
			\end{itemize}
	\end{alertblock}
	\vspace{-2mm}
	\begin{block}{Example}
		Let $X_1,...,X_n$ be a random sample from $double$-$exponential (\mu , 1)$, $\mu \in {\cal R}$,
		i.e., the joint pdf of $X$ is \vspace{-2mm}
		\[ f_\mu (x) = \frac{1}{2^n} \exp \left(- \sum_{i=1}^n |x_i - \mu | \right) \vspace{-2mm} \] 
		Consider ${\cal P}_0 = \{ f_0,f_1,...,f_n\}$, where each $f_j$ is the pdf with $\mu = j$, $j=0,1,...,n$. 
		
		By Theorem 6.6.5a, a minimal sufficient statistic for ${\cal P}_0$ is \vspace{-2mm} 
		\[ T = \left( \exp \left( \sum_{i=1}^n |X_i|- \sum_{i=1}^n |X_i - j| \right) , \ \ j=1,...,n \right) \vspace{-2mm}\]
		which is equivalent to 
		\end{block}}			
\frame{\vspace{-2mm} 
	\begin{block}{}
		\[ U = \left( \sum_{i=1}^n |X_i|- \sum_{i=1}^n |X_i - j|  , \ \ j=1,...,n \right) \vspace{-2mm}\]
		We can further show that $U$ is equivalent to $S$, the set of order statistics. 
		
		Since $S$ is sufficient, by Theorem 6.6.5b, $S$, $U$, and $T$ are all minimal sufficient. 
	\end{block}\vspace{-2mm}
		\begin{alertblock}{Definition 6.6.3 (Necessity).}
A statistic is said to be necessary if it can be written as a function of every sufficient statistic. 
\end{alertblock}\vspace{-2mm}
\begin{block}{}
$T$ is necessary if, for every sufficient $S$, there is a function $\psi $ such that $T= \psi (S)$. \vspace{1mm}

From the definitions of the necessity and minimal sufficiency, we can reach the following conclusion. 
\end{block}\vspace{-2mm}
\begin{alertblock}{Theorem 6.6.4.}
	A statistic is minimal sufficient if and only if it is necessary and sufficient. 
\end{alertblock}
}
\end{document}
