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\title[Stat 609 Lecture 24]{Stat 609: Mathematical Statistics \\ Lecture 24}
\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 24: Completeness}
\vspace{-1mm}
    \begin{alertblock}{Definition 6.2.16 (ancillary statistics)}
    A statistic $V(X)$ is ancillary iff its distribution
does not depend on any unknown quantity.
A statistic $V(X)$ is first-order ancillary iff $E[V(X)]$
does not depend on any unknown quantity.
\end{alertblock}\vspace{-2mm}
\begin{alertblock}{}
A trivial ancillary statistic is $V(X) \equiv $ a constant.
\vspace{1mm}

The following examples show that there exist many nontrivial ancillary statistics (non-constant ancillary statistics). 
\end{alertblock}\vspace{-2mm}
\begin{block}{Examples 6.2.18 and 6.2.19 (location-scale families)}
\begin{itemize}
\item
If $X_1,...,X_n$ is a random sample from a location family with location parameter $\mu \in {\cal R}$, then, for any 
pair $(i,j)$, $1 \leq i, j \leq n$, $X_i - X_j$ is ancillary, because $X_i-X_j = (X_i-\mu )-(X_j-\mu) $ and the distribution of 
$(X_i-\mu, X_j-\mu)$ does not depend on any unknown parameter.  
\vspace{1mm}

Similarly, $X_{(i)}-X_{(j)}$ is ancillary, where $X_{(1)},...,X_{(n)}$ are the order statistics, and the sample variance $S^2$ is ancillary. 
\end{itemize}
\end{block}}
\frame{\vspace{-2mm}
\begin{block}{}
\begin{itemize}
\item
Note that we do not even need to obtain the form of the distribution of $X_i-X_j$.
\item
If $X_1,...,X_n$ is a random sample from a scale family with scale parameter $\sigma >0$, then by the same argument we can show that, for any
pair $(i,j)$, $1 \leq i, j \leq n$, $X_i / X_j$ and $X_{(i)}/X_{(j)}$ are ancillary. 
\item
If $X_1,...,X_n$ is a random sample from a location-scale family with  parameters $\mu \in {\cal R}$
and $\sigma >0$, then, for any $(i,j,k)$, $1 \leq i,j,k \leq n$, 
$(X_i- X_k)/(X_j-X_k)$ and 
$(X_{(i)}- X_{(k)})/(X_{(j)}-X_{(k)})$ are ancillary. 
\end{itemize}
     \end{block}\vspace{-2mm}
\begin{alertblock}{}
\begin{itemize}
\item
If $V(X)$ is a non-trivial ancillary statistic, then
the the set  $\{ x: V(x )=v \}$
does not contain any information about $\theta$.
\item
If $T(X)$ is a statistic and $V(T(X))$ is a non-trivial ancillary
statistic, it indicates that the reduced data set by $T$ contains
a non-trivial part that does not contain any information about
$\theta$ and, hence, a further simplification of $T$ may still be needed.
\item
A sufficient statistic $T(X)$ appears to be most successful in reducing the
data if no nonconstant function of $T(X)$ is ancillary or even first-order
ancillary, which leads to the following definition. 
\end{itemize}
 \end{alertblock}}
\frame{\vspace{-2mm}
\begin{block}{Definition 6.2.21 (completeness)} 
Let $X$ be a sample from a family indexed by $\theta \in \Theta$ (the parameter space) and let
$E_\theta$ and $P_\theta$ be the expectation and probability, respectively, calculated with respect to a $\theta \in \Theta$.
\begin{itemize}
\item
A statistic $T(X )$ is complete iff for any function $g$ not depending on $\theta$,
$E_\theta [g(T)]=0$
for all $\theta \in \Theta $ implies $P_\theta ( g(T)=0)=1$
for all $\theta \in \Theta $.
\item
A statistic $T$ is boundedly complete iff the previous
statement holds for any bounded $g$.
\item
The family of distributions corresponding to a statistic $T$ is complete (or boundedly complete) iff 
$T$ is complete (or boundedly complete). 
\end{itemize}
\end{block}\vspace{-2mm}
\begin{alertblock}{}
 \begin{itemize}
   \item A complete statistic is boundedly complete.
   \item If $T$
is complete (or boundedly complete) and $S=\psi (T)$ for
a measurable $\psi$, then $S$ is complete (or boundedly complete).
   \item It can be shown that a complete
and sufficient statistic is minimal sufficient (Theorem 6.2.28).
   \item A minimal sufficient statistic is not necessarily complete. 
 \end{itemize}
\end{alertblock}
}
\frame{\vspace{-2mm}
\begin{block}{Example 6.2.15.}
In this example, $X_1,...,X_n$ is a random sample from $uniform(\theta , \theta +1)$, $\theta \in {\cal R}$, and we showed that 
$T=(X_{(1)},X_{(n)})$ is the minimal sufficient
statistic for $\theta$. 

We now show that $T$ is not complete. 

Note that $V(T) = X_{(n)}-X_{(1)}= (X_{(n)}-\theta )-( X_{(1)}- \theta )$
is in fact ancillary. 

Its distribution can be obtained using the result in Example 5.4.7, but we do not need that for arguing that $T$ is not complete. 

It is easy to see that $E_\theta (V)$ exists and it does not depend on $\theta$ since $V$ is ancillary. 

Letting $c= E(V)$, we see that $E_\theta (V-c) =0$ for all $\theta $. 

Thus, we have a function $g(x,y) = x-y-c$ such that $E_\theta [g(X_{(1)},X_{(n)})] = E_\theta (V-c) =0$ for all $\theta $ but 
  $P_\theta ( g(X_{(1)},X_{(n)}) =0 ) = P_\theta  ( V = c) \neq 0$.

This shows that $T$ is not complete. 
\end{block}\vspace{-2mm}
\begin{alertblock}{}
If a minimal sufficient statistic $T$ is not complete, then \vspace{-0.5mm}
\begin{itemize}
\item there is a non-trivial first order ancillary statistic $V(T)$;\vspace{-0.5mm}
\item
there does not exist any complete statistic. 
\end{itemize}
\end{alertblock}
}
\frame{\vspace{-2mm} 
\begin{block}{Example 6.2.22.} 
Let $T \sim binomial(n,\theta )$, $0< \theta < 1$, (note that $T$ is a sufficient statistic based on a random sample of size $n$ from $binomial(1,\theta )$).

If $g$ is a function such that $E_\theta [g(T)] = 0 $ for all $\theta$,  then \vspace{-2mm} 
\beqrs
 0 = E_\theta [g(T)] &=& \sum_{t=0}^n g(t) {n \choose t} \theta^t (1-\theta )^{n-t} \\
&=& (1-\theta )^n \sum_{t=0}^n g(t) {n \choose t} \left( \frac{\theta}{1-\theta }\right)^{t} 
\quad \quad \mb{all $\theta \in (0,1)$} \vspace{-2mm} 
\eeqrs
Since the factor $(1-\theta )^n \neq 0$, we must have \vspace{-2mm} 
\[ 0 = \sum_{t=0}^n g(t) {n \choose t} \varphi^t 
\quad \quad \mb{all $\varphi >0$}, \ \varphi = \frac{\theta}{1-\theta}  \vspace{-2mm} \]
The last expression is a polynomial in $\varphi$ of degree $n$.  

For this polynomial to be 0 for all $\varphi >0$, it must be true that
the coefficient of $\varphi^t $, which is $ g(t){n \choose t} $, is 0 for every $t$.

This shows that $g(t) =0 $ for $t=0,1,...,n$ and hence $P_\theta (g(T)=0 ) =1$ for all $\theta$, i.e., $T$ is complete.  
\end{block}}
\frame {\vspace{-2mm}
\begin{alertblock}{Example 6.2.23.}
Let $X_1,...,X_n$ be iid from
$uniform(0,\theta )$, $\theta >0$, with pdf 
\vspace{-2mm}
\[ f_\theta (x ) = \left\{ \br{ll}
1 & \quad 0<x_i<\theta , i=1,...,n \\
0 & \quad \mbox{otherwise} \er \right. 
= \left\{ \br{ll}
1 & \quad 0<x_{(n)}<\theta  \\
0 & \quad \mbox{otherwise} \er \right. \vspace{-2mm} \]
By the factorization theorem, $X_{(n)}$ is sufficient for $\theta$.

Using the result for the order statistics in Chapter 5, $X_{(n)}$ has pdf\vspace{-2mm} 
\[ f_{X_{(n)}}(t) =  \left\{ \br{ll}
nt^{n-1}\theta^{-n} & \quad 0<t<\theta  \\
0 & \quad \mbox{otherwise} \er \right. \vspace{-2mm} \]
For any $g$, \vspace{-2mm} 
\[ E_\theta [g(X_{(n)})]= n\theta^{-n} \int_0^\theta g(t) t^{n-1}dt =0 \quad \quad \mb{all $\theta >0$} \vspace{-2mm} \]
implies  that \vspace{-2mm} 
\[ 0 = \int_0^\theta g(t) t^{n-1}dt  
\quad \mb{and} \quad 0 = \frac{d}{d\theta } \int_0^\theta g(t) t^{n-1}dt
= g(\theta ) \theta^{n-1} \quad \quad \mb{all $\theta >0$} \vspace{-2mm} \]
Thus,  $g(t)=0$ for all $t >0$, which means that
$P_\theta (g(T)=0)=1$ for all $\theta$.

Hence, $X_{(n)}$ is also complete.
     \end{alertblock}
}
\frame { \vspace{-2mm} 
\begin{alertblock}{}
The relationship between an ancillary statistic and a complete and
sufficient statistic is characterized in the following result.
\end{alertblock}\vspace{-2mm}
\begin{block}{Theorem 6.2.24  (Basu's theorem)}
Let $V$ and $T$ be two statistics of $X$ from a population indexed by $\theta \in \Theta$.

If $V$ is ancillary and
$T$ is boundedly complete and sufficient for $\theta $, then $V$ and $T$ are independent with respect to $P_\theta $ for any $\theta \in \Theta$.
     \end{block}\vspace{-2mm} 
    \begin{block}{Proof.}
Let $B$ be an event on the range of $V$ and 
 $A$  an event on the range of $T$. 
 
 From the 3rd definition of the independence of random variables, we only need to show that 
 \vspace{-2mm} 
\[ P_\theta ( T^{-1}(A) \cap V^{-1}(B)) =P_\theta (T^{-1}(A))P_\theta (V^{-1}(B)), \quad  \quad \theta \in \Theta \vspace{-2mm} \]

Since $V$ is ancillary,
$P_\theta (V^{-1}(B)) = P_B$ does not depend on $\theta $. 

As $T$ is sufficient,
$E_\theta [I_B(V)|T] = h_B(T) $  is a function of $T$ (not depending on $\theta $), where $I_B( V)$ is the indicator function of $\{V \in B\}$. 

Since \vspace{-2mm}
\[ E_\theta [h_B(T)] = E_\theta \{ E[I_B(V)|T]\} = E_\theta \{ I_B(V)\} 
= P_\theta (V^{-1}(B))  = P_B \quad  \theta  \in \Theta , \]
\end{block}}
\frame{
\begin{block}{}
by the bounded completeness of $T$, \vspace{-2mm}
\[ P_\theta ( h_B(T) = P_B ) =1 \quad \quad   \theta  \in \Theta \vspace{-2mm}  \]
Then the result follows from \vspace{-2mm}
\beqrs
 P_\theta (T^{-1}(A)\cap V^{-1}(B)) & = &
E_\theta \{E_\theta [I_A(T)I_B(V)|T]\}\\
&= & E_\theta \{I_A(T)E_\theta [I_B(V)|T]\} \\
& = & E_\theta \{I_A(T)P_B\}=P_B E_\theta \{ I_A(T)\}\\
& =&  P_\theta (T^{-1}(A))P(V^{-1}(B)) \quad \quad \theta \in \Theta 
\eeqrs
     \end{block}
\vspace{-2mm}
    \begin{alertblock}{}
\begin{itemize}
\item
If a minimal sufficient statistic $T$ is not complete, then there may be an ancillary statistic $V$ such that $V$ and $T$ are not independent.

An example is in Example 6.2.15, $T=(X_{(1)}, X_{(n)})$ is minimal sufficient but not complete, and $T$ and the ancillary statistic $V = X_{(n)}-X_{(1)}$ 
is not independent. 
\item
     Basu's theorem is useful in proving the independence of two statistics.
\end{itemize}
     \end{alertblock} \vspace{-2mm}
     \begin{block}{}
     We first state without proof the following useful result. 
     \end{block}}
\frame{\vspace{-2mm}
\begin{block}{Theorem 6.2.25 (complete statistics in exponential families)}
Let $X=(X_1,...,X_n)$ be a random sample with pdf \vspace{-2mm}
\[{f}_\theta ( x )
= h(x) c(\theta ) \exp \left(  \sum_{j=1}^k w_j(\theta )t_j(x )\right) \quad \quad \theta \in \Theta   , \vspace{-2mm} \]
Then the statistic \vspace{-2mm} 
\[ T(X) = \left( \sum_{i=1}^n t_1(X_i),..., \sum_{i=1}^n t_k(X_i) \right) \vspace{-2mm} \]
 is complete as long as $\Theta $ contains an open set in ${\cal R}^k$
 (i.e., the family of distributions is of full rank).
\end{block}\vspace{-2mm}
\begin{alertblock}{}
\begin{itemize}
\item
Note that $T$ is also sufficient for $\theta$ (without requiring any condition on $\Theta$). 
\item
Compared with the result of minimal sufficient statistics in curved exponential families,
the condition on $\Theta$ in this theorem is stronger.  
\end{itemize}
\end{alertblock}\vspace{-2mm}
\begin{block}{}
We illustrate the application of Basu's theorem and Theorem 6.2.25 in the normal distribution family. 
\end{block} 
}
\frame { 
\vskip -.2cm
\begin{alertblock}{Example (the normal family).} 
Suppose that $X_1,...,X_n$ are iid from  $N(\mu,
\sigma^2)$, $\mu \in {\cal R}$, $\sigma >0$, $\theta = (\mu, \sigma^2)$.

The joint pdf of $X_1,...,X_n$ is
\vspace{-2mm}
\[  (2\pi)^{-n/2} \exp\left\{
 \eta_1 T_1+  \eta_2 T_2 - n \zeta(\eta) \right\},\vspace{-2mm}\]
where $T_1=\sum_{i=1}^n X_i$, $T_2=-\sum_{i=1}^n X_i^2$,
$\eta=(\eta_1,\eta_2)= \left(
 \frac{\mu}{\sigma^2} , \frac{1}{2\sigma^2} \right)$, and $\zeta ( \eta )$ is a function of $\eta$.

This is an exponential family of full rank, since 
 ${\cal R}\times
(0,\infty )$ is open.

By Theorem 6.2.25, $T(X)= (T_1, T_2 )$ is
complete and sufficient for $\theta $.

It can be shown that any one-to-one
function of a complete and sufficient statistic is also
complete and sufficient.

Thus, $(\bar{X}, S^2)$ is complete and sufficient for $\theta$. 
\vspace{2mm}

We now apply Basu's theorem to show that $\bar{X}$ and $S^2$ are independent for any $\theta$. 

For this purpose, we consider a sub-family with unknown $\mu \in {\cal R}$ and a known (fixed) $\sigma^2 >0$. 

Note that we only need to show that $\bar{X}$ and $S^2$ are independent for every fixed $(\mu, \sigma^2)$.  
     \end{alertblock}
}
\frame { 
\begin{alertblock}{}

If $X_1,...,X_n$ are iid from $N(\mu,
\sigma^2)$ with $\mu \in {\cal R}$ and a known $\sigma >0$, then 
it can be easily shown that the family  is an exponential family of full rank with
parameter $\mu \in {\cal R}$.

By Theorem 6.2.25,  $\bar{X}$
is complete and sufficient for $\mu$.

Since $S^2 = (n-1)^{-1} \sum_{i=1}^n (Z_i - \bar{Z})^2 $,
where $Z_i=X_i - \mu$ is $N(0,\sigma^2)$
and $\bar{Z}= n^{-1}\sum_{i=1}^n Z_i$,
$S^2$ is an ancillary statistic ($\sigma^2$ is known).

By Basu's theorem, $\bar{X}$ and $S^2$ are independent with respect to 
$N(\mu, \sigma^2)$ for any
$\mu \in {\cal R}$ and $\sigma^2>0$.
\vspace{2mm}

Note that this proof is simpler than the proof we gave in Chapter 5. 
\end{alertblock}\vspace{-2mm}
   \begin{block}{Example}
Let $X_1,...,X_n$ be 
   	a random sample from the family of pdf's of the form \vspace{-2mm}
   	\[ f_\theta (x)=C(\theta_1,...,\theta_n) \exp \{ - x^{2n} + \theta_1 x +
   	\theta_2 x^2 + \cdots + \theta_n x^n \} ,  \vspace{-2mm}\]
   	where $\theta_j \in {\cal R}$, $\theta = (\theta_1,...,\theta_n)$, and
   	$C(\theta_1,...,\theta_n)$ is a normalizing constant such that
   	$\int f(x) dx =1$.
   	
   	This family of pdf's is an exponential family of full rank.
 \vspace{1mm}
   	
By Theorem 6.2.25,
   	$U=(U_1,...,U_n)$ is a complete and sufficient 
   \end{block}
}
\frame{\vspace{-2mm}
	\begin{block}{}
statistic for $\theta$,	where
   	$U_j = \sum_{i=1}^n X_i^j$, $j=1,...,n$.
   	\vspace{1mm} 
   	
   	We want to show that $T(X)=(X_{(1)},...,X_{(n)})$, the vector of order statistics, 
   	is also complete and sufficient for $\theta$.  	   	
   	  	\vspace{1mm} 
   	  	
The result follows if we can show that there is a one-to-one correspondence
		between $T(X) $ and $U(X)$.
		  	\vspace{1mm} 
		  	
		Let $V_1 = \sum_{i=1}^n X_i$, $V_2 =
		\sum_{i < j} X_iX_j$, $V_3 = \sum_{i<j<k} X_iX_jX_k$,...,
		$V_n = X_1\cdots X_n$.
		
		From the identities \vspace{-2mm}
		\[ U_k -V_1 U_{k-1} + V_2 U_{k-2} - \cdots + (-1)^{k-1} V_{k-1}U_1
		+ (-1)^k k V_k =0, \vspace{-2mm}\]
		$ k=1,...,n$, there is a one-to-one correspondence
		between $U(X)$ and $V(X)  =  (V_1,...,V_n)$.
  	\vspace{1mm} 
		
		From the identity   	\vspace{-2mm} 
		\[ (t-X_1)\cdots (t-X_n) = t^n - V_1 t^{n-1} + V_2 t^{n-2} - \cdots
		+ (-1)^n V_n , \vspace{-2mm}\]
		there is a one-to-one correspondence
		between $V(X)$ and $T(X)$.
		  	\vspace{1mm} 
		  	
		  	Hence, $T(X)$ is sufficient and complete for $\theta$.
		\vspace{1mm} 
		
		In fact, $V(X)$ is also sufficient and complete for
		$\theta$.
		
		   	
	\end{block}	
}
\frame{\vspace{-2mm}
\begin{alertblock}{}
The relationship between minimal sufficiency and sufficiency with completeness is given by the following theorem.
\end{alertblock}\vspace{-2mm}
\begin{alertblock}{Theorem 6.2.28 (modified).}
	Suppose that $S$ is a sufficient statistic and $T$ is a complete and sufficient statistic. Then $T$ must be minimal sufficient and $S$ must be complete.  
\end{alertblock}
\vspace{-2mm}
\begin{alertblock}{Proof.}
	Since $S$ is minimal sufficient and $T$
	is sufficient, there exists a Borel function $h$ such that
	$S=h(T)$.
\vspace{0.5mm}
	
	 Since $h$ cannot be a constant function and $T$ is complete,
	we conclude that $S$ is complete. 
\vspace{0.5mm}
	
	Consider $T-E(T|S)=T-E[T|h(T)]$, which is a Borel
	function of $T$ and hence can be denoted as $g(T)$. 
\vspace{0.5mm}
	
	Note that
	$E[g(T)]=0$. 
\vspace{0.5mm}
	
	By the completeness of $T$,
	$g(T)=0$ a.s., that is, $T=E(T|S)$ a.s. 
\vspace{0.5mm}
	
	This means that $T$ is
	also a function of $S$ and, therefore, $T$ is minimal sufficient.	
\end{alertblock}	
	}
\frame{\vspace{-2mm}
	\begin{block}{Example 6.2.20 (ancillary precision)}
		Let $X_1$ and $X_2$ be iid from the discrete uniform distribution on three points
		$\{ \theta , \theta + 1 , \theta +2 \}$, where $\theta \in \Theta = \{ 0, \pm 1 , \pm 2 ,... \}$.
		\vspace{1mm}
		
		Using the same argument as in Example 6.2.15, we can show that the order statistics $(X_{(1)},X_{(2)})$ is minimal sufficient for $\theta$.
		\vspace{1mm}
		
		Let  $M= (X_{(1)}+X_{(2)})/2$ and $R=X_{(2)}-X_{(1)}$ (mid-range and range).
		\vspace{1mm}
		
		Since $(M,R)$ is a one-to-one function of $(X_{(1)},X_{(2)})$, it is also minimal sufficient for $\theta $.
		\vspace{1mm}
		
		Consider the estimation of $\theta$ using $(M,R)$.
		\vspace{1mm}
		
		Note that $R= (X_{(2)}-\theta )-(X_{(1)}-\theta )$
		is the range of the two order statistics from the uniform distribution on $\{0, 1, 2 \}$ and, hence the distribution of $R$
		does not depend on $\theta$, i.e., $R$ is ancillary.
		\vspace{1mm}
		
		One may think $R$ is useless in the estimation of $\theta$ and only $M$ is useful.
		\vspace{1mm}
		
		Suppose we observe $(M,R) = (m,r)$ and $m$ is an integer.
		\vspace{1mm}
		
		From the observation $m$, we know that $\theta$ can only be one of the 3 values $m$, $m-1$, and $m-2$;
		however, we are not certain which of the 3 values is $\theta$.
	\end{block}
}
\frame{
	\begin{block}{}
		We can know more if $r=2$, which must be the case that $X_{(1)} = m-1$ and $X_{(2)} = m+1$.
		\vspace{1mm}
		
		With this additional information, the only possible value for $\theta $ is $m-1$.
		\vspace{1mm}
		
		When $m$ is an integer, $r$ cannot be 1. If $r =0$, then we know that $X_1=X_2$ and we are not certain
		which of the 3 values is $\theta$.
		\vspace{1mm}
		
		
		The knowledge of the value of the ancillary statistic $R$ increases our knowledge about $\theta$, although $R$ alone gives us no information about $\theta$.
	\end{block}\vspace{-2mm}
	\begin{alertblock}{What we learn from the previous example?}
		\begin{itemize}
			\item
			An ancillary statistic that is a function of a minimal sufficient statistic $T$
			may still be useful for our knowledge about $\theta$.\\
			(Note that the ancillary statistic is still a function of $T$.)
			\item
			This cannot occur to a sufficient and complete statistic $T$, since, if $V(T)$ is ancillary, then by the completeness of $T$,
			$V$ must be a constant and is useless.
			\item
			Therefore, the sufficiency and completeness together is a much desirable (and strong) property.
		\end{itemize}
	\end{alertblock}
}


\end{document}
