title: Sufficient Statistic
author: Keith A. Lewis
institution: KALX, LLC
email: [email protected]
classoption: fleqn
abstract: Sufficient for what?
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\newcommand\RR{\bm{R}}
\newcommand\CC{\bm{C}}
\newcommand\FF{\bm{F}}
\newcommand\ker{\operatorname{ker}}
\newcommand\ran{\operatorname{ran}}
The set of bounded function on a set $X$, $B(X)$,
is a Banach space with norm $|f| = \sup_{x\in X}|f(x)|$.
The characteristic function of $A\subseteq X$ is $1_A$ where
$1_A(x) = 1$ if $x\in A$ and $1_A(x) = 0$ if $x\not\in A$.
The power set of $X$ is $\mathcal{P}(X) = {E\subseteq X}$.
Let $\mathcal{O}(X) = {1_A:A\subseteq X}$.
Exercise. $B(X)$ is the norm closed algebra generated by $\mathcal{O}(X)$.
If $f\colon X\to Y$ is a function define
$f^{\dashv}\colon\mathcal{P}(Y)\to\mathcal{P}(X)$ by
$f^{\dashv}(B) = {x\in X:f(x)\in B}\subseteq X$, $B\subseteq Y$.
We also write $f^{\dashv}1_B = 1_{f^{\dashv}(B)}$.
This can be extended to a map $B(Y)\to B(X)$.
An algebra of sets on $X$ is a subset of $\mathcal{P}(X)$ closed
under complement, union, and contains the empty set.
If $\mathcal{A}$ is an algebra of sets on $X$ let $B(X,\mathcal{A})$
be the norm closed algebra generated by ${1_A:A\in\mathcal{A}}$.
Exercise. If $\mathcal{A}\subseteq\mathcal{P}(X)$ then
the set of finite sums $\sum a_j 1{A_j}$, $a_j\in\RR$, $1_{A_j}\in\mathcal{A}$
is an algebra if and only if $\mathcal{A}$ is an algebra_.
Solution
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Let $ba(X,\mathcal{A})$ be the collection of finitely additive set
functions on $\mathcal{A}$, $α\colon\mathcal{A}\to\RR$ such
that $α(A\cup B) = α(A) + α(B) - α(A\cap B)$ and $α(\emptyset) = 0$, $A,B\in\mathcal{A}$.
Lemma. The vector space dual of $B(X,\mathcal{A})$ is $ba(X,\mathcal{A})$.
Hint: Given $α^\in B(X,\mathcal{A})^$ define $α\in ba(X,\mathcal{A})$ by $α(A) = α^(1_A)$.
Given $α\in ba(X,\mathcal{A})$ define $α^(1_A) = α(A)$ and use
Exercise. If $f = \sum_j a_j 1{A_j}$ is a finite sum then
$f = \sum_k b_k 1_{B_k}$ where the $B_k$ are pairwise disjoint_.
A statistic is a measurable function $T\colon X\to Y$ be measurable.
When does a transformation $T\colon B(Y,\mathcal{B})\to B(X,\mathcal{A})$
come from a function $t\colon X\to Y$?