The Annals of Mathematical Statistics

Sufficiency in the Undominated Case

D. L. Burkholder

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Abstract

In this paper the concept of statistical sufficiency is studied within a general probability setting. It is not assumed that the family of probability measures is dominated. That is, it is not assumed that there is a $\sigma$-finite measure $\mu$ such that each probability measure in the family is absolutely continuous with respect to $\mu$. In the dominated case, the theory of sufficiency has received a thorough-going and elegant treatment by Halmos and Savage [6], Bahadur [2], and others. Although many families of probability measures of importance for statistical work are dominated, many others are not. Nonparametric statistical work, especially, abounds with undominated families. It seems appropriate, therefore, to see what can be learned about sufficiency in the undominated case. Let $X$ be a set, $\mathbf{A}$ a $\sigma$-field of subsets of $X$, and $P$ a family of probability measures $p$ on $\mathbf{A}$. The probability structure $(X, \mathbf{A}, P)$ is to be kept in mind throughout the paper and is unrestricted except where specifically stated to the contrary. Any subfield (= sub-$\sigma$-field) entering the discussion is implicitly assumed to be a subfield of $\mathbf{A}$. If $H$ is a collection of subfields, let $\bigvee H$ denote the smallest $\sigma$-field containing each member of $H$. If $\mathbf{A}_1, \mathbf{A}_2, \cdots$ are subfields, write $\mathbf{A}_1 \bigvee \mathbf{A}_2$ for $\bigvee\{\mathbf{A}_1, \mathbf{A}_2\}, \mathbf{\bigvee}^\infty_{n = 1} \mathbf{A}_n$ for $\bigvee\{\mathbf{A}_1, \mathbf{A}_2, \cdots\}$, and so forth. A set $N$ is $P$-null if $N$ is $p$-null for each $p$ in $P$, that is, if $N$ is in $\mathbf{A}$ and $p(N) = 0, p \varepsilon P$. If $f$ and $g$ are $\mathbf{A}$-measurable functions, write $f = g\lbrack p\rbrack$ if the set $\{x\mid f(x) \neq g(x)\}$ is $p$-null and write $f = g\lbrack P\rbrack$ if this set is $P$-null. Let $\mathbf{N}$ be the smallest $\sigma$-field containing the $P$-null sets. If $\mathbf{A}_1$ and $\mathbf{A}_2$ are subfields, write $\mathbf{A}_1 \subset \mathbf{A}_2\lbrack P\rbrack$ if $\mathbf{A}_1 \subset \mathbf{A}_2 \bigvee \mathbf{N}$, and so forth. A subfield $\mathbf{B}$ is sufficient if, for each bounded $\mathbf{A}$-measurable function $f$, there is a $\mathbf{B}$-measurable function $g$ such that $\int_Bf dp = \int_Bg dp, B \varepsilon \mathbf{B}, p \varepsilon P$, that is, such that $g = E_p(f \mid\mathbf{B})\lbrack p\rbrack, p \varepsilon P$. Equivalent definitions are obtained if "bounded $\mathbf{A}$-measurable function" is replaced by $\mathbf{A}$-measurable characteristic function" or by "$P$-integrable function." Of course, $f$ is $P$-integrable if $f$ is $\mathbf{A}$-measurable and $\int _X|f| dp$ is finite for each $p$ in $P$. A subfield $\mathbf{B}$ is separable if it contains a countable subcollection such that $\mathbf{B}$ is the smallest $\sigma$-field containing the subcollection. In Section 2, we give an example of a nonsufficient subfield containing a sufficient subfield. This solves a problem posed by Bahadur (Problem 1 on page 441 of [2]). In fact, we show that often the collection of such nonsufficient subfields is much larger than the collection of sufficient subfields. Analogous results hold for statistics. Some of these and later results depend on Theorem 1 which gives a necessary condition for a subfield to be sufficient in the case that $\mathbf{A}$ is separable. Let $\mathbf{A}_1, \mathbf{A}_2, \cdots$ be a sequence of sufficient subfields. Are the subfields $\mathbf{A}_1 \bigcap \mathbf{A}_2, \mathbf{A}_1 \bigvee \mathbf{A}_2, \mathbf{\bigcap}^\infty_{n = 1}\mathbf{A}_n$, and $\mathbf{\bigvee}^\infty_{n = 1}\mathbf{A}_n$, necessarily sufficient? This question is investigated in Sections 3 and 4. Using martingale theory, we show that, if the sequence is decreasing (increasing), then $\mathbf{\bigcap}^\infty_{n = 1}\mathbf{A}_n(\mathbf{V}^\infty_{n = 1}\mathbf{A}_n)$ is sufficient. If the sequence is not necessarily monotone, it is still possible to show that $\mathbf{A}_1 \bigcap \mathbf{A}_2$ and $\bigcap^\infty_{n = 1}\mathbf{A}_n$ are sufficient under a small extra assumption involving $\mathbf{N}$. This result rests on a theorem proved in [3] regarding iterates of conditional expectation operators. One consequence of this result is of interest in connection with the theory of minimal sufficient subfields. It is not necessarily true that $\mathbf{A}_1 \bigvee \mathbf{A}_2$ is sufficient. This is shown in Example 4. Conditions under which $\mathbf{A}_1 \bigvee \mathbf{A}_2$ is sufficient are examined. The main result of Section 5 is related to Theorem 1 and indicates that if $\mathbf{A}$ is separable then each sufficient subfield is essentially equal to one of a very special type.

Article information

Source
Ann. Math. Statist., Volume 32, Number 4 (1961), 1191-1200.

Dates
First available in Project Euclid: 27 April 2007

Permanent link to this document
https://projecteuclid.org/euclid.aoms/1177704859

Digital Object Identifier
doi:10.1214/aoms/1177704859

Mathematical Reviews number (MathSciNet)
MR131287

Zentralblatt MATH identifier
0221.62003

JSTOR
links.jstor.org

Citation

Burkholder, D. L. Sufficiency in the Undominated Case. Ann. Math. Statist. 32 (1961), no. 4, 1191--1200. doi:10.1214/aoms/1177704859. https://projecteuclid.org/euclid.aoms/1177704859


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