Halmos-サベージ定理が優勢統計モデルのことを言う$(\Omega, \mathscr A, \mathscr P)$統計量$T: (\Omega, \mathscr A, \mathscr P)\to(\Omega', \mathscr A')$で十分であるすべてのための（及び場合のみ）であれば$\{P \in \mathscr{P} \}$が存在する$T$ラドンNikodym誘導体の-measurableバージョン$\frac{dP}{dP*}$ここで$dP*$、特権尺度であるように、$P*=\sum_{i=1}^\infty P_i c_i$するための$c_i >0, \sum _{i=1}^\infty c_i =1$と$P_i \in \mathscr P$。

定理が真である理由を直感的に把握しようとしましたが、成功しませんでしたので、定理を理解する直感的な方法があるかどうかが私の質問です。

技術補題

これがどれほど直感的かはわかりませんが、ハルモス・サベージの定理の声明の根底にある主な技術的結果は次のとおりです。

補題。 してみましょう$\mu$可能$\sigma$上-finite対策$(S, \mathcal{A})$。仮定し$\aleph$上の措置の集まりである$(S, \mathcal{A})$など、すべてのためのことを$\nu \in \aleph$、$\nu \ll \mu$。次いで、非負数のシーケンスが存在する$\{c_i\}_{i=1}^\infty$の要素と配列$\aleph$、$\{\nu_i\}_{i=1}^\infty$よう$\sum_{i=1}^\infty c_i = 1$と$\nu \ll \sum_{i=1}^\infty c_i \nu_i$ごとに1つずつ$\nu \in \aleph$。

これは、シェルビッシュの統計理論（1995）の定理A.78から逐語的に取られています。その中で、彼はそれをリーマンの統計統計仮説の検定（1986）（第3版へのリンク）に帰し、その結果はHalmosとSavage自身に帰せられます（補題7を参照）。別の参考資料として、Shaoの数学統計（2003年第2版）があります。関連する結果は補題2.1と定理2.2です。

上記の補題は、$\sigma$有限メジャーによって支配されるメジャーのファミリーで開始する場合、実際には、ファミリー内からメジャーのカウント可能な凸の組み合わせで支配メジャーを置き換えることができると述べています。シェルビッシュは、定理A.78を述べる前に、

「統計アプリケーションでは、多くの場合、単一の$\sigma$有限メジャーに関して絶対に連続するメジャーのクラスがあります。単一の支配メジャーが元のクラスにあるか、またはクラス。次の定理はこの問題に対処します。」

具体例

未知のθ > 0に対して区間[ 0 、θ ]に均一に分布すると考えられる量$X$測定を行うと仮定します。この統計問題では、[ 0 、θ ]の形式のすべての区間の均一分布からなるR上のボレル確率測度の集合Pを暗黙的に考慮しています。場合、あるλ示しルベーグ測度と、のためのθ > 0、P θ表すユニフォーム（[$[0, \theta]$$\theta > 0$$\mathcal{P}$$\mathbb{R}$$[0, \theta]$$\lambda$$\theta > 0$$P_\theta$$\operatorname{Uniform}([0, \theta])$分布（すなわち、

$$P_\theta(A) = \frac{1}{\theta} \lambda(A \cap [0, \theta]) = \int_A \frac{1}{\theta} \mathbf{1}_{[0, \theta]}(x) \, dx$$ すべてのボレルのために$A \subseteq \mathbb{R}$）、我々は単に持って $$\mathcal{P} = \{P_\theta : \theta > 0\}.$$ これは、測定値$X$候補分布のセットです。

家族$\mathcal{P}$明らかルベーグ測度によって支配される$\lambda$（ある$\sigma$（と上記補題ので、-finite）$\aleph = \mathcal{P}$）配列の存在を保証$\{c_i\}_{i=1}^\infty$非負に合計数の$1$及びAをシーケンス$\{Q_i\}_{i=1}^\infty$で一様分布の$\mathcal{P}$そのようなこと

$$P_\theta \ll \sum_{i=1}^\infty c_i Q_i$$ 各$\theta > 0$。この例では、このようなシーケンスを明示的に構築できます！

まず、聞かせて$(\theta_i)_{i=1}^\infty$正の有理数の列挙（もこれは明示的に行うことができます）、および聞かせて$Q_i = P_{\theta_i}$各$i$。次は、聞かせて$c_i = 2^{-i}$、だから$\sum_{i=1}^\infty c_i = 1$。私はこの組み合わせと主張$\{c_i\}_{i=1}^\infty$および$\{Q_i\}_{i=1}^\infty$ works.

To see this, fix $\theta > 0$ and let $A$ be a Borel subset of $\mathbb{R}$ such that $\sum_{i=1}^\infty c_i Q_i(A) = 0$. We need to show that $P_\theta(A) = 0$. Since $\sum_{i=1}^\infty c_i Q_i(A) = 0$ and each summand is non-negative, it follows that $c_i Q_i(A) = 0$ for each $i$. Moreover, since each $c_i$ is positive, it follows that $Q_i(A) = 0$ for each $i$. That is, for all $i$ we have

$$Q_i(A) = P_{\theta_i}(A) = \frac{1}{\theta_i} \lambda(A \cap [0, \theta_i]) = 0.$$ Since each $\theta_i$ is positive, it follows that $\lambda(A \cap [0, \theta_i]) = 0$ for each $i$.

Now choose a subsequence $\{\theta_{i_k}\}_{k=1}^\infty$ of $\{\theta_i\}_{i=1}^\infty$ which converges to $\theta$ from above (this can be done since $\mathbb{Q}$ is dense in $\mathbb{R}$). Then $A \cap [0, \theta_{\theta_{i_k}}] \downarrow A \cap [0, \theta]$ as $k \to \infty$, so by continuity of measure we conclude that

$$\lambda(A \cap [0, \theta]) = \lim_{k \to \infty} \lambda(A \cap [0, \theta_{i_k}]) = 0,$$ and so $P_\theta(A) = 0$. This proves the claim.

Thus, in this example we were able to explicitly construct a countable convex combination of probability measures from our dominated family which still dominates the entire family. The Lemma above guarantees that this can be done for any dominated family (at least as long as the dominating measure is $\sigma$-finite).

The Halmos-Savage Theorem

So now on to the Halmos-Savage Theorem (for which I will use slightly different notation than in the question due to personal preference). Given the Halmos-Savage Theorem, the Fisher-Neyman factorization theorem is just one application of the Doob-Dynkin lemma and the chain rule for Radon-Nikodym derivatives away!

Halmos-Savage Theorem. Let $(\mathcal{X}, \mathcal{B}, \mathcal{P})$ be a dominated statistical model (meaning that $\mathcal{P}$ is a set of probability measures on $\mathcal{B}$ and there is a $\sigma$-finite measure $\mu$ on $\mathcal{B}$ such that $P \ll \mu$ for all $P \in \mathcal{P}$). Let $T : (\mathcal{X}, \mathcal{B}) \to (\mathcal{T}, \mathcal{C})$ be a measurable function, where $(T, \mathcal{C})$ is a standard Borel space. Then the following are equivalent:

1. $T$ is sufficient for $\mathcal{P}$ (meaning that there is a probability kernel $r : \mathcal{B} \times \mathcal{T} \to [0, 1]$ such that $r(B, T)$ is a version of $P(B \mid T)$ for all $B \in \mathcal{B}$ and $P \in \mathcal{P}$).
2. There exists a sequence $\{c_i\}_{i=1}^\infty$ of nonnegative numbers such that $\sum_{i=1}^\infty c_i = 1$ and a sequence $\{P_i\}_{i=1}^\infty$ of probability measures in $\mathcal{P}$ such that $P \ll P^*$ for all $P \in \mathcal{P}$, where $P^* = \sum_{i=1}^\infty c_i P_i$, and for each $P \in \mathcal{P}$ there exists a $T$-measurable version of $dP/dP^*$.

Proof. By the lemma above, we may immediately replace $\mu$ by $P^* = \sum_{i=1}^\infty c_i P_i$ for some sequence $\{c_i\}_{i=1}^\infty$ of nonnegative numbers such that $\sum_{i=1}^\infty c_i = 1$ and a sequence $\{P_i\}_{i=1}^\infty$ of probability measures in $\mathcal{P}$.

(1. implies 2.) Suppose $T$ is sufficient. Then we must show that there are $T$-measurable versions of $dP/dP^*$ for all $P \in \mathcal{P}$. Let $r$ be the probability kernel in the statement of the theorem. For each $A \in \sigma(T)$ and $B \in \mathcal{B}$ we have

$$\begin{aligned} P^*(A \cap B) &= \sum_{i=1}^\infty c_i P_i(A \cap B) \\ &= \sum_{i=1}^\infty c_i \int_A P_i(B \mid T) \, dP_i \\ &= \sum_{i=1}^\infty c_i \int_A r(B, T) \, dP_i \\ &= \int_A r(B, T) \, dP^*. \end{aligned}$$ Thus $r(B, T)$ is a version of $P^*(B \mid T)$ for all $B \in \mathcal{B}$.

For each $P \in \mathcal{P}$, let $f_P$ denote a version of the Radon-Nikodym derivative $dP/dP^*$ on the measurable space $(\mathcal{X}, \sigma(T))$ (so in particular $f_P$ is $T$-measurable). Then for all $B \in \mathcal{B}$ and $P \in \mathcal{P}$ we have

$$\begin{aligned} P(B) &= \int_{\mathcal{X}} P(B \mid T) \, dP \\ &= \int_{\mathcal{X}} r(B, T) \, dP \\ &= \int_{\mathcal{X}} r(B, T) f_P \, dP^* \\ &= \int_{\mathcal{X}} P^*(B \mid T) f_P \, dP^* \\ &= \int_{\mathcal{X}} E_{P^*}[\mathbf{1}_B f_P \mid T] \, dP^* \\ &= \int_B f_P \, dP^*. \end{aligned}$$ Thus in fact $f_P$ is a $T$-measurable version of $dP/dP^*$ on $(\mathcal{X}, \mathcal{B})$. This proves that the first condition of the theorem implies the second.

(2. implies 1.) Suppose one can choose a $T$-measurable version $f_P$ of $dP/dP^*$ for each $P \in \mathcal{P}$. For each $B \in \mathcal{B}$, let $r(B, t)$ denote a particular version of $P^*(B \mid T = t)$ (e.g., $r(B, t)$ is a function such that $r(B, T)$ is a version of $P^*(B \mid T)$). Since $(T, \mathcal{C})$ is a standard Borel space, we may choose $r$ in a way that makes it a probability kernel (see, e.g., Theorem B.32 in Schervish's Theory of Statistics (1995)). We will show that $r(B, T)$ is a version of $P(B \mid T)$ for any $P \in \mathcal{P}$ and any $B \in \mathcal{B}$. Thus, let $A \in \sigma(T)$ and $B \in \mathcal{B}$ be given. Then for all $P \in \mathcal{P}$ we have

$$\begin{aligned} P(A \cap B) &= \int_A \mathbf{1}_B f_P \, dP^* \\ &= \int_A E_{P^*}[\mathbf{1}_B f_P \mid T] \, dP^* \\ &= \int_A P^*(B \mid T) f_P \, dP^* \\ &= \int_A r(B, T) f_P \, dP^* \\ &= \int_A r(B, T) \, dP. \end{aligned}$$ This shows that $r(B, T)$ is a version of $P(B \mid T)$ for any $P \in \mathcal{P}$ and any $B \in \mathcal{B}$, and the proof is done.

Summary. The important technical result underlying the Halmos-Savage theorem as presented here is the fact that a dominated family of probability measures is actually dominated by a countable convex combination of probability measures from that family. Given that result, the rest of the Halmos-Savage theorem is mostly just manipulations with basic properties of Radon-Nikodym derivatives and conditional expectations.