# Necessary condition for pairwise sufficient statistic [duplicate]

This question already has an answer here:

I'm struggling to prove the following.

If $T:\left(X,\mathbf{A}\right)\rightarrow\left(Y,\mathbf{B}\right)$ is a pairwise sufficient statistic for a set $\left\{\mu_0,\mu_1,\mu_2\right\}$ of three measures on $\left(X,\mathbf{A}\right)$, then $\frac{d\mu_0}{d\left(\mu_0+\mu_1+\mu_2\right)}$ (the Radon-Nikodym derivative) is $T^{-1}\left(\mathbf{B}\right)$-measurable modulo $\mu_0+\mu_1+\mu_2$.

It is supposedly proved in the otherwise accessible and irreproachable article "Application of the Radon-Nikodym Theorem to the Theory of Sufficient Statistics" by Halmos and Savage (Lemma 9, page 238), but i'm dissatisfied with the proof, since in my opinion it justifies the claim modulo $\mu_0$ only.

I'd appreciate help in either understanding Halmos & Savage's proof or proving it from scratch.

-

## marked as duplicate by Evan Aad, Davide Giraudo, Martin, Jim, Henry T. HortonApr 22 '13 at 22:16

Some years ago I took a course taught by Morris L. Eaton in which he said there is a gap in the argument in this paper. He said they had failed to prove that certain sets have measure $0$. I haven't looked at this enough to know whether you might be talking about the same thing. He also said someone else published a good proof later. –  Michael Hardy Mar 3 '13 at 0:21
@MichaelHardy: Thanks, Michael. I believe you are talking about the same thing, as the problem indeed boils down to proving that certain sets have measure $0$. I can't tell you what a relief it is to know that there's a known fault in this argument. I would be very obliged to you if you could let me know the name of the article where this "gap" is filled. –  Evan Aad Mar 3 '13 at 0:23
This notation is different from what I am accustomed to in thinking about sufficiency, and that's probably as it should be considering what they're trying to do. Just to be clear: when you say $T$ is sufficient, am I right in thinking that you mean that the conditional distribution of $x$ given $T(x)$ is the same regardless of which of the three measures is used? –  Michael Hardy Mar 3 '13 at 0:24
@MichaelHardy: Not exactly. What you describe is sufficiency, whereas the notion they refer to in this lemma is pairwise sufficiency, which means that $T$ is sufficient for every pair of measures, but not necessarily for all three of them. –  Evan Aad Mar 3 '13 at 0:28
I'll have to look at the paper . . . . –  Michael Hardy Mar 3 '13 at 0:31

Also, Bahadur's version of the factorization theorem supposes $\sigma$-finiteness of the dominating measure, while Halmos & Savage's article deals strictly with finite measures (though i believe all their arguments can be readily extended to the $\sigma$-finite case, but i have not checked that it is so). –  Evan Aad Apr 24 '13 at 20:20