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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.

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

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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
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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
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I'll have to look at the paper . . . . –  Michael Hardy Mar 3 '13 at 0:31

1 Answer 1

I'm pretty sure the author of the paper I mentioned in the comments was R. R. Bahadur. Here's his CV: http://www.stat.uchicago.edu/faculty/InMemoriam/bahadur/BahadurCV.pdf

Possibly this (?): http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.aoms/1177728715

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Thanks, Michael. I checked out the Bahadur article as well as a few others from the list you linked to. I doesn't look like the issue i mentioned is being addressed there, but i've only scanned these papers and didn't read them in depth. I'm therefore leaving this thread open. –  Evan Aad Mar 3 '13 at 6:21
    
I'd like to add, for future reference, that the gap that Bahadur corrected in said article was not a gap in argumentation, but a gap in the generality of the Fisher-Neyman factorization theorem. Where Halmos and Savage require integrability, Bahadur requires mere measurability. Bahadur proves the general, now standard version of this theorem in Corollary 6.1. He follows it with an example in which Halmos & Savage's version falls short, but his version applies. –  Evan Aad Apr 21 '13 at 18:18
    
In the end of the proof of Corollary 6.1, Bahadur indicates that sufficiency can alternatively be proved using Corollary 4 of Halmos and Savage's article. Corollary 4 is a consequence of Lemma 9, whose proof i questioned in my original post. Therefore, unfortunately, Bahadur's article does not solve my original question. However, Prof. Richard Dudley has proved the contented result in his MIT Open course notes on mathematical statistics (Theorem 2.1.4). I haven't read it yet, though. –  Evan Aad Apr 21 '13 at 18:26
    
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

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