Let $T_{\theta}(\mathbf{x})$ be a complete, sufficient statistic $T_{\theta}: \Omega \mapsto \mathbb{R}$, where $T_{\theta}$ is indexed by the parameter $\theta \in \mathbb{R}^n$.

Is it true that the sampling distribution of $T_{\theta}$ is free of relevant subsets. I.e., $P(T_{\theta}(\mathbf{x}) \in S \subset \mathbb{R}|\mathbf{x} \in Q \subset \Omega) = P(T_{\theta}(\mathbf{x}) \in S \subset \mathbb{R})$?

Example: We use a pivotal statistic to define a $(1-\alpha)\%$ confidence interval. If the distribution of this statistic has relevant subsets, then after we see the data $\mathbf{x}$ we realize that $\mathbf{x}$ falls into a subset such that confidence intervals formed using data sampled from this subset have a different confidence coverage than the unconditional confidence $(1-\alpha)\%$.

If this were true, that means that frequentist inferences (especially confidence intervals) will have a conditional confidence level that is the same as the unconditional confidence level, among other benefits.

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    $\begingroup$ What is $\Omega$ here? Is it just an arbitrary subset of $S$? If so, then taking $\Omega = \{ x_0 \}$ as a singleton set would reduce the left-hand-side to $\mathbb{I}(x_0 \in S)$, which would then require the statistic to have a point-mass distribution at $x_0$. Since this could be done for any $x_0 \in S$ that would mean that the requirement could never be satisfied for $|S| > 1$. Can you clarify why this doesn't happen? $\endgroup$ – Ben Jan 11 '19 at 1:40
  • $\begingroup$ @Ben good catch -- I meant $\Omega$ to be the sample space and that $\mathbf{x} \in Q \subset \Omega$ $\endgroup$ – Bey Jan 11 '19 at 3:16
  • $\begingroup$ Okay, so then my same question now applies to $Q$. Is there any restriction on this set, or can it just be a singleton? $\endgroup$ – Ben Jan 11 '19 at 3:28
  • $\begingroup$ @Ben: Usually it is via an ancillary statistic (subset defined by the preimage of the statistsic at its observed value). However, it can be a singleton, but in that case you assign confidence using a confidence function: web.uvic.ca/~dgiles/blog/Berger_and_Wolpert.pdf $\endgroup$ – Bey Jan 11 '19 at 15:37

Can you clarify the question? Even if your claim were true, I don't think it will follow that frequentist inferences (especially confidence intervals) will have a conditional confidence level that is the same as the unconditional confidence level. That because there do exist counterexamples, also for confidence interval procedures with optimal properties (and based on sufficient statistics), like $t$-based confidence intervals in the usual iid normal model.

That interval is based on a pivot, that is, $$ T=\frac{\bar{x}-\mu}{\sqrt{S^2/n}} $$ which, under the usual assumptions, has a known distribution not depending on the unknown parameters. $T$ is a function of the sufficient statistics $(\bar{X},S^2)$, but is not itself sufficient.

References for this result and discussion of relevant subsets can be found in the Likelihood Principle, second edition. Some relevant papers is here and here

  • $\begingroup$ How should I clarify? What was confusing? Also, I am familiar with that example, and so I was trying to formalize the notion of "relevant subset" -- in the case you cite, you can define subsets of the sample space where we know the sampling distribution of $T$ conditional on that subspace, has a different coverage. However, as you state, $T$ is not sufficient, so it doesn't apply here. $\endgroup$ – Bey Jan 10 '19 at 23:52
  • $\begingroup$ You could. At least add some relevant definitions and examples! Relevant subsets is not a well-known concept, with some more motivation more people could get interested (I will look at this again tomorrow, in bed now) $\endgroup$ – kjetil b halvorsen Jan 11 '19 at 0:03
  • $\begingroup$ I added an example and I link to a paper that goes into depth on what these are. There's not a simple definition. $\endgroup$ – Bey Jan 11 '19 at 3:23
  • $\begingroup$ I will loook at it again, but without clear definitions you cannot expect theorems ... $\endgroup$ – kjetil b halvorsen Jan 11 '19 at 13:23
  • $\begingroup$ I understand -- honestly, after reading this nice discussion (and references) some time back (web.uvic.ca/~dgiles/blog/Berger_and_Wolpert.pdf -- section 2.2) one is left with the feeling that the idea, while initially sensible, has serious conceptual/philosophical issues, so your confusion is not unwarranted. Fisher was a great thinker, but some of his ideas have generated more confusion than insight (this and Fiducial inference being two of the latter). Nancy Reid is currently working these issues (conditional frequentist inference). $\endgroup$ – Bey Jan 11 '19 at 15:39

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