I read this statement on the Wikipedia article about Credible Intervals and it seems dubious to me. Although it has a reference associated with it, the book is not available to read online.

It is possible to frame the choice of a credible interval within decision theory and, in that context, an optimal interval will always be a highest probability density set.

Wikipedia: Credible Intervals

If it's correct, can someone explain how? Maybe I'm just misunderstanding one of the terms.

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    Optimal in what sense? If optimal is not defined, it is automatically true. The highest probability density set will always be of shortest length - so it is optimal in that sense. – Samuel Benidt Jan 23 '14 at 19:49
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    I think if the terms of the statement are not clear, it's not really appropriate to stand on its own in an encyclopedia. That said, my interpretation (based on "within decision theory") was that a decision function based on the interval must minimize the expected loss with respect to an arbitrary loss function... which is why I found it dubious. – Fabio Beltramini Jan 23 '14 at 19:53
  • I see what you mean. If you can find a loss function where this doesn't hold, perhaps you can bring this issue up on the talk page of the article. – Samuel Benidt Jan 23 '14 at 20:00
  • My hope was that someone more knowledgeable would feel "bold" enough to edit it directly :) Does anyone even watch the talk page of obscure wikipedia articles? – Fabio Beltramini Jan 23 '14 at 20:04
up vote 5 down vote accepted

The "will always be" part is a little confusing, because that depends on your loss function. For instance, if the action space is the set of Borel sets of $\mathbb{R}$, the posterior density is $\pi(\theta\mid x)$, and $\lambda$ denotes Lebesgue measure, then, for the loss function $$ L(\theta,B) = \lambda(B) + c\,(1 - I_B(\theta)) $$ the Bayes decision is an HPD set of the form $B=\{\theta:\pi(\theta\mid x)\geq 1/c\}$. But for different loss functions the Bayes decision may not be an HPD set.

  • That was more or less my reasoning (but being unexperienced with applying loss functions, I came up with a rather artificial loss function myself). So if the statement does not stand with any loss function, and in particular with some commonly used loss functions, would you agree the statement is not actually true? – Fabio Beltramini Jan 23 '14 at 20:02
  • The statement is not true for every loss function. – Zen Jan 23 '14 at 20:03
  • Also, ideally, your loss function is as personal and subjective as your prior is. – Zen Jan 23 '14 at 20:19

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