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In one dimension it can be shown that the highest posterior density (HPD) interval is the shortest; I found a proof in Subjective and Objective Bayesian Statistics (Section 8.4) by S. James Press where a suitable Lagrangian is built and used to find the solution.

Does anyone know of a proof for the higher dimensional case?

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    $\begingroup$ The proof is the same: take a subset out of the HPD and you need more volume to achieve the same coverage. $\endgroup$
    – Xi'an
    Nov 5 '17 at 16:41
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    $\begingroup$ @Xi'an thanks! in the 1-d case, is your suggestion equivalent to the proof above? $\endgroup$
    – Valerio
    Nov 10 '17 at 10:25
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Answered in comments:

The proof is the same: take a subset out of the HPD and you need more volume to achieve the same coverage. – Xi'an

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