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Consider a misspecified model: $$P \sim \text{SegmentedUniform}(0,1) \\ Y \mid P \sim \text{Binomial}(N,P).$$ Where SegmentedUniform has uniform density on intervals $$(0.1, 0.2), (0.3, 0.4), (0.5, 0.6), (0.7, 0.8), (0.9,1.0)$$ and $0$ otherwise. Suppose I obtain posterior samples such that they concentrate around the segments $(0.3, 0.4), (0.5, 0.6)$.

Is it "valid" to derive a Bayes estimate $\hat p$ for $p$ that is not supported by the model? For example, $\hat p = 0.45$.

Here's my guess: A Bayes estimate by definition, is a minimizer of its expected posterior loss for some loss function. Although the expected loss is under the posterior, the minimizer itself is not constrained to the support, so $\hat p$ in this sense in the above example is valid. However, with this $\hat p$ being unsupported by the model, the resulting joint density would evaluate to $0$, and thus it is an "invalid" Bayes estimate.

Edit: (see second comment)

As a follow-up to this, consider the model $$P \sim \text{Uniform}(0,1) \\ Y | P \sim \text{Binomial}(N,P)$$ whose posterior is inferred with a "misspecified sampler", such that it does **not** produce samples from the true posterior but instead from the SegmentedUniform model's posterior.

In this case, the Bayes estimate would be supported, but derived from an incorrect posterior. Is there any merit to an analysis performed as such? Is there any sense in studying this incorrectly-inferred posterior?

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    $\begingroup$ It all depends on the definition of an optimal Bayesian procedure. Or of the decision space. That the posterior density at this procedure is equal to zero is not paradoxical as such. $\endgroup$
    – Xi'an
    Commented Aug 28, 2020 at 19:14
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    $\begingroup$ Furthermore, MCMC has nothing to state about misspecified models. It aims at simulating a given distribution. $\endgroup$
    – Xi'an
    Commented Aug 28, 2020 at 19:16
  • $\begingroup$ Continuing with the first example, suppose the decision space is (0,1). How does one interpret the decision $\hat p = 0.45$? $\endgroup$
    – fool
    Commented Aug 28, 2020 at 22:09
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    $\begingroup$ Once again, it depends on the definition and role of the decision procedure. If a decision outside the prior support is unacceptable, the loss should penalise it as such. $\endgroup$
    – Xi'an
    Commented Aug 29, 2020 at 15:04
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    $\begingroup$ "If a decision ... the loss should penalise it as such" this cleared things up, thank you! I also found this question and your answer there helpful: stats.stackexchange.com/questions/343171/… $\endgroup$
    – fool
    Commented Aug 29, 2020 at 18:32

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