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Suppose I've found that the Bayes risk is of the form $$r(\theta) = \int_{-a}^a \theta^2 \pi(\theta)d\theta $$

I want to show that the following distribution, $\pi(a)=\pi(-a)=0.5$, maximizes this quantity, i.e. that it's a least favorable prior distribution.

This seems intuitively right to me, as any distribution over a range of values, will surely reduce the variance, compared to a distribution that allocates only the extreme values - but I'm not sure how to prove this rigorously.

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  • $\begingroup$ Notation correction: $r(\theta)$ should be $r(\pi)$ since $\theta$ is integrated out. Any prior fully supported by $\{-a,a\}$ is an acceptable solution. Any prior that gives positive weight to $\{-a,a\}^c$ has a lower risk. $\endgroup$
    – Xi'an
    Commented Jan 16, 2021 at 18:07
  • $\begingroup$ @Xi'an what is $\{-a,a\}^c$? $\endgroup$
    – Maverick Meerkat
    Commented Jan 29, 2021 at 8:47
  • $\begingroup$ The complementary set of $\{-a,a\}^c$, i.e., $(-a,a)$. $\endgroup$
    – Xi'an
    Commented Jan 29, 2021 at 9:20
  • $\begingroup$ Thanks. Any formal proof of what you wrote? i.e. how to answer the original question $\endgroup$
    – Maverick Meerkat
    Commented Jan 29, 2021 at 9:41

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