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Suppose $X$ given $\theta$ has pdf $f(x\mid \theta)=e^{-(x-\theta)}I(x>\theta)$ and there is a standard Cauchy prior on $\theta$. As part of an exercise, I am trying to find a Bayes estimator of $\theta$ when the loss function is $L(\theta,a)=I(|\theta -a|>\delta)$ for some specified $\delta>0$.

So posterior density of $\theta$ given the data $x$ is

$$\pi(\theta\mid x) \propto \frac{e^{\theta}}{1+\theta^2}I(\theta<x)$$

Now a Bayes estimator $d=d(x)$ is such that the following is minimized:

$$\int I(|\theta -d|>\delta)\pi(\theta\mid x)\,d\theta=1-\int_{d-\delta}^{d+\delta}\pi(\theta\mid x)\,d\theta$$

This is same as maximizing $\int_{d-\delta}^{d+\delta}\pi(\theta\mid x)\,d\theta$ with respect to $d$. I think $d$ must satisfy $\pi(d+\delta\mid x)=\pi(d - \delta \mid x)$. But is there a general solution of $d$ from here?

The loss function is increasing in $|\theta-a|$, for which I know that Bayes estimator is posterior median as long as the posterior is symmetric and unimodal. But I don't think the posterior here is symmetric. Any hint would be great.

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    $\begingroup$ Basically, the question is equivalent to finding an interval $[ \theta - d , \theta + d ]$ that maximizes the area under the curve of $ \pi ( x | \theta ) $, I doubt a nice general solution can be found for all potential density function. $\endgroup$ Sep 13, 2022 at 15:33
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    $\begingroup$ It seems to me, intuitively, that if the posterior is unimodal with a central mode (as opposed to a mode at an endpoint), even if not symmetric, there would be only one $d$ such that $\pi(d+\delta|x) = \pi(d-\delta|x)$, and if the posterior doesn't have a nice enough closed form you'd probably just have to search for it - a one-dimensional root finding routine would do it - lacking elegance, I admit. $\endgroup$
    – jbowman
    Sep 13, 2022 at 17:18

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The posterior density $$\pi(\theta|x)\propto \frac{e^\theta}{1+\theta^2}\mathbb I_{\theta<x}$$ is an increasing function (in $\theta$): $$ \frac{\text d}{\text d\theta} \{\theta-\log(1+\theta^2)\}=1-\frac{2\theta}{1+\theta^2}=\frac{(1-\theta)^2}{1+\theta^2}\ge 0$$ Therefore, maximising the surface $$\int_{d-\delta}^{\min(d+\delta,x)} \frac{e^\theta}{1+\theta^2}\,\text d\theta$$ should prove straightforward. If not, note that the surface decreases when $d+\delta$ overshoots $x$ (hence $d+\delta\le x$) and, on the opposite, that it increases with $d$ when $d+\delta\le x$. Hence $$d^\star(x) = x-\delta$$is the optimal Bayesian decision.

Note: This exercise is somewhat related to Example 4.2 in my book, which points out the surprising phenomenon that, under a double exponential prior on $\theta$ and a Cauchy observation $x$ with median $\theta$, the MAP is always zero.

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    $\begingroup$ Got it. Thanks. $\endgroup$ Sep 14, 2022 at 21:22
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I think, as per my comment, it is worth noting that a maximum interval must exist and please someone critiques my proof.

WLOG,by extreme value theorem. Since integral must be continuous for a well-defined density, there exist an interval of width $2 \delta$ for each of set of intervals $ \{[-\delta , \delta],[-2 \delta,2 \delta],...\} $ that maximize the area under the curve of each interval

Since density function converges to zero as $ \theta \rightarrow \pm \infty $ , so does $ \int_{d - \delta }^{d + \delta} \pi (\theta | x ) d\theta $ as $ d \rightarrow \pm \infty $

This means there exists a K such that for all $d \in R \backslash [-K\delta,K\delta] $ , $ \int_{d - \delta }^{d + \delta} \pi (\theta | x ) d\theta < \int_{ - \delta }^{ + \delta} \pi (\theta | x ) d\theta$

Or in other words, one of the intervals contains the maximum $2\delta$ interval that maximizes the area.

Since there are a maximum interval, the method you suggest should work, maybe need some intuitive minor tweaking to handle the cases where the density is not a continuous function

Have to say I suddenly realized the answer before sleep and so not sure if there are error in my proof

Edit: The assumption that density function must tend to zero is flawed.

However, its conclusion that integral must converge to zero is still valid using Monotone convergence theorem.

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