What is Bayes estimator of $\theta$ when loss function is $L(\theta,a)=I(|\theta -a|>\delta)$? 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.

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.
