# 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

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.

• 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. Sep 13, 2022 at 15:33
• 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. Sep 13, 2022 at 17:18

The posterior density $$\pi(\theta|x)\propto \frac{e^\theta}{1+\theta^2}\mathbb I_{\theta is an increasing function (in $$\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.

• Got it. Thanks. Sep 14, 2022 at 21:22

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.