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.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

2 Answers 2


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.

  • 1
    $\begingroup$ Got it. Thanks. $\endgroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.