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Let $c > 0$ and \begin{equation} L(\theta,a)=\left\{ \begin{array}{@{}ll@{}} c|\theta-a|, & \text{if}\ \theta < a \\ |\theta-a|, & \text{if}\ \theta \ge a \quad. \end{array}\right. \end{equation} We assume here that $\theta$ has a continuous distribution. How do I show that the Bayes estimator of $\theta$ is the $\frac{1}{1+c}$th quantile of the posterior distribution of $\theta$?

Can anyone please give me any hints regarding this? Thanks.

Dwaipayan Gupta
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    Please explain what $L$ and $a$ are intended to represent. Likelihood? If $L$ is loss, then your question is answered at https://stats.stackexchange.com/questions/251600 (and in several other threads). – whuber May 25 '22 at 16:17
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    The solution is found in most textbooks, incl. mine. – Xi'an May 26 '22 at 04:59
  • @Xi'an can you please provide me the name of your testbook? Thanks! – Dwaipayan Gupta May 28 '22 at 06:11

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