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Let a prior $\pi(\theta)=\frac{1}{3}(\mathbb{I}_{[0,1]}(\theta)+\mathbb{I}_{[2,3]}(\theta)+\mathbb{I}_{[4,5]}(\theta))$ and $f(x\mid\theta)=\theta e^{-\theta x}$. Taking the multilinear loss $$L_{k_1,k_2}(\theta,\delta) = \begin{cases} k_2(\theta-\delta), &\theta>\delta, \\ k_1(\theta-\delta), & \theta\leq\delta \end{cases}$$ show that the Bayes estimator is not unique.

I know that $f(x\mid\theta)\sim \exp(\theta)$ and $\pi(\theta)$ is a sum of Uniform distributions such that $$\pi(\theta)\in (0,1) $$ and

Proposition: A Bayes estimator associated with prior $\pi$ and multilinear loss is the $\frac{k_1}{k_1+k_2}$ fractile of $\pi(\theta\mid x)$

$$\pi(\theta\mid x)\propto\theta e^{-\theta x}\mathbb{I}_{[0,1]}(\theta)+\theta e^{-\theta x}\mathbb{I}_{[2,3]}(\theta)+\theta e^{-\theta x}\mathbb{I}_{[4,5]}(\theta)$$

I don't understand right this proposition, I don't need to find $E[\pi(\theta\mid x)$? The estimator would be $$\frac{k_1}{k_1+k_2}E[\pi(\theta\mid x)]?$$

To show that is not unique I need to construct two estimators, or there is another way?

I'm trying to use this theorem

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but I don't understand this well , is just the fractile of posterior?

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  • $\begingroup$ If by "$E[\pi(\theta\mid x)]$" you mean $E[\theta\mid x]$, that would be the Bayes estimator under squared-error loss. You need to find the posterior expected loss, using the loss function that is given. The value of $\theta$ that minimizes that would be the Bayes estimator. $\qquad$ $\endgroup$ – Michael Hardy Mar 31 '16 at 23:11
  • $\begingroup$ BTW, I would say "piecewise linear" rather than "multilinear". $\qquad$ $\endgroup$ – Michael Hardy Mar 31 '16 at 23:12
  • $\begingroup$ Did you mean $\theta e^{-\theta x} \mathbb I_{2,3}(\theta)$ where you have $e^{-\theta x} \mathbb I_{2,3}(\theta)$? $\qquad$ $\endgroup$ – Michael Hardy Mar 31 '16 at 23:14
  • $\begingroup$ @MichaelHardy Yes, is a typo sorry. If possible take a look again $\endgroup$ – Roland Mar 31 '16 at 23:15

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