# Multilinear loss in Exponential-Uniform model

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

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

• What does notaation $\pi(\theta) \in (0,1)$ mean? And just to be clear, you are looking for a Bayes estimator for $\theta$? Apr 2, 2016 at 0:49
• @Greenparker Since we have a sum of three uniform random variables it just mean the range of possible values for $\pi(\theta)$.
– user72621
Apr 2, 2016 at 1:22
• $\pi(\theta) = 0$ or $\pi(\theta) = 1/3$. Using notation $\pi(\theta) \in (0,1)$ is very awkward. Apr 2, 2016 at 1:53

The loss in this problem is a piecewise linear function (not squared error loss), and thus the Bayes estimator is not the posterior expectation. (You use notation $E[\pi(\theta|x)]$ by which I assume you mean $E(\theta|x)$).

The proposition tells you exactly what the Bayes estimator is.

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

Thus the Bayes estimator is $\delta^*$ such that $$P(\theta \leq \delta^*|x) = \frac{k_1}{k_1 + k_2}.$$

You need to show that such a $\delta^*$ is not necessarily unique, or in other words, you need to find two values of $\delta^*$ such that the evaluation of the CDF at those points is the same.

First notice that the support of $\theta$ is $[0,1] \cup [2,3] \cup[4,5]$, thus the points of discontinuity in the CDF are $\{1, 2, 3, 4\}$. For 1 and 2, the CDF will take the same value and for 3 and 4, the CDF will take the same value. I will show that the Bayes estimator is not unique when $k_1/(k_1 + k_2)$ is equal to the CDF value at 1 (you can also do this for CDF valu at 3). Let

$$c = \int_{0}^{1} \pi(\theta|x)d\theta = \int_{0}^{1} \theta e^{-\theta x}d\theta.$$

You can find the exact form of $c$ in terms of $x$ by solving that integral. I have left that part as exercise. However, you can find values of $k_1$ and $k_2$ such that, $c = k_1/(k_1 + k_2)$. But the $c$th fractile is not unique since $$P(\theta \leq 2|x) = \int_0^2 \theta e^{-\theta x}d \theta = \int_0^1 \theta e^{-\theta x} d\theta + P(\theta = 2|x) = P(\theta \leq 1|x).$$

(I have left out some of the details, since this is tagged as self-study).

• I don't understand well, why you considered just the support $[0,1]$?
– user72621
Apr 2, 2016 at 12:20
• I added some more explanation, see if it makes sense now. Apr 2, 2016 at 13:02
• Don't is necessary find the posterior to show that?
– user72621
Apr 2, 2016 at 14:06
• I don't know what the previous comment means. Please edit the comment for more clarity. Apr 2, 2016 at 14:09
• You can just ignore $\pi(\theta)\in (0,1)$. To show that bayes estimator with respect to multlinear loss is not unique, I don't need to find the estimator explicity?
– user72621
Apr 2, 2016 at 19:14