# Finding Bayes estimator for $\theta$ of Unif$(0,\theta)$

Finding Bayes estimator for $$\theta$$ of Unif$$(0,\theta)$$

Let $$Y = \max{X_i}$$ where $$(X_1,\ldots,X_n)$$ is a random sample from Unif$$(0,\theta)$$. $$Y$$ is sufficient for $$\theta$$. Find the Bayes estimator $$w(Y)$$ for $$\theta$$ based on $$Y$$ using the loss function $$L(\theta,a) = \lvert a- \theta\rvert$$ The prior density of $$\theta$$ is $$\displaystyle \pi(\theta) = \frac{2}{\theta^3}1_{(1 < \theta < \infty)}$$

I am pretty unfamiliar with Bayesian inference.

From what I understand the posterior is given by $$\displaystyle p(\theta \mid \underline{x}) = \frac{L(\theta \mid \underline x)\pi(\theta)}{\int L(\theta \mid \underline x)\pi(\theta) \, d\theta }\, ;$$ where

$$L(\theta \mid \underline{x})\pi(\theta) = \frac{1}{\theta^n}1_{(0 \le \min(x_i))}1_{(y \le \theta)}\frac{2}{\theta^3}1_{(1<\theta<\infty)}$$

Aside from this I am not sure how I set this up to solve or where I use the loss function or how I base it off $$Y$$.

\begin{align} L(\theta) & = \begin{cases} \dfrac {\text{constant}} {\theta^n} & \text{if } \theta>y, \\[8pt] \,\,0 & \text{if } 0<\theta1, \\[8pt] \,\,0 & \text{if } \theta<1. \end{cases} \\[12pt] \text{Therefore } \pi(\theta\mid y)\, d\theta & \propto \begin{cases} \dfrac{\text{constant}\cdot d\theta}{\theta^{n+3}} & \text{if } \theta> \max\{1,y\}, \\[8pt] \,\,\,0 & \text{otherwise.} \end{cases} \end{align} (Here I have written $$\text{“}{>}\text{”}$$ and $$\text{“}{<}\text{”}$$ rather than $$\text{“}{\ge}\text{”}$$ and $$\text{“}{\le}\text{”}$$ whereas if we have been doing maximum-likelihood estimation then I would have written $$\theta\ge y.$$) $$\int_{\max\{1,y\}}^{+\infty} \frac{d\theta}{\theta^{n+3}} = \frac 1{(n+1)(\max\{ 1,y \})^{n+2}}.$$ So the posterior probability distribution is $$\frac{ (n+1)(\max\{ 1,y \})^{n+2}}{\theta^{n+3}} \, d\theta \qquad \text{ for } \theta > \max\{1,y\}.$$ Theorem: With absolute-error loss, the Bayes estimator is the posterior median.

It you know the theorem above, then what remains is to solve the equation below for $$m{:}$$ $$\int_{\max\{1,y\}}^m \frac{ (n+1)(\max\{ 1,y \})^{n+2}}{\theta^{n+3}} \, d\theta = \frac 1 2.$$

If you don't know the theorem above, then maybe that's the question you need to post.

Bayes estimator under absolute error loss is the posterior median (see here for example).

The prior you are given has a Pareto distribution, which is known to be a conjugate prior for $$\theta$$ when $$X_i\sim U(0,\theta)$$. This means the posterior distribution is also a Pareto distribution, which you can show by simply writing the posterior density as

\begin{align} \pi_{\theta\mid \boldsymbol X}(\theta\mid \boldsymbol x)&\propto f(\boldsymbol x\mid \theta)\pi(\theta) \\&=\frac1{\theta^n}\mathbf1_{01} \\&=\frac{2}{\theta^{n+3}}\mathbf1_{\theta>\max(1,x_{(n)})} \end{align}

Here $$x_{(n)}=\max\{x_1,\ldots,x_n\}$$ as usual.

The Bayes estimator of $$\theta$$ is nothing but the median of this Pareto distribution.