Find the posterior density of $\log \theta$ and Bayes estimator under 0-1 loss

Question

Let $X_1,...,X_n$ be iid from the uniform distribution on $(0,\theta)$ where $\theta>0$ is unknown. Suppose the prior distribution of $\theta$ is log-normal with parameters $\mu, \sigma^2$, which are known constants.

1. Find the posterior density of $\log \theta$.

2. Suppose the loss function is $L(\delta, \theta)=\begin{cases}0 & if \delta=0\\1&otherwise\end{cases}$. Find the Bayes estimator of $\theta$ under this loss function. Hint: Part (1) is related. With 0-1 loss the Bayes estimator would be the Maximum A Posteriori (MAP) estimator...in this case is the mode of the posterior.

We have $$\begin{split}\xi(\log\theta|x)&\propto f(x|\log\theta)\xi(\log\theta)\\ &=\frac 1 {\theta^n}\frac 1 {\sigma\sqrt{2\pi}}e^{-\frac 1 2\left(\frac{\theta-\mu}{\sigma}\right)^2}\end{split}$$

Is this correct; what distribution is this?

Answer 1

reading you posts I think you are a clever guy thus with a sketch I think you can do the rest by yourself.... I am just an amateur, but this is what I would do in this case...

Find the posterior density of $log \theta$.

1. The model is uniform thus

$$p(\mathbf{x}|\theta)=\frac{1}{\theta^n}\cdot\mathbb{1}_{(0;\theta)}(x_{(n)})=\frac{1}{\theta^n}\cdot\mathbb{1}_{(x_{(n)};+\infty)}(\theta)$$

2. The Prior is lognormal

$$\pi(\theta)\propto \frac{1}{\theta}e^{-(\log\theta-\mu)^2/(2\sigma^2)}$$

First I find the posterior of $\theta$

3. $$\pi(\theta|\mathbf{x})\propto \frac{1}{\theta^{n+1}}e^{-(\log\theta-\mu)^2/(2\sigma^2)}\cdot\mathbb{1}_{(x_{(n)};+\infty)}(\theta)$$

4. Now I transform the posterior getting the density of its log with the standard transformation theorem

$$\lambda=\log\theta$$

$$\theta=e^{\lambda}$$

$$\theta'=e^{\lambda}$$

Thus

$$\pi(\lambda|\mathbf{x})\propto e^{-\lambda n}e^{-(\lambda-\mu)^2/(2\sigma^2)}$$

5. Let's focus on the exponent

$$-\frac{1}{2\sigma^2}[2\sigma^2 n \lambda+\lambda^2+\mu^2-2\lambda\mu]$$

6. Waste the element $\mu^2$ because not depending on $\lambda$ (it will be part of the normalizing constant) do some algebraic manipulation completing the square inside the bracket and you will find the kernel of a certain Gaussian density... truncated in the support

$$(\lambda|\mathbf{x}) >\log x_{(n)}$$