Let the observation $x$ has the following density: \begin{equation} p(x|\theta)= \begin{cases} \frac{1}{\theta},& \text{if } 0 < x \leq \theta\\ 0, & \text{otherwise} \end{cases} \end{equation} and the r.v $\theta$ has: \begin{equation} p(\theta)= \begin{cases} \theta \exp(-\theta),& \text{if } 0 \leq \theta\\ 0, & \text{otherwise} \end{cases} \end{equation} I try to find MAP and MMSE estimation of the parameter $\theta$.
MAP:
$$ \hat{\theta} = \operatorname{argmax}_{\theta}$ $p(\theta|x) = \operatorname{argmax}_{\theta}, $$ $$ p(x|\theta)p(\theta) = \operatorname{argmax}_{\theta} \exp(-\theta), $$ which gives $\hat{\theta} = 0$. But it seems $p(x|\theta)$ is not well defined at $\theta = 0$. Am I missing something here?
MMSE:
$$\hat{\theta} = \mathbb{E}[\theta | x] = \int_{0}^{\infty} \theta \exp(-\theta) \,\mathrm d\theta = 1.$$
I want to verify if it's correct.
The posterior density is given by $$\begin{split}\xi(\theta|x)&\propto f(x|\theta)\xi(\theta)\\ &=\frac 1 \theta\textbf 1\{\theta\ge x\}\theta e^{-\theta}=e^{-\theta}\textbf 1\{\theta \ge x\}\end{split}$$
The normalizing constant is given by $$\int_x^\infty e^{-\theta}d\theta=e^{-x}$$
Thus the posterior pdf is $e^{-(\theta-x)}$ for $\theta\ge x$. You could have also gone to this directly after noting that the posterior is $\text{Exponential}(1)$ with a support starting from $x$, along with the fact that the exponential distribution is "memoryless."
This gives the mode or MAP estimate to be $x$ because $e^{-(\theta-x)}$ is decreasing in $\theta$.
The Bayes estimate (MMSE) will then be the mean of this distribution, which is $x+1$ either by direct calculation or noting that it is the mean of $\text{Exponential}(1)$ added to $x$, by similar memoryless shenanigans.