Bayesian improper prior for discrete data There is one remark in my lecture notes:
if $X$ (data) is a discrete distribution then one cannot use an improper prior. 
Can anyone prove it?
 A: The reason for not being able to use an improper prior in the case of a finite distribution is that, if the distribution $p_\theta$ of $X$ has support in $\{a_1,\ldots,a_N\}$, then the posterior$$\pi(\theta|x)=\pi(\theta)p_\theta(x)\big/ m(x)$$is not defined as
$$m(x)=\int p_\theta(x)\pi(\theta)\text{d}\theta$$cannot be finite for all $x$'s in $\{a_1,\ldots,a_N\}$:
$$\sum_{i=1}^N m(a_i) = \sum_{i=1}^N \int p_\theta(a_i)\pi(\theta)\text{d}\theta=\int\sum_{i=1}^Np_\theta(a_i)\pi(\theta)\text{d}\theta=\infty$$implies that at least one of the $m(a_i)$'s is infinite, hence that the posterior(s) do(es) not exist for the corresponding $i$'s.
The argument does not extend to the infinitely countable case though: take for instance a Poisson $\mathcal{P}(\lambda)$ distribution and an improper prior $\pi(\lambda)=1/\sqrt\lambda$. Then$$\pi(\lambda|x)\propto \lambda^{x-\frac{1}{2}}\exp\{-\lambda\}$$is well-defined as a Gamma$(x+\frac{1}{2},1)$ for all $x\in\mathbb{N}$.
A: Just a pointer about the $Beta(0,0)$ prior. If $X\sim Binomial(n,p)$, with $n\geq 2$ fixed, then the $Beta(0,0)$ prior for $p$, pointed out by @marmle and which is independent of the sample (opposite to what @Xi'an says), leads to a proper posterior distribution only if $y\neq 0, n$. That is, this is an improper prior that leads to a proper posterior distribution under some conditions. 
This prior is often used in practice since the posterior mean coincides with the MLE. Also, in practice, people tend to prefer a $Beta(\epsilon,\epsilon)$ prior, with $\epsilon\approx 0$, in order to obtain a posterior where the posterior mean (which is most often reported) is close to the MLE, while avoiding problems of impropriety of the posterior when $y=0,n$.
