Will Jeffrey's Prior Always be Improper? If our prior distribution has a support which ranges to infinity, then will the Jeffrey's prior necessarily be improper? For example, with a Gamma or Normal prior, the Jeffrey's prior is improper. But, with a Beta prior, we get an actual distribution when calculating the Jeffrey's prior. The support for the Gamma and Normal distributions range to infinity, whereas the support for the Beta distribution is $[0,1]$. Does this generalize to all priors whose support ranges to infinity?
EDIT: Also, if possible, it would be really interesting to see this claim proven (although I'm not sure something like that is even possible).
 A: The Gaussian distribution $f(x,\theta)=\frac{1}{\sqrt{2\pi}}\exp(-(x-\mu)^2)$ has Jeffrey's prior $p(\mu)\propto 1 $, that is to say the Jeffrey's prior is the uniform distribution. 
The uniform distribution on $(-\infty,\infty)$ is an improper prior.
But...
Let $f(x,\theta)=\frac{1}{\sqrt{2\pi}}\exp\left(-(x-\frac{1}{1+\theta^2})^2\right)$. 
Since Jeffrey's prior is invariant under reparamaterisation, the Jeffrey's prior for this distribution is $p(\frac{1}{1+\theta^2})\propto 1$, the uniform distribution. Because $\frac{1}{1+\theta^2}\in [0,1]$ for all $\theta\in(-\infty,\infty)$, our prior is that $\frac{1}{1+\theta^2}$ has the uniform distribution on $[0,1]$. This IS a proper prior.
A: Another counterexample:
Let $f(y|\theta) = \exp\{-(e^{-\theta} + \theta y)\} / y!$, where $\theta > 0$. The Fisher information for $\theta$ is $I(\theta)=e^{-\theta}$, so the Jeffreys prior is proportional to $e^{-\theta/2}$, i.e. an Exponential with rate 1/2.
[Yes, this is an extremely contrived example. $Y$ is just a Poisson with rate $e^{-\theta}$.]
