Can a proper prior and exponentiated likelihood lead to an improper posterior? (This question is inspired by this comment from Xi'an.)
It is well known that if the prior distribution $\pi(\theta)$ is proper and the likelihood $L(\theta | x)$ is well-defined, then the posterior distribution $\pi(\theta|x)\propto \pi(\theta) L(\theta|x)$ is proper almost surely.
In some cases, we use instead a tempered or exponentiated likelihood, leading to a pseudo-posterior
$$\tilde\pi(\theta|x)\propto \pi(\theta) L(\theta|x)^\alpha$$
for some $\alpha>0$ (for example, this can have computational advantages).
In this setting, is it possible to have a proper prior but an improper pseudo-posterior?
 A: For $\alpha \leq 1$, perhaps this is an argument to show that it is impossible to construct such a posterior?
We'd like to find out if it's possible for $\int \tilde \pi(\theta|x)d\theta = \infty$.
On the RHS:
$$ \int \pi(\theta) L^{\alpha}(\theta|x) d\theta = E_{\theta}(L^{\alpha}(\theta|x))$$
If $\alpha \leq 1$, $x^{\alpha}$ is a concave function, so by the Jensen inequality:
$$ E_{\theta}(L^{\alpha}(\theta|x)) \leq E^{\alpha}_{\theta}(L(\theta|x)) = m(x)^\alpha < \infty $$
... where $m(x)$ as Xi'an pointed out, is the normalising constant (the evidence).
A: It's possible to use the result in @InfProbSciX's answer to prove the result in general.
Rewrite $L(\theta\mid x)^\alpha\pi(\theta)$ as $$L(\theta\mid x)^{\alpha-1}L(\theta\mid x)\pi(\theta).$$
If $1 \leq \alpha \leq 2$, we have the Jensen's inequality case above, since we know that $L(x|\theta)\pi(\theta)$ is normalisable.
Similarly, if $2 \leq \alpha \leq 3$, we can write $$ L(x|\theta)^{\alpha-p} L(x|\theta)^p\pi(\theta),$$
with $1 \leq p \leq 2$, again falling into the same case, since we know that $L(x|\theta)^{p}\pi(\theta)$ is normalisable.
Now one can use (strong) induction to show the case in general.
Old comments
Not sure if this is super useful, but since I can't comment I will leave this in an answer. In addition to @InfProbSciX's excellent remark about $\alpha \leq 1$, if one makes the further assumption that $L(\theta \mid x) \in L^p$, then it is impossible to have a proper prior but an improper pseudo-posterior for $ 1 < \alpha \leq p$. For instance, if we know that the second ($p$-th) moment of $L(\theta \mid x)$ exists, we know it is in $L^2$ ($L^p$) and hence the pseudo-posterior will proper for $0 \leq \alpha \leq 2$. Section 1 in these notes goes into a bit more detail, but unfortunately it is not clear how broad the class of, say, $L^{10}$ pdfs is.
I apologise if I'm speaking out of turn here, I really wanted to leave this as a comment.
