mjhjmtu

(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

$$tildepi(theta|x)propto pi(theta) L(theta|x)^alpha$$
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?

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 $int tilde pi(theta|x)dtheta = infty$.

On the RHS:

$$ int pi(theta) L^alpha(theta|x) dtheta = E_theta(L^alpha(theta|x))$$

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).

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.