Many thanks. I wonder what kind of properties one could invoke here. Some form of distance-regularity?

@Ashok: in your example, write en as $$\pi_2e^{−bn}(\frac{\pi_1}{\pi_2}e^{−(a−b)n}+1)$$ then take the $-\frac 1 n \log$: the limit is $b$, not $a+b$. This is not the bug in the reasoning: see my post and the discussion at stats.stackexchange link.

As I said, this is Bayesian h.t. and what you look for is the error probability averaged over all $x$'s and $H$'s. So there is no such thing as the "true distribution" of $x$ (like, say, in the non-Bayesian case). So the expectation $E[\cdot]$ in the paper is just $\sum_x p(x)...$, with $p(x)=p(x|A)p(A)+p(x|B)p(B)$.

There is no difference. The formula in the cited paper (instantiated to the case $M=2$, binary h.t.) is $$ P_e = 1-E_x[\max_{i=1,2} p(\theta_i|x)] $$ where, in our case, $\theta_1=A$ and $\theta_2=B$. If you use Bayes, the probabilities $p(x)$ of the expectation and in $p(\theta_i|x)= p(x|\theta_i)p(\theta_i)/p(x)$ cancel out, and you get the same as my (1). M.

Many thanks. NB: edited my post to rectify formula (3): the previous formulation was probability of success, not of error. M.

Ashok, I found the paper, man thanks. Michele

Surely it is not always the case that if $(P_n,Q_n)\rightarrow (P,Q)$ then $\lim_{n\rightarrow \infty} D(P_n||Q_n)= D(P||Q)$, as you say. If $Q$ is the frontier of the PMF simplex (that is, $Q(x)=0$ for some $x$), consider a sequence $(P_n,Q_n)$ with $Q_n=Q$ and $P_n\rightarrow P$, where $P(x)=0$ whenever $Q(x)=0$. Moreover, assume all the $P_n$'s are in the interior of the simplex, that is they are all strictly positive on every $x$. Then, for each $n$, $D(P_n||Q_n)=+\infty$, hence $\lim_{n\rightarrow \infty} D(P_n||Q_n)=+\infty$, whereas $D(P||Q)<+\infty$.