Martingales: Why must expected posterior equal prior? For a posterior distribution to be plausible in the Bayesian sense (Bayes' Plausible), it is said that: 
$\mathbb{E}(\mu_{t+1} | \mu_t) = \mu_t$
where $\mu_t$ is the posterior distribution at time $t$, and consequently the prior at $t+1$ and $\mu_{t+1}$ is the posterior distribution at time $t+1$. 
So, for example $\mu_t(\omega)$ is the probability associated with the state being $\omega$ at time $t$. ( In other words, the belief).
Could someone explain why this should be the case? 
I think I understand why intuitively: on receiving no new informations, other than $\mu_t$, you can anticipate the new distribution to average to the current information you have. 
But if this intuition is incorrect or incomplete, I'd really appreciate it if someone could point that out. 
 A: I never heard the term before, but quick googling leads to papers by Kamenica and Gentzkow (2011a, 2011b) on "Bayesian persuasion", who define it as follows (p. 10 in the linked pdf to the 2011a paper):

A distribution of posteriors is Bayes-plausible if the expected
  posterior probability of each state equals its prior probability:
$$ \int \mu d\tau(\mu) = \mu_0. $$

So the name seems to be chosen to be self-explanatory: a Bayesian choosing informative prior would expect the posterior to be close to the prior (otherwise, why would he choose the obviously wrong prior?). Such outcome would be the most plausible for the Bayesian before seeing the data.
It does not imply that "on receiving no new informations, other than $\mu_t$, you can anticipate the new distribution to average to the current information you have", it simply defines the kind of distributions that follow this property as "Bayes-plausible". So if distribution is Bayes-plausible, then it follows this property by definition.
More precisely: it seems that it's not a property of any particular disruption, but rather of a prior-posterior pair.
It is not my area of expertise, but you can see the original papers for more details on game-theoretic implications and further discussion.

Kamenica, E., & Gentzkow, M. (2011a). Bayesian persuasion. The American Economic Review, 101(6), 2590-2615.
Gentzkow, M., & Kamenica, E. (2011b). Competition in persuasion (No. w17436). National Bureau of Economic Research.
A: I believe the answer to what you are looking is:
$$E \{ \pi(\omega|S) \}= \sum_s \pi(\omega|s) \pi(s) = \sum_s \frac{\pi(s|\omega) \pi(\omega)}{\pi(s)} \pi(s) = \pi(\omega)\underbrace{\sum_s 
 \pi(s|\omega)}_{=1} $$
