5
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ Please define $\mu_t$ and $\mu_{t+1}$ $\endgroup$
    – Juho Kokkala
    Nov 30 '16 at 19:02
  • $\begingroup$ It is the definition of a martingale, but the why here is referring to the property that Bayesian beliefs (ie beliefs formed by updating using Bayes Theorem) form a martingale. That is not true by definition. $\endgroup$
    – Fatsho
    Nov 30 '16 at 19:48
  • $\begingroup$ What is the expectation of a distribution?! I never heard of this before. $\endgroup$
    – Xi'an
    Nov 30 '16 at 21:47
2
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ The notion is rather weak as it does not extend to continuous probability distributions or to quantities related with this distribution. For instance, the variance of the expectation differs from the expectation of the variance. $\endgroup$
    – Xi'an
    Dec 1 '16 at 7:23
  • 1
    $\begingroup$ @Xi'an the papers are about "Bayesian persuasion" and updating "beliefs" rather then dealing with data and they have pretty informal approach, so this is pretty unconventional understanding of the concepts in general. $\endgroup$
    – Tim ♦
    Dec 1 '16 at 10:24
  • 1
    $\begingroup$ You might be reading the Kamenica and Gentzcow from the perspective of a statistician choosing a prior to deal with data. In game theory, we don't choose priors. We think of priors as parameters that represent the subjective attitudes of the agents in our models. Kamenica and Gentzcow prove in the paper that the equation you cited holds, no matter what your prior is, as long as you are able to derive your posterior from Bayes rule (you don't observe events that were null with respect to your prior). $\endgroup$
    – brunosalcedo
    Feb 24 '19 at 13:50
1
$\begingroup$

I believe the answer to what you are looking is:

$$E \{ \pi(\omega|S) \}= \sum_s \pi(\omega|s) \pi(s) = \sum_s \frac{\pi(s|\omega) \pi(\omega)}{\pi(s)} \pi(s) = \pi(\omega)\underbrace{\sum_s \pi(s|\omega)}_{=1} $$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.