I'm looking at Aumann's work "Agreeing to Disagree", and trying to understand the very first numerical example.

So, the paper starts with definitions

[L]et $(\Omega, \mathcal{B}, p)$ be a probability space, $\mathscr{P}_1$ and $\mathscr{P}_2$ [information] partitions of $\Omega$. [I]f the true state of the world is $\omega$, then $i$ is informed of that element $P_i(\omega)$ of $\mathscr{P}_i$ that contains $\omega$.

Let $A$ be an event, and let $q_i$ denote the posterior probability $p(A | \mathscr{P}_i)$ of $A$ given $i$'s information; i.e. if $\omega \in \Omega$, then $q_i(\omega) = p(A \cap P_i(\omega)) / p(P_i(\omega))$.

Then comes a simple example, that I fails to understand.

Suppose $\Omega$ has $4$ elements $\alpha, \beta, \gamma, \delta$ of equal (prior) probability, $\mathscr{P}_1 = \{\alpha\beta, \gamma\delta\}$, $\mathscr{P}_2 = \{\alpha\beta\gamma, \delta\}$, $A = \alpha\delta$, and $\omega = \alpha$. Then $1$ knows that $q_2$ is $\frac{1}{3}$, and $2$ knows that $q_1$ is $\frac{1}{2}$; but $2$ thinks that $1$ may not know what $q_2$ is ($\frac{1}{3}$ or $1$).

My problem starts with notation, what does $\alpha\beta\gamma$ mean? Is it $\{\alpha, \beta, \gamma\}$?

After I suppose this, I'm able to understand $\frac{1}{3}$ and $\frac{1}{2}$ part, then "but ..." sentence is totally cryptic. Why $2$ thinks this?

  • $\begingroup$ I agree to disagree this way to introduce bayes' theorem :) I share your difficulties to grasp the meaning of the example. $\endgroup$ – Peter Mar 7 at 10:13

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