# Agreeing to disagree “simple” example

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?

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