Probability distributions in Bayesian games

Question

In a Bayesian game, each player $i$ learns his own type, $\theta_i$, which is his private information, and then uses his prior $\phi_i$ to form posterior beliefs over the other types of players, using Bayes' rule:

$$\phi_i(\theta_{-i}|\theta_i)=\frac{\mathbb{P}(\theta_i\cap\theta_{-i})}{\sum_{t_{-i}\in\Theta_{-i}}\mathbb{P}(\theta_i\cap t_{i})}.\quad (1)$$

The belief function $\phi_i$ is a mapping from $\Theta_i$ into $\Delta(\Theta_{-i})$, the set of probability distributions over $\Theta_{-i}$. That is, for any possible type $\theta_i\in\Theta_i$, $\phi_i$ specifies a probability distribution $\phi_i(\cdot|\theta_i)$ over the set $\Theta_{-i}$ representing player $i$'s beliefs about the types of the other players if his own type were $\theta_i$.

First of all, can we say that $\Delta(\Theta_{-i})=[0,1]$ since $\Delta(\cdot)$ represents a set of probability distributions?

Also, I am not quite sure what we mean by "probability distribution" in this case. Most authors in game theory do not specify the term "probability distribution" (or at least provide more details), and it is unclear to me how it actually works.

For example, if we have $\theta_i\in\Theta_i=\{a,b\}$, then a common prior probability distribution assigns probabilities to each element of $\Theta_i$? That is, $\mathbb{P}(a)=p$ and $\mathbb{P}(b)=1-p$? Do we also have to assign the probability intersections, $\mathbb{P}(a\cap b)$? If not, then how can we use Bayes' rule (1) to compute the conditional probabilities?

Answer 1

Think abou the following. Whether you prefer apple to orange is your private information. So you have two types, $a$ and $o$. Type $a$ means you prefer apple to orange. Type $o$ means the converse. Your type space is $\Theta_1 = \{a, o\}$. Your girlfriend's preference is her private information. She also has two types $a$ and $o$ with the same interpretation. Her type space is $\Theta_2$.

The type space of the game is then $\Theta = \Theta_1 \times \Theta_2$. The common prior about types is $\mathbb{P}$, which is a probability distribution over $\Theta$. Then when you know your preference, what is your poterior belief about your girlfirend's preference? It is $\mathbb{P}(p_2 | p_1)$ where $p_1 \in \Theta_1$ and $p_2 \in \Theta_2$. So, given your own type, $\mathbb{P}( \cdot |p_1)$ is a probability distribution over $\Theta_2$. It is just Bayes' rule.

You used the notation $p_1 \cap p_2$. It should be understood as follows. Here $p_1$ actual means $\{p_1\} \times \Theta_2$ and $p_2$ means $\Theta_1 \times \{p_2\}$. Therefore $p_1 \cap p_2 = \{p_1\} \times \{p_2\}$, the event that your prerference is $p_1$ and your girlfriend's preference is $p_2$.