Bayesian game and the set of types

Question

In a finite Bayesian game, in most textbooks, a type of player $i$ is defined as $\theta_i\in\Theta_i$. Little is said about the "nature" of the set of types.

For example, could we have a two-element set $\Theta_i=\{(a,b),(c,d)\}$? If I am not mistaken this means that $\Theta_i$ is a subset of $\mathbb{R^2}$, right?

For the common prior then, we could have, for example, $\mathbb{P}((a,b))=p$ and $\mathbb{P}((c,d))=1-p$. Both probabilities describe the joint probability distribution that Nature assigns the types to the players.

Is this allowed in the definition of a Bayesian game?

I'd appreciate any help. Thank you.

Answer 1

Sure.

In the two type case there is an alternate solution that is frequently used. If $a \neq c$ and $b \neq d$, that is the types are really different in both attributes, then you can define a function $f$ for which $$ b = f(a), \hskip 20pt d = f(c). $$ Whatever you need the second attribute for, this way it becomes a function of the first one, and the values of the first attribute completely define the type space.

Example:
In Spence's model of job market signaling, people have a type that shows how much they are worth to a company, how hard working they are. It is assumed that the cost associated with getting a degree is a function of this type as well, and harder workers get a degree more easily. One way to show this would be to include both worth to the employer and cost of getting a degree in the type attributes, but Spence only defined the types and gave some functions that mapped to these values.