<p>Sure.</p>

<p>In the two type case there is an alternate solution that is frequently used. If <span class="math-container">$a \neq c$</span> and <span class="math-container">$b \neq d$</span>, that is the types are really different in both attributes, then you can define a function <span class="math-container">$f$</span> for which
<span class="math-container">$$
b = f(a), \hskip 20pt d = f(c).
$$</span>
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.</p>

<p>Example:
<br> In <a href="https://www.jstor.org/stable/1882010?seq=1#page_scan_tab_contents" rel="nofollow noreferrer">Spence's model of job market signaling</a>, 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.</p>

CC BY-SA 4.0
2019-01-13T09:10:18.770Z
2019-01-13T09:10:18.770Z
Giskard
<p>In a finite Bayesian game, in most textbooks, a type of player <span class="math-container">$i$</span> is defined as <span class="math-container">$\theta_i\in\Theta_i$</span>. Little is said about the "nature" of the set of types.</p>

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

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

<p>Is this allowed in the definition of a Bayesian game?</p>

<p>I'd appreciate any help. Thank you.</p>

<game-theory><bayesian-game>
Bayesian game and the set of types
johnny09
# Bayesian game and the set of types

(Asked by: username_0 on 2019-01-13)

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.

**Tags:** game-theory, bayesian-game


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## 1 answers

### Answer by username_1 on 2019-01-13. Score: 4 | Accepted answer

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](https://www.jstor.org/stable/1882010?seq=1#page_scan_tab_contents), 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.


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