Tadelis ‘trading places’ Bayesian game

Question

My question concerns the following problem: two players, $1$ and $2$, each owns a house. Each player $i$ values his own house at $v_{i}$. The value of player $i$'s house to the other player, i.e. to player $j\neq i$, is $\frac{3}{2}v_{i}$. Each player $i$ knows the value $v_{i}$ of his own house to himself, but not the value of the other player's house. The values $v_{i}$ are drawn independently from the interval $[0,1]$ with uniform distribution.

Payoffs and actions are defined as follows: each player announces simultaneously whether they want to exchange their houses. If both players agree to an exchange, the exchange takes place. Otherwise no exchange takes place.

As far as finding a Bayesian equilibrium is concerned, I have reached the following stage:

Given $j$ exchanges, $i$ exchanges so long as $v_{i}\leq\frac{3}{2}v_{j}$ and given $i$ exchanges, $j$ exchanges so long as $v_{j}\leq\frac{3}{2}v_{i}$. This means that $j$'s expected utility from exchanging (given $i$ exchanges) is $\frac{9}{8}v_{j}$ and $i$'s is $\frac{9}{8}v_{i}$. From this I do not know how to proceed.

Answer 1

This is a static Bayesian game. The two players share a common prior over the type space which is characterized by the uniform distribution. For Bayesian games, you have to specify the belief of each player given his/her type, and the belief of each player determines their action. Basically, the strategy is a mapping from one's type to an action through belief specification.

What is player $i$'s belief about $v_j$? It is independently drawn from the uniform distribution over $[0,1]$, so $E(v_j) = 0.5$ for now. And he will exchange only when $v_i \leqslant \frac{3}{2}v_j$, but he does not know exactly the value of $v_j$, so he takes an expectation over $v_j$, then as long as $v_i \leqslant \frac{3}{2} \times 0.5 = \frac{3}{4}$, he will choose to exchange.

The same goes for player $j$ because they are symmetric, so player $j$ will only exchange when $v_j \leqslant \frac{3}{4}$.

Given this, $E(v_j)$ now decreases to $\frac{3}{8}$, player $i$'s expected payoff from exchange decreases to $\frac{3}{2}\times \frac{3}{8} = (\frac{3}{4})^2$. Therefore, player $i$ will only choose exchange when $v_i \leqslant (\frac{3}{4})^2$.

Since $\frac{3}{4} < 1$, this process will go on and on such that no one will choose exchange in the end which is the only equilibrium.