# Exchange Houses game - Bayesian Game

Can anyone help me understand how to solve this type of asymmetric information Bayesian game? So the game is a different version of the Tadelis 'trading house games'. It involves 2 players that each own a house. Each player values his own house at $$v_i$$. The value of player $$i$$’s house to the other player, i.e. to player $$j \ne i$$, is $$\alpha v_i − c$$ where $$\alpha > 1$$. Each player knows the value $$v_i$$ of his own house to himself, but not the value of the opponent’s house. Both players know $$\alpha$$. The values $$v_i$$ are distributed uniformly on the interval [0, 1] and are independent across players.

I suppose that player $$i$$ will agree to exchange only if $$v_i \leq \alpha v_j - c$$ but I don't know how to go from there to find the Bayesian Nash equilibrium and how to assign probabilities. Are we suppose to use the expected value of $$v_i$$ through the uniform distribution, i.e. $$E(v_i)=1/2$$ ?

Would be very grateful for some help!!

• Is this homework? What if the probability distribution was discrete such that $v_i, v_j$ can take only two values (say $0,1$) with probability $0.5$ each. Can you now write this game in normal-form now? Nov 26 '20 at 11:10
• I think the probability distribution has to be continuous Nov 26 '20 at 11:12
• What is known to whom about $c$? Nov 26 '20 at 11:21
• Nothing is said about c :( Nov 26 '20 at 11:22
• since there is no index on $c$ it may be safe to assume that both know $c$ as well. Nov 26 '20 at 11:27

The trick with Bayesian games is to recognize that the players learn in equilibrium from the strategies of the other players. That is, you cannot simply take the unconditional expectation of $$v_i$$ when calculating strategies.
For example, you may be tempted to say that both players decide to trade whenever, $$v_i\leq E(\alpha v_j-c)=0.5\alpha-c$$ and $$v_j\leq E(\alpha v_i - c)=0.5\alpha-c$$. However, these strategies might not be mutual best responses because suppose player $$i$$ believes that $$j$$ will use that strategy and assume that $$0.5\alpha-c=0.8<1$$. That means that player $$i$$ should not expect $$v_j$$ to be greater than $$0.8$$, so a profitable deviation is to instead trade whenever $$v_i\leq E(\alpha v_j-c| v_j\leq 0.8)= 0.4\alpha-c$$.
A common trick to solve these games is to assume something about how the equilibrium will look like and then confirm that it indeed looks like that. For example, given the example above, it is not crazy to think that players will only trade if $$v_i\leq \bar v$$ (for some threshold that you will have to characterize later). It is also reasonable to assume that both players will use the same strategy since they face the same underlying uncertainty and the valuations for each other houses are symmetric.
Therefore, it must be that player I trades whenever $$v_i\leq E(\alpha v_j-c | v_j\leq \bar v)= \alpha\frac{\bar v}2-c$$. Since we assume that the threshold for trading or non trading is $$\bar v$$, it must be that $$\bar v = \alpha\frac{\bar v}2-c$$ which implies that $$\bar v=\frac{c}{\left(\frac{\alpha}{2}-1\right)}$$. This means that (assuming $$c>0$$), $$\alpha$$ must be strictly larger than 2 for some trade to exist since $$\bar v <0$$ means that players will never trade, and if $$\bar v>1$$, then trade always occurs.