How to find BNE of the exchange game?

External question — Economics Stack Exchange Author: Hãru2you Original post: https://economics.stackexchange.com/questions/57862 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. Each of two players receives a ticket t on which there is a number in [0,1]. The number on a players ticket is the size of a prize that he may receive. The two prizes are identically and independently distributed according to a uniform distribution. Each player is asked independently and simultaneously whether he wants to exchange his price for the other players prize. If both players agree than the prizes are exchanges; otherwise each player receives his own prize. I did it till best response of player 2 given player 1 is playing 'Exchange when T1<k' will be: Don't exchange, if T2>k Exchange if , T2<k But it turns out to be T2<k/2 in second part. Why is it so?

External answer — Economics Stack Exchange Author: tdm Original post: https://economics.stackexchange.com/a/57863 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. Assume that the decision rule (in equilibrium) for player 1 is to ask for an exchange iff $T_1 \le k_1$ . Now assume that player 1 does indeed ask for an exchange. Then in equilibrium, player 2 knows that $T_1 \le k_1$ . Given that the prior of $T_1$ is the uniform distribution over $[0,1]$ the posterior of $T_1$ for player 2 will now be the uniform over $[0,k_1]$ . This means that, if she agrees to exchange, then the expected payoff for player 2 equals $\frac{k_1}{2}$ . As such, Player 2 will agree with the exchange if and only if $T_2 \le \frac{k_1}{2}$ as $\frac{k_2}{2}$ is the expected value of a uniformly distributed random variable over $[0,k_1]$ .