<p>Assume that the decision rule (in equilibrium) for player 1 is to ask for an exchange iff <span class="math-container">$T_1 \le k_1$</span>.</p>
<p>Now assume that player 1 does indeed ask for an exchange. Then in equilibrium, player 2 knows that <span class="math-container">$T_1 \le k_1$</span>. Given that the prior of <span class="math-container">$T_1$</span> is the uniform distribution over <span class="math-container">$[0,1]$</span> the posterior of <span class="math-container">$T_1$</span> for player 2 will now be the uniform over <span class="math-container">$[0,k_1]$</span>.</p>
<p>This means that, if she agrees to exchange, then the expected payoff for player 2 equals  <span class="math-container">$\frac{k_1}{2}$</span>. As such, Player 2 will agree with the exchange if and only if <span class="math-container">$T_2 \le \frac{k_1}{2}$</span> as <span class="math-container">$\frac{k_2}{2}$</span> is the expected value of a uniformly distributed random variable over <span class="math-container">$[0,k_1]$</span>.</p>

CC BY-SA 4.0
2024-03-08T13:55:56.643Z
2024-03-09T08:41:51.747Z
2024-03-09T08:41:51.747Z
tdm
<p>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.</p>
<p>I did it till best response of player 2 given player 1 is playing 'Exchange when T1&lt;k' will be:</p>
<p>Don't exchange, if T2&gt;k</p>
<p>Exchange if , T2&lt;k</p>
<p>But it turns out to be T2&lt;k/2 in second part. Why is it so? </p>

CC BY-SA 4.0
2024-03-08T11:34:37.390Z
"*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*" Why do you think they should have the same $k$? Like if player $1$ wants to trade under $1/3$, why should player $2$ want to trade exactly under $1/3$ as well, why not $1/4$ or something else?
Giskard
<nash-equilibrium><bayesian-game><game-theory>
How to find BNE of the exchange game?
Hãru2you
# How to find BNE of the exchange game?

(Asked by: username_0 on 2024-03-08)

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?

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

> **Comments:**
> **username_1**: "*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*" Why do you think they should have the same $k$? Like if player $1$ wants to trade under $1/3$, why should player $2$ want to trade exactly under $1/3$ as well, why not $1/4$ or something else?  

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## 1 answers

### Answer by username_2 on 2024-03-08. Score: 1

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 $[0,k\_1]$.


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