<p>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 <span class="math-container">$v_i$</span> when calculating strategies.</p>
<p>For example, you may be tempted to say that both players decide to trade whenever,  <span class="math-container">$v_i\leq E(\alpha v_j-c)=0.5\alpha-c$</span> and <span class="math-container">$v_j\leq E(\alpha v_i - c)=0.5\alpha-c$</span>. However, these strategies might not be mutual best responses because suppose player <span class="math-container">$i$</span> believes that <span class="math-container">$j$</span> will use that strategy and assume that <span class="math-container">$0.5\alpha-c=0.8&lt;1$</span>. That means that player <span class="math-container">$i$</span> should not expect <span class="math-container">$v_j$</span> to be greater than <span class="math-container">$0.8$</span>, so a profitable deviation is to instead trade whenever <span class="math-container">$v_i\leq E(\alpha v_j-c| v_j\leq 0.8)= 0.4\alpha-c$</span>.</p>
<p>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 <span class="math-container">$v_i\leq \bar v$</span> (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.</p>
<p>Therefore, it must be that player I trades whenever <span class="math-container">$v_i\leq E(\alpha v_j-c | v_j\leq \bar v)= \alpha\frac{\bar v}2-c$</span>. Since we assume that the threshold for trading or non trading is <span class="math-container">$\bar v$</span>, it must be that <span class="math-container">$\bar v = \alpha\frac{\bar v}2-c$</span> which implies that <span class="math-container">$\bar v=\frac{c}{\left(\frac{\alpha}{2}-1\right)}$</span>. This means that (assuming <span class="math-container">$c&gt;0$</span>), <span class="math-container">$\alpha$</span> must be strictly larger than 2 for some trade to exist since <span class="math-container">$\bar v &lt;0$</span> means that players will never trade, and if <span class="math-container">$\bar v&gt;1$</span>, then trade always occurs.</p>

CC BY-SA 4.0
2021-02-03T07:47:01.120Z
2021-02-03T07:47:01.120Z
Regio
<p>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 <span class="math-container">$v_i$</span>. The value of player <span class="math-container">$i$</span>’s house to the other player, i.e. to player <span class="math-container">$j \ne i$</span>, is <span class="math-container">$\alpha v_i − c$</span> where <span class="math-container">$\alpha &gt; 1$</span>. Each player knows the value <span class="math-container">$v_i$</span> of his own house to himself, but not the value of the opponent’s house. Both players know <span class="math-container">$\alpha$</span>. The values <span class="math-container">$v_i$</span> are distributed uniformly on the interval [0, 1] and are independent across players.</p>
<p>I suppose that player <span class="math-container">$i$</span> will agree to exchange only if <span class="math-container">$v_i \leq \alpha v_j - c$</span> 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 <span class="math-container">$v_i$</span> through the uniform distribution, i.e. <span class="math-container">$E(v_i)=1/2$</span> ?</p>
<p>Would be very grateful for some help!!</p>

CC BY-SA 4.0
2020-11-26T11:10:03.330Z
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?
Dayne
CC BY-SA 4.0
2020-11-26T11:12:54.620Z
I think the probability distribution has to be continuous
peninette
CC BY-SA 4.0
2020-11-26T11:21:50.240Z
What is known to whom about $c$?
VARulle
CC BY-SA 4.0
2020-11-26T11:22:48.893Z
Nothing is said about c :(
peninette
CC BY-SA 4.0
2020-11-26T11:27:03.027Z
since there is no index on $c$ it may be safe to assume that both know $c$ as well.
Dayne
CC BY-SA 4.0
2020-11-26T11:28:04.683Z
Yes for sure but how should I go about finding the Bayesian Nash equilibrium?
peninette
CC BY-SA 4.0
2020-11-26T13:45:05.163Z
Does [this](https://economics.stackexchange.com/a/36928/24967) answer your question?
Dayne
CC BY-SA 4.0
2020-11-26T14:13:05.027Z
The Nash Equilibrium would depend on the bargaining process that the players use to determine whether to trade/whether there should be some payment.
H Rogers
CC BY-SA 4.0
2020-11-26T16:06:37.750Z
@Dayne I saw that but wasn't too sure if that helped because since I am not given a clear value for alpha, I cannot fully determine whether (keep,Keep) is the `Nash equilibrium
peninette
<game-theory><bayesian-game><asymmetric-information>
Exchange Houses game - Bayesian Game
peninette
# Exchange Houses game - Bayesian Game

(Asked by: username_0 on 2020-11-26)

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!!

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

> **Comments:**
> **username_1**: 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?  
> **username_0**: I think the probability distribution has to be continuous  
> **username_2**: What is known to whom about $c$?  
> **username_0**: Nothing is said about c :(  
> **username_1**: since there is no index on $c$ it may be safe to assume that both know $c$ as well.  
> **username_0**: Yes for sure but how should I go about finding the Bayesian Nash equilibrium?  
> **username_1**: Does [this](https://economics.stackexchange.com/a/36928/24967) answer your question?  
> **username_3**: The Nash Equilibrium would depend on the bargaining process that the players use to determine whether to trade/whether there should be some payment.  
> **username_0**: @username_1 I saw that but wasn't too sure if that helped because since I am not given a clear value for alpha, I cannot fully determine whether (keep,Keep) is the `Nash equilibrium  

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## 1 answers

### Answer by username_4 on 2021-02-03. Score: 3

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.


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