<h1>Hint:</h1>
<p>You can convert the game into the following normal form:</p>
<p><span class="math-container">\begin{equation}
\begin{array}{|c|c|c|}
\hline
&amp; Steal &amp; Split \\\hline 
Steal &amp; 100\beta, 0&amp; 100(1-\alpha),0 \\\hline 
Split &amp; 50(\alpha+\beta),50+50(1-\alpha-\beta) &amp; 50,50 \\\hline
\end{array}
\end{equation}</span></p>
<p>Note that Steal is a weakly dominant strategy for player 2, so (Split, Split) can never be an equilibrium.</p>

CC BY-SA 4.0
2019-01-01T22:35:38.063Z
Thanks for the answer, it was a big help realizing I could just write the game up in a bimatrix! I have one question however: Consider $\alpha = \beta = 0.5$. Wouldn't all four possibilities be equilibria in that case, including (Split, Split)?
Greased Dog
CC BY-SA 4.0
2019-01-02T02:14:21.427Z
@GreasedDog: You're right. Note that $\alpha=\beta=0.5$ implies state $N$ never occurs, so player 2's choice doesn't matter. The probability also implies that player 1 is indifferent between split and steal.
Herr K.
CC BY-SA 4.0
2018-11-20T22:37:45.670Z
2018-11-20T22:37:45.670Z
2020-06-17T09:40:54.643Z
Herr K.
<p>What are the perfect Bayesian equilibria in the following game?</p>

<p><a href="https://i.sstatic.net/wQ9fR.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wQ9fR.png" alt="enter image description here"></a></p>

<p>The part that is confusing me is the right-hand side, where player 2 is of type N.</p>

<p>Thanks!</p>

<p>EDIT: My workings out so far are given below:</p>

<p>We start off by noticing that if player 2 is type <span class="math-container">$N$</span>, it can be rational for him to play both <span class="math-container">$Steal$</span> and <span class="math-container">$Split$</span>, as he does not know what player 1 plays.</p>

<p>Let us therefore first look at the case where player 2 plays <span class="math-container">$Steal$</span> if he is type <span class="math-container">$N$</span>. Then player 1's expected payoff from stealing is <span class="math-container">$100\beta$</span> and his expected payoff from splitting is <span class="math-container">$50\alpha + 50\beta$</span>. In this case player 1 steals when <span class="math-container">$100\beta &gt; 50\alpha + 50\beta \Leftrightarrow \alpha &lt; \beta$</span>, and therefore splits if <span class="math-container">$\alpha &gt; \beta$</span> and is indifferent when <span class="math-container">$\alpha = \beta$</span>. This means we get the following potential PBE's in this case:
<span class="math-container">\begin{align*}
    (Steal, Steal; \alpha \leq \beta) \text{ and } (Split, Steal; \alpha \geq \beta).
\end{align*}</span></p>

<p>Let us next look at the case where player 2 plays <span class="math-container">$Split$</span> if he is type <span class="math-container">$N$</span>. Then player 1's expected payoff from stealing is <span class="math-container">$100\beta + 100(1-\alpha-\beta)=100(1-\alpha)$</span> and his expected payoff from splitting is <span class="math-container">$50$</span>. In this case player 1 steals when <span class="math-container">$100(1-\alpha) &gt; 50 \Leftrightarrow \alpha &lt; 1/2$</span>, and therefore splits if <span class="math-container">$\alpha &gt; 1/2$</span> and is indifferent if <span class="math-container">$\alpha = 1/2$</span>. This means that we get the following potential PBE's in this case:
<span class="math-container">\begin{align*}
    (Steal, Split; \alpha \leq 1/2) \text{ and } (Split, Split; \alpha \geq 1/2).
\end{align*}</span></p>

<p>As such, we have can write up the set of pure-strategy PBE:
<span class="math-container">\begin{alignat*}{1}
    \text{PBE} = \{&amp;(Steal, Steal; \alpha \leq \beta),(Split, Steal; \alpha \geq \beta), \\ &amp;(Steal, Split; \alpha \leq 1/2),(Split, Split; \alpha \geq 1/2)\}.
\end{alignat*}</span></p>

CC BY-SA 4.0
2018-11-19T17:39:10.773Z
What exactly is your confusion?
Herr K.
CC BY-SA 4.0
2018-11-20T18:17:30.330Z
I have added my working outs to the post. I guess my confusion is regarding if I can just say that it is rational for player 2 to play both Steal and Split and then leave it at that.
Greased Dog
CC BY-SA 4.0
2018-11-20T19:08:52.580Z
Please edit your tex code into the question rather than the image.
Giskard
CC BY-SA 4.0
2018-11-20T19:58:07.770Z
Okay, I have added tex code now.
Greased Dog
<microeconomics><game-theory>
Perfect Bayesian Equilibria
Greased Dog
# Perfect Bayesian Equilibria

(Asked by: username_0 on 2018-11-19)

What are the perfect Bayesian equilibria in the following game?

[![enter image description here](https://i.sstatic.net/wQ9fR.png)](https://i.sstatic.net/wQ9fR.png)

The part that is confusing me is the right-hand side, where player 2 is of type N.

Thanks!

EDIT: My workings out so far are given below:

We start off by noticing that if player 2 is type $N$, it can be rational for him to play both $Steal$ and $Split$, as he does not know what player 1 plays.

Let us therefore first look at the case where player 2 plays $Steal$ if he is type $N$. Then player 1's expected payoff from stealing is $100\beta$ and his expected payoff from splitting is $50\alpha + 50\beta$. In this case player 1 steals when $100\beta > 50\alpha + 50\beta \Leftrightarrow \alpha < \beta$, and therefore splits if $\alpha > \beta$ and is indifferent when $\alpha = \beta$. This means we get the following potential PBE's in this case:
\begin{align\*}
(Steal, Steal; \alpha \leq \beta) \text{ and } (Split, Steal; \alpha \geq \beta).
\end{align\*}

Let us next look at the case where player 2 plays $Split$ if he is type $N$. Then player 1's expected payoff from stealing is $100\beta + 100(1-\alpha-\beta)=100(1-\alpha)$ and his expected payoff from splitting is $50$. In this case player 1 steals when $100(1-\alpha) > 50 \Leftrightarrow \alpha < 1/2$, and therefore splits if $\alpha > 1/2$ and is indifferent if $\alpha = 1/2$. This means that we get the following potential PBE's in this case:
\begin{align\*}
(Steal, Split; \alpha \leq 1/2) \text{ and } (Split, Split; \alpha \geq 1/2).
\end{align\*}

As such, we have can write up the set of pure-strategy PBE:
\begin{alignat\*}{1}
\text{PBE} = \{&(Steal, Steal; \alpha \leq \beta),(Split, Steal; \alpha \geq \beta), \\ &(Steal, Split; \alpha \leq 1/2),(Split, Split; \alpha \geq 1/2)\}.
\end{alignat\*}

**Tags:** microeconomics, game-theory

> **Comments:**
> **username_1**: What exactly is your confusion?  
> **username_0**: I have added my working outs to the post. I guess my confusion is regarding if I can just say that it is rational for player 2 to play both Steal and Split and then leave it at that.  
> **username_2**: Please edit your tex code into the question rather than the image.  
> **username_0**: Okay, I have added tex code now.  

---

## 1 answers

### Answer by username_1 on 2018-11-20. Score: 1

# Hint:

You can convert the game into the following normal form:

\begin{equation}
\begin{array}{|c|c|c|}
\hline
& Steal & Split \\\hline
Steal & 100\beta, 0& 100(1-\alpha),0 \\\hline
Split & 50(\alpha+\beta),50+50(1-\alpha-\beta) & 50,50 \\\hline
\end{array}
\end{equation}

Note that Steal is a weakly dominant strategy for player 2, so (Split, Split) can never be an equilibrium.

> **Comments:**
> **username_0**: Thanks for the answer, it was a big help realizing I could just write the game up in a bimatrix! I have one question however: Consider $\alpha = \beta = 0.5$. Wouldn't all four possibilities be equilibria in that case, including (Split, Split)?  
> **username_1**: @GreasedDog: You're right. Note that $\alpha=\beta=0.5$ implies state $N$ never occurs, so player 2's choice doesn't matter. The probability also implies that player 1 is indifferent between split and steal.  

---

