<p>Let's take a generic symmetric <span class="math-container">$2\times2$</span> game as below:
<span class="math-container">$$
\begin{array}{|c|c|c|}\hline
&amp;A&amp;B\\\hline
A&amp;a,a &amp; b,c \\\hline 
B&amp;c,b &amp; d,d \\\hline
\end{array}
$$</span>
where <span class="math-container">$a,b,c,d \in \mathbb R$</span> are distinct. Further suppose that <span class="math-container">$(A,A)$</span> and <span class="math-container">$(B,B)$</span> are pure strategy NEs. By definition, <span class="math-container">$(A,A)$</span> risk dominates <span class="math-container">$(B,B)$</span> if and only if
<span class="math-container">$$a-c&gt;d-b\,, \tag{1}$$</span>
and the direction of domination reverses with the inequality.</p>
<p>Now consider the statement</p>
<blockquote>
<p>it is more costly to mistakenly think we are playing <span class="math-container">$B$</span> than to mistakenly think we are playing <span class="math-container">$A$</span>.</p>
</blockquote>
<p>Suppose player 2 plays <span class="math-container">$A$</span>, but player 1 <em>mistakenly</em> believes that player 2 is playing <span class="math-container">$B$</span>. That erroneous belief would lead player 1 to play <span class="math-container">$B$</span>, which causes player 1 to deviate from the equilibrium <span class="math-container">$(A,A)$</span> and entails a cost of <span class="math-container">$a-c$</span>. Likewise, if player 1 mistakenly thinks that player 2 would play <span class="math-container">$A$</span> while the latter is playing <span class="math-container">$B$</span>, the cost of deviating from the <span class="math-container">$(B,B)$</span> equilibrium would be <span class="math-container">$d-b$</span>.</p>
<p>Plug in the numbers from your first example, the above statement corresponds to condition <span class="math-container">$(1)$</span>, namely,
<span class="math-container">$$a-c=80-0&gt;100-80=d-b$$</span>
thus making <span class="math-container">$(A,A)$</span> the risk dominant NE.</p>
<p>In the second example, the numbers turn out to be
<span class="math-container">$$a-c=80-0&lt;100-0=d-b$$</span>
which reverses the inequality in <span class="math-container">$(1)$</span>, making <span class="math-container">$(B,B)$</span> risk dominant. The interpretation is symmetric to the one given for the first example:</p>
<blockquote>
<p>it is more costly to mistakenly think we are playing <span class="math-container">$A$</span> than to mistakenly think we are playing <span class="math-container">$B$</span>.</p>
</blockquote>

CC BY-SA 4.0
2022-11-09T04:08:11.253Z
2022-11-09T04:08:11.253Z
Herr K.
<p>I am trying to wrap my head around the idea of risk dominance but I am not confident whether or not what I am thinking is correct.</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th></th>
<th>A</th>
<th>B</th>
</tr>
</thead>
<tbody>
<tr>
<td>A</td>
<td>80, 80</td>
<td>80, 0</td>
</tr>
<tr>
<td>B</td>
<td>0, 80</td>
<td>100, 00</td>
</tr>
</tbody>
</table>
</div>
<p>The solution says that the risk dominance here is (A, A), and explains that it is more costly to mistakenly think we are playing B than to mistakenly think we are playing A. Here's my thought process so far.</p>
<p>Player 1 would prefer to stick with A to get a payoff of 80, rather than going for B with a chance to get 100 but also with a chance to get nothing with 0. Same thing for player 2 - he also wants to stick with A to get a payoff of 80 rather than risking for 100 or nothing.</p>
<p>Now here comes this example, which is where I am truly confused.</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th></th>
<th>A</th>
<th>B</th>
</tr>
</thead>
<tbody>
<tr>
<td>A</td>
<td>80, 80</td>
<td>0, 0</td>
</tr>
<tr>
<td>B</td>
<td>0, 0</td>
<td>100, 100</td>
</tr>
</tbody>
</table>
</div>
<p>The solution says that the answer is (Bottom, Right), which is (100, 100). Here's my thought process for this example. Suppose we are player 1. We will be playing A and we think that player 2 is going for A too, but he ends up going for B. That means we are losing out for 80 - 0 = 80. Now suppose that we will play B and we think that player 2 is going for B too, but he goes for A. That means we are losing out for 100 - 0 = 100. This is symmetric for player 2 too. So losing out for 80 and 100...obviously, losing 80 sucks less than losing 100.</p>
<p>Is my thought process correct? Thanks a lot!</p>

<game-theory><self-study>
Risk dominance in NE
kyrie3
# Risk dominance in NE

(Asked by: username_0 on 2022-11-06)

I am trying to wrap my head around the idea of risk dominance but I am not confident whether or not what I am thinking is correct.

|  | A | B |
| --- | --- | --- |
| A | 80, 80 | 80, 0 |
| B | 0, 80 | 100, 00 |

The solution says that the risk dominance here is (A, A), and explains that it is more costly to mistakenly think we are playing B than to mistakenly think we are playing A. Here's my thought process so far.

Player 1 would prefer to stick with A to get a payoff of 80, rather than going for B with a chance to get 100 but also with a chance to get nothing with 0. Same thing for player 2 - he also wants to stick with A to get a payoff of 80 rather than risking for 100 or nothing.

Now here comes this example, which is where I am truly confused.

|  | A | B |
| --- | --- | --- |
| A | 80, 80 | 0, 0 |
| B | 0, 0 | 100, 100 |

The solution says that the answer is (Bottom, Right), which is (100, 100). Here's my thought process for this example. Suppose we are player 1. We will be playing A and we think that player 2 is going for A too, but he ends up going for B. That means we are losing out for 80 - 0 = 80. Now suppose that we will play B and we think that player 2 is going for B too, but he goes for A. That means we are losing out for 100 - 0 = 100. This is symmetric for player 2 too. So losing out for 80 and 100...obviously, losing 80 sucks less than losing 100.

Is my thought process correct? Thanks a lot!

**Tags:** game-theory, self-study

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## 1 answers

### Answer by username_1 on 2022-11-09. Score: 2 | Accepted answer

Let's take a generic symmetric $2\times2$ game as below:
$$
\begin{array}{|c|c|c|}\hline
&A&B\\\hline
A&a,a & b,c \\\hline
B&c,b & d,d \\\hline
\end{array}
$$
where $a,b,c,d \in \mathbb R$ are distinct. Further suppose that $(A,A)$ and $(B,B)$ are pure strategy NEs. By definition, $(A,A)$ risk dominates $(B,B)$ if and only if
$$a-c>d-b\,, \tag{1}$$
and the direction of domination reverses with the inequality.

Now consider the statement

> it is more costly to mistakenly think we are playing $B$ than to mistakenly think we are playing $A$.

Suppose player 2 plays $A$, but player 1 *mistakenly* believes that player 2 is playing $B$. That erroneous belief would lead player 1 to play $B$, which causes player 1 to deviate from the equilibrium $(A,A)$ and entails a cost of $a-c$. Likewise, if player 1 mistakenly thinks that player 2 would play $A$ while the latter is playing $B$, the cost of deviating from the $(B,B)$ equilibrium would be $d-b$.

Plug in the numbers from your first example, the above statement corresponds to condition $(1)$, namely,
$$a-c=80-0>100-80=d-b$$
thus making $(A,A)$ the risk dominant NE.

In the second example, the numbers turn out to be
$$a-c=80-0<100-0=d-b$$
which reverses the inequality in $(1)$, making $(B,B)$ risk dominant. The interpretation is symmetric to the one given for the first example:

> it is more costly to mistakenly think we are playing $A$ than to mistakenly think we are playing $B$.


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