<p>(I only parsed what you did, perhaps I did not get it.)</p>
<p>You seem to be overcomplicating things a bit. A good starting point seems to be whether player 2 will mix or not.</p>
<ol>
<li><p>If not, you can see that player 1 will have a clear best response in each matrix, so you will have a pure equilibrium. You only have to validate that player 2's not mixed strategy is a best response to the expected strategy that player 1's types are playing.</p>
</li>
<li><p>If player 2 is mixing, then you can write down an equation showing their expected payoff as a function of player 1's strategy:
<span class="math-container">\begin{align*}
EU_2(\text{P1 strategy},x_2) &amp; =
(p \cdot q_{11x} + (1-p) \cdot q_{12x}) \cdot  1 + (p \cdot q_{11y} + (1-p) \cdot q_{12y}) \cdot  (-1) = 0.   
\end{align*}</span>
It equals zero because
<span class="math-container">$$EU_2(\text{P1 strategy},x_2) = - EU_2(\text{P1 strategy},y_2),$$</span>
so if they are mixing both <span class="math-container">$x_2$</span> and <span class="math-container">$y_2$</span> than their expected payoff is 0. Unless player 2 is mixing with 50-50% probability, player 1 will not mix. So this gives you the unlikely case that <br>
a. player 2 is mixing, player 1 is not, but <span class="math-container">$p$</span> is such that things work out, and the more likely case that <br>
b. player 2 is mixing with 50-50% and you can adjust player 1's strategy however you like, as long as player 2's expected payoff is 0.</p>
</li>
</ol>

CC BY-SA 4.0
2024-04-23T20:42:09.973Z
2024-04-24T04:37:47.267Z
2024-04-24T04:37:47.267Z
Giskard
<h2>Question</h2>
<p>I want to compute the Bayesian equilibria for the following Bayesian game:
<a href="https://i.sstatic.net/6xrCd.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6xrCd.png" alt="enter image description here" /></a>
With probability <span class="math-container">$p$</span>, player 1 would be of type 1.1. With probability <span class="math-container">$1-p$</span>, player 1 would be of type 1.2. Player 2 does not have any private information and she only has one type 2.0.</p>
<p>I know two methods to compute Bayesian equilibria. The first one, with which I think I am more familiar, is based on the presentation from Roger Myerson's <em>Game Theory: Analysis of conflict</em>. The other one is presented by William Spaniel's <em>Game Theory 101</em> Youtube course. I want to use both method to solve this question and, of course, I expect to get the same result. But I got some trouble when proceeding William Spaniel's method. I would really appreciate it if someone could help me figure out where my problem is!</p>
<hr />
<h2>Method 1<span class="math-container">$\quad$</span>[Roger Myerson]</h2>
<p>This is the same answer I worte for <a href="https://math.stackexchange.com/questions/4899648/characterize-the-mixed-strategy-bayesian-nash-equilibria-for-this-game/4900405#4900405">this MSE post</a>. But I will present it here as well.</p>
<p>Note that a Bayesian equilibrium of the game <span class="math-container">$\Gamma^b$</span> is any randomized-strategy profile <span class="math-container">$\sigma$</span> such that, for every player <span class="math-container">$i$</span> in <span class="math-container">$N$</span> and every type <span class="math-container">$t_i$</span> in <span class="math-container">$T_i$</span>,
<span class="math-container">\begin{align*}
\sigma_i(\cdot|t_i) \in \text{argmax}_{\tau_i \in \Delta(C_i)} \sum_{t_{-i} \in T_{-i}} p_i(t_{-i}|t_i) \sum_{c\in C}\left(\prod_{j \in N-i}\sigma_j(c_j|t_j)\right)\tau_i(c_i)u_i(c,t).
\end{align*}</span>
As for our game, we make the following definition:</p>
<blockquote>
<p><span class="math-container">$C_1 = \{x_1,y_1\}$</span>, where <span class="math-container">$x_1$</span> means &quot;up&quot; and <span class="math-container">$y_1$</span> means &quot;down&quot;.</p>
<p><span class="math-container">$C_2 = \{x_2,y_2\}$</span>, where <span class="math-container">$x_2$</span> means &quot;left&quot; and <span class="math-container">$y_2$</span> means &quot;right&quot;.</p>
<p><span class="math-container">$C = C_1\mathop{\Large\times}C_2$</span></p>
<p><span class="math-container">$\Delta(C)$</span> is the set of all probability distributions over the set <span class="math-container">$C$</span></p>
<p><span class="math-container">$T_1 = \{1.1,1.2\}$</span>, where 1.1 stands for player 1's type of coorperative and 1.2 stands for player 1's type of uncooperative.</p>
<p><span class="math-container">$T_2 = \{2.0\}$</span>, where 2.0 stands for player 2's only possible type.</p>
<p><span class="math-container">$T = T_1\mathop{\Large\times}T_2$</span></p>
<p><span class="math-container">$p_2(1.1|2.0) = p$</span> and <span class="math-container">$p_2(1.2|2.0) = 1-p$</span>, where <span class="math-container">$p_2(\cdot|2.0)$</span> represents what player 2 would believe about player 1's types given his own type being 2.0.</p>
</blockquote>
<p>First, consider the matrix for <span class="math-container">$t_1=1.1$</span>. Clearly, there are two pure strategy Nash equilibria:
<span class="math-container">\begin{align*}
\sigma_1(\cdot|1.1)=[x_1],\ \sigma_2=[x_2]\quad \text{and}\quad \sigma_1(\cdot|1.1)=[y_1],\ \sigma_2=[y_2],
\end{align*}</span>
and one mixed strategy Nash equilibrium:
<span class="math-container">\begin{align*}
\sigma_1(\cdot|1.1)=\frac{1}{2}[x_1]+\frac{1}{2}[y_1],\ \sigma_2 = \frac{1}{2}[x_2] + \frac{1}{2}[y_2].
\end{align*}</span></p>
<p>Next, consider the matrix for <span class="math-container">$t_1 = 1.2$</span>. Clearly, there is only one Nash equilibrium:
<span class="math-container">\begin{align*}
\sigma_1(\cdot|1.2)=\frac{1}{2}[x_1]+\frac{1}{2}[y_1],\ \sigma_2 = \frac{1}{2}[x_2] + \frac{1}{2}[y_2].
\end{align*}</span></p>
<p>Therefore, we have the following three cases:</p>
<p><em><strong>Case 1:</strong></em><span class="math-container">$\quad$</span><em><strong>[Mixed strategy Nash equilibria for both matrices]</strong></em></p>
<p>Since
<span class="math-container">\begin{align*}
\begin{alignedat}{4}
       &amp;p_2(1.1|2.0) \cdot (&amp;&amp;\sigma_1(x_1|1.1) \cdot \tau_2(x_2) \cdot u_1((x_1,x_2),(1.1,2.0)) &amp;\ +\\
       &amp;                    &amp;&amp;\sigma_1(y_1|1.1) \cdot \tau_2(x_2) \cdot u_1((y_1,x_2),(1.1,2.0)) &amp;\ +\\
       &amp;                    &amp;&amp;\sigma_1(x_1|1.1) \cdot \tau_2(y_2) \cdot u_1((x_1,y_2),(1.1,2.0)) &amp;\ +\\
       &amp;                    &amp;&amp;\sigma_1(y_1|1.1) \cdot \tau_2(y_2) \cdot u_1((y_1,y_2),(1.1,2.0)))&amp;\\
    =\ &amp;0                   &amp;&amp;                                                                   &amp;
\end{alignedat}
\end{align*}</span>
and
<span class="math-container">\begin{align*}
\begin{alignedat}{4}
       &amp;p_2(1.2|2.0) \cdot (&amp;&amp;\sigma_1(x_1|1.2) \cdot \tau_2(x_2) \cdot u_1((x_1,x_2),(1.2,2.0)) &amp;\ +\\
       &amp;                    &amp;&amp;\sigma_1(y_1|1.2) \cdot \tau_2(x_2) \cdot u_1((y_1,x_2),(1.2,2.0)) &amp;\ +\\
       &amp;                    &amp;&amp;\sigma_1(x_1|1.2) \cdot \tau_2(y_2) \cdot u_1((x_1,y_2),(1.2,2.0)) &amp;\ +\\
       &amp;                    &amp;&amp;\sigma_1(y_1|1.2) \cdot \tau_2(y_2) \cdot u_1((y_1,y_2),(1.2,2.0)))&amp;\\
    =\ &amp;0,                  &amp;&amp;                                                                   &amp;
\end{alignedat}
\end{align*}</span>
it follows that
<span class="math-container">\begin{align*}
&amp;\text{argmax}_{\tau_2 \in \Delta(C_2)} \sum_{t_{1} \in T_{1}} p_2(t_{1}|t_2) \sum_{c\in C}\left(\prod_{j \in \{1\}}\sigma_j(c_j|t_j)\right)\tau_2(c_2)u_2(c,t)\\
=\ &amp;\text{argmax}_{\tau_2 \in \Delta(C_2)} 0\\
=\ &amp;0.
\end{align*}</span>
Thus, <span class="math-container">$\sigma_2(\cdot|2.0) = \alpha[x_2] + (1-\alpha)[y_2]$</span> for all <span class="math-container">$\alpha \in [0,1]$</span>, <span class="math-container">$\sigma_1(\cdot|1.1) = \frac{1}{2}[x_1]+\frac{1}{2}[y_1]$</span>, and <span class="math-container">$\sigma_1(\cdot|1.2) = \frac{1}{2}[x_1] + \frac{1}{2}[y_1]$</span> is a Bayesian equilibrium.</p>
<p><em><strong>Case 2:</strong></em><span class="math-container">$\quad$</span><em><strong>[The first pure equilibrium for matrix <span class="math-container">$\mathbf{t_1=}\ $</span>1.1]</strong></em></p>
<p>Since
<span class="math-container">\begin{align*}
p_2(1.1|2.0)\cdot(1\cdot\tau_2(x_2)\cdot1+0+1\cdot\tau_2(y_2)\cdot(-1)+0) = p(\tau_2(x_2)-\tau_2(y_2)),
\end{align*}</span>
it follows that
<span class="math-container">\begin{align*}
&amp;\text{argmax}_{\tau_2 \in \Delta(C_2)} \sum_{t_{1} \in T_{1}} p_2(t_{1}|t_2) \sum_{c\in C}\left(\prod_{j \in \{1\}}\sigma_j(c_j|t_j)\right)\tau_2(c_2)u_2(c,t)\\
=\ &amp;\text{argmax}_{\tau_2 \in \Delta(C_2)} p\cdot(\tau_2(x_2)-\tau_2(y_2))\\
=\ &amp;(\tau_2(x_2)=1,\tau_2(y_2)=0)\\
=\ &amp;[x_2].
\end{align*}</span>
Thus, <span class="math-container">$\sigma_2(\cdot|2.0)=[x_2]$</span>, <span class="math-container">$\sigma_1(\cdot|1.1)=[x_1]$</span>, and <span class="math-container">$\sigma_1(\cdot|1.2) = \frac{1}{2}[x_1]+\frac{1}{2}[y_1]$</span> is a Bayesian equilibrium.</p>
<p><em><strong>Case 3:</strong></em><span class="math-container">$\quad$</span><em><strong>[The second pure equilibrium for matrix <span class="math-container">$\mathbf{t_1=}\ $</span>1.1]</strong></em></p>
<p>Since
<span class="math-container">\begin{align*}
p_2(1.1|2.0)\cdot(0+1\cdot\tau_2(x_2)\cdot(-1)+0+1\cdot\tau_2(y_2)\cdot1) = p(\tau_2(y_2)-\tau_2(x_2)),
\end{align*}</span>
it follows that
<span class="math-container">\begin{align*}
&amp;\text{argmax}_{\tau_2 \in \Delta(C_2)} \sum_{t_{1} \in T_{1}} p_2(t_{1}|t_2) \sum_{c\in C}\left(\prod_{j \in \{1\}}\sigma_j(c_j|t_j)\right)\tau_2(c_2)u_2(c,t)\\
=\ &amp;\text{argmax}_{\tau_2 \in \Delta(C_2)} p\cdot(\tau_2(y_2)-\tau_2(x_2))\\
=\ &amp;(\tau_2(x_2)=0,\tau_2(y_2)=1)\\
=\ &amp;[y_2].
\end{align*}</span>
Thus, <span class="math-container">$\sigma_2(\cdot|2.0)=[y_2]$</span>, <span class="math-container">$\sigma_1(\cdot|1.1)=[y_1]$</span>, and <span class="math-container">$\sigma_1(\cdot|1.2) = \frac{1}{2}[x_1]+\frac{1}{2}[y_1]$</span> is a Bayesian equilibrium.</p>
<hr />
<h2>Method 2<span class="math-container">$\quad$</span> [William Spaniel]</h2>
<p>You may refer to this short <a href="https://www.youtube.com/watch?v=E0_CA9TwZ8c&amp;list=PLKI1h_nAkaQoDzI4xDIXzx6U2ergFmedo&amp;index=67" rel="nofollow noreferrer">video</a> for his presentation. The idea of this method is to transfer the original type-agent representation to an &quot;averaged&quot; game matrix, and find the Nash equilibria for the new matrix.</p>
<p>The new &quot;averaged&quot; game matrix I got is the following:
<a href="https://i.sstatic.net/TEikN.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/TEikN.png" alt="enter image description here" /></a>
But how should I proceed next? What bothers me is the variable <span class="math-container">$p$</span>. For example, when <span class="math-container">$p=1$</span>, then there are four pure strategy Nash equilibria: <span class="math-container">$(x_1x_1,x_2), (x_1y_1,x_2), (y_1x_1,y_2), (y_1y_1,y_2)$</span>. But I don't think any of the Bayesian equilibria computed in Method 1 is compatible with them. I am baffled.</p>
<hr />
<h2>My Problem</h2>
<p>It might be that my Method 1 has some problems (I'm not sure). But I just didn't think I can get the same result using the two methods. And I don't know what to do with method 2. Could someone please help me out?</p>
<hr />
<h2>Related Definition from Myerson's Text</h2>
<blockquote>
<p><strong>Definition of A Bayesian Game</strong><span class="math-container">$\quad$</span> To define a <em>Bayesian game</em> (or a <em>game in Bayesian form</em>), we must specify a set of players <span class="math-container">$N$</span> and, for each player <span class="math-container">$i$</span> in <span class="math-container">$N$</span>, we must specify a set of possible <em>actions</em> <span class="math-container">$C_i$</span>, a set of possible <em>types</em> <span class="math-container">$T_i$</span>, a probability function <span class="math-container">$p_i$</span>, and a utility function <span class="math-container">$u_i$</span>. We let
<span class="math-container">\begin{align*}
    C = \mathop{\Large\times}_{i \in N} C_i,\quad T = \mathop{\Large\times}_{i \in N} T_i.
\end{align*}</span>
That is, <span class="math-container">$C$</span> is the set of all possible profiles or combinations of actions that the players may use in the game, and <span class="math-container">$T$</span> is the set of all possible profiles or combinations of types that the player may have in the game. For each player <span class="math-container">$i$</span>, we let <span class="math-container">$T_{-i}$</span> denote the set of all possible combinations of types for the players other than <span class="math-container">$i$</span>, that is,
<span class="math-container">\begin{align*}
    T_{-i} = \mathop{\Large\times}_{j \in N-i} T_j,
\end{align*}</span></p>
<p>The probability function <span class="math-container">$p_i$</span> in the Bayesian game must be a function from <span class="math-container">$T_i$</span> into <span class="math-container">$\Delta(T_{-i})$</span>, the set of probability distributions over <span class="math-container">$T_{-i}$</span>. That is, for any possible type <span class="math-container">$t_i$</span> in <span class="math-container">$T_i$</span>, the probability function must specify a probability distribution <span class="math-container">$p(\cdot|t_i)$</span> over the set <span class="math-container">$T_{-i}$</span>, representing what player <span class="math-container">$i$</span> would believe about the other players' types if his own type were <span class="math-container">$t_i$</span>. Thus, for any <span class="math-container">$t_{-i}$</span> in <span class="math-container">$T_{-i}$</span>, <span class="math-container">$p_i(t_{-i}|t_i)$</span> denotes the subjective probability that <span class="math-container">$i$</span> would assign to the event that <span class="math-container">$t_{-i}$</span> is the actual profile of types for the other players, if his own type were <span class="math-container">$t_i$</span>.</p>
<p>For any player <span class="math-container">$i$</span> in <span class="math-container">$N$</span>, the utility function <span class="math-container">$u_i$</span> in the Bayesian game must be a function from <span class="math-container">$C \times T$</span> into the real numbers <span class="math-container">$\mathbb{R}$</span>. That is, for any profile of actions and types <span class="math-container">$(c,t)$</span> in <span class="math-container">$C \times T$</span>, the function <span class="math-container">$u_i$</span> must specify a number <span class="math-container">$u_i(c,t)$</span> that represents the payoff that player <span class="math-container">$i$</span> would get, in some von Neumann-Morgenstern utility scale, if the players' actual types were all as in <span class="math-container">$t$</span> and the players all chose their actions as specified in <span class="math-container">$c$</span>.</p>
<p>These structures together define a Bayesian game <span class="math-container">$\Gamma^b$</span>, so we may write
<span class="math-container">\begin{equation}
    \Gamma^b = \left(N, (C_i)_{i \in N}, (T_i)_{i \in N}, (u_i)_{i \in N}\right).
\end{equation}</span>
<span class="math-container">$\Gamma^b$</span> is <em>finite</em> if and only if the sets <span class="math-container">$N$</span> and <span class="math-container">$C_i$</span> and <span class="math-container">$T_i$</span> (for every <span class="math-container">$i$</span>) are all finite.</p>
</blockquote>
<blockquote>
<p><strong>Definition of Bayesian Nash Equilibrium</strong><span class="math-container">$\quad$</span> To formally state the definition of a Bayesian equilibrium, let <span class="math-container">$\Gamma^b$</span> be a Bayesian game, so
<span class="math-container">\begin{align*}
    \Gamma^b = (N, (C_i)_{i \in N}, (T_i)_{i \in N}, (p_i)_{i \in N}, (u_i)_{i \in N}).
\end{align*}</span>
A <em>randomized-strategy profile</em> for the Bayesian game <span class="math-container">$\Gamma^b$</span> is any <span class="math-container">$\sigma$</span> in the set <span class="math-container">$\mathop{\Large\times}_{i \in N} \mathop{\Large\times}_{t_i \in T_i} \Delta(C_i)$</span>, that is, any <span class="math-container">$\sigma$</span> such that
<span class="math-container">\begin{align*}
    &amp;\sigma = \left((\sigma_i(c_i|t_i))_{c_i \in C_i}\right)_{t_i \in T_i,i \in N},\\
    \\
    &amp;\sigma_i(c_i|t_i) \geq 0,\quad \forall c_i \in C_i,\quad \forall t_i \in T_i,\quad \forall i \in N,\quad \text{and}\\
    \\
    &amp;\sum_{c_i \in C_i} \sigma_i(c_i|t_i) = 1,\quad \forall t_i \in T_i,\quad \forall i \in N.
\end{align*}</span>
In such a randomized-strategy profile <span class="math-container">$\sigma$</span>, the number <span class="math-container">$\sigma_i(c_i|t_i)$</span> represents the conditional probability that player <span class="math-container">$i$</span> would use action <span class="math-container">$c_i$</span> if his type were <span class="math-container">$t_i$</span>. In the randomized-strategy profile <span class="math-container">$\sigma$</span>, the randomized strategy for type <span class="math-container">$t_i$</span> of player <span class="math-container">$i$</span> is
<span class="math-container">\begin{align*}
    \sigma_i(\cdot|t_i) = (\sigma_i(c_i|t_i))_{c_i \in C_i}.
\end{align*}</span>
A Bayesian equilibrium of the game <span class="math-container">$\Gamma^b$</span> is any randomized-strategy profile <span class="math-container">$\sigma$</span> such that, for every player <span class="math-container">$i$</span> in <span class="math-container">$N$</span> and every type <span class="math-container">$t_i$</span> in <span class="math-container">$T_i$</span>,
<a href="https://i.sstatic.net/Sm6df.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Sm6df.png" alt="enter image description here" /></a></p>
</blockquote>
<hr />
<h2>An Example from William Spaniel's Lecture</h2>
<p>Here is the game matrix he presented in his lecture:
<a href="https://i.sstatic.net/iTKJt.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iTKJt.png" alt="enter image description here" /></a>
With probability <span class="math-container">$0.2$</span>, player 2 would be of type 2.1. With probability <span class="math-container">$0.8$</span>, player 2 would be of type 2.2.</p>
<p>In order to compute the Bayesian equilibria, William converted the original type-agent representation into a &quot;weighted-averaged&quot; game matrix. We got rid of the different types of player 2 and combined them into just a single player. So we have the following four columns: <span class="math-container">$LL$</span>, <span class="math-container">$LR$</span>, <span class="math-container">$RL$</span>, <span class="math-container">$LL$</span>. The interpretation is, for example, the <span class="math-container">$LL$</span> column stands for both type 2.1 and type 2.2 are going to choose &quot;Left&quot;; the <span class="math-container">$LR$</span> column stand for just type 2.1 is going to choose &quot;Left&quot; and just type 2.2 is going to choose &quot;Right&quot;. In order to fill in the payoff matrix, we calculate the expected payoff for each player in each cell, and we will get the following matrix:
<a href="https://i.sstatic.net/3dRA1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3dRA1.png" alt="enter image description here" /></a>
Notice that the first column is strongly dominated by the third column, and that the second column is strongly dominated by the fourth column. So, we can reduce this <span class="math-container">$2\times 4$</span> matrix to the following <span class="math-container">$2 \times 2$</span> matrix:
<a href="https://i.sstatic.net/wad8aiY8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wad8aiY8.png" alt="enter image description here" /></a>
Compute the Nash equilibrium for it, we have</p>
<ol>
<li><span class="math-container">$(\text{[Up]}, \text{[RL]})$</span></li>
<li><span class="math-container">$(\text{[Down]}, \text{[RR]})$</span>, and</li>
<li><span class="math-container">$(\frac{1}{2}\text{[Up]}+\frac{1}{2}\text{[down]}, \frac{5}{8}\text{[RL]}+\frac{3}{8}\text{[RR]})$</span>.</li>
</ol>
<p>The interpretation of the three Bayesian Nash equilibria is the following, respectively:</p>
<ol>
<li>There is a Bayesian Nash equilibrium in which player 1 chooses &quot;<span class="math-container">$\text{Up}$</span>&quot;,  player 2 as type 2.1 chooses &quot;<span class="math-container">$\text{Right}$</span>&quot;, and player 2 as type 2.2 chooses &quot;<span class="math-container">$\text{Left}$</span>&quot;.</li>
<li>There is a Bayesian Nash equilibrium in which player 1 chooses &quot;<span class="math-container">$\text{Down}$</span>&quot; and player 2 as both type 2.1 and 2.2 choose &quot;<span class="math-container">$\text{Right}$</span>&quot;.</li>
<li>There is a mixed strategy Bayesian Nash equilibrium in which player 1 chooses &quot;<span class="math-container">$\text{Up}$</span>&quot; with probability <span class="math-container">$\frac{1}{2}$</span> and &quot;<span class="math-container">$\text{Down}$</span>&quot; with probability <span class="math-container">$\frac{1}{2}$</span>, player 2 as type 2.1 chooses &quot;<span class="math-container">$\text{Right}$</span>&quot; with probability <span class="math-container">$1$</span> (because <span class="math-container">$\frac{5}{8}+\frac{3}{8}=1$</span>), and player 2 as type 2.2 chooses &quot;<span class="math-container">$\text{Left}$</span>&quot; with probability <span class="math-container">$\frac{5}{8}$</span> and &quot;<span class="math-container">$\text{Right}$</span>&quot; with probability <span class="math-container">$\frac{3}{8}$</span>.</li>
</ol>
<hr />
<p><em><strong>Please help! Pretty Please! Thank you very much!</strong></em></p>

CC BY-SA 4.0
2024-04-23T14:19:34.143Z
I updated the game table in the beginning. It should be good now.
Beerus
CC BY-SA 4.0
2024-04-23T14:38:19.320Z
I added the definition of a Bayesian game as well as a Bayesian Nash equilibrium as presented in Myerson's text.
Beerus
CC BY-SA 4.0
2024-04-23T15:37:58.497Z
I added an example to show how William Spaniel's method works.
Beerus
CC BY-SA 4.0
2024-04-23T20:29:44.763Z
Some notes: 1. Your question is very long, I would try to get rid of the unnecessary stuff. It seems there is even a video? 2. In "An Example from William Spaniel's Lecture" the type probabilities are given as specific numbers, not as parameters. This enables him to do elimination of dominated strategies.
Giskard
CC BY-SA 4.0
2024-04-23T20:47:42.713Z
@Giskard Thanks a lot for your comment and answer below! I realized that my question is very long, I apologize for that. This is because I feel (1) Myerson's presentation of Bayesian game and Bayesian equilibrium may not be the most popular one taught in school, and (2) yes the Spaniel's method does come from his video lecture. That is why I added the example in his lecture at the bottom so you don't have to go through his video.
Beerus
<microeconomics><mathematical-economics><game-theory><nash-equilibrium><bayesian-game>
About Two Methods of Computing Bayesian Equilibria
Beerus
# About Two Methods of Computing Bayesian Equilibria

(Asked by: username_0 on 2024-04-17)

## Question

I want to compute the Bayesian equilibria for the following Bayesian game:
[![enter image description here](https://i.sstatic.net/6xrCd.png)](https://i.sstatic.net/6xrCd.png)
With probability $p$, player 1 would be of type 1.1. With probability $1-p$, player 1 would be of type 1.2. Player 2 does not have any private information and she only has one type 2.0.

I know two methods to compute Bayesian equilibria. The first one, with which I think I am more familiar, is based on the presentation from Roger Myerson's *Game Theory: Analysis of conflict*. The other one is presented by William Spaniel's *Game Theory 101* Youtube course. I want to use both method to solve this question and, of course, I expect to get the same result. But I got some trouble when proceeding William Spaniel's method. I would really appreciate it if someone could help me figure out where my problem is!

---

## Method 1$\quad$[Roger Myerson]

This is the same answer I worte for [this MSE post](https://math.stackexchange.com/questions/4899648/characterize-the-mixed-strategy-bayesian-nash-equilibria-for-this-game/4900405#4900405). But I will present it here as well.

Note that a Bayesian equilibrium of the game $\Gamma^b$ is any randomized-strategy profile $\sigma$ such that, for every player $i$ in $N$ and every type $t\_i$ in $T\_i$,
\begin{align\*}
\sigma\_i(\cdot|t\_i) \in \text{argmax}\_{\tau\_i \in \Delta(C\_i)} \sum\_{t\_{-i} \in T\_{-i}} p\_i(t\_{-i}|t\_i) \sum\_{c\in C}\left(\prod\_{j \in N-i}\sigma\_j(c\_j|t\_j)\right)\tau\_i(c\_i)u\_i(c,t).
\end{align\*}
As for our game, we make the following definition:

> $C\_1 = \{x\_1,y\_1\}$, where $x\_1$ means "up" and $y\_1$ means "down".
>
> $C\_2 = \{x\_2,y\_2\}$, where $x\_2$ means "left" and $y\_2$ means "right".
>
> $C = C\_1\mathop{\Large\times}C\_2$
>
> $\Delta(C)$ is the set of all probability distributions over the set $C$
>
> $T\_1 = \{1.1,1.2\}$, where 1.1 stands for player 1's type of coorperative and 1.2 stands for player 1's type of uncooperative.
>
> $T\_2 = \{2.0\}$, where 2.0 stands for player 2's only possible type.
>
> $T = T\_1\mathop{\Large\times}T\_2$
>
> $p\_2(1.1|2.0) = p$ and $p\_2(1.2|2.0) = 1-p$, where $p\_2(\cdot|2.0)$ represents what player 2 would believe about player 1's types given his own type being 2.0.

First, consider the matrix for $t\_1=1.1$. Clearly, there are two pure strategy Nash equilibria:
\begin{align\*}
\sigma\_1(\cdot|1.1)=[x\_1],\ \sigma\_2=[x\_2]\quad \text{and}\quad \sigma\_1(\cdot|1.1)=[y\_1],\ \sigma\_2=[y\_2],
\end{align\*}
and one mixed strategy Nash equilibrium:
\begin{align\*}
\sigma\_1(\cdot|1.1)=\frac{1}{2}[x\_1]+\frac{1}{2}[y\_1],\ \sigma\_2 = \frac{1}{2}[x\_2] + \frac{1}{2}[y\_2].
\end{align\*}

Next, consider the matrix for $t\_1 = 1.2$. Clearly, there is only one Nash equilibrium:
\begin{align\*}
\sigma\_1(\cdot|1.2)=\frac{1}{2}[x\_1]+\frac{1}{2}[y\_1],\ \sigma\_2 = \frac{1}{2}[x\_2] + \frac{1}{2}[y\_2].
\end{align\*}

Therefore, we have the following three cases:

***Case 1:***$\quad$***[Mixed strategy Nash equilibria for both matrices]***

Since
\begin{align\*}
\begin{alignedat}{4}
&p\_2(1.1|2.0) \cdot (&&\sigma\_1(x\_1|1.1) \cdot \tau\_2(x\_2) \cdot u\_1((x\_1,x\_2),(1.1,2.0)) &\ +\\
& &&\sigma\_1(y\_1|1.1) \cdot \tau\_2(x\_2) \cdot u\_1((y\_1,x\_2),(1.1,2.0)) &\ +\\
& &&\sigma\_1(x\_1|1.1) \cdot \tau\_2(y\_2) \cdot u\_1((x\_1,y\_2),(1.1,2.0)) &\ +\\
& &&\sigma\_1(y\_1|1.1) \cdot \tau\_2(y\_2) \cdot u\_1((y\_1,y\_2),(1.1,2.0)))&\\
=\ &0 && &
\end{alignedat}
\end{align\*}
and
\begin{align\*}
\begin{alignedat}{4}
&p\_2(1.2|2.0) \cdot (&&\sigma\_1(x\_1|1.2) \cdot \tau\_2(x\_2) \cdot u\_1((x\_1,x\_2),(1.2,2.0)) &\ +\\
& &&\sigma\_1(y\_1|1.2) \cdot \tau\_2(x\_2) \cdot u\_1((y\_1,x\_2),(1.2,2.0)) &\ +\\
& &&\sigma\_1(x\_1|1.2) \cdot \tau\_2(y\_2) \cdot u\_1((x\_1,y\_2),(1.2,2.0)) &\ +\\
& &&\sigma\_1(y\_1|1.2) \cdot \tau\_2(y\_2) \cdot u\_1((y\_1,y\_2),(1.2,2.0)))&\\
=\ &0, && &
\end{alignedat}
\end{align\*}
it follows that
\begin{align\*}
&\text{argmax}\_{\tau\_2 \in \Delta(C\_2)} \sum\_{t\_{1} \in T\_{1}} p\_2(t\_{1}|t\_2) \sum\_{c\in C}\left(\prod\_{j \in \{1\}}\sigma\_j(c\_j|t\_j)\right)\tau\_2(c\_2)u\_2(c,t)\\
=\ &\text{argmax}\_{\tau\_2 \in \Delta(C\_2)} 0\\
=\ &0.
\end{align\*}
Thus, $\sigma\_2(\cdot|2.0) = \alpha[x\_2] + (1-\alpha)[y\_2]$ for all $\alpha \in [0,1]$, $\sigma\_1(\cdot|1.1) = \frac{1}{2}[x\_1]+\frac{1}{2}[y\_1]$, and $\sigma\_1(\cdot|1.2) = \frac{1}{2}[x\_1] + \frac{1}{2}[y\_1]$ is a Bayesian equilibrium.

***Case 2:***$\quad$***[The first pure equilibrium for matrix $\mathbf{t\_1=}\ $1.1]***

Since
\begin{align\*}
p\_2(1.1|2.0)\cdot(1\cdot\tau\_2(x\_2)\cdot1+0+1\cdot\tau\_2(y\_2)\cdot(-1)+0) = p(\tau\_2(x\_2)-\tau\_2(y\_2)),
\end{align\*}
it follows that
\begin{align\*}
&\text{argmax}\_{\tau\_2 \in \Delta(C\_2)} \sum\_{t\_{1} \in T\_{1}} p\_2(t\_{1}|t\_2) \sum\_{c\in C}\left(\prod\_{j \in \{1\}}\sigma\_j(c\_j|t\_j)\right)\tau\_2(c\_2)u\_2(c,t)\\
=\ &\text{argmax}\_{\tau\_2 \in \Delta(C\_2)} p\cdot(\tau\_2(x\_2)-\tau\_2(y\_2))\\
=\ &(\tau\_2(x\_2)=1,\tau\_2(y\_2)=0)\\
=\ &[x\_2].
\end{align\*}
Thus, $\sigma\_2(\cdot|2.0)=[x\_2]$, $\sigma\_1(\cdot|1.1)=[x\_1]$, and $\sigma\_1(\cdot|1.2) = \frac{1}{2}[x\_1]+\frac{1}{2}[y\_1]$ is a Bayesian equilibrium.

***Case 3:***$\quad$***[The second pure equilibrium for matrix $\mathbf{t\_1=}\ $1.1]***

Since
\begin{align\*}
p\_2(1.1|2.0)\cdot(0+1\cdot\tau\_2(x\_2)\cdot(-1)+0+1\cdot\tau\_2(y\_2)\cdot1) = p(\tau\_2(y\_2)-\tau\_2(x\_2)),
\end{align\*}
it follows that
\begin{align\*}
&\text{argmax}\_{\tau\_2 \in \Delta(C\_2)} \sum\_{t\_{1} \in T\_{1}} p\_2(t\_{1}|t\_2) \sum\_{c\in C}\left(\prod\_{j \in \{1\}}\sigma\_j(c\_j|t\_j)\right)\tau\_2(c\_2)u\_2(c,t)\\
=\ &\text{argmax}\_{\tau\_2 \in \Delta(C\_2)} p\cdot(\tau\_2(y\_2)-\tau\_2(x\_2))\\
=\ &(\tau\_2(x\_2)=0,\tau\_2(y\_2)=1)\\
=\ &[y\_2].
\end{align\*}
Thus, $\sigma\_2(\cdot|2.0)=[y\_2]$, $\sigma\_1(\cdot|1.1)=[y\_1]$, and $\sigma\_1(\cdot|1.2) = \frac{1}{2}[x\_1]+\frac{1}{2}[y\_1]$ is a Bayesian equilibrium.

---

## Method 2$\quad$ [William Spaniel]

You may refer to this short [video](https://www.youtube.com/watch?v=E0_CA9TwZ8c&list=PLKI1h_nAkaQoDzI4xDIXzx6U2ergFmedo&index=67) for his presentation. The idea of this method is to transfer the original type-agent representation to an "averaged" game matrix, and find the Nash equilibria for the new matrix.

The new "averaged" game matrix I got is the following:
[![enter image description here](https://i.sstatic.net/TEikN.png)](https://i.sstatic.net/TEikN.png)
But how should I proceed next? What bothers me is the variable $p$. For example, when $p=1$, then there are four pure strategy Nash equilibria: $(x\_1x\_1,x\_2), (x\_1y\_1,x\_2), (y\_1x\_1,y\_2), (y\_1y\_1,y\_2)$. But I don't think any of the Bayesian equilibria computed in Method 1 is compatible with them. I am baffled.

---

## My Problem

It might be that my Method 1 has some problems (I'm not sure). But I just didn't think I can get the same result using the two methods. And I don't know what to do with method 2. Could someone please help me out?

---

## Related Definition from Myerson's Text

> **Definition of A Bayesian Game**$\quad$ To define a *Bayesian game* (or a *game in Bayesian form*), we must specify a set of players $N$ and, for each player $i$ in $N$, we must specify a set of possible *actions* $C\_i$, a set of possible *types* $T\_i$, a probability function $p\_i$, and a utility function $u\_i$. We let
> \begin{align\*}
> C = \mathop{\Large\times}\_{i \in N} C\_i,\quad T = \mathop{\Large\times}\_{i \in N} T\_i.
> \end{align\*}
> That is, $C$ is the set of all possible profiles or combinations of actions that the players may use in the game, and $T$ is the set of all possible profiles or combinations of types that the player may have in the game. For each player $i$, we let $T\_{-i}$ denote the set of all possible combinations of types for the players other than $i$, that is,
> \begin{align\*}
> T\_{-i} = \mathop{\Large\times}\_{j \in N-i} T\_j,
> \end{align\*}
>
> The probability function $p\_i$ in the Bayesian game must be a function from $T\_i$ into $\Delta(T\_{-i})$, the set of probability distributions over $T\_{-i}$. That is, for any possible type $t\_i$ in $T\_i$, the probability function must specify a probability distribution $p(\cdot|t\_i)$ over the set $T\_{-i}$, representing what player $i$ would believe about the other players' types if his own type were $t\_i$. Thus, for any $t\_{-i}$ in $T\_{-i}$, $p\_i(t\_{-i}|t\_i)$ denotes the subjective probability that $i$ would assign to the event that $t\_{-i}$ is the actual profile of types for the other players, if his own type were $t\_i$.
>
> For any player $i$ in $N$, the utility function $u\_i$ in the Bayesian game must be a function from $C \times T$ into the real numbers $\mathbb{R}$. That is, for any profile of actions and types $(c,t)$ in $C \times T$, the function $u\_i$ must specify a number $u\_i(c,t)$ that represents the payoff that player $i$ would get, in some von Neumann-Morgenstern utility scale, if the players' actual types were all as in $t$ and the players all chose their actions as specified in $c$.
>
> These structures together define a Bayesian game $\Gamma^b$, so we may write
> \begin{equation}
> \Gamma^b = \left(N, (C\_i)\_{i \in N}, (T\_i)\_{i \in N}, (u\_i)\_{i \in N}\right).
> \end{equation}
> $\Gamma^b$ is *finite* if and only if the sets $N$ and $C\_i$ and $T\_i$ (for every $i$) are all finite.

> **Definition of Bayesian Nash Equilibrium**$\quad$ To formally state the definition of a Bayesian equilibrium, let $\Gamma^b$ be a Bayesian game, so
> \begin{align\*}
> \Gamma^b = (N, (C\_i)\_{i \in N}, (T\_i)\_{i \in N}, (p\_i)\_{i \in N}, (u\_i)\_{i \in N}).
> \end{align\*}
> A *randomized-strategy profile* for the Bayesian game $\Gamma^b$ is any $\sigma$ in the set $\mathop{\Large\times}\_{i \in N} \mathop{\Large\times}\_{t\_i \in T\_i} \Delta(C\_i)$, that is, any $\sigma$ such that
> \begin{align\*}
> &\sigma = \left((\sigma\_i(c\_i|t\_i))\_{c\_i \in C\_i}\right)\_{t\_i \in T\_i,i \in N},\\
> \\
> &\sigma\_i(c\_i|t\_i) \geq 0,\quad \forall c\_i \in C\_i,\quad \forall t\_i \in T\_i,\quad \forall i \in N,\quad \text{and}\\
> \\
> &\sum\_{c\_i \in C\_i} \sigma\_i(c\_i|t\_i) = 1,\quad \forall t\_i \in T\_i,\quad \forall i \in N.
> \end{align\*}
> In such a randomized-strategy profile $\sigma$, the number $\sigma\_i(c\_i|t\_i)$ represents the conditional probability that player $i$ would use action $c\_i$ if his type were $t\_i$. In the randomized-strategy profile $\sigma$, the randomized strategy for type $t\_i$ of player $i$ is
> \begin{align\*}
> \sigma\_i(\cdot|t\_i) = (\sigma\_i(c\_i|t\_i))\_{c\_i \in C\_i}.
> \end{align\*}
> A Bayesian equilibrium of the game $\Gamma^b$ is any randomized-strategy profile $\sigma$ such that, for every player $i$ in $N$ and every type $t\_i$ in $T\_i$,
> [![enter image description here](https://i.sstatic.net/Sm6df.png)](https://i.sstatic.net/Sm6df.png)

---

## An Example from William Spaniel's Lecture

Here is the game matrix he presented in his lecture:
[![enter image description here](https://i.sstatic.net/iTKJt.png)](https://i.sstatic.net/iTKJt.png)
With probability $0.2$, player 2 would be of type 2.1. With probability $0.8$, player 2 would be of type 2.2.

In order to compute the Bayesian equilibria, William converted the original type-agent representation into a "weighted-averaged" game matrix. We got rid of the different types of player 2 and combined them into just a single player. So we have the following four columns: $LL$, $LR$, $RL$, $LL$. The interpretation is, for example, the $LL$ column stands for both type 2.1 and type 2.2 are going to choose "Left"; the $LR$ column stand for just type 2.1 is going to choose "Left" and just type 2.2 is going to choose "Right". In order to fill in the payoff matrix, we calculate the expected payoff for each player in each cell, and we will get the following matrix:
[![enter image description here](https://i.sstatic.net/3dRA1.png)](https://i.sstatic.net/3dRA1.png)
Notice that the first column is strongly dominated by the third column, and that the second column is strongly dominated by the fourth column. So, we can reduce this $2\times 4$ matrix to the following $2 \times 2$ matrix:
[![enter image description here](https://i.sstatic.net/wad8aiY8.png)](https://i.sstatic.net/wad8aiY8.png)
Compute the Nash equilibrium for it, we have

1. $(\text{[Up]}, \text{[RL]})$
2. $(\text{[Down]}, \text{[RR]})$, and
3. $(\frac{1}{2}\text{[Up]}+\frac{1}{2}\text{[down]}, \frac{5}{8}\text{[RL]}+\frac{3}{8}\text{[RR]})$.

The interpretation of the three Bayesian Nash equilibria is the following, respectively:

1. There is a Bayesian Nash equilibrium in which player 1 chooses "$\text{Up}$", player 2 as type 2.1 chooses "$\text{Right}$", and player 2 as type 2.2 chooses "$\text{Left}$".
2. There is a Bayesian Nash equilibrium in which player 1 chooses "$\text{Down}$" and player 2 as both type 2.1 and 2.2 choose "$\text{Right}$".
3. There is a mixed strategy Bayesian Nash equilibrium in which player 1 chooses "$\text{Up}$" with probability $\frac{1}{2}$ and "$\text{Down}$" with probability $\frac{1}{2}$, player 2 as type 2.1 chooses "$\text{Right}$" with probability $1$ (because $\frac{5}{8}+\frac{3}{8}=1$), and player 2 as type 2.2 chooses "$\text{Left}$" with probability $\frac{5}{8}$ and "$\text{Right}$" with probability $\frac{3}{8}$.

---

***Please help! Pretty Please! Thank you very much!***

**Tags:** microeconomics, mathematical-economics, game-theory, nash-equilibrium, bayesian-game
> **Comments:**
> **username_0**: I updated the game table in the beginning. It should be good now.  
> **username_0**: I added the definition of a Bayesian game as well as a Bayesian Nash equilibrium as presented in Myerson's text.  
> **username_0**: I added an example to show how William Spaniel's method works.  
> **username_1**: Some notes: 1. Your question is very long, I would try to get rid of the unnecessary stuff. It seems there is even a video? 2. In "An Example from William Spaniel's Lecture" the type probabilities are given as specific numbers, not as parameters. This enables him to do elimination of dominated strategies.  
> **username_0**: @Giskard Thanks a lot for your comment and answer below! I realized that my question is very long, I apologize for that. This is because I feel (1) Myerson's presentation of Bayesian game and Bayesian equilibrium may not be the most popular one taught in school, and (2) yes the Spaniel's method does come from his video lecture. That is why I added the example in his lecture at the bottom so you don't have to go through his video.  

---

## 1 answers

### Answer by username_1 on 2024-04-23. Score: 1

(I only parsed what you did, perhaps I did not get it.)

You seem to be overcomplicating things a bit. A good starting point seems to be whether player 2 will mix or not.

1. If not, you can see that player 1 will have a clear best response in each matrix, so you will have a pure equilibrium. You only have to validate that player 2's not mixed strategy is a best response to the expected strategy that player 1's types are playing.
2. If player 2 is mixing, then you can write down an equation showing their expected payoff as a function of player 1's strategy:
   \begin{align\*}
   EU\_2(\text{P1 strategy},x\_2) & =
   (p \cdot q\_{11x} + (1-p) \cdot q\_{12x}) \cdot 1 + (p \cdot q\_{11y} + (1-p) \cdot q\_{12y}) \cdot (-1) = 0.
   \end{align\*}
   It equals zero because
   $$EU\_2(\text{P1 strategy},x\_2) = - EU\_2(\text{P1 strategy},y\_2),$$
   so if they are mixing both $x\_2$ and $y\_2$ than their expected payoff is 0. Unless player 2 is mixing with 50-50% probability, player 1 will not mix. So this gives you the unlikely case that   
   a. player 2 is mixing, player 1 is not, but $p$ is such that things work out, and the more likely case that   
   b. player 2 is mixing with 50-50% and you can adjust player 1's strategy however you like, as long as player 2's expected payoff is 0.


---

