<p>If the capacities <span class="math-container">$q_1$</span> and <span class="math-container">$q_2$</span> are the same, say <span class="math-container">$\overline{q}$</span> you can rule out some cases.</p>
<p>The best response functions are given by:
<span class="math-container">$$
\begin{align*}
\hat q_1 &amp;= \min\left\{\frac{M - (1-\theta) \hat q_2^L - \theta \hat q_2^H - m}{2}, \overline{q}\right\},\\
\hat q_2^L &amp;= \min\left\{\frac{M - \hat q_1 - m}{2}, \overline{q}\right\},\\
\hat q_2^H &amp;= \min\left\{\frac{\alpha M - \hat q_1 - m}{2}, \overline{q}\right\}
\end{align*}
$$</span></p>
<p>First we can show that <span class="math-container">$\hat q_2^H \ge \hat q_2^L$</span>.</p>
<p><strong>Proposition:</strong> <span class="math-container">$\hat q_2^H \ge \hat q_2^L$</span></p>
<blockquote>
<p><em>Proof:</em> If <span class="math-container">$\hat q_2^H = \overline{q}$</span> the proof is obvious. Else, from <span class="math-container">$\alpha &gt; 0$</span>, we have:
<span class="math-container">$$
\hat q_2^H = \frac{\alpha M - \hat q_1 - m}{2} \ge \frac{M - \hat q_1 - m}{2} \ge q_2^L.
$$</span></p>
</blockquote>
<p>Next, we can also show that <span class="math-container">$\hat q_2^L \ge \hat q_1$</span>.</p>
<p><strong>Proposition:</strong> <span class="math-container">$\hat q_2^L \ge \hat q_1$</span>.</p>
<blockquote>
<p><em>Proof:</em> if <span class="math-container">$\hat q_2^L = \overline{q}$</span> the proof is obvious. If not, then:
<span class="math-container">$$
\begin{align*}
\hat q_1 - \hat q_2^L &amp;\le \frac{M - (1-\theta) \hat q_2^L - \theta \hat q_2^H - m}{2} - \hat q_2^L,\\
&amp;= \frac{M - (1-\theta) \hat q_2^L - \theta \hat q_2^H - m}{2} - \frac{M - \hat q_1 - m}{2},\\
&amp;= \frac{-(1-\theta) \hat q_2^L - \theta \hat q_2^H + \hat q_1}{2},\\
\to 2 \hat q_1 - 2 \hat q_2^L &amp;\le -(1- \theta) \hat q_2^L - \theta \hat q_2^H + \hat q_1 \\
\to \hat q_1 &amp;\le \hat q_2^L - \theta(\hat q_2^H - \hat q_2^L) \le \hat q_2^L.
\end{align*}
$$</span>
where the last line follows from <span class="math-container">$\hat q_2^H \ge \hat q_2^L$</span>.</p>
</blockquote>
<p>The two propositions show that we only have to consider the following cases:</p>
<ol>
<li><span class="math-container">$\hat q_2^H &lt; \overline{q}$</span> and <span class="math-container">$\hat q_2^L &lt; \overline{q}$</span> and <span class="math-container">$\hat q_1 &lt; \overline{q}$</span></li>
<li><span class="math-container">$\hat q_2^H = \overline{q}$</span> and <span class="math-container">$\hat q_2^L &lt; \overline{q}$</span> and <span class="math-container">$\hat q_1 &lt; \overline{q}$</span>.</li>
<li><span class="math-container">$\hat q_2^H = \hat q_2^L = \overline{q}$</span> and <span class="math-container">$\hat q_1 &lt; \overline{q}$</span>.</li>
<li><span class="math-container">$\hat q_2^H = \hat q_2^L = \hat q_1 = \overline{q}$</span>.</li>
</ol>
<p>For each of them you can compute the equilibrium values and then you need to check if they indeed correspond to the best responses.
This reduces the number of cases from 8 to 4. I don't think you can do better.</p>

CC BY-SA 4.0
2021-05-05T00:26:56.170Z
Thanks for your detailed answer. I agree with the two propositions your derived and it intuitively makes sense. Unfortunately, in my problem setting, q_1 and q_2 are not the same. This is a two stage problem and I am using a backward solution approach. The setup I described is firms' second stage problem. In the first they determine production quantities in a Nash equilibrium setting.
EagleEdge0423
CC BY-SA 4.0
2021-05-05T00:28:41.800Z
One more question. What do you mean by saying that I need to check if they indeed correspond to the best responses?
EagleEdge0423
CC BY-SA 4.0
2021-05-05T05:13:03.837Z
I meant that: once you solved for $\hat q_1, \hat q_2^L$ and $\hat q_3^L$, you need to verify that they are indeed smaller than the capacities $\overline{q}$.
tdm
CC BY-SA 4.0
2021-05-04T18:15:46.963Z
2021-05-04T18:15:46.963Z
tdm
<p>I am having a hard time to solve a Bayesian Nash equilibrium game in a duopoly cournot competition setting. So, I have two firms with <strong>given</strong> production quantities, let's say <span class="math-container">$q_1$</span> and <span class="math-container">$q_2$</span> (i.e., not a decision). Firm 1's product is low quality whereas Firm 2's product is high quality with probability <span class="math-container">$\theta$</span> and low quality with <span class="math-container">$1-\theta$</span>. As a result, different quality products have different market potential for Firm 2. The way I set up the price function for low quality is
<span class="math-container">$$M-q_1-q_2$$</span>
For high quality products, there is a price premium such that the price for high quality is given by <span class="math-container">$$\alpha M-q_1-q_2$$</span> where <span class="math-container">$\alpha&gt;1$</span>. Another different aspect of the problem is that companies first sell products in the primary market under the price mechanism specified above. There exist a secondary market. It works in a way such that if any firm sells his (remaining) products there, he gets a guaranteed price <span class="math-container">$m$</span> (<span class="math-container">$M &gt; m$</span>).
Currently my solution involves 8 different cases. I solve Bayesian Nash equilibrium by solving three profit functions simultaneously to find the quantities to supply in the primary market. Since <span class="math-container">$q_1$</span> and <span class="math-container">$q_2$</span> are treated fixed, I need to have a <span class="math-container">$\min \{q_i, \hat{q}_i\}$</span> for <span class="math-container">$i=1, 2H, 2L$</span>, where <span class="math-container">$\hat{q}_i$</span> is the equilibrium solution for the primary market. But I am trying to narrow the cases down to a small set so that I can continue my other analysis with the solution. Accordingly, I have two questions: <strong>1. Is my current solution correct? 2. How can I narrow cases down to a manageable number?</strong></p>
<p>If you wonder what my equilibrium solution looks like for primary market, here it is.
<span class="math-container">$$\hat{q}_1 =\frac{M-m}{3}-\frac{\theta (\alpha M-M)}{3}$$</span>
<span class="math-container">$$\hat{q}_{2L} =\frac{M-m}{3}+\frac{\theta (\alpha M-M)}{6}$$</span>
<span class="math-container">$$\hat{q}_{2H}^M = \frac{\alpha M-m}{3}+\frac{(1+\theta)(\alpha M-M)}{6}$$</span></p>

<game-theory><nash-equilibrium><bayesian-game>
Bayesian Nash Equilibrium in a Duopoly Cournot Competition
EagleEdge0423
# Bayesian Nash Equilibrium in a Duopoly Cournot Competition

(Asked by: username_0 on 2021-04-28)

I am having a hard time to solve a Bayesian Nash equilibrium game in a duopoly cournot competition setting. So, I have two firms with **given** production quantities, let's say $q\_1$ and $q\_2$ (i.e., not a decision). Firm 1's product is low quality whereas Firm 2's product is high quality with probability $\theta$ and low quality with $1-\theta$. As a result, different quality products have different market potential for Firm 2. The way I set up the price function for low quality is
$$M-q\_1-q\_2$$
For high quality products, there is a price premium such that the price for high quality is given by $$\alpha M-q\_1-q\_2$$ where $\alpha>1$. Another different aspect of the problem is that companies first sell products in the primary market under the price mechanism specified above. There exist a secondary market. It works in a way such that if any firm sells his (remaining) products there, he gets a guaranteed price $m$ ($M > m$).
Currently my solution involves 8 different cases. I solve Bayesian Nash equilibrium by solving three profit functions simultaneously to find the quantities to supply in the primary market. Since $q\_1$ and $q\_2$ are treated fixed, I need to have a $\min \{q\_i, \hat{q}\_i\}$ for $i=1, 2H, 2L$, where $\hat{q}\_i$ is the equilibrium solution for the primary market. But I am trying to narrow the cases down to a small set so that I can continue my other analysis with the solution. Accordingly, I have two questions: **1. Is my current solution correct? 2. How can I narrow cases down to a manageable number?**

If you wonder what my equilibrium solution looks like for primary market, here it is.
$$\hat{q}\_1 =\frac{M-m}{3}-\frac{\theta (\alpha M-M)}{3}$$
$$\hat{q}\_{2L} =\frac{M-m}{3}+\frac{\theta (\alpha M-M)}{6}$$
$$\hat{q}\_{2H}^M = \frac{\alpha M-m}{3}+\frac{(1+\theta)(\alpha M-M)}{6}$$

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


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## 1 answers

### Answer by username_1 on 2021-05-04. Score: 4

If the capacities $q\_1$ and $q\_2$ are the same, say $\overline{q}$ you can rule out some cases.

The best response functions are given by:
$$
\begin{align\*}
\hat q\_1 &= \min\left\{\frac{M - (1-\theta) \hat q\_2^L - \theta \hat q\_2^H - m}{2}, \overline{q}\right\},\\
\hat q\_2^L &= \min\left\{\frac{M - \hat q\_1 - m}{2}, \overline{q}\right\},\\
\hat q\_2^H &= \min\left\{\frac{\alpha M - \hat q\_1 - m}{2}, \overline{q}\right\}
\end{align\*}
$$

First we can show that $\hat q\_2^H \ge \hat q\_2^L$.

**Proposition:** $\hat q\_2^H \ge \hat q\_2^L$

> *Proof:* If $\hat q\_2^H = \overline{q}$ the proof is obvious. Else, from $\alpha > 0$, we have:
> $$
> \hat q\_2^H = \frac{\alpha M - \hat q\_1 - m}{2} \ge \frac{M - \hat q\_1 - m}{2} \ge q\_2^L.
> $$

Next, we can also show that $\hat q\_2^L \ge \hat q\_1$.

**Proposition:** $\hat q\_2^L \ge \hat q\_1$.

> *Proof:* if $\hat q\_2^L = \overline{q}$ the proof is obvious. If not, then:
> $$
> \begin{align\*}
> \hat q\_1 - \hat q\_2^L &\le \frac{M - (1-\theta) \hat q\_2^L - \theta \hat q\_2^H - m}{2} - \hat q\_2^L,\\
> &= \frac{M - (1-\theta) \hat q\_2^L - \theta \hat q\_2^H - m}{2} - \frac{M - \hat q\_1 - m}{2},\\
> &= \frac{-(1-\theta) \hat q\_2^L - \theta \hat q\_2^H + \hat q\_1}{2},\\
> \to 2 \hat q\_1 - 2 \hat q\_2^L &\le -(1- \theta) \hat q\_2^L - \theta \hat q\_2^H + \hat q\_1 \\
> \to \hat q\_1 &\le \hat q\_2^L - \theta(\hat q\_2^H - \hat q\_2^L) \le \hat q\_2^L.
> \end{align\*}
> $$
> where the last line follows from $\hat q\_2^H \ge \hat q\_2^L$.

The two propositions show that we only have to consider the following cases:

1. $\hat q\_2^H < \overline{q}$ and $\hat q\_2^L < \overline{q}$ and $\hat q\_1 < \overline{q}$
2. $\hat q\_2^H = \overline{q}$ and $\hat q\_2^L < \overline{q}$ and $\hat q\_1 < \overline{q}$.
3. $\hat q\_2^H = \hat q\_2^L = \overline{q}$ and $\hat q\_1 < \overline{q}$.
4. $\hat q\_2^H = \hat q\_2^L = \hat q\_1 = \overline{q}$.

For each of them you can compute the equilibrium values and then you need to check if they indeed correspond to the best responses.
This reduces the number of cases from 8 to 4. I don't think you can do better.

> **Comments:**
> **username_0**: Thanks for your detailed answer. I agree with the two propositions your derived and it intuitively makes sense. Unfortunately, in my problem setting, q_1 and q_2 are not the same. This is a two stage problem and I am using a backward solution approach. The setup I described is firms' second stage problem. In the first they determine production quantities in a Nash equilibrium setting.  
> **username_0**: One more question. What do you mean by saying that I need to check if they indeed correspond to the best responses?  
> **username_1**: I meant that: once you solved for $\hat q_1, \hat q_2^L$ and $\hat q_3^L$, you need to verify that they are indeed smaller than the capacities $\overline{q}$.  

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