Bayesian Nash Equilibrium in a Duopoly Cournot Competition

Question

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. $$\hat{q}_1 =\frac{M-m}{3}-\frac{\theta (\alpha M-M)}{3}$$ $$\hat{q}_{2L} =\frac{M-m}{3}+\frac{\theta (\alpha M-M)}{6}$$ $$\hat{q}_{2H}^M = \frac{\alpha M-m}{3}+\frac{(1+\theta)(\alpha M-M)}{6}$$

Answer 1

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.