A few game theory exercises and answers

Introduction

I want to work out a couple of exercises from Game Theory: an Introduction, by Tadelis. Since the exercises from later chapters are not available online, and since they represent some of the more creative and fun challenges, I just wanted to post a few exercises which I worked out, as well as some (unasked) variations.

Exercise 12.2: Cournot Revisited

This exercise asks the following:

Consider the Cournot duopoly model in which two firms, 1 and 2, simultaneously choose the quantities they supply, \(q_1\) and \(q_2\). The price each will face is determined by the market demand function \(p(q_1, q_2) = a - b(q_1 + q_2)\). Each firm has a probability \(\mu\) of having a marginal unit cost of \(c_L\) and a probability \(1-\mu\) of having a marginal unit cost of \(c_H\). These probabilities are common knowledge, but the true type is revealed only to each firm individually. Solve for the Bayesian Nash equilibrium.

The exercise makes it clear that there is uncertainty about the marginal cost parameter of both firms, but I will first start with a more simple question, which is that firm 2’s marginal cost parameter is uncertain (for player 1, but observable for themselves), but not vice versa: player 1 has a fixed marginal cost parameter, say \(c_1\).

Then, the profit function of player 2 is:

$$ \Pi_2 = \begin{cases} (a - b(q_1 + q_2^L))q_2^L - c_L q_2^L \hspace{2em} &\text{ if } c = c_L \newline (a - b(q_1 + q_2^H))q_2^H - c_H q_2^H \hspace{2em} &\text{ if } c = c_H \end{cases} $$

And the profit function for player 1 is:

$$ \Pi_1 = \mu((a - b(q_1 + q_2^L))q_1 - c_1 q_1) + (1-\mu) ((a - b(q_1 + q_2^H))q_1 - c_1 q_1) $$

From these profit functions, the best responses can be derived by taking the derivative with respect to the control variables. Player 1’s best response will be \(q_1 = f(q_2^L, q_2^H)\). Substituting player 2’s best responses \(q_2^L = f(q_1)\) and \(q_2^H = f(q_1)\) into player 1’s best responses yields a solution for \(q_1\). Substituting this solution in player 2’s best responses then yields a solution for both incarnations of player 2.

From this solution, it is also easy to see what the solution for the actual problem would look like: The profit function for player \(i\), given that their marginal costs are low and high respectively, is:

$$ \Pi_i^L = \begin{cases} (a - b(\mathbb{E}[q_j] + q_i^L)) \cdot q_i^L - c_L q_i^L &\text{ if MC = Low} \newline (a - b(\mathbb{E}[q_j] + q_i^H)) \cdot q_i^L - c_H q_i^H &\text{ if MC = High} \end{cases} $$

The expected value of the other player’s production \(\mathbb{E}[q_j]\) equals \(\mu q_j^L + (1-\mu)q_j^H\). Substituting this in the profit function, and then differentiating to find the Best Responses for player \(i\) gives:

$$ BR_i = \begin{cases} q_i^L = \frac{a - b(\mu q_j^L + (1-\mu) q_j^H) - c_L}{2b} \newline q_i^H = \frac{a - b(\mu q_j^L + (1-\mu) q_j^H) - c_H}{2b} \end{cases} $$

Now, recognizing that the problems are symmetric, we can impose that \(q_j^L = q_i^L\) and \(q_j^H = q_i^H\). Alternatively, we could just make explicit the best responses for player 2, and then realize that in a Bayesian Nash Equilibrium, each player’s best response should be a best response to the others. That would give us a system of 4 equations, and 4 unknowns. But, recognizing the symmetric nature of the problem, it is easy to make do with a system of only two equations and two unknowns. Substituting \(q_j^L = q_i^L\) and \(q_j^H = q_i^H\) into the best responses gives:

$$ BR_i = \begin{cases} q_i^L = \frac{a - b(\mu q_i^L + (1-\mu) q_i^H) - c_L}{2b} \newline q_i^H = \frac{a - b(\mu q_i^L + (1-\mu) q_i^H) - c_H}{2b} \end{cases} $$

After a bit of algebra, the solution to this problem is:

$$ \begin{bmatrix} q_i^L \newline q_i^H \end{bmatrix} = \begin{bmatrix} \frac{2a + c_L (\mu-3) - (\mu-1)c_H}{6b} \newline \frac{2a + c_L \mu - c_H (\mu+2)}{6b} \end{bmatrix} $$

This solution has an elegant interpretation: suppose \(\mu = 1\), so that there is no uncertainty and all firms have low marginal costs. Then, the problem’s solution reduces to the solution of a standard Cournot problem with:

$$ q_i^L = \frac{2a + c_L (1-3) - (1-1) c_H}{6b} = \frac{2a - 2c_L}{6b} = \frac{a - c_L}{3b} $$

That is to say, this is the solution to the Cournot problem were \(\Pi_i = (a-b(q_i + q_j))q_i - c_L q_i\) where \(q_i = q_j\). If \(\mu = 0\), the solution for \(q_i^H\) yields a similar result, but then for \(c = c_H\).