Asymmetric Cournot with discrete quantities

05 Oct 2015

In standard Cournot models, firms are symmetric and choose continuous quantities (\(q_i \in \mathbb{R_+}\)). Here we consider a model of Cournot duopoly with asymmetric firms limited to discrete quantities (\(q_i \in \mathbb{Z_+}\)). Because the set we are optimizing over is no longer continuous, we can’t use the Kuhn-Tucker approach used in Cournot models with continuous quantities. Instead, we set up the problem as a game in normal form and find the pure- and mixed-strategy Nash equilibria by using iterated elimination of dominated strategies.

The setting

Let firm 1 have marginal cost \(c_1 = 4\) and firm 2 have marginal cost \(c_2 = 5\). Let the inverse demand function be given by \(\begin{align} P(Q)=\begin{cases} 10 - Q \ \ & if \ \ Q \le 10 \cr 0 \ \ & if \ \ Q \gt 10 \end{cases} \end{align}\)

and \(Q = q_1 + q_2\).

To start, we can eliminate any quantities greater than the firm’s monopoly quantities. That’s what the firm would do if it wasn’t constrained by its competitor, so it can’t be profit-maximizing to play anything above those quantities. To find the monopoly quantities, we can take the standard Cournot duopoly best-response function, \(q_i^* = \frac{a - bq_j^* - c_i}{2b}\), and plug in \(q_j^* = 0\). This gives us that firm 1’s monopoly quantity is 3, and firm 2’s is 2.5. We can eliminate any quantity choice greater than 3 for both firms, since they are strictly dominated.

How do we handle 2.5? Firm 2 has to choose 2 or 3, not 2.5. We can see that \(\pi_2(q_1=0,q_2=3) = 6\) and \(\pi_2(q_1=0,q_2=2) = 6\), so we can’t eliminate 3 from firm 2’s choices by strict domination.

The game matrix below shows the firms’ profits. Firm 1 is on the columns, and firm 2 is on the rows. The entries are “\(\pi_1, \pi_2\)”.

\[\begin{array}{c|lcr} & 0 & 1 & 2 & 3 \cr \hline 0 & 0,0 & 5,0 & 8,0 & 9,0 \cr 1 & 0,4 & 4,3 & 6,2 & 6,1 \cr 2 & 0,6 & 3,4 & 4,2 & 3,0 \cr 3 & 0,6 & 2,3 & 2,0 & 0,-3 \end{array}\]

Iterated elimination:

• Firm 1: 1 strictly dominates 0 \(\implies\) remove 0-column
• Firm 2: 2 strictly dominates 3 \(\implies\) remove 3-row
• Firm 1: 2 strictly dominates 1 \(\implies\) remove 1-column
• Firm 2: 1 strictly dominates 0 \(\implies\) remove 0-row

This leaves us with the following 2x2 game matrix:

\[\begin{array}{c|lr} & 2 & 3 \cr \hline 1 & 6,2 & 6,1 \cr 2 & 4,2 & 3,0 \cr \end{array}\]

There are 3 pure-strategy Nash equilibria, \((q_1,q_2,\pi_1,\pi_2)\):

1. \[(2,1,6,2)\]
2. \[(2,2,4,2)\]
3. \[(3,1,6,1)\]

Letting \(p_1 = Pr(q_2=1)\), \(p_2 = Pr(q_2=2)\), \(k_2=Pr(q_1=2)\), \(k_3=Pr(q_1=3)\), we can get their expected utilities:

\[\begin{align} EU_{1}(2) & = p_1(6) + p_2(4) = 6p_1 + 4p_2 \cr EU_{1}(3) & = p_1(6) + p_2(3) = 6p_1 + 3p_2 \cr EU_{2}(1) & = k_2(2) + k_3(1) = 2k_2 + k_3 \cr EU_{2}(2) & = k_2(2) + k_3(0) = 2k_2 \cr \end{align}\]

By applying expected-utility indifference, we can construct a mixed NE.

\[\begin{align} EU_1(2) & = EU_1(3) \cr EU_2(1) & = EU_2(2) \cr \implies p_2&=0,p_1=1 \cr k_3&=0,k_2=1 \cr \end{align}\]

This is odd: The mixed NE that’s coming out is a degenerate one, pure NE #1. I think I remember something to the effect that the total number of pure + mixed NE in a matrix game is supposed to be odd, so this might imply that there are actually no mixed NE? I was told that this game actually has infinitely many mixed NE, but I don’t see it.