Mixed Cournot/Bertrand Duopoly [closed]

I am learning basic Oligopoly models.

I know that :

• In Cournot model firms set output - output is the strategic variable.
• In Bertrand model firms set prices - price is the strategic variable.

Simple problems I see are usually given demand/cost fucntion for a differentiated duopoly and let us determine the Bertrand or Cournot equilibrium.

Here however is a mixed market:

Assume the following duopoly (firms are denoted by 1 and 2) with the following demand functions:

$$P_1=100-0.5Q_1-0.4Q_2 \\ P_2=100-0.5Q_2-0.4Q_1$$

Firm 1 plays Cournot while firm 2 plays Bertrand. Assuming that there are no costs, find the optimal Nash quantity-price solution.

How can I solve for the equilibrium?

closed as off-topic by Giskard, Herr K., Kitsune Cavalry♦Dec 30 '18 at 1:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

• I'm voting to close this question as off-topic because no effort toward the solution was shown on part of the questioner. The question already has a valid answer, so this action has little importance except to serve as a qualifier for what we regard as sufficient effort. – Giskard Aug 12 '15 at 21:26
• Have a look at meta.economics.stackexchange.com/questions/1420/…, meta.economics.stackexchange.com/questions/24/… and other threads on the meta about homework-like questions. It might help you improve your question so that it does not get closed. – Martin Van der Linden Aug 13 '15 at 10:57

Each player seeks to maximize his or her revenue. As you said, in the Cournot model, the player varies the output. In the Bertrand model, the player varies the price.

So we first solve for $Q_{2}$ as a function of $P_{2}$:

$$Q_{2} = 200 - 2P_{2} - 0.8Q_{1}$$

We then substitute $Q_{2}$ into $P_{1}$ to get:

$$P_{1} = 100 - 0.5Q_{1} - 0.4 (200 - 2P_{2} - 0.8Q_{1}) = 50 - 0.18Q_{1} + 0.8P_{2}$$

Player $1$'s maximization problem is:

$$\max_{Q_{1}} 50Q_{1} - 0.18Q_{1}^{2} + 0.8P_{2}Q_{1}$$

Which yields the first order conditions:

$$0.36Q_{1} = 50 + 0.8P_{2} \implies Q_{1}^{*} = \frac{1250}{9} + \frac{20}{9}P_{2}$$

Now we can substitute $Q_{1}^{*}$ into $Q_{2}$ to get:

$$Q_{2} = 200 - 2P_{2} - \frac{4}{5}(\frac{1250}{9} + \frac{20}{9}P_{2})$$

Now maximize $Q_{2}$ with respect to $P_{2}$. Can you take it from here?

• That's very kind of you!Many thanks!It really helps a lot. – ChocolateU Aug 12 '15 at 10:05