<p>You have solved the Cournot part correctly, but then you've gone completely off the road, by mistaking economics for mathematics. This usually happens.</p>

<p>First of all, you shouldn't assume just any value for $q^*_2$ unless you want to show some contradiction by a counterexample. Moreover, even if it was somehow correct, you have used the most unreasonable difficult to solve value $\frac{a-c}{2b}$ which isn't convenient at all.</p>

<p>Second of all, you weren't asked to solve for the SPNE, you were just asked to show whether that particular strategy is SPNE or not.</p>

<p>And lastly, you have mixed strategy profile for action profile, when you gave two answers for the question (a). You said YES and NO, while it only had one strategy. In fact, the number $q_2$ is not a strategy, it's action. The function $q_2(q_1)$ is a strategy.</p>

<p>Here is how you solve this kind of problems:</p>

<ol>
<li>Solve Cournot and obtain $q_c = q^*_1 = q^*_2 = \frac{a-c}{3b}$.</li>
</ol>

<p>Now, you should understand the strategy economically before solving it analytically. When Firm 1 has the first-mover advantage, it can (and will) dominate all of the market by producing more and not leaving a space for the Firm 2, which enters after Firm 1. So, Firm 2 has two options:</p>

<p>(a) Don't adjust and sell nothing and make zero profits</p>

<p>or</p>

<p>(b) Try to threaten Firm 1 against dominating all of the market and possibly make some positive profit.</p>

<p>Obviously, it chooses (b). What the strategy (in the question) says is basically the following: If you deviate from Cournot equilibrium, I will produce a lot more and make production so high, that the prices will drop to zero and both of us won't make any profits. Now, we need to see whether this threat is credible or not.</p>

<p>Before we proceed, please remember the following shortcut: Non-credible threats usually are Nash equilibria, and never are Subgame-Perfect equilibria.</p>

<p>2.a. Show that producing less than Cournot is worse than producing exactly Cournot:</p>

<p>Remember, that $q_1 = q_2 = q_c$ is already a Nash equilibrium. Now, let Firm 1 deviate from the equilibrium and produce less:</p>

<p>$ \pi_1 = (a-b( \frac{a-c}{3b} + \frac{a-c}{3b} - \epsilon) - c) \cdot (\frac{a-c}{3b} - \epsilon = \frac{(a-c)^2}{9b} - b\epsilon^2 &lt; \pi_{cournot} $,</p>

<p>So, it will not make any sense for Firm 1 to produce less.</p>

<p>2.b. Show that producing more than Cournot is worse than Cournot, because Firm 2 will respond with higher quantity:</p>

<p>$\pi_1 = (a-b(\frac{a-c}{3b} + \epsilon + \frac{a-c}{3b} + \delta) - c)(\frac{a-c}{3b} + \epsilon) = (\frac{a-c}{3} - b\epsilon - b\delta)(\frac{a-c}{3b} + \epsilon) = \frac{(a-c)^2}{9b} - b \epsilon^2 - \frac{a-c}{3b} b \delta - b\delta\epsilon&lt; \frac{(a-c)^2}{9b} = \pi_{cournot}$ </p>

<p>So, we conclude that the following strategy profile:</p>

<p>{ $q_1 = \frac{a-c}{3b}$; $ q_2(q_1) = \begin{cases}
q_c, &amp; q_1 = q_c \\
q_c+\epsilon, \forall\epsilon&gt;0&amp;q_1\neq q_c
\end{cases} \} $</p>

<p>is a Nash equilibrium. That is, a non-credible threat by a Firm 2, which threatened to bankrupt both firms if Firm 1 din't play by the rules, worked and forced Firm 1 to play by Cournot rules, even though it had a first-mover advantage.</p>

<p>The answer to part (A) is YES.</p>

<p>Now, we have to check, whether this strategy profile is Subgame-Perfect. (Recall, that we are <strong>not</strong> asked to solve for all subgame-perfect equilibria, just check the existing).</p>

<ol start="3">
<li>Show that in a proper subgame, this strategy is not a Nash equilibrium.</li>
</ol>

<p>Now, we will show that this threat is non-credible. To do that, identify the only proper subgame: the one where Firm 2 chooses the quanity given Firm 1's choice has already been made.</p>

<p>Assume, that Firm 1 didn't listen to Firm 2's threats and decided to produce more than Cournot $q_1 = q_c + \epsilon$. Now, if Firm 2 keeps its promise and goes by its own rules, it will have to produce more than Cournot as well: $q_2 = q_c + \delta$. But then we repeat the 2.b's equation and see that it's less profitable for Firm 2 to keep its promise and produce more than it would be to produce $q_c$. And since Firm's only objective is to maximize proifts, we conclude that Firm 2 breaks its promise and doesn't bankrupt both firms. That is, Firm 1 enjoys first mover advantage and Firm 2 obeys its fate.</p>

<p>The answer to (b) is NO. Firm 2's strategy is not a Nash equilibrium in the subgame, where it chooses the quantity.</p>

CC BY-SA 4.0
2018-05-18T03:07:28.053Z
2018-05-18T10:57:53.767Z
2018-05-18T10:57:53.767Z
Ravshan S.K.
<p>Consider an oligopoly between two identical firms producing a homogenous good with constant marginal cost where firms face linear market demand. $B_i(q_j)$ denotes firm i’s best response given the output by firm j for i,j=1,2 and $i\not= j$. Let $q_c$ denote the Cournot equilibrium output choice where firms choose output simultaneously. Assume now a Stackelberg setup, where firm 2 observes firm 1’s choice before making its own output choice. </p>

<p>Consider the following pair of strategies in the Stackelberg setup: firm 1 chooses $q_c$. Firm 2 chooses $q_c$ if firm 1 chooses $q_c$ otherwise firm 2 chooses $q^*_2 &gt;&gt;q_c$. </p>

<p>(A) show whether or not the pair of strategies described above constitutes a Nash equilibrium in the game. </p>

<p>(B) show whether or not the pair of strategies described above constitutes a subgame perfect equilibrium in the game. </p>

<hr>

<p><strong>What I did is</strong></p>

<p>Leaks inverse demand function $p=a-bq$ And $q=q_1+q_2$</p>

<p>And since marginal cost is constant c, then cost function is $cq_i$</p>

<p>First of all, I find Cournot equilibrium outputs $q_c$</p>

<p>I obtain by the profit maximization problem</p>

<p>$$q_c=\frac{a-c}{3b}$$</p>

<p>for both firms. </p>

<p>Secondly, since firm 2 also has a choice $q_2^* &gt;&gt; q_c$</p>

<p>Then let’s assume that $q_2^* = \frac{a-c}{2b}&gt;&gt;\frac{a-c}{3b}=q_c$</p>

<p>Then in order to obtain Stackelberg setup, by backward induction method </p>

<p>Let’s calculate $q^*_1$ which is the firm 1’s quantity when firm 2 chooses $q_2^*=\frac{a-c}{2b}$. </p>

<p>$$\pi_1=(a-bq_1-b(\frac{a-c}{2b}))q_1-cq_1$$</p>

<p>By FOC,</p>

<p>$$q_1^*=\frac{a-c}{4b}$$</p>

<p>And $$q^*_1&lt;&lt; q_c$$</p>

<p>Thirdly I calculated profits for four cases</p>

<p>Case 1: firm1’s quantity and firm 2’s quantity are both $q_c$ </p>

<p>Then both firms’ profits are equal and equal to $\frac{(a-c)^2}{9b}$</p>

<p>Case 2: firm1’s quantity is $q_1^*$ firm 2’s quantity is $q_c$ </p>

<p>Then,</p>

<p>Firm 1’s profit =$\frac{5(a-c)^2}{48b}$</p>

<p>Firm 2’s profit =$\frac{5(a-c)^2}{36b}$</p>

<p>Case 3: firm1’s quantity is $q_c$ firm 2’s quantity is $q_2^*$ </p>

<p>Then,</p>

<p>Firm 1’s profit =$\frac{(a-c)^2}{18b}$</p>

<p>Firm 2’s profit =$\frac{5(a-c)^2}{12b}$</p>

<p>Case 4: firm1’s quantity is $q_1^*$ firm 2’s quantity is $q_2^*$ </p>

<p>Then,</p>

<p>Firm 1’s profit =$\frac{(a-c)^2}{16b}$</p>

<p>Firm 2’s profit =$\frac{(a-c)^2}{8b}$</p>

<p>Now let’s look at the part (A)</p>

<p>I construct the following table</p>

<p><a href="https://i.sstatic.net/wRqoG.jpg" rel="nofollow noreferrer">Table</a></p>

<p>So as it is seen, only $(q_c, q_c)$ is Nash equilibrium. But $(q_1^*, q_2^*)$ is not Nash equilibrium. </p>

<p>Next let’s look at the part (B)</p>

<p>I construct tree in this case. </p>

<p><a href="https://i.sstatic.net/IFpWA.jpg" rel="nofollow noreferrer">Tree</a></p>

<p>When we look at the table,</p>

<p>First of all, it is for firm 2,</p>

<p>In the left hand side, since the payoff of $q_c$ is greater than the payoff of $q_2^*$, the firm 2 will choose $q-c$.</p>

<p>In the right hand side, since the payoff of $q_2^*$ is greater the payoff of $q_c$ for firm 2, then the firm 2 will choose $q^*_2$. </p>

<p>Now, as for firm 1, if firm 1 choose $q_c$ then it knows that its payoff will be $(a-c)^2/9b$. But if firm 1 choose $q^*_1$ the its payoff will be $(a-c)^2/16b$. So the firm 1 will decide to choose $q_c$.</p>

<p>That’s SPNE is $(q_c,q_cq_c)$.</p>

<p>That is, $(q_1^*,q_2^*)$ is not SPNE. </p>

<hr>

<p>I solve this question in this way. But I am exactly not sure about my way. Please share your ideas with me. I will be happy if say something abou my solution.  Thank you. </p>

<microeconomics><game-theory><self-study><oligopoly>
Question on oligopoly.
studentp
# Question on oligopoly.

(Asked by: username_0 on 2018-05-13)

Consider an oligopoly between two identical firms producing a homogenous good with constant marginal cost where firms face linear market demand. $B\_i(q\_j)$ denotes firm i’s best response given the output by firm j for i,j=1,2 and $i\not= j$. Let $q\_c$ denote the Cournot equilibrium output choice where firms choose output simultaneously. Assume now a Stackelberg setup, where firm 2 observes firm 1’s choice before making its own output choice.

Consider the following pair of strategies in the Stackelberg setup: firm 1 chooses $q\_c$. Firm 2 chooses $q\_c$ if firm 1 chooses $q\_c$ otherwise firm 2 chooses $q^\*\_2 >>q\_c$.

(A) show whether or not the pair of strategies described above constitutes a Nash equilibrium in the game.

(B) show whether or not the pair of strategies described above constitutes a subgame perfect equilibrium in the game.

---

**What I did is**

Leaks inverse demand function $p=a-bq$ And $q=q\_1+q\_2$

And since marginal cost is constant c, then cost function is $cq\_i$

First of all, I find Cournot equilibrium outputs $q\_c$

I obtain by the profit maximization problem

$$q\_c=\frac{a-c}{3b}$$

for both firms.

Secondly, since firm 2 also has a choice $q\_2^\* >> q\_c$

Then let’s assume that $q\_2^\* = \frac{a-c}{2b}>>\frac{a-c}{3b}=q\_c$

Then in order to obtain Stackelberg setup, by backward induction method

Let’s calculate $q^\*\_1$ which is the firm 1’s quantity when firm 2 chooses $q\_2^\*=\frac{a-c}{2b}$.

$$\pi\_1=(a-bq\_1-b(\frac{a-c}{2b}))q\_1-cq\_1$$

By FOC,

$$q\_1^\*=\frac{a-c}{4b}$$

And $$q^\*\_1<< q\_c$$

Thirdly I calculated profits for four cases

Case 1: firm1’s quantity and firm 2’s quantity are both $q\_c$

Then both firms’ profits are equal and equal to $\frac{(a-c)^2}{9b}$

Case 2: firm1’s quantity is $q\_1^\*$ firm 2’s quantity is $q\_c$

Then,

Firm 1’s profit =$\frac{5(a-c)^2}{48b}$

Firm 2’s profit =$\frac{5(a-c)^2}{36b}$

Case 3: firm1’s quantity is $q\_c$ firm 2’s quantity is $q\_2^\*$

Then,

Firm 1’s profit =$\frac{(a-c)^2}{18b}$

Firm 2’s profit =$\frac{5(a-c)^2}{12b}$

Case 4: firm1’s quantity is $q\_1^\*$ firm 2’s quantity is $q\_2^\*$

Then,

Firm 1’s profit =$\frac{(a-c)^2}{16b}$

Firm 2’s profit =$\frac{(a-c)^2}{8b}$

Now let’s look at the part (A)

I construct the following table

[Table](https://i.sstatic.net/wRqoG.jpg)

So as it is seen, only $(q\_c, q\_c)$ is Nash equilibrium. But $(q\_1^\*, q\_2^\*)$ is not Nash equilibrium.

Next let’s look at the part (B)

I construct tree in this case.

[Tree](https://i.sstatic.net/IFpWA.jpg)

When we look at the table,

First of all, it is for firm 2,

In the left hand side, since the payoff of $q\_c$ is greater than the payoff of $q\_2^\*$, the firm 2 will choose $q-c$.

In the right hand side, since the payoff of $q\_2^\*$ is greater the payoff of $q\_c$ for firm 2, then the firm 2 will choose $q^\*\_2$.

Now, as for firm 1, if firm 1 choose $q\_c$ then it knows that its payoff will be $(a-c)^2/9b$. But if firm 1 choose $q^\*\_1$ the its payoff will be $(a-c)^2/16b$. So the firm 1 will decide to choose $q\_c$.

That’s SPNE is $(q\_c,q\_cq\_c)$.

That is, $(q\_1^\*,q\_2^\*)$ is not SPNE.

---

I solve this question in this way. But I am exactly not sure about my way. Please share your ideas with me. I will be happy if say something abou my solution. Thank you.

**Tags:** microeconomics, game-theory, self-study, oligopoly

---

## 1 answers

### Answer by username_1 on 2018-05-18. Score: 5 | Accepted answer

You have solved the Cournot part correctly, but then you've gone completely off the road, by mistaking economics for mathematics. This usually happens.

First of all, you shouldn't assume just any value for $q^\*\_2$ unless you want to show some contradiction by a counterexample. Moreover, even if it was somehow correct, you have used the most unreasonable difficult to solve value $\frac{a-c}{2b}$ which isn't convenient at all.

Second of all, you weren't asked to solve for the SPNE, you were just asked to show whether that particular strategy is SPNE or not.

And lastly, you have mixed strategy profile for action profile, when you gave two answers for the question (a). You said YES and NO, while it only had one strategy. In fact, the number $q\_2$ is not a strategy, it's action. The function $q\_2(q\_1)$ is a strategy.

Here is how you solve this kind of problems:

1. Solve Cournot and obtain $q\_c = q^\*\_1 = q^\*\_2 = \frac{a-c}{3b}$.

Now, you should understand the strategy economically before solving it analytically. When Firm 1 has the first-mover advantage, it can (and will) dominate all of the market by producing more and not leaving a space for the Firm 2, which enters after Firm 1. So, Firm 2 has two options:

(a) Don't adjust and sell nothing and make zero profits

or

(b) Try to threaten Firm 1 against dominating all of the market and possibly make some positive profit.

Obviously, it chooses (b). What the strategy (in the question) says is basically the following: If you deviate from Cournot equilibrium, I will produce a lot more and make production so high, that the prices will drop to zero and both of us won't make any profits. Now, we need to see whether this threat is credible or not.

Before we proceed, please remember the following shortcut: Non-credible threats usually are Nash equilibria, and never are Subgame-Perfect equilibria.

2.a. Show that producing less than Cournot is worse than producing exactly Cournot:

Remember, that $q\_1 = q\_2 = q\_c$ is already a Nash equilibrium. Now, let Firm 1 deviate from the equilibrium and produce less:

$ \pi\_1 = (a-b( \frac{a-c}{3b} + \frac{a-c}{3b} - \epsilon) - c) \cdot (\frac{a-c}{3b} - \epsilon = \frac{(a-c)^2}{9b} - b\epsilon^2 < \pi\_{cournot} $,

So, it will not make any sense for Firm 1 to produce less.

2.b. Show that producing more than Cournot is worse than Cournot, because Firm 2 will respond with higher quantity:

$\pi\_1 = (a-b(\frac{a-c}{3b} + \epsilon + \frac{a-c}{3b} + \delta) - c)(\frac{a-c}{3b} + \epsilon) = (\frac{a-c}{3} - b\epsilon - b\delta)(\frac{a-c}{3b} + \epsilon) = \frac{(a-c)^2}{9b} - b \epsilon^2 - \frac{a-c}{3b} b \delta - b\delta\epsilon< \frac{(a-c)^2}{9b} = \pi\_{cournot}$

So, we conclude that the following strategy profile:

{ $q\_1 = \frac{a-c}{3b}$; $ q\_2(q\_1) = \begin{cases}
q\_c, & q\_1 = q\_c \\
q\_c+\epsilon, \forall\epsilon>0&q\_1\neq q\_c
\end{cases} \} $

is a Nash equilibrium. That is, a non-credible threat by a Firm 2, which threatened to bankrupt both firms if Firm 1 din't play by the rules, worked and forced Firm 1 to play by Cournot rules, even though it had a first-mover advantage.

The answer to part (A) is YES.

Now, we have to check, whether this strategy profile is Subgame-Perfect. (Recall, that we are **not** asked to solve for all subgame-perfect equilibria, just check the existing).

3. Show that in a proper subgame, this strategy is not a Nash equilibrium.

Now, we will show that this threat is non-credible. To do that, identify the only proper subgame: the one where Firm 2 chooses the quanity given Firm 1's choice has already been made.

Assume, that Firm 1 didn't listen to Firm 2's threats and decided to produce more than Cournot $q\_1 = q\_c + \epsilon$. Now, if Firm 2 keeps its promise and goes by its own rules, it will have to produce more than Cournot as well: $q\_2 = q\_c + \delta$. But then we repeat the 2.b's equation and see that it's less profitable for Firm 2 to keep its promise and produce more than it would be to produce $q\_c$. And since Firm's only objective is to maximize proifts, we conclude that Firm 2 breaks its promise and doesn't bankrupt both firms. That is, Firm 1 enjoys first mover advantage and Firm 2 obeys its fate.

The answer to (b) is NO. Firm 2's strategy is not a Nash equilibrium in the subgame, where it chooses the quantity.


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