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Suppose the market demand is $P(Q) = a-bQ$ and the manufacturing cost of an item is $c_1q_1$ for firm $1$ and $c_2q_2$ for firm $2$.

In the Cournot model, we maximize the profit functions and plug in the best response function of one firm into the other's best response function.

Why can't we do this instead: If $\pi_1 = (a-q_1-q_2)q_2 - c_1q_1$ is the profit of firm $1$, why don't we plug in $q_2^* = \text{BR}_2(q_1)$ and solve for firm $1$'s BR function. That is, we differentiate $\pi_i = (a-q_i-\text{BR}_{-i}(q_i))q_i - c_iq_i$ wrt $q_i$ for both $i=1,2$.

I tried this and it didn't work. This is also how the Stackelberg model is solved. So my question is, why does plugging in the BR function before and after create a difference when in the end both firms are choosing their best response functions? I don't get the math behind how plugging in the BR later creates the "first-mover advantage" while it doesn't if we do it the usual Cournot way.

Notation: If $i=1$, BR$_{-i}(q_1)$ corresponds to BR$_2(q_1)$. Similarly for $i=2$.

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  • $\begingroup$ Does this answer your question? Question on oligopoly. $\endgroup$
    – Giskard
    Commented Oct 5, 2022 at 18:29
  • $\begingroup$ Your question is much better formulated than the question there; but the answer there is very good. $\endgroup$
    – Giskard
    Commented Oct 5, 2022 at 18:30
  • $\begingroup$ @Giskard I read it. It doesn't answer my question. My question is, how does plugging in the BR function give the first move advantage? We do this in Stackelberg but not in Cournot. In Cournot, the players are playing the best responses, so why not assume the other player's going to respond with BR and plug in their BR directly in the profit function (before differentiation)? $\endgroup$
    – Rick_Morty
    Commented Oct 5, 2022 at 23:49
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    $\begingroup$ @Giskard If the second firm, say, goes for the NE move, its profits will be less but firm 1 will make profits lesser than what it would have made had it played the NE move. So this lack of information will force both players to play the NE move in Cournot while in the case of Stackelberg, we have more information (that is, the leader knows the choice of the follower) and exploit that. $\endgroup$
    – Rick_Morty
    Commented Oct 6, 2022 at 0:42
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    $\begingroup$ This might be helpful: qr.ae/pG0AkU $\endgroup$
    – Amit
    Commented Oct 7, 2022 at 13:38

1 Answer 1

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In a nutshell: in Cournot (in any simultaneous game), each firm chooses his best response given the other players' RESPONSE, whereas in Stackelberg one firm is choosing his best response given the other players' RESPONSE (the Follower), whereas the other is choosing the best response given the other player's BEST RESPONSE (the Leader), namely, she is "choosing herself" the S.P.N.E. .

In Cournot, each firm is choosing a function, their intersection is the N.E. (and they can't choose the N.E. themselves), whereas in Stackelberg one firm is choosing a function, the other firm is choosing the point of this function that maximizes her own payoffs.

If the question is "Why can't I optimize for the best responses of the two players in Cournot", the answer is simple: you are basically treating them as monopolists, you're choosing two points of two different functions (that happen to be the same just because the players are symmetric), you're not looking for the point where the two best-responses cross.

For example, if you have a monopolist that maximizes $\Pi(p_1,p_2,p_3)$, there's no difference whether you solve simultaneously or you choose sequentially first $p_1$ (and you find $p_1(p_2,p_3)$, then $p_2$ (so that you find $p_2(p_3)$, and then $p_3$: you're always choosing a vector in $R^3$, not a function in $R^3$.

I hope I was clear...

P.S. Note that the Leader "doesn't know" the choice of the Follower (She infers it! Like a chess player! She knows which moves he will play optimally for a given combination of parameters, let's say), but the Follower does know the choice of the Leader. The problem is not asymmetric information, is that this mathematical trick (choice of point vs choice of a function) "simulates" a strategic advantage.

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  • $\begingroup$ Yes yes, I was under the impression that when the moves are simultaneous (or non-sequential when one player doesn't know what the other player is going to choose), if one of the players goes for a non-NE allocation, then the other won't be (except for a $0 \%$ chance) reacting with the best response function. For instance, the NE response in that case won't be the BR. $\endgroup$
    – Rick_Morty
    Commented Oct 7, 2022 at 19:25

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