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For example, in the case of Bertrand competition, is the Stackelberg leader worse off relative to the follower? If so, why? Is this because the first mover can always undercut until marginal cost is reached?

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It depends on what you assume the firms know about each other. Under full information / rationality assumptions, with identical firms, second-mover advantage disappears because the first-mover will recognize the credible threat of undercutting, and set P = MC and that will be that.

If the firms are not identical (e.g., different marginal costs) then turn order is irrelevant, because only one firm has the ability to make a credible threat of undercutting. That firm will capture the market, whether it moves first or second.

Only if you drop the assumption of full information / rationality is it feasible that turn order would matter for Bertrand competition, because then the assessment of credible threat becomes a probabilistic one.

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  • $\begingroup$ This assumes that products offered by both firms are homogeneous. Differentiation in products allows firms to price above marginal cost. This may add some clarity. $\endgroup$ – lunar_props Nov 27 '19 at 19:43
  • $\begingroup$ @lunar_props I'm familiar with the classic Bertrand model. My point is that relaxing those assumptions is the only way to yield a practical turn-order advantage. This is to be contrasted with, e.g., a Cournot duopoly, in which there is a clear first-mover advantage even under the classic assumptions. $\endgroup$ – heh Nov 27 '19 at 19:51
  • $\begingroup$ I take your point. This is one way to yield a turn-order advantage. However, with differentiated products, price competition will yield a turn-order advantage for the follower even in the presence of full information. $\endgroup$ – lunar_props Nov 27 '19 at 20:06
  • $\begingroup$ Not under full info and rationality assumption. In that case, firms know each others' production functions and best-response functions. In particular, the stronger firm (by virtue of having a lower $MC = p_1$) will know that the weaker firm cannot make a credible threat to reduce price below $p_1$. Since $p_2 > p_1$, whether the weaker firm moves first or second, it will not be able to capture the market on price. $\endgroup$ – heh Nov 27 '19 at 20:14
  • $\begingroup$ I am having difficulties understanding this last comment. I am not sure what "lower $MC = p_1$" means, since $MC$ is a function of quantity sold, and that is also dependent on $p_2$, so it seems like something is missing here. Also, w.r.t. the last sentence: is capturing the entire market necessary for second mover advantage? Cannot I just be better off compared to the simultaneous equilibrium without capturing the entire market? $\endgroup$ – Giskard Nov 27 '19 at 21:26

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