Consider a game with two players with profit functions : $ \pi_1 (p_1; p_2) $ and $ \pi_2 (p_2; p_1) $, where $p_1 $ and $p_2 $ are the players' actions (say prices).

Is there any theorem that establishes conditions, when the pure strategy equilibrium will lead to the same prices, in the following two cases? :

  1. If the prices are decided simultaneously
  2. If the player 1 sets the price first and then player 2 observes the price of player 1 and then sets their own price

A different way to phrase your question is to ask for sufficient conditions under which the Nash equilibrium for a price competition game yields the same payoffs under simultaneous versus sequential choices.

A partial negative result is in: E. Gal-Or, "First Mover and Second Mover Advantages", International Economic Review, Vol. 26, No. 3 (Oct., 1985), pp. 649-653.

Suppose that the game is symmetric and has a unique symmetric equilibrium under simultaneous choices: then the players' payoffs are the same. This paper shows that, under sequential choices, this is never the case when the best replies are increasing (the Leader earns less than the Follower) or decreasing (the Follower earns less than the Leader). So in this simple context, you would need to have non-monotonic best replies to achieve the desired result.

On the side, let me suggest that you add more context to your question. It seems you are implicitly assuming unique equilibria (how are we supposed to compare multiple equilibria?) and it is unclear whether $\pi_1$ and $\pi_2$ should be supposed to be symmetric.

  • $\begingroup$ Thanks for your answer and the reference. To clarify on the points you asked, the game that is of interest to me is not symmetric ($\pi_1$ and $\pi_2$ are different), but has a unique equilibrium (in both simultaneous move and sequential move). $\endgroup$ – user917983 Aug 29 '17 at 13:22
  • $\begingroup$ The literature has results more general than Gal-Or about non-symmetric games. Do a reverse bibliographic search. Most of these results are going to be negative, I'm afraid. $\endgroup$ – mlc Aug 29 '17 at 15:21

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