Consider a game in which, simultaneously, player $1$ selects any real number $x$ and player $2$ selects any real number $y$. The payoffs are given by:

$u_1 (x, y) = 2x − x^2 + 2xy$

$u_2 (x, y) = 10y − 2xy − y^2.$

(a) Calculate each player’s best-response function as a function of the opposing player’s pure strategy.

(b) Find and report the Nash equilibria of the game.

(c) Determine the rationalizable strategy profiles for this game.

I got the first two part

$\partial u_1 / \partial X = 0 \implies 2 – 2X + 2Y = 0$ $\implies$ Best Response of Player $1 = 1 + Y$

$\partial u_2/ \partial Y = 0 \implies 10 – 2Y – 2X = 0$ $\implies$ Best Response of Player $2 = 5 – X$

Nash Equilibrium

$BR_2 = 5-(1+Y)$

so that

$Y^* = 2$ and $X^* = 3$.

The Nash equilibrium is $(3,2)$.

I am unsure abut rationalizable strategy. I think it is rationalizable strategy for both players are all real numbers because there is no single number that is strictly dominated by another number. For example, if player 2 chooses 10, 11 is the best strategy for player 1 and dominated other number. But, if player 2 chooses 11, 12 is the best strategy for player 1 and 12 dominated 11 in this case. This logic then applies to all real numbers. I am unsure if this logic is correct. Any help is greatly appreciated.

  • $\begingroup$ I rolled back your last edit, since the materials you deleted contained the real question to which Michael's answer is. Doing so is for the benefit of future visitors to this site, who may share a similar question. $\endgroup$
    – Herr K.
    Commented Feb 11, 2018 at 3:25
  • 1
    $\begingroup$ Also, please avoid cross-posting on different SE platforms. $\endgroup$
    – Herr K.
    Commented Feb 11, 2018 at 3:27

1 Answer 1


Your calculation gives you all Nash equilibria in pure strategies. The unique Nash equilibrium in pure strategies is indeed the only Nash equilibrium, but this requires an argument. A careful look at the payoff functions shows that it is irrelevant for player 1 whether player 2 plays the pure strategy $y$ or a mixed strategy with expectation $y$. A similar argument applies to the other side. But then the calculation gives us that every player has a unique best response in pure strategies to every mixed strategy of the other. Moreover, mixed best responses must randomize over pure best responses, so all best responses must be in pure strategies. In particular, every Nash equilibrium must be in pure strategies.

Now rationalizable strategies are the strategies that survive iterated elimination of never best responses. Every strategy is a best response here and does never get eliminated, so every pure strategy profile is rationalizable. If you define rationalizability in terms of mixed strategies, the argument above shows you that you still get only the pure strategies as rationalizable strategies. There are some subtleties in how one defines rationalizability with infinite action spaces, but they are all irrelevant here.

  • $\begingroup$ Is there a standard reference for defining rationalisability on infinite action spaces? $\endgroup$ Commented Feb 11, 2018 at 13:07
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    $\begingroup$ @TheoreticalEconomist The problem has mostly to do with the meaning of "iterative deletion". This note by Dov Samet provides a good overview for the case of the nonprobabilistic case. For beliefs, look at this paper by Itai Arieli. $\endgroup$ Commented Feb 11, 2018 at 19:56

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