Each player can contribute to the project with non-negative effort.

Player 1's utility is $u_1=e_1(1+e_2-s\cdot e_1)$ where $s\in [0,1]$.

Player 2's utility is $u_2=e_2(1+e_1-e_2)$

For case 1, each player contributes to the project simultaneously.

For case 2, player 1 moves first then play 2 moves next

I cannot find a Nash equilibria (SPNE for case 2) if $s$ is less than $1/4$.

If I draw best a response diagram, the two best response functions do not intersect. But I don't think that $e_1=0$, $e_2=0$ is a Nash equilibrium, since they have would have an incentive to exert more than $0$ effort.

Can $e_1=\infty$, $e_2=\infty$ be a nash equilibrium?


1 Answer 1


You are right, there is no equilibrium for $s\leq \frac{1}{4}$. Note that Nash's theorem only guarantees the existence of an equilibrium in finite games (which this game is not). (Also note that $0$ is not in the best-response set of any of the two players, so we know that, if there were an equilibrium, it would not contain zero effort by any player.)

To get an intuition of what is going on, consider the plot below, which shows the best response functions for the two players, with $s=\frac{1}{4}$. As you point out, the lines do not intersect. Suppose we start at point $(0,0)$, and let each player revise their action in turn, holding the other player's action constant (dashed line). Player 1 will update their action to $e_1=2$, the best response to $e_2=0$. But then, player 2 will update their action to $e_2=\frac{3}{2}$. This goes on ad infinitum, never reaching an equilibrium.

Best response diagram

  • $\begingroup$ As @Frederic has mentioned this game does not have an equilibrium in pure strategies. $e_1=e_2=\infty$ will not make a lot of sense. Though technically speaking it all depends if the game was defined as effort defined in the extended real and non-negative line. To my knowledge, there are no theorems that would guaranty existence in this game. The main complications are that action spaces are not finite, and more importantly, they are not compact. You could try to look for mixed strategies, but the space of mixed strategies is quite large, so beware. $\endgroup$
    – Regio
    Commented Jun 14, 2019 at 16:33
  • 2
    $\begingroup$ If you were willing to assume a maximum effort level (plausible assumption in most cases) so that $e_i\in[0,\bar e_i]$ for some $\bar e_i<\infty$ and for each $i\in\{1,2\}$, then an equilibrium in pure strategies is guaranteed to exists. In fact, it looks like the equilibrium would be $e_i^*=\bar e_i$ for all $i=1,2$ i.e. both players would exert the maximum effort. $\endgroup$
    – Regio
    Commented Jun 14, 2019 at 16:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.