<p>You are right, there is no equilibrium for <span class="math-container">$s\leq \frac{1}{4}$</span>. Note that Nash's theorem only guarantees the existence of an equilibrium in finite games (which this game is not). (Also note that <span class="math-container">$0$</span> 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.)</p>

<p>To get an intuition of what is going on, consider the plot below, which shows the best response functions for the two players, with  <span class="math-container">$s=\frac{1}{4}$</span>. As you point out, the lines do not intersect. Suppose we start at point <span class="math-container">$(0,0)$</span>, 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 <span class="math-container">$e_1=2$</span>, the best response to <span class="math-container">$e_2=0$</span>. But then, player 2 will update their action to <span class="math-container">$e_2=\frac{3}{2}$</span>. This goes on ad infinitum, never reaching an equilibrium.</p>

<p><a href="https://i.sstatic.net/YHGQx.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YHGQx.png" alt="Best response diagram"></a></p>

CC BY-SA 4.0
2019-06-14T16:33:03.933Z
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.
Regio
CC BY-SA 4.0
2019-06-14T16:37:41.257Z
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.
Regio
CC BY-SA 4.0
2019-06-14T10:54:30.017Z
2019-06-14T13:18:00.417Z
2019-06-14T13:18:00.417Z
Frederic Schneider
<p>Each player can contribute to the project with non-negative effort.</p>

<p>Player 1's utility is <span class="math-container">$u_1=e_1(1+e_2-s\cdot e_1)$</span> where <span class="math-container">$s\in [0,1]$</span>. </p>

<p>Player 2's utility is <span class="math-container">$u_2=e_2(1+e_1-e_2)$</span> </p>

<p>For case 1, each player contributes to the project simultaneously.</p>

<p>For case 2, player 1 moves first then play 2 moves next</p>

<p>I cannot find a Nash equilibria (SPNE for case 2) if <span class="math-container">$s$</span> is less than  <span class="math-container">$1/4$</span>.</p>

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

<p>Can <span class="math-container">$e_1=\infty$</span>, <span class="math-container">$e_2=\infty$</span> be a nash equilibrium?</p>

<game-theory><nash-equilibrium>
I cannot find for this Simple shared effort level game
user23421
# I cannot find for this Simple shared effort level game

(Asked by: username_0 on 2019-06-14)

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?

**Tags:** game-theory, nash-equilibrium

---

## 1 answers

### Answer by username_1 on 2019-06-14. Score: 2

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](https://i.sstatic.net/YHGQx.png)](https://i.sstatic.net/YHGQx.png)

> **Comments:**
> **username_2**: 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.  
> **username_2**: 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.  

---

