Player $i$ chooses an effort level, $e_i \in [0, 10]$. Let player $i$ have the following payoff function: $90 -e_i$ if $e_i > e_j$ and $80 -e_i$ if $e_i \leq e_j$. What is the Nash Equilibrium (NE) of this game? (I actually got confused at mixed-strategy part.)

My approach:

For the pure-strategy NE - Let player $i$ choose an arbitrary effort level, say $8$, then he/she gets $72$ if the other player chooses the same effort level or less. Then the player $i$ is better off by choosing an higher effort level, say $9$, but indeed here we have the same situation. Hence, the player is better off by choosing the highest effort, $10$, which allows him to get $80$ at most. However, choosing $0$ effort is superior to this because he can get $80$ and there is no risk for getting $70$. This procedure continues on and we conclude that there is no pure-strategy NE.

Now we need to consider mixed-strategy NE. Here is the part where I got confused. Player $i$ is mixing over infinitely many strategies. So, we should consider a distribution where $\int F(.) =1.$ Thus, what should be the probability distribution that allows us to be indifferent among strategies?


Let's first check if there's a mixed equilibrium with both strategies having support on all of $[0,10]$. Player $1$ ramping up her effort $e_1$ by $\mathrm de_1$ costs her $\mathrm de_1$ in effort and gains her $10\rho(e_1)\mathrm de_1$ in expected payoff from the comparison, where $\rho$ is player $2$'s probability density function. In an equilibrium with full support, these must be equal everywhere, so $\rho\equiv1/10$.

Thus, both players choosing uniformly randomly yields a Nash equilibrium, whose value for both players can be read off at either end of the interval as $80$.

Note that this result depends on the length of the interval being equal to the value of winning the comparison. If the upper effort limit were higher, all efforts above $10$ would be dominated by effort $0$, so the strategies would still be uniform with support on $[0,10]$. If the upper effort limit were lower, say, $b$, then a finite probability $1-b/10$ would be concentrated at the upper limit and the remaining $b/10$ would be uniformly spread with density $1/10$ over $[0,b]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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