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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?

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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 $[0,b]$.

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