samples/texts/2875771/page_33.md

The above condition implicitly defines the symmetric equilibrium effort function $e(\theta)$.

Note that since

$$z^{\mathrm{\prime }}(\theta )=\frac{dg(\theta ,e(\theta ))}{d\theta }=\frac{\mathrm{\partial }g}{\mathrm{\partial }\theta }+\frac{\mathrm{\partial }g}{\mathrm{\partial }e}\frac{de(\theta )}{d\theta },z\text{'}(\backslash theta)\; =\; \backslash frac\{dg(\backslash theta,\; e(\backslash theta))\}\{d\backslash theta\}\; =\; \backslash frac\{\backslash partial\; g\}\{\backslash partial\; \backslash theta\}\; +\; \backslash frac\{\backslash partial\; g\}\{\backslash partial\; e\}\; \backslash frac\{de(\backslash theta)\}\{d\backslash theta\},$$

we have that the first-order condition can be written as:

$$f^{(n-1)}(\theta )\frac{\frac{\mathrm{\partial }g}{\mathrm{\partial }e}}{\frac{\mathrm{\partial }g}{\mathrm{\partial }\theta }+\frac{\mathrm{\partial }g}{\mathrm{\partial }e}{e}^{\mathrm{\prime }}(\theta )}V=c^{\mathrm{\prime }}(e(\theta ))\mathrm{.}(9)f^\{(n-1)\}(\backslash theta)\; \backslash frac\{\backslash frac\{\backslash partial\; g\}\{\backslash partial\; e\}\}\{\backslash frac\{\backslash partial\; g\}\{\backslash partial\; \backslash theta\}\; +\; \backslash frac\{\backslash partial\; g\}\{\backslash partial\; e\}\; e\text{'}(\backslash theta)\}\; V\; =\; c\text{'}(e(\backslash theta)).\; \backslash quad\; (9)$$

Condition (9) has an intuitive interpretation. The LHS is the marginal probability of winning times the prize V in a symmetric equilibrium from the perspective of a player who knows that his or her skill is $\theta$. Given that a player only has a marginal incentive to exert effort when the strongest opponent (the highest order statistic) has the same skill, $f^{(n-1)}(\theta)$ is the “likelihood” of this situation. There are two main differences with respect to the corresponding condition for the case of symmetric uncertainty (equation (4)). First, because players know their own skill level, there is no need to integrate over all possible realizations of a considered player’s own skill. Second, instead of $a_e(x) =$

$$\frac{\mathrm{\partial }g(x,e)}{\mathrm{\partial }e}\mathrm{/}\frac{\mathrm{\partial }g(x,e)}{\mathrm{\partial }x}\text{ (which appeared inside the integral of (4)), we now have the factor }\frac{\frac{\mathrm{\partial }g}{\mathrm{\partial }e}}{\frac{\mathrm{\partial }g}{\mathrm{\partial }\theta }+\frac{\mathrm{\partial }g}{\mathrm{\partial }e}{e}^{\mathrm{\prime }}(\theta )}\backslash frac\{\backslash partial\; g(x,e)\}\{\backslash partial\; e\}\; /\; \backslash frac\{\backslash partial\; g(x,e)\}\{\backslash partial\; x\}\; \backslash text\{\; (which\; appeared\; inside\; the\; integral\; of\; (4)),\; we\; now\; have\; the\; factor\; \}\; \backslash frac\{\backslash frac\{\backslash partial\; g\}\{\backslash partial\; e\}\}\{\backslash frac\{\backslash partial\; g\}\{\backslash partial\; \backslash theta\}\; +\; \backslash frac\{\backslash partial\; g\}\{\backslash partial\; e\}\; e\text{'}(\backslash theta)\}$$

which includes the new term $\frac{\partial g}{\partial e}e'(\theta)$ in the denominator. This new term arises because effort is a function of skill in the private information case.

Recall that when we discussed the intuition behind $a_e(x)$ in equation (4), we explained that the purpose of a marginal effort increase is to beat rivals who have marginally higher skill. In the current setting, the output advantage of marginally more able rivals is not only determined by $\frac{\partial g}{\partial \theta}$ (which is positive) but also by the additional term $\frac{\partial g}{\partial e}e'(\theta)$ which generally has an ambiguous sign. If $\frac{\partial g}{\partial e}$ and $e'(\theta)$ are both strictly positive, more highly skilled rivals are harder to beat not only because of their skill advantage, but also because they exert higher effort, reducing the marginal incentive to exert effort by any player.

In the following example, we compute the equilibrium effort for a specific skill distribution and production function.²⁵

Example 7. Consider a contest with n symmetric players with privately known skills independently drawn from the uniform distribution on [0, 1]. The production function is given by $g(\theta, e) = \theta e$, and the cost function is $c(e) = \frac{e^2}{2}$. Then, the symmetric equilibrium

²⁵The detailed derivations for this example are provided in Appendix B.2.