samples/texts/2875771/page_13.md

$\theta_1 e_1 / e_2$. The probability of that event, and its first derivative with respect to $e_1$, are

$$P_1(e_1,e_2)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{F}_2\left(x\frac{e_1}{e_2}\right)f_1(x)dx,P\_1(e\_1,\; e\_2)\; =\; \backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; F\_2\backslash left(x\; \backslash frac\{e\_1\}\{e\_2\}\backslash right)\; f\_1(x)\; dx,$$

$$\frac{\mathrm{\partial }{P}_1(e_1,e_2)}{\mathrm{\partial }{e}_1}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{f}_2\left(x\frac{e_1}{e_2}\right)\left(\frac{x}{e_2}\right)f_1(x)dx\mathrm{.}\backslash frac\{\backslash partial\; P\_1(e\_1,\; e\_2)\}\{\backslash partial\; e\_1\}\; =\; \backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; f\_2\backslash left(x\; \backslash frac\{e\_1\}\{e\_2\}\backslash right)\; \backslash left(\backslash frac\{x\}\{e\_2\}\backslash right)\; f\_1(x)\; dx.$$

The first-order condition of player 1's maximization problem is

$$\frac{\mathrm{\partial }{P}_1(e_1,e_2)}{\mathrm{\partial }{e}_1}V=e_1\mathrm{.}\backslash frac\{\backslash partial\; P\_1(e\_1,\; e\_2)\}\{\backslash partial\; e\_1\}\; V\; =\; e\_1.$$

In a symmetric equilibrium with $e_1 = e_2 = e$, this can now be written as

$$V{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{f}_2(x)x{f}_1(x)dx=e^2\mathrm{.}V\; \backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; f\_2(x)\; x\; f\_1(x)\; dx\; =\; e^2.$$

For player 2 we obtain the same expression. Using our distributional assumptions, the left-hand side becomes

$$V{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{f}_2(x)x{f}_1(x)dx=V{\int }_1^2\frac{x}{\pi (1+x^2)}dx=V\frac{1}{2\pi }\log \left(\frac{5}{2}\right)\mathrm{.}V\; \backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; f\_2(x)\; x\; f\_1(x)\; dx\; =\; V\; \backslash int\_\{1\}^\{2\}\; \backslash frac\{x\}\{\backslash pi(1+x^2)\}\; dx\; =\; V\; \backslash frac\{1\}\{2\backslash pi\}\; \backslash log\; \backslash left(\backslash frac\{5\}\{2\}\backslash right).$$

We thus have a symmetric equilibrium, and the corresponding effort is $e^* = \sqrt{\frac{V \log(\frac{5}{2})}{2\pi}} \approx 0.38\sqrt{V}$.

5 Generalizations

In our previous analysis, the symmetry of the equilibrium was derived under the assumption that the production functions, prizes and cost functions were the same for the competing players. We show next that these assumptions can, under certain conditions, be relaxed. This allows us to deal with heterogeneity between players beyond heterogeneity in skill distributions in a surprisingly simple manner. At the end of the subsection, we will also provide an illustrative example that combines heterogeneity in skill distributions, prizes, cost functions, and production functions.

We begin with the possibility to allow for heterogeneity in production technologies, using the observation that, in some situations, different production functions can be reh interpreted as different skill distributions. This allows us to show that a symmetric equilibrium exists using the results of Theorem 1.

Corollary 1. Suppose that the production functions are different for the two competing