samples/texts/2875771/page_10.md

4 Equilibrium Characterization

We focus on pure-strategy Nash equilibria in which both players choose the same level of effort. The following lemma provides a sufficient condition for such a symmetric equilibrium to exist.

Lemma 1. A sufficient condition for a symmetric equilibrium to exist is that $\frac{\partial P_i(e_i, e_k)}{\partial e_i}|_{e_i=e_k=e}$ is the same for $i, k \in {1,2}, i \neq k$, and all $e \in \text{int } E$.

Proof. See Appendix A.1. □

We will make use of Lemma 1 to prove the existence of a symmetric equilibrium by checking the sufficient condition. Since this condition depends on the winning probability, we need to specify this probability first. For each $e > 0$, we define the function $g_e: \mathbb{R} \to \mathbb{R}$ by $g_e(x) = g(x,e)$. The function $g_e(x)$ is strictly increasing in $x$ and thus invertible, and we denote the (strictly increasing) inverse by $g_e^{-1}$. This notation can be motivated by the fact that the event of player $i$ winning over player $k$ can be written as

Considering all potential realizations of $\Theta_i$ and $\Theta_k$, the winning probability of player $i$ is

$$P_i(e_i,e_k)={\int }_{\mathbb{R}}{F}_k(g_{e_k}^{-1}(g_{e_i}(x)))f_i(x)dx\mathrm{.}P\_i(e\_i,\; e\_k)\; =\; \backslash int\_\{\backslash mathbb\{R\}\}\; F\_k(g\_\{e\_k\}^\{-1\}(g\_\{e\_i\}(x)))\; f\_i(x)\; dx.$$

By symmetry, the winning probability of player $k$ is

$$P_k(e_i,e_k)={\int }_{\mathbb{R}}{F}_i(g_{e_i}^{-1}(g_{e_k}(x)))f_k(x)dx\mathrm{.}P\_k(e\_i,\; e\_k)\; =\; \backslash int\_\{\backslash mathbb\{R\}\}\; F\_i(g\_\{e\_i\}^\{-1\}(g\_\{e\_k\}(x)))\; f\_k(x)\; dx.$$

The derivative of player $i$'s winning probability with respect to $e_i$ is given by:¹⁴

$$\frac{\mathrm{\partial }{P}_i(e_i,e_k)}{\mathrm{\partial }{e}_i}={\int }_{\mathbb{R}}{f}_k(g_{e_k}^{-1}(g_{e_i}(x)))\frac{d}{d{e}_i}(g_{e_k}^{-1}(g_{e_i}(x)))f_i(x)dx\mathrm{.}(1)\backslash frac\{\backslash partial\; P\_i(e\_i,\; e\_k)\}\{\backslash partial\; e\_i\}\; =\; \backslash int\_\{\backslash mathbb\{R\}\}\; f\_k(g\_\{e\_k\}^\{-1\}(g\_\{e\_i\}(x)))\; \backslash frac\{d\}\{de\_i\}\; (g\_\{e\_k\}^\{-1\}(g\_\{e\_i\}(x)))\; f\_i(x)\; dx.\; \backslash quad\; (1)$$

The derivative of player $k$'s winning probability with respect to $e_k$ is given by:

$$\frac{\mathrm{\partial }{P}_k(e_i,e_k)}{\mathrm{\partial }{e}_k}={\int }_{\mathbb{R}}{f}_i(g_{e_i}^{-1}(g_{e_k}(x)))\frac{d}{d{e}_k}(g_{e_i}^{-1}(g_{e_k}(x)))f_k(x)dx\mathrm{.}(2)\backslash frac\{\backslash partial\; P\_k(e\_i,\; e\_k)\}\{\backslash partial\; e\_k\}\; =\; \backslash int\_\{\backslash mathbb\{R\}\}\; f\_i(g\_\{e\_i\}^\{-1\}(g\_\{e\_k\}(x)))\; \backslash frac\{d\}\{de\_k\}\; (g\_\{e\_i\}^\{-1\}(g\_\{e\_k\}(x)))\; f\_k(x)\; dx.\; \backslash quad\; (2)$$

¹⁴Notice that $F_k$ is differentiable almost everywhere, since it is the cdf of the absolutely continuous random variable $\Theta_k$ with $f_k$ as the corresponding pdf.