Appendix
A Proofs
A.1 Proof of Lemma 1
Suppose that $\frac{\partial P_i(e_i, e_k)}{\partial e_i}|{e_i=e_k=e}$ is the same for both $i \in {1,2}$ and all $e \in \text{int } E$. Then we have $\frac{\partial P_1(e_1, e_2)}{\partial e_1}|{e_1=e_2=e} V - c'(e) = \frac{\partial P_2(e_2, e_1)}{\partial e_2}|{e_1=e_2=e} V - c'(e)$ for all $e \in \text{int } E$. Since $\pi_i(e_i, e_k)$ is continuously differentiable, $\frac{\partial P_i(e_i, e_k)}{\partial e_i}|{e_i=e_k=e} V - c'(e)$ is a continuous function of $e$. Furthermore, recall that there exist $\bar{e}_i, \tilde{e}i \in \text{int } E$ such that $\frac{\partial \pi_i(e_i, e_k)}{\partial e_i}|{e_i=e_k=\bar{e}i} < 0$ and $\frac{\partial \pi_i(e_i, e_k)}{\partial e_i}|_{e_i=e_k=\tilde{e}_i} > 0$. Hence, by the Intermediate Value Theorem, there is some $e^* \in \text{int } E$ such that $\frac{\partial P_i(e_i, e_k)}{\partial e_i}|{e_i=e_k=e^*} V - c'(e^*) = 0$. By Assumption 1, $e_1 = e_2 = e^*$ is a Nash equilibrium.
A.2 Proof of Theorem 1
Since we wish to apply the sufficient condition from Lemma 1, we restrict attention to $e_i > 0$. Then, the function $g_e : \mathbb{R} \to \mathbb{R}$ defined by $g_e(x) = g(x,e)$ is strictly increasing and, thus, invertible. The inverse, $g_e^{-1}$, is strictly increasing as well. For the two (different) players $i,k \in {1,2}$, we observe
Player k thus wins with probability
Differentiating with respect to $e_k$, we obtain
According to Lemma 1, and noting that $g_{e_k}^{-1}(g_{e_i}(x)) = g_{e_i}^{-1}(g_{e_k}(x)) = x$ if $e_i = e_k$, a sufficient condition for a symmetric equilibrium to exist is that