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$\begin{align*} & \frac{(n-1)}{e} \exp(-\lambda \Delta t) \lambda^2 \left( \int^0 x \exp(n\lambda x) dx + t \int^0 \exp(n\lambda x) dx \right) \\ = & \frac{(n-1)}{e} \exp(-\lambda \Delta t) \lambda^2 \frac{(-1+n\lambda t)}{n^2\lambda^2} \\ = & \frac{(n-1)}{e} \exp(-\lambda \Delta t) \frac{(-1+n\lambda t)}{n^2}. \end{align*}$

Notice that the last expression is positive if and only if $n\lambda t - 1 > 0$. Taking the derivative of the expression w.r.t. $n$ results in an expression that is positive if $t > 0$, which is implied by $n\lambda t - 1 > 0$.

B Other Computations and Derivations

B.1 Additional Derivations for Section 7.1.3.

Player i outperforms player k iff

$\begin{align*} g_{e_i}(t_i + \varepsilon_i) &> g_{e_k}(t_k + \varepsilon_k) \\ \Leftrightarrow \quad & \varepsilon_k < g_{e_k}^{-1}(g_{e_i}(t_i + \varepsilon_i)) - t_k. \end{align*}$

Recall that the $E_i$ are i.i.d., following the reflected exponential distribution on $(-\infty, 0]$. The cdf is denoted by $H$ and the pdf by $h$. Hence, player $i$ wins the contest with probability

$\int \prod_{k \neq i} H(g_{e_k}^{-1}(g_{e_i}(t_i+x)) - t_k) h(x) dx.$

In a symmetric equilibrium with $e_1^* = \dots = e_n^* =: e^*$, the marginal effect of effort on the probability of winning,

$\int \left( \prod_{k \neq i} H(t_i + x - t_k) \right) \left( \sum_{k \neq i} \left( \frac{d}{de_i} g_{e_k}^{-1}(g_{e_i}(t_i + x)) \right) \Big|_{e_1^* = \dots = e_n^* = e^*} \frac{h(t_i + x - t_k)}{H(t_i + x - t_k)} \right) h(x)dx,$

must be the same for all i. Denote $\Delta t = t_1 - t > 0$. For player 1, we have,

$\int \left( \prod_{k \neq 1} H(\Delta t + x) \right) \left( \sum_{k \neq 1} \left( \frac{d}{de_1} g_{e_k}^{-1}(g_{e_1}(t_1 + x)) \right) \Big|_{e_1^* = \dots = e_n^* = e^*} \frac{h(\Delta t + x)}{H(\Delta t + x)} \right) h(x) dx.$