Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 42 / 4 / Solution

First exclude boundary profiles. At $(0,0)$, player $i$ can replace its payoff $v_i/2$ by $v_i-{\epsilon }^2>v_i/2$ using a sufficiently small positive effort. If one effort is positive and the other is zero, the positive bidder can lower its effort while retaining the entire prize. Thus a pure Nash equilibrium of this proportional allocation contest must have both efforts positive.

Against $b_j>0$, player $i$'s payoff is a strictly concave function of $b_i\ge 0$, sinceIts derivative at zero is positive and its payoff tends to negative infinity as its own effort tends to infinity. Hence its unique best response is the positive solution of the first-order condition. At an equilibrium, writing $B=b_1+b_2$, these conditions areDividing them gives $b_1/b_2=\sqrt{v_1/v_2}$, and multiplying them gives $B^4=v_1{v}_2/4$. Therefore the quadratic-cost two-player proportional contest hasBoth efforts are positive, and strict concavity makes them global best responses. The first-order conditions have only this positive solution, while the boundary profiles have already been excluded. This proves both existence and uniqueness. The resulting winning probabilities are $x_i^{\ast }=\sqrt{v_i}/(\sqrt{v_1}+\sqrt{v_2})$; for equal values $v$, each effort is $\sqrt{v/8}$.