Question

2. Two players participate in a contest where the winner gets R and the loser gets P-R, 0 ≤ R ≤ P. The two players have identical convex cost of effort functions C(e). The chance player i wins is ei/(e1+e2). Show that, at equilibrium, effort increases in R.

Answer 1

the game is such that both the players maximize their winning amounts.

Expected Payoff to player 1 = pr(winning)*amountfrom winning + pr(losing)*amount from losing - cost of effort

= e1/(e1+e2) * R + (1 - [e1/ e1+ e2] )*(P - R) - C(e)

Taking the first order condition with respect to e1 and equating to 0

=> Re2/(e1 + e2)² + (P-R)*(-e2 /(e1+e2)²) - C'(e)...........................(1)

Now by symmetry of the game, player 2 has the exact some payoff and FOC

thus, we get the condition for playr 2 as Re1/(e1 + e2)² + (P-R)*(-e1 /(e1+e2)²) - C'(e)

and thus, e1 = e2 in equilibrium. putting this in (1) we get.

R/2e1 - P/4e1 = C'(e1)

=> (2R - P)/4C'(e1) = e

=> de/dR = 1/2C'(e1) > 0

Thus, we see that effort is, indeed, increasing in R (since C'(e1) is also > 0)