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
e_{i}/(e_{1}+e_{2}). Show that, at
equilibrium, effort increases in R.

Answer #1

Both the players will maximize their winning amounts.

Expected payoff to player 1 = pr(winning) * amount from 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 rquating to 0.

=> R e2/(e1+e2)^{2} + (P-R) * (-e2 / (e1 +
e2)^{2}) - C'(e1) ......eq (1)

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

so, R e1/ (e1 + e2)^{2} + (P-R) * (-e1 /
(e1+e2)^{2}) -C'(e2)

and thus e1 = e2 in equilibrium

Putting this in equation (1)

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

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

=> de1/dR = 1/2 C'(e1) >0

Thus, we see that effort is increasing in R.

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

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