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And so we can see that with <math>r=1\,</math> the rents will be fullly dissapated as <math>n \to \infty\,</math>.
 
===Additive Noise===
 
Consider a problem with two risk neutral contestants where the observation of effort is based on the actual effort plus some idiosyncratic noise, that is <math>x_i + \epsilon_i \forall i is observed. Letting <math>\epsilon = \epsilon_2 - \epsilon_1 we can write the contest success function as:
 
<math>
p_1(x_1,x_2) =
\begin{cases}
1 & \mbox{ if } x_1-x_2 > \epsilon \\
\frac{1}{2} & \mbox{ if } x_1-x_2 = \epsilon \\
0 & \mbox{ if } x_1-x_2 < \epsilon
\end{cases}
</math>
 
 
Suppose that <math>\epsilon\,</math> is distributed according to <math>G\,</math> on the interval <math>[-e,e]\,</math>, with <math>e>0\,</math>, and that the designer creates two prizes <math>b_w\,</math> and <math>b_l\,</math> for the winner and loser respectively. Each contestant then maximizes:
 
:<math>p_i(x_1,x_2)(v_i(b_w) - v_i(b_l)) - C_i(x_i) + v_i(b_l)\,</math>
 
 
Assume that <math>C(\cdot)\,</math> satisfies Inada conditions. If this is sufficiently convex then there is an interior solution for player 1 (and likewise for 2) given by the first order condition:
 
 
:<math>\frac{\partial G(x_1 - x_2)}{\partial x_1)(v_1(b_w) - v_1(b_l)) = C_i'(x_1)\,</math>
 
The IR constraint can be satisfied by making <math>b_l\,</math> sufficiently large.
 
 
If <math>v_i(b)=b\,</math> and <math>C_i = C \forall i\,</math> then there is a symmetric equilibrium with <math>x_1=x_2\,</math> which reduces the FOC to:
 
 
:<math>\frac{\partial G(0)}{\partial x_1)(b_w -b_l) = C'(x_1)\,</math>
 
 
With <math>\epsilon ~ U[e,e]\,</math> we have that:
 
 
:<math>G'(0) = g(0) = \frac{1}{b-a} = \frac{1}{2e}\,</math>
 
:<math>\therefore C'(x_i) = \frac{b_w - b_l}{2e}\,</math>
 
 
It is important to note that the prizes and be adjusted so that the contest can be designed to be optimal for either of the 'short' sides of the market (i.e. to give the rents to the players or to the mechanism designer).
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