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Konrad (2007) - Strategy In Contests-An Introduction (view source)
Revision as of 19:29, 25 April 2010
, 19:29, 25 April 2010→All-pay auctions
\end{cases}
</math>
Contestant 2 will never expend effort greater than <math>v_2,\</math>, and contestant 1 can win for sure is he expends an epsilon more, so this constraints the effort of both parties to lie on the range <math>[0,v_2],\</math>. If contestant 1 expends an epsilon more than <math>v_2,\</math> he gaurantees a payoff of <math>v_1 - v_2,\</math>; this is his equilibrium payoff. If he does this then player 2's payoff is zero; this is his equilibrium payoff! Under mixed strategies a player must be indifferent at all points, we use this indifference to solve the game:
:<math>\pi_1(x_1) = F_2(x_1)v_1 - x_1 = v_1-v_2,\</math>
and
:<math>\pi_2(x_2) = F_1(x_2)v_2 - x_2 = 0,\</math>
Together this can be used to solve for <math>F_1,\</math> and <math>F_2,\</math>, yielding the results above. Using the standard expected value formula (and differentiating the CDFs to get PDFs first):
:<math>\mathbb{E}(X) = \int_{-\infty}^\infty x f(x)\, \dx,\</math>
We get that:
<center><math>\mathbb{E}x_1 = \frac{v_2}{2} and \mathbb{E}x_2=\frac{v_2^2}{2v_1},\</math></center>
