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Konrad (2007) - Strategy In Contests-An Introduction (view source)
Revision as of 19:57, 25 April 2010
, 19:57, 25 April 2010no edit summary
So not only is the auction in efficient in that it doesn't extract the value of the second highest player, but it is also inefficient in the sense that it does not necessarily allocate the good to the individual with the highest valuation.
===Tullock Contests===
The Tullock contest success function is:
<math>
p_i(x) =
\begin{cases}
\frac{x_i^r}{\sum_{j=1}^n x_j^r & \mbox{ if } \max\{x\} > 0 \\
\frac{1}{N} & \mbox{otherwise}
\end{cases}
</math>
For <math>r=1\,</math> this is called the lottery contest. Generally it is assumed that the cost function <math>C_i(x_i)=x_i\,</math>. With <math>r=1\,</math> adding constants to the function or changing <math>C_i(x_i)\,</math> for some <math>i\,</math> introduces handicaps.
There exists a pure strategy Nash equilibrium iff <math>r \le \frac{n}{n-1}\,</math>, though this generally leads to rent dissapation. For <math>r > \frac{n}{n-1}\,</math> there are mixed strategy NE. In this case it is possible that <math>\sum x_i > B\,</math>, that is that there is excess total effort, but <math>\mathbb{E}(\sum x_i) \le B\,</math>.
For two players the first order conditions give:
:<math>\frac{rx_i^{r-1}x_j^r}{(x_i^r +x_j^r)}\cdot v_1 = 1\,</math>
Plugging in for <math>v_i\,</math>, <math>v_j\,</math> and <math>r\,</math> provides a set of reaction functions.
For any number of players one can solve for the symmetry pure strategy Nash equilibrium by assuming <math>x_i = x_j \forall i\,</math> and rearranging to get:
:<math>x_i = \frac{r(n-1)}{n^2}\cdot B \,</math>
Taking the summation gives:
:<math>\sum x_i = \frac{r(n-1)}{n}\cdot B\,</math>
And so we can see that with <math>r=1\,</math> the rents will be fullly dissapated as <math>n \to \infty\,</math>.
