0
$\begingroup$

I know that the bidding strategy for an all-pay auction is $ \frac{n-1}{n} v^n$, where $n$ is the number of bidders and $v$ is a bidder's value or type. Therefore, the expected revenue should be $\Sigma_{i=1}^n\left(\frac{n-1}{n} v_i^n\right)$. By the revenue equivalence theorem, this should be equal to $\frac{n-1}{n+1}$, the expected revenue for first and second price closed bid auctions. How do I show this equivalence?

Assume values are independent and uniformly distributed over $[0,1] $.

Thanks!

$\endgroup$
1
  • 1
    $\begingroup$ You should add the assumption that each $v_i$ is independently distributed according to a uniform distribution on $[0,1]$. $\endgroup$
    – mlc
    Dec 13, 2016 at 21:01

1 Answer 1

0
$\begingroup$

Hint:

You want to calculate

$$ \sum_{i=1}^nE\left[\frac{n-1}{n}v_i^n\right] = (n-1) \int_0^1 x^n dx $$

To understand what happens with the summation, recall that the $v_i$'s are iid.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .