1. week 81 COS 444 Internet Auctions: Theory and Practice Spring 2009 Ken Steiglitz ken@cs.princeton.edu

2. week 82 Some other auctions in A rs Consider simple case with n=2 and uniform iid values on [0,1]. We also choose the optimal v * = ½. How do we choose b 0 to achieve this v * ? All-pay with reserve

3. week 83 Standard argument: If your value is v * you win if and only if you have no rival bidder. (This is the point of indifference between bidding and not bidding, and your expected surplus is 0.) Therefore, bid as low as possible! Therefore, b(v * ) = b 0. And so b 0 = v * 2 = ¼. Notice that this is an example where the reserve is not equal to the entry value v *. All-pay with reserve

4. week 84

5. 5 Loser weeps auction, n=2, uniform v Winner gets item for free, loser pays his bid! Auction is in A rs. The expected payment is therefore and therefore, choosing v * = ½ as before (optimally), To find b 0, set E[surplus] = 0 at v = v *, and again argue that b(v * ) = b 0. This gives us ( goes to ∞ !)

6. week 86

7. 7 Santa Claus auction, n=2, uniform v Winner pays her bid Idea: give people their expected surplus and try to arrange things so bidding truthfully is an equilibrium. Pay bidders To prove: truthful bidding is a SBNE

8. week 88 Santa Claus auction, con’t Suppose 2 bids truthfully and 1 has value v and bids b. Then because F(b) = prob. winning. For equil.: □ (use reserve b 0 = v * )

9. week 89 Matching auction: not in A rs Bidder 1 may tender an offer on a house, b 1 ≥ v * Bidder 2 currently leases house and has the option of matching b 1 and buying at that price. If bidder 1 doesn’t bid, bidder 2 can buy at v * if he wants to

10. week 810 To compare with optimal auctions, choose v * = ½; uniform iid IPV’s on [0,1] Bidder 2’s best strategy: If 1 bids, match b 1 iff v 2 ≥ b 1 ; else bid ½ iff v 2 ≥ ½ Bidder 1 should choose b 1 ≥ ½ so as to maximize expected surplus. This turns out to be b 1 = ½. To see this Matching auction: not in A rs

11. week 811 Choose v * = ½ for comparison. Bidder 1 tries to max (v 1 − b 1 )·{ prob. 2 chooses not to match } = (v 1 − b 1 )·b 1  b 1 = 0 if v 1 < ½ = ½ if v 1 ≥ ½ Matching auction: not in A rs □

12. week 812 Notice: When ½ < v 2 < v 1, bidder 2 gets the item, but values it less than bidder 1  inefficient! E[revenue to seller] turns out to be 9/24 (optimal in A rs is 10/24; optimal with no reserve is 8/24) BTW, …why is this auction not in A rs ? Matching auction: not in A rs

13. week 813 Intuition: Suppose you care more about losses than gains. How does that affect your bidding strategy in SP auctions? First-price auctions? Risk aversion recall

14. week 814 Utility model

15. week 815 Risk aversion & revenue ranking Result: Result: Suppose bidders’ utility is concave. Then with the assumptions of A rs, R FP ≥ R SP Proof: Proof: Let γ be the equilibrium bidding function in the risk-averse case, and β in the risk-neutral case.

16. week 816 Revenue ranking Let the bidder bid as if her value is z, while her actual value is x. In a first-price auction, her expected surplus is where W(z) = F(z) n-1 is the prob. of winning. As usual, to find an equilibrium, differentiate wrt z and set the result to 0 at z = x: where w(x) = W΄(x).

17. week 817 Revenue ranking In the risk neutral case this is just: The utility function is concave:

18. week 818 Revenue ranking Using this, Now we can see that γ΄(0) > β΄(0). If not, then there would be an interval near 0 where γ ≤ β, and then which would be a contradiction.

19. week 819 Revenue ranking Also, it’s clear that γ(0) = β(0) = 0. So γ starts out above β at the origin. To show that it stays above β, consider what would happen should it cross, say at x = x* : A contradiction. □

20. week 820 Constant relative risk aversion (CRRA) Defined by utility In this case we can solve [MRS 03]MRS 03 for  Very similar to risk-neutral form, but bid as if there were (n−1)/ρ instead of (n−1) rivals!