1. Market Design and Analysis Lecture 5 Lecturer: Ning Chen ( 陈宁 ) Email: ningc@ntu.edu.sg

2. 2 Worst Case Optimal Auction Design  Consider an auction setting where there are  one item (unlimited supplies)  n buyers who desire one copy each at value v i  Goal: Design a truthful auction that maximizes the profit.  What if VCG?

3. 3 Characterization of Truthfulness  An auction is called bid-independent if for each buyer i, the auction computes t i  f(b -i )  if t i < b i, i wins at price p i = t i  if t i > b i, i does not win  if t i = b i, i either wins or not (doesnot matter)  Theorem. An auction is truthful if and only if it is equivalent to a bid-independent auction.

4. 4 Characterization of Truthfulness  Proof. Any bid-independent auction is truthful (trivial). On the other direction, consider any truthful auction A.  Let b i (x) = (b 1,…,b i-1,x,b i+1,…,b n )  If there is a value x* s.t. in A(b i (x*)), buyer i wins and pays p (≤ x*), then define f(b -i ) = p

5. 5 Characterization of Truthfulness  Consider any value x and b i (x) if buyer i wins, payment must be p. (otherwise, there is y s.t. the payment is q ≠ p in A(b i (y)). Assume wlog q>p. then the auction is not truthful when v i = y.) buyer i wins for any bid x > p. (otherwise, there is y > p s.t. i does not win when bidding y. Again the auction is not truthful when v i = y.)  Hence, the auction is equivalent to bid-independent auction f(∙) where f(b -i ) = p.

6. 6 A Bad News..  Consider a special case where there is only one buyer.  By the above theorem, any truthful auction is bid- independent  The value t i offered to the buyer must be a constant  The real value of the buyer, however, can be arbitrarily large…  In general, truthful auctions can work arbitrarily bad in the worst case.

7. 7 Analysis Approach  How we can arrive at a rigorous theoretical framework to determine the optimality of some auctions?  Answer: moving from absolute optimality to relative optimality. (E.g. in data compression, whenever there is an information theoretic obstacle preventing us from obtaining an absolute optimal solution, we try to approximate.)

8. 8 Analysis Approach  In our setting, the obstacle is the game theoretic constraint that an auction does not know the true values in advance and must solicit them in a truth- inducing manner.  The approach to study relative optimality of auctions is to find a suitable benchmark, and show that the auction is always within a small factor of the benchmark.

9. 9 Benchmark  Benchmarks. Assume the values are v 1 ≥ v 2 ≥ ∙∙∙ ≥ v n. The following are the natural candidates:  T(v) = ∑ i v i, i.e. the sum of values of all buyers  F(v) = max i i∙v i, i.e. the optimal profit given by a single price.  The example above, however, shows that both T(v) and F(v) are not good benchmark, as they can be arbitrarily larger than the profit of any truthful auction.

10. 10 Benchmark  Definition (F (2) ). The optimal single priced profit with at least two winners is F (2) (v) = max i≥2 i∙v i  Definition (competitive ratio). The competitive ratio of auction A is defined to be max v F (2) (v) / A(v), where A(v) is the (expected) profit of A on input v.  Goal: design a truthful auction A to minimize the competitive ratio (worst-case analysis), i.e. min max v F (2) (v) / A(v)

11. 11 Benchmark  Examples:  There are 50 bids $10 and 50 bids $1. Then F (2) = 50 ∙ 10 = $500  There are 5 bids $10 and 95 bids $1. Then F (2) = 1 ∙ 100 = $100

12. 12 An Impossibility Result  Theorem. Roughly speaking, no deterministic truthful auction is c-competitive for any given constant c > 0.  Therefore, we have to look for randomized auctions.

13. 13 Profit Extraction  Assume the goal of the seller is to raise a total profit R.  The auction profit extractor with target profit R, ProfitExtract(R), sells to the largest set of k buyers that can equally share R and change each R/k.  Example. Consider b = (5,4,3,2,1) and R = 9.  offer R/5 to all buyers, b 5 = 1 drops out  offer R/4 to first four buyers, b 4 = 2 drops out  offer R/3 to first three buyers, all accept at price R/3 = 3.

14. 14 Profit Extraction  Lemma 1. For any given value R, ProfitExtract(R) is truthful.  Proof. assignment.  Lemma.  If R ≤ F(v), then ProfitExtract(R) always obtains a profit of R.  If R > F(v), then ProfitExtract(R) obtains nothing.  Proof. assignment. F(v) = max i i∙v i

15. 15 Random Sampling Profit Extraction  The random sampling profit extraction auction (RSPE) works as follows:  randomly partition the bids b = (b 1,…,b n ) into two groups g 1 and g 2.  compute R 1 = F(g 1 ) and R 2 = F(g 2 ), the optimal profit given by a single price for each group.  run ProfitExtrac(R 1 ) on g 2  run ProfitExtrac(R 2 ) on g 1 F(v) = max i i∙v i

16. 16 Random Sampling Profit Extraction  Theorem. Random sampling profit extraction is truthful.  Proof. Consider any buyer i.  Assume wlog that i is in group g 1.  Then we will run ProfitExtrac(R 2 ) on g 1, where R 2, the optimal profit given by a single price for group g 2, is independent to the bid of i.  By Lemma 1, we know that ProfitExtrac(R 2 ) is truthful for buyer i.

17. 17 Random Sampling Profit Extraction  Theorem. Random sampling profit extraction is 4- competitive.

18. 18 Random Sampling Profit Extraction  Lemma 2. If we flip a coin uniformly k ≥ 2 times, then E[min {# heads, # tails}] ≥ k/4  Proof. Let M i = min{# heads, # tails} be a random variable after i coin flips. Observe that  E[M 1 ] = 0  E[M 2 ] = 1/2  E[M 3 ] = 3/4 Let X i = M i – M i-1. Thus,  E[X 1 ] = 0  E[X 2 ] = 1/2  E[X 3 ] = 1/4

19. 19 Random Sampling Profit Extraction  By linearity of expectation, E[M k ] = ∑ k i=1 E[X i ] Thus, to compute E[M k ], it suffices to compute E[X i ].  Case 1. i is even. Hence i – 1 is odd. Thus, prior to the i-th coin, # heads ≠ # tails. Assume wlog that # heads < # tails. Now we flip the i-th coin:  with probability 1/2, it is head and we increase the minimum by one;  with probability 1/2, it is tail and we do not increase the minimum. Thus, E[X i ] = 1/2.

20. 20 Random Sampling Profit Extraction  Case 2. i is odd. We use the trivial lower bound E[X i ] ≥ 0.  Therefore, where recall that X 1 = 0, X 2 = 1/2, X 3 = 1/4.

21. 21 Random Sampling Profit Extraction  Theorem. Random sampling profit extraction is 4- competitive.  Proof. Assume that F (2) (b) = kp, where k ≥ 2 is the number of winners and p is the price. Of these k winners,  let k 1 be the number of them that are in g 1  R 1 = F(g 1 ) ≥ k 1 p  let k 2 be the number of them that are in g 2  R 2 = F(g 2 ) ≥ k 2 p F(v) = max i i∙v i F (2) (v) = max i≥2 i∙v i

22. 22 Random Sampling Profit Extraction  Hence,  That is, for any possible input vector, we always have F (2) (v) / RSPE(v) ≤ 4, i.e. max v F (2) (v) / RSPE(v) ≤ 4

23. 23 Random Sampling Profit Extraction  Theorem. Random sampling profit extraction is truthful and 4-competitive.  Theorem. No randomized truthful auction can be better than 2.42-competitive.  Theorem. There is a truthful auction for n buyers with ratio 3.12.

24. 24 Reading Assignment  R. Myerson, Optimal Auction Design, Mathematics of Operations Research, V.6(1), 1981.  N. Nisan, A. Ronen, Algorithmic Mechanism Design, STOC 1999, 129-140.  A. Archer, E. Tardos, Truthful Mechanisms for One-Parameter Agents, FOCS 2001, 482-491.  A. Karlin, D. Kempe, T. Tamir, Beyond VCG: Frugality of Truthful Mechanisms, FOCS 2005, 615-626.  A. Goldberg, J. Hartline, A. Karlin, M. Saks, xxxx, Competitive auctions, Games and Economic Behavior, 2009.  X. Bei, N. Chen, N. Gravin, P. Lu, Budget Feasible Mechanism Design: From Prior-Free to Bayesian, STOC 2012.