Market Design and Analysis Lecture 5 Lecturer: Ning Chen ( 陈宁 )

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

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