1. Unlimited Supply Infinitely many identical items. Each bidder wants one item. –Corresponds to a situation were we have no marginal production cost. –Very common in “digital goods” – songs sold over the Internet, software, etc. Bidder i has private value v i for the item. –We assume throughout that bidders are ordered so that v 1 >v 2 >…>v n

2. Known probability distribution If the values of the players are i.i.d from a probability distribution F, the we can use the optimal auction we already know: –Since we have unlimited supply, we need to determine, for every bidder separately, if she wins or loses. –So we need an optimal auction for one bidder. –We already know that this is a take-it-or-leave-it auction, with a reservation price p such that p=(1-F(p))/f(p)

3. Worst-case analysis What if we do not know, and cannot estimate, the underlying probability distribution? We will design an auction with revenue close to the following benchmark auction: Suppose we know the values of the players, but must sell all items in the same price. The price will then be: F(v) = max i i·v i Remark: a non-discriminatory monopoly chooses this price. First attemp: find a small constant c and a truthful auction with revenue R(v) > F(v)/c for any input v. Remark: this is called a worst-case analysis.

4. First difficulty Reminder: a truthful auction must be monotone, and the price of a winner must be exactly her threshold value for winning. Problem: suppose two players with v 1 =v 2 =1. Then F(v)=2. suppose we find a truthful auction with R(v)=2. –If we increase v 1 to 100 · c, 1’s price cannot change, so R(v)=2. => R(v) < 100 = F(v)/c –Conclusion: situations with no competition are hopeless. A fix: define F (2) (v) = max i>2 i·v i –For “most” cases F(v)= F (2) (v), and we want to avoid the extreme situations for which this is not true. Second attempt: find a small constant c and a truthful auction with revenue R(v) > F (2) (v)/c for any input v.

5. A warm-up mechanism Let A={1…n/2} and B={(n/2)+1,…,n} Let p A and p B be the prices that achieve F (2) (v 1,…,v n/2 ) and F (2) (v n/2+1,…,v n ), respectively. Each bidder i has a take-it-or-leave-it price: p A if i  B and p B if i  A. Problem: suppose v 1 =v 2 =…=v 50 =2c >> 1, and v 51 =…=v 100 =1. What will happen? –p A = 2c and p B = 1. –Players 51..100 lose, and players 1,…50 win and pay $1 each –Overall R(v)=50 while F (2) (v)=50·2c, so R(v) < F (2) (v)/c

6. Allowing randomization If the input contains 50 low bidders and 50 high bidders then “most” partitions will find the correct optimal price. The problem is that we deterministically chose one fixed partition. A fix: allow the mechanism to toss coins. Third (and final) attempt: find a small constant c and a truthful auction with expected revenue E[R(v)] > F (2) (v)/c for any v. Remarks: –This is different than a randomization over the input v as we had up to now. In particular, we do not assume any known distribution. Instead, we create it. –Two truthfulness notions. Universal truthfulness: the auction is truthful for all coin tosses. Truthfulness-in-expectation: the player’s expected utility is maximized by being truthful.

7. A building block: truthful cost sharing We want a revenue of exactly R (or cancel the entire auction). The mechanism CostShare(R): collect bids b 1,…,b n, and repeat: –Suppose we have k bids left (initially we have k=n). –If there exists a bid lower than R/k, delete it, and repeat. –Otherwise all remaining bidders win, and each one pays R/k. Example: bids (5,4,3,2,1), R=9 –In the 1 st round the tentative price is 9/5. Bid 5 is deleted. –In the 2 nd round the tentative price is 9/4. Bid 4 is deleted. –In the 2 nd round the tentative price is 9/3. All three remaining bids win, each one pays 9/3=3. Total revenue is 9.

8. A building block: truthful cost sharing Claim: 1.CostShare is truthful 2.The revenue is exactly R. Proof: left as exercise. Observe that if F(v) > R then the item will indeed be sold, and the revenue will therefore be R.

9. Our auction 1.Randomly partition the bids to two sets (A, B) by tossing a fair coin for each player and associating her to A if the coin came out “head”, and otherwise to B. 2.Compute F A =F(A) and F B =F(B). 3.Run CostShare(A, F B ) and CostShare(B, F A ) to determine winners and prices. Claim: This auction is universally truthful proof: For every result of the coin toss, a player is associated to a fixed set (say A), and cannot affect F B. Therefore, truthfulness follows from the truthfulness of CostShare(A, F B )

10. Revenue Analysis (1) Claim: R(v) > min{F(A), F(B)} Proof: Suppose F(A) < F(B). CostShare(B, F(B)) succeeds by definition Therefore CostShare(B, F(A)) will also succeed, and will provide a profit of F(A).

11. A helpful fact Flip a fair coin k>1 times. Let A k,B k denote the number of times it fell head,tail, respectively, and let M k = min(A k,B k ). Lemma: E[min(A k,B k )] > k/4.

12. A helpful fact Flip a fair coin k>1 times. Let A k,B k denote the number of times it fell head,tail, respectively, and let M k = min(A k,B k ). Lemma: E[min(A k,B k )] > k/4. Proof:For any 1<i<k, define X i = M i – M i-1. Observe: M k = X 1 +…+X k

13. A helpful fact Flip a fair coin k>1 times. Let A k,B k denote the number of times it fell head,tail, respectively, and let M k = min(A k,B k ). Lemma: E[min(A k,B k )] > k/4. Proof:For any 1<i<k, define X i = M i – M i-1. Observe: M k = X 1 +…+X k for all i, X i > 0 => E[X i ] > 0

14. A helpful fact Flip a fair coin k>1 times. Let A k,B k denote the number of times it fell head,tail, respectively, and let M k = min(A k,B k ). Lemma: E[min(A k,B k )] > k/4. Proof:For any 1<i<k, define X i = M i – M i-1. Observe: M k = X 1 +…+X k for all i, X i > 0 => E[X i ] > 0 E[M 2 ]=1/2 ; E[M 3 ]=3/4=> E[X 3 ]=E[M 3 ]-E[M 2 ]=1/4

15. A helpful fact Flip a fair coin k>1 times. Let A k,B k denote the number of times it fell head,tail, respectively, and let M k = min(A k,B k ). Lemma: E[min(A k,B k )] > k/4. Proof:For any 1<i<k, define X i = M i – M i-1. Observe: M k = X 1 +…+X k for all i, X i > 0 => E[X i ] > 0 E[M 2 ]=1/2 ; E[M 3 ]=3/4=> E[X 3 ]=E[M 3 ]-E[M 2 ]=1/4 If i is even then E[X i ]=1/2: –i-1 is odd, so A k ≠ B k. Assume A k > B k. With prob. 1/2 the i’th coin is A and the minimun stays the same (so X i =0) and with probability 1/2 it’s and the minimum increases by 1 (so X i =1).

16. A helpful fact Flip a fair coin k>1 times. Let A k,B k denote the number of times it fell head,tail, respectively, and let M k = min(A k,B k ). Lemma: E[min(A k,B k )] > k/4. Proof:For any 1<i<k, define X i = M i – M i-1. Observe: M k = X 1 +…+X k for all i, X i > 0 => E[X i ] > 0 E[M 2 ]=1/2 ; E[M 3 ]=3/4=> E[X 3 ]=E[M 3 ]-E[M 2 ]=1/4 If i is even then E[X i ]=1/2: –i-1 is odd, so A k ≠ B k. Assume A k > B k. With prob. 1/2 the i’th coin is A and the minimun stays the same (so X i =0) and with probability 1/2 it’s and the minimum increases by 1 (so X i =1).  E[M k ]=E[X 1 ] +…+E[X k ] > 0 + 1/2 + 1/4 + 1/2 + 0 + 1/2 + 0 +…

17. A helpful fact Flip a fair coin k>1 times. Let A k,B k denote the number of times it fell head,tail, respectively, and let M k = min(A k,B k ). Lemma: E[min(A k,B k )] > k/4. Proof:For any 1<i<k, define X i = M i – M i-1. Observe: M k = X 1 +…+X k for all i, X i > 0 => E[X i ] > 0 E[M 2 ]=1/2 ; E[M 3 ]=3/4=> E[X 3 ]=E[M 3 ]-E[M 2 ]=1/4 If i is even then E[X i ]=1/2: –i-1 is odd, so A k ≠ B k. Assume A k > B k. With prob. 1/2 the i’th coin is A and the minimun stays the same (so X i =0) and with probability 1/2 it’s and the minimum increases by 1 (so X i =1).  E[M k ]=E[X 1 ] +…+E[X k ] > 0 + 1/2 + 1/4 + 1/2 + 0 + 1/2 + 0 +… = =1/4 + (  k  /2)·(1/2) > 1/4 + (k-1)/4 = k/4.

18. Revenue Analysis (2) Lemma: E[R(v)] > F (2) (v)/4 Proof: Recall that R(v) > min{F (2) (A), F (2) (B)} Suppose F (2) (v) = y·p (p is the monopolistic price). Let y A,y B denote the number of bids larger than p that belong to A,B, respectively. Note that y= y A +y B. Observe that F(A) > y A ·p, F(B) > y B ·p E[R(v)] > E[min{F(A), F(B)}] > E[min{y A ·p, y B ·p}] = = p·E[min{y A, y B }] > p·y/4 = F (2) (v)/4.

19. Remarks The factor c is called the “approximation factor” (or approximation ratio). This is a notion from computer science. The best approximation factor known for this setting is 3.25. It is also known that no truthful auction can approximation factor better than 2.42.

20. Bibliographic notes Worst-case analysis is the standard in computer science (as opposed to all other sciences…) This work indeed came from computer scientists (Fiat, Goldberg, Hartline, and Karlin, 2002). The interaction of CS and Economics is quite new, and it is not clear what performs better in practice –Designing an auction based on distributional knowledge we don’t have (and then estimating it somehow), or –Designing a worst-case auction that gives worst-case bounds (that are of-course lower than the bounds of above). In particular, an experimental study that evaluates this would be interesting and useful.