Market Design and Analysis

Lecture 5

Lecturer: Ning Chen (陈宁 )

Email: ningc@ntu.edu.sg

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 vi

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 ti f(b-i)

if ti < bi, i wins at price pi = ti

if ti > bi, i does not win

if ti = bi, 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 bi(x) = (b1,…,bi-1,x,bi+1,…,bn)

If there is a value x* s.t. in A(bi(x*)), buyer i wins and pays p (≤ x*), then define f(b-i) = p

5

Characterization of Truthfulness

Consider any value x and bi(x) if buyer i wins, payment must be p.

(otherwise, there is y s.t. the payment is q ≠ p in A(bi(y)). Assume wlog q>p. then the auction is not truthful when vi = 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 vi = 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 ti 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 arev1 ≥ v2 ≥ ∙∙∙ ≥ vn. The following are the natural candidates: T(v) = ∑i vi, i.e. the sum of values of all buyers

F(v) = maxi i∙vi, 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) = maxi≥2 i∙vi

Definition (competitive ratio). The competitive ratio of auction A is defined to be maxv 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 maxv 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, b5 = 1 drops out

offer R/4 to first four buyers, b4 = 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) = maxi i∙vi

15

Random Sampling Profit Extraction The random sampling profit extraction auction

(RSPE) works as follows: randomly partition the bids b = (b1,…,bn) into two

groups g1 and g2.

compute R1 = F(g1) and R2 = F(g2), the optimal profit given by a single price for each group.

run ProfitExtrac(R1) on g2

run ProfitExtrac(R2) on g1

F(v) = maxi i∙vi

16

Random Sampling Profit Extraction Theorem. Random sampling profit extraction is

truthful.

Proof. Consider any buyer i. Assume wlog that i is in group g1.

Then we will run ProfitExtrac(R2) on g1, where R2, the optimal profit given by a single price for group g2, is independent to the bid of i.

By Lemma 1, we know that ProfitExtrac(R2) 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 Mi = min{# heads, # tails} be a random variable

after i coin flips. Observe that E[M1] = 0

E[M2] = 1/2

E[M3] = 3/4

Let Xi = Mi – Mi-1. Thus, E[X1] = 0

E[X2] = 1/2

E[X3] = 1/4

19

Random Sampling Profit Extraction By linearity of expectation,

E[Mk] = ∑ki=1 E[Xi]

Thus, to compute E[Mk], it suffices to compute E[Xi].

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[Xi] = 1/2.

20

Random Sampling Profit Extraction Case 2. i is odd. We use the trivial lower bound

E[Xi] ≥ 0.

Therefore,

where recall that X1 = 0, X2 = 1/2, X3 = 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 k1 be the number of them that are in g1

R1 = F(g1) ≥ k1p

let k2 be the number of them that are in g2

R2 = F(g2) ≥ k2p

F(v) = maxi i∙vi F(2)(v) = maxi≥2 i∙vi

22

Random Sampling Profit Extraction Hence,

That is, for any possible input vector, we always have F(2)(v) / RSPE(v) ≤ 4, i.e. maxv 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), 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.