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Chapter 6: Prior-free Mechanisms Roee and Ofir (Also from “Envy Freedom and Prior-free Mechanism Design” by Devanur, Hartline, Yan) 1.

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Presentation on theme: "Chapter 6: Prior-free Mechanisms Roee and Ofir (Also from “Envy Freedom and Prior-free Mechanism Design” by Devanur, Hartline, Yan) 1."— Presentation transcript:

1 Chapter 6: Prior-free Mechanisms Roee and Ofir (Also from “Envy Freedom and Prior-free Mechanism Design” by Devanur, Hartline, Yan) 1

2 Talk overview 1.Introduction to prior free mechanisms and comparison with prior independent mechanisms. 2.Theorem: No anonymous, deterministic digital good auction is better than an n- approximation to the envy-free benchmark. Solution 1: Random Sampling Solution 2: Profit Extraction 2

3 The trouble with priors The prior can be inaccurate- For example, agents can lie during a market survey if they know that the results will affect their prices in the future. Prior dependent mechanisms are non robust- A mechanism that was designed to work on one distribution will probably not work on another. 3

4 Prior free vs. Prior independent Prior-independent mechanism can rely on there being a distribution where as the prior-free mechanism cannot. ↓ The class of good prior-free mechanisms should be smaller than the class of good prior- independent mechanisms. 4

5 What is a good mechanism? A good mechanism approximates the optimal mechanism for the distribution if there is a distribution; moreover, when there is no distribution this mechanism still performs well. 5

6 Tradeoff The goal of prior-free mechanism design and this work therein is to sacrifice optimality to obtain prior freedom. 6

7 What is the objective? The objective is profit maximization- we characterize (p,x) that gives the highest total revenue. (p- payments vector, x – allocation vector.) 7

8 How do we evaluate prior free mechanisms? Envy-free optimal revenue benchmark: An outcome, allocation and payments, (x,p), is envy-free if no agent prefers to swap outcome (allocation and payment) with another agent. (Similar in structure to incentive compatible mechanisms). 8

9 Incentive Compatibility versus Envy Freedom A mechanism is incentive compatible if no agent prefers the outcome when misreporting her value to the outcome when reporting the truth. ∀ i, z, v. v i x i (v) − p i (v) ≥ v i x i (z, v -i ) − p i (z, v -i ) An allocation x with payments p is envy free for valuation profile v if no agent prefers the outcome of another agent to her own. ∀ i, j. v i x i − p i ≥ v j x j − p j IC ≈ EF 9

10 Digital Good Environment There are n unit-demand agents denoted N = {1,..., n} and any subset of them can be served. I.e., X = 2 n 10

11 Optimal mechanism given a i.i.d distribution (in digital environment) Post the monopoly price as a take-it-or-leave-it offer to each agent. ↓ v i < monopoly price → x i = 0 v i > monopoly price → x i = 1 This mechanism is envy free. 11

12 Optimal mechanism not given an i.i.d distribution (in digital environment) 12

13 Example- Selling a song to 5 people: AgentValue iiv (i) ↓ argmax i iv (i) = 3 → 13

14 Why is max i iv (i) not a good benchmark? Not a good benchmark when the maximization is obtained at i=1. ↓ The envy-free (optimal) benchmark for digital goods is defined as EFO (2) (v) = max i≥2 iv (i) This will be the benchmark. ivivi

15 Intro to designing prior-free auctions Deterministic auctions cannot give good prior- free approximation. We will describe two approaches for designing prior-free auctions for digital goods. 15

16 Deterministic Auctions When figureing out a price to offer agent i we can use statistics from the values of all other agents v -i 16

17 Deterministic optimal price auction The deterministic optimal price auction offers each agent i the take-it-or-leave-it price of τ i equal to the monopoly price for v –i. This mechanism is prior independent but not prior free. 17

18 Example AgentValue 110 …… 111 …… 1001 ↓ The price that will be offered for agents 1-10 is 1, and the value that will be offered to agents is 10. (Derived from v –i for each i.) → The revenue will be 10, which is much less then EFO (2) (v) = EFO(v) = 100. (Sell to the first 10 agents for 10.) AgentValuev –i Price Offered Profit high valued 90 low valued high valued 89 low valued

19 Theorem No anonymous, deterministic digital good auction is better than an n-approximation to the envy-free benchmark. 19

20 Proof 20

21 Proof cont. 21

22 22

23 Proof cont. 23

24 Proof cont. 24

25 Motivation The problem with the deterministic optimal price auction is that it sometimes offers high- valued agents a low price and low-valued agents a high price. Either of these prices would have been good if only it offered consistently to all agents. 25

26 First solution: Random Sampling 1. Randomly partitions the agents into S′ and S′′ (by flipping a fair coin for each agent) 2. Compute (empirical) monopoly prices η′ and η′′ for S′ and S′′ respectively 3. Offers η′ to S′′ and η′′ to S′ S S S′ 26

27 Example v = (1.1, 1) → EFO (2) (v) = 2 With probability ½ both agents are in the same partition → revenue is 0. With probability ½ each agent is in a different partition → revenue is 1. So the expected revenue is ½, which is a 4- approximation to the benchmark. 27

28 Theorem For all valuation profiles, the random sampling auction is at least a 15-approximation to the envy-free benchmark. 28

29 Proof First, we assume that v (1) ∈ S′ and we call S’ the Market and a S’’ the sample. Now we want to prove two main theorems: Show that EFO(v S’’ ) is close to EFO (2) (v). Show that revenue from price η′′ on S′ is close to EFO(v S’’ ). 29

30 Proof cont. Define: 1. v (i) represent the i-th largest valued agent 2. X i is an indicator to the event that i ∈ S′′ 3.Define S i = ∑ j

31 Proof cont. We will prove that with good probability EFO(v S’’ ) is close to the benchmark, EFO (2) (v): Define the event B that S k ≥ k/2 Notice that EFO(v S’’ ) ≥ S k v k (Optimal revenue≥ Revenue from price v k ) Now from Event B we can see: S k v k ≥ v k k/2 (multiply by v k ) 31

32 Proof cont. EFO(v S’’ ) ≥ S k v k and S k v k ≥ v k k/2 ↓ (v k k/2 = EFO (2) (v)/2 by definition.) EFO(v S’’ ) ≥ EFO (2) (v)/2 32

33 Proof cont. But, we didn’t prove that event B happens in a good probability. Therefore we now want to show that Pr(B)=1/2. (proof will be for even k, proof for odd k omited) 33

34 Proof cont. We assume in the beginning that v (1) ∈ S′. Therefore, we need to divide k-1 (odd number) agents into the market and the sample (S’ and S’’). At least one partition receives at least k/2 of these agents and half the time it is the sample; therefore, Pr[B] = 1/2. 34

35 Proof cont. Now we want to prove the second part, that with good probability, revenue from price η′′ on S′ is close to EFO(v S’’ ). 35

36 Proof cont. Like in the first part we define an event: ε= “ ∀ i, (i − S i ) ≥ S i /3” Let k′′ be index of the agent whose value is the monopoly price for the sample. ↓ v k’’ = η′′ and EFO(v S’’ )= S k’’ v k’’ (by definition) ↓ EFO(v S’’ )/3 = S k’’ v k’’ /3 (Divide the 2nd equation by 3) 36

37 Proof cont. We combine the theorems: EFO(v S’’ ) ≥ EFO (2) (v)/2 and EFO(v S’’ )/3 = S k’’ v k’’ /3 ↓ If B and ε holds, then the expected revenue is at least EFO (2) (v)/ 6. 37

38 Proof cont. From the Balanced Sampling Lemma (no proof) we assume Pr[ε] ≥ 0.9 B and ε holds = Pr[ε ∧ B] = 1−Pr[ ¬ ε]−Pr[ ¬ B] ≥ 0.4. Therefore, the random sampling auction is a 15-approximation to the envy-free benchmark. 38

39 Solution 2: Profit extractor We design a mechanism that obtains profit at least R on any input v with EFO(v) ≥ R. We call this mechanism a profit extractor. 39

40 Profit extractor The digital good profit extractor for target R and valuation profile v finds the largest k such that v(k) ≥ R/k, sells to the top k agents at price R/k, and rejects all other agents. If no such set exists, it rejects all agents. 40

41 Profit extractor The digital good profit extractor is dominant strategy incentive compatible. 41

42 Proof: We need to design a mechanism that makes profit R from n agents by selling them a digital good. Try #1: 1.Offer price R to the agents- sell if 1 agent accepts. 2.If not, offer price R/2- sell if 2 agents accept. 3.And so on… This mechanism is not DSIC! 42

43 Proof cont. (Ascending price mechanism) Solution: 1.Offer price R/n to all agents. Sell if all n agents accept. 2. If not, offer price R/(n-k) to the n-k agents who accepted the last offer. Sell if all n-k agents left accept. 3.And so on… This mechanism is DSIC. (Agents drop out when the price rises above their valuation.) 43

44 Profit extractor For all valuation profiles v, the digital good profit extractor for target R obtains revenue R if R ≤ EFO(v) and zero otherwise. 44

45 Proof EFO(v) = kv (k) for some k If R <= EFO(v) → R/k <= v (k) and the profit extractor will find this k. If R > EFO(v) → R > EFO(v) = max k kv (k) then there is no k for which R/k <= v (k) → The mechanism has no winners and no revenue. 45

46 Approximate Reduction to Decision Problem We use random sampling to approximately reduce the mechanism design problem of optimizing profit to profit extraction. 46

47 Random Sampling profit extraction auction The random sampling profit extraction auction works as follows: 1. Randomly partition the agents by flipping a fair coin for each agents and assigning her to S′ or S′′. 2. Calculate R′= EFO(v s’ ) and R′′= EFO(v s’’ ), the benchmark profit for each part. 3. Profit extract R′′ from S′ and R′ from S′′ 47

48 Random Sampling profit extraction auction The revenue of this mechanism is: min(R’, R’’). ↓ (Profit extractor is DSIC.) ↓ Random sampling profit extraction auction is DSIC. 48

49 Lemma Flip k > 1 coins then: E[min{#heads,#tails}] >= k/4 49

50 For digital good environments and all valuation profiles, the revenue of the random sampling profit extraction auction is a 4-approximation to the envy-free benchmark. 50

51 Proof Define: REF: Envy-free benchmark and its revenue. APX: Random sampling profit extraction auction and its expected revenue. (= E[min(R′,R′′)]) Assume k >=2 51

52 Proof cont. Assume that the envy-free benchmark sells to k agents at price p. → REF = kp Of the k Winners in REF let k’ be the number of them that are in S’, and k’’ the number in S’’. ↓ R’ >= k’p, R’’ >= k’’p ↓ 52

53 Proof cont. APX/REF = E[min(R′,R′′)]/kp ≥ E[min(k′p,k′′p)]/kp = E[min(k′,k′′)]/k ≥ ¼ (from the lemma) 53

54 The End 54


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