[PPT] - Announcements HW1 grading will be out next Tuesday, and sample PowerPoint Presentation

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Announcements

ØHW1 grading will be out next Tuesday, and sample solution is out

n Collab

ØHW 2 is due next Tuesday

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CS6501: T

pics in Learning and Game Theory

(Fall 2019) Mechanism Design from Samples

Instructor: Haifeng Xu

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Outline

Ø Optimal Auction and its Limitations Ø The Sample Mechanism and its Revenue Guarantee

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Recap: Optimal Auction for Single Item

Theorem. For single-item allocation with regular value distribution

𝑤" ∼ 𝑔

" independently, the following auction is BIC and optimal:

1. Solicit buyer values 𝑤%, ⋯ , 𝑤( 2. Transform 𝑤" to “virtual value” 𝜚"(𝑤") where 𝜚" 𝑤" = 𝑤" −

%./0(10) 20(10)

3. If 𝜚" 𝑤" < 0 for all 𝑗, keep the item and no payments 4. Otherwise, allocate item to 𝑗∗ = arg max

"∈[(] 𝜚"(𝑤") and charge him

the minimum bid needed to win, i.e., 𝜚"

.% max max ?@"∗ 𝜚?(𝑤?) , 0

.

Ø Recall, “regular” means 𝜚" 𝑤" is monotone non-decreasing Ø Will always assume distributions are regular and “nice” henceforth

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Recap: Optimal Auction for Single Item

An important special case: 𝑤" ∼ 𝐺 i.i.d.

ØThe second-price auction with reserve 𝜚.%(0) is optimal

1. Solicit buyer values 𝑤%, ⋯ , 𝑤(
2. If 𝑤" < 𝜚.%(0) for all 𝑗, keep the item and no payments
3. Otherwise, allocate to 𝑗∗ = arg max

"∈[(] 𝑤" and charge him the minimum

bid needed to win, i.e., max( max

?@"∗ 𝑤? , 𝜚.% 0

)

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Recap: Optimal Auction for Single Item

An important special case: 𝑤" ∼ 𝐺 i.i.d.

ØThe second-price auction with reserve 𝜚.%(0) is optimal

1. Solicit buyer values 𝑤%, ⋯ , 𝑤(
2. If 𝑤" < 𝜚.%(0) for all 𝑗, keep the item and no payments
3. Otherwise, allocate to 𝑗∗ = arg max

"∈[(] 𝑤" and charge him the minimum

bid needed to win, i.e., max( max

?@"∗ 𝑤? , 𝜚.% 0

)

Intuitions about why second-price auction with reserve is good

à second highest bid is pretty much the best choice Ø Incentive compatibility requires payment to not depend on bidder’s

wn bid

Ø Use the reserve to balance between “charging a higher price” and “disposing the item”

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Recap: Optimal Auction for Single Item

Myerson’s Lemma is central to the proof

Lemma. Consider any BIC mechanism 𝑁 with interim allocation 𝑦

and interim payment 𝑞, normalized to 𝑞" 0 = 0. The expected revenue of 𝑁 is equal to the expected virtual welfare served ∑"F%

(

𝔽10∼20 𝜚" 𝑤" 𝑦"(𝑤")

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Drawbacks of the Optimal Auction

ØIn this lecture, we will keep Assumption 1, but relax Assumption 2

1. Buyer’s value 𝑤" is assumed to be drawn from a distribution 𝑔

"

2. The precise distribution 𝑔

" is assumed to be known to seller

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Drawbacks of the Optimal Auction

ØIn this lecture, we will keep Assumption 1, but relax Assumption 2 ØThis is precisely the machine learning perspective

ML assumes data drawn from distributions
The precise distribution is unknown; instead samples are given

1. Buyer’s value 𝑤" is assumed to be drawn from a distribution 𝑔

"

2. The precise distribution 𝑔

" is assumed to be known to seller

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Task and Goal of This Lecture

ØWill focus on setting with 𝑜 buyer, i.i.d. values ØBuyer value 𝑤" is drawn from regular distribution 𝑔, which is

unknown to the seller Goal: design an auction that has revenue close to the optimal revenue when knowing 𝑔 Ø Optimal auction is a second-price auction with reserve 𝜚.%(0) Ø “Closeness” will be measured by guaranteed approximation ratio

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Task and Goal of This Lecture

ØWill focus on setting with 𝑜 buyer, i.i.d. values ØBuyer value 𝑤" is drawn from regular distribution 𝑔, which is

unknown to the seller Goal: design an auction that has revenue close to the optimal revenue when knowing 𝑔 Ø Optimal auction is a second-price auction with reserve 𝜚.%(0) Ø “Closeness” will be measured by guaranteed approximation ratio But wait . . . we cannot have any guarantee without assumptions

n bidder values – is this a contradiction?
No, we assumed 𝑤" ∼ 𝑔

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A Natural First Attempt

ØSince 𝑤"’s are all drawn from 𝑔, these 𝑜 i.i.d. samples can be used

to estimate 𝑔

ØThis results in the following “empirical Myerson” auction

Empirical Myerson Auction

1. Solicit buyer values 𝑤%, ⋯ , 𝑤(
2. Use 𝑤%, ⋯ , 𝑤( to estimate an empirical distribution ̅

𝑔

3. Run second-price auction with reserve J

𝜚.%(0) where J 𝜚 is calculated using ̅ 𝑔 instead Q: does this mechanism work?

No, may fail in multiple ways

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Issues of Empirical Myerson

ØNot incentive compatible – reserve depends on bidder’s report

This is a crucial difference from standard machine learning tasks

where samples are assumed to be correctly given

Empirical Myerson Auction

1. Solicit buyer values 𝑤%, ⋯ , 𝑤(
2. Use 𝑤%, ⋯ , 𝑤( to estimate an empirical distribution ̅

𝑔

3. Run second-price auction with reserve J

𝜚.%(0) where J 𝜚 is calculated using ̅ 𝑔 instead problematic

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Issues of Empirical Myerson

ØNot incentive compatible – reserve depends on bidder’s report

This is a crucial difference from standard machine learning tasks

where samples are assumed to be correctly given

ØEven bidders report true values, ̅

𝑔 may not be regular

ØEven ̅

𝑔 is regular, J 𝜚.%(0) may not be close to 𝜚.%(0)

Depend on how large is 𝑜, and shape of 𝑔

Empirical Myerson Auction

1. Solicit buyer values 𝑤%, ⋯ , 𝑤(
2. Use 𝑤%, ⋯ , 𝑤( to estimate an empirical distribution ̅

𝑔

3. Run second-price auction with reserve J

𝜚.%(0) where J 𝜚 is calculated using ̅ 𝑔 instead problematic

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Outline

Ø Optimal Auction and its Limitations Ø The Sample Mechanism and its Revenue Guarantee

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The Basic Idea

ØWant to use second-price auction with an estimated reserve ØLesson from previous example – if a bidder’s bid is used to

estimate the reserve, we cannot use this reserve for him

ØMain idea: pick a “reserve buyer” à use his bid to estimate the

reserve but never sell to this buyer

I.e., we give up any revenue from the reserve buyer

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The Basic Idea

ØWant to use second-price auction with an estimated reserve ØLesson from previous example – if a bidder’s bid is used to

estimate the reserve, we cannot use this reserve for him

ØMain idea: pick a “reserve buyer” à use his bid to estimate the

reserve but never sell to this buyer

I.e., we give up any revenue from the reserve buyer

Q: why only pick one reserve buyer, not two or more?

We have to give up revenue from reserve buyers, better not too many

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The Basic Idea

ØWant to use second-price auction with an estimated reserve ØLesson from previous example – if a bidder’s bid is used to

estimate the reserve, we cannot use this reserve for him

ØMain idea: pick a “reserve buyer” à use his bid to estimate the

reserve but never sell to this buyer

I.e., we give up any revenue from the reserve buyer

Q: why only pick one reserve buyer, not two or more?

We have to give up revenue from reserve buyers, better not too many

Q: which buyer to choose as the reserve buyer?

A-priori, they are the same à pick one uniformly at random

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The Basic Idea

ØWant to use second-price auction with an estimated reserve ØLesson from previous example – if a bidder’s bid is used to

estimate the reserve, we cannot use this reserve for him

ØMain idea: pick a “reserve buyer” à use his bid to estimate the

reserve but never sell to this buyer

I.e., we give up any revenue from the reserve buyer

Q: how to use a single buyer’s value to estimate reserve?

Not much we can do . . . just use his value as reserve

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The Mechanism

Second-Price auction with Random Reserve (SP-RR)

1. Solicit buyer values 𝑤%, ⋯ , 𝑤(
2. Pick 𝑘 ∈ [𝑜] uniformly at random as the reserve buyer
3. Run second-price auction with reserve 𝑤? but only

among bidders in 𝑜 ∖ 𝑘 .

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The Mechanism

Second-Price auction with Random Reserve (SP-RR)

1. Solicit buyer values 𝑤%, ⋯ , 𝑤(
2. Pick 𝑘 ∈ [𝑜] uniformly at random as the reserve buyer
3. Run second-price auction with reserve 𝑤? but only

among bidders in 𝑜 ∖ 𝑘 .

Claim. SP-RR is dominant-strategy incentive compatible.

For any bidder 𝑗 Ø If 𝑗 is picked as reserve, his bid does not matter to him, so truthful bidding is an optimal strategy Ø If 𝑗 is not picked, he faces a second-price auction with

reserve. Again, truthful bidding is optimal

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The Mechanism

Theorem. Suppose 𝐺 is regular. In expectation, SP-RR achieves

at least

% M ⋅ (.% ( fraction of the optimal expected revenue.

Remarks

Ø

% M ⋅ (.% ( is a worst-case guarantee

ØThe first time we use approximation as a lens to analyze

algorithms in this class

ØIt is possible to have a good auction even without knowing 𝐺

But we still assumed 𝑤" ∼ 𝐺 i.i.d.

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The Mechanism

Theorem. Suppose 𝐺 is regular. In expectation, SP-RR achieves

at least

% M ⋅ (.% ( fraction of the optimal expected revenue.

ØEquivalently, SP-RR is a second-price auction for (𝑜 − 1) i.i.d.

bidders, with a reserve 𝑠 drawn from 𝐺.

ØTo prove its revenue guarantee, we have to argue

1.

Discarding one buyer does not hurt revenue much (the

(.% (

term)

2.

Using a random 𝑤 ∼ 𝐺 as an estimated reserve is still good (the

% M

term)

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The Mechanism

Theorem. Suppose 𝐺 is regular. In expectation, SP-RR achieves

at least

% M ⋅ (.% ( fraction of the optimal expected revenue.

Next, we will give a formal proof

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Step 1: discarding a buyer does not hurt revenue much

Proof: use Myerson’s Lemma

ØExpected revenue for 𝑜 buyers is ∑"F%

(

𝔽10∼20 𝜚" 𝑤" 𝑦"

( 𝑤"

𝑦"

(() = interim allocation of the optimal auction for 𝑜 buyers

ØBy symmetry of the auction and buyer values, each buyer’s interim

allocation must be the same, i.e., 𝑦"

( (𝑤) = 𝑦(()(𝑤) for some 𝑦(()

ØThus, optimal revenue with 𝑜 bidders is

Lemma 1. The expected optimal revenue for an environment with (𝑜 − 1) buyers is at least

(.% (

fraction of the optimal expected revenue for 𝑜 buyers. 𝑆 𝑜 = ∑"F%

(

𝔽10∼20 𝜚" 𝑤" 𝑦"

( 𝑤"

= 𝑜 ⋅ 𝔽1∼2 𝜚 𝑤 𝑦 ( 𝑤

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Step 1: discarding a buyer does not hurt revenue much

Proof: use Myerson’s Lemma

ØDue to less competition, we have 𝑦((.%) 𝑤 ≥ 𝑦(()(𝑤)

They face the same reserve 𝜚.%(0), but with 𝑜 − 1 buyers, bidder 𝑗

has more chance to win

ØTherefore,

Lemma 1. The expected optimal revenue for an environment with (𝑜 − 1) buyers is at least

(.% (

fraction of the optimal expected revenue for 𝑜 buyers. 𝑆 𝑜 − 1 = 𝑜 − 1 ⋅ 𝔽1∼2 𝜚 𝑤 𝑦 (.% 𝑤 ≥ 𝑜 − 1 ⋅ 𝔽1∼2 𝜚 𝑤 𝑦 ( 𝑤 ≥ 𝑜 − 1 𝑜 𝑆(𝑜)

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Step 2: using random reserve is not bad

Lemma 2. Rev(SP-RR) ≥

% M Rev(SP-OR) for any 𝑜 and regular 𝐺.

Consider the following two auctions for i.i.d. bidders with 𝑤" ∼ 𝐺 Ø SP-OR: second price auction with optimal reserve 𝑠∗ = 𝜚.%(0) Ø SP-RR: second price auction with random reserve 𝑠 ∼ 𝐺 Note: this completes our proof of the theorem

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Lemma 2. Rev(SP-RR) ≥

% M Rev(SP-OR) for any 𝑜 and regular 𝐺.

Step 1: characterize how much revenue 𝑗 contribute in each auction What about SP-RR? Ø Similar argument, but use a random reserve 𝑠 instead Ø In expectation, 𝑗 contributes 𝔽Y∼/𝔽1X0[ W 𝑆 max 𝑢, 𝑠 ] In expectation, 𝑗 contributes 𝔽1X0[ W 𝑆 max 𝑢, 𝑠∗ ] in SP-OR Step 2: prove 𝔽Y∼/[ W 𝑆 max 𝑢, 𝑠 ≥

% M W

𝑆 max 𝑢, 𝑠∗ for any 𝑢

This proves Lemma 2

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ØNote: this is really the fundamental reason for why using uniform

reserve is not bad

ØProof is based on an elegant geometric argument

Claim. 𝔽Y∼/[ W

𝑆 max 𝑢, 𝑠 ] ≥

% M W

𝑆 max 𝑢, 𝑠∗ for any 𝑢.

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ØNote: this is really the fundamental reason for why using uniform

reserve is not bad

ØProof is based on an elegant geometric argument ØRecall W

𝑆 𝑞 = 𝑞 ⋅ (1 − 𝐺(𝑞)). The (not so) magic step: change variable for function W 𝑆 𝑞

Define new variable 𝑟 = 1 − 𝐺(𝑞), so 𝑞 = 𝐺.%(1 − 𝑟)
Define 𝑆 𝑟 = 𝑟 ⋅ 𝐺.%(1 − 𝑟)
Note: value of 𝑆 𝑟 equals value of W

𝑆 𝑞 (when 𝑟 = 1 − 𝐺(𝑞))

ØIt turns out that 𝑆(𝑟) is concave if and only if 𝐺 is regular

This is also the intrinsic interpretation of the regularity assumption
Claim. 𝔽Y∼/[ W

𝑆 max 𝑢, 𝑠 ] ≥

% M W

𝑆 max 𝑢, 𝑠∗ for any 𝑢.

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Calculating derivative of 𝑆 𝑟 = 𝑟 ⋅ 𝐺.%(1 − 𝑟):

Claim. 𝔽Y∼/[ W

𝑆 max 𝑢, 𝑠 ] ≥

% M W

𝑆 max 𝑢, 𝑠∗ for any 𝑢. 𝑒 𝑆(𝑟) 𝑒 𝑟 = 𝐺.% 1 − 𝑟 + 𝑟 ⋅ 𝑒 𝐺.%(1 − 𝑟) 𝑒 𝑟 = 𝐺.% 1 − 𝑟 − 𝑟 ⋅ 1 𝑔(𝐺.%(1 − 𝑟)) Derive on the board

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Calculating derivative of 𝑆 𝑟 = 𝑟 ⋅ 𝐺.%(1 − 𝑟):

Claim. 𝔽Y∼/[ W

𝑆 max 𝑢, 𝑠 ] ≥

% M W

𝑆 max 𝑢, 𝑠∗ for any 𝑢. 𝑒 𝑆(𝑟) 𝑒 𝑟 = 𝐺.% 1 − 𝑟 + 𝑟 ⋅ 𝑒 𝐺.%(1 − 𝑟) 𝑒 𝑟 = 𝐺.% 1 − 𝑟 − 𝑟 ⋅ 1 𝑔(𝐺.%(1 − 𝑟)) = 𝑞 − (1 − 𝐺(𝑞)) ⋅ 1 𝑔(𝑞) Use the equation 1 − 𝐺 𝑞 = 𝑟

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Calculating derivative of 𝑆 𝑟 = 𝑟 ⋅ 𝐺.%(1 − 𝑟):

Claim. 𝔽Y∼/[ W

𝑆 max 𝑢, 𝑠 ] ≥

% M W

𝑆 max 𝑢, 𝑠∗ for any 𝑢. 𝑒 𝑆(𝑟) 𝑒 𝑟 = 𝐺.% 1 − 𝑟 + 𝑟 ⋅ 𝑒 𝐺.%(1 − 𝑟) 𝑒 𝑟 = 𝐺.% 1 − 𝑟 − 𝑟 ⋅ 1 𝑔(𝐺.%(1 − 𝑟)) = 𝑞 − (1 − 𝐺(𝑞)) ⋅ 1 𝑔(𝑞) = 𝜚(𝑞) Use the equation 1 − 𝐺 𝑞 = 𝑟 Ø Regularity means 𝜚(𝑞) is increasing in 𝑞 Ø Moreover, 𝑞 is decreasing in 𝑟, so 𝑆′(𝑟) is decreasing in 𝑟 Ø This implies 𝑆(𝑟) is concave

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Claim. 𝔽Y∼/[ W

𝑆 max 𝑢, 𝑠 ] ≥

% M W

𝑆 max 𝑢, 𝑠∗ for any 𝑢. 𝑠∗ satisfies 𝜚 𝑠∗ = 0, i.e., the point where derivative of 𝑆 𝑟 is 0

Here: 1 − 𝐺 𝑠∗ = 𝑟∗

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Claim. 𝔽Y∼/[ W

𝑆 max 𝑢, 𝑠 ] ≥

% M W

𝑆 max 𝑢, 𝑠∗ for any 𝑢. First, prove the 𝑢 = 0 case. Claim (when 𝒖 = 𝟏). 𝔽Y∼/[ W 𝑆 𝑠 ] ≥

% M W

𝑆 𝑠∗ . Proof Ø 𝔽Y∼/ W 𝑆 𝑠 = 𝔽`∼a[b,%][𝑆 𝑟 ] by variable change 𝑟 = 1 − 𝐺(𝑠)

If 𝑠 ∼ 𝑔, then 𝐺 𝑠 ∼ 𝑉[0,1]

Ø 𝔽`∼a[b,%][𝑆 𝑟 ] is precisely the area under the 𝑆(𝑟) curve Ø W 𝑆 𝑠∗ = 𝑆(𝑟∗) is precisely the area of the rectangle Ø By geometry, 𝔽Y∼/[ W 𝑆 𝑠 ] ≥

% M W

𝑆 𝑠∗

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Claim. 𝔽Y∼/[ W

𝑆 max 𝑢, 𝑠 ] ≥

% M W

𝑆 max 𝑢, 𝑠∗ for any 𝑢. For general 𝑢 Ø If 𝑢 ≤ 𝑠∗, left-hand side increases, right-hand side no change Ø If 𝑢 > 𝑠∗, W 𝑆 max 𝑢, 𝑠∗ = W 𝑆 𝑢 𝔽Y∼/ W 𝑆 max 𝑢, 𝑠 = Pr 𝑠 ≤ 𝑢 ⋅ W 𝑆 𝑢 + Pr 𝑠 > 𝑢 ⋅ 𝔽Y∼/|hij W 𝑆 𝑠 Similar geometric argument shows 𝔽Y∼/|hij W 𝑆 𝑠 ≥

% M W

𝑆(𝑢) ≥ Pr 𝑠 ≤ 𝑢 ⋅ W 𝑆 𝑢 + Pr 𝑠 > 𝑢 ⋅ 1 2 W 𝑆(𝑢) ≥ 1 2 W 𝑆(𝑢)

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Remarks

ØApproximation ratio can be improved to

% M (i.e. without the (.% (

term)

Idea: don’t discard the reserve buyer; instead randomly choose

another buyer’s bid as the reserve for him

Ø

% M approximation is the best possible guarantee for SP-RR

The worst case is precisely when 𝑆(𝑟) curve is a triangle

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Remarks

ØIf we have sufficiently many bidders (more than Θ(𝜗.n ln 𝜗.%) many),

can obtain 𝜗-optimal auction

Idea: pick many reserve bidders and use their values to estimate a better

reserve

The estimation is tricky, not simply using the empirical distribution of the

reserve bidders’ values

ØThese results can all be generalized to “single-parameter” settings

E.g., selling 𝑙 identical copies of items to 𝑜 buyers

ØMany open questions in this broad field of learning optimal auctions

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Thank You

Haifeng Xu

University of Virginia hx4ad@virginia.edu