Yang Cai Oct 01, 2014. An overview of today’s class Myerson’s Auction Recap Challenge of Multi-Dimensional Settings Unit-Demand Pricing.

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Yang Cai Oct 01, 2014

An overview of today’s class Myerson’s Auction Recap Challenge of Multi-Dimensional Settings Unit-Demand Pricing

Myerson’s Auction Recap [Myerson ’81 ] For any single-dimensional environment. Let F= F 1 × F 2 ×... × F n be the joint value distribution, and (x,p) be a DSIC mechanism. The expected revenue of this mechanism E v~F [Σ i p i (v)]=E v~F [Σ i x i (v) φ i (v i )], where φ i (v i ) := v i - (1-F i (v i ))/f i (v i ) is called bidder i’s virtual value (f i is the density function for F i ).

1 1  Bidders report their values; transformed  The reported values are transformed into virtual - values;  the virtual-welfare maximizing allocation is chosen.  Charge the payments according to Myerson’s Lemma.  Transformation = depends on the distributions; deterministic function (the virtual value function); Myerson’s Auction Recap  Myerson’s auction looks like the following

Nice Properties of Myerson’s Auction  DSIC, but optimal among all Bayesian Incentive Compatible (BIC) mechanisms!  Deterministic, but optimal among all possibly randomized mechanisms!  Central open problem in Mathematical Economics: How can we extend Myerson’s result to Multi-Dimensional Settings?  Important progress in the past a few years.  See the Challenges first!

Challenges in Multi-Dimensional Settings

Example 1:  A single buyer, 2 non-identical items  Bidder is additive e.g. v({1,2}) = v 1 +v 2.  Further simplify the setting, assume v 1 and v 2 are drawn i.i.d. from distribution F = U{1,2} (1 w.p. ½, and 2 w.p. ½).  What’s the optimal auction here?  Natural attempt: How about sell both items using Myerson’s auction separately?

Example 1:  Selling each item separately with Myerson’s auction has expected revenue $2.  Any other mechanism you might want to try?  How about bundling the two items and offer it at $3?  What is the expected revenue?  Revenue = 3 × Pr[v 1 +v 2 ≥ 3] = 3 × ¾ = 9/4 > 2!  Lesson 1: Bundling Helps!!!

Example 1:  The effect of bundling becomes more obvious when the number of items is large.  Since they are i.i.d., by the central limit theorem (or Chernoff bound) you know the bidder’s value for the grand bundle (contains everything) will be a Gaussian distribution.  The variance of this distribution decreases quickly.  If set the price slightly lower than the expected value, then the bidder will buy the grand bundle w.p. almost 1. Thus, revenue is almost the expected value!  This is the best you could hope for.

Example 2:  Change F to be U{0,1,2}.  Selling the items separately gives $4/3.  The best way to sell the Grand bundle is set it at price $2, this again gives $4/3.  Any other way to sell the items?  Consider the following menu. The bidder picks the best for her. -Buy either of the two items for $2 -Buy both for $3

Example 2:  Bidder’s choice:  Expected Revenue = 3 × 3/9 + 2 × 2/9 =13/9 > 4/3! v 1 \v $0 $2 1$0 $3 2$2$3

Example 3:  Change F 1 to be U{1,2}, F 2 to be U{1,3}.  Consider the following menu. The bidder picks the best for her. -Buy both items with price $4. -A lottery: get the first item for sure, and get the second item with prob. ½. pay $2.50.  The expected revenue is $2.65.  Every deterministic auction — where every outcome awards either nothing, the first item, the second item, or both items — has strictly less expected revenue.  Lesson 2: randomization could help!

Unit-demand Bidder Pricing Problem

Unit-Demand Bidder Pricing Problem (UPP) 1 i n … …  A fundamental pricing problem v 1 ~ F 1 v i ~ F i v n ~ F n  Bidder chooses the item that maximizes v i - p i, if any of them is positive.  Revenue will be the corresponding p i.  Focus on pricing only, not considering randomized ones.  It’s known randomized mechanism can only get a constant factor better than pricing.

Our goal for UPP  Goal: design a pricing scheme that achieves a constant fraction of the revenue that is achievable by the optimal pricing scheme.  Assumption: F i ’s are regular. Theorem [CHK ‘07]: There exists a simple pricing scheme (poly-time computable), that achieves at least ¼ of the revenue of the optimal pricing scheme. Remark: the constant can be improved with a better analysis.

What is the Benchmark???  When designing simple nearly-optimal auctions. The benchmark is clear.  Myerson’s auction, or the miximum of the virtual welfare.  In this setting we don’t know what the optimal pricing scheme looks like.  We want to compare to the optimal revenue, but we have no clue what the optimal revenue is?  Any natural upper bound for the optimal revenue?

 (a) UPP  One unit-demand bidder  n items  Bidder’s value for the i-th item v i is drawn independently from F i  (b) Auction  n bidders  One item  Bidder I’s value for the item v i is drawn independently from F i Two Scenarios 1 i n … v 1 ~ F 1 v i ~ F i v n ~ F n Item 1 i n … … Bidders v 1 ~ F 1 v i ~ F i v n ~ F n

Benchmark Lemma 1: The optimal revenue achievable in scenario (a) is always less than the optimal revenue achievable in scenario (b). - Proof: See the board. - Remark: This gives a natural benchmark for the revenue in (a).

An even simpler benchmark  In a single-item auction, the optimal expected revenue E v~F [max Σ i x i (v) φ i (v i )] = E v~F [max i φ i (v i ) + ] (the expected prize of the prophet)  Remember the following mechanism RM we learned in Lecture 6. 1.Choose t such that Pr[max i φ i (v i ) + ≥ t] = ½. 2.Set a reserve price r i =φ i -1 (t) for each bidder i with the t defined above. 3.Give the item to the highest bidder that meets her reserve price (if any). 4.Charge the payments according to Myerson’s Lemma.  By prophet inequality: ARev(RM) = E v~F [Σ i x i (v) φ i (v i )] ≥ ½ E v~F [max i φ i (v i ) + ] = ½ ARev(Myerson)  Let’s use the revenue of RM as the benchmark.

Inherent loss of this approach  Relaxing the benchmark to be Myerson’s revenue in (b)  This step might lose a constant factor already.  To get real optimal, a different approach is needed.