Revenue optimal auctions Recap Last week: Vickrey-Clarke-Groves - PowerPoint PPT Presentation

Algorithmic game theory Ruben Hoeksma December 10, 2018 Revenue optimal auctions

Recap Last week: ◮ Vickrey-Clarke-Groves auctions ◮ Revenue equivalence Today: ◮ Optimal revenue single item auction (Myerson’s auction)

Single item, single bidder Auctioneer wants to get the maximum price (revenue) for their item. ◮ Single item ◮ Single bidder with private value v ∼ ϕ Q: What is the revenue of the Vickrey auction? ◮ Vickrey auction results in revenue of 0 Q: Can we do better? ◮ Revenue equivalence says that always allocating the item needs to result in revenue of 0! Q: How do we maximize revenue? Maximize expected revenue: max r · P [ v ≥ r ] r

Single item, two bidders ◮ Single item ◮ Two bidders, Alice and Bob with private values a , b ∼ U (0 , 1) The Vickrey auction has expected revenue: E [min { a , b } ] = 1 3 Q: Can we do better? We have to change the allocation rule ◮ Reserve price r ◮ If a < r and b < r nobody gets the item ◮ If one > r and one < r highest bid wins with price r ◮ Otherwise, highest bid wins and pays second highest bid Q: What is the expected revenue for reserve price r ?

Single item, two bidders - second price with reserve Let X be the revenue of the auction E [ X ] = E [ X | a < r , b < r ] · P [ a < r ] · P [ b < r ] + 2 · E [ X | a ≥ r , b < r ] · P [ a ≥ r ] · P [ b < r ] + E [ X | a ≥ r , b ≥ r ] · P [ a ≥ r ] · P [ b ≥ r ] Note: ( a | a ≥ r ) ∼ U ( r , 1) ◮ Still dominant strategy to bid true value [homework exercise] ◮ If a , b < r ; probability r 2 ; revenue 0 ◮ If a ≥ r , b < r ; probability r (1 − r ); revenue r ◮ If a , b ≥ r ; probability (1 − r ) 2 ; revenue min { a , b } E [ X ] = 0 · r 2 + 2 · r 2 (1 − r ) + E [min { a , b } | a , b ≥ r ] · (1 − r ) 2 · (1 − r ) 2 = 1 + 3 r 2 − 4 r 3 = 2 · r 2 (1 − r ) + 1 + 2 r 3 3

Model ◮ Bidders with private valuations v i ∈ V i ◮ Known discrete distribution ϕ of the valuations ( ϕ ( v j i ) = P [ v i = v j i ]) ◮ Players’ strategies: bid b i ∈ V i (depending on true v i ) ◮ Auction output: allocation ( a ( b )) + payments ( π ( b )) ◮ Payments: individual rational (IR) ◮ Truthful auction: Dominant strategy incentive compatible (DSIC) For any bids v ∈ V , let a i ( v ) ∈ { 0 , 1 } denote if player i is allocated the item (wins). (IR) v i a i ( v i , v − i ) − π i ( v i , v − i ) ≥ 0 (DSIC) v i a i ( v i , v − i ) − π i ( v i , v − i ) ≥ v i a i ( b i , v − i ) − π i ( b i , v − i )

Feasible allocation Q: How can we tell that an allocation rule can be used in a truthtelling auction? A: If there exist feasible payments such that (IR) and (DSIC) hold Q: How can we compute feasible payments for a given allocation rule? A: Use (DSIC) and (IR) constraints (DSIC) v i a i ( v i , v − i ) − π i ( v i , v − i ) ≥ v i a i ( b i , v − i ) − π i ( b i , v − i ) ⇔ π i ( v i , v − i ) − π i ( b i , v − i ) ≤ v i a i ( v i , v − i ) − v i a i ( b i , v − i ) ⇔ π i ( v i , v − i ) − π i ( b i , v − i ) ≤ v i ( a i ( v i , v − i ) − a i ( b i , v − i )) (DSIC) gives the maximum difference between payments if player j bids v i and when she bids b i , given v − i .

How to compute payments visually – type graph For a given allocation rule (given by a ( v )) and fixed v − i : We omit the dependence on v − i in the notation from here. (DSIC) π i ( v i ) ≤ π i ( b i ) + v i ( a i ( v i ) − a i ( b i )) v 1 i ( a i ( v 1 i ) − a i ( v 2 i )) v 1 v 2 i i Length of the edge ( v i , v ′ i ): the amount that the payment in ( v i , v − i ) can be larger than in ( v ′ i , v − i ).

How to compute payments visually – type graph · · · dummy v 1 v 2 v k i i i Steps to compute payments: ◮ For each player i and each v − i , create a complete digraph D with a node for each valuation (type). ◮ Any node potential on D is a payment rule that satisfies (BNIC). Lemma There is a feasible node potential on D iff it has no negative cycles.

How to compute payments visually – type graph Lemma There is a feasible node potential on D iff it has no negative cycles. Proof. “ ⇒ ”: If there is a negative cycle, no labeling of the nodes is feasible, since any node on the negative cycle needs to have value higher than itself. “ ⇐ ”: If there is no negative cycle: ◮ Set all node labels to zero. ◮ Increase the value of each label such that the constraints are satisfied given the previous assignment. ◮ Repeat until nothing changes. Since there are no negative cycles, each iteration fixes one label.

How to compute payments visually – type graph · · · dummy v 1 v 2 v k i i a i ( dummy ) = 0 i π i ( dummy ) = 0 Steps to compute payments: ◮ Create a complete digraph D with a node for each type. ◮ Any node potential on D is a payment rule that satisfies (BNIC). No upper bound on the node potentials: all payments can be arbitrarily large. (IR) v i ( t i )( a i ( t i ) − a i ( dummy )) − π i ( t i ) + π i ( dummy ) ≥ 0

Graph representation – type graph · · · dummy v 1 v 2 v k i i i Compute maximal payments by finding the largest value node potential. Minimum length paths from any node to the dummy node give maximum payment. Order the valuations s.t. v 1 i > v 2 i > . . . > v k i . Lemma D contains no negative cycles iff a is monotone in v i , i.e. a ( v j i ), for all v j i ) ≥ a ( v ℓ i > v ℓ i .

Lemma D contains no negative cycles iff a is monotone in v i , i.e. a ( v j i ), for all v j i ) ≥ a ( v ℓ i > v ℓ i . Proof. “ ⇒ ” If not monotone, a ( v j i ), for some v j i ) < a ( v ℓ i > v ℓ i , and v j i ( a ( v j i ) − a ( v j i ) − a ( v ℓ i )) + v ℓ i ( a ( v ℓ i )) = ( v j i )( a ( v j i − v ℓ i ) − a ( v ℓ i )) “ ⇐ ” Suppose monotone, let C be minimum length negative cycle. i be its largest value and ( v j Let v ℓ i , v ℓ i ) , ( v ℓ i , v k i ) ∈ C with j , k < ℓ . Then, v j i ( a ( v j i ) − a ( v ℓ i )) + v ℓ i ( a ( v ℓ i ) − a ( v k i )) � � � � = v j a ( v j i ) − a ( v ℓ +1 ) + a ( v ℓ +1 ) − a ( v ℓ + v ℓ a ( v ℓ i ) − a ( v k i ) i ) i i i i � � � � ≥ v j a ( v j i ) − a ( v j +1 + v j +1 a ( v j +1 ) − a ( v ℓ ) i ) . i i i i

Proof. “ ⇐ ” Suppose monotone, let C be minimum length negative i be its largest value and ( v j cycle. Let v ℓ i , v ℓ i ) , ( v ℓ i , v k i ) ∈ C with j , k < ℓ . Then, v j i ( a ( v j i ) − a ( v ℓ i )) + v ℓ i ( a ( v ℓ i ) − a ( v k i )) = v j � a ( v j � i ) − a ( v ℓ +1 ) + a ( v ℓ +1 ) − a ( v ℓ i ) i i i � � + v ℓ a ( v ℓ i ) − a ( v ℓ +1 ) + a ( v ℓ +1 ) − a ( v k i ) i i i � � � � = v j a ( v j + v j i ) − a ( v ℓ +1 a ( v ℓ +1 ) − a ( v ℓ ) i ) i i i i � � � � + v ℓ a ( v ℓ i ) − a ( v ℓ +1 + v ℓ a ( v ℓ +1 ) − a ( v k ) i ) i i i i � � � � = v j a ( v j i − v j i ) − a ( v ℓ +1 + ( v ℓ a ( v ℓ i ) − a ( v ℓ +1 ) i ) ) i i i � � + v ℓ a ( v ℓ +1 ) − a ( v k i ) i i � � � � > v j a ( v j i ) − a ( v ℓ +1 + v ℓ +1 a ( v ℓ +1 ) − a ( v k ) i ) . i i i i

The reduced type graph Lemma D contains no negative cycles iff a is monotone in v i , i.e. a ( v j i ), for all v j i ) ≥ a ( v ℓ i > v ℓ i . Corollary There is a payment rule π , such that ( a , π ) is a DSIC IR auction, iff a is monotone. Corollary If a is monotone, the shortest ( v j i , dummy )-path in D is ( v j i , v j +1 , . . . , v k i , dummy ). i · · · dummy . . . v 1 v 2 v k > > > i i i

The optimal auction Given a feasible allocation rule ( a ), we now know how to compute the payments. · · · dummy . . . v 1 v 2 v k > > > i i i Thus, we can compute the expected revenue we get from player i : � ϕ ( v j i ) π ( v j ϕ ( v j v j i ( a ( v j i ) − a ( v j +1 � � i ) = i ) )) i v j v j i ∈ V i i ∈ V i � + v j +1 ( a ( v j +1 ) − a ( v j +2 )) + . . . + v k i a ( v k i ) i i i � � a ( v j ϕ ( v j i ) v j i − P [ v i > v j i ]( v j − 1 − v j � = i ) i ) i v j i ∈ V i ϕ ( v j i ) v j i a ( v j � = i ) v j i ∈ V i

The optimal auction Maximize the total expected revenue � � � � max ϕ ( v ) π i ( v ) = max ϕ ( v ) v i a i ( v ) , v ∈ V i ∈ N v ∈ V i ∈ N where the virtual valuations are equal to i − P [ v i > v j i ] v j i = v j ( v j − 1 − v j v 1 i = v 1 and i ) . i i ϕ ( v j i ) The optimal allocation rule solves the optimization problem for every bid vector. Optimal auction? 1. Compute virtual values 2. For any given vector of bids assign the item to the player with the highest positive virtual valuation.

Regular distributions and ironing What if the optimal allocation rule is not monotone? Definition A distribution ϕ is regular if v j i ⇔ v j i ≥ v ℓ i > v ℓ i . Lemma If the distribution ϕ is regular, the allocation rule is monotone. Ironing If the distribution ϕ is not regular, we iron the virtual weights as follows: = ϕ ( v j i ) v j i + ϕ ( v j +1 ) v j +1 v j , j +1 i i i ϕ ( v j i ) + ϕ ( v j +1 ) i