Optimal mechanisms with simple menus
OOptimal mechanisms with simple menus
Zihe WangIIIS, Tsinghua [email protected] Pingzhong TangIIIS, Tsinghua [email protected] 9, 2018
Abstract
We consider optimal mechanism design for the case with one buyer and two items. The buyer’svaluations towards the two items are independent and additive. In this setting, optimal mechanism isunknown for general valuation distributions. We obtain two categories of structural results that shedlight on the optimal mechanisms. These results can be summarized into one conclusion: under certainconditions, the optimal mechanisms have simple menus.The first category of results state that, under certain mild condition, the optimal mechanism hasa monotone menu. In other words, in the menu that represents the optimal mechanism, as paymentincreases, the allocation probabilities for both items increase simultaneously. This theorem complementsHart and Reny’s recent result regarding the nonmonotonicity of menu and revenue in multi-item settings.Applying this theorem, we derive a version of revenue monotonicity theorem that states stochasticallysuperior distributions yield more revenue. Moreover, our theorem subsumes a previous result regardingsufficient conditions under which bundling is optimal [Hart and Nisan 2012]. The second category ofresults state that, under certain conditions, the optimal mechanisms have few menu items. Our firstresult in this category says, for certain distributions, the optimal menu contains at most 4 items. Thecondition admits power (including uniform) density functions. Our second result in this category worksfor a weaker condition, under which the optimal menu contains at most 6 items. This condition includesexponential density functions. Our last result in this category works for unit-demand setting. It statesthat, for uniform distributions, the optimal menu contains at most 5 items. All these results are in sharpcontrast to Hart and Nisan’s recent result that finite-sized menu cannot guarantee any positive fraction ofoptimal revenue for correlated valuation distributions.
Optimal mechanism design has been a topic of intensive research over the past thirty years. The generalproblem is, for a seller, to design a revenue-maximizing mechanism for selling k items to n buyers, giventhe buyers’ valuations distributions but not the actual values. A special case of the problem, where thereis only one item ( k = 1 ) and buyers have independent valuation distributions towards the item, has beenresolved by Myerson’s seminal work [Myerson 1981]. Myerson’s approach has turned out to be quitegeneral and has been successfully applied to a number of similar settings, such as [Maskin and Riley 1989,Jehiel et al. 1996, Levin 1997, Ledyard 2007, Deng and Pekeˇc 2011], just to name a few.While this line of work has flourished, it does not deepen our understanding of the cases with more thanone items ( k > ). In fact, even for the simplest multi-item case, where there are two independent items(k=2) and one buyer (n=1) with additive valuations, a direct characterization of the optimal mechanism isstill open for general, especially continuous, valuation distributions.When the distributions are discrete, Daskalakis and Weinberg [2011], Cai et al. [2012a,b] show thatthe general optimal mechanism ( k > ) is the solution of a linear program. They provide different meth-ods to solve the linear program efficiently. For continuous distributions, Chawla et al. [2010], Cai and1 a r X i v : . [ c s . G T ] A p r uang [2013] study the possibility of using simple auctions to approximate optimal auctions. In addition,Daskalakis and Weinberg [2012], Cai and Huang [2013] provide PTAS of the optimal auction under variousassumptions on distributions.Zoom in and look at the case with two independent items and a single buyer, much progress has beenmade in this particular setting. Hart and Nisan [2012] investigate two simplest forms of auctions: sellingthe two items separately and selling them as a bundle. They prove that selling separately obtains at leastone half of the optimal revenue while bundling always returns at least one half of separate sale revenue.They further extend these results to the general case with k independent items: separate sale guaranteesat least a clog k fraction of the optimal revenue; for identically distributed items, bundling guarantees atleast a clog k fraction of the optimal revenue. Li and Yao [2013] tighten these lower bounds to clog k and c respectively. Under some technical assumptions, Daskalakis et al. [2013] show close relation betweenmechanism design and transport problem and use techniques there to solve for optimal mechanisms in afew special distributions. Hart and Nisan [2013] investigate how the “menu size” of an auction can effectthe revenue and show that revenue of any finite menu-sized auction can be arbitrarily far from optimal (thusconfirm an earlier consensus that restricting attention to deterministic auctions, which has an exponentially-sized (at most) menu, indeed loses generality). On the economic front, Manelli and Vincent [2006, 2007],Pavlov [2011a,b] obtain the optimal mechanisms in several specific distributions (such as both items aredistributed according to uniform [0,1]). We will discuss these results in much more detail as we proceed torelevant sections.In this paper, we study the case with one buyer and two independent items, in hopes of a direct character-ization of exact optimal mechanisms. We obtain several exciting structural results. Our overall conclusionis that, under fairly reasonable conditions, optimal mechanisms has “simple” menus.
We summarize ourresults below into two parts, based on the conditions under which the results hold, as well as differentinterpretations of simplicity.For ease of presentation, we need the following definition: for a density function h , the power rate of h is P R ( h ( x )) = xh (cid:48) ( x ) h ( x ) . • Part I (Section 4). If density functions f and f satisfy P R ( f ( x )) + P R ( f ( y )) ≤ − , ∀ x, y . Theoptimal mechanism has a monotone menu – sort the menu items in ascending order of payments, theallocation probabilities of both items increase simultaneously – a desirable property that fails to holdin general (cf [Hart and Reny 2012]). Our result complements this observation in general and has twoimportant implications in particular.1. [Hart and Nisan 2012, Theorem 28]. Hart and Nisan show that, if two item distribution arefurther identical (i.e., f = f ), bundling sale is optimal. Our result subsumes this theorem as acorollary.2. A revenue monotonicity theorem. Based on menu monotonicity theorem, we are able to provethat, stochastically superior distributions yield higher revenue, another desirable property thatfails to hold in general.Our proof is semi-constructive in the sense that we fix part of the buyer utility function (for this part,relation between revenue and buyer utility is unknown/undesirable) and construct the remainder ofthe utility function (for this part, relation between revenue and buyer utility is known/desirable). Thistechnique might be of independent interest. • Part II. (Section 5). If the density functions f and f satisfy P R ( f ( x )) + P R ( f ( y )) ≥ − , ∀ x, y .The optimal mechanisms often contain few menu items. In particular,1. If both P R ( f ( x )) and P R ( f ( y )) are constants, the optimal mechanism contains at most 4menu items. The result is tight. Constant power rate is satisfied by a few interesting classesof density functions, including power functions h ( x ) = ax b and uniform density as a special2ase. This is consistent with earlier results for uniform distributions [Manelli and Vincent 2006,Pavlov 2011a]: the optimal mechanisms indeed contain four menu items.2. If P R ( f ( x )) ≤ y A f ( y A ) − and P R ( f ( y )) ≤ x A f ( x A ) − , where x A and y A are thelowest possible valuations for item one and two respectively, the optimal mechanism contains 2menu items.3. If we relax the condition to be that both P R ( f ( x )) and P R ( f ( y )) are monotone, then theoptimal mechanism contains at most 6 menu items. This condition is general and admits densityfunctions such as exponential density and polynomial density.4. Our last result requires the buyer demands at most one item. Under this condition, for uniformdensities, the optimal mechanism contains at most 5 menu items. The result is tight.These results are in sharp contrast to Hart and Nisan’s recent result that there is some distributionwhere finite number of menu items cannot guarantee any fraction of revenue [Hart and Nisan 2013].Here we show that, for several wide classes of distributions, the optimal mechanisms have a finite andeven extremely simple menus.Our proofs for this part are based on Pavlov’s characterization and careful analyses of how the revenuechanges as a function of the buyer’s utility. A rough line of reasoning is as follows, the “extremepoints” in the set of convex utility functions on the boundary values are piecewise linear functions.Since the utility on the boundaries contains only few linear pieces and the utility on inner values arelinearly related to that on the boundary, it must be the case that the utility function on the inner pointscontains only few linear pieces as well. In other words, the mechanism only contains few menu items.Our results not only offer original insights of “what do optimal mechanisms look like”, but are also inline with the “simple versus optimal” literature (cf [Hartline and Roughgarden 2009, Hart and Nisan 2012]):in our case, simple mechanisms are exactly optimal. We consider a setting with one seller who has two distinct items for sale, and one buyer who has privatevaluation x for item 1, y for the item 2, and x + y for both items. The seller has zero valuation for any subsetof items.As usual, x and y are unknown to the seller and are treated as independent random variables according todensity functions f on [ x A , x B ] ⊂ R and f on [ y A , y C ] ⊂ R respectively. The valuation (aka. type) spaceof the buyer is then V = [ x A , x B ] × [ y A , y C ] . To visualize, we sometimes refer to V as rectangle ABDC ,where A represents the lowest possible type ( x A , y A ) and D represents the highest possible type ( x B , y C ) .Let f ( x, y ) = f ( x ) f ( y ) be the joint density on V . We assume the f and f are positive, bounded, anddifferentiable densities.The seller sells the items through a mechanism that consists of an allocation rule q and a payment rule t . In our two-item setting, an allocation rule is conveniently represented by q = ( q , q ) , where q i is theprobability that buyer gets item i ∈ { , } . Given valuation ( x, y ) , buyer’s utility is u ( x, y ) = xq ( x, y ) + yq ( x, y ) − t ( x, y ) In other words, buyer has a quasi-linear, additive utility function. It is sometimes convenient to viewa mechanism as a (possibly infinite) set of menu items { ( q ( x, y ) , q ( x, y ) , t ( x, y )) | ( x, y ) ∈ [ x A , x B ] × [ y A , y C ] } . Given a mechanism, the expected revenue of the seller is R = E ( x,y ) [ t ( x, y )] .A mechanism must be Individually Rational (IR) : ∀ ( x, y ) , u ( x, y ) ≥ . In other words, a buyer cannot get negative utility by participation.3y revelation principle, it is without loss of generality to focus on the set of mechanisms that are
Incen-tive Compatible (IC) : ∀ ( x, y ) , ( x (cid:48) , y (cid:48) ) , u ( x, y ) ≥ xq ( x (cid:48) , y (cid:48) ) + yq ( x (cid:48) , y (cid:48) ) − t ( x (cid:48) , y (cid:48) ) . This means, it is the buyer’s (weak) dominant strategy to report truthfully. Equivalently, an IC mechanismpresents a set of menu items and let the buyer do the selection (aka. the taxation principle). As a result, u ( x, y ) = sup ( x (cid:48) ,y (cid:48) ) { xq ( x (cid:48) , y (cid:48) ) + yq ( x (cid:48) , y (cid:48) ) − t ( x (cid:48) , y (cid:48) ) } , which is the supremum of a set of linear functions of ( x, y ) . Thus, u must be convex.Fixing y , by IC , we have u ( x (cid:48) , y ) − u ( x, y ) − q ( x, y )( x (cid:48) − x )= x (cid:48) q ( x (cid:48) , y ) + yq ( x (cid:48) , y ) − t ( x (cid:48) , y ) − xq ( x, y ) − yq ( x, y ) + t ( x, y ) − x (cid:48) q ( x, y ) + xq ( x, y )= x (cid:48) q ( x (cid:48) , y ) + yq ( x (cid:48) , y ) − t ( x (cid:48) , y ) − ( x (cid:48) q ( x, y ) + yq ( x, y ) − t ( x, y )) ≥ Substitute x (cid:48) twice by x − = x − (cid:15) and x + = x + (cid:15) respectively, for any arbitrarily small positive (cid:15) , we have pu x ( x − , y ) ≤ q ( x, y ) ≤ u x ( x + , y ) , where u x denotes the partial derivative of u on the x dimension. The inequality above implies u is differ-entiable almost everywhere on x and u x = q ( x, y ) . Similarly, u is differentiable almost everywhere on y and u y = q ( x, y ) . As a result u x and u y must be within interval [0 , . This means, the seller can neverallocate more than one pieces of either item. Now, payment function t can be represented by utility function u , t ( x, y ) = xu x ( x, y ) + yu y ( x, y ) − u ( x, y ) .The seller’s problem is to design a non-negative, convex utility function, whose partial derivatives onboth x and y are within [0 , , that maximizes expected revenue R (cf. [Hart and Nisan 2012, Lemma 5]). Let Ω denote any area in V and R Ω be the revenue obtained within Ω . Let z = ( x, y ) T and T ( z ) = zu ( z ) f ( z ) .By Green’s Theorem, we have (cid:82) Ω ∇ · T dz = (cid:72) ∂ Ω T · ˆn ds. ∇ · T = 2 u ( z ) f ( z ) + ( ∇ u ( z )) T zf ( z ) + u ( z ) z T ∇ f ( z )= [( ∇ u ( z )) T z − u ( z )] f ( z ) + [3 f ( z ) + z T ∇ f ( z )] u ( z )= t ( z ) f ( z ) + (cid:52) ( z ) u ( z ) where (cid:52) ( x, y ) = 3 f ( x ) f ( y ) + xf (cid:48) ( x ) f ( y ) + yf (cid:48) ( y ) f ( x ) .Seller’s revenue formula within Ω is as follows: R Ω = (cid:90) Ω t ( z ) f ( z ) dz = (cid:90) Ω ( ∇ · T − (cid:52) ( z ) u ( z )) dz = (cid:73) ∂ Ω T · ˆn ds − (cid:90) Ω (cid:52) ( z ) u ( z ) dz Set Ω to be the rectangle ABDC , the seller’s total revenue R ABDC is (cid:90) y C y A x B u ( x B , y ) f ( x B ) f ( y ) dy + (cid:90) x B x A y C u ( x, y C ) f ( x ) f ( y C ) dx − (cid:90) y C y A x A u ( x A , y ) f ( x A ) f ( y ) dy − (cid:90) x B x A y A u ( x, y A ) f ( x ) f ( y A ) dx − (cid:90) x B x A (cid:90) y C y A u ( x, y ) (cid:52) ( x, y ) dydx (1)4ormula (1) consists of 5 terms. The first term represents the part of seller’s revenue that depends onutilities on edge BD only. Moreover, this part is increasing as utilities on edge BD increase. Similarly,the second term represents the part of seller’s revenue that depends positively on utilities on edge CD . Thethird and fourth terms represent respectively the parts of seller’s revenue that depend negatively on utilitieson edges AC and AB . The fifth term represents the part of revenue that depends on the utilities on theinner points of the rectangle. Under different conditions, (cid:52) ( x, y ) can be either positive or negative, whichsuggests this part can either increase or decrease as the utilities on inner points increases. We now definethese conditions. Definition 3.1
For any density h ( x ) , let P R ( h ( x )) = xh (cid:48) ( x ) h ( x ) be the power rate of h . Consider the following two conditions regarding power rate.
Condition 1:
P R ( f ( x )) + P R ( f ( y )) ≤ − , ∀ ( x, y ) ∈ V. Condition 2:
P R ( f ( x )) + P R ( f ( y )) ≥ − , ∀ ( x, y ) ∈ V. Clearly, under Condition 1, we have (cid:52) ( x, y ) ≤ . This means seller’s revenue depends positively onutilities of the inner points. Similarly, under Condition 2, seller’s revenue depends negatively on utilities ofthe inner points.Based on the two conditions above, we obtain two parts of results: under Condition 1, the optimalmechanisms have simple menus in the sense that their menus are monotone – allocation probabilities andpayment are increasing in the same order; under Condition 2, the optimal mechanisms also have simplemenus, but in a different sense, that their menus only contain a few items.Daskalakis et al. [2013] consider the same problem but restrict to the case where y A f ( y A ) = 0 , x A f ( x A ) = 0;lim x → x B x f ( x ) = 0 , lim y → y C y f ( y ) = 0 . These assumptions ignore the effect of utilities on the edges of the rectangle. While we do not have anyof these constraints. As a result, their techniques (such as reduction to optimal transport) do not apply toour more general case. In fact, one of our main techniques is to show how the optimal revenue changes as afunction of the utilities on the edges.
In this section, we consider the case where power rates of both density functions satisfy Condition 1. Whenthis condition is not met, Hart and Reny [2012] give several interesting counter-examples of revenue mono-tonicity: the optimal revenue for stochastically inferior valuation distributions may be greater than that ofstochastically superior distributions. When this condition is met, for identical item distributions, Hart andNisan [2012] prove that, bundling sale is the optimal mechanism. In this section, we show that, under Con-dition 1, the optimal menu can be sorted so that both allocations as well as payment monotonically increase.We coin this result menu monotonicity theorem . The theorem has two immediate consequences. First, ityields a version of revenue monotonicity theorem, thus complements the Hart-Reny result above. Second,it subsumes the above Hart-Nisan result as a corollary.Our analysis starts from a simple observation: any optimal mechanism must extract all of the buyer’svaluation when he is in the lowest type.
Lemma 4.1
In the optimal mechanism, u ( x A , y A ) = 0 . roof. Suppose otherwise that u ( x A , y A ) > , one can revise every menu item from ( q ( x, y ) , q ( x, y ) , t ( x, y )) to ( q ( x, y ) , q ( x, y ) , t ( x, y ) + u ( x A , y A )) and obtain a mechanism with strictly higher revenue, contradic-tion. (cid:117)(cid:116) Theorem 4.2
Menu Monotonicity
Under Condition 1, each menu item of the optimal mechanism can be labeled by a real number s : ( q ( s ) , q ( s ) ,t ( s )) , such that q ( s ) , q ( s ) is weakly increasing and t ( s ) are strictly increasing in s . Roughly speaking, Theorem 4.2 suggests that, among the menu items of the optimal mechanism, higher t corresponds to higher q and q . Note that allocation and payment monotonicity are well understood insingle-item optimal auction (i.e., Myerson auction) but in general fail to hold in two item settings [Hart andReny 2012].In the following, we give a semi-constructive proof. By Formula (1), under Condition 1, we know thatseller’s revenue is increasing as the utilities of the buyer increases on V , only except on edges AB and AC . Our idea is then, to fix the optimal utility function on AB and AC and construct the (largest possible)remainder of the utility function subject to convexity. Proof.
By Lemma 4.1, u ( x A , y A ) = 0 . In the following, we start from the optimal utility function u . Let u ( AB ) denote the part of u on edge AB . Similar for u ( AC ) .Let u ( x ,y ) , ( x, y ) = xq ( x , y )+ yq ( x , y ) − t ( x , y ) for any ( x , y ) . In other words, u ( x ,y ) , ( x, y ) is the buyer’s utility at type ( x, y ) but chooses menu item ( q ( x , y ) , q ( x , y ) , t ( x , y )) . By IC, u ( x ,y ) , ( x, y ) ≤ u ( x, y ) . To visualize, think of u ( x ,y ) , ( x, y ) as a plane that is always weakly below u ( x, y ) but touches u at the point of ( x , y ) .Apply the following two-step operation:Figure 1: A menu item in the optimal mechanism • Step 1. Translation. Rise up the plane of u ( x ,y ) , uniformly for every ( x, y ) until it touches oneof u ( AB ) and u ( AC ) , say u ( AB ) . Denote the plane after Step 1 as u ( x ,y ) , and notice the part of u ( x ,y ) , on AB , which is a straight line. • Step 2. Rotation. Fix u ( x ,y ) , ( AB ) , rotate the plane u ( x ,y ) , until it touches u ( AC ) .(We must alsoconsider the case where the slope of u ( x ,y ) , ( AC ) reaches 1 but still does not touches u ( AC ) . If thisoccurs, simply return the current plane.) Denote the plane after Step 2 as u ( x ,y ) , . In Fig. 1, the reddashed plane represents plane u ( x ,y ) , , it touches both u ( AB ) and u ( AC ) .We have u ( x ,y ) , ( x , y ) ≥ u ( x ,y ) , ( x , y ) ≥ u ( x ,y ) , ( x , y ) = u ( x , y ) . Repeat the sameprocedure for all ( x , y ) and define another function u ∗ = sup ( x ,y ) { u ( x ,y ) , } . In other words, u ∗ is thesupremum of all the planes after the two-step operation.6e claim that, when the u is optimal, u ∗ must be optimal as well. In fact, by the construction of u ∗ ,we know that u ∗ ( x, y ) ≥ u ( x,y ) , ( x, y ) ≥ u ( x, y ) for any ( x, y ) . Also, by the construction of u ( x,y ) , , forany ( x , y ) on AB or AC , u ( x ,y ) , ( x , y ) = u ( x , y ) and for any ( x, y ) , u ( x,y ) , ( x , y ) ≤ u ( x , y ) since any plane u ( x,y ) , resulted from the two-step operation must weakly remain below u ( AB ) and u ( AC ) .Thus, by the definition of u ∗ , u ∗ and u are identical on AB and AC . So u ∗ will lead to a weakly greaterrevenue than u according to Formula(1). Thus, u ∗ must be optimal as well. This is sufficient to establishthat u ∗ must consist of a set of planes.Pick one arbitrary plane ˆ u among u ∗ , it intersects with the u-axis at some point k = (0 , , k ) , we assumeat the same time it intersects with vertical line ( x = x A , y = y A ) at some point ( x = x A , y = y A , u = u k ) ( u k is zero or negative).We now show that ˆ u is the unique plane among u ∗ that goes through point ( x A , y A , u k ) . Consider ˆ u ∩ ( y = y A ) (intersection line of two plane ˆ u and plane y = y A ), this line is unique. Moreover, it eitherhas gradient 1 or touches u ( AB ) . So the gradient, say q , of this line is unique. Recall that q is also thegradient of plane ˆ u in the x -dimension. So the gradient of plane ˆ u in the x -direction is unique. For samereason, q , the gradient of plane ˆ u in y direction is also unique. Because there is only one plane that goesthrough point ( x A , y A , u k ) can satisfy the two gradient conditions, plane ˆ u can be uniquely determined by u k . Finally, lower u k results in weakly larger q and q (by convexity of u ( AC ) and u ( AB ) ), strictly lower k = z k − q x A − q y A , and hence strictly larger payment t = | k | . In other words, larger t indeed correspondsto steeper ˆ u ( AB ) and ˆ u ( AC ) , hence larger q and q . This completes the proof of menu monotonicity. (cid:117)(cid:116) Theorem 4.2 implies the aforementioned Hart-Nisan result as a corollary.
Corollary 4.3 [Hart and Nisan 2012, Theorem 28] For two i.i.d. items,
P R ( f ) = P R ( f ) ≤ − ,bundling sale is optimal. Proof.
It is without loss to restrict attention to symmetric mechanisms [Maskin and Riley 1984]. Thus, u ( AB ) is identical to u ( AC ) . u ( AB ) and u ( AC ) have the same slope (in fact, plane u (cid:48) touches both u ( AB ) and u ( AC ) simultaneously). So, q ( v ) = q ( v ) ∀ v ∈ V . In other words, the two items are alwayssold with the same probability. The seller’s revenue of this optimal mechanism is equivalent to a mechanismthat sells two items as a bundle with the same probability. So bundling is optimal as well. (cid:117)(cid:116) As another application of Theorem 4.2, we obtain a revenue monotonicity theorem in this setting.
Theorem 4.4 (Revenue Monotonicity)
Under Condition 1, optimal revenue is monotone: let F i , G i be the cumulative distribution function ofdensity functions f i , g i , i = 1 , , respectively. If G and G first-order stochastically dominate F and F respectively, optimal revenue obtained for ( G , G ) is no less than that of ( F , F ) . Proof.
Consider any two points ( x , y ) and ( x , y ) , where x > x . If q ( x , y ) < q ( x , y ) , by Theo-rem 4.2, we must have t ( x , y ) < t ( x , y ) . If q ( x , y ) = q ( x , y ) , then q ( x, y ) = q ( x , y ) , ∀ x ∈ [ x , x ] . u ( x , y ) = u ( x , y ) + q ( x , y )( x − x ) , which can be achieved by choosing the same menuas ( x , y ) chooses. While buyer at type ( x , y ) has several menu items that all achieve the highest utility,we can assume the buyer chooses the menu with the highest payment ([Hart and Reny 2012]).Thus there is an optimal choice guarantees t ( x , y ) ≤ t ( x , y ) .To sum up, t ( x , y ) ≤ t ( x , y ) when x ≤ x . For same reason, t ( x , y ) ≤ t ( x , y ) when y ≤ y .Hence t ( x, y ) is a weakly monotone function in both directions. Suppose G and G first-order stochas-tically dominates F and F respectively. Let R ( F × F ) denote the optimal revenue when item and distributes independently according to F and F . When distribution G × G chooses the same mechanismas F × F does, let the revenue be R ∗ ( G × G ) . We have R ∗ ( G × G ) ≥ R ( F × F ) , since t is weaklymonotone. By transitivity, R ( G × G ) ≥ R ∗ ( G × G ) ≥ R ( F × F ) . (cid:117)(cid:116) G i first-order stochastically dominates F i if G i ( x ) ≤ F i ( x ) for all x and G i ( x ) > F i ( x ) for some x . In this section, we investigate optimal mechanisms under Condition 2. We obtain several results saying thatthe optimal mechanism contains only few menu items. All these results are built upon Pavlov’s characteri-zation [Pavlov 2011a] and an important lemma introduced in the next subsection.
If both f and f satisfy Condition 2, Pavlov [2011a, Proposition 2] states that, in the optimal mechanism,the seller either keeps both items (i.e., q = (0 , ), or sells one of the items at probability 1 (i.e., q = 1 or q = 1 ).For graphic representation, let the buyer’s valuation be within rectangle ABDC , we have the followinglemma.
Lemma 5.1
Graph Representation Lemma
Under Condition 2, the optimal mechanism can be represented by one of the rectangles shown in Fig. 2 orFig. 3. More precisely, the optimal mechanism divides the valuation space into four regions, where1. in the bottom left region (region
ASM E in both figures), it assigns q = (0 , and u ( x, y ) = 0 to anypoint ( x, y ) in the region. Furthermore, region ASM E is convex.2. in the top right region, it assigns q = (1 , to any point in the region.3. in the top left region, it assigns q = ( ∗ , to any point in the region, where ∗ is a variable. Thus thisregion represents a set of menu items, each of which is a vertical slice.4. Symmetrically, in the bottom right region, it assigns q = (1 , ∗ ) to any point in the region. This regionrepresents a set menu items, each of which is a horizontal slice.5. The boundary between the top left and right regions is vertical ( QL in both figures); the boundary ofthe top right and bottom right regions is horizontal ( M N in Fig. 2 or LI in Fig. 3). The boundarybetween (1 , ∗ ) region and ( ∗ , region is in the upper right direction. To describe our first theorem under Condition 2, we need the following condition on density functions.
Condition 3:
P R ( f i ( x )) , i = 1 , , is constant. Theorem 5.2
Under Conditions 2 and 3, optimal mechanism has at most 4 menu items. M and it intersects BD at N . Then draw avertical line through M crossing CD at G . We have the following two lemmas. Lemma 5.3
Optimal utility function on BN is piecewise linear with at most two pieces. Lemma 5.4
The optimal utility function on
N D is piecewise linear with at most 2 pieces.
With these two lemmas, we are able to prove Theorem 5.2 (in the appendix). Hart and Nisan [2012,Theorem 1 and Lemma 14] state that bundling 4-approximates optimal revenue for general two-item setting.As an application of Theorem 5.2, we obtain a better lower bound for bundling sale.
Corollary 5.5
Under Conditions 2 and 3, bundling 3-approximates optimal revenue.
Proof.
Revenue of an optimal mechanism with 3 non-zero menu items is less than or equal to the sum ofrevenues of 3 mechanisms, each of which has only 1 non-zero menu items. Since bundling is optimal amongall mechanisms that contains only 1 non-zero menu item, thus no worse than any of these three mechanisms.Consequently bundling gives a 3-approximation of the optimal revenue. (cid:117)(cid:116)
Following a similar proof of Theorem 5.2, we obtain another condition under which 3 menu items areenough. Note that, this condition does not impose constant power rate, thus is not a special case of Condition3.
Condition 4: − ≤ P R ( f ( x )) ≤ y A f ( y A ) − , ∀ x and − ≤ P R ( f ( y )) ≤ x A f ( x A ) − , ∀ y. Theorem 5.6
Under Conditions 2 and 4, there is an optimal mechanism such that it contains at most 3menu items, thus bundling gives a 2-approximation of the optimal revenue.
The requirement of power rate to be constant might be restrictive. If one relaxes this requirement to be monotone power rate , one only needs to add two more menu items.
Condition 5:
P R ( f i ( x )) , i = 1 , , is weakly monotone. Theorem 5.7
Under Conditions 2 and 5, if f = f , optimal mechanism consists of at most 6 menu items. AD intersects SE at point M . In region ASM E , sellerkeeps both items. Item 2 is sold deterministically in
CSM D and item 1 is sold determinately in
M EBD .Let the allocation rule on point ( x, y ) in CSM D be ( q ( x ) , . Similar to the proof of Theorem 5.2, westart with the following lemma. Lemma 5.8
Optimal utility function on
N D is piecewise linear with at most 2 pieces.
Similarly, we show that u ( CN ) is piecewise linear as well. Lemma 5.9
Optimal utility function on CN is piecewise linear with at most 2 pieces. With the two lemmas above, we are able to prove Theorem 5.7.We say h ( x ) are nonnegative-coefficient polynomial if h ( x ) = a n x n + a n − x n − + ... + a x + a , a i ≥ , i = 0 , ..., n . Let h and h are nonnegative-coefficient polynomials, it is easy to show that h ( x ) e h ( x ) satisfies Condition 5. In particular, this expression includes nonnegative-coefficient polynomial density andexponential density as special cases. A buyer has unit-demand if q ( x, y ) + q ( x, y ) ≤ . Under unit-demand model, Pavlov [2011b, Propo-sition 2] states that, if distribution functions satisfy Condition 2, it is without loss to restrict attention onmechanisms such that q ( x, y ) + q ( x, y ) ∈ { , } ∀ ( x, y ) Pavlov solves the optimal mechanism for two items with identical uniform distributions. The resultingmechanism contains 5 menu items for uniform distribution on [ c, c + 1] ∗ [ c, c + 1] , c ∈ (1 , ¯ c ) (where ¯ c ≈ . ). We show that in nonidentical settings, the optimal mechanism also contains at most 5 menuitems. It follows trivially that our result is tight. Theorem 5.10
In unit-demand model, if both f and f are uniform distributions, the optimal mechanismconsists of at most 5 menu items. Let
ASE denote the zero utility region and
CSEBD the non-zero utility region. For the same reason inLemma 5.1,
ASE is convex. For points in
ASE , allocation (0 , is the best. For ( x, y ) ∈ CSEBD , ( q ( x, y ) , q ( x, y )) (cid:54) = (0 , , so q ( x, y ) + q ( x, y ) = 1 . The mechanism is shown in Fig. 6. Draw a degree line across E , intersecting BD or CD at W . Draw a degree line across S , intersecting BD or CD at G . We consider here the case that W is on BD and G is on CD . Other cases follow from similararguments.The theorem can be similarly proved via the following two lemmas. Lemma 5.11
Optimal utility function on BW is piecewise linear with at most 2 pieces. Lemma 5.12
Optimal utility function on
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A Appendix: Proofs of Section 5
Proof. of Lemma 5.1
We first determine the relative positions of the four possible regions.If the seller keeps both items, the buyer’s utility is zero. Since u ( x, y ) is an increasing function, it assigns q = (0 , in the bottom left region, i.e. ASM E . Since u ( x, y ) is convex, the convex combination of anytwo zero-utility points must also be zero. Therefore, ASM E is a convex region.If for a type ( x , y ) with q ( x , y ) = q ( x , y ) = 1 , for any point ( x , y ) , IC requires that x + y − t ( x , y ) ≥ x q ( x , y ) + y q ( x , y ) − t ( x , y ) ,x q ( x , y ) + y q ( x , y ) − t ( x , y ) ≥ x + y − t ( x , y ) . Summing the two inequalities, we get ( q ( x , y ) − x − x ) + ( q ( x , y ) − y − y ) ≥ . If x > x , y > y , we must have q ( x , y ) = q ( x , y ) = 1 .Let ( x , y ) be a point where some positive proportions of the items are sold, then according to Pavlov’scharacterization [Pavlov 2011a], one of the items must be sold deterministically. Consider two types ( x , y ) and ( x , y ) where q ( x , y ) = 1 , q ( x , y ) < and q ( x , y ) < , q ( x , y ) = 1 . By IC , we musthave x + y q ( x , y ) − t ( x , y ) ≥ x q ( x , y ) + y − t ( x , y ) ,x q ( x , y ) + y − t ( x , y ) ≥ x + y q ( x , y ) − t ( x , y ) . Summing up the two inequalities, we get (1 − q ( x , y ))( x − x ) + (1 − q ( x , y ))( y − y ) ≥ . So,one of x < x and y > y does not hold. This implies the second part of (5).To sum up, (1 , must be assigned to the upper right corner, (1 , q ( x, y )) is assigned to the bottom rightcorner, ( q ( x, y ) , is assigned to the upper left corner, and (0 , is assigned to the bottom left corner,(some regions may be empty).Let the allocation vector at ( x , y ) be (1 , q ( x , y )) . For any x ∈ [ x , x B ] , by IC, we must have q ( x, y ) = 1 , so u ( x, y ) = u ( x , y ) + x − x . It is still in the buyer’s best interest to choose menu item (1 , q ( x , y )) at ( x, y ) . This immediately implies the first part of (5): the boundary between for different q in (1 , q ( x, y )) is horizontal. In particular, in Fig. 2, M N is horizontal. Similarly, LQ is vertical.If there is a point on curve SE that chooses menu item (1 , , the mechanism is of the form shown inFig. 2, otherwise it is of the form shown in Fig. 3. (cid:117)(cid:116) Proof. of Lemma 5.3
For any point K on BN (see Fig.7), draw a horizontal line that intersects curve SE at W . Because u x = 1 for all the types in region EBN M , we can represent the utility of any point ( x, y K ) on line KW using u ( K ) . Formally, u ( x, y K ) = u ( K ) + x − x K , ∀ x ∈ [ x W , x K ] .12igure 7: The optimal allocation that there is no point on curve SME chooses allocation menu (1,1).The revenue obtained in EBN M is R EBNM = (cid:73) EBNM T · ˆn ds − (cid:90) EBNM (cid:52) ( z ) u ( z ) dz = (cid:90) BN T · ˆn ds − (cid:90) y M y A (cid:90) x B x B − u ( x B ,y ) (cid:52) ( x, y ) u ( x, y ) dxdy + (cid:90) NMEB T · ˆn ds = (cid:90) y M y A [ x B u ( x B , y ) f ( x B ) f ( y ) − (cid:90) x B x B − u ( x B ,y ) (cid:52) ( x B , y )( u ( x B , y ) + x − x B ) dx ] dy + c = (cid:90) y M y A R ( u ( x B , y ) , y ) dy + cc = (cid:82) NMEB T · ˆn ds depends on u ( N M ) , u ( M E ) ,and u ( EB ) . We fix these utility and study u ( BN ) . Let R ( u ( x B , y ) , y ) = x B u ( x B , y ) f ( x B ) f ( y ) − (cid:82) x B x B − u ( x B ,y ) (cid:52) ( x B , y )( u ( x B , y )+ x − x B ) dx and R u ( u ( x B , y ) , y ) denote the partial derivative on u ( x B , y ) dimension. R u ( u ( x B , y ) , y ) = f ( y )[ x B f ( x B ) − (cid:90) x B x B − u ( x B ,y ) f ( x )(3 + P R ( f ( x )) + P R ( f ( y ))) dx ] Let v ( l ) = (cid:82) x B x B − l f ( x )(3 + P R ( f ( x )) + P R ( f ( y ))) dx . Because P R ( f ( x )) + P R ( f ( y )) ≥ is a constant, v is an increasing function of l . When x B f ( x B ) − v ( u ( x B , y )) > , R EBNM is increasingas u ( x B , y ) increases. From the Fig.7 we can see, when ( x B , y ) moves from B to N , u ( x B , y ) weaklyincreases, v ( u ( x B , y )) weakly increases and x B f ( x B ) − v ( u ( x B , y )) weakly decreases.There are 3 possibilities: 1. x B f ( x B ) − v ( u ( x B , y )) > , y ∈ [ y B , y N ] , 2. x B f ( x B ) − v ( u ( x B , y )) < , y ∈ [ y B , y N ] , 3. ∃ y (cid:48) ∈ [ y B , y N ] , x B f ( x B ) − v ( u ( x B , y )) = 0 and x B f ( x B ) − v ( u ( x B , y )) ≤ , y ∈ [ y B , y (cid:48) ] , x B f ( x B ) − v ( u ( x B , y )) ≥ , y ∈ [ y (cid:48) , y N ] . 1 can be regarded as a special case of 3 by setting y (cid:48) = y B , because u ( b, y ) , y ∈ [ y B , y N ] must be also as large as possible. 2 can be regarded as an specialcase of 3 by setting y (cid:48) = y N , because u ( x B , y ) , y ∈ [ y B , y N ] must be as small as possible.So it is without loss to restrict attention on case 3. Since K is randomly chosen, set y (cid:48) = y K . Revenue isincreasing as u ( KN ) decrease and revenue is increasing as u ( KB ) increase. By convexity of u , with fixed u ( B ) , u ( K ) , and u ( N ) , optimal u ( BN ) comprises of two lines: the straight line(with extended line) acrosspoints ( y B , u ( B ))( y K , u ( K )) , and straight line across point ( y N , u ( N )) with slope q ( N ) . (cid:117)(cid:116) roof. of Lemma 5.4 The revenue obtained in region
M N DG is R MNDG = (cid:73) MNDG T · ˆn ds − (cid:90) MNDG (cid:52) ( z ) u ( z ) dz = (cid:90) DGMN T · ˆn ds + (cid:90) ND T · ˆn ds − (cid:90) y D y N (cid:90) x N x M (cid:52) ( x, y ) u ( x, y ) dxdy = c + (cid:90) y D y N [ x B u ( x B , y ) f ( x B ) f ( y ) − (cid:90) x N x M (cid:52) ( x, y ) u ( x, y ) dx ] dyc = (cid:82) DGMN T · ˆn ds depends on u ( DG ) , u ( GM ) , and u ( M N ) . We fix these utilities and study u ( DN ) .Let R ( u ( x B , y ) , y ) = x B u ( x B , y ) f ( x B ) f ( y ) − (cid:82) x N x M (cid:52) ( x, y ) u ( x, y ) dx . Pick a random point P on N D ,draw a horizontal line across P that intersects curve M LQ at J , intersects segment M G at R . In region CSM LID point ( x, y ) gets item 2 deterministically and we assume it gets item 1 with probability q ( x ) . R ( u ( x B , y J ) , y J )= u ( x B , y J ) x B f ( x B ) f ( y J ) − (cid:90) x J x M u ( x, y J ) (cid:52) ( x, y J ) dx − (cid:90) x B x J u ( x, y J ) (cid:52) ( x, y J ) dx = [ u ( R ) + (cid:90) x J x M q ( x ) dx + x B − x J ] x B f ( x B ) f ( y J ) − (cid:90) x J x M [ u ( R ) + (cid:90) xx R q ( l ) dl ] (cid:52) ( x, y J ) dx − (cid:90) x B x J [ u ( R ) + (cid:90) x J x M q ( l ) dl + x − x J ] (cid:52) ( x, y J ) dx Because u ( x B , y J ) = u ( R ) + (cid:82) x J x M q ( x ) dx + x B − x J = u ( R ) + x B − (cid:82) x J x M (1 − q ( x )) dx , we have ∂u∂x J ( x B , y J ) ≤ ∂R∂x J ( u ( x B , y J ) , y J )= ( q ( x J ) − x B f ( x B ) f ( y J ) − (cid:90) x B x J ( q ( x J ) − (cid:52) ( x, y J ) dx = ( q ( x J ) − f ( y J )[ x B f ( x B ) − (cid:90) x B x J f ( x )(3 + P R ( f ( x )) + P R ( f ( y J ))) dx ]= ( q ( x J ) − f ( y J ) v ( x J ) In the last equality, we reset v ( x J ) = x B f ( x B ) − (cid:82) x B x J f ( x )(3 + P R ( f ( x )) + P R ( f ( y J ))) dx . Since q ( x J ) − ≤ , sgn ( v ( x J )) = − sgn ( ∂R∂x J ) . Thus sgn ( v ( x J )) = − sgn ( ∂R∂u ∂u∂x J ) = sgn ( ∂R∂u ) Since P R ( f ( x ))+ P R ( f ( y )) ≥ , v ( x J ) is weakly monotone in x J . By (3) and (4) of Lemma 5.1,while P moves up, the intersection J moves towards right, i.e., x J weakly increases. Then v ( x J ) weaklyincreases. So if v ( x J ) switches sign, it must switch sign from negative to positive. So R u ( u ( x B , y J ) , y J ) can only switch sign from negative to positive.There are three cases for the sign of R u ( u ( x B , y J ) , y J ) . It’s similar to the proof in Lemma 5.3. WLOG,we can assume that v ( x J ) = 0 . Thus v ( x ) ≥ , x ∈ [ x J , x D ] and v ( x ) ≤ , x ∈ [ x A , x J ] . There-fore R u ( u ( x B , y J ) , y J ) ≥ , y ∈ [ y P , y D ] , R u ( u ( x B , y J ) , y J ) ≤ , y ∈ [ y N , y P ] . So R MNDG is in-creasing as u ( N P ) decrease and is increasing as u ( P D ) increase. With fixed u ( N ) , u ( P ) , and u ( D ) ,14igure 8: Utility function on BD.by convexity, optimal u ( N D ) comprises of two lines: the straight line (with extended line) across points ( y P , u ( P ))( y D , u ( D )) , and straight line across point ( y N , u ( N )) with slope q ( N ) . (cid:117)(cid:116) Proof. of Theorem 5.2
We sum up the conclusions drawn on different segments of BD and settle final shape of u ( BD ) , subjectto the convexity constraints and fixed u ( B ) , u ( K ) , u ( P ) , and u ( D ) . In Fig. 8, the black solid line is anarbitrary convex utility function. By Lemma 5.3 and 5.4, the optimal u ( BD ) consists of at most 3 pieces:the straight line (with extended line) across points ( y B , u ( B ))( y K , u ( K )) , the straight line (with extendedline) across points ( y P , u ( P ))( y D , u ( D )) , and the straight line across point ( y N , u ( N )) with slope q ( N ) .In fact, we can prove an even stronger result: it turns out that the line across point ( y N , u ( N )) is unnecessary.The remainder of the proof is to confirm this claim.In Fig. 8, the red dashed utility consists of two parts: straight line across points ( y B , u ( B )) and ( y K , u ( K )) ,and straight line across points ( y P , u ( P )) and ( y D , u ( D )) . We denote ”original” utility function to be theblack arbitrary convex utility function. We denote ”new” utility function to be the red dashed utility function.We prove that the revenue based on new utility function is greater than or equal to the revenue based on theblack original utility. Therefore when u ( B ) , u ( K ) , u ( P ) , u ( D ) and their coordinates on the y -dimension y B , y K , y P , y D are fixed, the optimal utility function must be of the shape portrayed as the red dashed line.First, we study what the graph representation looks like. Since u ( Q ) , u ( J ) , u ( M ) , u ( W ) remain thesame, they are still on the boundaries (see Fig. 7): • M is on the boundary between ( ∗ , region and (0 , region. • W is on the boundary between (1 , ∗ ) region and (0 , region. • J is on the boundary between (1 , ∗ ) region and ( ∗ , region. • Q is on the boundary between ( ∗ , region and (1 , ∗ ) region.Let M (cid:48) denote the new intersection of the three parts: (0 , , ( ∗ , , and (1 , ∗ ) . The new boundary between (1 , ∗ ) and ( ∗ , is the dashed line QJ M (cid:48) . The new boundary between ( ∗ , and (0 , is the dashed line SM M (cid:48) . The new boundary between (1 , ∗ ) and (0 , is the dashed line M (cid:48) W E . Since u ( P D ) and u ( KB ) weakly increase, and u ( KP ) weakly decreases, M (cid:48) must lie in M W KN region, dashed QJ lies in GM LQ region, dashed
W E lies in
SM W E region. Draw horizontal line through M (cid:48) and it intersects BD at N (cid:48) ,intersects the extended line of GM at O .With fixed u ( CD ) , given the boundary of (1 , ∗ ) , we can calculate u ( BD ) as follows: say ( x, y ) is onthe boundary between (1 , ∗ ) amd ( ∗ , , then u ( x B , y ) = u ( x, y ) + x B − x = u ( x, y C ) + y − y C + x B − x .15o we can define u ( BD ) according to their boundary. Let the original utility and revenue function basedon the boundary QLJ M W E is u and R . Let the new utility and revenue function based on the boundary QJ M (cid:48) W and dashed W E is u and R . Our goal is to prove R ≤ R . R DGMW EB ≤ R DGM (cid:48)
W EB
Since R DGMW EB = R DGMW K + R W EBK (here
W E is the solid line) and R DGM (cid:48)
W EB = R DGMM (cid:48)
W K + R W EBK (here,
W E is the dashed line). It suffices to prove R DGMW K ≤ R DGMM (cid:48)
W K and R W EBK ≤ R W EBK separately. R W EBK = (cid:90) W K T · ˆn ds + (cid:90) EB T · ˆn ds + (cid:90) y K y B u ( x B , y ) x B f ( x B ) f ( y ) dy − (cid:90) y K y B (cid:90) x B x B − u ( x B ,y ) u ( x, y ) (cid:52) ( x, y ) dxdy = (cid:90) W K T · ˆn ds + (cid:90) EB T · ˆn ds + (cid:90) y K y B R ( u ( x B , y ) , y ) dy ≤ (cid:90) W K T · ˆn ds + (cid:90) EB T · ˆn ds + (cid:90) y K y B R ( u ( x B , y ) , y ) dy = R W EBK
Here T i = zu i f ( z ) and we use the same R ( u ( x B , y ) , y ) as that in Lemma 5.3. What still remains to showis R DGMW K ≤ R DGMM (cid:48)
W K .In order to prove this claim, we introduce an intermediate utility and revenue as follows. Let the in-termediate utility and revenue function based on the boundary
QLJ M M (cid:48) W and dashed W E is u and R . Consider the intermediate case, M M (cid:48) is the boundary shared by all three regions (1 , ∗ ) , ( ∗ , and (0 , . For any point ( x, y ) on M M (cid:48) , we can calculate the utility of corresponding point ( x B , y ) as u ( x B , y ) = u ( x, y ) + x B − x = u ( x, y C ) + y − y C + x B − x . In fact, according to Lemma 5.1, thiscase cannot happen and u ( BD ) does not retain convexity any more. It is important to note that, here,we are only concerned with R , not the feasibility of u . That is, we use R as a number to facilitate thecomparison between R and R .Now we show R DGMW K ≤ R DGMM (cid:48)
W K . R DGMW K = R DGMN + R MW KN ≤ R DGMN + R MM (cid:48) W KN = R DGMN + R MM (cid:48) N (cid:48) N + R M (cid:48) W KN (cid:48) = R DGMM (cid:48) N (cid:48) + R M (cid:48) W KN (cid:48) = R DGMM (cid:48) N (cid:48) + R M (cid:48) W KN (cid:48) ≤ R DGMM (cid:48) N (cid:48) + R M (cid:48) W KN (cid:48) = R DGMM (cid:48)
W K
The first inequality follows from a similar proof of that in R W EBK ≤ R W EBK .In order to show the second inequality, introduce u and u as follows: u ( x, y ) = (cid:26) u ( x, y ) ( x, y ) ∈ DGM M (cid:48) N (cid:48) u ( x, y C ) + y − y C ( x, y ) ∈ M M (cid:48) O ( x, y ) = (cid:26) u ( x, y ) ( x, y ) ∈ DGM M (cid:48) N (cid:48) u ( x, y C ) + y − y C ( x, y ) ∈ M M (cid:48) OR DGMM (cid:48) N (cid:48) − R DGMM (cid:48) N (cid:48) = (cid:90) DGMM (cid:48) N (cid:48) T · ˆn ds + (cid:90) N (cid:48) D T · ˆn ds − (cid:90) y D y N (cid:48) (cid:90) x N (cid:48) x M (cid:52) ( x, y ) u ( x, y ) dxdy + (cid:90) MM (cid:48) O (cid:52) ( z ) u ( z ) dz − R DGMM (cid:48) N (cid:48) = (cid:90) y D y N (cid:48) [ x B u ( x B , y ) f ( x B ) f ( y ) − (cid:90) x N x M (cid:52) ( x, y ) u ( x, y ) dx ] dy − (cid:90) y D y N (cid:48) [ x B u ( x B , y ) f ( x B ) f ( y ) − (cid:90) x N x M (cid:52) ( x, y ) u ( x, y ) dx ] dy = (cid:90) y D y N (cid:48) [ R ( u ( x B , y ) , y ) − R ( u ( x B , y ) , y )] dy (2)In the last equality, we use the same R ( u ( x B , y ) , y ) as in Lemma 5.4.In the following, we show Equation (2) is non-positive. According to Lemma 5.4, R u ( u ( x B , y ) , y ) ≥ y ∈ [ y J , y Q ] , u ( x B , y ) ≥ u ( J ) R u ( u ( x B , y ) , y ) ≥ y ∈ [ y M (cid:48) , y J ] , u ( x B , y ) ≤ u ( J ) Because u ( x B , y ) ≤ u ( x B , y ) y ∈ [ y P , y D ] and u ( x B , y ) ≥ u ( x B , y ) y ∈ [ y N (cid:48) , y P ] , wehave R ( u ( x B , y ) , y ) ≤ R ( u ( x B , y ) , y ) , y ∈ [ y N (cid:48) , y P ] . So Equation (2) ≤ , then R DGMM (cid:48) N (cid:48) ≤ R DGMM (cid:48) N (cid:48) .Up to now, we have proved that red dashed utility function on BD segment indeed yield the highestrevenue subject to fixed u ( B ) , u ( K ) , u ( P ) , and u ( D ) (Fig. 8). In other words, u ( BD ) is piecewise linearwith two pieces. Same for u ( CD ) .We have showed region CSEBD consists of 4 menu items. In other words, the whole mechanismconsists of 5 menu items. Say, points on BD segments choose menus (1 , α, t α ) , (1 , γ, t γ ) , α ≤ γ . Pointson CD choose menus ( β, , t β ) , ( θ, , t θ ) , β ≤ θ . What remains to show is that the top right two regionsboth allocate with probabilities (1,1) thus are in fact a unique region.Manelli and Vincent [2007, Theorem 16] state that in the optimal mechanism, there must exist a segmentchooses allocation (1 , . By Lemma 5.1, the (1 , region is on the top right corner of the rectangle. So inthe optimal mechanism there are slopes of utility lines equal to 1 in both BD and CD . Hence, γ = θ = 1 . BD and CD share the same menu item (1 , .To sum up, the optimal mechanism menu consists of at most 4 menu-items: q q t α t α β t β t (cid:117)(cid:116) Proof. of Theorem 5.6
SM J DC represents the ( ∗ , region. When y ∈ [ y B , y N ] , we have u ( x B , y ) ≤ x B − x A , Lemma 5.3 implies R u ( u ( x B , y ) , y ) ≥ f ( y )[ x B f ( x B ) − (cid:90) x B x A f ( x )[3 + P R ( f ( x )) + P R ( f ( y ))] dx ] . When y ∈ [ y N , y D ] , we have x J ≥ x A , Lemma 5.4 implies v ( x J ) ≥ x B f ( x B ) − (cid:90) x B x A f ( x )[3 + P R ( f ( x )) + P R ( f ( y ))] dx. We have x B f ( x B ) − (cid:90) x B x A f ( x )[3 + P R ( f ( x )) + P R ( f ( y ))] dx = x A f ( x A ) + (cid:90) x B x A [ xf (cid:48) ( x ) + f ( x )] dx − (cid:90) x B x A f ( x )[3 + P R ( f ( x )) + P R ( f ( y ))] dx = x A f ( x A ) (cid:90) x B x A f ( x ) dx − (cid:90) x B x A f ( x )[2 + P R ( f ( y ))] dx = (cid:90) x B x A f ( x )[ x A f ( x A ) − − P R ( f ( y ))] dx ≥ Thus R u ( u ( x B , y ) , y ) ≥ , y ∈ [ y B , y N ] and v ( x J ) ≥ , py ∈ [ y N , y D ] . Because sgn ( R u ( u ( x B , y ) , y )) = sgn ( v J ) ≥ , y ∈ [ y N , y D ] , R u ( u ( x B , y ) , y ) ≥ , y ∈ [ y N , y D ] . By Lemma 5.3 and 5.4, with fixed u ( B ) , u ( N ) , and u ( D ) , optimal u ( N D ) and u ( BN ) are straight lines. Using similar method as shown inthe Theorem 5.2, we can prove that with fixed u ( B ) and u ( D ) , the optimal utility function is u ( x B , y ) = u ( B ) + q ( y − y B ) y ∈ [ y B , y D ] , where q = u ( D ) − u ( B ) y D − y B . But we should notice that, this operation maychange u ( AC ) . This leads to two further cases: (1) the operation changes u ( AC ) ; (2) otherwise. It can beseen case (1) is strictly more general, so we only need to consider case (1).Draw a vertical line through S across BD at P , we fix u ( B ) and u ( D ) . Instead of letting u ( BD ) be astraight line, we let u ( BD ) be piece linear with one break point u ( P ) in order not to change u ( AC ) . Let u ( P ) = x B − x A , u ( x B , y ) = u ( P ) + y − y P y D − y P · ( u ( D ) − u ( P )) y ∈ [ y P , y D ] u ( x B , y ) = u ( B ) + y − y B y P − y B · ( u ( P ) − u ( B )) y ∈ [ y B , y P ] u ( S ) , u ( P ) , u ( D ) are in a new plane. So the boundary of ( ∗ , region, curve M J D , moves into
CSD region. Using similar method as is shown in the Theorem 5.2, total revenue willweakly increase.Then we set u ( x, y C ) = u ( C ) + x − x C x D − x C · ( u ( D ) − u ( C )) x ∈ [ x C , x D ] After this operation, points u ( C ) , u ( S ) , u ( D ) are in a new plane. The boundary of ( ∗ , region, originallycurve M J D , becomes the straight line SD . So this operation will not affect the utilities on segment AB .Using same argument above, total revenue will weakly increase. Now there are 2 non-zero menu itemschosen in region CSP D and 1 non-zero menu item chosen in region
SM EBP . Manelli and Vincent[2007, Theorem 16] says in the optimal mechanism, there must exist a segment chooses (1 , allocation.So these 2 non-zero menus chosen by region CSP D must be the same menu (1 , , t ) . So there are twonon-zero menus in total: (1 , , t ) and (1 , α, t ) .Case 2: When there is no ( ∗ , region.Now Condition ” P R ( f ( x )) ≤ y A f ( y A ) − , ∀ x ” is not enough, because when we do the same ma-nipulation for u ( BD ) , u ( CS ) will change at the same time. If a point have chosen non-zero menu item, itmust choose first item with probability 1. According to this, we have the following equation: R SP DC = (cid:90) SP T · ˆn ds + (cid:90) CD T · ˆn ds + (cid:90) P D T · ˆn ds + (cid:90) SC T · ˆn ds − (cid:90) SP DC (cid:52) ( z ) u ( z ) dz = c + (cid:90) y C y S u ( x B , y ) x B f ( x B ) f ( y ) dy − (cid:90) y C y S ( u ( x B , y ) + x A − x B ) x A f ( x A ) f ( y ) dy − (cid:90) y C y S (cid:90) x B x A (cid:52) ( x, y )( u ( x B , y ) + x − x B ) dxdy = c + (cid:90) y C y S R ( u ( x B , y ) , y ) dy Here, c = (cid:82) SP T · ˆn ds + (cid:82) CD T · ˆn ds . When we fix u ( D ) and u ( P ) , c is constant. Let R = u ( x B , y ) x B f ( x B ) f ( y ) − ( u ( x B , y ) + x A − x B ) x A f ( x A ) f ( y ) − (cid:82) x B x A (cid:52) ( x, y )( u ( x B , y ) + x − x B ) dx and R u ( u ( x B , y ) , y ) denote the partial derivative wrt. u ( x B , y ) . R u ( u ( x B , y ) , y )= [ x B f ( x B ) − x A f ( x A ) − (cid:90) x B x A f ( x )[3 + P R ( f ( x )) + P R ( f ( y ))] dx ] · f ( y )= [ − − P R ( f ( y ))] · f ( y ) ≤ This means R SP DC is increasing as utilities on segment
P D decreases and is decreasing as utilities onsegment
P D increases. Using same argument as in Case 1, we know total revenue is increasing as utilitieson segment
P B increases and is decreasing as utilities on segment
P D decreases.19igure 10: Utility function on BD, the red dashed line including point u(D) has derivative 1.As shown in Fig. 10, the new mechanism with red dashed line u ( BD ) yields higher revenue than theoriginal mechanism with black solid line u ( BD ) . So there are two non-zero menu items in total: (1 , , t ) and (1 , α, t ) . (cid:50) (cid:117)(cid:116) Proof. of Lemma 5.8
Since point M is on (0 , part, so u ( M ) = 0 . For point ( x, y ) in region N M D , u ( x, y ) = u ( M ) + y − y M + (cid:82) xx M q ( l ) dl = y − y M + (cid:82) xx M q ( l ) dl .Rewrite the revenue formula for region N M GD as follows, R NMGD = (cid:73) NMGD T · ˆn ds − (cid:90) NMGD (cid:52) ( z ) u ( z ) dz = 2 (cid:90) x D x N y C u ( x, y C ) f ( y C ) f ( x ) dx − (cid:90) NM T · ˆn ds − (cid:90) x D x N (cid:90) y C x u ( x, y ) (cid:52) ( x, y ) dydx = 2 (cid:90) x D x N x B [ y C − y M + (cid:90) xx N q ( l ) dl ] f ( x B ) f ( x ) dx − (cid:90) x D x N (cid:90) y C x [ y − y M + (cid:90) xx N q ( l ) dl ] (cid:52) ( x, y ) dydx − (cid:90) NM T · ˆn ds = 2 (cid:90) x D x N q ( x )( (cid:90) x D x [ x B f ( x B ) f ( l ) − (cid:90) x D l (cid:52) ( l, y ) dy ] dl ) dx + C = 2 (cid:90) x D x N q ( x ) v ( x ) dx + C The rest terms are denoted by C which only depends on y M . When y M is fixed, C is a constant. Inthe last equality, let v ( x ) = (cid:82) x D x [ x B f ( x B ) f ( l ) − (cid:82) x D l (cid:52) ( l, y ) dy ] dl , thus it is independent from q .To find out the optimal q , divide [ x N , x D ] into several intervals according to v ( x ) . Since v ( x ) is acontinuous function, we can assume in intervals v ( x ) ≤ , x ∈ [ n i , d i ] , ∀ ≤ i ≤ l , and v ( x ) ≥ , x ∈ [ d i , n i +1 ] , ∀ ≤ i ≤ l − , and m = x N , b n = x D . There is always a rational number in any interval, so20he number of intervals is countable. It is possible that d = x N , n l = x D and l may be infinite. (cid:90) x D x N q ( x ) v ( x ) dx = l (cid:88) i =1 (cid:90) d i n i q ( x ) v ( x ) dx + l − (cid:88) i =1 (cid:90) n i +1 d i q ( x ) v ( x ) dx ≤ l (cid:88) i =1 q ( n i ) (cid:90) d i n i v ( x ) dx + l − (cid:88) i =1 q ( n i +1 ) (cid:90) n i +1 d i v ( x ) dx (3) = q ( n ) (cid:90) x D x N v ( x ) dx + l (cid:88) i =2 ( q ( n i ) − q ( n i − )) (cid:90) d l d i − v ( x ) dx ≤ q ( n ) (cid:90) x D x N v ( x ) dx + l (cid:88) i =2 ( q ( n i ) − q ( n i − )) (cid:90) x D x (cid:48) v ( x ) dx (4) ≤ q ( x N ) (cid:90) x (cid:48) x N v ( x ) dx + 1 × (cid:90) x D x (cid:48) v ( x ) dx Let x (cid:48) denote the point in [ x N , x D ] where g ( s ) = (cid:82) x D s v ( x ) dx achieves the maximum value in (4).The inequality (3) are only based on the facts that q ( x ) is a weakly monotone function and they becomeequalities by setting q ( x ) = q ( x N ) , x ∈ [ x N , x (cid:48) ) , q ( x ) = 1 , x ∈ [ x (cid:48) , x D ] . So there are at most two piecesin u ( BD ) . (cid:117)(cid:116) Proof. of Lemma 5.9
The proof is similar to Lemma 5.3. Rewrite the revenue formula of region
CSM N as follows, R CSMN = (cid:73) CSMN T · ˆn ds − (cid:90) CSMN (cid:52) ( z ) u ( z ) dz = (cid:90) CSMN T · ˆn ds + (cid:90) x N x C y C u ( x, y C ) f ( y C ) f ( x ) dx − (cid:90) x N x C (cid:90) y C y C − u ( x,y C ) (cid:52) ( x, y ) u ( x, y ) dydx = (cid:90) x N x C [ y C u ( x, y C ) f ( y C ) f ( x ) − (cid:90) y C y C − u ( x,y C ) ( u ( x, y C ) − y C + y ) (cid:52) ( x, y ) dy ] dx + C = (cid:90) x N x C R ( u ( x, y C ) , x ) dx + C We let C = (cid:82) CSMN T · ˆn ds , it only depends on u ( C ) and u ( N ) . Let R ( u ( x, y C ) , x ) = y C u ( x, y C ) f ( y C ) f ( x ) − (cid:82) y C y C − u ( x,y C ) ( u ( x, y C ) − y C + y ) (cid:52) ( x, y ) dx . Pick point K in CN , we have, ∂R∂u ( u ( x K , y C ) , x K ) = f ( x K )[ y C f ( y C ) − (cid:90) y C y C − u ( x K ,y C ) f ( y )[3 + P R ( f ( x K )) + P R ( f ( y ))] dy ] While x K increases, u ( x K , y C ) and P R ( f ( x K )) increase, y C f ( y C ) − (cid:82) y C y C − u ( x K ,y C ) f ( y )[3+ P R ( f ( x K ))+ P R ( f ( y ))] dy decreases. WLOG, we can assume ∂R∂u ( u ( x, y C ) , x ) ≥ , x ∈ [ x C , x K ] and ∂R∂u ( u ( x, y C ) , x ) ≤ , x ∈ [ x K , x N ] . Revenue is increasing as u ( KN ) decrease. Revenue is increasing as u ( CK ) increase. Let u ( x, y C ) = u ( C ) + u ( K ) − u ( C ) x K − x C ( x − x C ) , u ( x, y C ) = u ( N ) + q ( x N , y N )( x − x N ) , then u ( x, y C ) = max ( u ( x, y C ) , u ( x, y C )) , x ∈ [ x C , x N ] gives the optimal revenue with fixed u ( C ) , u ( K ) , u ( N ) and q ( x N , y C ) . So optimal u ( CN ) comprises at most two pieces. (cid:117)(cid:116) Proof. of Theorem 5.7
When q ( x N ) are fixed, q ( x )( x ∈ ( x N , x D ]) is irrelevant to R CSMN . When u ( C ) and u ( N ) are fixed, u ( x, y B )( x ∈ ( x C , x M )) is irrelevant to R NMGD .Let’s look at Fig. 11. When u ( C ) , u ( K ) , u ( N ) , q ( x N ) are fixed, the buyer utility on the red dashedline is greater than that on the black solid line between CK . The buyer utility on the red dashed line is lessthan that on the black solid line between KN . According to Lemma 5.8 and 5.9, R CSMN and R NMGD arelarger, so the total revenue is larger. Furthermore, optimal utility function on
N D is piecewise linear withtwo slopes: q ( x N ) and . To sum up, the optimal utilities on CD is piecewise linear with at most threepieces. Therefore, the optimal mechanism is of the following form: q q t α t α t α t α t t (cid:117)(cid:116) Proof. of Lemma 5.11
For ( x, y ) in CSEBD , the utility of choosing ( q, − q ) is xq + y (1 − q ) − t ( q ) =( x − y ) q + y − t ( q ) . The buyer will choose the menu item that achieves max q { ( x − y ) q + y − t ( q ) } = y + max q { ( x − y ) q − t ( q ) } . This means that which menu item will be chosen depends entirely on the valueof x − y . For two points ( y + l, y ) and ( y + l, y ) in CSEBD , they must share the same allocation. Use q ( y ) as a shortcut of q ( x B , y ) .Fixing u ( W ) and q ( y W ) , for any point ( x, y ) in region EBW , the buyer’s utility is u ( x, y ) = u ( W ) − ( u ( W ) − u ( x B , y + x B − x )) − ( u ( x B , y + x B − x ) − u ( x, y ))= u ( W ) − (( x B − x )(1 − q ( y + x B − x )) + ( x B − x ) q ( y + x B − x )) − (cid:90) y W x B + y − x q ( l ) dl = u ( W ) − ( x B − x ) − (cid:90) y W x B + y − x q ( l ) dl R EBW = (cid:73) EBW T · ˆn ds − (cid:90) EBW (cid:52) ( z ) u ( z ) dz = (cid:90) W E T · ˆn ds + (cid:90) BW T · ˆn ds + (cid:90) EB T · ˆn ds − (cid:90) EBW (cid:52) ( z ) u ( z ) dz The forms of the second and third terms in the expression above are as follows. Here, a i , b i , i = 1 , ..., and C j , v j , j = 1 , ..., are expressions independent of q ( y ) , y ∈ [ y B , y W ] . (cid:90) BW T · ˆn ds = (cid:90) y W x B y B u ( x B , y ) dy = (cid:90) b a h ( y )[ (cid:90) b ( y ) a ( y ) q ( y ) dx + g ( y )] dy = (cid:90) y W y B q ( y ) v ( y ) + C (cid:90) EB T · ˆn ds = (cid:90) x B x E − y A u ( x, y A ) dx = (cid:90) b a h ( x )[ (cid:90) b ( x ) a ( x ) q ( y ) dy + g ( x )] dx = (cid:90) y W y C q ( y ) v ( y ) + C Since f and f are uniform distributions, (cid:52) ( x, y ) = 3 f ( x ) f ( y )+ xf (cid:48) ( x )+ yf (cid:48) ( y ) = 3 f is a constant. (cid:90) EBW (cid:52) ( z ) u ( z ) dz = (cid:90) y W y C (cid:90) x B x B − y + y C f u ( x, y ) dxdy = (cid:90) b a (cid:90) b ( y ) a ( y ) [ (cid:90) b ( x,y ) a ( x,y ) q ( l ) dl + g ( x, y )] dxdy = (cid:90) y W y C q ( y ) v ( y ) + C Thus, R EBW = (cid:90) W E T · ˆn ds + (cid:90) y W y C q ( y )( v ( y ) + v ( y ) + v ( y )) dy + C + C + C = (cid:90) y W y C q ( y ) v ( y ) dy + C The following proof is similar to Lemma 5.8, optimal q ( y ) y ∈ [ y C , y W ] comprises two parts. Thereis y (cid:48) ∈ [ y C , y W ] such that q ( y ) = (cid:26) y ∈ [ y C , y (cid:48) ) q ( y W ) y ∈ [ y (cid:48) , y W ] The utility function is u ( x B , y ) = (cid:26) u ( W ) + ( y − y W ) q ( y W ) y ∈ [ y (cid:48) , y W ] u ( W ) + ( y (cid:48) − y W ) q ( y W ) y ∈ [ y B , y (cid:48) ] Hence u ( x B , y ) ≥ u ( W ) + ( y (cid:48) − y W ) q ( y W ) ≥ u ( W ) + ( y B − y W ) . Because W E is a 45 degree line, x B − x E = y W − y B . Then u ( W ) = ( y W − y B ) q ( y W ) + ( x B − x E )(1 − q ( y W )) = y W − y B . We get u ( x B , y ) ≥ . This new utility function satisfies the convexity and nonnegative property, so it’s feasible.Therefore, u ( BW ) comprises at most two pieces. (cid:117)(cid:116) Proof. of Lemma 5.12 R GSMEW D = (cid:90) GSMEW T · ˆn ds + (cid:90) GD T · ˆn ds + (cid:90) W D T · ˆn ds − (cid:90) GSMD f u ( z ) dz − (cid:90) DMEW f u ( z ) dz = C + (cid:90) x D x G y C u ( x, y C ) f dx − f (cid:90) x D x G (cid:90) y C y C − u ( x,y C ) ( y − y C + u ( x, y C )) dydx + (cid:90) y D y W x B u ( x B , y ) f dy − f (cid:90) y D y W (cid:90) x W x W − u ( x W ,y ) ( x − x W + u ( x W , y )) dxdy = C + (cid:90) x D x G f u ( x, y C )( y C − u ( x, y C )) dx + (cid:90) y D y W f u ( x B , y )( x B − u ( x B , y )) dy So, for fixed u ( D ) and u ( W ) , when u ( x B , y ) > x B , x ∈ ( y W , y D ) , revenue is decreasing in u ( x B , y ) ;when u ( x B , y ) < x B , y ∈ ( y W , y D ) , revenue is increasing in u ( x B , y ) . Similar to the proof in Lemma 5.3,WLOG, we can assume there is a point P on W D segment, such that u ( x B , y P ) = x B . Since u ( P D ) ≥ u ( P ) , revenue is increasing as u ( P D ) decrease. Since u ( W P ) ≤ u ( P ) , revenue is increasing as u ( W P ) increase. Let u ( x B , y ) = u ( W ) + u ( P ) − u ( W ) y P − y W ( y − y W ) , u ( x B , y ) = u ( D ) + q ( y D )( y − y D ) . Then u ( x B , y ) = max ( u ( x B , y ) , u ( x B , y )) y ∈ [ y W , y D ] gives the optimal revenue with fixed u ( W ) , u ( P ) , u ( D ) ,and q ( y D ) . So optimal u ( W D ) comprises at most two pieces. (cid:117)(cid:116) Proof. of Theorem 5.10
We can now settle the optimal utilities on BD , subject to fixed values of u ( W ) and u ( P ) as well asthe convexity of u . Let α = u ( P ) − u ( W ) y P − y W and t α = (1 − α ) x P + αy P − u ( P ) . Adding a new menu item (1 − α, α, t α ) , the utility of choosing this new item is u α , which is denoted by the red dashed line shown inFig. 12. Let u D be the utility obtained by choosing the same menu item as point D : (1 − q ( y C ) , q ( y C ) , t D ) .Thus, for any point ( x, y C ) on CD , we have u ( x, y C ) ≥ u D ( x, y C ) = (1 − q ( y C )) x + q ( y C ) y C − t D ≥ (1 − q ( y C )) x D + (1 − q ( y C ))( x − x D ) + q ( y C ) − t D = u ( D ) + (1 − q ( y C ))( x − x D ) ≥ u α ( D ) + (1 − α )( x − x D )= (1 − α ) x D + αy C − t α + (1 − α )( x − x D ) = u α ( x, y C ) Thus, after adding this new menu item, u ( CD ) does not change and u ( P W ) will weakly increase.24n the optimal mechanism with fixed q ( y W ) , according to Lemma 5.11, types on BW choose only 2menu items: (1 − α, α, t α ) and (1 , , t ) for some t . Using same arguments as CD , for BD , we haveanother two menu items: (1 − β, β, t β ) and (0 , , t ) for some t .We now consider a brand new menu with only the above four menu items and (0 , , . Comparedto the old menu, u ( P D ) weakly decreases while u ( P W ) weakly increase. According to Lemma 5.12, R GSMEW D weakly increases. For fixed q ( y W ) and u ( W ) , R EBW is maximized. For fixed q ( x G , y G ) and u ( G ) , R CSG is maximized. The total revenue weakly increase. The optimal menu consists of at most 5menu items: q q t t − α α t α t β − β t β (cid:117)(cid:116)(cid:117)(cid:116)