Efficient cost sharing with a cheap residual claimant Hervé Moulin∗ Rice University, Houston,Texas, USA September 2006

Abstract ∗

The residual mechanism shares efficiently a one-commodity convex technology and generates a small budget imbalance. The mechansim may subsidize a user with a null demand. If the cost function is totally monotone (e.g., polynomials with positive coefficients) it guarantees Voluntary Participation and generates a budget surplus. The ratio of this surplus to the overall surplus in the economy is no larger than min{ log2 n , 1}, where n is the number of potential users. For cost functions of the form C(a) = ap , p > 1 and their positive linear combinations, the mechanism may generate a budget deficit, however the above ratio converges faster to zero. Voluntary Participation appears to be true. For analytic cost functions, the ratio of the budget imbalance to the efficient surplus converges to zero exponentially when the set of users increases. All properties above are lost if the cost function is not smooth.

1

The residual claimant approach

The residual claimant is a simple way to align monetary incentives and efficiency. Its most familiar application is the rich class of Vicrey-Clarke-Groves (VCG) mechanisms. If individual preferences are quasilinear in money (or any other numeraire) a VCG mechanisms take the "efficient" decision, except for the fact that the budget imbalance must be covered by a residual claimant ([25, 7]). Thus we achieve full efficiency (Pareto optimality) in the augmented economy where the claimant’s preferences are taken into account, and he only cares about cash transfers. In this case we speak of residual efficiency in the initial economy. ∗ I am greatly indebted to Fedor Nazarov at Michigan State University, for the crucial insight into the proof of Theorem 1, and to Doug Hensley at Texas A&M, who provided numerical simulations in support of a key conjecture. This work is supported by the NSF under grant SES-0414543.

1

I submit that the realism of the residual claimant idea hinges around the size of the cash transfer he or she receives. If the residual claimant (thereafter RC) pockets a surplus commensurate to -or larger than- that available in the economy, the cost of generating the efficient decision is prohibitive: the RC receives a significant rent, and the choice of the entity playing the role of RC is a matter of dispute. These difficulties are amplified if the RC must finance a substantial deficit, in effect paying out an additional rent to the participants: now it may not even be feasible to find an entity willing to play the RC role. I propose a canonical mechanism, called residual ∗ , achieving residual efficiency while running a one commodity convex technology. This mechanism is simpler than a VCG mechanism, in the sense that individual messages are one-dimensional "demands", namely a request for a certain amount of output, which the mechanism must meet. Its incentives properties are weaker: no individual demand is a dominant strategy for most users; however for any profile of convex quasi-linear preferences, the non cooperative analysis of the demand game is quite straightforward. This game is a potential game ([14]), therefore Nash equilibrium behavior is a convincing prediction even when information on preferences is entirely private: a Nash equilibrium is achieved by fictitious play, best reply dynamics, and many other learning algorithms (??). For many smooth cost functions, in particular for all analytic functions, the residual∗ mechanism generates a budget transfer to the RC that is vanishingly small relative to the overall surplus in the economy, when the set of potential users of the commons grows. Thus the introduction of a RC to run the mechanism is not controversial, whether this entity will receive a small surplus from, or pay a small subsidy to the participants. The benefit of aligning efficiency with incentive-compatibility obtains far outweighs the cost of establishing an independent Residual Claimant. One of the oldest1 normative requirements of a resource allocation mechanism is that agents should have no objection to participate. In many contexts, this results from private property rights (e.g., in any trading mechanism); in other contexts (provision of a local public good), it reflects a concern for fairness: an agent who is forced to participate subsidizes other agents, she does not get a share of the collective surplus. For the rich class of totally monotone cost functions, the residual∗ mechanism guarantees voluntary participation. In particular a user with a null or small demand may reseive a cash subsidy. This makes good sense when marginal costs increase: refraining from using the technology benefits all active users. In our mechanism inactive users in effect have a claim on the surplus generated by the active users. The manager of the mechanism must monitor entry, lest many newcomers show up to demand no output and receive some cash. Relation to the literature Recent applications of demand mechanisms include resource allocation in networks with elastic supply,such as sharing the cost of a bandwidth, assigning service priorities, etc.. ([6, 8, 9, 10, 23, 27, 3]). 1 It

goes back to the early literature on public finance. See e.g. Wicksell’s entry in [20].

2

In our model we can construct fully efficient (in particular, budget balanced) mechanisms only when the cost function is a polynomial of low enough degree, in all other cases a RC is required to balance the budget (Lemma 4). This result is technically similar to the classic impossibility of balancing the budget with a VCG mechanism, and in particular to the ‘cubical array’ Lemma in [26]. Our asymptotic efficiency results (Theorems 2,3,4) contrast with other results of the same flavour, such as those in the literature on double auctions (??). The latter are always probabilistic statements about the expected convergence of the equilibrium surplus to the efficient surplus. Thus they are predicated on the existence of common priors on individual types. By contrast, our convergence results bear on the worst case configuration of types, they are completely independent of any prior belief on individual types. Evaluating the worst relative surplus loss in equilibrium, over the entire domain of preference profiles, is a promising systematic approach to mechanism design. It is inspired by the recent literature on the price of anarchy in congestion games (??).

2

Overview of the results

The residual∗ mechanism introduced here allows each user i of the commons to demand an arbitrary quantity xi of "service" (output). Given the increasing, convex, and differentiable cost function C, for each demand profile x the mechanism computes monetary (input) charges yi . Thus each individual demand is served in full by the mechanism, who can only adjust the monetary charges to the various users, including those who choose not to consume. This is the property called Consumer Sovereignty in [19], and the defining feature of simple cost sharing mechanisms (see references above). Given quasi-linear utility functions ui (xi , yi ) = vi (xi ) − yi for each user, the predicted outcome is some Nash equilibrium of the resulting demand game. To guarantee residual efficiency ( Pareto optimality in the economy augmented by theP residual claimant) at all equilibria for all utility profiles, we must choose yi = C( i xi ) − hi (x−i , C) where the function hi does not depend on xi but is otherwise arbitrary (Lemma 1). By Nash’s theorem if vi is concave the demandPgame has at least one P Nash equilibrium, and at each equilibrium vi0 (xi ) = C 0 ( i xi ) 0 0 (or P vi (xi ) ≤ C ( i xi ) if xi = 0) for all i. Therefore thePequilibrium demand P i xi is optimal, i.e., it maximizes total surplus P (x) = i vi (xi ) − C( i xi ). Moreover the function P is a potential for the demand game, thus ensuring strong convergence properties of the classic learning algorithms ([14, 15, 12]). Finally, when n potential users the technology, the P budget imbalance (surP share P P plus or deficit) ∆(x, C) = i yi − C( i xi ) = (n − 1)C( i xi ) − i hi (x−i , C) is transferred to the RC. |∆(x,C)|S At the equilibrium demand profile x, rn (x, C, v) = S vi (x meai )−C( i i xi ) sures the relative inefficiency of the equilibrium outcome. We call this ratio the residual cost of the mechanism at equilibrium x. A natural choice of the transfers functions is hi (x−i , C) = C(xNÂ i ), defining the incremental mechanism where each user pays the incremental cost of adding 3

her own demand to that of other users2 . Note that these transfers are defined by the normatively appealing property that a null demand, xi = 0, results in a zero charge, yi = 0. Convexity of C ensures that the RC receives a positive surplus, however this surplus may exhaust all but a n1 -th fraction of the efficient surplus: we can choose for all n a profile v of utilities and a corresponding Nash equilibrium x, such that rn (x, C, v) ≥ 1 − K n , where the constant K does not depend on n. See section 4. P With the notation xS = i∈S xi for any subset S of agents, our residual∗ mechanism is defined as follows 1 X 1 X C(xNÂ i,j ) + C(xN Â i,j,k )− h∗i (x−i ) = (n − 1){C(xN Â i ) − 2 3 j∈N Â i

(−1)n−2 X C(xj )} n−1 S∈NÂ i,|S|=k j∈N Â i P The first property of the cost shares yi∗ = C( i xi ) − h∗i (x−i ) is that they cover exactly the budget whenever the cost function C is a polynomial of degree at most n − 1. In this case every Nash equilibrium of the demand game is fully efficient and there is no need to introduce a residual claimant. Polynomials of degree n−1 or less are in fact the only cost functions for which we can choose the hi to balance the budget for any demand profile (Lemma 3). For all other cost functions, residual efficiency implies budget imbalance at some demand profile; our hope is make it small. Second property: its budget is balanced if not all users are active. Ppotential P For any cost function C, xi = 0 for some i implies yi∗ = C( xi ). In combination with the symmetric treatment of all individual demands, this property defines the residual∗ mechanism (Lemma 4). A third remarkable property is that for many cost functions, the sign of the budget transfer to the RC is independent of the utility profile: if n users share the commons, and the n-th derivative C (n) is everywhere non negative (resp. everywhere non positive), the residual∗ mechanism always generates a budget surplus (resp. a deficit): Lemma 6. Our two main results, in section 6, apply to totally monotone cost functions (k) C, namely indefinitely differentiable with all derivatives P∞C ,kk = 1, · · · non negative on R+ . Such functions are analytic C(a) = 1 λk a with λk ≥ 0. Here the residual∗ mechanism is compelling. On the one hand (Theorem 1) it never generate a budget deficit, and its residual cost is at most min{ log2 n , 1}, irrespective of C, the utility profile v = (vi ), and the Nash equilibrium x. On the other hand (Theorem 2) it guarantees the two critical normative requirements Ranking (RKG) and Voluntary Participation (VP). Ranking is a classic test of fairness (e.g., Moulin and Sprumont), requiring cost shares to be co-monotonic with demands: xi ≤ xj ⇒ yi ≤ yj . Absent this property, some users have a very strong claim that their charge is unfair, i.e., they pay more for less service! ··· +

2 This

(−1)k k+1

X

j,k∈NÂ i

C(xNÂ i,S ) + · · · +

idea has a long history in the cost sharing literature: see [24, 2]

4

Voluntary Participation is an even more common3 normative test, with a stronger incentives flavour than RKG. If users can declime to participate in the mechanism, no outcome where they are worse off than receiving no output at no charge is sustainable. In our model, VP means that a null demand is never taxed: xi = 0 ⇒ yi∗ ≤ 0 (Definition 2). Note that this does not preclude to subsidize a null demand, xi = 0 and yi∗ < 0. Unlike the incremental one, our residual∗ mechanism routinely hads out cash to inactive users; for instance Lemma 5 shows that with many more inactive than active users, the rent of the former is quite large. Recall that if there is at least one inactive user, the charges will exactly cover the costs: that is because inactive users are, in effect, sharing the residual claim. Thus the right to participate in the residual∗ mechanism must be carefully monitored. In section 7, we examine the performance of our mechanism for several other classes of smooth cost functions. The first PK one consists of the positive generalized polynomials, of the form C(a) = k=1 λk apk where pk > 1 and λk > 0. Although it is no longer possible to sign the budget imbalance, the asymptotic efficiency property of the residual∗ mechanism remains strong: the worst case residual cost (over all utility profiles and all Nash equilibria) converges to zero 1 as ninf k {p (Theorem 3). Numerical evidence strongly suggests that both k }−1 Voluntary Participation and Ranking still hold for these functions, but I have been unable to prove or disprove this conjecture. P k We turn next to the class of analytic functions C(a) = ∞ where the 1 λk a P K sign of λk is arbitrary, and to the generalized polynomials C(a) = 1 λk apk with pk > 1 and λk ∈ R. Here Voluntary Participation and Ranking may fail. However a weaker form of asymptotic efficiency still holds: the residual cost converges to zero as the set of users increases (as opposed to an increase in the number of users). The convergence is exponential in the former case, and hyper-polynomial in the latter: Theorem 4. We show finally in section 8 that the regularity of the cost function is critical to the good behavior of our mechanism. For a piecewise linear cost function, and more generally for a non differentiable cost function, the residual∗ mechanism generates surpluses and deficits, growing exponentially with n. It grossly violates VP and RK. In fact for such cost function„ P any residually efficient mechanism (i.e., with cost shares of the form yi = C( i xi ) − hi (x−i , C)) has a residual cost of at least 100%.

3

Residually efficient demand mechanisms

A finite set N of agents share a technology C producing a units of output at cost C(a). We assume that C is convex, non decreasing, and differentiable on R+ , with C(0) = 0, C 0 (∞) = ∞. Each user i ∈ N requests the quantity xi , xi ≥ 0, of output and is charged $yi : this charge can be negative, namely user i can be subsidized by the mech3A

formal statement in the cost sharing context is in [19].

5

anism (for instance as a reward for a null demand). User i’s allocation is zi = (xi , yi ) ∈ R+ × R. A profile of individual allocations, one for each user i, is denoted z ∈ (R+ × R)N and is called an outcome. P If S is a subset of N , we use the notations xS = i∈S xi , and similarly for yS ; we use the convention x∅ = 0. The budget imbalance of outcome z given C is ∆(z, C) = yN − C(xN ) which we call a surplus if ∆(z, C) ≥ 0, and a deficit if ∆(z, C) ≤ 0. Preferences of user i are represented by a quasi-linear utility function vi (xi )− yi , where vi is concave, nondecreasing, and normalized by vi (0) = 0. We write D for this preference domain, and we abuse language by speaking of vi as an element of D. Definition 1 The outcome z is efficient in the economy (N, C, v) if it is budget balanced, ∆(z, C) = 0, and Pareto optimal in the set of budget balanced allocations. The outcome z is residually efficient in (N, C, v) if the augmented outcome (z, zrc ), where zrc = (0, −∆(z, C)), is Pareto optimal in the augmented economy (N ∪ {rc}, C, (v, vrc )) where the residual claimant rc only cares about money: vrc (z) = −y. P We write es(C, v) = maxx∈RN { N vi (xi )−C(xN )} for the efficient surplus of + the economy (N, C, v) above the null outcome (recall the normalizations C(0) = 0 and vi (0) = 0). Our assumptions guarantee that this maximum is achieved, and that the set of efficient outcomes is non P empty and compact. Clearly, outcome z is residually efficient if and only if N vi (xi ) − C(xN ) = es(C, v). In other words, residual efficiency is ‘efficiency except for budget balance’. Given a set of potential users N and a cost function C, a demand mechanism is a game form associating to each profile of demands x ∈ RN + a profile of charges y = ξ(x) ∈ RN . Each utility profile v generates a non cooperative demand game (N, v, ξ), where xi is user i’s strategy and vi (xi ) − ξ i (x) is her payoff. The mechanisms achieving residual efficiency at every Nash equilibrium outcome are easy to describe. Lemma 1 For any N and cost function C as above, the two following statements are equivalent: i) for any utility profile v in DN the demand game (N, v, ξ) has at least one Nash equilibrium, and all Nash equilibria are residually efficient. ii) the demand mechanism takes the form yi = ξ i (x) = C(xN ) − hi (x−i , C) for all i ∈ N and x ∈ RN +

(1)

NÂi . where the function hi is arbitrary over R+ A mechanism meeting these properties is called residually efficient. Proof Statement P ii) ⇒ i). The P normal form game (N, v, ξ) is a potential game for P (x) = i vi (xi ) − C( i xi ). Our assumptions guarantee that the set of the

6

maximizers of P over RN + is non empty and compact. That a maximizer of P is a Nash equilibrium is clear. Conversely pick a Nash equilibrium x, and an 0 arbitrary x0 ∈ RN + . For each i, C (xN ) supports vi at xi : vi (x0i ) − vi (xi ) ≤ C 0 (xN )(x0i − xi ) for all x0i , xi ≥ 0

(2)

Convexity of C gives C(x0N ) − C(xN ) ≥ C 0 (xN )(x0N − xN ), and P (x0 ) ≤ P (x) follows. Statement i) ⇒ ii). We fix a mechanism ξ as statement i), and an arbitrary 0 x ∈ RN + . If C (xN ) > 0, we chose for each i a strictly concave and smooth function vi ∈ D such that vi0 (xi ) = C 0 (xN ), and vi0 (∞) = 0. If C 0 (xN ) = 0, we take vi ∈ D strictly concave and smooth on [0, xN ] and such that vi0 (xi ) = 0. We claim that x is the only residually efficient demand profile at v. This is clear if C 0 (xN ) = 0, so we assume C 0 (xN ) > 0. Note that vi0 is continuous and strictly decreasing, therefore invertible on ]0, vi0 (0)]. We extend the definition of (vi0 )−1 by setting (vi0 )−1 (α) = 0 for α ≥ vi0 (0). Now the first order conditions characterizing a maximizer x of P over RN + are simply the existence of some α ≥ 0 such that X {C 0 (xN ) = α and xi = (vi0 )−1 (α) for all i} ⇔ C 0 ( (vi0 )−1 (α)) = α N

The equation in α has a unique solution because its left hand side is continuous and non increasing from ∞ to C 0 (0). By assumption the demand game (N, v, ξ) has at least one residually efficient equilibrium, which can only be x. Consider now agent i’s opportunity set O(x−i ) = {zi = (ai , bi ) ∈ R+ × R|bi = ξ i (ai , x−i )} when other strategies are fixed at x−i . The equilibrium property is that O(x−i ) is to the south-east of the indifference curve through (xi , ξ i (x)); the latter is the curve with equation bi − ξ i (x) = vi (ai ) − vi (xi ). For appropriate choices of vi , this indifference curve can be arbitrarily close to the line with slope C 0 (xN ): this implies the following inequality: ξ i (ai , x−i ) − ξ i (x) ≥ C 0 (xN )(ai − xi ) for all ai ≥ 0 Fix now x−i and write for all a ≥ 0 : f (a) = ξ i (a, x−i ) and γ(a) = C 0 (a+xN Â i ). We have just proved f (b) − f (a) ≥ γ(a)(b − a) for all a, b ≥ 0 Hence f (b) = supa≥0 {f (a) + γ(a)(b − a)} is convex, and its subgradient contains γ. Therefore f is a primitive of γ, namely f (a) = C(a + xNÂ i ) + hi , where hi depends only upon x−i . This concludes the proof. ¥ In the above proof, differentiability of C is important: we show in section 8 that, if C has even a single kink, a Nash equilibrium in a demand mechanism of the form (1) may not be residually efficient. For any residually efficient mechanism, user i’s cost share yi is non decreasing in her demand xi , however the sign of this cost share is not restricted: a null 7

demand xi = 0 may well result in a payment to i, i.e., yi < 0. In that case we think of user i as a partial claimant of the surplus generated by the other, active users of the technology. This is in sharp contrast to most of the literature on cost sharing mechanisms(e.g., [17, 18, 19]), where a null demand is neither charged nor subsidized (xi = 0 ⇒ yi = 0). Subsidizing an inactive user is justifiable under increasing marginal costs, because such a user helps the active users. However what is not desirable is the configuration xi = 0 and yi > 0, in which user i would rather stay away entirely, so that participation in the mechanism is not voluntary. Definition 2 The residually efficient mechanism ξ satisfies Voluntary Participation (VP) if it meets the following two equivalent properties i) for all demand profile x and all i ∈ N : xi = 0 ⇒ yi = ξ i (x) ≤ 0 ii) for any v∈DN and any Nash equilibrium outcome z, vi (xi ) − yi ≥ 0 for all i ∈ N (recall vi (0) = 0). Under assumption i), an agent guarantees the null utility by demanding no output, which implies ii). Conversely, if there exists x and i such that xi = 0 and yi > 0, we define vi ≡ 0 and choose vj so that vj0 > C 0 (xN ) on [0, xj ], after which vj is flat. Then x is a Nash equilibrium outcome. If VP has a dual interpretation in terms of incentives and equity, our next property is a straight test of equity. We say that the mechanism ξ satisfies Ranking (RKG) if for all demand profile x and all i, j ∈ N : xi ≤ xj ⇒ yi ≤ yj Ranking rules out a very coarse form of inequity, whereby a user is charged more for demanding less output4 . The most natural residually efficient mechanism is the incremental + mechanism: for all i ∈ N and x ∈ RN + yi+ = C(xN ) − C(xN Â i )

(3)

Within residually efficient mechanisms, it is characterized by the property xi = 0 ⇒ yi = 0. By convexity of C, and C(0) = 0, it never generates a budget + deficit: yN ≥ C(xN ) for all x. A ‘dual’ residually efficient mechanism never generates a budget surplus; we call it incremental − : yi− = C(xN ) −

n−1 n ) C( x n n − 1 NÂ i

− ≤ C(xN ) is a simple consequence of the convexity of C. The inequality yN Both mechanisms meet Ranking and VP, moreover yi− ≤ yi+ for all x and i. Yet the resulting social cost of the residual claimant is too high. We show in the next section that under incremental+ , the RC can swallow almost the entire efficient surplus in the economy; under incremental− , the size of the deficit may be arbitrarily larger than the efficient surplus. 4 A stronger property is No Envy : no user prefers another user’s equilibrium outcome to his own. In fact No Envy is out of reach for residually efficient mechanisms, with the exception of the two person problem with quadratic costs. Proof available upon request.

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4

The residual cost of a residually efficient mechanism

We measure the (social) cost of the residual claimant, first at a specific outcome, then for a given mechanism by evaluating the ‘worst case scenario’, over all equlibrium outcomes at all utility profiles. Definition 3 Given are the set of agents N and a cost function C. i) The residual cost of the residually efficient outcome z at utility profile v ∈ DN is |yN − C(xN )| |∆(z, C)| =P r(z, C, v) = es(C, v) N vi (xi ) − C(xN )

ii ) The residual cost of the residually efficient mechanism ξ is r(C, ξ) =

sup

r(z, C, v) where z = (x, ξ(x))

v∈DN ,x∈Neq(N,v,ξ)

(the notation N eq stands for the set of Nash equilibrium demands) Note that we count surplus and deficit alike. As explained in the introduction, both represent a liability, and a good mechanism is one where the budget imbalance is small, irrespective of its sign. A useful observation: if the residually efficient mechanism ξ never generates a budget deficit, and satisfies Voluntary Participation, then r(C, ξ) ≤ 1 for all C. An example is the incremental+ mechanism. Proof: at any equilibrium outcome z, VP implies vi (xi ) ≥ yi for all i. Summing up X vi (xi ) ≥ yN = C(xN ) + ∆(z, C) N

∆(z, C) ≤ 1 =⇒ r(C, ξ) ≤ 1 N vi (xi ) − C(xN )

=⇒ P

The following key result is a variant of the more general "linearity lemma" introduced in [8]. It reduces the computation of the residual cost of a mechanism to a tractable maximization problem. We write the budget imbalance ∆(z, C) = ξ N (x) − C(xN ) at the outcome z = (x, ξ(x)) as ∆(x, C, ξ). Lemma 2 Given N and C, and a residually efficient mechanism ξ, we have P |(n − 1)C(xN ) − N hi (x−i , C)| |∆(x, C, ξ)| r(C, ξ) = sup = sup (4) Γ(xN , C) Γ(xN , C) RN RN + + where Γ(a, C) = aC 0 (a) − C(a) for a ≥ 0. Proof Pick an arbitrary utility profile v in D and a Nash equilibrium x in the corresponding demand game. Write zi = (xi , ξ i (x)). The concave function 9

i (x) x0i → vi (x0i ) − ξ i (x0i , x−i ) reaches its maximum at xi , and ∂ξ∂x = C 0 (xN ), i 0 0 implying (2). Applying the latter at xi = 0 gives vi (xi ) ≥ C (xN ) · xi . Next consider the unanimous profile v ∗ of linear utilities vi∗ (a) = C 0 (xN ) · a for all i and a ≥ 0. Clearly x is still a Nash equilibrium in (N, v ∗ , ξ), and we have X X vi (xi ) − C(xN ) ≥ vi∗ (xi ) − C(xN ) = Γ(xN , C)

N

¥

N

|∆(x, C, ξ)| |∆(x, C, ξ)| ≤ = r(z, C, v ∗ ) v (x ) − C(x ) Γ(x , C) N N N i i

⇒ r(z, C, v) = P

The proof shows that for any demand profile x, the unanimous linear utilities with slope C 0 (xN ) achieve the worst residual cost over all utility profiles such that x is a Nash equilibrium. At this utility profile, x0 is a Nash equilibrium if and only if x0N = xN 5 . To illustrate the power of formula (4), we evaluate the residual cost of the two incremental mechanisms. Starting with incremental+ , recall ∆(x, C, ξ + ) = P (n − 1)C(xN ) − N C(xN Â i ) ≥ 0. Fixing a = xN , the maximum over x of ∆(x, C, ξ + ) obtains for xi = xj all i, j. Therefore (4) reads rn (C, ξ + ) = (n − 1) sup a>0

n C(a) − n−1 C( n−1 n a) aC 0 (a) − C(a)

(5)

The discussion after Definition 3 proved rn (C, ξ + ) ≤ 1 which is also easy to check directly from (5). Fixing a in the right hand side of (5), the ratio goes to 1 as n → ∞. We conclude limn→∞ rn (C, ξ + ) = 1: when n grows large, the RC may claim almost the whole surplus for himself. The discussion of the incremental− mechanism is similar, but the conclusion P n is worse. There |∆(x, C, ξ − )| = (n − 1){ n1 N C( n−1 xN Â i ) − C(xN )}, and the maximum for a = xN fixed obtains whenever all but one demands are null, therefore n n−1 C( n−1 a) − C(a) rn (C, ξ − ) = (n − 1) sup n (6) 0 (a) − C(a) aC a>0

n 1 a) ≥ C(a)+ n−1 aC 0 (a) in the right hand side of equation Using inequality C( n−1

(6), we get rn (C, ξ − ) ≥ (n−1) n , so for large n, the worst case deficit is barely below the entire surplus. For a power function Cp (a) = ap , check limn rn (Cp , ξ − ) = 1; on the other hand for E(a) = ea − 1, pick a = n(n − 1) in (6) to conclude limn rn (E, ξ − ) = ∞.

5

The residual∗ mechanism

Can a residually efficient mechanism be budget balanced for every profile of demands ? The answer to this question singles out a small class of cost functions: 5 If C 0 increases strictly at x . In general any x0 such that C 0 (x0 ) = C 0 (x ) is an N N N N equilibrium.

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Lemma 3 i)If C is a polynomial of degree at most n − 1, there is a n-person residually efficient demand mechanism (1) that is budget balanced for all demands x ∈ Rn+ . ii) If C is bounded on every interval (e.g., continuous), and is not a polynomial of degree at most n − 1, there is no n-person residually efficient mechanism achieving budget balance at all demands. A short-cut to the proof of statement ii) assumes that C is n times differentiable. The budget balance property for the mechanism (1) amounts to X (n − 1)C(xN ) = hi (x−i , C) for all x ∈ RN (7) + N

(n)

∂ Taking the n-th derivative ∂x1 ∂x on both sides gives C (n) ≡ 0, so that C 2 ···∂xn is a polynomial of degree at most n − 1. We give in the appendix a proof of statement ii) when we do not assume the differentiability of C. The proof of statement i) is combined with that of statement i) in Lemma 4 below. There we solve equation (7) in the unknown functions hi under the additional assumption that the mechanism is anonymous, namely treats symmetrically the n demands. It turns out that there is a unique function h∗ on Rn−1 + , symmetric in its n − 1 variables, and such that hi (x−i , C) = h(x−i ) solves (7). In turn h∗ defines our canonical mechanism for any cost function. Recall the convention x∅ = 0. Lemma 4 i)If C is a polynomial of degree at most n − 1, the only budget-balanced, anonymous, residually efficient demand mechanism is the following

h∗i (x−i , C) = h∗ (x−i , C) = (n − 1)

n−1 X 0

(−1)k k+1

X

T ⊂N Â i,|T |=k

C(xN Â {i}∪T )

(8)

ii)If C is an arbitrary cost function, (8) defines the n-residual∗ mechanism. It is residually efficient, anonymous, and budget balanced whenever at least one user demands no output, xi = 0 for some i. Together these three properties characterize the n-residual∗ mechanism. Proof Step 1: preliminary result. Fix N and a set function V from P subsets P of N , including N and the empty set, into R. Define B(N, V ) = n0 (−1)k T :|T |=k V (N ÂT ). Call agent i ∈ N a dummy in (N, V ) if V (T ∪ {i}) = V (T ) for all T ⊆ N Â{i}. Then if some agent i is a dummy, B(N, V ) = 0. Indeed T → T ∪ {i} is a one-to-one mapping from 2N Â {i} to 2N Â2N Â {i} , and in the sum defining B, the terms V (T ) and V (T ∪ {i}) cancel each other. Now let G be an arbitrary function from x ∈ RN + to G(x) ∈ R. For any P Pn k x, define BN (x, G) = T :|T |=k G(xN Â T , 0T ). Applying the above 0 (−1) remark to the set function V (T ) = G(xN Â T , 0T ), we see that BN (x, G) = 0 whenever at least one coordinate xi is null. 11

Step 2: proof of statement ii). The budget imbalance of the residual∗ mechanism (8) is easily computed: X h∗ (x−i , C) = (n − 1)Bn (x, C) (9) ∆∗n (x, C) = (n − 1)C(xN ) − N

where Bn (x, C) =

n X X (−1)k C(xNÂ T ) 0

T :|T |=k

By step 1, Bn (x, C) = 0 whenever at least one coordinate xi is null. Next we claim that h∗ isP the only function h symmetric in its n−1 variables, such that (n − 1)C(xN ) − N h(x−i , C) = 0 whenever at least one coordinate of x is null. This is the second part of statement ii). Pick such a function h(x−i ) (we omit C in h(x−i , C) for simplicity). We prove the claim by induction on the number e ∈ P of non-zero coordinates of x n Rn−1 + . For x = 0 ∈ R+ , budget balance N h(x−i ) = (n − 1)C(xN ) gives h(e 0) = C(0). Assume h(e x) is determined for each x e with at most k − 1 non zero x, 0) ∈ coordinates, and consider x e = (x1 , · · · , xk , 0, · · · , 0) ∈ Rn−1 + . Write x = (e P x) = (n − 1)C(x{1,··· ,k} ), Rn+ ; budget balance at x gives k1 h(x−i ) + (n − k)h(e which determines h(e x). Step 3: proof of statement i) in Lemma 4 (and Lemma 3). We only need to check that our mechanism is budget-balanced P everywhere if C is a polynomial of degree at most n−1. Define Ank (x, C) = T ⊂N,|T |=k C(xN Â T ), and for any a ≥ 0, Ca (b) = C(a + b). We check first the following identity, for all i, all k = 0, · · · , n − 1, all C and x: Ank (x, C) − Ank ((0, x−i ), C) = An−1 (x−i , Cxi − C) k

(10)

For any T containing i, the two corresponding terms in the LHS cancel; for T ⊆ N Â{i}, they are C(xN Â T ) − C(xN Â T ∪{i} ) = (Cxi − C)(x(N Â {i})Â T ), establishing (10). As Bn (x, C) is a linear combination of Ank (x, C), and Ann (x, C) = Ann ((0, x−i ), C) = C(0), (10) gives Bn (x, C)−Bn ((0, x−i ), C) = Bn−1 (x−i , Cxi − C). Step 2 gives Bn ((0, x−i ), C) = 0, hence we have for all i, C,and x: Bn (x, C) = Bn−1 (x−i , Cxi − C)

(11)

Statement i) follows by an easy induction on n. Start with B1 (x, C) = C(x) − C(0), which is null if C is a constant. Next if C is a polynomial of degree at most n − 1, then Cxi − C is a polynomial of degree at most n − 2.¥ With two potential users, n = 2, h∗ (x−1 ) = C(x2 ) hence the residual∗ and incremental + mechanisms coincide: y1∗ = y1+ = C(x1 + x2 ) − C(x2 ). ForP n ≥ 3, the computation is more involved. For instance, with the notation x = xi , n = 3 : y1∗ = C(x) − 2C(x2 + x3 ) + C(x2 ) + C(x3 )

n = 4 : y1∗ = C(x) − 3C(x2 + x3 + x4 ) + 12

X 3X C(xi + xj ) − C(xi ) 2

The defining property of our mechanism is budget balance whenever at least one demand is null (statement ii) above). However a null demand xi = 0 typically does not mean a null charge for the inactive user i, who often has a partial claim on the surplus (or deficit). As a first illustration of this important point, suppose among n potential users, only the first two are active, x1 , x2 > 0, x3 = · · · = xn = 0. Then (8) implies yi∗ = yi+ for i = 1, 2; yj = −

1 (C(x1 + x2 ) − C(x1 ) − C(x2 )) for j ≥ 3 (12) n−2

+ In words, the inactive users share the budget surplus left by Pthe incremental n mechanism.To check this claim, use the notation Ak (x, C) = T ⊂N,|T |=k C(xN Â T ) introduced in the above proof, so that

h∗ (x−1 , C) = (n − 1)

n−1 X 0

(−1)k n−1 (x−1 , C) A k+1 k

(13)

¡ ¢ and for the demand profile where only x1 , x2 are positive, Ank (x−1 , C) = n−2 k C(x2 ). k¡ ¢ Pm m 1 The claim follows from the identity 0 (−1) k+1 k = m+1 . This result generalizes. Lemma 5 Let A be the set of active users, xi > 0, and I that of inactive agents, xi = 0, with cardinality nA , nI respectively. As we fix C, A, and the demands xi , i ∈ A, while the number nI grows, the total subsidy to inactive agents converges to X lim (−yI ) = (nA − 1)C(xA ) − C(xAÂ {i} ) (14) nI →∞

i∈A

The proof is in the Appendix. By convexity of C, in the limit inactive agents receive a positive subsidy, and limnI →∞ (−yI ) ≤ xA C(xA ) − C 0 (xA ) = Γ(xA ), the efficient surplus in the economy. Moreover as nA grows, this inequality converges to an equality. We check now that the budget imbalance generated by the residual∗ mechanism, when all demands are positive, can be a deficit or a surplus. This contrasts with incremental + and incremental − , where we have respectively always a surplus and always a deficit. The smooth, convex, increasing function 3 C(a) = ea − a3 −a−1, is an example where the sign of ∆∗3 (x, C), i.e., of B3 (x, C), varies with x. Figure 1 shows B3 (x, C) for x = ( a3 , a3 , a3 ) and a ≤ 2. However in many cases we can predict if the residual∗ mechanism will produce a budget surplus or a deficit. Lemma 6 Suppose the function C is n times differentiable on R+ Â{0}. i) If C (n) ≥ 0 (resp. > 0) on R+ Â{0}, then ∆∗n (x, C) is non-decreasing (resp. strictly increasing) in x and ∆∗n (x, C) ≥ 0 on Rn+ (resp. ∆∗n (x, C) > 0 on Rn+ Â{0}). 13

ii) If C (n) ≤ 0 (resp. < 0) on R+ Â{0}, then ∆∗n (x, C) is non-increasing in x (resp. strictly decreasing), and ∆∗n (x, C) ≤ 0 on Rn+ (resp. ∆∗n (x, C) < 0 on Rn+ Â{0}). Note that there are no sign constraint on the first derivatives of C, in particular C need not be convex. Proof By linearity of B in C, both statements are equivalent. We prove first by induction on n the following property P (n) {C (n) ≥ 0 on R+ Â{0}} ⇒ Bn (x, C) non-decreasing in x Property P (1) is clear from B1 (x, C) = C(x) − C(0). If C is n times differentiable, Cx1 − C is n − 1 times, and C (n) ≥ 0 implies (Cx1 − C)(n−1) ≥ 0. The inductive argument follows at once from (11). From P (n) and Bn ((0, x−1 ), C) = 0 (Lemma 4), we see that Bn (x, C) is non negative. A similar argument shows {C (n) > 0 on R+ Â{0}} ⇒ Bn (x, C) strictly increasing in x ¥ Consider a power function Cp (a) = ap , p ≥ 1. If p is an integer, all its derivatives are non negative so the residual∗ mechanismP generates a surplus for K k any n. The same is true of any polynomial C(a) = 0 λk a with positive coefficients λk . If p is not an integer, the sign of the transfer to the residual claimant depends on the parity of n. Write dpe for the smallest integer larger than p: the first dpe derivatives of Cp are positive, then they are alternatively negative (for dpe + 1, dpe + 3, . . .) and positive (for dpe + 2, dpe + 4, . . .). In the Appendix (section 10.4.1) we prove an additionalproperty of the budget imbalance ∆∗n (x, C), when the sign of C (n) is constant. For any given a = xN , the largest budget transfer (in absolute value) over all choices of x is achieved for identical demands.

6

Totally monotone cost functions

We now define a class of cost functions for which the efficiency, incentives and fairness properties of the residual∗ mechanism are compelling. Definition 4 The function C is totally monotone if it is indefinitely differentiable on R+ , and all its derivatives are non negative. Totally monotone functions are increasing and convex, as well as analytic6 , stable by positive linear combinations, multiplication, and composition. Examples include polynomials with positive coefficients, the functions ea , ba for any 6 because

a ≥ 0.

their Taylor development at 0 implies C(a) ≥

14

SK 0

k

C (k) (0) ak! for all K and all

b > 0, eP (a) + e−P (a) , eP (a) − eQ(a) if P, Q, and P − Q are polynomials with positive coefficients, etc.. We already know (Lemma 6) that for such functions our mechanism yields a budget surplus. Our next two results deal successively with its equity and efficiency properties. Theorem 1 i) If C is n − 1 times differentiable and C (k) ≥ 0 for k = 1, · · · , n − 1, the residual ∗ mechanism satisfies RKG, and VP; the latter is equivalent to yi∗ ≤ yi+ = C(xN ) − C(xN Â i ) for all x ∈ Rn+ , all i ∈ N

ii) If C is n times differentiable and C (k) ≥ 0 for k = 1, · · · , n

n−1 n ) for all x ∈ Rn+ , all i ∈ N C( x n n − 1 NÂ i The following notation is used in our next result, and in subsequent sections: an = Ω(bn ) (resp. an = O(bn )) means that abnn remains bounded away from zero (resp. infinity). And an = θ(bn ) means that an is both Ω(bn ) and O(bn ). Theorem 2 If C is totally monotone, we have 2 rn∗ (C) ≤ min{ , 1} for all n (15) log n yi∗ ≥ yi− = C(xN ) −

If E(a) = ea − 1 (a totally monotone function), rn∗ (E) = θ( log1 n ) The proof of both Theorems is in the Appendix. Recall from section 4 that the bound rn∗ (C) ≤ 1 follows from VP and the fact that the the residual∗ mechanism can only generate a surplus. The bound rn∗ (C) ≤ log2 n starts to bite for n ≥ 8, for instance ∗ ∗ ∗ ∗ r10 (C) ≤ 0.87; r25 (C) ≤ 0.62; r50 (C) ≤ 0.51; r100 (C) ≤ 0.43

The bound (15) is nevertheless remarkable, because it does not depend on the particular cost function in the totally monotone class. Of course for specific cost functions, the convergence rate O( log1 n ) can often be improved. For instance, if C is a polynomial, rn∗ (C) = 0 as soon as n exceeds its degree. Also, for the exponential function E n 2 5 10 25 50 rn∗ (E) 0.5 0.22 0.20 0.17 0.15

7

Other smooth cost functions

Which properties of the residual∗ mechanism are preserved when C is not totally monotone? We discuss successively the familiar power functions, then all analytic functions. On the negative side, the budget imbalance cannot be signed any more, and the Voluntary Participation and Ranking properties become hard to crack. On the other hand a form of asymptotic efficiency remains true for positive linear combinations of power functions, as well as all analytic cost functions. 15

7.1

Power cost functions

These are the cost functions Cp (a) = ap , where p > 1 is not an integer. They are derived from Cobb-Douglas production functions, hence are the most popular one dimensional family of cost functions in stylized models. See Mas-Colell et al.. Recall from Lemma 6 and the discussion following it, that the sign of the budget imbalance depends on the parity of n: it is a surplus if n = 2, · · · , dpe, or n = dpe + 2, dpe + 4, · · · ; a deficit if n = dpe + 1, dpe + 3, · · · . Theorem 3 1 i)For any p, p > 1, we have rn∗ (Cp ) = O( np−1 ); P pk with pk > 1 and ii) If C is a positive generalized polynomial C(a) = K 0 λk a 1 ∗ ∗ λk > 0, we have rn (C) = O( np∗ −1 ), where p = mink pk . Proof in the Appendix. Unlike Theorem 2 for totally monotone cost functions, the convergence of the relative cost of the RC is only in the asymptotic sense, it is not uniform in p. We only claim the existence of a constant Kp deKp . Some numerical examples for p = 1.5 pending upon p such that rn∗ (Cp ) ≤ np−1 1 and p = 2.5 suggest that the convergence is faster than np−1 : n 2 5 10 25 50 rn∗ (C1.5 ) 0.58 0.1 0.04 0.017 0.009 n 2 5 10 25 50 . rn∗ (C2.5 ) 0.43 1.0 × 10−2 1.2 × 10−3 1.2 × 10−4 2.5 × 10−5 Although one checks easily that ρn = supp>1 rn∗ (Cp ) is finite for all p, it is unclear whether it converges to zero as n grows. Numerically one checks that for small values of n, rn∗ (Cp ) is largest over p ∈]1, ∞[ when p approaches 1. Here is ρn for such values of n: n 2 5 10 25 50 rn∗ 0.69 0.47 0.36 0.28 0.17 Do the Voluntary Participation and Ranking properties hold for power functions ? By Theorem 1 they do if p ≥ n − 2, in particular for any p if n = 3. For larger n I conjecture that VP and RKG hold for all power functions; 1140 numerical tests in the case n = 7 did not disprove the conjecture7 . On the other hand the inequality yi∗ ≥ yi− certainly does not hold for n ≥ 7. A counterexample8 is p = 1.2, x = (0, 3, 6, 10, 12, 14, 20), for which y1∗ = −38.73 < −3.38 = y1− .

7.2

Analytic cost functions and generalized polynomials

Recall that totally monotone functions are simply analytic functions C(a) = P ∞ k 0 λk a where the coefficients λk are non negative. When we remove the sign restriction on the λk , the residual∗ mechanism is still asymptotically efficient, but in a weaker sense than in Theorems 2 and 3. Specifically, we think of 7I

am grateful to Doug Hensley for running these random tests. provided by Doug Hensley.

8 again

16

the technology C as an increasingly crowded ‘commons’: we evaluate the limit residual cost as the set of potential participants increases indefinitely9 . Definition 5 Fix the cost function C and for all n a residually efficient mechanism ξ n . For any utility profile v n ∈ Dn , the residual cost rn (C, v n , ξ n ) is rn (C, v, ξ n ) =

sup x∈N eq(v,ξn )

r(z, C, v) where z = (x, ξ n (x))

We call the mechanism {ξ n } asymptotically efficient (at C) if for any sequence of utilities (vn , n = 1, 2, · · · ) in D, we have lim rn (C, (v1 , v2 , · · · , vn ), ξ n ) = 0

n→∞

(16)

We illustrate this definition for the incremental+ mechanism. Fix a > 0, and consider the sequence of identical linear preferences with slope C 0 (a) (vi = C 0 (a)xi − yi ). With n users, the unanimous demand profile xi = na is the worst Nash equilibrium, hence as in (5) we get +

)= rn (C, (v1 , v2 , · · · , vn ), ξ inc n

(n − 1)C(a) − nC( n−1 n a) 0 aC (a) − C(a)

and this ratio goes to 1 as n goes to infinity. The same conclusion holds for the incremental− mechanism. Lemma 7 The residual∗ mechanism is asymptotically efficient at C if for all a > 0: lim

sup |∆∗n (x, C)| = lim n · sup |Bn (x, C)| = 0

n→∞ x:xN ≤a

n→∞

(17)

x:xN ≤a

Proof Fix C satisfying (17), and the sequence vn . Then pick for each n an arbitrary Nash equilibrium xn of the demand game for v n = (v1 , v2 , · · · , vn ). We must show limn rn∗ (z n , C, v n ) = 0. In the proof below, we assume for simplicity that C has no flat part, namely C 0 is strictly increasing on R+ . The details of the proof for the general case are omitted for brevity. If for some n, at least one coordinate of xn is null, then z n = (x, ξ ∗n (x)) is budget balanced (Lemma 4) so rn∗ (z n , C, v n ) = 0. Thus we focus on the subsequence of v n such that xn À 0, and we need only consider the case where an infinite subsequence exists. We denote it v n for simplicity, keeping in mind that from v n to v n+1 we may add more than one new user. Let an = xn{1,··· ,n} be the total demand at xn ; we check first that the sequence an is bounded above. Consider user 1: xn1 > 0 implies v10 (xn1 ) ≥ C 0 (an ), therefore C 0 (an ) ≤ v10 (0), implying an ≤ b a where b a is defined by C 0 (b a) = v10 (0). 9 Reference [4] develops a related analysis for two simple budget balanced demand games, but for a given distribution of preferences.

17

Next we show that the sequence an is non decreasing. Suppose an > an+1 . Then for each i = 1, · · · , n, we have by Lemma 1 v10 (xn+1 ) ≤ C 0 (an+1 ) < C 0 (an ) ≤ vi0 (xni ) ⇒ xn+1 ≥ xni i i a contradiction. Finally the inequalities vi (xni ) ≥ vi0 (xni ) · xni ≥ C 0 (an ) · xni imply X vi (xni ) − C(an ) ≥ C 0 (an ) · an − C(an ) = Γ(an , C) s(z n , C, v n ) = N

where the sequence Γ(an , C) is strictly positive and non decreasing. We now have rn∗ (z n , C, v n ) ≤

supx:xN ≤ea |∆∗n (x, C)| supx:xN ≤ea |∆∗n (x, C)| |∆∗n (xn , C)| ≤ ≤ n n Γ(a , C) Γ(a , C) Γ(a1 , C)

and assumption (17) implies limn rn∗ (z n , C, v n ) = 0 at once.¥ For totally monotone functions, (15) in Theorem 2 implies the following: lim rn∗ (C) = lim {sup

n→∞

n→∞ a>0

supx:xN =a |∆∗n (x, C)| }=0 Γ(a, C)

which is stronger than asymptotic efficiency because the convergence is uniform in a, instead of pointwise as in (16). I do not know if for an arbitrary analytic cost function C, rn∗ (C) converges to zero as n grows large. However the weaker property of Definition 5 holds true. Theorem 4 i) If C is analytic on R+ , the residual∗ mechanism is asymptotically efficient. For any sequence (vn , n = 1, 2, · · · ) in D, the convergence of the residual cost is exponential: rn∗ (C, (v1 , v2 , · · · , vn )) = O( 21n ). PK ii) If C is a generalized polynomial, C(a) = 0 λk apk , with λk ∈ R, pk > 1, the residual∗ mechanism is weakly asymptotically efficient, and the convergence to zero is geometric: rn∗ (C, (v1 , v2 , · · · , vn )) = O( np∗1−1 ), where p∗ = mink pk .

8

Non smooth cost functions

When the cost function is not smooth, our initial characterization of residually efficient mechanisms (Lemma 1) breaks down. If the demand mechanism takes the form (1), the corresponding demand games are still potential games (for the same potential function), but at some utility profiles not all Nash equlibrium outcomes are residually efficient. Consider the two-piece linear cost function C(a) = max{a, 2a − 1}, with 2 users and linear utilities v1 (a) = a, v2 (a) = 32 a. Residual efficiency requires to produce 1 unit, entirely consumed by user 1, yet x1 = x2 = 12 is a Nash equlibrium in any demand game (1). We can nevertheless define the residual∗ mechanism with the same equations (8), or the equivalent properties in Lemma 4, but the appealing properties of asymptotic budget balance, VP and RKG are entirely lost, as we now show. 18

The function C(a) = max{a − 1, 0} and any other two-piece linear cost function (as above) have the same residual cost rn∗ (C)10 , thus the choice of C is without loss of generality. en (a, C) = Bn (( a , a , · · · , a ), C) for the budget imbalance when all deWrite B n n n ¡ ¢ en (a, C) = Pn (−1)k n max{ n−k a − 1, 0} is mands are equal11 .Check that B 0 n k n n linear for n−k ≤ a ≤ n−k−1 and compute en ( B

n , C) = n−k =

µ ¶ µ ¶ k k X 1 X n n (−1)j (−1)j (n − j) − j j n − k j=0 j=0 µ ¶ µ ¶ µ ¶ 1 n−2 n−1 (−1)k−1 n − 1 } − {(−1)k }= {n(−1)k k k n−k n−1 k−1

en (a, C) for n = 10, 20. Figure 2 shows the function a → B As a varies between 1 and n, the budget imbalance ∆∗n ( na 1, C) varies wildly between a large surplus and a large deficit. For a = 2, k is close to n2 and ¡ ¢ |∆∗n ( n2 1, C)| ∼ n−1 k−1 . Comparing this to the surplus Γ(2, C) = 1, we see that the relative cost of the residual claimant £ ¤ is 100% for n = 2, 200% for n = 3, and grows exponentially with n. Let n2 be the largest integer bounded above by n2 , we have µ ¶ en ( £ n ¤ , C)| = £ n ¤− 1 ∼ 2n rn∗ (C) ≥ (n − 1)|B n n 2 −1 2 ¡ n−1 ¢ where the asymptotic behavior of [ n ]−1 follows from Stirling formula. 2 We let the reader check that VP and RKG fail soundly, e.g., for n = 4 and x = (0, 1, 1, 1). We concude with an impossibility result. Within the class of demand games (1), where the functions hi are arbitrary, we cannot avoid the situation where the transfer to the RC captures the whole surplus. Lemma 8 For any n and any mechanism ξ of the form (1), we have rn (C, ξ) ≥ 1. Proof Pick x such that xN Â {i} ≤ 1 for all i, and consider the unanimous utility profile viε = εxi −yi , with ε < 1. Here es(C, v ε ) = ε, so if ξ does not balance the budget at z = (x, ξ(x)), then rn (C, ξ) ≥ |∆(z,C)| ⇒ rn (C, ξ) = ∞ and we are done. ε P Thus we may assume N hi (x−i , C) = 0. Next consider the demand profile 1 x0i = n−1 for all i, and the unanimous linear utility profile of slope 1. The P efficient surplus is 1; we just saw N hi (x0−i , C) = 0, therefore the imbalance of ξ at x0 is ∆(x0 , C, ξ) = (n − 1)C(x0N ) = 1.¥ 1 0 This follows from the linearity of B (·, C) and Γ(·, C) in C, and the fact that B (·, C ) = n n 0 Γ(·, C0 ) = 0 for a linear function C0 (a) = λa. 1 1 We show in the Appendix (lemma 9) that for many smooth cost functions, B hn (a, C) achieves the largest imbalance in absolute value. But we make no such claim for the function C.

19

Notice that the incremental+ mechanism achieves rn (C, ξ) = 1 (recall rn (C, ξ) ≤ 1 holds because it satisfies VP and never yields a deficit).

9

Concluding comments

1. Our residual∗ mechanism can also be used as a VCG mechanism, in those problems where individual types is one-dimensional and total surplus depends only upon the aggregate type. Consider the provision of a public good problem where the convex cost of producing z units is γ(z), and user i’s utility is ui = xi · z − ti when z units are produced and he is charged $ti . The non negative marginal utility parameter xi is only known to user i. In a VCG mechanism (refce) each user reports xi ; then the optimal P level of public good (at the reported utilities) z ∗ is produced (z ∗ maximizes ( xi ) · z − γ(z) over R+ ), and user i is charged ti = γ(z ∗ ) − xN Â i · z ∗ + hi (x−i ) where the function hi (x−i ) is arbitrary. To see the connection with our cost sharing problem, write the efficient surplus at x as C(xN ) = max{xN · z − γ(z)} z≥0

and check that P the net utility Ui of user i is Ui = C(x∗N ) − hi (x−i ). First best efficiency is Ui = C(xN ), or equivalently tN = γ(z ): users pay exactly the cost of the efficient level of output. By Lemma 3 we can achieve this for all x if and only if C is a polynomial of the right degree. If γ(z) = z q , q > 1, then up to q m a constant C(z) = z q−1 which is a polynomial only if q = m−1 for some integer m, which requires q ≤ 2. For any q > 1, the choice of hi (x−i ) = h∗ (x−i ) as in (8) gives an asymptotically efficient VCG mechanism (Theorem 3), generating alternatively a budget surplus or a deficit depending on the parity of n (Lemma 6). 2. Many of our results are valid under very general convex and monotonic preferences. Definition 1 of residually efficient outcomes and Definition 2 generalize in a straightforward manner, and so does Lemma 1. We can define the residual cost of an outcome or a mechanism (Definition 3) using money as the numeraire, then the key (linearity) Lemma 2 still holds and so do all subsequent results12 . One big difference however is that the demand game generated by cost shares of the form (1) is no longer a potential game, so the interpretation of the Nash equilibrium. behavior under incomplete information is less compelling.

References [1] E. Anshelevich, A. Dasgupta, E. Tardos and T. Wexler, Near-optimal network design with selfish agents. proceedings of the 35th Annual ACM Symposium on the Theory of Computing, 2004. 1 2 Details

available upon request from the author.

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[2] W. Baumol, J. Panzar and r. Willig, Contestable markets and the theory of industry structure, New York: Harcourt Brace Jovanovich, 1982. [3] Y. Chen and J. Zhang, Design of price mechanisms for network resource allocation via price of anarchy, mimeo, New york University, 2005. [4] H. Cres and H. Moulin, Commons with increasing marginal cost: random priority versus average cost, International Economic Review, 44, 3, 10971115, 2003. [5] M. Cripps and J. Swinkels, Depth and efficiency of large double auctions, Econometrica, 74, 1,47-92,2006. [6] A. Demers, S. Keshaw and S. Shenker, Analysis and simulation of a fair queuing algorithm, Internetworking: Research and experience, 1, 3-26, 1990. [7] J. Green, and J.J. Laffont , Incentives in public decision making, Amsterdam: North-Holland, 1979. [8] R. Johari and J. Tsitsiklis Efficiency loss in a network resource allocation game, Maths of Operations Research, 29, 3, 407-435, 2004. [9] R. Johari and J. Tsitsiklis A scalable network resource allocation mechanism, mimeo, Stanford and MIT, 2005. [10] R. Johari, S. Mannor and J. Tsitsiklis Efficiency loss in a network resource allocation game: the case of elastic supply, 2004. [11] E. Koutsoupias and C. Papadimitriou Worst case equilibria Proceedings of the 16th Symposium on Theoretical aspects of computer science, 404-413, 1999. [12] N. Kukushkin, Best response dynamics in finite games with additive aggregation, Games and Economic Behavior, 48, 94-110,2004. [13] R. P. McAfee, A dominant strategy double auction, Journal of Economic Theory, 56, 434-50, 1992. [14] D. Monderer and L. Shapley, Potential games, Games and Economic Behavior, 14, 124-143,1996. [15] D. Monderer and L. Shapley, Fictitious play property for games with identical interests, Journal of Economic Theory, 68, 258-265, 1996. [16] H. Moulin, The price of anarchy of serial, average, and incremental cost sharing, Rice University, 2006. [17] H.Moulin and S. Shenker, Serial cost sharing, Econometrica,60, 1009-1037, 1992.

21

[18] H.Moulin and S. Shenker, Average cost pricing versus serial cost sharing: an axiomatic comparison, Journal of Economic Theory, 64, 178-201, 1994. [19] H.Moulin and S. Shenker, Strategy-proof sharing of submodular costs: budget balance versus efficiency, Economic Theory, 18, 3, 511-533, 2001. [20] R. Musgrave and A. Peacock, Classics in the theory of public finance, London: McMillan, 1958. [21] T. Roughgarden The price of anarchy is independent of the network topology, STOC, 2002. [22] T. Roughgarden and E. Tardos How bad is selfish routing ? Journal of the ACM, 49, 2, 236-259, 2002. [23] S. Sanghavi and B. Hajek Optimal allocation of a divisible good to strategic buyers, Proceedings of the 43d IEEE conference on Decision and Control, 2004. [24] P.D. Straffin and J.P. Heaney, Game Theory and the Tennessee Valley Authority, International Journal of Game Theory,10, 1,35-43,1981. [25] N. Tideman and G. Tullock, A new and superior principle for collective choice, Journal of Political Economy, 84,1145-59, 1976. [26] M. Walker, On the non-existence of dominant strategy mechanisms for making optimal public decisions, Econometrica, 48, 1521-40, 1980. [27] S. Yang and B. Hajek Revenue and stability of a mechanism for efficient allocation of a divisible good, mimeo, University of Illinois at Urbana Champaign, 2005.

10 10.1

Appendix: remaining proofs Proof of statement ii) in Lemma 3

Fixn and C, and assume there is a residually efficient mechanism achieving budget balance for all demands. By Lemma 1 there exists some functions hi such that equation (7) holds(Lemma 3). By step 1 in the proof of Lemma 4, Bn (x, hi ) = 0 for all x, and by linearity of Bn (x, C) with respect to C, Bn (x, C) = 0 for all x. We write P(m) for the set of polynomials of degree at most m, and consider the two properties A(n) and B(n) of a function C bounded on every interval of R+ : A(n) : {Bn (x, C) = 0 for all x ∈ Rn+ } ⇒ C ∈ P(n − 1)

(18)

B(n) : {Ca − C ∈ P(n − 1) for all a > 0} ⇒ C ∈ P(n)

(19)

Assume B(n) is true for all n. Then A(n) follows easily by induction on n and (11). Property A(1) is clear; next Bn (x, C) = 0 for all x implies Bn−1 (x−i , Cxi − 22

C) = 0 for all x, and Cxi − C is bounded on every interval if C is. By the inductive assumption A(n − 1), we get Ca − C ∈ P(n − 2), then C ∈ P(n − 1) by B(n − 1). We prove B(n) by induction on n = 0, 1, · · · . Start with B(0): if Ca (b)−C(b) is constant in b, we have C(a + b) = C(a) + f (b) for some function f , and the conclusion C ∈ P(1) follows from Cauchy equation, given our assumption that C is bounded on every interval(see Aczel). Assume now B(m) up to m = n − 1 and consider C such that Ca − C ∈ P(n − 1). There exist functions f k , k = 0, · · · , n − 1, defined on R+ such that for all a, x ≥ 0, we have C(a + x) − C(x) =

n−1 X

f k (a)xk

(20)

k=0

Notice that f k (0) = 0, and f k is bounded on every interval. To check the latter claim, fix n distinct values xk , and use the corresponding equations (20), a non singular linear system in (f 0 (a), · · · , f n−1 (a)), to compute f k as a linear combination of functions Cxk . Applying (20) successively to C(a+b+x)−C(a+x) then to C(a+b+x)−C(x), we get n−1 n−1 n−1 X X X f k (a + b)xk = f k (a)xk + f k (b)(a + x)k k=0

k=0

k=0

k

Identify the coeeficients of x on both sides of this equation: we see that fbk − f k ∈ P(n − 1 − k) for k = 0, · · · , n − 1. The inductive assumption gives f k ∈ P(n − k) for k = 1, · · · , n − 1, then for k = 0 the identification gives f 0 (a + b) = f 0 (a) + f 0 (b) + R(a, b) for all a, b ≥ 0

(21)

where R is a symmetric polynomial in a, b, a linear combination of terms ap bq where p + q ≤ n and p, q ≥ 1 (the latter because f k (0) = 0). If f and g are two solutions of (21), both bounded on every interval, f − g satisfies the Cauchy equation ϕ(a+b) = ϕ(a)+ϕ(b) and is therefore in P(1). Now for any polynomial R as above, there is a solution f of (21) in P(n) (we omit the straightforward details). This establishes that f 0 ∈ P(n), and by (20), so does C. The proof of B(n) is complete.

10.2

Proof of Lemma 5

Fix C and a demand P profile x as in the premises of Lemma 5. Set m = nI − 1. We write Φk = T ⊂A,|T |=k C(xAÂ T ) for k = 0, · · · , nA , so that Φ0 = C(xA ) and ΦnA = C(0) = 0. We develop the cost share of inactive agent 1 by means

23

of (8) −y1

µ ¶ ¸ ¸ ∙µ ¶ ∙µ ¶ 1 m 1 m m = −Φ0 + (n − 1){Φ0 − Φ1 + Φ2 + · · · Φ0 + Φ1 + Φ0 + 1 2 1 3 2 µ ¶ µ ¶ ¸ ∙µ ¶ (−1)m m m m + Φ1 + Φ2 + · · · Φ0 + m−1 m−1 m+1 m µ ¶ µ ¶ ¸ ∙µ ¶ m m m (−1)m+1 Φ2 + Φ3 + · · · + · · · Φ1 + + m−1 m−1 m+2 m µ ¶ ∙µ ¶ ¸ m+nA −1 m m (−1) ΦnA } ΦnA −1 + + m−1 m m + nA

where the sum inside the k−th bracket with coefficient (−1)k k+1 ends with Φk if ¡ m ¢ if n ≤ k. Regrouping with respect to the Φk -s: Φ nA ≥ k, with k−n nA A A

−y1 = −Φ0 + (n − 1){β 1 Φ0 − β 2 Φ1 + β 3 Φ2 − · · · + (−1)nA −1 β nA ΦnA −1 } (22)

where βt =

µ ¶ m X (−1)k m 0

k+t

k

=

Z

0

1

z t−1 (1 − z)m dz =

(t − 1)!m! for t = 1, · · · , nA (m + t)! (23)

in particular β 1 = n1I and β 2 = nI (n1I +1) . Now we fix nA and let nI grow. The total subsidy (−nI y1 )to inactive agents (we make no claim about the sign of this number) is a linear combination of the Φt , in which for t ≥ 3 the coefficient of Φt−1 is (up to a sign) (n − 1)β t = (nA +nI −1)(t−1)! nI (nI +1)(nI +2)···(nI +t−1) and converges to zero. Therefore the limit subsidy is computed from the terms in Φ0 and Φ1 lim (−nI y1 ) = lim{nI (

nI →∞

n−1 nI (n − 1) − 1)Φ0 − Φ1 = (nA − 1)Φ0 − Φ1 nI nI (nI + 1)

as was to be proved.

10.3

Proof of Theorem 1

Step 1 The two inequalities in statements i) and ii) reduce to the left and right part of C(xN Â i ) ≤ h∗ (x−1 , C) ≤

n−1 n ) C( x n n − 1 NÂi

Changing n into n + 1 and using the notations in (13), this writes n

X (−1)k 1 1 n+1 C(xN ) ≤ Ank (x, C) ≤ C( xN ) n k + 1 n + 1 n 0 24

(24)

For any function f on R+ , not necessarily such that f (0) = 0, we consider two slightly more general inequalities, denoted (L25) and (R25) for left and right n

X (−1)k 1 1 1 n+1 f (xN )− f (0) ≤ Rn (x, f ) = An (x, f ) ≤ f( xN ) n n(n + 1) k + 1 n−k n+1 n k=0 (25) We show in step 2 that (L25) holds if f is n times differentiable, and f (k) ≥ 0 on R+ for k = 1, · · · , n; then in step 3 we show (R25) if f is n + 1 times differentiable, and f (k) ≥ 0 on R+ for k = 1, · · · , n + 1. This establishes (24) under the premises of statements i) and ii) respectively. As for Ranking, it clearly is the property that Rn (x, f ) is non decreasing in all variables xi . Using the identity 0 ∂1 Ank (x, f ) = An−1 k−1 (x−1 , fx1 )

(26)

we have ∂1 Rn (x, f ) = Rn−1 (x−1 , fx0 1 ) and an obvious induction argument: if the first n derivatives of f are non negative, the first n − 1 derivatives of fx0 1 are non negative as well. Step 2 proof of (L25). Write (L25) as P (n, q), where q is the number of non-zero coordinates in x1 , · · · , xn . We prove P (n, q) by a double induction on n and q. Both P (1, q) and P (n, 0) are identities, the latter because (23) for t = 1 is µ ¶ n X n 1 (−1)k = (27) k+1 n−k n+1 k=0

Assume P (n, q) holds whenever n + q ≤ r and prove it for n, q s.t. n + q = r + 1. Identity (10) implies Rn (x, f ) = Rn ((0, x−1 ), f ) + Rn−1 (x−1 , fx1 − f ) By induction (L25) holds for Rn ((0, x−1 ), f ) and Rn−1 (x−1 , fx1 − f ), the latter because (fx1 − f )(k) ≥ 0 for k = 1, · · · , n − 1.Therefore: Rn ((0, x−1 ), f ) ≥ Rn−1 (x−1 , fx1 − f ) ≥ ⇒ Rn (x, f ) ≥

1 1 {f (xN Â 1 ) − f (0)} n n+1 1 1 {f (xN ) − f (xN Â 1 ) − (f (x1 ) − f (0))} n−1 n

1 1 1 {f (xN )− f (0)}+ {f (xN )−f (xNÂ 1 )−f (x1 )+f (0)} n n+1 n(n − 1)

Convexity of f implies f (xN ) − f (xN Â 1 ) − f (x1 ) + f (0) ≥ 0, and step 1 is complete. Step 3 proof of (R25) Substep 3.1 We show that for any b > 0, the maximum of Rn (x, f ) over all x such that xN = b is achieved for b∗ = ( nb , · · · , nb ). This is clear for n = 1 and for n = 2 by 25

convexity of f , so we assume n ≥ 3. As Rn (x, f ) is symmetric in the variables xi it is enough to show for all x x1 < x2 =⇒ ∂1 Rn (x, f ) ≥ ∂2 Rn (x, f ) where we use the notation ∂i =

∂ ∂xi

(28)

. For all k = 1, · · · , n − 1, check the identity

∂1 Ank (x, f ) − ∂2 Ank (x, f ) = An−1 (x−2 , f 0 ) − An−1 (x−1 , f 0 ) k k whereas ∂1 Ann (x, f ) − ∂2 Ann (x, f ) = ∂1 An0 (x, f ) − ∂2 An0 (x, f ) = 0. Therefore ∂1 Rn (x, f ) − ∂2 Rn (x, f ) = ∂2 Rn (x, f ) − ∂1 Rn (x, f ) =

n−1 X k=1 n−2 X k=0

(−1)k n−1 0 (x−2 , f 0 ) − An−1 (A n−k (x−1 , f )) k + 1 n−k (−1)k n−1 0 (x−2 , f 0 ) − An−1 (A n−k−1 (x−1 , f )) k + 2 n−k−1

k Pn−2 n−1 0 We claim now that k=0 (−1) k+2 An−k−1 (x−i , f ) is non decreasing in x−i , which completes the proof of (28) and of substep 3.1. The claim is obvious for n = 1, and for n ≥ 2 we use (26) to reduce it to the following inequality

n−2 X k=0

(−1)k n−2 (2) (e x, fb ) ≥ 0 A k + 2 n−2−k

(29)

and b ≥ 0. We set m = n − 2, write x e simply as x, and rewrite where x e ∈ Rn−2 + (29) as the inequality Tm (x, g) =

m X (−1)k

k=0

k+2

Am m−k (x, g) ≥ 0

whenever g is m times differentiable, g and g (k) ≥ 0 on R+ for k = 1, · · · , m. We prove a stronger statement namely, under the same premises for g Tm (x, g) ≥

1 2 (g(xM ) − g(0)) m(m + 1) m+2

(30)

where M = {1, · · · , m}. The proof of (30) parallels that of (L25) in step 1. Refer to (30) as Q(m, q), where q is the number of non-zero coordinates in x1 , · · · , xm and use a double induction on m + q. Both Q(1, q) and Q(m, 0) are actually equalities, the latter because µ ¶ m X (−1)k m 1 = k+2 m−k (m + 1)(m + 2) k=0

For m ≥ 2 (10) implies as above Tm (x, g) = Tm ((0, x−1 ), g) + Tm−1 (x−1 , gx1 − g) 26

We can apply the inductive assumption because gx1 − g ≥ 0 and (gx1 − g)(k) ≥ 0 for k = 1, · · · , m − 1, hence Tm (x, g) is bounded below by 1 2 2 1 (g(xM Â 1 )− g(0))+ (g(xM )−g(xMÂ 1 )− (g(x1 )−g(0))) m(m + 1) m+2 (m − 1)m m+1 =

1 2 2 (g(xM )− g(0))+ (g(xM )−g(xMÂ 1 )−g(x1 )+g(0)) m(m + 1) m+2 (m − 1)m(m + 1)

The desired conclusion follows by convexity of g. Substep 3.2 1 By substep 3.1 it remains to prove Rn (b∗ , f ) ≤ n+1 f ( n+1 n b) for any b > 0 and b b ∗ b = ( n , · · · , n ), namely µ ¶ n X n n−k 1 n+1 (−1)k Rn (b , f ) = f( b) ≤ f( b) k+1 n−k n n+1 n k=0 µ ¶ n+1 X n+1−k k n+1 (−1) f( ⇔ b) ≥ 0 k n ∗

(31)

k=0

Define

µ ¶ n X n−k k n Bn (b, f ) = (−1) f( b) k n k=0

Our last claim, denoted S(n), is that for any n, {f (k) ≥ 0 for k = 1, · · · , n} implies Bn (b, f ) ≥ 0. Then (31) is simply Bn+1 ( n+1 n b, f ) ≥ 0 and we are done. Define g = f b − f and compute n

Bn−1 (

µ ¶ n−1 X n−1 n−1 n−k n−k−1 (−1)k (f ( b, g) = b) − f ( b)) = Bn (b, f ) k n n n k=0

Property S(1) is clear, and S(n) follows from the above equality by induction on n.

10.4 10.4.1

Proof of Theorems 2, 3 and 4 Preliminary result

If the n-th derivative of C is of constant sign, the largest budget transfer (in absolute value) is achieved for identical demands. For any a ≥ 0, we write ¡ ¢ n−k Pn a a a a a k n e 0 (−1) k C( n a). n 1 = ( n , n , · · · , n ), and Bn (a, C) = Bn ( n 1, C) = Lemma 9 Suppose the function C is continuous on R+ and n times differentiable on R+ Â{0}. If the sign of C (n) is constant on R+ Â{0}, then for all a > 0 en (a, C)| max |Bn (x, C)| = |B

x:xN =a

27

(32)

rn∗ (C) = (n − 1) sup a>0

en (a, C)| |B Γ(a, C)

(33)

Proof By Lemma 5, (32) holds if we show that for all a > 0, na 1 is a solution of one of the programs maxx:xN =a Bn (x, C) and minx:xN =a Bn (x, C), depending on the sign of C (n) . Both proofs are identical so we only discuss the case C (n) ≥ 0. Then (33) follows at once from Lemma 2. The proof parallels that of substep 3.1, where we established that Rn (x, C), a different linear combination of the Ank (x, C) than Bn (x, C), reaches its maximum over {x : xN = a} at na 1. We want to prove (28) for Bn (x, C) instead of Rn (x, f ), and by exactly the same argument, this reduces to the analog of (29) namely n−2 X k=0

(2)

(2)

(−1)k An−2 x, Cb ) ≥ 0 ⇔ Bn−2 (e x, Cb ) ≥ 0 for all x e ∈ Rn−2 and b ≥ 0 + n−2−k (e

The desired conclusion follows now from Lemma 6. Incidentally if C (n) is strictly positive everywhere, the proof can be adapted to show that na 1 is the unique maximizer of Bn (x, C) over the relevant simplex. 10.4.2

Theorem 2 statement ii)

We consider the exponential function E(a) = ea − 1. ¡ ¢ en (a, E) = Pn (−1)k n e n−k n a = We apply (33) with Γ(a, E) = ea (a−(1−e−a )) and B 0 k a a −n n e (1 − e ) : a

rn∗ (E) = (n − 1) sup a≥0

(1 − e− n )n (1 − e−a )n ∼ sup −a a − (1 − e ) a≥0 a − n1 (1 − e−na )

(34)

For n ≥ 3 the RHS function to maximize is zero at a = 0 and a = ∞, with a single critical point an solving ean = nan + e−(n−1)an It follows easily that an → ∞ and

this back into (34) gives rn∗ (E) ∼ 10.4.3

e−1 log n

ean nan

→ 1, then that

as desired.

an log n

→ 1. Plugging

Theorem 2, statement i)

For any integers n, m define σ n,m

µ ¶ n n−k m n e n X k n ( = Bn (1, Cm ) = (−1) ) k m m 0 n

Recall σ n,m = 0 if m < n (Lemma 4). We claim σ n,m ≤

2 for all m log n 28

(35)

Compute for all a ≥ 0 ∞ ∞ X X m am 1 e en (a, E) = (e na − 1)n σ n,m = Bn (a, Cm ) = B n m! m! m=1 m=1

(36)

a

In the development of e n − 1 as a series, the m-th term is no larger than the a m-th term in the development of e n . Because all such terms are non negative, a it follows that the m-th term in the development of (e n − 1)n is no larger than a the m-th term in the development of (e n )n = ea . Thus m n σ n,m ≤ 1 for all m. n . It remains to deal with those m such that This gives (35) for m ≥ n log 2 1≤

m log n ≤ n 2

(37)

Equation (36) implies mm m m m σn,m ≤ (e n − 1)n = em (1 − e− n )n n m! combine this with the upper bound for m! given by Stirling’s formula m! ≤

√ √ m 1 m 2 2πm( )m e 12m ≤ 3 2 m( )m e e

1

because m ≥ 2 ⇒ e 12m ≤ 1.1. Combining the two inequalities √ m n 1 n σ n,m ≤ 3 √ (1 − e− n )n ≤ 3 2 n(1 − √ ) 2 2 m n where the last inequality uses the two bounds in (37). It is now a simple matter to check (e.g., numerically) √ 1 n 0.6 3 2 n(1 − √ ) ≤ 2 n log n completing the proof of (35). P∞ We fix now an arbitrary totally monotone function C(a) = k=0 λk ak , with h

n (a,C) that does not depend λk ≥ 0. Inequality (35) gives an upper bound for |BΓ(a,C) on a or C: P∞ P∞ k k en (a, Ck ) en (a, C) λk B B k=n λk n σ n,k a k=n = P∞ = P∞ k Γ(a, C) k=0 λk Γ(a, Ck ) k=0 (k − 1)λk a P∞ kλ ak 2 2 P∞ k=n k ≤ ≤ n log n k=n (k − 1)λk ak (n − 1) log n

where the last inequality follows from complete.

k k−1

29

≤

n n−1 ,

and the proof of (15) is

10.4.4

Theorem 3

Statement i) Fix p, p > 1 and not an integer, and n, n ≥ dpe + 1 = k∗ . Apply (11) at x = and use the linearity of Bn−1 in C:

a n1

en (a, Cp ) = B en−1 ( n − 1 a, (Cp ) a ) − B en−1 ( n − 1 a, Cp ) B n n n

en−1 (a0 , Cp ) Recall from the discusion immediately after Lemma 6 that for n ≥ k ∗ , B 0 e e and Bn−1 (a , (Cp )a00 ) have the same sign, and the opposite sign of Bn (a, Cp ) (for en (a, Cp )| ≤ |B en−1 ( n−1 a, Cp )|. Repeating this arguany a, a0 , a00 ). Therefore |B n ∗ ment for n − 1, · · · , k + 1 gives en (a, Cp )| ≤ |B ek∗ ( |B

k∗ a e ∗ a, Cp )| = ( )p |B k∗ (k , Cp )| n n

(38)

en (a, Cp )| is a constant multiple of ( a )p . The desired estimation of Thus |B n follows at once from Γ(a, Cp ) = (p − 1)ap and (33). Statement ii) Fix n and two n−times differentiable functions Ci , i = 1, 2. We claim that if (n) (n) the sign of C1 and that of C2 are constant, though not necessarily identical rn∗ (Cp )

rn∗ (λ1 C1 + λ2 C2 ) ≤ max{rn∗ (C1 ), rn∗ (C2 )} for all λ1 , λ2 ≥ 0

(39)

Lemma 9, and the linearity of Bn and Γ in C imply |λ1 Bn (x, C1 ) + λ2 Bn (x, C2 )| λ1 Γ(x, C1 ) + λ2 Γ(x, C2 )

≤ ≤ PK

λ1 |Bn (x, C1 )| + λ2 |Bn (x, C2 )| λ1 Γ(x, C1 ) + λ2 Γ(x, C2 ) en (xN , C1 ) + λ2 B en (xN , C2 ) λ1 B λ1 Γ(x, C1 ) + λ2 Γ(x, C2 )

pk with pk > 1 and λk > 0, we hence the claim. If now C(a) = 0 λk a set C = C1 + C2 , where C1 (resp. C2 ) collects all terms of which the n−th derivative is positive (resp. negative). We can apply (39) repeatedly to the C1 -sum because any partial sum has a positive n−th derivative. In view of statement i), this gives rn∗ (C1 ) = O( p∗11−1 ), where p∗1 is the smallest exponent n in the C1 -sum. Similarly rn∗ (C2 ) = O( p∗21−1 ), and we use (39) one last time n between C1 and C2 .

10.5

Theorem 4

Proof Statement i) P∞ Fix an analytic function C(a) = 0 λk ak . From Bn (x, Ck ) = 0 if k < n, we P∞ get Bn (x, C) = n λk Bn (x, Ck ). By (32) and the proof of Statement ii) in Theorem 2 k en (xN , Ck ) = xk B e Bn (x, Ck ) ≤ B N n (1, Ck ) ≤ xN 30

(the last inequality is just xN ≤ a ⇒ |Bn (x, C)| ≤

k n σ n,k ∞ X n

≤ 1 and it holds for all n, k). Therefore

|λk |Bn (x, Ck ) ≤ P∞

∞ X n

|λk |ak ≤

∞ 1 X |λk |(3a)k 3n n

Because C is analytic on R+ , n |λk |(3a)k converges to zero when n is large. Thus n · supx:xN ≤a |Bn (x, C)| = O( 21n ). From the the proof of Lemma 7, we conclude as desired that the sequence rn∗ (C, (v1 , v2 , · · · , vn )) is also O( 21n ). Statement ii) The proof is essentially the same as for Theorem 3. Inequality (38) gives for all p>1 1 n · sup |Bn (x, Cp )| = O( p−1 ) n x:xN ≤a We conclude by linearity of Bn in C. The sign of the coefficients λk doers not matter any more because we are not dividing Bn (x, C) by the efficient surplus.

31

example where B3 > 0 and < 0 with C smooth and C (3) not of constant sign. 3 C(z) = ez − z3 − z 3 2x x x 3 3 3 B3 = ex − x3 − 3e 3 + ( 2x 3 ) + 3e − ( 3 ) − 1

y

0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 0.2

0.4

0.6

0.8

1.0

-0.02

FIGURE 1

1

1.2

1.4

1.6

1.8

2.0

x

C(z) = (z − 1)+ , comput of Bn (C, a · e); recall ∆ = (n − 1)B n = 10

y

14 12 10 8 6 4 2 0 0.1

-2

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

x

-4 -6 -8 -10 -12 -14

n = 20

y

5000 4000 3000 2000 1000 0 0.1

0.2

0.3

0.4

0.5

-1000 -2000 -3000 -4000 -5000

1

0.6

0.7

0.8

0.9

1.0

x