Nash Equilibria for a Quadratic Voting Game

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Aug 31, 2014 - ... grant DMS-1106669. †University of Chicago and Microsoft Research New England. ... there will be the possibility of recounts, litigation, etc. 2 ...
arXiv:1409.0264v1 [cs.GT] 31 Aug 2014

Nash Equilibria for a Quadratic Voting Game Steven P. Lalley∗and E. Glen Weyl† September 2, 2014

Abstract Voters making a binary decision purchase votes from a centralized clearing house, paying the square of the number of votes purchased. The net payoff to an agent with utility u who purchases v votes is Ψ(Sn+1 )u − v 2 , where Ψ is a monotone function taking values between -1 and +1 and Sn+1 is the sum of all votes purchased by the n + 1 voters participating in the election. The utilities of the voters are assumed to arise by random sampling from a probability distribution FU with compact support; each voter knows her own utility, but not those of the other voters, although she does know the sampling distribution FU . Nash equilibria for this game are described. These results imply that the expected inefficiency of any Nash equilibrium decays like 1/n.

1 Introduction 1.1 Quadratic voting The principle of one man, one vote (or perhaps we should say one person, one vote, or one shareholder, one vote) has been a cornerstone of democratic institutions since antiquity. But although it is the most equitable way of distributing influence in group decision-making, it has obvious drawbacks, perhaps the most notable being the tyranny of the majority and the noise of the rabble. It is often the case that a minority of a population have a much greater stake in the outcome of an election than the majority (consider, for instance, the issue of gay marriage, where the economic and social costs of prohibition to gays can be orders of magnitude higher than the social costs accruing to those opposed to legalization); in such cases one-person-one-vote elections will often lead to outcomes at variance with the aggregate social utility. Related to this problem of tyranny of the majority is the distinct problem of the noise of the rabble, which occurs when such a large fraction of the voting population has so little at stake – or such a poor understanding of the consequences of the choice being made – that their votes are essentially random. In such cases, the magnitude of the “noise” can be so large that it overwhelms the “signal” in the votes of the much smaller minority of voters who are well-informed, or have a real stake in the outcome. ∗ †

University of Chicago; research supported by NSF grant DMS-1106669. University of Chicago and Microsoft Research New England.

1

Various alternatives to one-person-one-vote democracy have been discussed in the economics literature (see the forthcoming paper [6] for a detailed discussion and bibliography). Recently, a simple mechanism designed to eliminate some of the problems that arise in one-person-one-vote democracy, Quadratic Voting, has been proposed by Weyl [11] (in part motivated by earlier work of Hylland and Zeckhauser [5]) and advocated by Posner and Weyl [8] for use in corporate governance (see also [7] and [3]). In quadratic voting, voters making a binary decision purchase votes from a centralized clearing house, paying the square of the number of votes purchased. The ultimate payoff to the voter is her utility minus the cost of the votes she purchases, in the event that her side wins the election, or minus the cost of her votes in the event that her side loses. It is not our purpose in this paper to discuss the relative advantages (and disadvantages) of quadratic voting or its competitors – this will be done in a subsequent article. Rather, the intent of this article is to elucidate the mathematical properties of the mechanism, in particular, the structure of the (Bayes-)Nash equilibria in a quadratic voting game modeled on the Weyl proposal.

1.2 Assumptions In this game there are n + 1 agents, whose utilities U1 , U2 , . . . , Un+1 are gotten by independent random sampling from a probability distribution FU on R with mean µU ≥ 0, variance σU2 > 0, and with C ∞ density fU (u) supported by a closed interval [−A, B] containing the point 0. (In particular, we assume that fU is C ∞ in the interior (−A, B), and that fU and all of its derivatives extend continuously to [−A, B].) To avoid pathologies, we shall assume that fU is strictly positive, and therefore bounded below, on [−A, B]. The distribution FU is known to the agents, as is the sample size n + 1, but the actual values Ui of the other voters are not. Agents buy votes subject to the quadratic cost rule: to buy ±v votes, an agent must pay v 2 . The payoff to an agent i with utility Ui who purchases Vi votes is (1) Ψ(Sn+1 )Ui − Vi2 Pn+1 where Sn+1 = j=1 Vj is the sum of all votes purchased by the n + 1 agents. Here Ψ : R → [−1, 1] is a nondecreasing, C ∞ function1 whose derivative ψ(x) := Ψ′ (x) satisfies the following conditions: (a) ψ has support [−δ, δ]; and (b) ψ/2 is a strictly positive probability density on [−δ, δ]. Thus, in particular, the payoff function Ψ is identically 1 on [δ, ∞) and identically −1 on (−∞, δ] and is strictly increasing on [−δ, δ]. In Theorem 7 (sec. 5) we will require in addition that Ψ is odd, but this is not necessary for any of the other main results. Agents are assumed to be rational, and so an agent will always buy votes of the same sign (±) as her utility, and 1 A better model for many elections might be the function Ψ = 1(0,∞) − 1(−∞,0) , which has a jump discontinuity at 0 and so is not C ∞ . The hypothesis in this paper that Ψ is infinitely differentiable is primarily for the sake of mathematical tractability. However, it can be argued that smoothed versions of the payoff function make sense on economic grounds, as well, at least in some contexts, since (for instance) in nearly tied elections there will be the possibility of recounts, litigation, etc.

2

no agent will pay more for votes than 2|u|, where u is the voter’s utility, since 2|u| is the maximum change in utility payoff that could possibly result.

1.3 Terminology √ √ A pure strategy is a Borel measurable function v : [−A, B] → [− 2A, 2B]; when a pure strategy v is adopted, each agent buys v(u) votes, where u is the agent’s utility. A mixed 2 strategy is a Borel measurable πV : [−A, B] → Π, where Π is the collection of Borel √ function √ probability measures on [− 2A, 2B]; when a mixed strategy πV is adopted, each agent i will buy a random number Vi of votes, where V1 , V2 , . . . are conditionally independent given the utilities U1 , U2 , . . . and Vi has conditional distribution πV (Ui ). Clearly, the set of mixed strategies contains the pure strategies. A best response for an agent with utility u to a strategy (either pure or mixed) is a value v such that EΨ(v + Sn )u − v 2 = sup EΨ(˜ v + Sn )u − v˜2 , (2) v˜

where Sn is the sum of the votes of the other n agents when these agents all play the specified strategy and E denotes expectation. (Thus, under E, the random variables Vi of the n other voters are distributed in accordance with the strategy and the sampling rule for utility values Ui described above.) Since Ψ is continuous and bounded, the equation (2) and the dominated convergence theorem imply that for each u the set of best responses is closed, and hence has well-defined maximal and minimal elements v+ (u), v− (u). A mixed strategy πV is a Nash equilibrium if for every u ∈ [−A, B] the measure πV (u) is supported by the set of best responses to πV for an agent with utility u.

1.4 Main Results The existence of Nash equilibria is guaranteed by general existence theorems in game theory, for instance those of Athey [1], or alternatively those of Reny [9], [10]. Details can be found in the forthcoming paper [6]. Nash equilibria need not be unique, as we will show, nor are they necessarily pure strategies. However, every Nash equilibrium is “essentially” a pure strategy, as the following theorem implies. Theorem 1. If a mixed strategy πV is a Nash equilibrium, then the set of utility values u ∈ [−A, B] for which there is more than one best response (and hence the set of values u such that πV (u) is not supported by just a single point v(u)) is at most countable. Since by hypothesis the utility values Ui are sampled from a distribution FU that is absolutely continuous with respect to Lebesgue measure, it follows that for every Nash equilibrium there is an equivalent pure-strategy Nash equilibrium v(u). √ √ The space of Borel probability measures on [− 2A, 2B] is given the topology of weak convergence; Borel measurability of a function with range Π is relative to the Borel field induced by this topology. Theorem 1 below implies that in the quadratic voting game only pure strategies are relevant, so measurability issues will play no role in this paper. 2

3

Theorem 2. Every pure-strategy Nash equilibrium v(u) is a strictly increasing function of u ∈ [−A, B], and therefore is continuous except at possibly countably many points. If πV is a mixedstrategy Nash equilibrium and v(u) the equivalent pure-strategy Nash equilibrium, then the utility values u ∈ [−A, B] for which there is more than one best response are precisely the points at which v(u) is discontinuous. Theorems 1 and 2 both follow √ √by relatively routine monotonicity arguments, using the fact that the action space [− 2A, 2B] and the type space [−A, B] are totally ordered. The details are laid out in section 2. The main results of the paper, to which we now turn, concern the nature of the pure-strategy Nash equilibria. First, we will prove that in all cases Nash equilibria are nearly linear functions, except in the extreme tails of the utility distribution. Theorem 3. For any ǫ > 0 there exist constants nǫ < ∞ and C < ∞ such that if n ≥ nǫ then for any Nash equilibrium v(u) and for all u ∈ [−A + Cn−3/2 , B − Cn−3/2 ], (1 − ǫ)Eψ(Sn )|u| ≤ |2v(u)| ≤ (1 + ǫ)Eψ(Sn )|u|.

(3)

Furthermore, for all sufficiently large n any Nash equilibrium v(u) with no discontinuities must satisfy (3) for all u ∈ [−A, B]. Thus, except for agents with extreme utilities, the number of votes v(u) that will be purchased will be roughly proportional to the agent’s utility u, with proportionality constant Eψ(Sn ). The size of this proportionality constant will vary from order 1/n to 1/n1/4 , depending on the properties of the utility distribution and the payoff function Ψ. There are two main cases, depending on whether the mean µU of the utility distribution is zero or non-zero; the case where the mean is non-zero splits into two sub-cases, depending on the relative “steepness” of the payoff function. Recall that Ψ is strictly increasing on the interval [−δ, δ] and flat on both [δ, ∞) and (−∞, −δ]. Theorem 4. Assume that µU > 0. If there is no 0 < v ≤ 2δ such that (1 − Ψ(δ − v))A ≥ v 2 , then for each ǫ > 0 and all sufficiently large values of the sample size n, (i) no Nash equilibrium has a discontinuity; and (ii) for every Nash equilibrium, P {|Sn − δ| ≥ ǫ} < ǫ. In conjunction with Theorem 3, this implies that for large n the proportionality constant in (3) will be approximately Eψ(Sn ) ≈ n−1 δ/(2µU ). This yields the following rough description of a Nash equilibrium when n is large: every agent will buy about δu/(2nµU ) votes, where u is her utility; since the mean µU is positive, the law of large numbers will guarantee that the vote total will be near δ. One might wonder why those agents with negative utility values would purchase votes at all, since they will, with overwhelmingly high probability, lose the election. The answer is that Ψ is strictly increasing on [−δ, δ], and so the expected payoff for an agent with negative utility is slightly increased by a small move of the vote total to the left. The case where µU > 0 and there exists w ∈ [−δ, δ] such that (1 − Ψ(w))A > (δ − w)2 is more interesting, but has several wrinkles that must be ironed out before a characterization 4

of Nash equilibria can be given. If such a w exists, then there exists α > δ such that (1 − Ψ(w))A = (α − w)2 ′

′ 2

(1 − Ψ(w ))A ≤ (α − w )

and

(4) ′

for all w 6= w.

For a given payoff function Ψ and A > 0 there might exist several pairs (α, w) for which the conditions (4) hold. However, for generic Ψ and A there will be only one such pair, and at this pair the second derivative of (α − w′ )2 + Ψ(w′ )A with respect to w′ will be positive: 2 + ψ ′ (w)A > 0.

(5)

This is the only sub-case of (4) that we will consider. Theorem 5. Assume that µU > 0 and that there is a unique pair (α, w) satisfying (4) and (5). Then there exists γ > 0 such that for any ǫ > 0, all sufficiently large values of the sample size n, and any Nash equilibrium v(u), (i) (ii) (iii) (iv)

v(u) has a single discontinuity at u∗ , where |u∗ + A − γn−2 | < ǫn−2 ; |v(u) − α + w| < ǫ for u ∈ [−A, u∗ ); α = γψ(w)fU (−A); and P {|Sn − α| > ǫ} < ǫ.

Under the hypotheses of Theorem 5 Nash equilibria are not unique, for a trivial reason: any Nash equilibrium v has a discontinuity u∗ , and both the left limit v(u∗ −) and the right limit v(u∗ +) are best responses at u∗ , so a Nash equilibrium can be obtained by setting v(u∗ ) to either (or to any convex combination, for a mixed-strategy Nash equilibrium). The case considered in Theorem 5 is especially interesting, in that the Nash equilibria are driven by the possible presence in the sample of an extremist (an agent with utility u < −A + γn−2 ) who will singlehandedly buy the election. Such an agent will occur only with probability ≈ n−1 fU (−A)γ, and so will affect the expected payoff to a non-extremist by an amount of order only 1/n, but this is enough for a non-extremist to push her vote purchase to the point where the aggregate vote of the non-extremists in the sample will be near the point α where only agents in the 1/n2 tail would find it beneficial to use the extremist strategy. The case µU = 0, where the utility distribution is (at least in a crude sense) balanced is more technically challenging, and here we have less complete results. First, there are no extremists and no discontinuities. Theorem 6. If µU = 0, then for all sufficiently large values of the sample size n no Nash equilibrium v(u) has a discontinuity in [−A, B]. Moreover, for any ǫ > 0, if n is sufficiently large then every Nash equilibrium v(u) satisfies kvk∞ ≤ ǫ. (6) Under additional hypotheses, the proportionality constant in (3) is of order n−1/4 , as the next theorem shows. Theorem 7. Assume that the function ψ = Ψ′ is even. If µU = 0 and EU 3 = 0 then for any ǫ > 0, if n is sufficiently large then every Nash equilibrium satisfies p (7) | πn/2(Eψ(Sn ))2 σU − 2| < ǫ, 5

and therefore, by the proportionality rule (3), for all u 6= 0, p √ √ | 2σU 4 πn/2v(u)/u − 2| < ǫ

(8)

We do not know if the hypotheses that ψ is even and EU 3 = 0 are necessary for the validity of the theorem. The proof of the theorem, in section 5 below, will use the Edgeworth expansion for the density of a sum of independent, identically distributed random variables (cf. Feller [2], Ch. XVI, sec. 2, Th. 2), and in order that the relevant terms in this expansion behave in the desired fashion we must have EU 3 = 0 and ψ even. The rest of the article is arranged as follows. In section 2, we prove Theorems 1 and 2, and we establish a necessary condition for a pure-strategy to be a Nash equilibrium. In section 3, we use the necessary condition to prove that pure-strategy Nash equilibria must be continuous and smooth except in small neighborhoods of the endpoints −A, B, and we prove the approximate proportionality rule (Theorem 3). Finally, we prove Theorems 4 and 5 in section 4, and Theorem 7 in section 5.

2 Necessary Conditions for Nash Equilibrium Let πV be a mixed-strategy Nash equilibrium, and let Sn be the sum of the votes of n agents with utilities Ui gotten by random sampling from Fu , all acting in accordance with the strategy πV . For an agent with utility u, a best response v must satisfy equation (2), and so in particular for every ∆ > 0, if u > 0 then E {Ψ(Sn + v + ∆) − Ψ(Sn + v)} u ≤ 2∆v + ∆2

E {Ψ(Sn + v − ∆) − Ψ(Sn + v)} u ≤ −2∆v + ∆

and

(9)

2

Similarly, if u < 0 and ∆ > 0 then E {Ψ(Sn + v − ∆) − Ψ(Sn + v)} u ≤ −2∆v + ∆2

E {Ψ(Sn + v + ∆) − Ψ(Sn + v)} u ≤ 2∆v + ∆

and

(10)

2

Since Ψ is C ∞ and its derivative ψ has compact support, differentiation under the expectation if permissible. Thus, we have the following necessary condition. Proposition 2.1. If πV is a mixed-strategy Nash equilibrium then for every u a best response v must satisfy Eψ(Sn + v)u = 2v. (11) Consequently, every pure-strategy Nash equilibrium v(u) must satisfy the functional equation Eψ(Sn + v(u))u = 2v(u).

6

(12)

Proposition 2.2. Let πV be a mixed-strategy Nash equilibrium, and let v, v˜ be best responses for agents with utilities u, u ˜, respectively. If u = 0 then v = 0, and if u < u ˜, then v ≤ v˜. Consequently, any pure-strategy Nash equilibrium v(u) is a nondecreasing function of u, and therefore has at most countably many discontinuities and is differentiable almost everywhere. Proof. It is obvious that the only best response for an agent with u = 0 is v = 0, and the monotonicity of the payoff function Ψ implies that a best response v for an agent with utility u must be of the same sign as u. If v, v˜ are best responses for agents with utilities 0≤u 0. But if the two expectations were equal then v could not possibly be a best response at u, because an agent with utility u could obtain the same expected payoff EΨ(v + Sn )u at a lower vote cost by purchasing v˜ votes. This proves that if 0 ≤ u < u ˜ then best responses v, v˜ for agents with utilities u, u ˜ must satisfy 0 ≤ v ≤ v˜. A similar argument shows that if u < u ˜ ≤ 0 then best responses v, v˜ for agents with utilities u, u ˜ must satisfy v ≤ v˜ ≤ 0. Proof of Theorem 1. For each u denote by v− (u) and v+ (u) the minimal and maximal best responses at u. Proposition 2.2 implies that if u < u ˜ then v+ (u) ≤ v− (˜ u). Consequently, for any ǫ > 0 the set of utilities values u at which v+ (u) − v− (u) ≥ ǫ must be finite, because otherwise v√ + (u) → ∞ √as u → B, which is impossible since best responses must take values between − 2A and 2B. Remark 2.3. It follows from Theorem 1 that a mixed-strategy Nash equilibrium is essentially equivalent to a pure-strategy Nash equilibrium, specifically, the pure strategy v(u) = v− (u). This is because by hypothesis the sampling distribution FU is absolutely continuous with respect to Lebesgue measure, and so with probability one the sample U1 , U2 , . . . , Un+1 will contain only values u at which v− (u) = v+ (u). This proves the second assertion of Theorem 2. Henceforth, we shall consider only pure-strategy Nash equilibria. Proposition 2.4. If v(u) is a Nash equilibrium, then v(u) 6= 0 for all u 6= 0. 7

Proof. If v(u) = 0 for some u > 0 then by Proposition 2.2 v(u′ ) = 0 for all u′ ∈ (0, u). Since the utility density fU (u) is strictly positive on [−A, B], it follows that there is positive probability p that every agent in the sample casts vote Vi = 0. But then an agent with utility u could improve her expectation by buying ε > 0 votes, where ε ≪ uψ(0)p, because then the expected utility gain would be at least uΨ(ε)p ∼ uψ(0)pε at a cost of ε2 . Since by hypothesis ψ(0) > 0, the expected utility gain would overwhelm the increased vote cost for small ε > 0. Corollary 2.5. Any Nash equilibrium v(u) is strictly monotone on [−A, B]. Proof. Propositions 2.1 and 2.4 imply that Eψ(Sn + v(u)) > 0 for every u 6= 0. Now differentiation of the necessary condition (12) gives Eψ(Sn + v(u)) = (2 − Eψ ′ (Sn + v(u)))v ′ (u) at every u where v(u) is differentiable. Since such points are dense in [−A, B], and since ψ and ψ ′ are C ∞ functions with compact support, it follows that v ′ (u) 6= 0 on a dense set. But v ′ (u) ≥ 0 at every point where the derivative exists, so it follows that v ′ (u) > 0 almost everywhere, and this implies that v is strictly monotone.

3 Continuity and Smoothness Properties of Nash Equilibria 3.1 Weak consensus bounds According to the necessary condition (12), in a Nash equilibrium the number of votes v(u) purchased by an agent with utility u must satisfy the equation 2v(u) = Eψ(v(u) + Sn )u. It is natural to expect that when the sample size n + 1 is large the effect of adding a single vote v to the aggregate total Sn should be small, and so the function v(u) should satisfy the approximate proportionality rule 2v(u) ≈ Eψ(Sn )u. As we will show later, this naive approximation can fail badly for utility values u in the extreme tails of the distribution FU , and even in the bulk of the distribution the relative error in the approximation can be significant. Nevertheless, the idea of approximate population consensus on the expectations Eψ(v(u) + Sn ) can be used to obtain weak bounds that we will find useful. The following lemma states, roughly, that if it is optimal for some agent in the bulk of the population to buy a moderately large number of votes, then most agents will be forced to buy a moderately large number of votes. Lemma 3.1. For every ǫ > 0 there exist constants α, β > 0 such that for all sufficiently large n and any Nash equilibrium v(u) v(u) ≥ α max(−v(−A + ǫ), v(B − ǫ)) − e−βn u 8

for all |u| > 2ǫ.

(13)

Proof. It suffices to establish the lower bound αv(B − ǫ) − e−βn, as the other half of (13) can be proved in virtually the same way. Set uǫ = B − ǫ and pǫ = 1 − FU (uǫ ) where FU is the cumulative distribution function of the utility distribution. Let N = Nǫ be the number of points in the sample U1 , U2 , . . . , Un that fall in the interval [B − ǫ, B], and let U = Un+1 be independent of U1 , U2 , . . . , Un . Then X n E(ψ(v(U ) + Sn ) | U < uǫ ) = pm (1 − pǫ )n−m E(ψ(v(U ) + Sn ) | U < uǫ , N = m), m ǫ m≥0 X n E(ψ(v(U ) + Sn ) | U ≥ uǫ ) = pm (1 − pǫ )n−m E(ψ(v(U ) + Sn ) | U ≥ uǫ , N = m) m ǫ m≥0

Now conditional on N = m, the sample U1 , U2 , . . . , Un is obtained by choosing m points at random according to the conditional distribution of U given U ≥ uǫ and n − m according to the conditional distribution of U given U < uǫ . Consequently, for each m ≥ 0, E(ψ(v(U ) + Sn ) | U ≥ uǫ , N = m) = E(ψ(v(U ) + Sn ) | U < uǫ , N = m + 1). Furthermore, for any small ǫ′ > 0 and for m in the range [npǫ − nǫ′ , npǫ + nǫ′ ], the ratio     n m n n−m p (1 − pǫ ) pm+1 (1 − pǫ )n−m−1 m ǫ m+1 ǫ is between 1/2 and 2. Since the binomial-(n, pǫ ) distribution puts only an exponentially small (in n) mass outside the interval [npǫ −nǫ′ , npǫ +nǫ′ ], it follows that for some constants α′ , β ′ depending on ǫ but not n, ′

E(ψ(v(U ) + Sn ) | U < uǫ ) ≥ α′ E(ψ(v(U ) + Sn ) | U ≥ uǫ ) − e−β n

(14)

for all sufficiently large n. A similar argument proves that for suitable constants α′′ , β ′′ > 0, and for any interval J ⊂ [−A, B] of length ǫ not overlapping [B − ǫ, B] E(ψ(v(U ) + Sn ) | U ∈ J) ≥ α′′ E(ψ(v(U ) + Sn ) | U ≥ uǫ ) − e−β

′′ n

.

(15)

To see this, let N be the number of points in the sample U1 , U2 , . . . , Un that fall in the interval [B−ǫ, B], and let N ′ be the number of points in the sample that fall in J. Decompose the conditional expectations E(ψ(v(U ) + Sn ) | U ∈ J) and E(ψ(v(U ) + Sn ) | U > uǫ ) according to the values of N and N ′ , and use the identity E(ψ(v(U )+Sn ) | U ∈ J, N = m+1, N ′ = m′ ) = E(ψ(v(U )+Sn ) | U > uǫ , N = m, N ′ = m′ +1). As in the proof of (14), the ratio P {N = m + 1, N ′ = m′ } P {N = m, N ′ = m′ + 1} is near one for all pairs (m, m′ ) except those in the tails of the joint distribution, and the tails are exponentially small, by standard estimates for the multinomial distribution. 9

Now recall that any Nash equilibrium v(u) is monotone, and satisfies the necessary condition 2v(u) = Eψ(v(u) + Sn )u. Since any u > 2ǫ is the right endpoint of an interval J = [u − ǫ, u] of length ǫ for which the ratio of the right to the left endpoint is less than 2, it follows that for any such u, Eψ(v(u) + Sn ) ≥ 2α′′ E(ψ(v(U ) + Sn ) | U ≥ uǫ ) − 2e−β ≥ α′′ v(uǫ )/B − 2e−β

′′ n

′′ n

and similarly for any u < −2ǫ. The assertion (13) now follows from another application of the necessary condition (12).

3.2 Concentration and size constraints Since the vote total Sn is the sum of independent, identically distributed random variables v(Ui ) (albeit with unknown distribution), its distribution is subject to concentration restrictions, such as those imposed by the following lemma. Proposition 3.2. For any ǫ > 0 there exists a constant γ = γ(ǫ) < ∞ such that for all sufficiently large values of the sampling rate n and any Nash equilibrium v(u), if √ (16) max(v(B − ǫ), −v(−A + ǫ) ≥ γ/ n, then P {|Sn + v| ≤ δ} < ǫ

for all v ∈ R

(17)

and therefore |v(u)| ≤ ǫkψk∞

for all u ∈ [−A, B].

(18)

We will deduce Proposition 3.2 from the following general fact about sums of independent, identically distributed random variables. Proposition 3.3. Fix δ > 0. For any ǫ > 0 and any C < ∞ there exists C ′ = C ′ (ǫ, C) > 0 and n′ = n′ (ǫ, C) < ∞ such that the following statement is true: if n ≥ n′ and Y1 , Y2 , . . . , Yn are independent random variables such that E|Y1 − EY1 |3 ≤ Cvar(Y1 )3/2

var(Y1 ) ≥ C ′ /n P then for every interval J ⊂ R of length δ or greater, the sum Sn = ni=1 Yi satisfies and

P {Sn ∈ J} ≤ ǫ|J|/δ.

(19)

(20)

The proof of the proposition, a routine exercise in the use of Fourier methods, is relegated to the appendix. Proof of Proposition 3.2. Lemma 3.1 implies that there are constants α, β > 0 such that for every u ∈ [−A, B]\[−2ǫ, 2ǫ] the ratio v(u)/u is at least αv(uǫ )−e−βn , where uǫ = B−ǫ. Since the utility density fU is bounded below, it follows that for a suitable constant 0 < C < ∞, 10

for every Nash equilibrium v(u) there is an interval [u′ , u′′ ] ⊂ [B/2, B) of probability p such √ that if v(uǫ ) ≥ γ/ n then v(uǫ ) ≤ Cv(u′′ ) ≤ C 2 v(u′ ). (21) Similarly, there exists an interval [u∗ , u∗∗ ] ⊂ [−A, 0] of probability p such that |v(u∗ )| ≤ C|v(u∗∗ )|.

(22)

Let N be the number of points Ui in the sample U1 , U2 , . . . , Un that fall in [u∗ , u∗∗ ] ∪ and let Sn∗ be the sum of the votes v(Ui ) for those agents whose utility values fall in this range. Observe that N has the binomial-(n, 2p) distribution, and that conditional on the event N = m and Sn − Sn∗ = w, the random variable Sn∗ is the sum of m independent random variables Yi whose variance is at least v(u′ )2 /4 and whose third moment obeys the restriction (67) (this follows from the inequalities (21)–(22)). Consequently, by Proposi√ tion 3.3, if v(uǫ ) n is sufficiently large then the conditional probability, given N = n ≥ np and Sn − Sn∗ = w, that Sn∗ lies in any interval of length δ is bounded above by ǫ/2. Since P {N ≤ np} is, for large n, much less than ǫ/2, the inequality (17) follows. [u′ , u′′ ],

Proposition 3.2 implies that for any ǫ > 0, if n is sufficiently large then for any Nash equilibrium v(u) the absolute value |v(u)| can assume large values only at utility values u within distance ǫ of one of the endpoints −A, B. The following proposition improves this to the extreme tails of the distribution. Proposition 3.4. For any 0 < C < ∞ there exists C ′ > 0 such that for all sufficiently large n and any Nash equilibrium v(u) satisfies the inequality |v(u)| ≤ C

for all u ∈ [−A + C ′ n−3/2 , B − C ′ n−3/2 ].

(23)

Proof. Fix C > 0, and suppose that 2v(u∗ ) ≥ C for some u∗ > 0. Since any Nash equilibrium v is monotone, we must have 2v(u) ≥ C for all u ≥ u∗ , and by the necessary condition (12) it follows that Eψ(v(u) + Sn )u ≥ C

=⇒

Eψ(v(u) + Sn ) ≥ C/B ∀ u ≥ u∗ .

(24)

Consequently, the distribution of Sn is concentrated: since the function ψ has support [−δ, δ], the probability that Sn + v(u) ∈ [−δ, δ] must be at least C/Bkψ ′ k∞ . Thus, Proposition 3.2 implies that for any ǫ > 0 there exists γǫ > 0 (depending on both ǫ and C, but not on n) such that √ (25) max(−v(−A + ǫ), v(B − ǫ)) ≤ γǫ / n

In particular, for all sufficiently large n, γB/2 v(B/2) ≤ √ n

=⇒ =⇒ =⇒

Eψ(v(B/2) + Sn ) ≤

2γB/2 √ B n

2γB/2 √ + kψ ′ k∞ v(B/2) B n CB/2 Eψ(Sn ) ≤ √ n Eψ(Sn ) ≤

11

(26)

for a constant CB/2 < ∞ that may depend on B/2 and C but not on either n or the particular Nash equilibrium. Fix C ′ large, and suppose that 2v(u∗ ) ≥ C for u∗ = B − C ′ n−3/2 . Let N∗ be the number of points Ui in the sample U1 , U2 , . . . , Un that fall in the interval [u∗ , B]; by our assumptions concerning the sampling procedure, the random variable N∗ has the binomial distribution with mean Z B

EN∗ = n

fU (u) du = C ′ Cf n−1/2

u∗

where Cf is the mean value of fU on the interval [u∗ , B] (which for large n will be close to fU (B) > 0). Since EN∗ is vanishingly small for large n, the assumption v(u∗ ) ≥ C implies that Eψ(v(u) + Sn )1{N∗ = 0} ≥ C/2B for all u ≥ u∗ . (27)

This expectation can be decomposed by partitioning the probability space into the event G = {Un ∈ [−A + ǫ, B − ǫ]} and its complement. On the event G, the contribution of v(Un ) √ to the vote total Sn is at most γǫ / n in absolute value, by (25). On the complementary event Gc the integrand is bounded above by kψk∞ . Therefore, Eψ(v(u) + Sn )1{N∗ = 0} ≤ P (Gc )kψk∞ + Eψ(v(u) + Sn )1{N∗ = 0}1G

√ ≤ P (Gc )kψk∞ + Eψ(v(u) + Sn−1 )1{N∗ = 0} + kψ ′ k∞ (γǫ / n)

≤ ǫ′ + Eψ(v(u) + Sn−1 )1{N∗ = 0}

where ǫ′ > 0 can be made arbitrarily small by choosing ǫ > 0 small and n large. This together with inequality (27) implies that for large n, Eψ(v(u) + Sn−1 )1{N∗ = 0} ≥ C/4B

for all u ≥ u∗ .

(28)

Now consider the conditional distribution of Sn given that N∗ = 1: this can be simulated by generating Sn−1 from the conditional distribution of Sn−1 given that N∗ = 0 and then adding an independent v(U ) where U = Un is drawn from the conditional distribution of U given that U ≥ u∗ . Consequently, by inequality (28), E(ψ(Sn ) | N∗ = 1) = E(ψ(Sn−1 + v(U )) | N∗ = 0) ≥ C/4B. But this implies that √ E(ψ(Sn )) ≥ (C/2B)P {N∗ ≥ 1} ≈ CC ′ Cf /(2B n). For large C ′ this is incompatible with inequality (26) when n is sufficiently large.

3.3 Discontinuities Since any Nash equilibrium v(u) is monotone in the utility u, it can have at most countably many p discontinuities. Moreover, since any Nash equilibrium is bounded in absolute value by 2 max(A, B) (as no agent will pay more pfor votes than she could gain in expected utility) the sum of the jumps is bounded by 2 max(A, B). We will now show that there is a lower bound on the size of |v| at a discontinuity. 12

Lemma 3.5. Let v(u) be a Nash equilibrium. If v is discontinuous at u ∈ (−A, B) then Eψ ′ (˜ v + Sn )u = 2

(29)

for some v˜ ∈ [v− , v+ ], where v− and v+ are the left and right limits of v(u′ ) as u′ → u. Proof. The necessary condition (12) holds at all u′ in a neighborhood of u, so by monotonicity of v and continuity of ψ, the equation (12) must hold when v(u) is replaced by either of v± , that is, 2v+ = Eψ(v+ + Sn )u

and

2v− = Eψ(v− + Sn )u. Subtracting one equation from the other and using the differentiability of ψ we obtain Z v+ Z v+ ′ Eψ ′ (t + Sn ) dt. ψ (t + Sn ) dt = u 2v+ − 2v− = uE v−

v−

The result then follows from the mean value theorem of calculus. Proposition 3.6. There is a constant ∆ > 0 such that for all sufficiently large n, at any point u of discontinuity of a Nash equilibrium, v(u+ ) ≥ ∆ if u ≥ 0,

and

(30)

v(u− ) ≤ −∆ if u ≤ 0.

Consequently, there is a constant β < ∞ not depending on the sample size n such that for all sufficiently large n no Nash equilibrium v(u) has a discontinuity at a point u at distance greater than βn−3/2 from one of the endpoints −A, B. Proof. Since the function ψ has support contained in the interval [−δ, δ], equation (29) implies that v can have a discontinuity only if the distribution of Sn is highly concentrated: specifically, 2 . (31) P {Sn + v˜ ∈ [−δ, δ]} ≥ kψ ′ k max(A, B) In fact, since ψ ′ vanishes at the endpoints of [−δ, δ], there exists 0 < δ′ < δ such that P {Sn + v˜ ∈ [−δ′ , δ′ ]} ≥

1 . kψ ′ k max(A, B)

(32)

Proposition 3.2 asserts that strong concentration of the distribution of Sn can occur only if |v(u)| is vanishingly small in the interior of the interval [−A, B]. In particular, if ǫ < √ (kψ ′ k max(A, B))−1 and n is sufficiently large then |v(u)| < γǫ / n for all u ∈ [−A+ ǫ, B − ǫ]. But v(u) must satisfy the necessary condition (12) at all such u, so √ Eψ(v(u) + Sn )|u| ≤ 2γǫ / n 13

for all u ∈ [−A + ǫ, B − ǫ]. Since the function ψ is positive and bounded away from 0 in any interval [−δ′′ , δ′′ ] where 0 < δ′′ < δ, it follows from (32) that for sufficiently large n, |˜ v | ≥ (δ − δ′ )/3 := ∆. Thus, by the monotonicity of Nash equilibria, at every point u of discontinuity we must have (30). Proposition 3.4 now implies that any such discontinuities can occur only within a distance βn−3/2 of one of the endpoints −A, B.

3.4 Smoothness Since Nash equilibria are monotone, by Proposition 2.2, they are necessarily differentiable almost everywhere. We will show that in fact differentiability must hold at every u, except near the endpoints −A, B. Lemma 3.7. If v(u) is a Nash equilibrium then at every u where v is differentiable, Eψ(Sn + v(u)) + Eψ ′ (Sn + v(u))uv ′ (u) = 2v ′ (u).

(33)

Proof. This is a routine consequence of the necessary condition (12) and the smoothness of the function ψ. Equation (33) can be rewritten as a first-order differential equation: v ′ (u) =

Eψ(Sn + v(u)) . 2 − Eψ ′ (Sn + v(u))u

(34)

This differential equation becomes singular at any point where the denominator approaches 0, but is regular in any interval where Eψ ′ (Sn + v(u))u ≤ 1. The following lemma implies that this will be the case on any interval where |v(u)| remains sufficiently small. Lemma 3.8. For any α > 0 there exists a constant β = βα > 0 such that for any strategy v(u), any v˜ ∈ R, any u ∈ [−A, B], and all n, E|ψ ′ (˜ v + Sn )u| ≥ α ′′

E|ψ (˜ v + Sn )u| ≥ α

=⇒

Eψ(˜ v + Sn )|u| ≥ β and

=⇒

(35)

Eψ(˜ v + Sn )|u| ≥ β

Proof. Recall that ψ/2 is a C ∞ probability density with support [−δ, δ] and such that ψ is strictly positive in the open interval (−δ, δ). Consequently, on any interval J ⊂ (−δ, δ) where |ψ ′ | (or |ψ ′′ |) is bounded below by a positive number, so is ψ.

Fix ǫ > 0 so small that ǫ max(−A, B) < α/2. In order that E|ψ ′ (˜ v + Sn )u| ≥ α, it must be the case that the event {|ψ ′ (˜ v + Sn )| ≥ ǫ} contributes at least α/2 to the expectation; hence, P {|ψ ′ (˜ v + Sn )| ≥ ǫ} ≥

2kψ ′ k∞

14

α . max(−A, B)

But on this event the random variable ψ(˜ v + Sn ) is bounded below by a positive number η = ηǫ , so it follows that Eψ(˜ v + Sn )|u| ≥

ηα 2kψ ′ k∞ max(−A, B)

.

A similar argument proves the corresponding result for ψ ′′ . Proposition 3.9. There exist constants C, α > 0 such that for all sufficiently large n, any Nash equilibrium v(u) is continuously differentiable on any interval where |v(u)| ≤ C (and therefore, by Proposition 3.4, on (−A + C ′ n−3/2 , B − C ′ n−3/2 )), and the derivative satisfies α≤

v ′ (u) ≤ α−1 . Eψ(v(u) + Sn )

(36)

Proof. The function v(u) is differentiable almost everywhere, by Proposition 2.2, and at every point u where v(u) is differentiable the differential equation (34) holds. By Proposition 3.6, the sizes of discontinuities are bounded below, and so if C > 0 is sufficiently small then a Nash equilibrium v(u) can have no discontinuities on any interval where |v(u)| ≤ C. Furthermore, if C > 0 is sufficiently small then by Lemma 3.8 and the necessary condition (12), we must have Eψ ′ (v(u) + Sn ) ≤ 1 on any interval where |v(u)| ≤ C. Since the functions v 7→ Eψ(Sn + v) and v 7→ Eψ ′ (Sn + v) are continuous (by dominated convergence), it now follows from equation (34) that if C > 0 is sufficiently small then on any interval where |v(u)| ≤ C the function v ′ (u) extends to a continuous function. Finally, since the denominator in equation (34) is at least 1 and no larger than 2 + kψ ′ k∞ , the inequalities (36) follow. Similar arguments show that Nash equilibria have derivatives of higher orders provided the sample size is sufficiently large. The proof of Theorem 7 in section 5 will require information about the second derivative v ′′ (u). This can be obtained by differentiating under the expectations in (34): Eψ(v(u) + Sn )(Eψ ′′ (v(u) + Sn )v ′ (u)u + Eψ ′ (v(u) + Sn ) Eψ ′ (v(u) + Sn )v ′ (u) + . 2 − Eψ ′ (Sn + v(u))u (2 − Eψ ′ (Sn + v(u)u))2 (37) A repetition of the proof of Proposition 3.9 now shows that for suitable constants C, β > 0 and all sufficiently large n, any Nash equilibrium v(u) is twice continuously differentiable on any interval where |v(u)| ≤ C and satisfies the inequalities v ′′ (u) =

β≤

v ′′ (u) ≤ β −1 . Eψ(v(u) + Sn )

(38)

3.5 Approximate proportionality The information that we now have about the form of Nash equilibria can be used to sharpen the heuristic argument given in paragraph 3.1 to support the “approximate proportionality rule”. Recall that in a Nash equilibrium the number of votes v(u) purchased 15

by an agent with utility u must satisfy the equation 2v(u) = Eψ(v(u) + Sn )u. We have shown in Proposition 3.4 that for any Nash equilibrium, v(u) must be small except in the extreme tails of the distribution (in particular, for all u at distance much more than n−3/2 from both endpoints −A, B). Since ψ is uniformly continuous, it follows that the expectation Eψ(v(u) + Sn ) cannot differ by very much from Eψ(Sn ). Unfortunately, this argument only shows that the approximation 2v(u) ≈ Eψ(Sn )u is valid up to an error of size ǫn |u| where ǫn → 0 as n → ∞. However, as n → ∞ the expectation Eψ(Sn ) → 0, and so the error in the approximation above might be considerably larger than the approximation itself. Theorem 3 (section 1) makes the stronger assertion that when n is large the relative error in the approximate proportionality rule is small. Proof of Theorem 3. Since ψ has compact support, it and all of its derivatives are uniformly continuous and uniformly bounded, and so the function v 7→ Eψ(v + Sn ) is differentiable with derivative Eψ ′ (v + Sn ). Consequently, by the mean value theorem, for every u there exists v˜(u) intermediate between 0 and v(u) such that 2v(u) = Eψ(v(u) + Sn )u = Eψ(Sn )u + Eψ ′ (˜ v (u) + Sn )v(u)u.

(39)

We will argue that for all C > 0 sufficiently small, if |v(u)| ≤ C then the expectation Eψ ′ (˜ v (u) + Sn ) remains below ǫ in absolute value, provided n is sufficiently large. Proposition 3.4 will then imply that there exists C ′ < ∞ such that (3) holds for all u ∈ (−A, B) at distance greater than C ′ n−3/2 from the endpoints −A, B, proving the first assertion of Theorem 3. If |2v(u)| ≤ C then |Eψ(v(u) + Sn )| ≤ C/ max(A, B), by the necessary condition (12). By Proposition 3.6, if C < ∆, where ∆ is the discontinuity threshold, then v(u) is continuous on any interval [0, uC ] where |v(u)| ≤ C, and so for each u in this interval there is a u′ ∈ [0, u] such that v˜(u) = v(u′ ). Consequently, |Eψ(˜ v (u) + Sn )| ≤ C/ max(A, B). But Lemma 3.8 implies that for any ǫ > 0, if C > 0 is sufficiently small then for all n and any Nash equilibrium v(u), |Eψ ′ (˜ v (u) + Sn )| < ǫ on any interval [0, uC ] where |v(u)| ≤ C. Thus, the error in the approximation (39) will be small when n is large and |v(u)| < C, for u > 0. A similar argument applies for u ≤ 0.

Finally, suppose that v(u) is a Nash equilibrium with no discontinuities. By Proposition 3.4, for any C > 0 there exists C ′ < ∞ such that |v(u)| ≤ C/2 except at arguments u within distance C ′ /n3/2 of one of the endpoints. Moreover, Proposition 3.9 implies that if C is sufficiently small then on any interval where |v(u)| ≤ C the function v is differentiable, with derivative v ′ (u) ≤ C ′′ for some constant C ′′ < ∞ not depending on n or on the particular Nash equilibrium. It then follows that v(B) ≤ C/2 + C ′ C ′′ n−3/2 ≤ C provided n is large. Since C > 0 can be chosen arbitrarily small, it follows that v(u) must satisfy the proportionality relations (3) on [0, B]. A smilar argument applies to the interval [−A, 0]. 16

3.6 Consequences of Theorem 3 Theorem 3 puts strong constraints on the distribution of the vote total Sn in a Nash equilibrium. According to Theorem 3, the approximate proportionality rule (3) holds for all u ∈ [−A, B] except those values u within distance Cn−3/2 of one of the endpoints −A, B. Call such values extremists, and denote by G the event that the sample U1 , U2 , . . . , Un co Moreover, for Nash equilibria with no discontinuities, (3) holds for all u ∈ [−A, B]. Thus, conditional on the event G (or, for continuous Nash equilibria, unconditionally) the random variables v(Ui ) are (at least for sufficiently large n) bounded above and below by Eψ(Sn )B and −Eψ(Sn )A, and so Hoeffding’s inequality [4] applies. Corollary 3.10. Let G be the event that the sample Ui contains no extremists. Then for all sufficiently large n and any Nash equilibrium v(u), P (|Sn − ESn | ≥ tEψ(Sn ) | G) ≤ exp{−2t2 /n max(A2 , B 2 )};

(40)

and for any Nash equilibrium with no discontinuities, P (|Sn − ESn | ≥ tEψ(Sn )) ≤ exp{−2t2 /n max(A2 , B 2 )}.

(41)

Theorem 3 also implies uniformity in the normal approximation to the distribution of Sn , because the proportionality rule (3) guarantees that the ratio of the third moment to the 3/2 power of the variance of v(Ui ) is uniformly bounded. Hence, by the Berry-Esseen theorem, we have the following corollary. Corollary 3.11. There exists κ < ∞ such that for all sufficiently large n and any Nash equilibrium v(u), the vote total Sn satisfies p (42) sup |P ((Sn − ESn ) ≤ t var(Sn ) | G) − Φ(t)| ≤ κn−1/2 ; and for any Nash equilibrium with no discontinuities, p sup |P ((Sn − ESn ) ≤ t var(Sn )) − Φ(t)| ≤ κn−1/2 .

(43)

Here Φ denotes the standard normal cumulative distribution function.

4 Unbalanced populations: The case µU > 0 4.1 Concentration of the Vote Total Proposition 4.1. If µU > 0 then for all large n no Nash equilibrium v(u) has a discontinuity at a nonnegative value of u. Moreover, if µU > 0 then for any ǫ > 0, if n is sufficiently large then in any Nash equilibrium the vote total Sn must satisfy √ (i) ESn ∈ [δ − ǫ, δ + ǫ + 2A] and (ii) P {|Sn − ESn | > ǫ} < ǫ. 17

Furthermore, there is a constant γ > 0 such that for any ǫ > 0, if n is sufficiently large and v(u) is a Nash equilibrium with no discontinuities, then (iii) P {|Sn − ESn | > ǫ} < e−γn . Proof. By Proposition 3.6, a Nash equilibrium v(u) can have no discontinuities at distance greater than Cn−3/2 of one of the endpoints −A, B. Agents with such utilities are designated extremists; the event G that the sample U1 , U2 , . . . , Un contains no extremists has probability 1 − O(n−1/2 ).

By Theorem 3, any Nash equilibrium v(u) obeys the approximate proportionality rule (3) except in the extremist regime. The contribution of extremists √ √ to ESn is vanishingly small c −1/2 for large n, since P (G ) = O(n ) and |v| ≤ max( 2A, 2B). Consequently, (3) implies that for any ǫ > 0, if n is large then Eψ(Sn )µU (1 − ǫ) ≤ ESn /n ≤ Eψ(Sn )µU (1 + ǫ).

(44)

Since µU > 0, this implies that ESn ≥ 0 for all sufficiently large n.

Suppose now that ESn < δ − 2ǫ′ for some small ǫ′ > 0. If ǫ > 0 is sufficiently small relative to ǫ′ then (44) implies that nEψ(Sn )µU ≤ δ − ǫ′ /2. But then Hoeffding’s inequality (40) (for this a weaker Chebyshev bound would suffice), together with the fact that P (Gc ) ≤ Kn−1/2 , implies that P {Sn ∈ [−δ/2, δ − ǫ′ /4]} ≥ 1 − ǫ

for large n. This is impossible, though, because we would then have Eψ(Sn ) ≥ (1 − ǫ)

min

v∈[−δ/2,δ−ǫ′ /4]

ψ(v),

and since ψ is bounded away from 0 on any compact sub-interval of (−δ, δ) this contradicts the fact that nEψ(Sn ) < δ − ǫ′ /2. This proves that for all large n and all Nash equilibria, ESn ≥ δ − 2ǫ′ . √ Next suppose that ESn > δ + 2A + 2ǫ′ , where ǫ′ > 0. The proportionality √ rule (3) (applied with some ǫ > 0 small relative to ǫ′ ) then implies that nEψ(Sn ) > δ + 2A + ǫ′ . Hence, by the Hoeffding inequality (40), there exists γ = γ(ǫ′ ) > 0 such that √ P (Sn ≤ δ + 2A | G) ≤ e−γn , √ because on the event Sn ≤ δ + 2A the √ sum Sn must deviate from its expectation by more than nEψ(Sn )ǫ′ . Hence, for all v ∈ [− 2A, 0] ≤ 0, Eψ(v + Sn ) ≤ e−γn kψk∞ + P (Gc )kψk∞ . Thus, |v(−A)| must be vanishingly small, and so by Proposition 3.6 there can be no discontinuities in [−A, 0]. But this implies that the proportionality rule (3) holds for all u ∈ [−A, B − Cn−3/2 ], and so another application of Hoeffding’s inequality (coupled with the

18

observation that v(u)/u ≥ (1−ǫ)Eψ(Sn ) holds for all u ∈ [−A, B] if v has no discontinuities at negative values of u) implies that √ P (Sn ≤ δ + 2A) ≤ e−γn =⇒ Eψ(Sn ) ≤ e−γn kψk∞ , which is a contradiction. This proves assertion (i). Since ESn is now bounded away from 0 and ∞, it follows as before that nEψ(Sn ) is bounded away from 0 and ∞, and so the proportionality rule (3) implies that the conditional variance of Sn given the event G is O(n−1 ). The assertion (ii) therefore follows from Chebyshev’s inequality and the bound P (Gc ) = O(n−1/2 ). Given (i) and (ii), we can now conclude that there can be no discontinuities at nonnegative values of u, because in view of Proposition 3.6, the monotonicity of Nash equilibria, and the necessary condition (12), this would entail that Eψ(v(B) + Sn )B ≥ 2∆, which is incompatible with (i) and (ii). Finally, if v is a Nash equilibrium with no discontinuities then Corollary 3.10 implies the exponential bound (iii).

4.2 Proof of Theorem 4 Proposition 4.1 implies that for large n the distribution of Sn is concentrated near ESn , and ESn ≥ δ − ǫ. Hence, the expected utility payoff for an agent with u > 0 is near u, and so when n is large it would be sub-optimal for an agent with utility u near B to buy more than ǫ votes. Since the size of a discontinuity is bounded below, by Proposition 3.6, it follows that for large n no Nash equilibrium will have a discontinuity in [0, B]. If a Nash equilibrium had a discontinuity at some u < 0 it would have to occur within distance O(n−3/2 ) of the endpoint −A. Thus, since Nash equilibria are monotone, v(−A) ≤ −∆, by Proposition 3.6. By Proposition 4.1, the expected payoff to an agent with utility −A would then be bounded above by Ψ(δ − v(−A) − ǫ)(−A) − v(−A)2 , provided n is sufficiently large. On the other hand, the expected payoff to such an agent buying 0 votes would be at least −A; since by hypothesis v 2 − Ψ(δ − v) is bounded below by a positive number ǫ′ , if ǫ < ǫ′ then buying v(−A) < −∆ would be suboptimal, contradicting the hypothesis that v is a Nash equilibrium. This proves that for large n, no Nash equilibrium can have a discontinuity. It remains to show that in a Nash equilibrium the distribution of Sn is, for large n, concentrated near δ. By Proposition 4.1, the distribution of Sn is concentrated near ESn , and since no Nash equilibrium has a discontinuity the probability that |Sn − ESn | ≥ ǫ is exponentially decaying in n. Consequently, if ESn were larger than δ + 2ǫ then Eψ(Sn ) would also be exponentially decaying in n. But this would contradict the proportionality rule (3), as this would imply that Eψ(Sn ) ≥ δn−1 for large n.

19

4.3 Proof of Theorem 5 Proposition 4.1 implies that for large n the distribution of Sn must be highly concentrated near ESn in any Nash equilibrium, and for any ǫ > 0 there exists γ > 0 such that for any Nash equilibrium with no discontinuities, P {|Sn − ESn | ≥ ǫ} ≤ e−γn . Hence, if ESn > δ + ǫ then Eψ(Sn ) < e−γn . But Proposition 3 asserts that if a Nash equilibrium v(u) has no discontinuities then the proportionality rule (3) holds for all u ∈ [−A, B], and so ESn ≤ (1 + ǫ′ )nEψ(Sn ) ≤ (1 + ǫ′ )ne−γn , contradicting the fact that ESn ≥ δ − ǫ. This proves that for large n, any Nash equilibrium v(u) with no discontinuities must satisfy ESn < δ + ǫ. Suppose that ESn < α − 2ǫ for some ǫ > 0. If ǫ > 0 is sufficiently small, then for some ǫ′ > 0 depending on ǫ, (1 − Ψ(w + ǫ)))(A − ǫ′ ) > (α − ǫ − w)2 . Consequently, if ESn ≤ α − 2ǫ, then an agent with utility u ∈ [−A, −A + ǫ′ ] purchasing α + ǫ − w votes would have expected payoff at least −Ψ(w + ǫ))(A − ǫ′ )P {Sn ≤ α − ǫ} − (α − ǫ − w)2 . This strictly dominates the expected payoff ≈ −AP {Sn ≥ α − 3ǫ} for buying votes in accordance with the approximate proportionality rule (3). But any Nash equilibrium must satisfy the rule (3) except in the extremist regime, so we have a contradiction. This proves that for all sufficiently large n, in any Nash equilibrium we must have ESn > α − 2ǫ. It follows that for all sufficiently large n, every Nash equilibrium has a discontinuity. The discontinuity must be located within distance Cn−3/2 of the endpoint −A, by Proposition 3.6. Now suppose that ESn > α + 3ǫ. Then, by Hoeffding’s inequality, P (Sn ≤ α + 2ǫ | G) is exponentially small for large n. Furthermore, since (α, w) is the unique pair satisfying (4), (1 − Ψ(w′ + ǫ)))A + 2Ae−γn < (α + 2ǫ − w′ )2

for all w′ ∈ [−δ, δ]

and so it would be suboptimal for an agent with utility value −A to buy more than α+2ǫ−δ votes. Clearly it would also be suboptimal to buy more than ∆ but no more than α + 2ǫ − δ votes, where ∆ is the discontinuity threshold (cf. Proposition 3.6), because this would leave the expected utility payoff below −A(1 − e−γn ). Consequently, if ESn > α + 3ǫ then for large n no Nash equilibrium would have a discontinuity; since we have shown that for large n every Nash equilibrium has a discontinuity it follows that ESn cannot exceed α + 3ǫ for large n. We have therefore proved that for any ǫ > 0, if n is sufficiently large then (a) every Nash equilibrium has a discontinuity in the extremist regime near −A; and (b) |ESn − α| < ǫ. Assertion (iv) of the theorem follows, by Proposition 4.1.

Let v(u) be a Nash equilibrium, and let u∗ be the rightmost point u∗ of discontinuity of v. Consider the strategy v(u) for an agent with utility value u < u∗ : since v is monotone, 20

v(u) ≤ −∆. Moreover, the expected payoff for an agent with utility u must exceed the expected payoff under the alternative strategy of buying no votes. The latter expectation is approximately −A, because Sn is highly concentrated near ESn > α−ǫ and so EΨ(Sn ) ≈ 1. On the other hand, the expected payoff at u for an agent playing the Nash strategy v is approximately Ψ(α − v(u))(−A) − v(u)2 . Consequently, since (α, w) is the unique pair such that relations (4) hold, we must have |v(u)| ≈ α − w. This proves assertion (ii). That v has only a single point of discontinuity u∗ follows from the hypothesis (5). Recall (cf. Lemma 3.5) that if v is discontinuous at u then Eψ ′ (˜ v + Sn ) = 2 for some v˜ intermediate between the right and left limits v(u+) and v(u−). But any discontinuity u must occur within distance βn−3/2 of −A, and if u < u∗ then v(u) ≈ −α + w. Hence, since the distribution of Sn is concentrated in a neighborhood of α, Eψ ′ (v(u±) + Sn ) ≈ ψ ′ (w), and so by (5), for u ∈ [−A, u∗ ) there cannot be a value v˜ ∈ [v(u−), v(u+)] satisfying the necessary condition Eψ ′ (˜ v + Sn ) = 2 for a discontinuity. Finally, since u∗ must be within distance Cn−3/2 of −A, the conditional probability that there are at least two extremists in the sample U1 , U2 , . . . , Un given that there is at least one is of order O(n−1/2 ). Consequently, Eψ(Sn ) = ψ(w)fU (−A)(u∗ + A) + O(n−1/2 (u∗ + A)). On the other hand, since ESn ≈ α, the proportionality rule (3) implies that nEψ(Sn ) ≈ α. Therefore, u∗ − A ∼ γn−2 where γ is the unique solution of the equation α = γψ(w)fU (−A). This proves assertions (i) and (iii).

5 Balanced Populations: µU = 0 5.1 Continuity of Nash Equilibra Proposition 5.1. If µU = 0, then for all sufficiently large values of the sample size n no Nash equilibrium v(u) has a discontinuity in [−A, B]. Moreover, for any ǫ > 0, if n is sufficiently large then every Nash equilibrium v(u) satisfies kvk∞ ≤ ǫ. 21

(45)

Proof. The size of any discontinuity is bounded below by a positive constant ∆, by Proposition 3.6, so it suffices to prove the assertion (45). By Proposition 3.2, for any ǫ > 0 there exists γ = γ(ǫ) such that if n is sufficiently large then any Nash equilibrium v(u) satisfy√ ing kvk∞ > ǫ must also satisfy |v(u)| ≤ γ/ n for all u not within distance ǫ of one of the endpoints −A, B. Hence, the approximate proportionality relation (3) implies that C Eψ(Sn ) ≤ √ n

(46)

for a suitable C = C(γ). Since v(u)/u is within a factor (1 + ǫ)±1 of Eψ(Sn ) for all u not within distance Cǫ′ n−3/2 of −A or B, it follows from Chebyshev’s inequality that for any α > 0 there exists β = β(α) such that P {|Sn − ESn | ≥ β} ≤ α. On the other hand, if kvk∞ ≥ ǫ, then by the necessary condition (12), there is some u such that ǫ . P {Sn + v(u) ∈ [−δ, δ]} ≥ kψk∞ max(A, B) Since Sn is concentrated around ESn , it follows that ESn must be at bounded distance from v(u), and so the Berry–Esseen bound (42) implies that P {Sn ∈ [−δ/2, δ/2]} is bounded below. But this in turn implies that Eψ(Sn ) is bounded below, which for large n is impossible in view of (46). Thus, if n is sufficiently large then no Nash equilibrium v(u) can have kvk∞ ≥ ǫ.

5.2 Proof of Theorem 7 The proof of Theorem 7 will use the Edgeworth expansion for the density of a sum of independent, identically distributed random variables (cf. Feller [2], Ch. XVI, sec. 2, Th. 2). The relevant summands here are the random variables v(Ui ), and since the function v(u) depends on the particular Nash equilibrium (and hence also on n), it will be necessary to have a version of the Edgeworth expansion in which the error is precisely quantified. The following variant of Feller’s Theorem 2 (which can be proved in the same manner as in [2]) will suffice for our purposes. Theorem 8. Let Y1 , Y2 , . . . , Yn be independent, identically distributed random variables with mean EY1 = 0, variance EY12 = 1, and finite 2rth moment E|Y1 |2r = µ2r ≤ m2r . Assume that the distribution of Y1 has a density f1 (y) whose Fourier transform fˆ1 satisfies |fˆ1 (θ)| ≤ g(θ), where g is a C 2r function such that g ∈ Lν for some ν ≥ 1 and such that for every ǫ > 0, sup g(θ) < 1.

(47)

|θ|≥ǫ

Then there ǫn → 0 depending only on m2r and on the function g such that the density Pis a sequence √ fn (y) of ni=1 Yi / n satisfies ! 2r −x2 /2 X e ǫn −(k−2)/2 fn (x) − √ 1+ n Pk (x) ≤ −r+1 (48) n 2πn k=3 22

for all x ∈ R, where Pk (x) = Ck Hk (x) is a multiple of the kth Hermite polynomial Hk (x), and Ck is a continuous function of the moments µ3 , µ4 , . . . , µk of Y1 . The following lemma ensures that in any Nash equilibrium the sums Sn = after suitable renormalization, meet the requirements of Theorem 8.

Pn

i=1 v(Ui ),

Lemma 5.2. There exist constants 0 < σ1 < σ2 < m2r < ∞ and a function g(θ) satisfying the hypotheses of Theorem 8 (with r = 4) such that for all sufficiently large n and any Nash equilibrium v(u) the following statement holds. If w(u) = 2v(u)/Eψ(Sn ) then (a) σ12 < var(w(Ui )) < σ22 ; (b) E|w(Ui ) − Ew(Ui )|2r ≤ m2r ; and (c) the random variables w(Ui ) have density fW (w) whose Fourier transform is bounded in absolute value by g. Proof. These statements are consequences of the proportionality relations (3) and the smoothness of Nash equilibria. By Proposition 5.1, Nash equilibria are continuous on [−A, B] and for large n satisfy kvk∞ < ǫ, where ǫ > 0 is any small constant. Consequently, by Proposition 3, the proportionality relations (3) hold on the entire interval [−A, B]. Since EU1 = 0, it follows that for any ǫ > 0, if n is sufficiently large then |Ew(Ui )| < ǫ, and so assertions (a)–(b) follow routinely from (3). The existence of the density fW (w) follows from the smoothness of Nash equilibria, which was established in section 3.4. In particular, by Proposition 3.9, inequalities (38), and the proportionality relations (3), if the sample size n is sufficiently large and v is any continuous Nash equilibrium then v is twice continuously differentiable on [−A, B], and there are constants α, β > 0 not depending on n or on the particular Nash equilibrium such that the derivatives satisfy α≤

v ′ (u) ≤ α−1 Eψ(Sn )

and

β≤

v ′′ (u) ≤ β −1 Eψ(Sn )

(49)

for all u ∈ [−A, B]. Consequently, if U is a random variable with density fU (u) then the random variable W := 2v(U )/Eψ(Sn ) has density fW (w) = fU (u)Eψ(Sn )/(2v ′ (u))

where w = 2v(u)/Eψ(Sn )

(50)

Furthermore, the density fW (w) is continuously differentiable, and its derivative ′ fW (w) =

satisfies

fU′ (u)(Eψ(Sn ))2 fU (u)(Eψ(Sn ))2 v ′′ (u) − 4v ′ (u)2 4v ′ (u)3 ′ |fW (w)| ≤ κ

(51)

where κ < ∞ is a constant that does not depend on either n or on the choice of Nash equilibrium. It remains to prove the existence of a dominating function g(θ) for the Fourier transform of fW . This will be done in three pieces: (i) for values |θ| ≤ γ, where γ > 0 is a small 23

fixed constant; (ii) for values |θ| ≥ K, where K is a large but fixed constant; and (iii) for γ < |θ| < K. Region (i) is easily dealt with, in view of the bounds (a)–(b) on the second and third moments and the estimate |Ew(U )| < ǫ′ , as these together with Taylor’s theorem imply that for all |θ| < 1, |fˆW (θ) − (1 + iθEw(U ) − θ 2 var(w(U ))/2| ≤ m3 |θ|3 . Next consider region (ii), where |θ| is large. Integration by parts shows that fˆW (θ) =

Z

w(B)

iθw

fW (w)e w(−A)

dw = −

Z

w(B) w(−A)

w(B) eiθw ′ eiθw fW (w) dw + fW (w) ; iθ iθ w(−A)

′ (w)| ≤ since fW (w) is uniformly bounded at w(−A) and w(B), by (49) and (50), and since |fW κ, by (51), it follows that there is a constant C < ∞ such that for all sufficiently large n and all Nash equilibria, |fˆW (θ)| ≤ C/|θ| ∀ θ 6= 0.

Thus, setting g(θ) = C/|θ| for all |θ| ≥ 2C, we have a uniform bound for the Fourier transforms fˆW (θ) in the region (ii). Finally, to bound |fˆW (θ)| in the region (iii) of intermediate θ−values, we use the proportionality rule once again to deduce that |w(u) − u| < ǫ. Therefore, fˆW (θ) =

Z

B

−A B

=

Z

eiθw(u) fU (u) du iθu

e

fU (u) du +

−A

= fˆU (θ) + R(θ)

Z

B

−A

(eiθw(u) − eiθu )fU (u) du

where |R(θ)| < ǫ′ uniformly for |θ| ≤ C and ǫ′ → 0 as ǫ → 0. Since fˆU is the Fourier transform of an absolutely continuous probability density, its absolute value is bounded away from 1 on the complement of [−γ, γ], for any γ > 0. Since ǫ > 0 can be made arbitrarily small (cf. Proposition 3), it follows that there is a continuous, positive function g(θ) that is bounded away from 1 on |θ| ∈ [γ, C] such that |fˆW )θ| ≤ g(θ) for all |θ| ∈ [γ, C]. The extension of g to the whole real line can now be done by smoothly interpolating at the boundaries of regions (i), (ii), and (iii). Proof of Theorem 7. We have already observed, in the proof of Lemma 5.2, that for any ǫ > 0, if n is sufficiently large then for any Nash equilibrium, |Ew(U )| < ǫ. It therefore follows from the proportionality rule that 4 var(v(U )) E|v(u) − Ev(u)|k (52) (Eψ(S ))2 σ 2 − 1 ≤ ǫ and (Eψ(Sn ))k E|U |k < ǫ ∀ k ≤ 8. n U Moreover, Lemma 5.2 and Theorem 8 imply that the distribution of Sn has a density with an Edgeworth expansion, and so for any continuous function ϕ supported by [−δ, δ], with

24

σV2 := var(v(U )), δ

2

e−y /2 ϕ(x) √ Eϕ(Sn ) = 2πnσV −δ Z

1+

8 X

n

−(k−2)/2

!

Pk (y)

k=3

dx + rn (ϕ)

(53)

p where y = y(x) = (x − ESn )/ var(Sn ) and Pk (y) = Ck Hk (y) is a multiple of the kth Hermite polynomial. The constants Ck depend only on the first k moments of w(U ), and consequently are uniformly bounded by constants Ck′ not depending on n or on the choice of Nash equilibrium. The error term rn (ϕ) satisfies ǫn |rn (ϕ)| ≤ 3 n

Z

δ

−δ

p

|ϕ(x)| dx. 2πvar(Sn )

(54)

The strategy of the proof will be to show that for every ǫ > 0, if n is sufficiently large then for every Nash equilibrium, |rn (ψ)| ≤ ǫEψ(Sn ) and p δ + |ESn | ≤ ǫ var(Sn ).

(55) (56)

It will then follows that in the expansion (53) the lead term dominates (because the integral will range over values of x for which |y(x)| ≤ ǫ), and so for large n, with ϕ = ψ, 1 Eψ(Sn ) = √ 2πnσV

Z

δ −δ

ψ(x) dx(1 ± ǫ).

Since 4 σV2 ≈ (Eψ(Sn ))2 σU2 this will imply that p

2

πn/2σU (Eψ(Sn )) =

Z

δ −δ

ψ(x) dx(1 ± ǫ) = 2 ± 2ǫ,

proving the assertion (7). Proof of (55). We will first argue that Eψ(Sn ) cannot be of smaller order of magnitude than n−1 . Fix C > 0, and suppose that Eψ(Sn ) ≤ Cn−1 . Then by the proportionality rule (3) and equation (12), for any ǫ > 0 and all sufficiently large n |ESn | ≤ nǫEψ(Sn )E|U | ≤ Cǫ. Thus, by Hoeffding’s inequality (Corollary 3.10), the distribution of Sn must be highly concentrated in a neighborhood of 0. But if this were so we would have, for all large n, Eψ(Sn ) ≈ ψ(0) > 0, which is a contradiction. This proves that for large n any Nash equilibrium must satisfy Eψ(Sn ) ≥ Cn−1 . 25

(57)

It now follows from (52) that var(Sn ) ≥ C 2 σU2 n−1 , and so by (54), ǫn kϕk1 |rn (ϕ)| ≤ √ . 2πCσU n5/2

(58)

Consequently, the error term rn (ϕ) in the Edgeworth expansion (53) is of smaller order of magnitude than n−3/2 Eψ(Sn ). This proves inequality (55). It follows, in particular, that the error in the expansion (53) for ϕ = ψ, is negligible, and hence for sufficiently large n every Nash equilibrium satisfies Z δ 1 ψ(x) dx + O(n−1 σV−1 ). Eψ(Sn ) ≤ √ 2πnσV −δ Since 4 σV2 ≈ (Eψ(Sn ))2 σU2 for large n, this implies that for a suitable constant κ < ∞, κ Eψ(Sn ) ≤ √ . 4 n

(59)

Proof of (56). p 2δ ≥ ǫ var(Sn ), or 2|ESn | ≥ p Inequality (56) could p fail in one of two ways: either ǫ var(Sn ). If 2δ ≥ ǫ var(Sn ), then var(v(U )) ≤ 4δ2 /(nǫ2 ), and so by (52), Eψ(Sn )2 ≤ 32δ2 σU2 /(nǫ2 ), at least for sufficiently large n. But in order that Eψ(Sn ) be this small, it would be necessary that |ESn | be large, because otherwise the Edgeworth expansion (53) with p ϕ = ψ would imply that Eψ(Sn ) is of order 1. Consequently, the inequality 2δ ≥ ǫ var(Sn ) would imply that |ESn | ≥ δ. This shows that to prove (56) it suffices to prove that for any ǫ > 0 if n is suffices large, p |ESn | ≤ ǫ var(Sn ), or equivalently, by (52),

√ |Ev(U )| ≤ ǫEψ(Sn )/ n.

(60)

Recall again the necessary condition 2v(u) = Eψ(v(u) + Sn )u for a Nash equilibrium; since ψ is infinitely differentiable and has compact support, the expectation can be expanded in a Taylor series: 2v(u) =

1 X k=0

1 v (u) + Sn )v(u)2 u. Eψ (k) (Sn )(v(u)k u) + Eψ ′′ (˜ 2

(61)

where v˜(u) is a point intermediate between 0 and v(u). Since EU = 0, the expectation of the k = 0 term in (61) is 0, and so 1 2Ev(U ) = Eψ ′ (Sn )E(v(U )U ) + Eψ ′′ (Sn + v˜(U ))E(v(U )2 U ). 2

(62)

To complete the proof of inequality (60) we will bound each of the two terms on the right side of (62) separately. The easier of the two terms is the second, so we dispose of it first. 26

Second order term. We will prove that for any ǫ > 0 and all large n, any Nash equilibrium must satisfy √ |Eψ ′′ (Sn + v˜(U ))v(U )2 U | ≤ ǫEψ(Sn )/ n. (63) First-order Taylor expansion gives Eψ ′′ (Sn + v˜(U ))v(U )2 U = Eψ ′′ (Sn )E(v(U )2 U ) + Eψ ′′′ (ˆ v (U ) + Sn )˜ v (U )v(U )2 U where vˆ(u) is a point intermediate between 0 and v˜(u) and v˜(u) is intermediate between 0 and v(u). By the proportionality rule, |v(u)| ≤ 2(A ∨ B)Eψ(Sn ) for large n, and so for a suitable C < ∞ |Eψ ′′′ (ˆ v (U ) + Sn )˜ v (U )v(U )2 U | ≤ Cρ(Eψ(Sn ))3 . √ where ρ = supv∈R P {Sn + v ∈ [−δ, δ]}. Since Eψ(Sn ) ≤ κ/ 4 n, by (59), this expression √ will be bounded by ǫEψ(Sn )/ n provided ρ < ǫ/C. By Proposition 3.3, this will be true for large n unless σV2 ≤ C ′ /n. But σV2 ≈ Eψ(Sn )2 σU2 /4 for large n, so if ρ ≥ ǫ/C then (Eψ(Sn ))2 ≤ C ′′ /n. Thus, for large n, √ (64) |Eψ ′′′ (ˆ v (U ) + Sn )˜ v (U )v(U )2 U | < ǫEψ(Sn )/ n. Next, consider the term Eψ ′′ (Sn )E(v(U )2 U ). Using the Taylor expansion (61) for v(u) together with the hypothesis that EU 3 = 0, we obtain Eψ ′′ (Sn )E(v(U )2 U ) =Eψ ′′ (Sn )(Eψ(Sn ))2 EU 3 + 2Eψ ′′ (Sn )Eψ(Sn )Eψ ′ (˜ v (U ) + Sn )v(U )U 3 + Eψ ′′ (Sn )(Eψ ′ (˜ v (U ) + Sn ))2 v(U )2 U 3 =2Eψ ′′ (Sn )Eψ(Sn )Eψ ′ (˜ v (U ) + Sn )v(U )U 3 + Eψ ′′ (Sn )(Eψ ′ (˜ v (U ) + Sn ))2 v(U )2 U 3 Using the proportionality rule as before, we have for large n |Eψ ′′ (Sn )Eψ(Sn )Eψ ′ (˜ v (U ) + Sn )v(U )U 3 | ≤ Cρ2 Eψ(Sn )2 ′′



2

2

3

3

and

2

|Eψ (Sn )(Eψ (˜ v (U ) + Sn )) v(U ) U | ≤ Cρ Eψ(Sn )

By Proposition 3.3 (using again the fact that σV2 ≈ (Eψ(Sn ))2 σU2 /4), there is a constant C ′ < ∞ such that for all large n, C′ , ρ≤ √ nEψ(Sn ) and so for large n, Cρ2 Eψ(Sn )2 ≤ ǫ

Eψ(Sn ) √ n

√ unless Eψ(Sn ) ≤ C ′′ / n. But if this is the case then by Lemma 3.8, there exist ǫn → 0 such that |Eψ ′′ (Sn )| ≤ ǫn ,

27

√ and so if Eψ(Sn ) ≤ C ′′ / n then

√ |Eψ ′′ (Sn )Eψ(Sn )Eψ ′ (˜ v (U ) + Sn )v(U )U 3 | ≤ Cρǫn Eψ(Sn )2 ≤ C ′′′′ ǫn Eψ(Sn )/ n and √ |Eψ ′′ (Sn )(Eψ ′ (˜ v (U ) + Sn ))2 v(U )2 U 3 | ≤ Cρ2 ǫn Eψ(Sn )2 ≤ C ′′′′ ǫn Eψ(Sn )/ n.

This proves that for any ǫ > 0, if n is large then √ |Eψ ′′ (Sn )E(v(U )2 U )| ≤ ǫEψ(Sn )/ n.

(65)

Together with inequality (64), this implies (63).

First order term. Our analysis of the second-order term in the expansion (62) proves that for any ǫ > 0 if n is sufficiently large then √ |2Ev(U ) − Eψ ′ (Sn )E(v(U )U )| ≤ ǫEψ(Sn )/ n. Consequently, to prove (60), it will suffice to show that for any ǫ > 0 if n is sufficiently large then √ Eψ ′ (Sn )E(v(U )U ) ≤ ǫEψ(Sn )/ n if Ev(U ) > 0, and √ Eψ ′ (Sn )E(v(U )U ) ≥ −ǫEψ(Sn )/ n if Ev(U ) < 0. Now v(U )U ≥ 0 and |v(U )| ≤ 2(A ∨ B)Eψ(Sn ), so it will suffice to prove that for any ǫ > 0 if n is large then Eψ ′ (Sn ) ≤ 1/n1/2+ǫ ′

Eψ (Sn ) ≥ −1/n

if Ev(U ) > 0 and

1/2+ǫ

(66)

if Ev(U ) < 0 .

Assume that Ev(U ) > 0; the opposite case can be handled in the same manner. To establish the inequality (66) we will use the Edgeworth series (53) with φ = ψ ′ . By inequality (58), the error term in the Edgeworth expansion (53) for Eψ ′ (Sn ) is bounded by a constant multiple of ǫn kψ ′ k1 /n5/2 , and hence this error can be ignored. The dominant term in the Edgeworth expansion (53) is Z δ 2 e−y /2 ′ √ ψ (x) dx 2πnσV −δ p where y = y(x) = ((x−ESn )/ var(Sn )). If Ev(U ) > 0 then ESn > 0 and so exp{−y(x)2 /2} is an increasing function on [−δ, δ]. But ψ ′ (x) is an odd function of x that is negative for x > 0; consequently, if Ev(U ) > 0 then the dominant term is negative. (Similarly, if Ev(U ) < 0 then the dominant term is positive.) Therefore, to prove the inequality (66) it suffices to show that the higher-order terms 3 ≤ k ≤ 8 in the Edgeworth expansion are bounded in absolute value by n−1/2+ǫ . The k = 3 term in (53) is T3 := √

1 2πnσV

Z

δ

ψ ′ (x)e−y

−δ

28

2 /2

P3 (y) dx

where P3 (y) = C3 H3 (y) and H3 (x) = x3 − x is the third Hermite polynomial. The constant Ck is a continuous function of the first three moments of w(U ), which by (52) are uniformly 2 bounded. Since the function H3 (y)e−y /2 is bounded on R, and since σV ≈ σU Eψ(Sn )/2, by (52), it follows that for a suitable constant κ < ∞, if n is sufficiently large then T3 ≤

κ . nEψ(Sn )

The terms 4 ≤ k ≤ 8 in the expansion (53) can be bounded in similar fashion: Tk ≤

κ n(k−2)/2 Eψ(S

n)

.

Thus, if Eψ(Sn ) ≥ n−1/2+2ǫ then the bound (66) will hold for all large n.

It remains to prove that (66) remains valid when Eψ(Sn ) ≤ n−1/2+2ǫ , and since the dominant term in the Edgeworth expansion is negative when Ev(U ) > 0, it suffices to consider the terms Tk for 3 ≤ k ≤ 8. Consider the Edgeworth expansion for Eψ(Sn ): ! Z δ 8 X 1 −y 2 /2 −(k−2)/2 Eψ(Sn ) = √ 1+ ψ(x)e n Pk (y) dx + rn (ψ). 2πnσV −δ k=3

Since Eψ(Sn ) ≥ Cn−1 , by (57), and since ψ ≥ 0, the terms 3 ≤ k ≤ 8 and the remainder term are of smaller order of magnitude than the dominant term, and so for large n, 1 Eψ(Sn ) ≥ √ 2 2πnσV

Z

δ

ψ(x)e−y

2 /2

−δ

dx ≥ (C/4)n−1 .

2

Hence, since ψ is even and e−y /2 is monotone in x on x ∈ [−δ, δ], there exists constants 0 < C ′ < C ′′ such that for all large n, 2

e−y(x) /2 C ′ n−1 ≤ √ ≤ C ′′ Eψ(Sn ) for all x ∈ [−δ, δ]. 4 2πnσV √ This implies that |y(x)| ≤ C ′′′ log n for all x ∈ [−δ, δ] (since nσV is bounded below), and so for an appropriate constant C ∗ , |Pk (y)| ≤ C ∗ (log n)4

for each 3 ≤ k ≤ 8 and x ∈ [−δ, δ]

where Pk (y) are the polynomials in the Edgeworth expansion (53). Therefore, for each 3 ≤ k ≤ 8, Z δ 1 2 ψ ′ (x)e−y /2 Pk (y) dx Tk = √ 2πn(k−2/2) σV −δ ≤ C ∗ C ′′′ n−1/2 (log n)4 Eψ(Sn ) ≤ C ∗ C ′′′ n−1+2ǫ

for all sufficiently large n. This proves (66). 29

6 Appendix: Proof of Proposition 3.3 Proposition 3.3. Fix δ > 0. For any ǫ > 0 and any C < ∞ there exists β = β(ǫ, C) > 0 and n′ = n′ (ǫ, C) < ∞ such that the following statement is true: if n ≥ n′ and Y1 , Y2 , . . . , Yn are independent random variables such that E|Y1 − EY1 |3 ≤ Cvar(Y1 )3/2

and var(Y1 ) ≥ β/n P then for every interval J ⊂ R of length δ or greater, the sum Sn = ni=1 Yi satisfies P {Sn ∈ J} ≤ ǫ|J|/δ.

(67)

(68)

Proof. It suffices to prove this for intervals of length δ, because any interval of length nδ can be partitioned into n pairwise disjoint intervals each of length δ. Without loss of generality, EY1 = 0 and δ = 1 (if not, translate and re-scale). Let g be a nonnegative, even, C ∞ function with kgk∞ = 1 that takes the value 1 on [− 12 , 12 ] and is identically zero outside [−1, 1]. It is enough to show that for any x ∈ R, Eg(Sn + x) ≤ ǫ. Since g is C ∞ and has compact support, its Fourier transform is real-valued and integrable, so the Fourier inversion theorem implies that Z 1 gˆ(θ)ϕ(−θ)n e−iθx dθ, Eg(Sn + x) = 2π where ϕ(θ) = EeiθY1 is the characteristic function of Y1 . Because EY1 = 0, the derivative of the characteristic function at θ = 0 is 0, and hence ϕ has Taylor expansion 1 1 |1 − ϕ(θ) − EY12 θ 2 | ≤ E|Y1 |3 |θ|3 . 2 6 Consequently, if the hypotheses (67) hold then for any γ > 0, if n is sufficiently large, |ϕ(θ)n | ≤ e−β

2 θ 2 /4

for all |θ| ≤ γ. This implies (since |ˆ g | ≤ 2) that Z Z 1 1 −β 2 θ 2 /4 e dθ + |ˆ g (θ)| dθ Eg(Sn + x) ≤ π |θ|