On the Distortion Value of the Elections with Abstention
OOn the Distortion Value of Elections with Abstention
Mohammad Ghodsi , Mohamad Latifian , and Masoud Seddighin Sharif University of Technology Institute for Research in Fundamental Sciences (IPM)August 4, 2020
Abstract
In Spatial Voting Theory, distortion is a measure of how good the winner is. It has been proved thatno deterministic voting mechanism can guarantee a distortion better than 3, even for simple metrics suchas a line. In this study, we wish to answer the following question: how does the distortion value changeif we allow less motivated agents to abstain from the election?We consider an election with two candidates and suggest an abstention model, which is a general formof the abstention model proposed by Kirchg¨assner [26]. Our results characterize the distortion value andprovide a rather complete picture of the model. The goal in Social Choice Theory is to design mechanisms that aggregate agents’ preferences into a collectivedecision. Voting is a well-studied method for aggregating preferences with many applications in artificialintelligence and multi-agent systems. Roughly, a voting mechanism takes the preferences of the agents overa set of alternatives and selects one of them as winner.One fruitful approach to estimate the quality of a voting mechanism is to use the utilitarian view whichassumes that each agent has cost over the alternatives [33, 11, 9, 7, 24, 6]. For example, spatial modelslocate the voters and the alternatives in a finite metric space M , and the cost of voter v i for Alternative x equals to their distance [4, 2, 3, 23, 8, 5, 28]. Considering these costs, the optimal candidate is definedto be the candidate that minimizes the social cost (the total cost of the voters). Ideally, we would like theoptimal candidate to be the winner; however, since voting mechanisms only take the ordinal preferences ofvoters as input, it is reasonable to expect that the winner is not always optimal. The question then arises:how good is the winner, i.e., what is the worst-case ratio of the social cost of the winner to the social costof the optimal candidate? This ratio is called the distortion value of a voting mechanism. It is known thatno deterministic voting mechanism can guarantee a distortion better than 3, even for simple metrics such asa line [4]. To see this, consider the example shown in Figure 1. In this example, candidate (cid:96) is the optimalcandidate, and under the plurality voting rule candidate r is the winner. Thus, the distortion value is0 . . − ε ) + 0 . · . . ε ) (cid:39) . However, the example of Figure 1 seems unrealistic in some ways. Although the voters located near thepoint 0 . r , they have a very low incentive to vote for r , since their costs for both candidatesare almost equal. On the other hand, agents located at 0 have a strong incentive to vote for (cid:96) . Indeed, if A preliminary version of this paper is accepted in AAAI 2019. For two candidates, all the well-known deterministic voting mechanisms (e.g. Borda, k -approval, Copeland, etc) turn intoplurality. a r X i v : . [ c s . G T ] A ug rε
49% 51% { Figure 1: An example with distortion value close to 3. In this example, 49% of the voters are located atpoint 0 and 51% of the voters are located at point 0 . (cid:15) . In addition, candidates L and R are respectivelylocated at points 0 and 1.voters are allowed to abstain, which is a natural assumption in many real-world elections, we expect (cid:96) to bethe winner rather than r . In this study, our goal is to tackle this problem:How does the distortion value change, if we allow less motivated agents to abstain? Scientists have long studied the factors affecting participation in an election. For example, Wolfinger andRosenstone [19] argue that more educated voters participate with a higher probability, or Lijphart [27]discusses that the voters on the left side of the political spectrum participate less frequently. Similarly, thedecision to vote may rely on variables such as income level or the sense of civic duty [37].Traditionally, both game-theoretic and decision-theoretic models of turnout have been proposed. At theheart of most of these models lies the assumption that there are costs for voting. These costs include thecosts of collecting and processing information, waiting in the queue and voting itself. Presumably, if a voterdecides to abstain, she does not have to pay these costs. Therefore, a rational voter must receive utility fromvoting. There is evidence suggesting that voters behave strategically when deciding to vote and take thecosts and benefits into account. For example, Riker and Ordeshook [35] show that the turnout is inverselyrelated to voting costs.Apart from social-psychological traits, other studies suggest that voters’ abstention may stem from theirideological distances from the candidates. The work of Downs [16] initiated this line of research. He arguesthat in a two-candidate election under the majority rule, the choice between voting and abstaining is relatedto the voter’s comparative evaluation of the candidates. Riker and Ordeshook [35] later improve this modelby reformulating the original equation to incorporate other social and psychological factors.Many empirical studies in spatial theory of abstention suggest that the voters are more likely to abstainwhen they feel indifferent toward the candidates or alienated from them [26]. The models introduced byDowns [16] and Riker and Ordeshook [35] are only capable of explaining the indifference-based abstentionwhich occurs when the difference between the costs of candidates for a voter is too small to justify votingcosts. On the other hand, these models cannot justify alienation-based abstention, which occurs when avoter is too distant from the alternatives to justify voting costs. To alleviate this, some studies argue thatthe relative ideological distance plays a more critical role than the absolute distance [26, 21]. Our model ofabstention in this paper a generalization of the model introduced by Kirchg¨assner [26] which incorporatesthe relative distances.
In this paper, we consider the effect of abstention on the distortion value. In our study, there are twocandidates, and the voters decide whether to vote or abstain based on a comparison between the cost (i.e.,distance) of their preferred alternative and the cost of the other alternative. We define the concepts of expected winner and expected distortion to evaluate the distortion of an election in our model. Our resultscharacterize the distortion value and provide a complete picture of the model. For the special case that our There are other decision theoretic explanations of abstention that do not rely on costs, e.g., see [22]. . α > α ) D ∗ , where D ∗ is the distortion of the expected winner.Finally, we generalize our results to include arbitrary metric spaces. We show that the same upper boundsobtained for the distortion value for the line metric also work for any arbitrary metric space. The utilitarian view, which assumes that the voters have costs for each alternative, is a well-known approachin welfare economics [36, 30] and has received attention from the AI community during the past decade[33, 9, 10, 32, 3, 12, 23, 25, 1, 11]. Procaccia and Rosenschein [33] first introduced distortion as a benchmarkfor measuring the efficiency of a social choice rule in utilitarian settings. The worst-case distortion of manysocial choice functions is shown to be high or even unbounded. However, imposing some mild constraintson the cost functions yields strong positive results. One of these assumptions which is reasonable in manypolitical and social settings, is the spatial assumption which assumes that the agent costs form a metricspace [17, 28, 20, 4, 2, 31, 29].Anshelevich, Bhardwaj and Postl [4] were first to analyze the distortion of ordinal social choice func-tions when evaluated for metric preferences. For plurality and Borda rules, they prove that the worst-casedistortion is 2 m −
1, where m is the number of alternatives. On the positive side, they show that for theCopeland rule, the distortion value is at most 5. They also prove the lower bound of 3 for any deterministicvoting mechanism and conjecture that the worst-case distortion of Ranked Pairs social choice rule meetsthis lower-bound. This conjecture is later refuted by Goel, Krishnaswamy, and Munagala [23]. Recently,Munagala and Wang [29] present a weighted tournament rule with distortion of 4 . α -decisive metric spaces any randomized rule has a lower-bound of 1+ α on the distortion value. For the case of two alternatives, theypropose an optimal algorithm with the expected distortion of at most 1 + α . Cheng et al. [14] characterizedthe positional voting rules with constant expected distortion value (independent of the number of candidatesand the metric space).Chen, Dughmi, and Kempe [13] consider the case that candidates are drawn randomly from the populationof voters. They prove the tight bound of 1 . embedding into votingrules introduced by Caragiannis and Procaccia [11]. An embedding is a set of instructions that suggestseach agent how to vote, based only on the agents own utility function. For example, when the votingmechanism is majority, one possible embedding is that voters vote for each candidate with a probabilitywhich is proportion to their utility for that candidate. Among other results, Caragiannis and Procaccia [11]show that this embedding results in constant distortion. Indeed, our abstention model can be seen as aembedding for elections with majority rule where voters are allowed to abstain. In an α -decisive metric, for every voter, the cost of her preferred choice is at most α times the cost of her second bestchoice. Preliminaries
In our study, every election ξ consists of four ingredients: • A set V ξ of n voters. We denote the i ’th voter by v i . • A set C ξ candidates. In this study, we suppose that there are only two candidates and denote thecandidates by (cid:96) (left candidate) and r (right candidate). • A finite metric space M ξ where the candidates and the voters are located. Unless explicitly statedotherwise, we suppose that M ξ is a line, and (cid:96) and r are located respectively at points 0 and 1. Inaddition, each voter is attributed a value x i ∈ ( −∞ , ∞ ) which shows her location on the line. Wedenote by d i,a , the distance between voter v i and alternative a ∈ { (cid:96), r } . • A mechanism by which the winner is selected. In this paper, we consider a simple scenario where thewinning candidate is elected via the majority rule (in case of a tie, the winner is determined by tossinga fair coin). Note that for two candidates, almost all the well-known deterministic voting mechanismsselect the candidate preferred by the majority as winner.
Definition 2.1.
For an election ξ and candidate a ∈ { (cid:96), r } , we define the social cost of a in ξ as sc ξ ( a ) = (cid:88) v i ∈V ξ d i,a . The optimal candidate of election ξ , denoted by o is the candidate that minimizes the social cost, i.e., o ξ = arg min a ∈{ (cid:96),r } sc ξ ( a ) . We suppose that each voter either abstains or votes for one of the candidates. In Section 2.1 we give aformal description of the voting behavior of the agents.
We employ a simple probabilistic model, where each voter independently decides whether to abstain orparticipate by evaluating her distances from the candidates. Fix an election ξ and a voter v i ∈ V ξ and let a ∈ { (cid:96), r } be the candidate closer to v i in M ξ and ¯ a be the other candidate. We suppose that v i votessincerely for her preferred candidate a with a probability p i where p i is a function of d i,a and d i, ¯ a , andabstains with probability 1 − p i .Denote by f the probability function from which p i is derived, i.e., p i = f ( d i,a , d i, ¯ a ) . Since f representsthe probability of voting, we expect f to satisfy certain axiomatic assumptions. Recall that in spatial votingmodels, there are two crucial sources of abstention [26]: • Indifference-based Abstention (IA): the smaller the difference between the distances of a voterfrom the candidates is, the less likely it is that she casts a vote. • Alienation-based Abstention (AA): the further a voter is located from his preferred candidate,the less likely it is that she casts a vote.To illustrate, for the voters in Figure 2, we have: • Voters v , v , and v prefer (cid:96) and voters v and v prefer r . • Voter v has a strong incentive to cast a vote since her cost for (cid:96) is zero. • Voter v always abstains, since her costs for both the candidates are equal (IA).4 rv v v v v v Figure 2: A simple election.Figure 3: f β for different values of β , when two candidates (cid:96) and r are located at points 0 and 1. For anypoint z , the curves show f β ( | z − | , | z − | ) for different β . • For voters v and v , we have p ≥ p , since v is more alienated (AA). • For voters v and v , we have p ≥ p , since d ,(cid:96) ≤ d ,(cid:96) , and d ,r − d ,(cid:96) ≥ d ,r − d ,(cid:96) (IA,AA).As mentioned, the models of Downs [16] and Riker and Ordeshook [35] are only capable of explainingthe Indifference-based abstention, since they only consider the absolute difference between the distances ofthe candidates to a voter. To resolve this, some recent studies argue that the relative distance, rather thanabsolute distance, is relevant. In this study, we follow the model of Kirchg¨assner [26] which is based onthe relative distances. The idea is that the probability that a voter casts a vote depends on her ability todistinguish between the candidates. By WeberFechner’s law (see [18]), the ability to distinguish between thecandidates depends on their relative distances to the voter. Formally, the probability p i that voter v i votesfor a is calculated via the following formula: p i = f ( d i,a , d i, ¯ a ) = | d i,a − d i, ¯ a | d i,a + d i, ¯ a . (1)Here we consider a more general form of Equation (1). We suppose that each voter v i in election ξ casts avote with probability p i , where p i = f β ( d i,a , d i, ¯ a ) = (cid:18) | d i,a − d i, ¯ a | d i, ¯ a + d i,a (cid:19) β , (2)where β is a constant in [0 , f β for different values of β and differentlocations on the line. As is clear from Figure 3, for the smaller values of β , voters are more eager toparticipate. Indeed, the exponent β can be seen as a quantitative measure of how much this ideologicaldistance matters. For the special case of β = 0, voters always participate in the election, regardless of theirlocation. We refer to β as the participation parameter. It can be easily observed that for any 0 ≤ β ≤ f β satisfies all the desired criteria. As mentioned, our assumption is that the winner is determined by the majority rule. However, according tothe stochastic behavior of the voters, the winner is not deterministic, i.e., each candidate has a probability In few cases, we also use (cid:96) (cid:48) and r (cid:48) to refer to the left and right candidates.
5f winning. Denote by a β , the expected number of voters who vote for candidate a , when the participationparameter is β . Furthermore, denote by P a,β , the probability that candidate a wins the election, when theparticipation probability is β . We define the expected winner of election ξ for participation parameter β ,denoted by ω ξ,β as the candidate with the maximum expected number of votes. ω ξ,β = arg max a ∈C ξ a β . Definition 2.2.
For election ξ and a candidate a ∈ { (cid:96), r } we define the distortion of a in election ξ , denotedby D ( a ) , as the ratio sc ξ ( a ) / sc ξ ( o ξ ) . By definition, the distortion of the optimal candidate is 1. We discuss two approaches to evaluate thedistortion of an election ξ . In the first approach, we evaluate election ξ by the distortion of its expectedwinner, i.e., D ( ω ξ,β ). Another approach is to define the distortion of election ξ as the expected distortion ofthe winner, over all possible outcomes, i.e.,¯ D β ( ξ ) = P (cid:96),β · D ( (cid:96) ) + P r,β · D ( r ) . (3)Finally, for any 0 ≤ β ≤
1, we define worst-case distortion values D ∗ β and ¯ D ∗ β as: D ∗ β = max ξ ∈ Ω D ( ω ξ,β ) , and ¯ D ∗ β = max ξ ∈ Ω ¯ D β ( ξ ) , where Ω is the set of all possible elections ξ . We dedicate two separate sections to analyze the value of D ∗ β and ¯ D ∗ β . Even though the value of D ∗ β and ¯ D ∗ β essentially depend on β , we provide necessary tools to analyzedistortion these values for any β ∈ [0 , β is fixed we drop the subscript ‘ β ’ and simply use P (cid:96) , a instead of P (cid:96),β , a β . Throughout this section, we analyze the worst-case distortion of the expected winner. Recall that theexpected winner is the candidate with a higher expected number of votes. There are two reasons why weconsider the distortion value of the expected winner. First, since the number of votes that each candidatereceives is concentrated around its expectation , in elections with a large number of voters, the expectedwinner has a very high chance of winning; especially when there is a non-negligible separation between theexpected number of votes that each candidate receives. Secondly, we use the tight upper-bound on thedistortion value of the expected winner to prove an upper bound on the expected distortion of the electionfor the second approach. Recall that the probability that a voter v i casts a vote for his favorite candidate inelection ξ is: f β = (cid:18) | d i,(cid:96) − d i,r | d i,(cid:96) + d i,r (cid:19) β . In this section, we suppose without loss of generality that candidate (cid:96) is the expected winner. Moreover,we assume that the optimal candidate is r ; otherwise the distortion equals 1. We also consider four regions A , B , C and D as in Figure 4.In Theorem 3.1 we state the main result of this section. We can show this claim using concentration bounds such as
Hoeffding . A simple form of this inequality states that for n independent random variables bounded by [0 , P ( S n − E [ S n ] > t ) ≤ exp( − nt ) , where S n is the sum of the variables. B C D ℓ r Figure 4: Regions A , B , C , and D Theorem 3.1.
For any β ∈ [0 , , there exists an election ξ , such that D ( ω ξ,β ) = D ∗ β and the voters in ξ are located at two different locations x b ∈ B and x d ∈ D . The basic idea to prove Theorem 3.1 is as follows: we prove that for every election ξ , there exists anelection ξ (cid:48) with D ( ω ξ (cid:48) ,β ) ≥ D ( ω ξ,β ), such that the voters in ξ (cid:48) are located in at most 2 different locations.To show this, we collect the voters in ξ by carefully moving them forward and backward via a sequence of valid displacements , as defined in Definition 3.2. Definition 3.2.
Define a displacement as the operation of moving a subset of the voters forward or backwardon the line to a new location. A displacement is valid if it does not alter the expected winner, and furthermore,does not decrease the distortion value of the expected winner.
In Lemmas 3.3, 3.4, and 3.5 we introduce three sorts of valid displacement which help us collect the voters.For convenience, here we only state the lemmas and defer the proofs to Section 3.2. Figure 5, illustrates asummary of the valid displacements introduced in these lemmas. Note that these displacements are validfor any β ∈ [0 , Lemma 3.3.
Moving a voter v i from x i ∈ A to is a valid displacement. Lemma 3.4.
Consider voters v i and v j respectively at x i ∈ B and x j ∈ C . Then, • If d i,(cid:96) ≤ d j,r , moving v i to x i + x j − / and v j to / is a valid displacement. • If d i,(cid:96) > d j,r , moving v i to x i − x j and v j to is a valid displacement. Lemma 3.5.
Consider voters v i , v j , where x i , x j ∈ B or x i , x j ∈ D . Then moving both the voters to ( x i + x j ) / is a valid displacement. We also state two simple and natural Corollaries of Lemmas 3.4, and 3.5.
Corollary 3.6 (of Lemma 3.4) . We can move each voter in region C to either or / by a sequence ofvalid displacements.Proof. Consider an arbitrary voter v j ∈ (1 / , (cid:96) is the expected winner, there exists at least onevoter, say v i , in region B . By Lemma 3.4, if d i,(cid:96) ≤ d j,r , we can move v j to 1 / d i,(cid:96) > d j,r , we canmove v j to 1. Corollary 3.7 (of Lemma 3.5) . We can collect all the voters of region B at some point x ∈ B via a sequenceof valid displacements. Furthermore, we can collect all the voters of region D at some point x (cid:48) ∈ D via asequence of valid displacements.Proof. By applying Lemma 3.5 iteratively to the furthest voters, the maximum distance between the votersin each region decreases. This procedure can be applied until all the voters gather at one point.Now, we are ready to prove Theorem 3.1. 7 r A v i ℓ r CB v j v i v ′ i v ′ j ℓ r B D
Figure 5: Valid displacements introduced in Lemmas 3.3, 3.4, 3.5.
Proof of Theorem 3.1.
First, we prove that by Lemma 3.3 and Corollaries 3.6, and 3.7, every election ξ canbe reduced to an election ξ (cid:48) , such that • ξ and ξ (cid:48) have the same expected winner. • D ( ω ξ,β ) ≤ D ( ω ξ (cid:48) ,β ). • All the agents in ξ (cid:48) are located at two points x b ∈ B and x d ∈ D .Consider an arbitrary election ξ . Using Lemma 3.3, we move all the voters in region A to 0. Afterwards,using Corollary 3.6 we move each voter in region C to one of the points 1 / B or D (we suppose that the voters located in the borderlines belong to bothregions). Finally, using Corollary 3.7, we collect all the voters in regions B and D at some points x b ∈ B ,x d ∈ D .Finally, let ξ be an arbitrary election such that D ( ω ξ,β ) = D ∗ β . Applying the above reduction on ξ , yieldsan election ξ ∗ with D ( ω ξ ∗ ,β ) = D ∗ β , and the desired structure.According to Theorem 3.1, for any β ∈ [0 , ξ ∗ with the maximum distortion,and the following structure (see Figure 6): the interior of regions A and C contain no voter. All the votersare located at two points x b ∈ B and x d ∈ D . Note that, the maximum distortion value and the location of x b and x d in the worst-case scenario depends on the value of β . ℓ rx b x d x m Figure 6: For any β ∈ [0 , ξ ∗ with D ( ω ξ ∗ ,β ) = D ∗ β , and the above structure.8 .1 A Tight Upper Bound on D ∗ β We now evaluate D ∗ β for different values of 0 ≤ β ≤
1. Let us start by the boundary case β = 0. For β = 0,the probability that a voter casts a vote is 1 , independent of her location. It is proved that for this case,we have D ∗ β = 3 [4]. Indeed, the same example we provided in Figure 1 is the scenario with the highestdistortion for β = 0.Now, consider β >
0, and let ξ ∗ be the election that maximizes D ( ω ξ ∗ ,β ). As discussed in the previoussection, we can assume without loss of generality that the voters in ξ ∗ are located at two points, namely, x b ∈ B and x d ∈ D . Suppose that q b voters are at x b and q d voters are at x d . We have: (cid:96) = (1 − x b ) β q b and r = (cid:18) x d − β (cid:19) q d . Since (cid:96) is the expected winner, we have(1 − x b ) β q b ≥ (cid:18) x d − β (cid:19) q d . On the other hand, we have sc ξ ∗ ( (cid:96) ) = q b x b + q d x d , and sc ξ ∗ ( r ) = q b (1 − x b ) + q d ( x d − . Thus, D ( (cid:96) ) = sc ξ ∗ ( (cid:96) ) sc ξ ∗ ( r )= q b x b + q d x d q b (1 − x b ) + q d ( x d − q b x b + ( n − q b ) x d q b (1 − x b ) + ( n − q b )( x d − q b x b + ( n − q b ) x d q b (1 − x b ) + ( n − q b )( x d − . t . (1 − x b ) β q b ≥ n − q b (2 x d − β , ≤ q b ≤ , ≤ x b ≤ / , ≤ x d . (4)Now consider another boundary case: β = 1. For β = 1 the answer to the above optimization problemis (1+ √ √ (cid:39) . q b = n √ , x b = 0, and x d = √ . A graphicalrepresentation of this construction is shown in Figure 7.In general for 0 < β <
1, the maximum distortion value equals the answer of Optimization Problem (4).In Figure 8, we show the answer of this program for different values of β . Interestingly, with β increasingfrom 0 to 1, D ∗ β initially decreases and then increases. As illustrated in Figure 8, it can be seen that theminimum possible value for D ∗ β is (cid:39) √ β (cid:39) . r √ n √ n √ √ Figure 7: A tight example for β = 1.Figure 8: Worst-case distortion for 0 ≤ β ≤ In this section, we prove Lemmas 3.3, 3.4, and 3.5. One important tool to prove these lemmas is Observation3.8.
Observation 3.8.
Let a, b, c, d > be four positive constants. We have : • If ab > cd then a + cb + d < ab and a − cb − d > ab . • If ab < cd then a + cb + d > ab and a − cb − d < ab . Lemma 3.3.
Moving a voter v i from x i ∈ A to is a valid displacement.Proof. Initially, v i votes for (cid:96) with probability (cid:16) − x i (cid:17) β . After moving v i to 0, she votes for (cid:96) with probability1. Therefore, if we move v i to 0, the value of (cid:96) does not decrease, and the expected winner does not change.Furthermore, by this movement both sc ξ ( (cid:96) ) and sc ξ ( r ) are decreased by − x i . Let c and c (cid:48) be the contributionof v − i (that is, all voters except v i ) to the social cost of (cid:96) and r respectively. Before moving v i to 0, we have D ( (cid:96) ) = c − x i c (cid:48) + 1 − x i , and after the movement we have D ( (cid:96) ) = cc (cid:48) + 1= ( c − x i ) − ( − x i )( c (cid:48) + 1 − x i ) − ( − x i ) , (5) For the second inequalities, we assume d < b .
10y applying Observation 3.8 on Equation (5), we have c − x i c (cid:48) + 1 − x i ≤ ( c − x i ) − ( − x i )( c (cid:48) + 1 − x i ) − ( − x i )which implies that moving v i to 0 is a valid displacement. Lemma 3.4.
Consider voters v i and v j respectively at x i ∈ B and x j ∈ C . Then, • If d i,(cid:96) ≤ d j,r , moving v i to x i + x j − / and v j to / is a valid displacement. • If d i,(cid:96) > d j,r , moving v i to x i − x j and v j to is a valid displacement.Proof. Initially, v i votes for (cid:96) with probability (1 − x i ) β , and v j votes for r with probability (2 x j − β .Since these movements do not change the regions where the voters belong, after the movement they vote forthe same candidate but with different probabilities. Let ∆ i be the difference between the contribution of v i to (cid:96) , before and after the displacement. Similarly, let ∆ j be the difference between the contribution of v j to r before and after the movement. We consider two cases. Case I ( d i,(cid:96) ≤ d j,r ): if we move v i to x i + x j − / v j to 1 / v i votes for (cid:96) with probability(2 − x i − x j ) β and v j votes for r with probability 0. Thus, we have∆ i = (2 − x j − x i ) β − (1 − x i ) β and ∆ j = 0 β − (2 x j − β . Since β ≤
1, by straightforward calculus we have:((1 − x i ) − (2 x j − β ≥ (1 − x i ) β − (2 x j − β ((1 − x i ) − (2 x j − β − (1 − x i ) β ≥ − (2 x j − β ∆ i ≥ ∆ j . Case II ( d i,(cid:96) > d j,r ): if we move v i to x i + x j − v j to 1, after the displacement, v i votes for (cid:96) withprobability (3 − x i − x j ) β , and v j votes for r with probability 1. Therefore, we have∆ i = (3 − x i − x j ) β − (1 − x i ) β and ∆ j = 1 β − (2 x j − β . Since β ≤
1, we have ((1 − x i ) − (2 x j − β ≥ (1 − x i ) β − (2 x j − β ((1 − x i ) − (2 x j − β − (1 − x i ) β ≥ − (2 x j − β ((1 − x i ) − (2 x j − β − (1 − x i ) β ≥ β − (2 x j − β ∆ i ≥ ∆ j . Hence the expected winner does not change. In addition, since we move two voters in Regions B and C equallyin the opposite directions in both cases, the distortion value of each candidate remains unchanged. Lemma 3.5.
Consider voters v i , v j , where x i , x j ∈ B or x i , x j ∈ D . Then moving both the voters to ( x i + x j ) / is a valid displacement.Proof. Let ε = | x i − x j | /
2. Recall the definition of ∆ i and ∆ j from the proof of Lemma 3.4. For the case of x i , x j ∈ D , we have: ∆ i = (cid:18) x i + 2 ε − (cid:19) β − (cid:18) x i − (cid:19) β , g and x i < x j , we have g ( x i + ε ) − g ( x i ) ≤ g ( x j ) − g ( x j − ε ) ≤ j = (cid:18) x j − ε − (cid:19) β − (cid:18) x j − (cid:19) β . Thus, we have∆ i + ∆ j = (cid:18) x i + 2 ε − (cid:19) β − (cid:18) x i − (cid:19) β + (cid:18) x j − ε − (cid:19) β − (cid:18) x j − (cid:19) β . Since x j > x i , these two inequalities imply ∆ i + ∆ j ≤ r does not increaseand the expected winner does not change.Similarly For the case of x i , x j ∈ B , we have:∆ i = (1 − x i − ε ) β − (1 − x i ) β , and ∆ j = (1 − x j + 2 ε ) β − (1 − x j ) β . Thus, we have ∆ i + ∆ j = (1 − x i − ε ) β − (1 − x i ) β + (1 − x j + 2 ε ) β − (1 − x j ) β . Note that since f (cid:48) β ( x ) = (1 − x i ) β is a decreasing and concave function, we have d (cid:0) f (cid:48) β (cid:1) dx ≤ , and d (cid:0) f (cid:48) β (cid:1) dx ≤ . Since x j > x i , these two inequalities imply ∆ i + ∆ j ≥
0. Thus, value of (cid:96) does not decrease and theexpected winner does not change. In addition, since the voters move in the opposite directions and by thesame distance, the distortion value of the candidates do not change. Therefore, this modification is a validdisplacement. 12
Expected Distortion
Recall that in our second approach, we define the distortion of an election as the expected distortion of thewinner, where the expectation is taken over the random behavior of the voters. Our main result in thisSection is Theorem 4.1.
Theorem 4.1.
For any α > , value of ¯ D β ( ξ ) for every election ξ whose candidates receive at least φ ( α ) = ( α + 1) α ( α − √ α + 1) expected number of votes is at most (1 + 2 α ) D ∗ β . In this section, we suppose without loss of generality that candidate r is the optimal candidate. Thus,Equation (3) can be rewritten as ¯ D β ( ξ ) = P (cid:96) sc ξ ( (cid:96) ) sc ξ ( r ) + P r . (6)In this case, if (cid:96) would be the expected winner, we have:¯ D β ( ξ ) = P (cid:96) D ( (cid:96) ) + P r D ( r ) ≤ P (cid:96) D ( (cid:96) ) + P r D ( (cid:96) ) ( D ( (cid:96) ) ≥ D ( r ))= D ( (cid:96) ) . (7)In addition, we know that the distortion of the expected winner is at most D ∗ β , which together with Equation(7) implies ¯ D ( ξ ) ≤ D ∗ β for the case that (cid:96) is the expected winner. Therefore, throughout this section wesuppose that r is both the optimal and the expected winner candidate.In Theorem 4.2, we prove that there is an election with the maximum distortion value and a simplestructure. Theorem 4.2.
For any β ∈ [0 , , there exists an election ξ ∗ such that ¯ D β ( ξ ∗ ) is maximum, and in ξ ∗ thereis no voter in the interior of regions A and C , and also all the voters in D are located at a single point x d ∈ D . The basic idea to prove Theorem 4.2 is as follows: we prove that for every election ξ , there exists anelection ξ (cid:48) with ¯ D β ( ξ (cid:48) ) ≥ ¯ D β ( ξ ) and the desired structure. To show this, we collect some of the the votersin ξ via a sequence of valid displacements , albeit with a new definition for valid displacement. Definition 4.3.
A displacement is valid, if it does not decrease ¯ D ( ξ ) . The process of proving that a displacement is valid for this case is relatively tougher than the previousmodel. The reason is that we do not even have a closed-form expression which represents the winningprobability of each candidate. In Lemmas 4.4 and 4.5 we explain our tools to discover valid displacements.For brevity, we defer the proofs to these lemmas to Section 4.2.
Lemma 4.4.
For each voter v i ∈ A , there is a point x (cid:48) i ∈ B such that moving v i to x (cid:48) i is a valid displacement.Furthermore, for each voter v j ∈ C , there is a point x (cid:48) j ∈ D such that moving v j to x (cid:48) j is a valid displacement. Lemma 4.5.
Let v i and v j be two voters located respectively at x i , x j ∈ D . Then, there exists a point x between x i and x j , such that moving both the voters to x is a valid displacement. Figure 10, shows a summary of the displacements described in Lemmas 4.4 and 4.5. Using these displace-ments, one can establish an election with the maximum expected distortion, and the following structure (seeFigure 11): the interior of regions A and C contain no voter. All the voters in D are located at point x d ∈ D . Proof of Theorem 4.2.
Consider an election with the maximum expected distortion. By Lemma 4.4 we cansuppose that the interior of regions A and C are empty. Furthermore, by iteratively applying Lemma 4.5 onthe farthest pair of points in Region D , we can collect all the voters of D into a single point and transformthe election into one with the maximum distortion, and the desired structure.13 rℓ r Figure 10: Valid displacements introduced in Lemmas 4.4 and 4.5. r x d . ℓ β ∈ [0 , ¯ D ∗ β In this section, we discuss the value of ¯ D ∗ β , for any β ∈ [0 , β = 0. By a similar argument as in Section3.1, for β = 0 all the voters vote for their preferred candidate and so we have ¯ D ∗ = 3. For β >
0, we proveTheorem 4.1 which provides an asymptotic upper bound on ¯ D ∗ β for any β ∈ (0 , Theorem 4.1.
For any α > , value of ¯ D β ( ξ ) for every election ξ whose candidates receive at least φ ( α ) = ( α + 1) α ( α − √ α + 1) expected number of votes is at most (1 + 2 α ) D ∗ β . To prove Theorem 4.1, we first prove Lemmas 4.6 and 4.7.
Lemma 4.6.
Let α be a constant, and ξ be an election, with the property that r is the optimal and theexpected winner candidate, and r/ (cid:96) ≤ α . Then ¯ D β ( ξ ) ≤ (1 + 2 α ) D ∗ β . Proof.
To prove this lemma we add sufficient number of agents at point 0 to alter the expected winner to (cid:96) . After this operation, since (cid:96) is the expected winner, we know that the expected distortion of (cid:96) is at most D ∗ β . Next, based on the number of voters added at point 0, we bound the value of ¯ D β ( ξ ).Let s be the minimum number of voters we need to add at point 0 to convert (cid:96) to the expected winner.Since r ≤ (1 + α ) · (cid:96) , and each voter at point 0 contributes 1 to (cid:96) , we have s ≤ α · (cid:96). Let ξ (cid:48) be the election,after adding the agents at point 0. Since the expected winner of ξ (cid:48) is (cid:96) , the expected distortion of (cid:96) is upperbounded by D ∗ β : sc ξ (cid:48) ( (cid:96) ) sc ξ (cid:48) ( r ) ≤ D ∗ β . (8)14oreover, since we add the agents at point 0, their cost for candidate (cid:96) is zero and hence, sc ξ (cid:48) ( (cid:96) ) = sc ξ ( (cid:96) ).Thus, we have sc ξ ( (cid:96) ) / sc ξ ( r ) sc ξ (cid:48) ( (cid:96) ) / sc ξ (cid:48) ( r ) = sc ξ (cid:48) ( r ) sc ξ ( r ) (9)Now, we show that the ratio sc ξ (cid:48) ( r ) / sc ξ ( r ) is upper bounded by 1 + 2 α . First, let us calculate the explicitformulas of sc ξ ( r ) and sc ξ (cid:48) ( r ). As discussed before, we can assume that the agents in ξ are located either inRegion B or at point x d ∈ D . Let q d be the population of the voters that are located at x d . We have sc ξ ( r ) = (cid:88) v ∈V ξ : x v ∈ B (1 − x v ) + q d ( x d − . Furthermore, we have sc ξ (cid:48) ( r ) = sc ξ ( r ) + s, where s ≤ α · (cid:96) = α (cid:88) v ∈V ξ : x v ∈ B (1 − x v ) β . Thus, we have sc ξ (cid:48) ( r ) sc ξ ( r ) ≤ α (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) β (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) + q d ( x d − ≤ α · (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) β (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) , and since for any x ≤ / (1 − x ) β − x ≤ sc ξ (cid:48) ( r ) sc ξ ( r ) ≤ α. (10)Inequality (10) together with Equations (8) and (9) implies: sc ξ ( (cid:96) ) / sc ξ ( r ) D ∗ β ≤ α. Thus, by Equation (6), we have ¯ D β ( ξ ) ≤ P (cid:96) (1 + 2 α ) D ∗ β + P r , and since P (cid:96) + P r = 1 we conclude that ¯ D β ( ξ ) ≤ (1 + 2 α ) D ∗ β . Lemma 4.7.
Let α be a constant, and ξ be an election, with the property that r is the optimal and theexpected winner candidate, and r/ (cid:96) > α . Then, if the number of candidates would be large enough, wehave ¯ D β ( ξ ) ≤ (1 + 2 α ) D ∗ β .Proof. To prove Lemma 4.7, we use the fact that the number of votes that a candidate receives is concentrated15round it’s expected value. By definition, we have¯ D β ( ξ ) = P (cid:96) sc ξ ( (cid:96) ) sc ξ ( r ) + (1 − P (cid:96) )= P (cid:96) ( sc ξ ( (cid:96) ) sc ξ ( r ) −
1) + 1= P (cid:96) (cid:80) v ∈V ξ : x v ∈ B (2 x v −
1) + q d (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) + q d ( x d −
1) +1 ≤ P (cid:96) q d (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) + 1 ( x v < / ≤ P (cid:96) q d (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) + 1 . (11)Let ˆ (cid:96) and ˆ r be two random variables indicating the number of votes that (cid:96) and r receive in ξ , respectively.These two variables are the sum of i.i.d. Bernoulli variables each indicating whether a voter casts a vote ornot. Note that all the voters that contribute to ˆ r are located at the same point, but voters contributing toˆ (cid:96) might have different locations. Using these facts we can calculate the expected value and the variance of ˆ (cid:96) and ˆ r . We have: E [ˆ r ] = r = q d (2 x d − β , Var(ˆ r ) = σ r = q d (2 x d − β × (1 − x d − β ) , E [ˆ (cid:96) ] = (cid:96) = (cid:88) v ∈V ξ : x v ∈ B (1 − x v ) β , Var(ˆ (cid:96) ) = σ (cid:96) = (cid:88) v ∈V ξ : x v ∈ B (1 − x v ) β × (1 − (1 − x v ) β ) . Let t = (cid:96) + r √ α . Since t ∈ [ (cid:96), r ], we have P (ˆ (cid:96) ≥ ˆ r ) ≤ P (ˆ (cid:96) ≥ t ) + P (ˆ r ≤ t ) . (12)Now, since we know both the expected value and the variance of ˆ r and ˆ (cid:96) we can use Chebyshev’s inequalityto provide an upper bound on P (ˆ (cid:96) ≥ ˆ r ).Chebyshev’s inequality states that for a random variable T with finite expected value µ and finite non-zerovariance σ , and for any real number k > P ( | T − µ | ≥ k ) ≤ σ k . (13)Therefore we have: P (ˆ (cid:96) ≥ t ) ≤ P ( | ˆ (cid:96) − (cid:96) | ≥ r √ α ) ≤ σ (cid:96) α r ≤ (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) β × (1 − (1 − x v ) β ) r × (cid:96) ≤ (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) β × (1 − (1 − x v ) β ) r × (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) β ≤ r . (14)16n the other hand, P (ˆ r ≤ t ) ≤ P ( | ˆ r − r | ≥ r − (cid:96) − r √ α )) ≤ P ( | ˆ r − r | ≥ r − r α − r √ α )) ≤ σ rα +1+ α − α √ α (1+ α ) r = q d (2 x d − β × (1 − x d − β ) ( α −√ α +1) (1+ α ) r = (1 − x d − β ) ( α −√ α +1) (1+ α ) r . (15)Let φ ( α ) = ( α − √ α + 1) (1 + α ) . Putting Equations (12), (14) and (15) together we have: P (ˆ (cid:96) ≥ ˆ r ) ≤ r + 1 − x d − β f ( α ) r , and by Equation (11) we have:¯ D β ( ξ ) ≤ (cid:32) r + 1 − x d − β f ( α ) r (cid:33) × q d (cid:80) v ∈V ξ : x v ∈ B (1 − x v ) + 1= (cid:18) (2 x d − β + (2 x d − β − f ( α ) (cid:19) × (cid:96) + 1 . (16)Note that since x d is the only location more distant to (cid:96) than r , even for q d = 1 the distortion of candidate (cid:96) and consequently the distortion of the election is upper-bounded by x d /x d −
1. Therefore, if x d ≥ α + 1,the distortion of the election is upper bounded by 1 + 2 α (i.e. ¯ D β ( ξ ) ≤ (1 + 2 α ) D ∗ β ). So here we assume x d < α + 1. If we substitute α + 1 for x d in (16) we have:¯ D β ( ξ ) ≤ (cid:32)(cid:18) α + 1 (cid:19) β + ( α + 1) β − f ( α ) (cid:33) × (cid:96) + 1 ≤ (cid:18) α + 1 (cid:19) × (cid:18) f ( α ) (cid:19) × (cid:96) + 1= 1 + αα × (cid:18) α ) ( α − √ α + 1) (cid:19) × (cid:96) + 1 ≤ α + 1) α ( α − √ α + 1) × (cid:96) + 1 , (17)where the last line is due to the fact that 1 ≤ (1 + α ) ( α − √ α + 1) . (cid:96) ≥ ( α + 1) α ( α − √ α + 1) . By Equation (17) we have: ¯ D β ( ξ ) ≤ α + 1) α ( α − √ α + 1) × α ( α − √ α + 1) ( α + 1) + 1 ≤ α ≤ (1 + 2 α ) D ∗ β . This completes the proof.Now, we are ready to prove Theorem 4.1.
Proof of Theorem 4.1.
Fix any α > β ∈ [0 , ξ ∈ Ω β be an arbitrary election whose candidatesreceive at least ( α + 1) α ( α − √ α + 1) expected number of votes. Recall that our assumption is that r is both the optimal and the expected winner.Now, based on the value of r/ (cid:96) , there are two cases: either r/ (cid:96) ≤ α or r/ (cid:96) > α . For the firstcase, by Lemma 4.6 value of ¯ D β ( ξ ) is upper bounded by (1 + 2 α ) D ∗ β . For the second case, since (cid:96) ≥ ( α + 1) α ( α − √ α + 1) , by Lemma 4.7, the expected distortion is upper bounded by (1 + 2 α ) D ∗ β . Combining these two cases yieldsthe upper-bound of (1 + 2 α ) D ∗ β on ¯ D β ( ξ ).As an example, for α = 0 .
1, Theorem 4.1 states that for every election ξ whose candidates receive atleast 148 expected number of votes, the expected distortion is upper bounded by 1 . D ∗ β which for β = 1 is (cid:39) . D ∗ β . Example 1.
Consider Optimization Problem 4, with an additional constraint that (cid:96) ≥ r (1 + ε ) for a fixedconstant ε , and let D ∗∗ β be the answer of this optimization problem and ξ ∗∗ be its corresponding election. ByChernoff bound, for a large enough value of (cid:96) , candidate (cid:96) wins the election with a high probability, i.e., lim (cid:96) →∞ ¯ D β ( ξ ∗∗ ) (cid:39) D ( (cid:96) ) (cid:39) D ∗ β . Note that, the bound provided by Theorem 4.1 is almost tight; as the election size grows, the upperbounds of Theorem 4.1 tends to the distortion value of Example 1. However, for elections with a smallnumber of voters, the distortion value might be larger. For example, consider a simple scenario where thereis one voter located at point 1 + ε ∈ D and β = 1 (see Figure 12). For this case, the distortion value is P (cid:96) · sc ξ ( (cid:96) ) sc ξ ( r ) + P r = P (cid:96) · εε + P r = ε ε · εε + 1 + ε ε = 2 + 2 ε ε , which tends to 2 as ε →
0. We conjecture that this example is the worst possible scenario and value of ¯ D ∗ β is upper bounded by 2 for any election with any size while β = 1.18 ℓ ε Figure 12: An example with maximum expected distortion. ¯ D β ( ξ ) for β = 1 tends to 2 as ε → In this section, we prove Lemmas 4.4 and 4.5.
Lemma 4.4.
For each voter v i ∈ A , there is a point x (cid:48) i ∈ B such that moving v i to x (cid:48) i is a valid displacement.Furthermore, for each voter v j ∈ C , there is a point x (cid:48) j ∈ D such that moving v j to x (cid:48) j is a valid displacement.Proof. We prove the statement of Lemma 4.4 for regions A and B . Similar arguments can be used to provethe lemma for regions C and D . Let x i be the current location of v i in region A ( x i < v i casts a vote with probability 1 / (1 − x i ) β . Now, consider point x = − x i / (1 − x i ). We claim that an agentat x , votes for (cid:96) with the same probability as v i . First, note that since x i < ≤ − x i / (1 − x i ) ≤ / . Hence, the preferred candidate of the voter located at x is (cid:96) . Furthermore, for any agent at x , the probabilityof casting a vote is (1 − x ) β = (cid:18) − − x i − x i (cid:19) β = (cid:18) − x i + 2 x i − x i (cid:19) β = (cid:18) − x i (cid:19) β . Therefore, by moving v i from x i ∈ A to x (cid:48) i = − x i − x i , the probability that v i votes for (cid:96) remains the same.Let ξ (cid:48) be the election, after moving v i to x . We have sc ξ (cid:48) ( (cid:96) ) sc ξ (cid:48) ( r ) = sc ξ ( (cid:96) ) − (cid:2) − x i − ( − x i / (1 − x i )) (cid:3) sc ξ ( r ) − (cid:2) (1 − x i ) − ( − x i / (1 − x i )) (cid:3) . Since we have − x i − ( − x i / (1 − x i ))(1 − x i ) − ( − x i / (1 − x i )) ≤ ≤ sc ( (cid:96) ) sc ( r ) , using Observation 3.8 we conclude that sc ξ (cid:48) ( (cid:96) ) sc ξ (cid:48) ( r ) ≥ sc ξ ( (cid:96) ) sc ξ ( r ) which in turn implies that moving v i to x is a validdisplacement. Lemma 4.5.
Let v i and v j be two voters located respectively at x i , x j ∈ D . Then, there exists a point x between x i and x j , such that moving both the voters to x is a valid displacement.Proof. Assume without loss of generality that x i < x j . We show that we can move both the voters to point t = (cid:112) (2 x i −
1) (2 x j −
1) + 12 . v k be a random variable which is equal to 1, if v k casts a vote and 0 otherwise. In addition, let A k = P ( r wins the election | ˆ v i + ˆ v j = k )where 0 ≤ k ≤
2. Trivially, we have A ≤ A ≤ A , and P r = A · P (ˆ v i + ˆ v j = 0)+ A · P (ˆ v i + ˆ v j = 1)+ A · P (ˆ v i + ˆ v j = 2) . (18)Furthermore, note that we have P (ˆ v i + ˆ v j = 0) = (cid:18) − x i − β (cid:19) (cid:18) − x j − β (cid:19) = 1 + (2 x i − β (2 x j − β − (2 x i − β − (2 x j − β (2 x i − β (2 x j − β , P (ˆ v i + ˆ v j = 2) = 1(2 x i − β (2 x j − β . Let ˆ v (cid:48) i and ˆ v (cid:48) j be variables indicating whether v i and v j cast a vote or not, after the displacement. Wehave P (cid:0) ˆ v (cid:48) i + ˆ v (cid:48) j = 0 (cid:1) = (cid:18) − t − β (cid:19) = (cid:32) − (cid:112) (2 x i −
1) (2 x j − β (cid:33) = 1 + (2 x i − β (2 x j − β − (cid:113) (2 x i − β (2 x j − β (2 x i − β (2 x j − β , P (cid:0) ˆ v (cid:48) i + ˆ v (cid:48) j = 2 (cid:1) = 1(2 t − β = 1(2 x i − β (2 x j − β . Thus, we have P (ˆ v i + ˆ v j = 2) = P (cid:0) ˆ v (cid:48) i + ˆ v (cid:48) j = 2 (cid:1) . Now, we show P (ˆ v i + ˆ v j = 0) ≤ P (ˆ v (cid:48) i + ˆ v (cid:48) j = 0) . We have P (ˆ v i + ˆ v j = 0) − P (ˆ v (cid:48) i + ˆ v (cid:48) j = 0) = 2 (cid:113) (2 x i − β (2 x j − β − (2 x i − β − (2 x j − β (2 x i − β (2 x j − β . Since (2 x i − β (2 x j − β > (cid:113) (2 x i − β (2 x j − β − (2 x i − β − (2 x j − β ≤ , which is trivial due to the fact that2 (cid:113) (2 x i − β (2 x j − β − (2 x i − β − (2 x j − β = − (cid:18)(cid:113) (2 x i − β − (cid:113) (2 x j − β (cid:19) . (cid:88) ≤ k ≤ P (cid:0) ˆ v (cid:48) i + ˆ v (cid:48) j = k (cid:1) = 1 , we have P (ˆ v i + ˆ v j = 1) > P (cid:0) ˆ v (cid:48) i + ˆ v (cid:48) j = 1 (cid:1) . Considering Equation (18), and the fact that A ≤ A we conclude that after this movement, the valueof P r decreases and the value of P (cid:96) increases.Finally, let C and C (cid:48) be the cost of the agents other than v i and v j for (cid:96) and r , respectively. By definition,before the displacement, we have D ( (cid:96) ) = C + [ x i + x j ] C (cid:48) + [ x i + x j − v i and v j to point t , we have: D ( (cid:96) ) = C + [ (cid:112) (2 x i −
1) (2 x j −
1) + 1] C (cid:48) + [ (cid:112) (2 x i −
1) (2 x j − − . Again, by straightforward calculus, one can easily verify that x i + x j ≥ (cid:113) (2 x i −
1) (2 x j −
1) + 1 , Thus, after this displacement, both D ( (cid:96) ) and P (cid:96) increases, and so does the value of ¯ D β ( ξ ). We now extend our results to general metric spaces. Suppose that the voters and candidates are located inan arbitrary metric M . By definition, for every voter v i and candidates (cid:96), r we have: • d i,(cid:96) , d i,r ≥ • d i,(cid:96) + d i,r ≥ d (cid:96),r (triangle inequality).We suppose without loss of generality that d (cid:96),r = 1. For this case, we prove Theorem 5.1, which states thatfor every 0 ≤ β ≤
1, the same upper bounds we obtained on the distortion value for the line metric alsoworks for any arbitrary metric space.
Theorem 5.1.
For every election ξ in an arbitrary metric space, there exists an election ξ (cid:48) in line metric,such that D ( ω ξ,β ) ≤ D ( ω ξ (cid:48) ,β ) and ¯ D β ( ξ ) ≤ ¯ D β ( ξ (cid:48) ) .Proof. Let ξ be an election in an arbitrary metric space M ξ . Assume w.l.o.g. that candidate r is the optimalcandidate and let V ξ be the set of voters in election ξ . For each voter v i ∈ V ξ , let γ i = d i,(cid:96) /d i,r . Based onthe value of c i , we partition the voters into two subsets V + ξ and V − ξ , where V − ξ = { v i | γ i ≤ D ( (cid:96) ) }V + ξ = { v i | γ i > D ( (cid:96) ) } . ξ (cid:48) as follows: consider a line and two candidates (cid:96) (cid:48) , r (cid:48) located respectively at 0and 1. For each voter v i ∈ V − , we consider a voter v (cid:48) i in ξ (cid:48) , located at point x (cid:48) i = c i c i +1 . Since (cid:18) | d i,(cid:96) (cid:48) − d i,r (cid:48) | d i,(cid:96) (cid:48) + d i,r (cid:48) (cid:19) β = (cid:18) | x (cid:48) i − | (cid:19) β = (cid:12)(cid:12)(cid:12)(cid:12) γ i γ i + 1 − (cid:12)(cid:12)(cid:12)(cid:12) β = (cid:12)(cid:12)(cid:12)(cid:12) d i,(cid:96) /d i,r d i,(cid:96) /d i,r + 1 − (cid:12)(cid:12)(cid:12)(cid:12) β = (cid:12)(cid:12)(cid:12)(cid:12) d i,(cid:96) − d i,R d i,(cid:96) + d i,R (cid:12)(cid:12)(cid:12)(cid:12) β , both v i and v (cid:48) i participate in their corresponding elections with equal probabilities. Similarly, for each voter v i ∈ V + , we consider a voter v (cid:48) i located at point x i = γ i γ i − . Again, it can be observed that (cid:18) | d i,(cid:96) (cid:48) − d i,r (cid:48) | d i,(cid:96) (cid:48) + d i,r (cid:48) (cid:19) β = (cid:18) | d i,(cid:96) − d i,r | d i,(cid:96) + d i,r (cid:19) β . In conclusion, for every i , voters v i and v (cid:48) i cast a vote in their corresponding elections with equal probabilities.Thus, expected winners of ξ (cid:48) and ξ are the same, and we have P (cid:96) (cid:48) = P (cid:96) , P r (cid:48) = P r . (19)Now, we prove D ( (cid:96) ) ≤ D ( (cid:96) (cid:48) ). For convenience, let A = (cid:88) v i ∈V − d i,(cid:96) A (cid:48) = (cid:88) v i ∈V − d i,(cid:96) (cid:48) B = (cid:88) v i ∈V − d i,r B (cid:48) = (cid:88) v i ∈V − d i,r (cid:48) C = (cid:88) v i ∈V + d i,(cid:96) C (cid:48) = (cid:88) v i ∈V + d i,(cid:96) (cid:48) D = (cid:88) v i ∈V + d i,r D (cid:48) = (cid:88) v i ∈V + d i,r (cid:48) . Note that for each v i ∈ V − , d i,(cid:96) (cid:48) = d i,(cid:96) / ( d i,r + d i,(cid:96) ), d i,r (cid:48) = d i,r / ( d i,r + d i,(cid:96) ), and d i,r + d i,(cid:96) ≥
1. Hence d i,(cid:96) ≥ d i,(cid:96) (cid:48) and d i,r ≥ d i,r (cid:48) . Therefore we have A − A (cid:48) ≥ B − B (cid:48) ≥
0. In addition, A − A (cid:48) B − B (cid:48) = (cid:80) i ∈V − d i,(cid:96) − γ i γ i +1 (cid:80) i ∈V − d i,r − γ i +1 = (cid:80) i ∈V − d i,(cid:96) + γ i d i,(cid:96) − γ i γ i +1 (cid:80) i ∈V − d i,r + γ i d i,r − γ i +1 ≤ max i ∈V − d i,(cid:96) + γ i d i,(cid:96) − γ i d i,r + γ i d i,r −
1= max i ∈V − γ i d i,r + γ i d i,(cid:96) − γ i d i,r + d i,(cid:96) − γ i d i,r = d i,(cid:96) ) ≤ max i ∈V − γ i ≤ D ( (cid:96) ) (20)22n the other hand, for each v i ∈ V + , d i,(cid:96) (cid:48) = d i,(cid:96) / ( d i,(cid:96) − d i,r ), d i,r (cid:48) = d i,r / ( d i,(cid:96) − d i,r ), and d i,(cid:96) − d i,r ≤ d i,(cid:96) ≤ d i,(cid:96) (cid:48) and d i,r ≤ d i,r (cid:48) . Therefore we have C (cid:48) − C ≥ D (cid:48) − D ≥
0. Furthermore, we have: C (cid:48) − CD (cid:48) − D = (cid:80) i ∈V + γ i γ i − − d i,(cid:96) (cid:80) i ∈V + γ i − − d i,r = (cid:80) i ∈V + γ i − γ i d i,(cid:96) + d i,(cid:96) γ i − (cid:80) i ∈V + − γ i d i,r + d i,r γ i − ≥ min i ∈V + γ i − γ i d i,(cid:96) + d i,(cid:96) − γ i d i,r + d i,r = min i ∈V + γ i − γ i d i,(cid:96) + γ i d i,r − d i,(cid:96) + d i,r ( γ i d i,r = d i,(cid:96) ) ≥ min i ∈V + γ i ≥ D ( (cid:96) ) . (21)By Equations (20) and (21), and using Observation 3.8 we have( C (cid:48) − C ) − ( A − A (cid:48) )( D (cid:48) − D ) − ( B − B (cid:48) ) ≥ D ( (cid:96) ) , (22)and D ( (cid:96) ) = A + CB + D ≤ ( A + C ) + ( C (cid:48) − C ) − ( A − A (cid:48) )( B + D ) + ( D (cid:48) − D ) − ( B − B (cid:48) ) (Observation 3.8 and Equation (22))= C (cid:48) + A (cid:48) D (cid:48) + B (cid:48) = D ( (cid:96) (cid:48) ) . (23)Since the expected winner is the same in ξ and ξ (cid:48) , Inequality (23) immediately implies that D ( ω ξ,β ) ≤ D ( ω ξ (cid:48) ,β ) . Furthermore, considering Equations (3) ,(19), and (23) we have ¯ D β ( ξ ) ≤ ¯ D β ( ξ (cid:48) ) . In this study, we analyzed the distortion value in a spatial voting model with two candidates, when thevoters are allowed to abstain. The set of results in this paper provides a rather complete picture of themodel. Nevertheless, some important open questions remain open. • The most immediate open question is to analyze the expected distortion value of the elections for asmall number of voters. The counter-example in Section 4.1 refutes the existence of an upper boundbetter than 2. We believe that this example is the worst possible scenario. However, we don’t have aformal proof for this claim. • Another direction is to provide a closed-form expression for the distortion of the expected winner.Currently, the maximum distortion is obtained via a mathematical program, which is not even convex .Beyond the above direct questions, this research also initiates an interesting line of work and opens afruitful direction for the future research. In the following, we discuss two of these directions: Of course, we have a short note on how to reduce this program into a convex one, by eliminating some of the variables. In this paper, we focused on a majority election between two candidates. When more than two candi-dates are running, vote aggregation becomes more complex. One interesting direction is to generalizethe models in this paper for elections with more than two candidates and analyze the performanceof different well-established voting mechanisms such as Borda, k -approval, Veto, Ranked pairs, andCopland under abstention assumption. One can also consider abstention in evaluating the distortionof different randomized mechanisms. • Similar to the elections with no abstention, it seems that high distortion scenarios stem from the issueof representativeness of candidates. Cheng et. al. [14] show that when the candidates are of thepeople (i.e., they have the same distribution as the voters), distortion ratio improves to a constantupper-bound strictly better than 2 for general metrics. The question is, how does the distortion valuechange if we allow abstention in the societies that voters and candidates have the same distribution?
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