Vote Delegation and Misbehavior
VVote Delegation and Misbehavior
Hans Gersbach
CER-ETH and CEPRZ¨urichbergstrasse 188092 Zurich, [email protected]
Akaki Mamageishvili
CER-ETHZ¨urichbergstrasse 188092 Zurich, [email protected]
Manvir Schneider
CER-ETHZ¨urichbergstrasse 188092 Zurich, [email protected]
Last updated: February 18, 2021
Abstract
We study vote delegation with ”well-behaving” and ”misbehaving” agents and com-pare it with conventional voting. Typical examples are validation or governance tasks onblockchains. There is a majority of well-behaving agents, but since voting is costly, theymay want to abstain or delegate their vote to other agents. Misbehaving agents always vote.We compare conventional voting allowing for abstention with vote delegation. Preferencesof voters are private information and a positive outcome is achieved if well-behaving voterswin. We provide three insights: First, if the number of misbehaving voters, denoted by X ,is high, both voting methods fail to deliver a positive outcome. Second, if X is moderate,conventional voting delivers a positive outcome, while vote delegation fails with probabilityone. Third, if X is low, delegation delivers a positive outcome with a higher probabilitythan conventional voting. Finally, our results allow us assessing the performance of votedelegation which is known as ”liquid democracy”. a r X i v : . [ c s . G T ] F e b Introduction
We study vote delegation when preferences are private information and voting is costly. Votedelegation plays a crucial role in two contexts—blockchain governance and so-called ”liquiddemocracy”, that is, a democracy in which the electorate can choose between voting itself ordelegating its right to vote before each collective decision.Let us look at the standard problem of blockchains with proof-of-stake. Stakeholders verifytransactions, whereas verification entails some cost for the stakeholders. Yet, they may alsoreceive some rewards when they participate in transaction verification. Besides monetary con-siderations, the stakeholders’ objectives depend on their type: Either they are well-behavingand seek verification and continuation of the blockchain or they are misbehaving and want toharm, by not validating transactions, validating them with delay, or engineering fake trans-actions with double-spending. In effect, these misbehaving individuals are similar to a groupof voters who wants to obtain a “negative” voting outcome, that is, an undesirable collectivedecision which is detrimental to the majority.One key question concerning blockchain management is whether stakeholders should be al-lowed to delegate their stake to other stakeholders, which is identical to vote delegation. Suchvote delegation is part of the basic procedures of some blockchain governance, see e.g. Good-man (2014) and the proposal in Damg˚ard et al. (2020) . With vote delegation, well-behavingstakeholders could avoid the verification cost and earn a small return if their vote goes to other,well-behaving stakeholders. However, they incur a risk that their vote goes to misbehavingstakeholders who aim at making false transactions and disrupting the blockchain. If misbehav-ing agents can accumulate a majority of votes, a negative outcome occurs.In the context of democracies—representative or direct—, suggestions are discussed thatcitizens should have the option to delegate their voting rights to other citizens. Citizens whooften abstain could then exercise their voting rights indirectly by delegating their votes, so thatthe electorate as a whole might be better represented. Yet, one drawback of vote delegation isthat it may open opportunities for minorities to win with vote delegation, while the majoritywould prevail under conventional voting. Set-up and Results
To delegate a vote means to transfer one’s voting right to another voter. Say, for example, avoter v i delegates his/her vote to a voter v j . Then, voter v j has his/her vote plus the delegatedvotes s/he received—in total 2 votes. In general, a voter can obtain and use any number ofvotes, if as many votes are delegated to him/her. When a voter votes, all of his/her votes countfor one alternative, that is, a voter can not split his/her votes between different alternatives.We study vote delegation with well-behaving and misbehaving agents under two assump-tions. First, voting is costly for well-behaving agents. That means, if a well-behaving individualabstains or delegates his/her vote, s/he is better off than with voting as long as his/her actiondoes not affect the voting outcome. Second, we assume that the minority composed of mis-behaving voters always votes. The rationale is that this minority is composed of determinedagents who have either a strong desire to disrupt the functioning of the system, or derive autility from enforcing their minority view that is larger than any cost of voting. The latterassumption also allows to study vote delegation under the most unfavorable circumstances.We compare vote delegation with conventional voting. In vote delegation, the well-behavingagents’ decision to delegate depends on an assessment of whether to vote or to abstain andwhether their vote delegation would allow misbehaving agents to obtain a majority of votes.Typically, some fraction of well-behaving voters will delegate, while the rest will vote. Sincepreferences are private information, delegated votes go uniformly at random to either thosewell-behaving voters who vote or the misbehaving voters who (always) vote. We model thetotal number of well-behaving voters as a Poisson random variable. This assumption simplifies See page 7 in Damg˚ard et al. (2020). c , there are thresholds f ∗ ( c ) and n ∗ ( f ) suchthat for any number of misbehaving voters f above f ∗ and an expected number of well-behavingvoters above n ∗ , misbehaving voters will have the majority of votes and will win. This meansthat if the cost of voting is close to zero, for example, there must be many misbehaving votersand the total number of voters must be of order f , so that misbehaving voters can win. Applications
Our results have immediate implications for blockchains, i.e. that vote delegation shouldonly be allowed if it is guaranteed that the absolute number of misbehaving agents is belowa certain threshold. Otherwise, the risk for negative outcomes increases. Our results can alsohelp assess the performance of vote delegation in democracy, a form that is known as ”liquiddemocracy”. We can view misbehaving voters as a determined minority who will vote, no matterthe costs. In the same setting, the well-behaving voters can be viewed as majority voters, whoincur a cost and/or analyze rationally. If there is a sufficiently large minority in absolute termswho is determined to engage in its cause, vote delegation can lower the likelihood that themajority wins.The paper is organized as follows: In Section 2, we discuss the related literature. In Section 3,we introduce our model. In Section 4, we analyze the equilibria and state our main result. Theproofs are in Appendix A.
The rational voting model is discussed in Palfrey and Rosenthal (1983) and Ledyard (1984). Votedelegation in a network and its dangers are studied in G¨olz et al. (2018) and Kahng et al. (2018).From an algorithmic perspective, Kahng et al. (2018) find that there is a delegation procedurethat outperforms direct voting, depending on the information that the voters have. In our paper,we focus on vote delegation from a game-theoretic perspective, when preferences of agents areprivate information. Bloembergen et al. (2019) study vote delegation in a directed graph, but donot study the case where voters aim to maximize the chance that their type wins. In our case,voters want their type to win. That is, all misbehaving voters vote for the same undesirablealternative and all well-behaving voters vote for the same desirable alternative. Leonardos et al.(2020) study weighted voting on blockchain and consider delegation of stakes as well. Finally, theliterature on Poisson games started with Myerson (1998). In the costly voting setting, Taylorand Yildirim (2010) justify the use of Poisson games, which makes the analysis of pivotalprobabilities easier. This setting is particularly useful when the size of the electorate is large.Vote delegation can take place over several voting rounds, as discussed in Casella and Palfrey(2019). In such settings, any pair of voters ( v i , v j ) can write a contract about a vote exchange.A voter v i gives a vote to a voter v j in round r in exchange for a vote in round w . So muchfreedom regarding vote delegation can entail an attack, even with a single misbehaving voter.A misbehaving voter can choose a voting round t in the future. If t is larger than half thenumber of all voters, the misbehaving voter can engineer vote exchange contracts with themajority of the other voters, such that at voting round t , s/he has the majority of votes, and3hus compromises the entire system. This simple yet efficient attack called ” t -period attack”shows that a free intertemporal vote exchange can be very dangerous. In this paper, we explorethe performance of vote delegation when there is a single delegation round. We consider a society consisting of well-behaving and misbehaving voters. In our setting, thereis a good and a bad alternative. A well-behaving voter, if s/he votes at all, will vote for thegood alternative and hence incur a cost 0 < c (cid:54)
1. A misbehaving voter will always turn outand vote for the bad alternative to harm the system. We assume that misbehaving voters donot have any cost of voting. The assumption of Poisson games is that the total number ofwell-behaving voters N is distributed as a Poisson random variable with parameter n , where n is some positive real number. Moreover, we assume that the number of misbehaving voters f ∈ N is common knowledge.Each voter has the same strategy set consisting of voting and delegation. Delegation meansthat the vote of the delegating voter goes to some other voter. We consider a totally mixedNash equilibrium solution concept, where well-behaving voters randomize between voting anddelegating.Moreover, we consider a symmetric Bayesian Nash equilibrium solution concept, that is,all voters have the same probability of delegating. Let γ ∈ [0 ,
1] denote the probability ofdelegation, i.e., a well-behaving voter delegates his/her vote with probability γ and votes withprobability 1 − γ . A value of γ characterizes an equilibrium if well-behaving voters are indifferentbetween delegating and voting.The procedure of delegation is performed by the pooling of delegated votes and giving themuniformly at random to those who are voting. We abstract from the network structure of thevote delegation process, since the identities of the misbehaving voters are not known. Further,we assume that delegation has no cost.From the decomposition property of Poisson games, we have that K , the number of well-behaving voters who are delegating , is distributed as a Poisson random variable with parameter nγ . On the other hand, S , the number of well-behaving voters who are voting , is distributed asa Poisson random variable with parameter n (1 − γ ). This means that K votes are delegated toa group of S + f voters consisting of the remaining S well-behaving voters and f misbehavingvoters. Let h denote the number of votes that are delegated to the well-behaving voters. Thenremaining K − h votes are delegated to the misbehaving voters. The following figure illustratesthe setting. KS fh K − h Figure 1: h out of K votes are delegated to S well-behaving voters and K − h votes are delegatedto f misbehaving voters.Note that by the environmental equivalence property of Poisson games, we can take the See Myerson (1998). See Myerson (1998). S to S + 1 or thevoter is delegating and hence increasing the number of the delegating voters’ group from K to K +1. In the first case, the random variable h follows the binomial distribution with parameters K and S +1 S +1+ f , because h out of K votes are delegated to a group of S + 1 well-behaving votersout of S + 1 + f voters. In the latter case, i.e., if the additional voter is delegating, h followsthe binomial distribution with parameters K + 1 and SS + f , because now h out of K + 1 votesare delegated to S well-behaving voters out of S + f .For later analysis, let us define the function g ( x, y ) := x > y, if x = y, x < y, where g ( x, y ) denotes the gain for well-behaving voters if they receive x votes, and misbehavingvoters receive y votes. If well-behaving voters have more votes, the gain is 1, whereas the gainis 0 if misbehaving voters have the majority of votes. Ties are broken by a fair coin toss, hencethe expected gain of when x and y are equal. We start our analysis by the equilibrium indifference condition. The equilibrium indifferencecondition consists of equating the cost of voting c with the difference of the expected utility ofvoting and the expected utility of delegating, that is, the expected utility gain from voting. Wecan write, c = E [ U (voting)] − E [ U (delegating)] . The expected utilities are calculated in the following way: First, we need to sample K and S , using the probability mass function of the Poisson distribution with parameter nγ , resp. n (1 − γ ). Then, based on the sampling, we sample the number of votes h that well-behavingvoters are being delegated using the binomial distribution with the parameters specified above.In the case where the additional voter is voting, we have a size of S + 1 of the well-behavingvoter group. Hence, the well-behaving voters have in total S + 1 + h votes versus f + K − h misbehaving votes, which will be used to calculate the gain function g . Similarly, in the casewhere the additional voter is delegating, we have a size of K + 1 of the delegating group. Thewell-behaving voters will have a total of S + h votes versus f + K + 1 − h misbehaving votes.The following equation explicitly states the indifference condition for the additional voter:5 = ∞ (cid:88) K =0 ∞ (cid:88) S =0 ( nγ ) K e nγ K ! ( n (1 − γ )) S e n (1 − γ ) S ! ×× (cid:20) K (cid:88) h =0 (cid:18) Kh (cid:19) (cid:18) S + 1 S + 1 + f (cid:19) h (cid:18) fS + 1 + f (cid:19) K − h g ( S + 1 + h, f + K − h ) − K +1 (cid:88) h =0 (cid:18) K + 1 h (cid:19) (cid:18) SS + f (cid:19) h (cid:18) fS + f (cid:19) K +1 − h g ( S + h, f + K + 1 − h ) (cid:21) . (1)Let us elaborate on the right-hand side of this equation. First, we call the right-hand side of (1)from now on ξ n,f ( γ ). The first two sums, together with the Poisson probability mass functions,account for the sampling of K and S . Whereas the other two sums, together with the binomialprobability mass functions, account for the sampling of h .Next, we state our main result. Theorem 1
For any c > , there exists f ∗ ( c ) ∈ Θ( δ c ) and n ∗ ( f ) ∈ Θ( f δ ) , for any δ ∈ N , so that, for all f ≥ f ∗ ( c ) and n ≥ n ∗ ( f ) , well-behaving voters lose. Well-behaving voters losing is equivalent to the fact that there exists no solution to theequation (1). In particular, the right-hand side of equation (1) is strictly smaller than c . Or,equivalently, every well-behaving voter is delegating, because the cost of voting is larger thanthe expected benefits for voting, and hence misbehaving voters have the majority (in fact,all) of the votes. From the theorem, we see that lowest value for f is for constant δ , and itis from the set of functions Θ (cid:0) c (cid:1) . In this case, however, n needs to be large as a functionof f , namely f . Verbally, the theorem states that vote delegation is dangerous if there aresufficiently many misbehaving voters and sufficiently many well-behaving voters. The intuitionbehind this result is the following. If the number of well-behaving voters is even slightly smallerthan the number of misbehaving voters, then with high probability, the number of delegatingvoters is large enough to guarantee that well-behaving voters will lose. That is, costly voting isnot a dominant strategy. If the number of well-behaving voters is much bigger than the numberof misbehaving voters, then for any number of delegating voters, well-behaved voters will winwith high probability. On the other hand, if the number of well-behaving voters is moderatelylarger than the number of misbehaving agents, then the number of delegating agents is largeenough to guarantee that well-behaved agents win with high probability. That is, in neither ofthe latter two cases a well-behaved player is motivated to vote. Last, the probability that thenumber of well-behaved voters is approximately equal to the number of misbehaving agents issufficiently small. It follows from a large enough expected number of misbehaving agents andfrom the Poisson random variable properties. To assess the efficiency of a voting rule, we look at two values. First, we consider the probabilitythat well-behaving voters win in equilibrium under a specific voting rule. Second, we considerthe value of per-capita social welfare, which consists of the expected benefits of the whole (well-behaving) society minus the costs incurred by voting. The probability p that the well-behaving6oters win is calculated by the following formula: p ( n, f, γ ) = ∞ (cid:88) K =0 ∞ (cid:88) S =0 K (cid:88) h =0 ( nγ ) K e nγ K ! ( n (1 − γ )) S e n (1 − γ ) S ! (cid:18) Kh (cid:19) (cid:18) SS + f (cid:19) h (cid:18) fS + f (cid:19) K − h g ( S + h, f + K − h ) . We note that K and S are sampled and then, given K , h is sampled. Therefore, we have S + h votes for the right alternative and f + K − h votes for misbehaving voters. This impliesthat well-behaving voters win with certainty if S + h > f + K − h and they win with probability when S + h = f + K − h . The probability that well-behaving voters win is described by the g function introduced in the last section.We calculate per-capita social welfare as the difference between the probability of winningand the voting costs. The former represents per capita expected utility. Per-capita welfare isthus given by W ( n, f, γ, c ) = ∞ (cid:88) K =0 ∞ (cid:88) S =0 r ( K + S ) K (cid:88) h =0 ( nγ ) K e nγ K ! ( n (1 − γ )) S e n (1 − γ ) S ! (cid:18) Kh (cid:19) (cid:18) SS + f (cid:19) h (cid:18) fS + f (cid:19) K − h (cid:18) ( K + S ) g ( S + h, f + K − h ) − Sc (cid:19) , where the function r is defined as follows: r ( x ) := (cid:40) x if x > , x = 0 . Note that the term − Sc in the above formula is the cost spent on voting by well-behavingvoters, and it is subtracted from the total benefits for the society. The latter is captured by theterm ( K + S ) g ( S + h, f + K − h ). With conventional voting, there is no option to delegate one’s voting right. Thus, the voters’strategy sets only consist of voting and abstaining. We look for symmetric Bayesian Nashequilibria under conventional voting. Let α ∈ [0 ,
1] be the probability of voting. To determine α , we next derive the indifference condition between voting and abstention. We have to considerthe cases where an additional vote would impact the probability that well-behaving voters win.Then, we have to equate the cost of voting, c , with the difference between the expected utilitiesof voting and abstaining, that is c = E [ U (voting)] − E [ U (abstaining)] . This indifference can only hold if well-behaving voters have either f or f − f votes, there is a draw ( f versus f ). Then, one additional vote from well-behaving voters will7in and creates a utility gain of . In the other case, where well-behaving voters have exactly f − . These expected utilities are equated with the cost of voting andwe obtain the following indifference relation between voting and abstaining: c = 12 ( nα ) f e nα f ! + 12 ( nα ) f − e nα ( f − . (2)For a given probability of voting α , the probability of the well-behaving voters winning in thebaseline game is calculated by the following formula: q ( n, f, α ) = ∞ (cid:88) k =0 ( nα ) k e nα k ! g ( k, f ) . The per-capita social welfare is calculated as the difference between the probability of winning,measuring the expected benefits, and the cost of voting. To calculate per-capita social welfare,we first sample the total number of citizens, denoted by N and then, out of those N individuals, k individuals will vote with probability α . That is, k is sampled as a Binomial random variablewith parameters N and α , which leads to the following formula: W ( n, f, α, c ) = ∞ (cid:88) N =0 n N e n N ! r ( N ) N (cid:88) k =0 (cid:18) Nk (cid:19) α k (1 − α ) N − k ( N g ( k, f ) − kc ) , where the r ( N ) term has been introduced above and measures the per-capita benefit whenthere are N well-behaving voters in total. The term N g ( k, f ) stands for the total benefits, while − kc stands for the costs incurred by k voters voting. In this section, to compare the performance of vote delegation with conventional voting and wedo this in two ways: the probability that the right alternative is winning and per-capital socialwelfare.From the proof of Theorem 1 and by taking δ = 1, we obtain the lower bound thresholdon f ∗ ( c ) for vote delegation, above which no well-behaving voter votes. We can compare thisto the corresponding threshold for conventional voting that has been derived in Gersbach et al.(2019). This leads to the following proposition. Proposition 1
There are constants < t < t , such that(i) If f > t c , misbehaving voters will win with probability 1 under vote delegation and underconventional voting.(ii) If f ∈ [ t c , t c ] , the probability that well-behaved voters win with vote delegation is zero,while the probability that well-behaved voters win with conventional voting is positive. Hence, with the above proposition, we obtain two insights: First, if the number of misbehav-ing voters is high, both voting methods fail to deliver a positive outcome. Second, if the number8f misbehaving voters is moderate, conventional voting delivers a positive outcome, while votedelegation fails with probability one. Hence, in such cases, conventional voting performs betterthan vote delegation,Next, we numerically compare the performance of two voting rules for small values of f and n . The analysis around Theorem 1 requires a large electorate, that is, n and f need to be largeenough numbers for the result to hold. Theorem 1 tells us that there are no mixed equilibriasolutions if n and f are large enough.By numerical simulations, we find that there are mixed equilibria solutions if n and f aresmall enough. We illustrate this by the following example. The cost of voting c is equal to0 .
14. The expected number of well-behaving voters, n , is equal to 30. Approximate numericalsolutions show that we have two different equilibria solutions for any f >
1, but f not too large.In Table 1, the probabilities p and p represent the equilibrium probabilities of well-behavingvoters winning in the delegation game. In Table 2, the values q and q represent the equilibriumprobabilities of well-behaving voters winning in the conventional voting game. In both tables, W and W are the per-capita social welfare values for the corresponding voting game.Figure 2: Delegation game: The x -axis repre-sents values of γ and the y -axis displays ex-pected utility gains from voting, with the bluehorizontal line representing cost. The graphsrepresent the right hand side (RHS) of equa-tion (1) for different values of f . f p p W W p , p of well-behaving voters win-ning with delegation and per-capitasocial welfare values of the corre-sponding equilibria states W , W for c = 0 .
14 and n = 30.Extensive numerical calculations for different parameter values correspond to the patternshown in the tables. They suggest that the two equilibrium probabilities of well-behaving voterswinning in the delegation game are higher than the probabilities in the conventional voting gamefor sufficiently low values of f . For moderate values of f , however, the conventional voting gameyields higher probabilities that the correct alternative is chosen and vote delegation may evenprevent the correct alternative from having any chance to win. Finally, for high values of f , misbehaving voters win with certainty in both, the delegation and the conventional votinggames. The latter observation is in line with our main result.We address the special case f = 1. In this case, with vote delegation, we have a totallymixed equilibrium for a large enough value of c , and the equilibrium is unique. Proposition 2
For f = 1 , equation (1) has a solution for any c ∈ [ e n ( + n + n ) , ] . We note that for n arbitrarily large, the left endpoint of the interval in the proposition converges9igure 3: Conventional game: The x -axis rep-resents values of α and the y -axis displays ex-pected utility gains from voting, with the bluehorizontal line representing cost. The graphsrepresent the RHS of equation (2) for differentvalues of f . f q q W W q , q of well-behaving voterswinning in the conventional votinggame and per-capita social welfarevalues of the corresponding equilib-ria states W , W for c = 0 .
14 and n = 30.to 0. Therefore, the interval of c for which there is a solution converges to [0 , ] and is anatural upper bound for the cost of voting.For the conventional voting game, we also obtain a totally mixed equilibrium for a largeenough value of c , and the equilibrium is unique. In particular, we obtain the following result: Corollary 1
For any f ≥ , equation (2) has a solution for any c ∈ [0 , e √ f ( f − (cid:18) √ f ( f − f f ! + √ f ( f − f − ( f − (cid:19) ] .For f = 1 , equation (2) has a solution for any c ∈ [ n +12 e n , ] . In the case f ≥
2, we note that the right endpoint of the interval converges towards 0 as f grows and reaches the critical value above which no well-behaving individual will vote anymore.In the case f = 1, the left endpoint of the interval converges to 0 if n grows unboundedly. We have studied the role of vote delegation in the costly voting setting. In particular, we showedthat with malicious parties, vote delegation is a risky procedure if the number of misbehavingvoters is not low. However, for a low number of misbehaving voters, we showed that votedelegation dominates conventional voting, i.e., it implements the right alternative with a higherprobability than conventional voting. Overall, our results suggest that one should be cautiouswith the implementation of vote delegation.Our setting of costly voting is orthogonal to the information acquisition setting studied in theliterature. We believe the cost-saving aspect of vote delegation makes it worth studying further.Our analysis can be extended in various ways. First, we could study caps on the number ofdelegated votes, that is, every voter is only allowed to cast at most a constant maximum numberof votes. Second, we could assume that there is a number of committed well-behaving voterswho will vote no matter the costs. Note that this case is not equivalent to considering fewermisbehaving voters, because having some number of well-behaving voters voting with certaintychanges the probability that a vote will be delegated to a misbehaving voter.10 eferences
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Proofs
Proof of Theorem 1.
We analyze ξ n,f ( γ ), the right-hand side of equation (1). To makenotation easier, we define the function G f ( K, S ) := K (cid:88) h =0 (cid:18) Kh (cid:19) (cid:18) S + 1 S + 1 + f (cid:19) h (cid:18) fS + 1 + f (cid:19) K − h g ( S + 1 + h, f + K − h ) − K +1 (cid:88) h =0 (cid:18) K + 1 h (cid:19) (cid:18) SS + f (cid:19) h (cid:18) fS + f (cid:19) K +1 − h g ( S + h, f + K + 1 − h ) . Note that by the property of a probability distribution and by the definition of g , we have | G f ( K, S ) | (cid:54)
1. Our goal is to show that c > ξ n,f ( γ ) for any γ ∈ [0 , f . Equivalently,we prove that the value ξ n,f ( γ ) is very small. That is, we show it is close enough to zero forlarge enough f .First recall the definition ξ n,f ( γ ) = ∞ (cid:88) K =0 ∞ (cid:88) S =0 ( nγ ) K e nγ K ! ( n (1 − γ )) S e n (1 − γ ) S ! G f ( K, S ) . Next, we derive upper bounds on G f ( K, S ). Let us rewrite G f ( K, S ) = K (cid:88) h = (cid:98) f + K − S +12 (cid:99) (cid:18) Kh (cid:19) (cid:18) S + 1 S + 1 + f (cid:19) h (cid:18) fS + 1 + f (cid:19) K − h (3)+ 12 (cid:18) K f + K − S − (cid:19) (cid:18) S + 1 S + 1 + f (cid:19) f + K − S − (cid:18) fS + 1 + f (cid:19) K − f + K − S − (4) − K +1 (cid:88) h = (cid:98) f + K +3 − S (cid:99) (cid:18) K + 1 h (cid:19) (cid:18) SS + f (cid:19) h (cid:18) fS + f (cid:19) K +1 − h (5) − (cid:18) K + 1 f + K +1 − S (cid:19) (cid:18) SS + f (cid:19) f + K +1 − S (cid:18) fS + f (cid:19) K − f + K +1 − S . (6)The above equation holds because g ( S + 1 + h, f + K − h ) = 1 ⇐⇒ h > f + K − S − , and g ( S + 1 + h, f + K − h ) = 12 ⇐⇒ h = f + K − S − . Analogously, for g ( S + h, f + K + 1 − h ), we have g ( S + h, f + K + 1 − h ) = 1 ⇐⇒ h > f + K + 1 − S g ( S + h, f + K + 1 − h ) = 12 ⇐⇒ h = f + K + 1 − S . We define a = f + K − S − , and a = f + K + 1 − S . (7)Further, we define random variables X , X , so that X is distributed as Bin ( K, S +1 S +1+ f ), abinomial random variable with parameters K and S +1 S +1+ f , and X is distributed as Bin ( K +1 , SS + f ). Then, equation (3) is equal to the probability P [ X > a ] and equation (5) is equal to − P [ X > a ]. Similarly, (3) + (4) is equal to P [ X > a ] + P [ X = a ] and (5) + (6) is equalto − P [ X > a ] − P [ X = a ]. Together we have, G f ( K, S ) = P [ X > a ] + 12 P [ X = a ] − P [ X > a ] − P [ X = a ] . (8)where P [ X = a ] and P [ X = a ] vanish if a , resp. a , are not integers. Let δ ( f ) , ˜ δ ( f ) , σ ( f )be some functions of f , determined later in the proof. We consider the following three cases: • S ≤ f − δ ( f ), low value of S . • S ≥ f + ˜ δ ( f ), intermediate value of S . • S ∈ ( f − δ ( f ) , f + ˜ δ ( f )), high value of S .We resolve all cases in the following. Case 1.
Let S ≤ f − δ ( f ). As S is small, K is large with high probability since S and K are distributed as Poisson random variables with parameters n (1 − γ ) and nγ , respectively.That is, if one value is small, the other is large with high probability.The larger K , the higher the chance that well-behaving voters obtain more votes. In theworst case for well-behaving voters, S = f − δ ( f ), i.e. S is as large as possible in Case 1 andhence K will be smaller with high probability than what it would be with high probability if S were even lower than f − δ ( f ).Let us consider equality, S = f − δ ( f ). Then p the probability that a vote is delegated to S + 1 well-behaving voters is p = S + 1 S + 1 + f = 12 − δ ( f ) − f − δ ( f ) + 1) , and a , as given in (7), is a = f + K − S −
12 = K + δ ( f ) − . We want to upper-bound the first two terms in (8), i.e. P [ X ≥ a ] < c . First we stateHoeffding’s inequality which gives us an exponential upper bound for some real-valued (cid:15)P [ X (cid:62) K ( p + (cid:15) )] (cid:54) exp( − (cid:15) K ) . (9) See Hoeffding (1963).
13n order to make use of (9), we need to find (cid:15) first. By setting K ( p + (cid:15) ) = a , we can solve thisfor (cid:15) : (cid:15) = a K − p = δ ( f ) −
12 ( 1 K + 12 f − δ ( f ) + 1 ) (10)Note that for K > f , we can upper bound (cid:15) to make the ensuing analysis easier, (cid:15) ≤ δ ( f )4 f . (11)By the inequality of Mitzenmacher and Upfal (2005) for 2 f < nγ we can lower-bound theprobability that K > f , P [ K > f ] ≥ − e − nγ ( enγ ) f (2 f ) f . As n is large, we see that this probability is high. That is, for any β > n ∗ , suchthat for all n ≥ n ∗ , we have e − nγ ( enγ ) f (2 f ) f ≤ β. The threshold n ∗ is calculated by solving the above inequality. Later in the proof, we will obtainmore thresholds for n . In the end we choose the maximum of all thresholds. Hence we canuse (11).Next, we make use of the upper bound in (9). We set exp( − (cid:15) K ) = c and solve it for K .But remember that now, we are using the upper bound on (cid:15) , (11):exp( − (cid:15) K ) = c ≥ exp( −
2( 3 δ ( f )4 f ) K ) . We end up with a lower bound on K which we call K ∗ , K ≥ f δ ( f ) 89 log( 3 c ) =: K ∗ . (12)Then, for any K ≥ K ∗ we have P [ X ≥ a ] < c and hence, G f ( K, S ) = P [ X > a ] + 12 P [ X = a ] − P [ X > a ] − P [ X = a ] ≤ P [ X ≥ a ] < c , (13)where the last inequality precisely follows from Hoeffding’s inequality.In the last step, it remains to show that the probability that K ≥ K ∗ is high and hence (13)holds. For this, we can use the inequality from Mitzenmacher and Upfal (2005) for K ∗ < nγ , P [ K ≤ K ∗ ] ≤ e − nγ ( enγ ) K ∗ ( K ∗ ) K ∗ . n large enough or by taking δ ( f ) > f (cid:113) nγ /c ) + 1 , we can ensure that K ∗ < nγ and hence by Mitzenmacher and Upfal (2005), P [ K ≥ K ∗ ] = 1 − P [ K ≤ K ∗ ] ≥ − e − nγ ( enγ ) K ∗ ( K ∗ ) K ∗ . We find parameters such that this event has a probability at least q , where q is close to 1.Note that ι ( n ) := e − nγ ( enγ ) K ∗ ( K ∗ ) K ∗ has only one extremum for γ, K ∗ , n (cid:54) = 0, which attains its maximum at n ∗ = K ∗ γ . Further, note that ι ( n ) → n → { , ∞} . This means that for any β >
0, there are some0 < n < n ∗ < n , so that for all n < n and all n > n , we have ι ( n ) < β .Let 0 < β <
1. We solve ι ( n ) = β . We know from above that there are two solutions: e − nγ ( enγ ) K ∗ ( K ∗ ) K ∗ = β (14) ⇐⇒ − n γK ∗ e − n γK ∗ = − β /K ∗ e . (15)To solve this, we use the Lambert W function. A short overview of the Lambert W functionis given in Johansson (2020). The Lambert W function is the inverse function of f ( w ) = we w .It is multi-valued and has a infinite number of branches W k ( z ) for k ∈ Z and z ∈ C . W k ( z ) iscomplex everywhere except for k = 0 and k = − W ( z ) is real-valued and monotone increasingfor R (cid:51) z ≥ − /e , with the image [ − , + ∞ ). W − ( z ) is real-valued and monotone decreasingfor real z ∈ [ − /e, −∞ , − W functionto equation (15). On the left hand side (LHS), we have we w for w = − n γK ∗ . On the RHS wehave a real number that is in the interval ( − /e, β ∈ (0 , W and W − to (15). Let k ∈ { , − } (15) ⇐⇒ W k ( − n γK ∗ e − n γK ∗ ) = W k ( − β /K ∗ e ) ⇐⇒ − n γK ∗ = W k ( − β /K ∗ e ) ⇐⇒ n = − K ∗ γ W k ( − β /K ∗ e ) . W ( − /e ) = W − ( − /e ) = − W (0) = 0, W ( − β /K ∗ e ) ∈ ( − , W − ( − β /K ∗ e ) ∈ ( −∞ , − . Hence, the two solutions to equation (15) are n := − K ∗ γ W ( − β /K ∗ e ) < − K ∗ γ W − ( − β /K ∗ e ) =: n . Clearly, we see that n < n ∗ < n .Therefore, we obtain two thresholds n , n using the Lambert W function, one for applying W and the other for applying W − : n , = − K ∗ γ W , − ( − (1 − q ) /K ∗ e ) . We choose the higher threshold.
Case 2.
Let S ≥ f + ˜ δ ( f ). Then we have to consider two cases: First, if S is much largerthan f , e.g. larger than f + σ ( f ). In this case well-behaving voters win for any K . The secondcase ist the one with S is larger than f + ˜ δ ( f ) and smaller than f + σ ( f ).Let S > f + σ ( f ). We then have to distinguish between two cases: K < σ ( f ) and K > σ ( f ). • If K < σ ( f ), then well-behaving voters win, because even if all K votes are being delegatedto misbehaving voters, these misbehaving voters still end up having less votes. • If K > σ ( f ), we define a binomial random variable Y that stands for the number of votes(out of K ) that are delegated to well-behaving voters. The probability that a vote goesto a well-behaving voter is p = SS + f . Note that p is an increasing function in S . Although S > f + σ ( f ), we consider the case where S = f + σ ( f ). Hence we consider a lowerprobability p than the actual p . Now, p = SS + f = f + σ ( f )2 f + σ ( f ) = 1 − f f + σ ( f ) . Note that Y ∼ Bin ( K, p ). The probability that well-behaving voters can at most create16 tie is P [ Y ≤ K − ( S − f )2 ] < P [ Y ≤ K − σ ( f )2 ] ≤ exp( −
2( 12 − f f + σ ( f ) + σ ( f )2 K ) K )= exp( − σ ( f )( 14 f + 2 σ ( f ) + 12 K ) K )= exp( − σ ( f )( K (4 f + 2 σ ( f )) + 14 f + 2 σ ( f ) + 14 K )) , where we used the fact that S − f > σ ( f ) for the first strict inequality. For the secondinequality, we solved K ( p − (cid:15) ) = K − σ ( f )2 and used Hoeffding’s inequality. We see thatthe last function of the expression converges to 0 as f increases. This means again thatfor sufficiently high f , well-behaving voters win.Taking these together, we find that in the case S > f + σ ( f ), well-behaving voters winfor sufficiently high f . The fact that well-behaving voters win means that they have morevotes than misbehaving voters, i.e. g ( · , · ) = 1. Hence, G f ( K, S ) = 0 in this case, as wesubtract to binomial sums which are both equal to 1.Let f + ˜ δ ( f ) ≤ S ≤ f + σ ( f ). • Let S = f + ˜ δ ( f ). Then, the probability p that a vote is being delegated to the well-behaving voters is p = SS + f = 12 + ˜ δ ( f )2(2 f + ˜ δ ( f ))and a , as given in (7), is a = f + K + 1 − S K + 1 + ˜ δ ( f )2 . We want to upper-bound the following: P [ X ≤ a ] < c/
3. Again we recall Hoeffding’sinequality for some real-valued (cid:15) , P [ X (cid:54) K ( p − (cid:15) )] (cid:54) exp( − (cid:15) K ) . (16)To use (16), we first need to find (cid:15) . By setting K ( p − (cid:15) ) = a , we can solve (cid:15) : (cid:15) = p − a K = ˜ δ ( f )4 f + 2˜ δ ( f ) − ˜ δ ( f ) + 12 K . (17)Note that for
K > f , we can upper-bound (cid:15) as we did in Case 1, (cid:15) ≤ ˜ δ ( f )4 f . (18)By the inequality from Mitzenmacher and Upfal (2005), for 8 f < nγ , we can lower-bound17he probability of the event K > f : P [ K > f ] ≥ − e − nγ ( enγ ) f (8 f ) f . With large n , this probability is high. That is, for any β >
0, there exists n ∗ , so that forall n ≥ n ∗ , e − nγ ( enγ ) f (8 f ) f ≤ β. The threshold n ∗ is calculated by solving the previous inequality. At this point we obtainanother threshold for n . Remember that in the end, we will choose the maximum of thesethresholds. Hence we can use (18).Next, we make use of the upper bound in (16). We set exp( − (cid:15) K ) = c and solve it for K . Again, remember that we will use the upper bound on (cid:15) , (18):exp( − (cid:15) K ) = c ≥ exp( −
2( ˜ δ ( f )4 f ) K ) . We end up with a lower bound on K , which we denote by K ∗ , K ≥ f ˜ δ ( f ) 8 log( 3 c ) =: K ∗ . (19)Then, for any K ≥ K ∗ , we have P [ X ≤ a ] < c and hence, G f ( K, S ) = P [ X > a ] + 12 P [ X = a ] − P [ X > a ] − P [ X = a ] < − P [ X > a ] < − (1 − c c , (20)where the last inequality precisely follows from Hoeffding’s inequality. As a last step, itremains to show that the probability that K ≥ K ∗ is high and hence (20) holds. Wecan use again the inequality of Mitzenmacher and Upfal (2005) for bounding a Poissonrandom variable, namely, for K ∗ < nγ , P [ K ≤ K ∗ ] ≤ e − nγ ( enγ ) K ∗ ( K ∗ ) K ∗ . Again, with n large enough or by taking˜ δ ( f ) > f (cid:113) nγ /c ) − ,
18e can make ensure that K ∗ < nγ and hence P [ K ≥ K ∗ ] = 1 − P [ K ≤ K ∗ ] > − e − nγ ( enγ ) K ∗ ( K ∗ ) K ∗ . We want to lower-bound the latter by q , where q is close to 1. By the property of theLambert function, we obtain two thresholds n , n , one for applying W and the other forapplying W − , n , = − K ∗ γ W , − ( − (1 − q ) /K ∗ e . We choose the higher threshold. • Let S = f + σ ( f ). This part is treated as the previous part, where S = f + ˜ δ ( f ). Then,the probability p that a vote is delegated to the well-behaving voters is p = SS + f = 12 + σ ( f )2(2 f + σ ( f ))and a , as given in (7), is a = f + K + 1 − S K + 1 + σ ( f )2 . We want to upper bound the following: P [ X ≤ a ] < c/
3. We use Hoeffdings’s inequalityfor some real-valued (cid:15) , P [ X (cid:54) K ( p − (cid:15) )] (cid:54) exp( − (cid:15) K ) . (21)To use (21), we need to find (cid:15) . By setting K ( p − (cid:15) ) = a , we can solve (cid:15) : (cid:15) = p − a K = σ ( f )4 f + 2 σ ( f ) − σ ( f ) + 12 K . (22)Again, for
K > f , we can upper-bound (cid:15) as we did it before, (cid:15) ≤ σ ( f )4 f . (23)By the Poisson random variable concentration bound, for 8 f < nγ , we can lower-boundthe probability that K > f , P [ K > f ] ≥ − e − nγ ( enγ ) f (8 f ) f . For large n , this probability is high. That is, for any β >
0, there exists n ∗ such that forall n ≥ n ∗ : e − nγ ( enγ ) f (8 f ) f ≤ β. n ∗ is calculated by solving the above inequality. We obtain one morethreshold for n . We will finally choose the maximum of all thresholds. Hence we canuse (23).Next, we make use of the upper bound in Hoeffding’s inequality, (21). We set exp( − (cid:15) K ) = c and solve it for K . We again use the upper bound on (cid:15) , (23):exp( − (cid:15) K ) = c ≥ exp( − σ ( f )4 f ) K ) . We end up with a lower bound on K , which we again call K ∗ , K ≥ f σ ( f ) log( 3 c ) =: K ∗ . (24)Then, for any K ≥ K ∗ we have P [ X ≤ a ] < c and hence, G f ( K, S ) = P [ X > a ] + 12 P [ X = a ] − P [ X > a ] − P [ X = a ] < − P [ X > a ] < − (1 − c c , (25)where the last inequality precisely follows from Hoeffding’s inequality. As the last step,it remains to show that the probability for K ≥ K ∗ is high and hence that (25) holds.For this, we can again use the concentration bound on the Poisson random variable, inparticular, for K ∗ < nγ , P [ K ≤ K ∗ ] ≤ e − nγ ( enγ ) K ∗ ( K ∗ ) K ∗ . Again, with n large enough or by taking σ ( f ) > f (cid:113) nγ /c ) − , we can ensure that K ∗ < nγ and hence, P [ K ≥ K ∗ ] = 1 − P [ K ≤ K ∗ ] > − e − nγ ( enγ ) K ∗ ( K ∗ ) K ∗ . We want the latter to be at least q , where q is close to 1.Therefore, we obtain two thresholds n , n , one by applying W and the other by applying W − , n , = − K ∗ γ W , − ( − (1 − q ) /K ∗ e ) . We choose the higher threshold.
Case 3.
Let S ∈ ( f − δ ( f ) , f + ˜ δ ( f )). We can bound the probability that S is in this20nterval by P [ S ∈ ( f − δ ( f ) , f + ˜ δ ( f ))] ≤ δ ( f ) · ( f − δ ( f )) f − δ ( f ) e f − δ ( f ) ( f − δ ( f ))! . (26)We want to obtain the following upper bound: P [ S ∈ ( f − δ ( f ) , f + ˜ δ ( f ))] < c . To obtain this upper bound, we first recall Stirling’s inequality. For any d , the following holds: √ πd d +1 / e − d +1 / (12 d +1) < d ! . Let us rewrite (26), using d := f − δ ( f ), and apply Stirling’s inequality: P [ S ∈ ( f − δ ( f ) , f + ˜ δ ( f ))] ≤ δ ( f ) d d e d d ! < δ ( f ) √ πde / (12 d +1) . Next, we set the RHS equal to c/ δ = πc
18 ( f − δ ) e / (12( f − δ )+1) . We can upper-bound δ by the following: δ ( f ) ≤ ce (cid:114) π (cid:112) f . The latter can be rewritten as f ≥ e π δ c .δ ( f ) should be at least 1. This is a requirement to have S strictly smaller than f . By takingthe maximum of lower bounds on K , (12), (19) and (24), we obtain a lower bound for the valueof n , too.Altogether, we have shown that in each case, G f ( K, S ) < c . This ends the proof of thetheorem. Proof of Proposition 2. If f = 1 and γ = 0, then the right-hand side of equation (1) is e n . The right-hand side of equation (1) for f = 1 reads ξ n, ( γ ) = ∞ (cid:88) K =0 ∞ (cid:88) S =0 ( nγ ) K e nγ K ! ( n (1 − γ )) S e n (1 − γ ) S ! ×× (cid:20) K (cid:88) h =0 (cid:18) Kh (cid:19) (cid:18) S + 1 S + 2 (cid:19) h (cid:18) S + 2 (cid:19) K − h g ( S + 1 + h, K − h ) − K +1 (cid:88) h =0 (cid:18) K + 1 h (cid:19) (cid:18) SS + 1 (cid:19) h (cid:18) S + 1 (cid:19) K +1 − h g ( S + h, K + 2 − h ) (cid:21) . γ = 1, only the terms remain where S = 0, ξ n, (1) = ∞ (cid:88) K =0 n K e n K ! (cid:20) K (cid:88) h =0 (cid:18) Kh (cid:19) (cid:18) (cid:19) K g (1 + h, K − h ) (cid:21) . Note that g (1 + h, K − h ) = if 2 h = K and g (1 + h, K − h ) = 1 if 2 h > K . Hence, wehave ξ n, (1) = ∞ (cid:88) K =0 n K e n K ! (cid:20)(cid:18) Kh (cid:19) (cid:18) (cid:19) K { h = K } + (cid:18) (cid:19) K K (cid:88) h = (cid:100) K +12 (cid:101) (cid:18) Kh (cid:19)(cid:21) . (27)We use the following property about the sum of binomial coefficients: K (cid:88) h =0 (cid:18) Kh (cid:19) = 2 K . (28)Recall the symmetry property of binomial coefficients for non-negative K and h : (cid:18) Kh (cid:19) = (cid:18) KK − h (cid:19) . (29)With the two properties (28) and (29) we end the proof.In equation (27), the first term in brackets vanishes for odd K . This means that for odd K ,the entire expression in brackets is the following: (cid:18) (cid:19) K K (cid:88) h = (cid:100) K +12 (cid:101) (cid:18) Kh (cid:19) = (cid:18) (cid:19) K K , where the first equality follows from (28) and (29), as we only sum over exactly half of thebinomial coefficients.If K is even, the expression in brackets in equation (27) is (cid:18) K K (cid:19) (cid:18) (cid:19) K
12 + (cid:18) (cid:19) K K (cid:88) h = (cid:100) K +12 (cid:101) (cid:18) Kh (cid:19) = (cid:18) (cid:19) K (cid:18) K K (cid:19)
12 + K (cid:88) h = (cid:100) K +12 (cid:101) (cid:18) Kh (cid:19) = (cid:18) (cid:19) K K , where again, the second equality follows from (28) and (29).22t follows that ξ n, (1) = 12 ∞ (cid:88) K =0 n K e n K ! = 12 . The right-hand side of equation (1) for f = 1 reads ξ n, ( γ ) = ∞ (cid:88) K =0 ∞ (cid:88) S =0 ( nγ ) K e nγ K ! ( n (1 − γ )) S e n (1 − γ ) S ! ×× (cid:20) K (cid:88) h =0 (cid:18) Kh (cid:19) (cid:18) S + 1 S + 2 (cid:19) h (cid:18) S + 2 (cid:19) K − h g ( S + 1 + h, K − h ) − K +1 (cid:88) h =0 (cid:18) K + 1 h (cid:19) (cid:18) SS + 1 (cid:19) h (cid:18) S + 1 (cid:19) K +1 − h g ( S + h, K + 2 − h ) (cid:21) . For γ = 0, only the terms remain where K = 0, ξ n, (0) = ∞ (cid:88) S =0 n S e n S ! (cid:20) g ( S + 1 , − (cid:88) h =0 (cid:18) h (cid:19) (cid:18) SS + 1 (cid:19) h (cid:18) S + 1 (cid:19) − h g ( S + h, − h ) (cid:21) = ∞ (cid:88) S =0 n S e n S ! (cid:20) g ( S + 1 , − S + 1 g ( S, − SS + 1 g ( S + 1 , (cid:21) . From the equation above, we see immediately that the term in brackets is 0 for S ≥
3, as all g ( · , · )-terms are 1. Therefore, we only need to consider ξ n, (0) for S = 0 , ,
2. It follows that ξ n, (0) = 12 e n + n e n + n e n = 1 e n ( 12 + n n
12 ) . Since ξ n, ( γ ) is a continuous function, equation (1) has a solution for any c ∈ [ e n ( + n + n ) , ]. Proof of Corollary 1.
The derivative w.r.t. α of the RHS of (2) is given by n f f ! f α f − − nα f e nα + n f − f − f − α f − − nα f − e nα . Setting this equation equal to 0, we obtain0 = n f ( f α f − − nα f ) + n f − f (( f − α f − − nα f − )= α f − ( f ( f − − n α ) . The only positive solution is α ∗ = √ f ( f − n . We can easily verify that the RHS of (2) has amaximum at α ∗ by inserting α ∗ into the second derivative of the RHS of (2). As a last step, wehave to insert α ∗ into the RHS of (2) to obtain the right endpoint of the interval. As for f ≥ α = 0, the RHS of (2) is 0, the left endpoint of the interval is 0.For f = 1, we simply insert α = 0 and αα