## Abstract

In open, anonymous settings such as the Internet, agents can participate in a mechanism multiple times under different identities. A mechanism is false-name-proof if no agent ever benefits from participating more than once. Unfortunately, the design of false-name-proof mechanisms has been hindered by a variety of negative results. In this paper, we show how some of these negative results can be circumvented by making the realistic assumption that obtaining additional identities comes at a (potentially small) cost. We consider arbitrary such costs and apply our results within the context of a voting model with two alternatives.

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## Notes

See, for instance, http://techcrunch.com/2011/01/09/quora-4/ and http://goo.gl/7JWyXE.

Conitzer et al. (2010) explicitly model optional costly verification of selected users in the context of social networks.

As is the case for strategy-proofness, false-name-proofness is a dominant-strategies criterion, that is, using only one identifier should be optimal regardless of one’s preferences and regardless of what the other agents do. A (weaker) Bayesian definition can also be given, but the dominant-strategies version is the one we study in this paper. For an overview of false-name-proofness, see Conitzer and Yokoo (2010).

This notation suggests that all agents have the same cost function and that there is no uncertainty about this cost function. However, these assumptions are not necessary: all of our analysis goes through if \(c(t)\) is the greatest lower bound on all realizable total costs for obtaining \(t\) identifiers.

As is common in the literature, voluntary participation fails in our model if there is a positive cost for the first identifier. However, voting more than once would still be a dominated strategy. That is, given that an agent is going to participate, the dominant strategy would still be to vote only once. One may also say that if the cost of the first identifier is only \(\epsilon \), then behaving truthfully is \(\epsilon \)-dominant. We do imagine settings where casting the first vote is relatively easy (

*e.g.*, when the agent comes upon the election, the agent is already logged into her existing account). It is also quite possible for the cost of the first identifier to be*negative*, i.e., getting the first account is actually enjoyable. For example, an agent could get a coupon after voting (ideally, a coupon of which it is useless to have more than one copy). Even if there is some effort cost to voting, this may be exceeded by the worth of the coupon, in which case all of our results go through.Since agents’ utilities are normalized to 1 for their preferred outcome and 0 for the other outcome, the units in this inequality are consistent.

One can argue about the precise definition of responsiveness. For example, the rule that chooses \(A\) if the total number of votes is odd and \(B\) otherwise is more “responsive” in the sense that each additional vote changes the outcome. However, such rules violate neutrality and voluntary participation. For our purposes, a rule is most responsive if, given some constraints (neutrality, false-name-proofness), it comes as close as possible to the outcome of the majority rule.

For instance, suppose that obtaining additional identifiers comes at the cost of other activities. If the time value of other activities is concave (i.e., exhibits decreasing marginal value), and if the marginal effort spent to get another identifier is roughly constant, then the opportunity cost per identifier is increasing (as it comes at the cost of ever more valuable alternative activities).

A preliminary investigation of this setting with linear costs appears in Wagman and Conitzer (2008).

To see this, note that \(c''(\cdot )\ge 0\) implies \(c(t+1)-c(t)\ge c(t)-c(t-1)\ge \ldots \ge c(2)-c(1)\). Now, since \(c(1)=0,\,c(k'+1)=c(1)+c(k'+1)\). By the above inequalities, \(c(k'+1+1)\ge c(1+1)+c(k'+1)\). Similarly, \(c(k'+1+2)\ge c(1+2)+c(k'+1),\ldots ,c(k'+1+(k-k'))\ge c(1+k-k')+c(k'+1)\). The final inequality gives \(c(k+1)\ge c(k-k'+1)+c(k'+1)\).

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## Acknowledgments

We thank conference participants and participants at the Duke University Microeconomic Theory workshops; the Rice University Microeconomic Theory workshops; and Kyushu University in Japan. We especially thank Makoto Yokoo’s research group, Hervé Moulin, Lirong Xia, and anonymous referees for helpful feedback and discussions. Wagman has benefited from support from the Program for Advanced Research in the Social Sciences and from the Yahoo! Faculty Research and Engagement Program. Conitzer is grateful for support through NSF CCF-1101659, IIS-0812113, and IIS-0953756, ARO W911NF-12-1-0550 and W911NF-11-1-0332, and an Alfred P. Sloan Research Fellowship.

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## Additional information

Early versions of this work were presented at the Twenty-Third Conference on Artificial Intelligence (AAAI 2008), where it was awarded one of two Outstanding Paper Awards; at the International Industrial Organization Conference (2009); and at the University of Chicago Institute on Computational Economics (2009). Parts were also presented at the INFORMS Annual Meeting (2008) and at the World Congress of the Game Theory Society (2008).

## Appendix

### Appendix

### 1.1 A omitted proofs

**Lemma 3**
\({\textit{FNP2}}\)
*satisfies voluntary participation and strategy-proofness.*

###
*Proof*

It suffices to show that \(P_A(x_A,x_B)\) is weakly increasing in \(x_A\). From Definition 3 and the fact that \(c(\cdot )\) is nondecreasing, it immediately follows that for any \(0\le x_B<x_A,\,P_A(x_A+1,x_B)\ge P_A(x_A,x_B)\), because \(P_A(x_A,x_B) = \min _{t\in \{1,\ldots ,x_A-1\}}\{P_A(x_A-t,x_B)+c(t+1),1\} = \min _{t' \in \{2,\ldots ,x_A\}}\{P_A(x_A+1-t',x_B)+c(t'),1\} = \min _{t' \in \{2,\ldots ,x_A\}}\{P_A(x_A,x_B)+c(1), P_A(x_A+1-t',x_B)+c(t'),1\} = \min _{t' \in \{1,\ldots ,x_A\}}\{P_A(x_A+1-t',x_B)+c(t'),1\} \le \min _{t' \in \{1,\ldots ,x_A\}}\{P_A(x_A+1-t',x_B)+c(t'+1),1\} = P_A(x_A+1,x_B)\).

There are two cases left to prove: **(i)** for \(x>1,\,P_A(x+1,x)-P_A(x,x)\ge 0\) (monotonicity at \((x,x)\) profiles; this is not immediately clear from Definition 3 because \(P_A(x,x)=.5\) is assigned separately from the recursion) ; and **(ii)** for \(x_A<x_B,\,P_A(x_A+1, x_B) - P_A(x_A,x_B) \ge 0\) (monotonicity at profiles where \(x_A<x_B\)). Actually, for this second case, we will prove the equivalent statement: for \(x_B<x_A,\,P_B(x_A,x_B+1)-P_B(x_A,x_B)=P_A(x_A,x_B)-P_A(x_A,x_B+1)\ge 0\).

**(i)** Consider profile \((x+1,x)\). By Definition 3, \(P_A(x+1,x)=\min _{t\in \{1,\ldots ,x\}}\{P_A(x+1-t,x)+c(t+1),1\}\). In addition, for all \(t\in \{1,\ldots ,x\},\,P_A(x+1-t,x)=1-P_A(x,x+1-t)\), and \(P_A(x,x+1-t)=\min _{k\in \{1,\ldots ,x\}}\{P_A(x-k,x+1-t)+c(k+1),1\}\). Thus, \(P_A(x,x+1-t)\le P_A(x+1-t,x+1-t)+c(t)=.5+c(t)\). But then \(P_A(x+1-t,x)=1-P_A(x,x+1-t)\ge 1-(P_A(x+1-t,x+1-t)+c(t))=.5 -c(t)\). Therefore, \(P_A(x+1,x)\ge \min _{t\in \{1,\ldots ,x\}}\{.5-c(t)+c(t+1),1\}\ge .5=P_A(x,x)\). Hence, for all \(x>1,\,P_A(x+1,x)-P_A(x,x)\ge 0\).

**(ii)** We prove this part by induction. First, \(P_A(2,2)=0.5\le P_A(2,1)=\min \{.5+c(2),1\}\). This is the base step. Now, hypothesize that for some \(k>1,\,P_A(x_A,x_B)\ge P_A(x_A,x_B+1)\) for all \(x_B<x_A < k\). By symmetry, this induction hypothesis implies that \(P_A(x_B+1,x_A)\ge P_A(x_B,x_A)\) for all \(x_B<x_A<k\) (i.e., voluntary participation holds with respect to alternative \(B\) in a square of size \(k-1\) and a southwest vertex at \((1,1)\) in a 2-dimensional grid with \(A\) on the horizontal axis and \(B\) on the vertical axis).

Consider profile \((k,x_B)\) where \(x_B<k\) (i.e., extending the square diagonal by one grid point). If \(k=x_B+1,\,P_A(k,x_B+1)= .5 \le P_A(k,k-1)=P_A(k,x_B)\) follows directly from part **(i)**. Suppose then without loss of generality that \(x_B+1<k\). From Definition 3, \(P_A(k,x_B+1)=\min _{t\in \{1,\ldots ,k-1\}}\{P_A(k-t,x_B+1)+c(t+1),1\}\) and \(P_A(k,x_B)=\min _{t\in \{1,\ldots ,k-1\}}\{P_A(k-t,x_B)+c(t+1),1\}\). By the induction hypothesis, for all \(t\in \{1,\ldots ,k-1\},\,P_A(k-t,x_B+1)\le P_A(k-t,x_B)\). Thus, \(P_A(k-t,x_B+1)+c(t+1) \le P_A(k-t,x_B) + c(t+1)\). But then

It follows that \(P_A(k,x_B+1)\le P_A(k,x_B)\), which completes the induction. Consequently, for \(x_B<x_A,\,P_B(x_A,x_B+1)\ge P_B(x_A,x_B)\), which proves **(ii)**. \(\square \)

**Lemma 4**
\({\textit{FNP2}}\)
*is false-name-proof.*

###
*Proof*

Part 1 of Lemma 2 follows from Lemma 3. In addition, for every profile where \(x_A>x_B,\,P_A(x_A,x_B)\le \min _{t\in \{1,\ldots ,x_A-1\}} \{P_A(x_A-t,x_B)+c(t+1)\}\) follows from Definition 3. It remains to prove that for \(1<x_A\le x_B,\,P_A(x_A,x_B)\le \min _{t\in \{1,\ldots ,x_A-1\}}\{P_A(x_A-t,x_B)+c(t+1)\}\) (we actually prove that for \(1<x_B\le x_A,\,P_B(x_A,x_B)\le \min _{t\in \{1,\ldots ,x_B-1\}}\{P_B(x_A,x_B-t)+c(t+1)\}\), which is equivalent).

We begin by considering a profile \((x_A,x_B)\) where \(1<x_B<x_A\) (we treat profiles where \(x_B=x_A\) next). Consider a profile \((x_A,x_B-k)\), where \(k\in \{1,\ldots ,x_B-1\}\). For false-name-proofness, we need \(P_B(x_A,x_B)\le P_B(x_A,x_B-k)+c(k+1)\). Note that

By Definition 3, we have \(P_A(x_A,x_B)=\min _{t\in \{1,\ldots ,x_A-1\}} \{P_A(x_A-t,x_B)+c(t+1),1\}\). The case where \(P_A(x_A,x_B)=1\) holds trivially because \(P_A(x_A,x_B-k)\le 1\). Hence, assume without loss of generality that \(P_A(x_A,x_B)<1\) and let \(t^{\star }\in \{1,\ldots ,x_A\}\) be such that \(P_A(x_A,x_B)=P_A(x_A-t^{\star },x_B)+c(t^{\star }+1)\) (i.e., it is the binding constraint of the minimum). Note that due to \(P_A(x_A,x_B-k)=\min _{t\in \{1,\ldots ,x_A-1\}}\{P_A(x_A-t,x_B-k) +c(t+1),1\}, \,P_A(x_A,x_B-k)\le P_A(x_A-t^{\star },x_B-k)+c(t^{\star }+1)\) holds. Then

If \(x_A-t^{\star }=x_B\), we can stop at this point, since \(P_A(x_A-t^{\star },x_B)=P_A(x_B,x_B)=.5\), and \(P_A(x_A-t^{\star },x_B-k)=P_A(x_B,x_B-k)\le P_A(x_B-k,x_B-k)+c(k+1)=.5+c (k+1)\). Combining this with (1) and (2), we obtain \(P_B(x_A,x_B)-P_B(x_A,x_B-k)\le c(k+1)\).

If \(x_A-t^{\star }\ne x_B\), we reiterate the above analysis. In particular, similarly to the above analysis, there exists \(t^{\star \star }\in \{1,\ldots ,x_A-t^{\star }-1\}\) such that \(P_A(x_A-t^{\star },x_B)= P_A(x_A-t^{\star }-t^{\star \star },x_B)+c(t^{\star }+t^{\star \star }+1)\). Also similarly, \(P_A(x_A-t^{\star },x_B-k)\le P_A(x_A-t^{\star }-t^{\star \star },x_B-k)+c(t^{\star }+t^{\star \star }+1)\). Thus,

Applying this process iteratively, we either reach profile \((x_B,x_B)\), in which case the above conclusion applies, or we obtain

Since \({\textit{FNP2}}\) is defined symmetrically, \( P_A(1,x_B-k)-P_A(1,x_B)=(1-P_A(x_B-k,1))-(1-P_A(x_B,1)) =P_A(x_B,1)-P_A(x_B-k,1)\le c(k+1)\), where the inequality follows from \(P_A(x_B,1)= \min _{t\in \{1,\ldots ,x_B-1\}} \{P_A(x_B-t,1)+c(t+1),1\}\). Combining all of the above observations, we have \(P_B(x_A,x_B)-P_B(x_A,x_B-k)\le c(k+1)\).

It remains to show that for \(x>1\) and \(k\in \{1,\ldots x-1\},\,P_B(x,x)-P_B(x,x-k)=.5-P_B(x,x-k)\le c(k+1)\). Since \(P_A(x,x-k)=\min _{t\in \{1,\ldots ,x-1\}}\{P_A(x-t,x-k) +c(t+1),1\},\,P_A(x,x-k)\le P_A(x-k,x-k)+c(k+1)=.5+c(k+1)\). Thus, \(P_B(x,x)-P_B(x,x-k)=.5-(1-P_A(x,x-k))=P_A(x,x-k)-.5\le c(k+1)\). This completes the proof that \({\textit{FNP2}}\) satisfies the conditions of Lemma 2, and is therefore false-name-proof. \(\square \)

Proposition 2 *For all*
\(0<x_B<x_A,\,{\textit{FNP2}}\)
*satisfies:*

** (i)** If \(c''(\cdot )\ge 0\) then

(The linear cost model in which \(c(t+1)=k\cdot t\) for \(t,k\ge 0\), whereby \(c(2)=k\), is a special case.)

** (ii)** Let \(\pi (\cdot )\) be defined (recursively) as follows: \(\pi (1)=0\) and \(\pi (x)=\min _{t\in \{1,\ldots ,x-1\}} \{\pi (x-t)+c(t+1)\}\) for \(x> 1\). If \(P_A(x_A,1)<1\), then \(P_A(x_A,x_B)=.5+\pi (x_A)-\pi (x_B)\)

** (iii)** If \(P_A(x_A,1)<1\) and \(c''(\cdot )<0\) then

###
*Proof*

**(i)** Consider \(0<x_B<x_A\) and \(k,k'\in \{1,\ldots ,x_A-1\}\) such that \(k'<k\). By false-name-proofness, \(P_A(x_A-k',x_B)\le P_A(x_A-k,x_B)+c(k-k'+1)\). Since \(c(1)=0\) and \(c''(\cdot )\ge 0,\,c(k+1)\ge c(k-k'+1)+c(k'+1)\).^{Footnote 12} But then \(P_A(x_A-k,x_B)+c(k+1)\ge P_A(x_A-k,x_B)+c(k-k'+1)+c(k'+1)\ge P_A(x_A-k',x_B)+c(k'+1)\). Thus, the false-name-proofness constraint that \(P_A(x_A,x_B)\le P_A(x_A-k,x_B)+c(k+1)\) is already implied by the constraint \(P_A(x_A,x_B)\le P_A(x_A-k',x_B)+c(k'+1)\), where \(k'<k\). Since \(k\) and \(k'\) were arbitrarily chosen in \(\{1,\ldots ,x_A-1\}\), it follows that \(P_A(x_A,x_B)=\min \{P_A(x_A-1,x_B)+c(2),1\}\). Similarly, \(P_A(x_A-1,x_B)=\min \{P_A(x_A-2,x_B)+c(2),1\},\ldots , P_A(x_B+1,x_B)=\min \{.5+c(2),1\}\). Combining these equalities, we obtain \(P_A(x_A,x_B)=\min \{0.5 +c(2)(x_A-x_B), 1\}\).

**(ii)** It is straightforward to check that \(P_A(x_A,1)=\min _{t\in \{1,\ldots ,x_A-1\}}\{P_A(x_A-t,1)+c(t+1),1\}= \min \{P_A(1,1)+\pi (x_A),1\} =\min \{.5+\pi (x_A),1\}\). If \(P_A(x_A,1)<1\), then \(P_A(x_A,1)=.5+\pi (x_A)\). It also follows from Lemma 3 that for any \(k<x_A,\,P_A(k,1)=.5+\pi (k)\).

Assume \(x_A>2\). By neutrality, \(P_A(1,2)=1-P_A(2,1)=.5-\pi (2)\). We also have that \(P_A(3,2)=\min \{1,P_A(1,2)+c(3),P_A(2,2)+c(2)\}\). However, \(P_A(2,2)=.5+\pi (2)-\pi (2)=P_A(1,2)+\pi (2)\). It follows that \(P_A(3,2)=\min \{P_A(1,2)+\pi (3),1\}=\min \{.5+\pi (3)-\pi (2),1\}\). Similarly, for any \(2<k\le x_A,\,P_A(k,2)=.5+\pi (k)-\pi (2)\) (where \(P_A(k,2)<1\) follows from Lemma 3). A similar process can be done for \(P_A(k,3)\) for \(3<k\le x_A\). Specifically, assuming \(x_A>3\), we have \(P_A(1,3)=1-P_A(3,1)=.5-\pi (3),\,P_A(2,3)=1-P_A(3,2)=.5+\pi (2)- \pi (3)=P_A(1,3)+\pi (2)\), and \(P_A(3,3)=.5+\pi (3)-\pi (3)=P_A(1,3)+\pi (3)\). It then follows that \(P_A(k,3)=P_A(1,3)+\pi (k)=.5 + \pi (k)-\pi (3)\).

The proof proceeds by induction (the above being the base step). Hypothesize that for \(t\in \{1,\ldots ,k_B\},\,k_B<x_B\), and for any \(k_A\le x_A,\,P_A(k_A,t)=.5+\pi (k_A)-\pi (k_B)\) (where \(P_A(k_A,t)<1\) follows from Lemma 3 and the assumption that \(P_A(x_A,1)<1\)). By the induction hypothesis, for \(x< k_B+1,\,P_A(x,k_B+1)=1-P_A(k_B+1,x)=.5+\pi (x)-\pi (k_B+1)\). In addition, from the definition of \({\textit{FNP2}},\,P_A(k_B+1,k_B+1)=.5=.5+\pi (k_B+1)-\pi (k_B+1)\). It follows that for \(k_B+1<k_A\le x_A,\,P_A(k_A,k_B+1)=.5+\pi (k_A)-\pi (k_B+1)\), which completes the induction. Therefore, if \(P_A(x_A,1)<1\), then \(P_A(x_A,x_B)=.5+\pi (x_A)-\pi (x_B)\).

**(iii)** For \(k>0,\,c''(\cdot )< 0\) implies that \(c(k+1)-c(k)< c(k)-c(k-1)< \ldots < c(2)-c(1)=c(2)\). Now, since \(c(1)=0,\,c(k)=c(1)+c(k)\). By the above inequalities, \(c(k+1)< c(1+1)+c(k)\). Similarly, \(c(k+2)< c(1+2)+c(k),\ldots ,c(k+t_1)< c(1+t_1)+c(k)\). Applying a similar set of inequalities, we can obtain \(c(k+t_1+\ldots +t_m)<c(1+t_1)+\ldots +c(1+t_m)+c(k)\). It follows that \(\pi (k)=c(k)\).

Consider any \(0<x_B<x_A\) such that \(P_A(x_A,1)<\). By Part **(ii)**, we have \(P_A(x_A,x_B)= .5+\pi (x_A)-\pi (x_B)\). Combining this with the above, we have \(P_A(x_A,x_B)= .5+c(x_A)-c(x_B)\). \(\square \)

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Wagman, L., Conitzer, V. False-name-proof voting with costs over two alternatives.
*Int J Game Theory* **43**, 599–618 (2014). https://doi.org/10.1007/s00182-013-0397-3

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DOI: https://doi.org/10.1007/s00182-013-0397-3