False-name-proof voting with costs over two alternatives

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.

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

$$\begin{aligned} \min _{t\in \{1,\ldots ,k-1\}}\{P_A(k-t,x_B+1)+c(t+1)\}\\ \le \min _{t\in \{1,\ldots ,k-1\}}\{P_A(k-t,x_B)+c(t+1)\} \end{aligned}$$

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

$$\begin{aligned}&P_B(x_A,x_B)-P_B(x_A,x_B-k) \nonumber \\&\quad =1-P_A(x_A,x_B)-(1-P_A(x_A,x_B-k)) \nonumber \\&\quad = P_A(x_A,x_B-k)-P_A(x_A,x_B) \end{aligned}$$

(1)

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

$$\begin{aligned}&P_A(x_A,x_B-k)-P_A(x_A,x_B) \nonumber \\&\quad \le P_A(x_A-t^{\star },x_B-k)+c(t^{\star }+1) - (P_A(x_A-t^{\star },x_B)+c(t^{\star }+1)) \nonumber \\&\quad =P_A(x_A-t^{\star },x_B-k)-P_A(x_A-t^{\star },x_B) \end{aligned}$$

(2)

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,

$$\begin{aligned}&P_A(x_A-t^{\star },x_B-k)-P_A(x_A-t^{\star },x_B) \\&\quad \le P_A(x_A-t^{\star }-t^{\star \star },x_B-k) - P_A(x_A-t^{\star }-t^{\star \star },x_B) \end{aligned}$$

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

$$\begin{aligned} P_A(x_A,x_B-k)-P_A(x_A,x_B)\le P_A(1,x_B-k)-P_A(1,x_B) \end{aligned}$$

(3)

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

$$\begin{aligned} P_A(x_A,x_B)=\min \{0.5 +c(2)(x_A-x_B), 1\} \end{aligned}$$

(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

$$\begin{aligned} P_A(x_A,x_B)=.5+c(x_A)-c(x_B) \end{aligned}$$

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 $c''(\cdot )\ge 0,\,c(k+1)\ge c(k-k'+1)+c(k'+1)$.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 $P_A(x_A,x_B)= .5+c(x_A)-c(x_B)$. $\square$