Existence of representative apportionments

´ G . Laffond, J. Laine Mathematical Social Sciences 39 2000 35 –53 43 Definition 9. A Condorcet-consistent solution S is sequentially neutral to appor- tionment if for any electorate N, ;[s][PN , there exists A[GN such that ST s S[T s ]5S[T s ]. Furthermore, S is weakly sequentially A 1k K k neutral to apportionment if ;N, ;[s][PN, A[GN such that [ST s S[T s ]] S[T s ] ± [. A 1k K k A solution is neutral or sequentially neutral under the sequential rule to an apportionment if it selects the same set of winners in both referendum and representa- tional systems. It is weakly neutral or sequentially neutral to an apportionment if the two sets of final outcomes intersect. Whenever uXu52, any Condorcet-consistent solution S is neutral to any representative apportionment. We will see below that this is no longer true for larger sets of candidates. Another sensitivity concept is defined as follows: Definition 10. Let S,GN . A Condorcet-consistent solution S is non-sensitive to S if 9 9 ;[s][PN, ;A5 hN ,...,N j, A95hN ,...,N j[S, ST s S[T s ]5 1 k 1 K 9 A 1k K k ST s S[T s ]. Furthermore, S is weakly non-sensitive to S A9 1k 9K 9 k 9 9 if ;A5 hN ,...,N j, A95 hN9 ,...,N ,...,N j[S, [ST s S[T s ]] 1 k 1 1 K 9 A 1k K k [ST s S[T s ]] ± [. A9 1k 9K 9 k 9 A solution S is non-sensitive to a set of apportionments if it provides the same set of winners whatever the actual apportionment that prevails among admissible ones. It is weakly non-sensitive if any pair of apportionments share at least one winning outcome. The informal idea which underlines this definition is that, even if the choice between a direct and an indirect democratic system does influence the finally elected candidates, the latter should not change in case of a reform of the apportionment method. Such a requirement allows for restricting the set of potentially admissible electoral reforms. Our approach to the representative voting systems naturally suggests the following questions, which are studied in the next two parts: • Is it true that for any preference profile, there exists a representative apportionment? If not, is it possible to define a non-obvious restriction on the set of profiles which ensures the existence of such a representative apportionment? • Does the representativeness requirement for apportionment matter? In particular, is it possible, for both collective choice methods defined above, to exhibit a profile for which a Condorcet-consistent solution is non-neutral to any non-representative apportionment? • Can we find a Condorcet-consistent solution which is neutral to the set of all representative apportionments?

3. Existence of representative apportionments

We begin our presentation of results with three propositions dealing with the existence ´ of representative apportionments. The first one, which is proved in Laffond and Laine 1996 is the existence of profiles admitting no such apportionment: ´ 44 G . Laffond, J. Laine Mathematical Social Sciences 39 2000 35 –53 Theorem 1. There exists a preference profile [s][PN such that G N,s5[. R However, the proof presented in the reference above involves few voters and a very large set of candidates. The next theorem states that the same inexistence problem may arise even if the ratio candidates voters is less than one: Theorem 2. There exists a profile [s][PN such that uXu uNu7 9 and G N,s5[. R Proof. Let N 5 h1,...,9j and X5h1,...,7j and let [s] be such that: • ;i [N, 1s 2 ⇔ i [N95 h1,...,5j i • ;i [N9, ;m[X2 h1,2j, ms 2 ⇔ i 5m 22 i • ;i [N2N9, ;m[X2 h1,2j, ms 2 i All other majority pairwise comparisons do not matter. It is obvious to check that T s is such that 2 is a Condorcet loser. Now let F 5 hT ,T ,T j be the set of sub-tournaments on X associated with any 3-apportionment 1 2 3 A5 hN ,N ,N j of N, where each constituency involves 3 voters. We claim that A is not 1 2 3 representative. Indeed, suppose that A[G N,s and w.l.g. that 1T 2 and 1T 2. Since R 1 2 ;m±1,2, mTs2, it follows that ;m±1,2, m22[N N otherwise 2 would get a 1 2 majority of votes against m in N and N , which would imply that 2 T s m, a 1 2 A contradiction. Thus one can suppose w.l.g. that N ,N9 and uN N9u52. Let i [N 1 2 1 and m 5i 12. Since i is the only individual in N who prefers m to 2, then 2T m . 1 1 Furthermore, every individual in N N9 prefers m to 2, hence 2T m .Thus m [ A2 2 2 h1,2j such that 2T m and 2T m , hence 2 T sm , which contradicts the representa- 1 2 A tiveness of A. h Theorem 2 points out a significant difference between the dichotomous and the general case: in the former case, there always exist a district map which yields agreement with direct democracy, whereas in the latter case, maybe no district map yields a social preference relation that coincides with overall majority preferences. An essential ingredient in the proof above is the existence of majority dominations with tiny margins. What about the possibility to design representative apportionments when the overall tournament involves strong majorities, i.e. widespread social agree- ments? Suppose that n 5Km, where K [ h3,n21j denotes the odd number of con- stituencies hence allowing for a perfect apportionment. Moreover, suppose that a majority of voters prefers some candidate y to some other one x. Then, in order for x to defeat y in a representational system, it must be the case that x is preferred to y by at least 0.25K 11m 11 voters this is a necessary requirement for x to get a majority of votes against y in a majority of constituencies. It follows that the minimal proportion q 21 of voters supporting x against y is [4Km] [Km 1K 1m 11]; since q converges to 0.25 as K and m both go to `, it follows that no reversal of a majority preference can prevail in representational systems when the preference profile is such that all majority margins are larger than 75. The next theorem generalizes this point to the case of non necessarily perfect apportionments: ´ G . Laffond, J. Laine Mathematical Social Sciences 39 2000 35 –53 45 Theorem 3. ;a 0.75, ;[s][PN such that Ts[Da, G N,s5GN. R Proof. Let A5 hN ,...,N j be a K-apportionment of N where n [h2p11,2p21j, 1 2k 11 k 2 1k K. Let for any x, y[X n resp. n 5 uhi [N: xs yjuresp. n 5uhi [N: ys xju. x y i y i 2 Now suppose that xT s y i.e. n .n and that yT s x. This implies that n pb 1 x y A y 1 2 1 p 11b , where b 5 uhh[h1,...,2k11j: uN u52p21 and yT sxju and b 5uhh[ h h 2 1 h1,...,2k11j: uN u52p11 and yT sxju, and with the condition b 1b k11. One h h then has: 2 1 1 1 • n p b 1b 1b pk 111b y 1 1 • n 2p 212k 112b 12p 11b 1 It follows that 4n 2n 2p 12k 1112b .0, hence that n n .0.25, a contradiction. y y h This result provides an interesting restriction on the set of profiles which ensures that majority preferences will not be biased through the representation process. It simply states that ‘popular willingness’ has to be strong enough. An immediate corollary of Theorem 3 is the existence of at least one non-representative apportionment when at least one majority margin is less than 75 as long as the number of constituencies as well as the size of the electorate are large enough. In fact, things may become even worse: indeed, we now prove that, as long as some pairwise majority margin is lower than 0.75, there exists at least one profile for which no representative apportionment 8 exists. Theorem 4. ;e. 0, there exists [s][PN such that Ts[D 0.752e and G N,s R 5[. Proof. Let X 5 hwN:uwu5mj, where 0,mn. Let P be a complete linear order on X. 2 For any i [N, i’s preferences on X are defined as follows: let w,w9[X , with w ±w9; then 1. if i [[w w9], w s w9 ⇔ wPw9 i 2. if i [[w 2w9], w9s w i 3. if i [[w92w], w s w9 i 4. if i [N 2[w w9], w s w9 ⇔ w9Pw. i It follows by definition that ;w,w9[X such that wPw9, uhi [N: ws w9ju5uw9u5m. i 2 Thus, if m .0.5n, T s coincides with P. We denote by P the reverse order of P. Let assume that n is such that A5 hN ,...,N j[GN such that n 52p11, 1 2k 11 h 1h 2k 11. The proof is organized in three steps: 8 It should be obvious that this result is not implied by Theorem 3: indeed, it is impossible to build a profile by considering mutually independent pairwise comparisons of candidates. ´ 46 G . Laffond, J. Laine Mathematical Social Sciences 39 2000 35 –53 Step 1. Let h [ h1,...,2k11j and w,w9[X. Let w resp. w9 5wN [resp. w9N ]. h h h h We claim that the preference profile ensures the following four simple statements: 9 1. If uw u p11, uw9 u p11 and wPw9, then wT w9: indeed, i [w ⇒ w s w9 if h h h h i 9 9 i [w w , this follows from Point 1 and wPw9, whereas if i [w 2w , this follows h h h h from Point 3 and thus it follows that at least p 11 individuals in N prefer w to w9. h 9 2. If uw u p11, uw u, p11 and w9Pw, then w9T w: note first that i [w ⇒ w9s w: h h h h i 9 9 indeed, i [w 2w ⇒ w9s w from Point 2, and i [w w ⇒ w9s w from Point 1 h h i h h i and w9Pw. Then uw u p11 implies that more than p11 individuals in N prefer h h w9 to w. 9 9 3. If uw u, p11 and wPw9, then w9T w: one has i [ ⁄ w ⇒ w9s w: indeed, i [w 2 h h h i h 9 9 w ⇒ w9s w from Point 2, and i [[N 2w w ] ⇒ w9s w from Point 4 and h i h h h i 9 wPw9. Thus uN 2w u p11 implies that more than p11 individuals in N prefer h h h w9 to w. Step 2. m 2p 11k 1pk 11 ⇒ G N,s5[. R Proof. Let w be the Condorcet-winner of T s 5P. We can deduce from Step 1 that if A is representative, it must be true that ;w [X 2 hwj, uhh[h1,...,2k11j: uw u p1 h 1 juk11: indeed, this ensures that wT sw;w±w. Now suppose that m2p1 A 1k 1pk 11. Let w9 be the alternative defined by w 5 hi , 1xmj, where x k hi ,...,i k jj5 N , and where V 5 hi ,...,i j is allocated among the 1 2p 11 h 51 h 2p 11 k 11 m remaining constituencies of A in such a way that ;h [ hk11,...,2k11j, uN Vup. It is h 9 obvious to check that uhh[h1,...,2k11j: uw u p11ju,k11, which implies that h w9T s w. Hence there is no representative apportionment. A Step 3. ;e .0, N can be chosen such that T s [D0.752e and G N,s 5[. R Proof. Let m 52p 11k 1pk 11. It is straightforward to check that ;k, p 0, m n , 0.75. The fact that lim [m n]50.75 ends the proof of Theorem 4. h k 5p → ` 4. Sensitivity of tournament solutions to representation 4.1. Representative systems under one-shot method We now investigate the sensitivity of tournament solution to apportionment when the final choice is made through the one-shot method. The existence of a representative apportionment ensures that direct and indirect democratic choices coincide. However, as seen above, there may exist no representative apportionment for some preference profile. Nevertheless, it might well be the case that some solution concept S may be non- sensitive to a non-representative apportionment. The next theorem states that this is not the case for the Uncovered Set: ´ G . Laffond, J. Laine Mathematical Social Sciences 39 2000 35 –53 47 Theorem 5. There exists a profile [s][PN such that uXu uNu7 9, and ;A[GN, UC [Ts]± UC[T s]. A Proof. Let N 5 h1,...,9j and X5h1,...,7j and let [s] be defined by voters’ names appear in the first row, and alternatives are listed in decreasing preference order: 1 2 3 4 5 6 7 8 9 1 2 a a a 3 4 1 2 a a 3 4 5 5 2 5 4 b b b b b 1 5 3 3 2 4 1 5 1 4 1 2 1 3 5 5 2 3 2 3 4 5 4 3 4 3 2 b b b b 5 1 2 1 4 a a a a It is easily checked that T s hereafter denoted by T is such that 1 T 2,4,a,b, 2 T 3,4,a,b, 3 T 1,4,a,b, 4 T 5,b, 5 T 1,2,3,b, and a T 3,4,5,b. Thus b is a Condorcet loser, and it is easily checked that UCT 5 ha,1,2,3,4,5j. Let A5N be a 3-apportionment of N and let S 5[T s ] be the p h p51,2,3j p p 51,2,3 associated set of tournaments on X. It follows from Theorem 2 that G N,s 5[. The R proof is organized according to the following strategy: since no representative apportion- 2 9 ment exists, x, y[X such that xTy and yT s x; each step of the proof is devoted to A a specific set of x, y-type pairs of candidates. Step 1. y 5b, x [ h1,...,5j. Case 1. x 51. It follows that one may assume w.l.g. that voters 2,3,4 and 5 form majorities in both N and N , and that either a is a Condorcet winner in T or 2 T a T 1,3,4,5,b. In the 1 2 A A A former case, UCT 5 haj±UCT . In the later case, UCT 5UCT ⇒ 1,3,4,5 T 2. A A A Since hi [N: 4s 2j5h3,4,6,7j, and hi [N: 3s 2j5h3,5,6,8j, it is obviously seen that 4 i i T 2 ⇔ 2 T 3, hence either 3[ ⁄ UCT , or 4[ ⁄ UCT . A A A A Case 2. x 52. Voters 1,3,4 and 5 form a majority in N and A in N , which implies that either a is a 1 2 Condorcet winner in T or 1 T a T 2,3,4,5,b. As above, UCT 5UCT ⇒ the later A A A A case prevails, and 2,3,4,5 T 1 [condition ]. Since hi [N: 4s 1j5h2,4,7,9j, one has to A i consider the following possible sub-cases: 9 To make notations simple, T s and T s will be respectively written T and T . A p p 51,2,3 A p p 51,2,3 ´ 48 G . Laffond, J. Laine Mathematical Social Sciences 39 2000 35 –53 Case 2.1. h2,4j,N and h7,9j,N : it must be the case that N is either h1,2,4j or h2,3,4j, 1 2 1 or h2,4,5j. N 5h1,2,4j and h1,3,4,5j have a majority in N and N imply with that 1 1 2 A5 hN ,N ,N j[h1,2,4,6,7,9,3,5,8jh1,2,4,6,7,8,3,5,9j ⇒ 2 T 1 2 3 A 3,4,5 ⇒ h3,4,5jUCT 5[. Moreover, N 5h2,3,4j, , 3 T 1 and b T 2 ⇒ A5 A 1 A A h2,3,4,7,6,9,1,5,8j ⇒ 4 T 2,3,5 ⇒ h2,3,5jUCT 5[. Finally, the reader may A A easily check that N 5 h2,4,5j, 5 T 1 and b T 2 ⇒ A5 h2,4,5,7,6,9,1,3,8j ⇒ 4 T 3,5, 1 A A A whereas 2 T 3 T 5 T 2 and 2 T 4 ⇒ 3[ ⁄ UCT , hence UCT ±UCT . A A A A A A Case 2.2. h2,7j,N and h4,9j,N : b T 2 ⇒ N [ h1,4,9jh3,4,9jh4,5,9j. In the first 1 2 A 2 case, b T 2 and 5 T 1 ⇒ A5 h2,7,8,1,4,9,3,5,6j ⇒ 2 T 3,4,5 ⇒ h3,4,5jUCT 5 A A A A [. Moreover, if N 5 h3,4,9j, then b T 2, 5 T 1 and 3 T 1 are mutually incompatible. 2 A A A Finally, if N 5 h4,5,9j, then b T 2 and ⇒ A5 hN ,N ,N j[h2,6,7,4,5,9,1,3,8j 2 A 1 2 3 h2,7,8,4,5,9,1,3,6j. In both cases, one get 2,4,5 T 3, hence 3[ ⁄ UCT . A A Case 2.3. h2,9j,N and h4,7j,N : b T 2 ⇒ N [ h1,4,7jh3,4,7jh4,5,7j. In the first 1 2 A 2 case, and b T 2 ⇒ A5 h2,8,9,1,4,7,3,5,6j ⇒ 2,4,5 T 3 ⇒ 3[ ⁄ UCT . In the A A A second case, and b T 2 ⇒ A[ h2,8,9,1,4,7,3,5,6jh2,8,9,3,4,7,1,5,6j, A which leads to the same conclusion. Finally, if N 5 h4,5,7j, b T 2 ⇒ either 1 T 5 or 1 2 A A T 3, hence either 5[ ⁄ UCT or 3[ ⁄ UCT . This proves that b T 2 implies that A A A A UCT ±UCT . A Case 3. x [ h3,4,5j. This means that voters 1,2,4 and 5 form a majority in both N and N . It follows from 1 2 [s] that a T 3,4,5,b. Suppose that h1,2j,N and h4,5j,N . Then if N 5h1,2,3j, a is a A 1 2 1 Condorcet winner in T , hence UCT ±UCT . Thus UCT 5UCT ⇒ N 5 h1,2,zj, A A A 1 where z [ h6,...,9j, which implies that 1,2 T a T 3,4,5,b. It is straightforward to check A A that the same conclusion prevails for all other way to allocate 1,2,4 and 5 among N and 1 N . Now assume that 1 T 2; UCT 5UCT ⇒ 3,4,5 T 1. Moreover, 1 T 2 and 4 T 1 2 A A A A A implies the existence of at least one constituency k [ h1,2,3j where 4 T 1 T 2. But this k k is impossible since hi [N: 4s 1j5h1,3,5,6,8jhi [N: 1s 2j5h2,4,7,9j5[. Thus 1 T i i A 4 and thus 4[ ⁄ UCT . Finally, suppose that 2 T 1. Similarly, it must be the case that A A 3,4,5 T 2. But this is impossible since hi [N: 2s 1j5h2,4,7,9jhi [N: 3s 2j5 A i i h3,5,6,8j5[. Thus, 3[ ⁄ UCT . This proves that if x 53, UCT ±UCT . The reader A A may check that if x [ h4,5j, the same argument applies. This concludes the proof of Step 1. Step 2. y 5b, x 5a. This implies that voters 6,7,8 and 9 form a majority in two constituencies. It follows that a is a Condorcet loser in T , hence a [ ⁄ UCT , thus UCT ±UCT . A A A Step 3. x 5a, y [ h3,4,5j. This implies that voters 6,7,8 and 9 form a majority in two constituencies. Hence we get the same conclusion as in Step 2. ´ G . Laffond, J. Laine Mathematical Social Sciences 39 2000 35 –53 49 Step 4. y 5a, x [ h1,2j. Suppose first that a T 1. It follows from [s] that voters 2,3,4 and 5 form a majority A in two constituencies. But this implies that b T 1, and thus UCT ±UCT from Step A A 1. Finally, if a T 2, voters 1,3,4 and form a majority in two constituencies, which A implies that b T 2, hence again UCT ±UCT from Step 1. This concludes the proof. A A h When no restriction bear upon the ratio between the number of voters and the number of candidates, things may become even worse: it is possible to exhibit a preference profile such that, for any Condorcet-consistent tournament solution, direct and repre- sentative voting systems always leads to mutually disjoint sets of winners. Theorem 6. No Condorcet-consistent tournament solution S is weakly neutral to apportionment . Proof. Consider the preference profile [s] defined in the proof of Theorem 4. Let 1 2 1 A5 hN ,...,N j[GN. Let X 5hw[X: uw u, p11j, and X 5X2X . It 1 2k 11 h h h h follows from the three assertions proved in Step 2 that ;h [ h1,...,2k11j, T s is a h linear order on X defined by: 1 2 • ;w,w9[X , wT s w9 ⇔ wP w9 h h 2 • ;w,w9[X , wT s w9 ⇔ wPw9 h h 1 2 • ;w[X , ;w9[X , wT s w9. h h h Let us examine the tournament among representatives T s . For w,w9[X, define a, A b, c and resp. by: 2 2 • a 5 uhh[h1,...,2k11j: w[X and w9[X ju h h 2 1 • b 5 uhh[h1,...,2k11j: w[X and w9[X ju h h 1 2 • c 5 uhh[h1,...,2k11j: w[X and w9[X ju h h 1 1 • d 5 uhh[h1,...,2k11j: w[X and w9[X ju. h h Then T s will behave according to the relative values of a, b, c and d. This leads to A the following possible cases: Case 1. b 2c. ua2du. This implies that wT sw9. But b2c.a2d ⇔ uhh[ A 1 2 h1,...,2k11j: w[X juk11 and b2c.d2a ⇔ uhh[h1,...,2k11j: w9[X juk11. h h Case 2. a 2d . uc2bu. Then wT sw9 ⇔ wPw9. Moreover a 2d .c 2b ⇔ uhh[ A 2 2 h1,...,2k11j: w9[X juk11 and a2d.b2c ⇔ uhh[h1,...,2k11j: w[X juk11. h h 2 Case 3. d 2a. uc2bu. Then wT sw9 ⇔ wP w9. Moreover d 2a.c 2b ⇔ uhh[ A 1 1 h1,...,2k11j: w[X juk11 and a2d.b2c ⇔ uhh[h1,...,2k11j: w9[X juk11. h h ´ 50 G . Laffond, J. Laine Mathematical Social Sciences 39 2000 35 –53 Case 4. c 2b. ua2du. We get w9T sw. Moreover c2b.a2d ⇔ uhh[h1,...,2k1 A 1 2 1 j: w9[X juk11 and a2d.b2c ⇔ uhh[h1,...,2k11j: w[X juk11. h h 1 2 1 1 Define for A the set X resp. X by X 5 hw[X: uhh[h1,...,2k11j: w9[X ju A A A h 2 1 k 11 j resp. X 5X2X . We deduce from the four cases above that T s is the A A A linear order defined on X by: 1 2 • ;w,w9[X , wT s w9 ⇔ wP w9 A A 2 • ;w,w9[X , wT s w9 ⇔ wPw9 A A 1 2 • ;w[X , ;w9[X , wT sw9. A A A Finally, for any apportionment A, any Condorcet-consistent solution S will select from 1 2 T s as unique winner the top-element of X for P , whereas S[T s ] is clearly A A reduced to the top-element of X for P. The theorem follows from the fact that 1 ;A[GN, X ± [. h A 4.2. Representative system under the sequential choice method We turn now to the comparison between the direct system and the representative system based on the sequential choice method. The next result points out that the same negative result prevails as for the one-shot method: Theorem 7. No Condorcet-consistent solution S is weakly sequentially neutral to apportionment . Proof. Consider again the profile introduced in the proof of Theorem 4. It is shown in the proof of Theorem 4 that ;A5 hN , 1h2k11j, T s is a linear order on X k h defined by: 1 2 • ;w,w9[X , wT s w9 ⇔ wP w9 h h 2 • ;w,w9[X , wT s w9 ⇔ wPw9 h h 1 2 • ;w[X , ;w9[X , wT s w9 h h h 1 2 1 1 where X 5 hw[X: uw u, p11j, and X 5X2X . Let x and x, respectively, h h h h h denote the Condorcet winner of T s and of T s . Since T s coincides with P, it h 1 follows that ;h 51,...,2k 11, [ x ] hxj5[. Hence, for any Condorcet-consistent h h 1 solution S, S[T s ]5 hxj[ST s x ]5[ h. A 1h 2k 11 h This result again rests heavily upon the non-existence of some representative apportionment. We may expect such apportionment to play the same role for the one-shot and the sequential choice methods. The next theorem shows that this conjecture is false, despite the very restrictive nature of the representativeness requirement: ´ G . Laffond, J. Laine Mathematical Social Sciences 39 2000 35 –53 51 Theorem 8. There exists a profile [s][PN such that GN5G N,s and, R for any Condorcet consistent solution S finer than UC , [STs][ST s S T s]5[ ;A hN ,1 k Kj. A 1 k K k k ´ Proof. See Laffond and Laine 1996. We just sketch the proof here. It involves a slight modification of the profile introduced in the proof of Theorem 4: let X 5 hw,N: uwu52p21j and P be a linear order on X; voter i’s preferences over a pair hw,w9j are now defined by: • if i [[w w9], w s w9 ⇔ w9Pw i • if i [[w 2w9], w s w9 i • if i [N 2[w w9], w s w9 ⇔ wPw9. i Let w be the maximal element of X for P. It is rather easy to check that P 5T s , hence that w is the Condorcet-winner of T s . Let A5 hN ,...,N j[GN where 1 K uN u[h2p21,2p11j, and let w,w9[X be such that wPw9. Suppose that w[ ⁄ A. Consider k constituency N . It follows from [s] that w9T s w ⇔ uN w9u.p. Thus uhk: k k k w9T s w ju1 since constituencies are mutually disjoint. This implies that A is k representative. Now suppose that w[ ⁄ A. Let w [X be s.t. wT s w. We have from 1 [s] that uN 2wu.0.5uN u. Moreover, ;i [N 2w, N s w. Since N s w;i [N , we 1 1 1 1 i 1 i 1 get that N → w in T s , hence w[ ⁄ UC[T s ]. In the case where w[ A, it is left to 1 k 1 the reader to prove that w → w in all T s , where w is the second top element of k X for P. Thus w[ ⁄ UC[T s ]. Hence w[ ⁄ S[T s ] for any S , 1k K k 1k K k UC. h This result may be given the following statistical-type interpretation, dealing with the accuracy of predictions about the election results: suppose that the prevailing system in a society N with 2m 112k 11 voters is direct democracy, and that the set of candidates is X 5 hw,N: uwu52p21j; therefore, the actual voting outcome is given by S[Ts]; moreover, suppose that T s has a Condorcet winner w. Now, consider an opinion research institute which tries to forecast the election outcome, using a sample N of size h 2k 11. The survey outcome is then given by S[T s ]. The above proof defines a profile k for which this outcome announce that, for any possible sampling design, w will be defeated. Our last result deals with the sensitivity of Condorcet-consistent solutions to the choice of a specific apportionment. It states that any such solution is sensitive to the set of representative apportionments. More precisely, the profile described in the proof of the next theorem allows for two representative apportionments leading to mutually disjoint sets of winners: Theorem 9. No Condorcet consistent solution S is weakly non-sensitive to G N,s. R Proof. Consider the profile defined in the proof of Theorem 8 and assume that N 5 h1,...,9j. It is already seen that GN5G N,s. Let A5hN ,N ,N j[GN. Define R 1 2 3 ´ 52 G . Laffond, J. Laine Mathematical Social Sciences 39 2000 35 –53 1 2 2 2 2 for k 51,2,3 the sets X and X by X 5 hw[X: uwN u2j and X 5X2X . k k k k k k Using the same arguments as in the proof of Theorem 6, one get that T has w as k k 2 Condorcet winner, where w is the worst element of X for P. Now suppose that P is k k 2 such that w 5 h1,2,3jPw 5h4,5,6jPw 5h7,8,9jPw 5h1,4,7j Pw 5h2,5,8j P w 5 1 2 3 4 5 6 2 h3,6,9j P w ;w±w , h51,...,6. Let A and A9 be the apportionments, respectively h 9 9 9 defined by A5 hN ,N ,N j5hw ,w ,w j and A95hN ,N ,N j5hw ,w ,w j. It fol- 1 2 3 1 2 3 1 2 3 4 5 6 9 lows that ST 5 hw ,w ,w j, whereas ST 5 hw ,w ,w j. Thus h 51,2,3 h 1 2 3 h 51,2,3 h 4 5 6 ST ST 5[. It is left to the reader to check that the proof is easily A k 51,2,3 k generalized to any number n of voters. h

5. Concluding comments