Divided Majority and Information Aggregation - Theory and Experiment Appendices

Appendices Appendix A1: Pivot Probabilities and Correcting Factors in AV AV The pivotal event pivAC is de…ned as follows: xA > xB 1 and xA + xAB 2 fnC xA = xB and xA + xAB = nC ; or xA = xB 1; nC g 1 and xB + xAB = nC : Therefore, the probability of event pivAC in state ! is: ! AC AV Pr pivAC j! = (n 1)! 1 2(nC X Xi) i=0 + + (n 1)! 3 (n 1)! 6 n ( ! nC A) i xAB (n 1) (nC ( ! ( ! AB ) B) i xAB )! xAB ! (n 1 nC +i)! xAB 2(nC xAB =0 2nC +1 n ! n 1 nC ( ! [ ! AB ) A B] 2 [(n 1 nC )!] (2nC n + 1)! n 1 nC 2nC n n nC ( ! ( ! ( ! A) AB ) B) (n i) 1 nC )! (n nC )! (2nC n)! : ! ! ! The pivot probabilities of event pivBC ; pivAB ; and pivBA can be de…ned and computed simi- larly. In particular: ! BC AV Pr pivBC j! = (n 1)! 1 2(nC X Xi) i=0 + + (n 1)! 3 (n 1)! 6 n ( ! (n A) (nC 1) 2(nC i) xAB nC i ( ! ( ! AB ) B) i xAB )! xAB ! (n 1+i nC )! xAB =0 2nC +1 n ! n 1 nC [ ! ( ! A B] AB ) 2 [(n 1 nC )!] (2nC n + 1)! n nC 2nC n n nC ( ! ( ! ( ! A) AB ) B) (n 1 nC )! (n nC )! (2nC 1 n)! ; xAB ! AB ! Pr (pivAB j!) = (n 1)! 1) 1 n+2(iX X nC ( ! k A) i n+(i 1) 2k ( ! ( AB ) 2k!(k i)!(n+i 1 2k)! ! k B) i=0 (n + 1)! 3 (n + 1)! 6 k=i 2nC +1 n ! ! n 1 nC ( ! [ A B] AB ) 2 [(n 1 nC )!] (2nC n + 1)! 2nC n n nC n 1 nC ( ! ( ! ( ! AB ) B) A) (n 1 nC )! (n nC )! (2nC n)! ; and ! BA AV Pr pivBA j! = (n 1)! 1) 1 n+2(iX X nC ( ! k ! n+(i A ) ( AB ) 1) 2k k ( ! B) 2k!(k i)!(n+i 1 2k)! i i=0 + + (n 1)! 3 (n 1)! 6 k=i 2nC +1 n ! ! n 1 nC ( ! [ A B] AB ) 2 [(n 1 nC )!] (2nC n + 1)! n nC n nC 2nC n ( ! ( ! ( ! B) A) AB ) (n 1 nC )! (n nC )! (2nC 1 n)! : Using the pivot probabilities and the expected gains of the di¤erent actions (i.e. (12) ; (13) ; and (14)), we can compute the correcting factors a = [Pr (XA = XB = nC a and b 44 : 1ja) (V + v) + Pr (XA = XB + 1 = nC ja) v + ::: ::: + Pr (XA + 1 = XB = nC ja) V + Pr (XA = XB = nC ja) (V + v)]=6; and b = [Pr (XA = XB = nC 1jb) (V + v) + Pr (XA = XB + 1 = nC jb) V + ::: ::: + Pr (XA + 1 = XB = nC jb) v + Pr (XA = XB = nC jb) (V + v)]=6: ! We directly see that 7.1 = 0 when ! AB 2 f0; 1g; or ! A = 0, or ! B = 0: Appendix A2: Plurality, Equilibrium Analysis Consider a distribution of signals such that r (sA ja) Proof of Proposition 3. r (sB jb) > (n; nc ), in which case sincere voting is not an equilibrium. That is, there exists a signal s 2 fsA ; sB g such that all the voters who received signal s strictly prefer to deviate from a strategy pro…le sincere f sA (A) ; sB (B)g = f1; 1g : Case 1: s = sA . In this case, sincere ) G (AjsA ) G (BjsA ) < 0. Now, consider a second strategy n o 1 pro…le 0 [r (sA ja) + r (sA jb)] ; 1 . With this pro…le, we have: aA = bB and bA = aB ; and thus pbBC = paAC > 0 and pbAC = paBC > 0 and, from (9): G (Ajs) where (i) [V paAC G (Bjs) = [V paAC vpaBC ] [q (ajs) vpaBC ] is positive, and (ii) [q (ajs) q (bjs)] ; q (bjs)] is positive for sA and negative for sB . In other words, all voters would strictly prefer to deviate from 44 Proof available upon request. (18) 0 by voting sincerely. This means that the value of G (AjsA ) [r (sA ja) + r (sA jb)] 1 G (BjsA ) changes sign when sA . (A) is increased from to 1. Since all pivot probabilities are continuous in continuous in sA sA , the di¤erential G (AjsA ) This implies that there must exist a value sA G (BjsA ) is also (A) 2 [r (sA ja) + r (sA jb)] 1 ;1 such that voters with signal sA are indi¤erent between playing A and B. Now, we prove that the strategy pro…le f rium. This pro…le implies: 2 > nC > n and hence: a A Since G (AjsA ) > b B sA (A) ; sB r(sA ja) r(sA ja)+r(sA jb) ; r (sA ja) a A b A > a B (B)g = and b B sA (A) ; 1 is an equilib- A ja) r (sB jb) ; r(sAr(s ja)+r(sA jb) 2 ; paAC > paBC and pbBC > pbAC . G (BjsA ) = 0 for that strategy pro…le, we know from (9) that: q (ajsA ) [V paAC vpaBC ] = q (bjsA ) [V pbBC vpbAC ]; (19) where both sides of the equality are strictly positive. Since q (ajsB ) < q (ajsA ), (19) implies: q (ajsB ) [V paAC vpaBC ] < q (bjsB ) [V pbBC vpbAC ]; (20) which means that a voter who received signal sB strictly prefers to play B. sincere Case 2: s = sB . In this case, another strategy pro…le 00 ) G (Ajs) f"; 1g, with " ! 0 (and hence 1, this strategy pro…le implies G (Ajs) payo¤s with respect to G (AjsA ) sA G (Bjs) > 0 for both signals. Now, consider (B) ! 1). From Proposition G (Bjs) < 0 for both signals. By the continuity of the (A) ; there must therefore exist a value G (BjsA ) = 0 and, by the same argument as in (19 Hence, the strategy pro…le f sA sA (A) ; sB (B)g = sA sA (A) 2 (0; 1) such that 20), G (AjsB ) G (BjsB ) < 0. (A) ; 1 is an equilibrium. Note that sincere stability is not a binding restriction, since all voters vote for their preferred alternative with a probability strictly larger than 0. Appendix A3: Approval Voting, Equilibrium Analysis Lemma 2 If there exists a signal s such that GAV (Ajs) GAV (ABjs) = 0 then GAV (AjsA ) GAV (ABjsA ) > GAV (AjsB ) GAV (ABjsB ) (21) GAV (Bjs) GAV (ABjs) = 0 then GAV (BjsB ) GAV (ABjsB ) > GAV (BjsA ) GAV (ABjsA ) , and (22) GAV (Ajs) GAV (Bjs) = 0 then GAV (AjsA ) GAV (BjsA ) > GAV (AjsB ) GAV (BjsB ) : (23) Proof. We detail the proof for (21). It is similar for the other two implications. Remember that the second term in (15) is necessarily negative. Thus GAV (Ajs) GAV (ABjs) = 0 implies that the …rst term must be strictly positive. It follows immediately that: GAV (Ajs) Thus, (21) follows from GAV (ABjs) q(ajsA ) q(bjsA ) 0 i¤ b AB (V a (V AB q (ajs) q (bjs) b BC V a v BC v) + v) b a + : q(ajsB ) q(bjsB ) . > Lemma 3 In any voting equilibrium under AV, neither A nor B can be approved by all voters. Proof. We prove the proposition by contradiction and for the limit case in which " = 0. By de…nition the results hold when " > 0. Policy A is approved by all voters if and only if In this case, we have: xA + xAB = n and hence pivot events are when xAB = n 1 or n sA ! AC (A) + sA (AB) = 1 = ! BC =0= ! and sB = q (ajs) a AB q (bjs) b AB (V v) ? 0 G (Bjs) G (ABjs) = q (bjs) b BA q (ajs) a BA (V v) ? 0: = ( n 1 ! AB ) , and 2 (AB) : 2. Hence: G (ABjs) ! AB sB = 0. The only possible G (Ajs) with: (A) + ! BA = ( n 2 ! AB ) 2 [(n 1) + (2 n) ! AB ] : Therefore, b BA a BA a AB b AB Now, we show that increasing in (n a AB b AB b BA a BA b AB a AB a AB b AB = = is increasing in 2) log b AB log a AB + log (n b AB n This is non-negative if and only if > 1) + (2 1) + (2 n) n) b AB a ; AB (24) n 1 (25) b AB a AB (from (25) ; it is straightforward that a AB b AB is also a AB , 1) + (2 n) b AB log [(n 1) + (2 n) a AB ] yields: 2 b AB b AB (n (n ). Taking logs, we have that the right-hand side of (24) is Di¤erentiating with respect to when n 2 b AB (n n 2 1) + (2 n) b AB : 1. Therefore, we have that b AB > a AB and b BA > a BA and conversely. We now use this result to prove that A cannot be approved by all voters. From Theorem 1, Lemma 2, and Lemma 2 (in this Appendix), there are 2 cases to check: (i) sA (A) = 1 and sB b AB (A) 2 [0; 1); and (ii) > a AB : (A) = 0 and sB Hence, we have that b BA > there cannot be any equilibrium in which (A) = 0, then either sB G (AjsA ) a AB > a AB G (ABjsA ) > 0: If sA (A) 2 (0; 1]. If a BA ; sA or a AB < b AB ; then (A) = 1 and which implies G (BjsB ) (A) = 1 and b AB sA < sB sB G (ABjsB ) > 0: Thus, (A) 2 [0; 1): If sA b a b a AB : If AB > AB ; then AB b a BA > BA ; and thus G (BjsB ) Therefore, there cannot be any equilibrium in which sB (A) = 0 and (A) 2 [0; 1), then sA (A) 2 (0; 1] and > b AB ; and thus G (ABjsB ) > 0: (A) 2 (0; 1]: Proof of Theorem 1. From McLennan (1998), a strategy that maximizes expected utility must be an equilibrium of such a common value game (and any …nite Bayesian game like ours must have an equilibrium). Now, conjecture some strategy pro…le That is, s (AB) = 0 for s = sA ; sB . In this case, Therefore, G ! AB AV (Ajs) G AV (ABjs) < 0 and G AV ! AB = (Bjs) that can be played under plurality. ! BA AV G = ! ! AC ; =0< ! BC ; ! = a; b. (ABjs) < 0; 8s. This means that = 0 cannot be part of an equilibrium under AV, and that the welfare-maximizing equilibrium under AV must produce strictly higher expected utility than plurality. It remains to show that this equilibrium is sincerely stable. We actually show the stronger statement that, to maximize expected welfare, a strategy must satisfy sA (A) ; sB (B) > 0. We show this by contradiction: suppose that ^ maximizes expected welfare and is such that ^ sA (A) = 0. By Lemma 3, we have other strategy 0 ! A; ! B; ! AB > 0 and hence ^ sB (A) > 0. Then, compare ^ with some in which sA -voters transfer some of their votes from B towards AB; whereas sB -voters adapt their voting strategy so as to maintain all vote shares unchanged in state b.45 As a result, the total vote share of A in state a must increase (i.e. a A (^ ) + a AB a A ( 0) + a AB ( 0) > (^ )), whereas the expected fraction of double votes increases (the total vote share of B remains unchanged). As a result, in state a, the probability that A wins must increase, whereas the probability that C wins decreases weakly. In state b, winning probabilities are unchanged. Hence, ^ cannot maximize expected welfare: a contradiction. Proof of Theorem 2. We prove the Theorem in two steps. First, we show that there is no interior equilibrium in which a voter strictly mixes across the three actions A; B; and AB. Second, we show that sA -voters never play B, nor sB -voters play A in an interior equilibrium: It follows that the only possible interior equilibrium is such that voters with signal sA mix between A and AB, and voters with signal sB mix between B and AB. First, conjecture an equilibrium in which sA (A) ; sA (B) ; sA (AB) > 0: This requires: G (AjsA ) = G (BjsA ) = G (ABjsA ) : In this case, by Lemma 2 (in this Appendix); sB -voters must be playing B with probability 1; i.e. sB (B) = 1: The equilibrium is therefore not interior, a contradiction. Similarly, sA -voters must play A with probability 1 if sB -voters strictly mix between A; B; and AB. 45 If ^ sA (AB) = 1, then one must consider a transfer of sA -votes from AB towards A, and sB -voters adapt their strategy to maintain all a unchanged. Second, imagine that sB -voters play A with strictly positive probability in equilibrium: (0; 1). This requires either (i) G (AjsB ) = G (ABjsB ) sB (A) 2 G (BjsB ) or (ii) G (AjsB ) = G (BjsB ) G (ABjsB ) : By Lemma 2; both (i) and (ii) imply that G (AjsA ) > G (ABjsA ) ; G (BjsA ) ; and hence that A’s strategy cannot be interior. By symmetry, interior equilibrium either. sA (B) 2 (0; 1) cannot be part of an Appendix A4: Sample Instructions (of the main part) This part consists of 100 rounds. The rules are the same for all rounds. At the beginning of this part, you will be randomly assigned to a group of 12 (including yourself). You will belong to the same group throughout the whole experiment. The earnings in this part will depend partly on your decisions, partly on the decisions of the other participants in your group and partly on chance. The Jar. There are two jars: the red jar and the blue jar. The red jar contains 6 red balls and 3 blue balls. The blue jar contains 6 blue balls and 3 red balls. At the beginning of each round, one of the two jars will be randomly selected. Each jar is equally likely to be selected, i.e., each jar is selected with a chance of 50%. You will not be told which jar has been chosen when making your decision.. The Ball. After a jar is selected for your group, the computer will show each of the participants in your group (including yourself) the color of one ball randomly drawn from that jar. Since you are 12 in your group, the computer performs this random draw 12 times. Each ball will be equally likely to be drawn for every member of the group. That is, if the color of the selected jar for your group were red, then all members of your group would draw their sample balls from a jar containing 6 red and 3 blue balls. If the color of your group’s jar were blue, then all members of your group would draw their sample balls from a jar containing 3 red and 6 blue balls. Therefore, if the selected jar is blue, each member of your group has a chance of 2 thirds of receiving a blue ball. And if the selected jar is red, each member of your group has a chance of 2 thirds of receiving a red ball. But you will only see the color of your own ball. This will be the only information you will have to help you guess which jar was selected by the computer. Your decision. Once you have seen the color of one of the balls, you can make your decision. [Treatment Plurality] You will have to vote for Blue, Red or Black. [Treatment Approval Voting] You will have to vote for Blue, Red, Black or any combination of them. That is, you can vote Blue, Red, Black, Blue and Red, Blue and Black, Red and Black, or Blue, Red and Black. You can vote for one of the colors by clicking below the color. You can unvote by clicking again below the color. After making your decision, please press the ‘OK’key. Group Decision. When all participants have taken their decision, the votes of all participants will be added up. On top of that, the computer will add 7 votes for Black. The group decision will depend on the …nal amount of votes that each color receives: If one color has strictly more votes than other colors, this color will be the group decision. If there is a tie between several colors with the most votes, one of the colors with the most votes will be selected randomly. Each color will have the same probability of being chosen. The chosen color will be the group decision. Payo¤ in Each Round. Your payo¤ depends on the group decision and the color of the selected jar. Your payo¤ is indicated in the following table: Jar Selected by the Computer Blue Red Group Decision Blue Red Black 200 110 20 110 200 20 The left part of the table indicates the group decision and the top part of the table indicates the color of the selected jar. If the selected jar is Blue and the group decision is Blue, you get 200 cents. _ _ _ _ _ _ _ _ _ _ Blue _ _ _ _ _ _ _ _ _ _ _ _ _ _ Red, you get 110 cents. _ _ _ _ _ _ _ _ _ _ Blue _ _ _ _ _ _ _ _ _ _ _ _ _ _ Black, you get 20 cents. _ _ _ _ _ _ _ _ _ _ Red _ _ _ _ _ _ _ _ _ _ _ _ _ _ Blue, you get 110 cents. _ _ _ _ _ _ _ _ _ _ Red _ _ _ _ _ _ _ _ _ _ _ _ _ _ Red, you get 200 cents. _ _ _ _ _ _ _ _ _ _ Red _ _ _ _ _ _ _ _ _ _ _ _ _ _ Black, you get 20 cents. To summarize, if the color of the group decision matches the color of the selected jar, your payo¤ is 200. If the color of the group decision is either blue or red, but does not coincide with the color of the selected jar, your payo¤ is 110. Finally, if Black is the color of the group decision, your payo¤ is 20. Information at the end of each Round. Once you and all the other participants have made your choices, the round will be over. At the end of each round, you will receive the following information about the round: Total amount of votes for Blue Total amount of votes for Red Total amount of votes for Black (including the 7/9 added by the computer) Group decision Color of the selected jar Your payo¤ Final Earnings of this Part. At the end of this part, the computer will randomly select four rounds and you will earn the payo¤s you obtained in these rounds. Each of the 100 rounds has the same chance of being selected. Control Questions. Before starting this part, you will have to answer some control questions in the computer terminal. Click Ok after you have answered all the questions of a page. Once you and all the other participants have answered all the questions, Part 1 will start.