Online Appendix for Obviously Strategy-Proof Mechanisms

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1 Onlne Appendx for Obvously Strategy-Proof Mechansms Shengwu L June 19, Proofs omtted from the man text 1.1 Proof of Theorem 1 Proof. Frst we prove the f drecton. Fx agent 1 and preferences 1. Suppose that S 1 s not obvously domnant n G = hh,, A, A,P, c, (I ) 2N,g. We need to demonstrate that there exsts G that s -ndstngushable from G, such that domnant n G. We proceed by constructon. G, G (S 1) s not weakly Let (S1 0,I 1,h sup,s sup 1,dsup c,h nf,s nf 1,dnf c ) be such that I 1 2 (S 1,S1 0 ), hnf 2 I 1, h sup 2 I 1, and u G 1 (h sup,s1,s 0 sup 1,dsup c, 1 ) >u G 1 (h nf,s 1,S nf 1,d nf c, 1 ) (1) Snce G s a game of perfect recall, we can pck (S nf 1,dnf c ) such that h nf z G (h ;,S 1,S nf 1,dnf c ), by specfyng that (S nf 1,dnf c ) plays n a way consstent wth h nf at any h h nf. Lkewse for h sup and (S sup 1,dsup c ). Suppose we have so done. We now defne another game G s 1-ndstngushable from G. Intutvely, we construct ths as follows: 1. We add a chance move at the start of the game; chance can play L or R. 2. Agent 1 does not at any hstory know whether chance played L or R. 3. If chance plays L, then the game proceeds as n G. 4. If chance plays R, then the game proceeds mechancally as though all players n N \ 1 and chance played accordng to S sup 1,dsup c shengwu l@fas.harvard.edu n G, wth one excepton: 1

2 5. If chance played R, we reach the nformaton set correspondng to I 1, and agent 1 plays S 1 (I 1 ), then the game henceforth proceeds mechancally as though all players n N \ 1 and chance played accordng to S nf 1,dnf c n G. Formally, the constructon proceeds as such: Ã = A [ {L, R}, wherea \ {L, R} = ;. There s a new startng hstory h ;, wth two successors ( h ; )={ h L, h R }, Ã( h L )=L, Ã( h R )=R, P ( h ; )=c. Thesubtree H L H startng from h L ordered by s the same as the arborescence (H, H. For j 6= 1, Ĩj s defned as on H. h ). (Ã, P, c, g) are defned on H L exactly as (A,P, c,g) are on We now construct the subtree startng from h R. Let h be such that h 2 (h sup ), z G (h sup,s 1,S sup 1,dsup c ). H 0 {h 2 H 9S 00 1 : h z G (h ;,S 00 1,S sup 1,dsup c )}} \[{h 2 H P (h) =1} [ {h 2 Z}] \{h 2 H h h} (2) In words, these are the hstores that can be reached by some S1 00 when facng, where ether agent 1 s called to play or that hstory s termnal, and such S sup 1,dsup c that those hstores are not h or ts successors. Let h be such that h 2 (h nf ), h z G (h nf,s 1,S nf 1,dnf c ). H 00 {h 2 H 9S 00 1 : h z G (h ;,S 00 1,S nf 1,d nf c )} \[{h 2 H P (h) =1} [ {h 2 Z}] \{h 2 H h h} (3) In words, these are the hstores that can be reached by some S 00 1 when facng Snf 1,dnf c, where ether agent 1 s called to play or that hstory s termnal, and such that those hstores are h or ts successors. We now paste these together. Let H R be the rooted subtree ordered by, for some bjecton : H R! H 0 [ H 00, such that for all h, h 0 2 H R, h h0 f and only f 1. EITHER: ( h), ( h 0 ) 2 H 0 and ( h) ( h 0 ) 2. OR: ( h), ( h 0 ) 2 H 00 and ( h) ( h 0 ) 3. OR: ( h) h and h ( h 0 ) 2

3 The root of ths subtree exsts and s unque; t corresponds to 1 (h), where h s the earlest hstory precedng z G (h ;,S 1,S sup 1,dsup c )} where 1 s called to play. Let h R be the root of H R. Ths completes the specfcaton of H. For all h 2 H R,wedefne: 1. g( h) =g( ( h)) f h s a termnal hstory. 2. P ( h) =1f h s not a termnal hstory. For all h 2 H R \ h R,wedefneÃ( h) =A(h), for the unque ( h 0,h) such that: 1. h 2 ( h 0 ) 2. h 2 ( ( h 0 )) 3. h ( h) We now specfy the nformaton sets for agent 1. Every h 2 H L corresponds to a unque hstory n H. Weuse as L on H L and on H R. L to denote the bjecton from H L to H. Let ˆ be defned 1 s nformaton partton Ĩ1 s defned as such: 8 h, h 0 2 H : f and only f 9Ĩ0 1 2 Ĩ1 : h, h 0 2 Ĩ0 1 (4) 9I I 1 :ˆ( h), ˆ( h 0 ) 2 Ĩ0 1 (5) All that remans s to defne c ; we need only specfy that at h ;, c plays R wth certanty. 1 G = h H,, Ã, Ã, P, c, (Ĩ) 2N, g s 1-ndstngushable from G. Every experence at some hstory n H L corresponds to some experence n G, and vce versa. Moreover, any experence at some hstory n H R could also be produced by some hstory n H L. Let G, G be the approprate bjecton from 1 s nformaton sets and actons n G onto 1 s nformaton sets and actons n G. Take arbtrary S 1. Observe that snce I 1 2 (S 1,S 0 1 ), G, G (S 1) and G, G(S 0 1 ) result n the same hstores followng h R,untlthey reach nformaton set G, G (I 1). Havng reached that pont, G, G (S 1) leads to outcome g(z G (h nf,s 1,S nf 1,dnf c )) and G, G(S 1 0 ) leads to outcome g(zg (h sup,s1 0,Ssup 1,dsup c )). Thus, 1 If one prefers to avod c wthout full support, an alternatve proof puts probablty on L, for close to 0, or for games wth N > 2assgns P ( h ; )=2. 3

4 E c[u G 1 ( h ;, G, G (S0 1), S 1, d c, 1 )] = u G 1 (h sup,s 0 1,S sup 1,dsup c, 1 ) >u G 1 (h nf,s 1,S nf 1,d nf c, 1 ) = E c[u G 1 ( h ;, G, G (S 1), S 1, d c, 1 )] (6) So G, G(S 1 ) s not weakly domnant n G. We now prove the only f drecton. Take arbtrary G. Suppose G, G (S 1) S 1 s not weakly domnant n G (gven type 1 ). We want to show that S 1 s not obvously domnant n G. There exst S 1 0 and S 0 1 such that: E c[u G 1 ( h ;, S 0 1, S 0 1, d c, 1 )] > E c[u G 1 ( h ;, S 1, S 0 1, d c, 1 )] (7) Ths nequalty must hold for some realzaton of the chance functon, so there exsts d c such that: Fx ( S 0 1, S 0 1, d c, 1 ). u G 1 ( h ;, S 0 1, S 0 1, d c, 1 ) >u G 1 ( h ;, S 1, S 0 1, d c, 1 ) (8) Defne: z G( h ;, S 0 1, S 0 1, d c ) 6= z G( h ;, S 1, S 0 1, d c ) (9) H { h 2 H h z G( h ;, S 0 1, S 0 1, d c ) and h z G( h ;, S 1, S 0 1, d c )} (10) h h 2 H such that 8 h 0 2 H : h 0 h (11) Snce the opponent strateges and chance moves are held constant across both sdes of Equaton 9, P ( h ) = 1 and h 2 Ĩ1, where S 1 (Ĩ1) 6= S 0 1 (Ĩ1). Moreover, Ĩ1 2 ( S 1, S 0 1 ) and G,G (Ĩ1) 2 (S 1,S 0 1 ), where we denote S0 1 G,G ( S 0 1 ). Snce G and G are 1-ndstngushable, consder the experences G,G ( 1 (z G( h ;, S 1, S 0 1, d c ))) and G,G ( 1 (z G( h ;, S 0 1, S 0 1, d c ))). In G, G,G ( 1 (z G( h ;, S 1, S 0 1, d c ))) could lead to outcome g(z G( h ;, S 1, S 0 1, d c )). We use (S nf 1,dnf c ) to denote the correspondng opponent strateges and chance realzatons that lead to that outcome. We denote h nf h 2 G,G (Ĩ1) :h z G (h ;,S 1,S nf 1,dnf c ). 4

5 In G, G,G ( 1 (z G( h ;, S 1 0, S 0 1, d c ))) could lead to outcome g(z G( h ;, S 1 0, S 0 1, d c )). We use (S sup 1,dsup c ) to denote the correspondng opponent strateges and chance realzatons that lead to that outcome. We denote h sup h 2 G,G (Ĩ1) :h z G (h ;,S1 0,Ssup 1,dsup c ). u G 1 (h sup,s 0 1,S sup 1,dsup c, 1 ) = u G 1 ( h ;, S 0 1, S 0 1, d c, 1 ) >u G 1 ( h ;, S 1, S 0 1, d c, 1 ) = u G 1 (h nf,s 1,S nf 1,d nf c, 1 ) (12) where h sup,h nf 2 G,G (Ĩ1) and G,G (Ĩ1) 2 (S 1,S 0 1 ). Thus S 1 s not obvously domnant n G. 1.2 Proof of Theorem 2 Proof. The key s to see that, for every G 2 G, there s a correspondng S 0, and vce versa. We use S 0 to denote the support of S0. In partcular, observe the followng somorphsm: Table 1: Equvalence between extensve game forms and Planner mxed strateges G S0 d c S0 2 S 0 c the probablty measure specfed by S 0 g(z) for z 2 Z the Planner s choce of outcome when she ends the game I ((m k,r k,r k ) t 1 k=1,m t,r t ) consstent wth some S 0 2 S 0 A(I ) R t (z) o c consstent wth some S 0 2 S 0 and S N Informaton sets n G are equvalent to sequences of past communcaton ((m k,r k,r k ) t 1 k=1,m t,r t ) under S 0. Avalable actons at some nformaton set A(I ) are equvalent to acceptable responses R t. Thus, for any strategy n some game G, we can construct an equvalent strategy gven approprate S 0, and vce versa. Furthermore, fxng a chance realzaton d c and agent strateges S N unquely results n some outcome. Smlarly, fxng a realzaton of the planner s mxed strategy S 0 2 S 0 and agent strateges S N unquely determnes some outcome. Consequently, for any G 2 G, there exsts S 0 wth the same strateges avalable for each agent and the same resultng 5

6 (probablty measure over) outcomes, and vce versa. 2 Table 1 summarzes. The next step s to see that a blateral commtment Ŝ 0 s equvalent to the Planner promsng to run only games n some equvalence class that s -ndstngushable. Suppose that there s some G that OSP-mplements f. Pck some equvalent S 0 wth support S 0. For each 2 N, specfy the blateral commtment Ŝ 0 1 ( ( S 0 )). These blateral commtments support f. To see ths, take any S Ŝ 0, wth support S 0 0. For any S S 0 0, for any S 0 N,there exsts S 0 2 S 0 and S N such that ( S 0 0, S 0 N )= ( S 0, S N ). By constructon, G s such that: There exsts z 2 Z where (z) and g(z) are equvalent to ( S 0, S N ). Thus, for G 0 that s equvalent to S 0 0, every termnal hstory n G0 results n the same experence for and the same outcome as some termnal hstory n G. Consequently, G and G 0 are -ndstngushable. Thus, by Theorem 1, the strategy assgned to agent wth type s weakly domnant n G 0, whch mples that t s a best response to S 0 0 and any S N\ n the blateral commtment game. Thus, f f s OSP-mplementable, then f can be supported by blateral commtments. Suppose that f can be supported by blateral commtments (Ŝ 0 ) 2N, wthrequste S 0 (wth support S 0 ) and S N. Wthout loss of generalty, let us suppose these are mnmal blateral commtments,.e. Ŝ 0 = 1 ( ( S 0 )). Pck G that s equvalent to S 0. G OSP-mplements f. To see ths, consder any G 0 such that G and G 0 are -ndstngushable. Let S 0 0 denote the Planner strategy that corresponds to G 0. At any termnal hstory z 0 n G 0, the resultng experence (z 0 ) and outcome g 0 (z 0 ) are equvalent to the experence (z) and outcome g(z) for some termnal hstory z n G. These n turn correspond to some observaton o 2 ( S 0 ). Thus S Ŝ 0. Snce f s supported by (Ŝj 0 ) j2n, S s a best response (for type )to S 0 0 and any S N\. Thus, the equvalent strategy S ( ) s weakly domnant n G 0. Snce ths argument holds for all -ndstngushable G 0,by Theorem 1, S ( ) s obvously domnant n G. Thus,ff can be supported by blateral commtments, then f s OSP-mplementable. 1.3 Proof of Proposton 2 Proof. We prove the contrapostve. Suppose ( G, S N ) does not OSP-mplement f. Then there exsts some (,, S 0, Ĩ) such that Ĩ 2 ( S ( ), S 0 ) and 2 Implctly, ths reles on the requrement that both G and S 0 have fnte length. If one had fnte length but the other could be nfntely long, the resultng outcome would not be well defned and the equvalence would not hold. 6

7 u G ( h, S ( ), S, d c,, ) <u G ( h 0, S, 0 S 0, d 0 c,, ) (13) for some ( h, S, d c ) and ( h 0, S 0, d 0 c). Notce that h and h 0 correspond to hstores h and h 0 n G. Moreover, we can defne S 0 = S 0 at nformaton sets contanng hstores that are shared by G and G, and specfy S 0 arbtrarly elsewhere. We do the same for ( S, d c ) and ( S 0, d 0 c), to construct (S,d c ) and (S,d c ). But, startng from h and h 0 respectvely, these result n the same outcomes as ther partners n G. Thus, u G (h, S ( ),S,d c,, ) <u G (h 0,S 0,S 0,d 0 c,, ) (14) h, h 0 2 I, for I 2 (S ( ),S 0 ). Thus (G, S N) does not OSP-mplement f. 1.4 Proof of Theorem 3 We splt ths proof nto two parts. Proposton 1. If (G, S N ) OSP-mplements f y, then P(G, S N ) s a personal-clock aucton. Proof. Throughout ths proof, we use the followng notaton: Gven some type-strategy S, S denotes the strategy assgned to by S. Take any (G, S N ) that mplements (f y,f t ). For any hstory h, wedefne h { N h z G (h ;, (S ) 2N} (15) For nformaton set I,wedefne h, { 9 :(, ) 2 h } (16) I [ h2i h (17) I, { 9 :(, ) 2 I } (18) 1 I, { 9 :(, ) 2 I and 2 f y (, )} (19) 7

8 0 I, { 9 :(, ) 2 I and /2 f y (, )} (20) Some observatons about ths constructon: 1. Snce player s type-strategy depends only on hs own type, I, = h, for all h 2 I. 2. I, = 1 I, [ 0 I, 3. Snce SP requres 1 2fy( ) weakly ncreasng n, 1 I, domnates 0 I, n the strong set order. Lemma 1. Suppose (G, S N ) OSP-mplements (f y,f t ), where G hh, For all, foralli 2 I,f:, A, A,P, c, (I ) 2N,g. 1. < I, I, a I. then S (I )=S 0 (I ). Equvalently, for any I, there exsts a I such that for all 2 1 I, \ 0 I,, S (I )= Suppose not. Take (, I,, 0 ) consttutng a counterexample to Lemma 1. Snce 2 1 I,, there exsts h 2 I and S such that 2 g y (z G (h, S,S )). Fx t g t, (z G (h, S,S )). Snce I,, there exsts h0 2 I and S 0 such that /2 g y (z G (h 0,S 0,S0 )). Fx t0 g t,(z G (h 0,S 0,S0 )). Snce S (I ) 6= S 0 (I ) and [ 0 I,, I 2 (S,S 0 ). Thus, OSP requres that u (,h,s,s ) u (,h 0,S 0,S0 ) (21) whch mples + t t 0 (22) and u ( 0,h,S,S ) apple u ( 0,h 0,S 0,S0 ) (23) 8

9 whch mples 0 + t apple t 0 (24) But 0 >,so a contradcton. Ths proves Lemma t >t 0 (25) Lemma 2. Suppose (G, S N ) OSP-mplements (f y,f t ) and P(G, S N )=G. Take any I such that 1 I, \ 0 I, 6= ;, and assocated a I. 1. If there exsts 2 0 I, such that S (I ) 6= a I, then there exsts t 0 such that: (a) For all 2 0 I, such that S (I ) 6= a I,forallh 2 I,forallS, g t, (z G (h, S,S )) = t 0. (b) For all 2 I, such that S (I ) = a I, for all h 2 I, for all S, f /2 g y (z G (h, S,S )), then g t, (z G (h, S,S )) = t If there exsts 2 1 I, such that S (I ) 6= a I, then there exsts t 1 such that: (a) For all 2 1 I, such that S (I ) 6= a I,forallh 2 I,forallS, g t, (z G (h, S,S )) = t 1. (b) For all 2 I, such that S (I ) = a I, for all h 2 I, for all S, f 2 g y (z G (h, S,S )), then g t, (z G (h, S,S )) = t 1. Take any type I, such that S 0 (I ) 6= a I. Take any type I, such that S 00 (I )=a I. (By 1 I, \ 0 I, 6= ; there exsts at least one such type.) Notce that I 2 (S 0,S 00 ). By Lemma 1, 0 /2 1 I,, and the game s pruned. Thus, Snce I,, 8h 2 I : 8S : /2 g y (z G (h, S 0,S )) (26) 9h 2 I : 9S : /2 g y (z G (h, S 00,S )) (27) OSP requres that type 0 does not want to (nf-sup) devate. Thus, 9

10 nf h2i,s g t, (z G (h, S 0,S )) sup {g t, (z G (h, S 00,S )) : /2 g y (z G (h, S 00,S ))} h2i,s (28) OSP also requres that type 00 does not want to (nf-sup) devate. Ths mples nf {g t, (z G (h, S 00,S )) : /2 g y (z G (h, S 00,S ))} h2i,s sup g t, (z G (h, S 0,S )) h2i,s (29) The RHS of Equaton 28 s weakly greater than the LHS of Equaton 29. The RHS of Equaton 29 s weakly greater than the LHS of Equaton 28. Consequently all four terms are equal. Moreover, ths argument apples to every I, such that S 0 (I ) 6= a I, and every 00 (1b) I, 2 0 I, such that S 00 (I )=a I. Snce the game s pruned, 00 satsfes and S 00 (I )=a I. Ths proves part 1 of Lemma 2. Part 2 follows by symmetry; we omt the detals snce they nvolve only small changes to the above argument. Lemma 3. Suppose (G, S N ) OSP-mplements (f y,f t ) and P(G, S N )=G. Take any I such that 1 I, \ 0 I, 6= ;, and assocated a I. Let t 1 and t 0 be defned as before. 1. If there exsts 2 0 I, such that S (I ) 6= a I, then for all (h 2 I,S,S ),f 2 g y (z G (h, S S )), then g t, (z G (h, S,S )) apple t 0 sup{ 2 0 I, : S (I ) 6= a I }. 2. If there exsts 2 1 I, such that S (I ) 6= a I, then for all (h 2 I,S,S ),f /2 g y (z G (h, S S )), then g t, (z G (h, S,S )) apple nf{ 2 1 I, : S (I ) 6= a I } + t 1. Suppose that part 1 of Lemma 3 does not hold. Fx (h 2 I,S,S ) such that 2 g y (z G (h, S S )) and g t, (z G (h, S,S )) >t 0 sup{ 2 0 I, : S (I ) 6= a I }.Snce G s pruned, we can fnd some 0 2 I, such that for every Ĩ 2 {I 0 2 I : I 2 (I 0)}, S 0 (Ĩ) =S (Ĩ). Fx that 0. Fx I, such that S (I ) 6= a I and 00 Snce G s pruned and 00 /2 1 I, By constructon, I 2 (S 0,S 00 ). OSP requres that, for all h 00 2 I,S 00 : sup{ 2 0 I, : S (I ) 6= a I }. (by Lemma 1), t must be that S 00 (I ) 6= S 0 (I ). 10

11 u ( 00,h 00,S 00,S 00 ) u ( 00,h,S 0,S ) (30) whch entals t g t, (z G (h, S,S )) (31) whch entals t 0 sup{ 2 0 I, : S (I ) 6= a I } + g t, (z G (h, S,S )) (32) But, by hypothess, t 0 sup{ 2 0 I, : S (I ) 6= a I } <g t, (z G (h, S,S )) (33) Snce ths argument holds for all > 0, we can pck small enough to create a contradcton. Ths proves part 1 of Lemma 3. Part 2 follows by symmetry. Lemma 4. Suppose (G, S N ) OSP-mplements (f y,f t ) and P(G, S N )=G. Take any I such that 1 I, \ 0 I, > 1 and assocated a I. 1. If there exsts 2 0 I, such that S (I ) 6= a I, then for all I,, S 0 (I )=a I. 2. (Equvalently) If there exsts 2 1 I, such that S (I ) 6= a I, then for all I,, S 0 (I )=a I. Suppose Part 1 of Lemma 4 does not hold. Fx I, and choose 0 < 00 such that { 0} [ { 00 } 1 I, \ 0 I,. Fx I, such that S 000 (I ) 6= a I. By Lemma 1, f ,thenS 000 I, (I )=a I, a contradcton. Thus, I, \ 0 I,, and snce 1 I, domnates 0 I, n the strong set order, 00 < 000. Snce I,, there exsts h0 2 I and 0 such that ( 0, 0 ) 2 I and 2 g y (z G (h 0,S 0,S 0 )). By Lemma 2, there exsts a 2 A (I ) such that a 6= a I and choosng a ensures /2 y and t = t 0.Thus,byGSP 0 + g t, (z G (h 0,S 0,S 0 )) t0 (34) By 00 Lemma I,,thereexstsh00 2 I and 00 such that /2 g y(z G (h 00,S 00,S 00 )). By g t, (z G (h 00,S 00,S 00 )) = t0 (35) 11

12 By G SP, 2 g y (z G (h 0,S 000 Notce that I 2 (S 00 devate to 000,S 0 )) and g t,(z G (h 0,S 000,S 0 )) = g t,(z G (h 0,S 0,S 0 )). does not want to (nf-sup),s 000 ). Thus, OSP requres that 00 s strategy, whch entals: g t, (z G (h 00,S 00,S 00 )) + g t, (z G (h 0,S 000,S 0 )) (36) 00 whch contradcts Equaton 34. t g t, (z G (h 0,S 000,S 0 )) (37) > 0 + g t, (z G (h 0,S 0,S 0 )) Part 2 s the contrapostve of Part 1. Ths proves Lemma 4. Lemma 5. Suppose (G, S N ) OSP-mplements (f y,f t ) and P(G, S N )=G. For all I,f 1 I, \ 0 I, apple 1 and A(I ) 1. For all 2 I,, h 2 I, S : (a) If /2 g y (z G (h, S,S )) then g t, (z G (h, S,S )) = t 0 (b) If 2 g y (z G (h, S,S )) then g t, (z G (h, S,S )) = t 1 2. If 1 I, > 0 and 0 I, > 0, then t1 = nf{ 2 1 I, } + t0 2, then there exsts t 1 and t 0 such that: By G pruned, I, 6= ;. By the Green-La ont-holmström Theorem, f t, (, )= 1 2fy( ) nf{ 0 2 f y ( 0, )} + r ( ) (38) Consder the case where 1 I, = ;. Pck some I, and some h0 2 I,S 0. Fx t 0 g t, (z G (h 0,S 0,S0 )). Suppose there exsts some ( 00, 00 ) 2 I such that f t, ( 00, 00 )=t0 00 6= t 0. Pck h 00 2 I such that h 00 z G (;,S 00,S 00 ). By Equaton 38, for all 2 I,, f t, (, 00 )=t0 00.ByGprunedand A(I ) 2, we can pck I, such that S 000 (I ) 6= S 0 (I ). Notce that I 2 (S 000,S 0 ). If t0 00 >t 0,then u (,h 0 0,S 0,S0 )=t 0 <t 0 00 = u (,h 0 00,S 000,S 00 ) (39) so S 0 s not obvously domnant for (, 0). If t0 00 <t 0,then u ( 000,h 0,S 0,S0 )=t 0 >t 0 00 = u ( 000,h 00,S 000,S 00 ) (40) 12

13 so S 000 s not obvously domnant for (, 000 ). By contradcton, ths proves Lemma 5 for ths case. A symmetrc argument proves Lemma 5 for the case where 0 I, = ;. Note that, f Lemma 5 holds at some nformaton set I, t holds at all nformaton sets I 0 that follow I. Thus, we need only consder some earlest nformaton set I at whch 1 I, \ 0 I, apple 1 and A(I ) 2. Now we consder the case where 1 I, 6= ; and 0 I, 6= ;. At every pror nformaton set I pror to I, 1 I, \ 0 I, > 1. Snce 1 I, 6= ; and 0 I, 6= ;, by Lemma 4, I s reached by some nterval of types all takng the same acton. Thus sup{ 2 0 I,} =nf{ 2 1 I,}. Fx ˆ 2 0 I, such that ˆ sup{ 2 0 I,}. Choose correspondng ĥ 2 I and ˆ 2 I, such that /2 g y (z G (ĥ, S ˆ,Sˆ )). Defne t0 g t,(z G (ĥ, S ˆ,Sˆ )). Fx ˇ 2 1 I, such that ˇ apple nf{ 2 1 I,} +. Choose correspondng ȟ 2 I and ˇ 2 I, such that 2 g y (z G (ȟ, S ˇ,Sˇ )). Defne t1 g t,(z G (ȟ, S ˇ,Sˇ )). Suppose there exsts some ( 0, 0 ) 2 I such that /2 f y ( 0, 0 ) and f t,( 0, 0 )= t 0 0 6= t 0. Snce sup{ 2 0 I,} = nf{ 2 1 I,}, t follows that for all 2 I,, nf{ : 2 f y (, )} = nf{ 2 1 I,}. Thus, by Equaton 38, for all 2 I,: f t, (, 0 ) = 1 2f y(, 0 ) nf{ 2 1 I,} + t0 0. Fx h 0 2 I such that z G (;,S 0,S 0 ). By A(I ) 2, we can pck some 00 2 I, such that S 00 (I ) 6= S ˆ (I ). Notce that I 2 (S 00,Sˆ ). Ether I or I \ 0 I. Suppose I. Suppose t0 0 >t 0. By OSP, h 0 u (ˆ, ĥ, S ˆ,Sˆ ) u (ˆ,h 0,S 00,S 0 ) (41) whch entals t 0 t fy( 00, 0 ) ( ˆ nf{ 2 1 I,}) t fy( 00, 0 ) ( ) t 0 0 (42) OSP and we can pck small enough to consttute a contradcton. Suppose t 0 0 <t 0.By u ( 00, ĥ, S ˆ,Sˆ ) apple u ( 00,h 0,S 00,S 0 ) (43) 13

14 whch entals whch s a contradcton. t 0 apple t fy( 00 = t fy( 00, 0 ) ( 00, 0 ) ( 00 nf{ 2 1 I,}) sup{ 2 0 I,}) apple (44) t0 0 The case that remans s I \ 0 I. Then 2 f y( 00, 0 ) and f t,( 00, 0 )= nf{ 2 1 I,} + t0 0. Suppose t 00 >t 0.OSPrequres: u (ˆ, ĥ, S ˆ,Sˆ ) u (ˆ,h 0,S 00,S 0 ) (45) whch entals t 0 ˆ nf{ 2 1 I,} + t0 0 t 0 0 (46) and we can pck small enough to consttute a contradcton. 0 Suppose t 0 < t 0. Snce S 00 (I ) 6= S ˆ (I ), ether S ˇ (I ) 6= S ˆ (I ) or S ˇ (I ) 6= S 00 (I ). Moreover, f t, (ˇ, 0)= 1 2f y(ˇ, 0 ) nf{ 2 1 I,} + t0 0. Suppose S ˇ (I ) 6= S ˆ (I ). OSP requres: u (ˇ,h 0,Sˇ,S 0 ) u (ˇ, ĥ, S ˆ,Sˆ ) (47) whch entals 1 2fy(ˇ, 0 )( ˇ nf{ I,})+t0 t 0 (48) whch entals 1 2fy(ˇ, 0 ) + t0 0 t 0 (49) and we can pck small enough to yeld a contradcton. Suppose S ˇ (I ) 6= S 00 (I ). By Equaton 38, f t, (ˇ, 0 )= 1 2f y(ˇ, 0 ) nf{ 2 1 I,} + t0 0, and ft, ( 00, ˆ )= nf{ 2 1 I,} + t0.osprequres u (ˇ,h 0,Sˇ,S 0 ) u (ˇ, ĥ, S 00,Sˆ ) (50) 14

15 whch entals 1 2fy(ˇ, 0 )(ˇ nf{ I,})+t0 (ˇ nf{ 2 1 I,})+t0 (51) whch entals 1 2fy(ˇ, 0 ) + t0 0 + t 0 (52) whch entals t 0 0 t 0 (53) a contradcton. By the above argument, for all I satsfyng the assumptons of Lemma 5, there s a unque transfer t 0 for all termnal hstores z passng through I such that /2 g y (z). Equaton 38 thus mples that there s a unque transfer t 1 for all termnal hstores z passng through I such that 2 g y (z). Moreover, t 1 = nf{ 2 1 I, } + t0. Ths proves Lemma 5. Now to brng ths all together. We leave showng parts (1.d.v) and (2.d.v) of Defnton 15 to the last. Take any (Ĝ, ŜN) that OSP-mplements (f y,f t ). Defne G P(Ĝ, ŜN) and S N as ŜN restrcted to G. By Proposton 2, (G, S N )OSP-mplements (f y,f t ). We now characterze (G, S N ). For any player, consder any nformaton set I such that A(I ) 2, and for all pror nformaton sets I0 I, A(I0 ) = 1. By Lemma 1, there s a unque acton a I taken by all types n 1 I, \ 0 I,. Ether 1 I, \ 0 I, > 1 or 1 I, \ 0 I, apple 1. If 1 I, \ 0 I, > 1, then by Lemma 4, G pruned and A(I ) 2, 1. EITHER: There exsts 2 0 I, such that S (I ) 6= a I, and for all I,, S 0 (I )=a I. 2. OR: There exsts 2 1 I, such that S (I ) 6= a I, and for all I,, S 0 (I )= a I. In the frst case, then by Lemma 2, there s some t such that, for all (S,S ), for all h 2 I,f/2 g y (z G (h, S,S ), then g t, (z G (h, S,S )= t. Moreover, we can defne a gong transfer at all nformaton sets I 0 such that I I 0 : t (I 0 ) mn I 00:I I 00 [ t sup{ 2 0 I 0 I 00, : S (I00 ) 6= a I 00 }] (54) 15

16 Notce that ths functon falls monotoncally as we move along the game tree; for any I 0,I00 such that I 0 I 00, t (I 0) t (I 00). Moreover, by constructon, at any I0,I00 such that I 0 I 00, t (I 0) > t (I 00 ), and there does not exst I000 such that I 0 I 000 I 00,then there exsts a 2 A(I 00) that yelds /2 y, and by Lemma 2 ths yelds transfer t. We defne A q to nclude all such quttng actons;.e. A q s the set of all actons such that: 1. a 2 I for some I 2 I 2. For all z such that a 2 (z): /2 g y (z) and g t, (z) = t Lemma 3 and SP together mply that, at any termnal hstory z, f 2 g y (z), then g t, (z) = nf t (I ) (55) I :I I z Ths holds because, f g t, (z) < nf I :I I z t (I ), then type such that t nf I :I I z t (I )) < < t g t, (z) could proftably devate to play a 2 A 0 at nformaton set I. In the second case, then by Lemma 2, there s some t such that, for all (S,S ), for all h 2 I,f 2 g y(z G (h, S,S ), then g t, (z G (h, S,S )= t. Moreover, we can defne a gong transfer at all nformaton sets I 0 such that I I 0 ): t (I 0 ) mn I 00:I I 00 We defne A 1 symmetrcally for ths second case. [ t +nf{ 2 0 I 0 I 00, : S (I00 ) 6= a I 00 }] (56) Part (1.d.) and (and ts analog n Clause 2) of Defnton 15 follow from Lemma 4. The above constructons su ce to prove Theorem 3 for cases where 1 I, \ 0 I, > 1. Cases where 1 I, \ 0 I, apple 1 are dealt wth by Lemma 5. Now for the last pece: We prove that parts (1.d.v) and (and ts analog n Clause 2) of Defnton 15 hold. The proof of part (1.d.v) s as follows: Suppose we are facng Clause 1 of Defnton 15, and for some I, A(I 0 ) \ A0 > 1. By part (1.d.), we know that the gong transfer t can fall no further. Snce G s pruned and A(I 0 ) \ A0 > 1, there exst two dstnct types of,, 0 2 I 0,, who do not qut at I 0, and take d erent actons. Snce nether quts at I 0 and the gong transfer falls no further, there exst, 0 2 I 0, such that 2 f y (, ) and 2 f y ( 0, 0 ). So there exst (h 2 I0,S ) and (h 0 2 I 0,S0 ) such that 2 g y (h, S,S ) (57) 2 g y (h 0,S 0,S0 ) (58) 16

17 g t, (h, S,S )=g t, (h 0,S 0,S0 )= t (I 0 ) (59) WLOG suppose < 0. Suppose that there does not exst a 2 A(I0 ) such that, for all z such that a 2 (z), 2 g y (z). Then there must exst (h 00 2 I 0,S00 ) such that /2 g y (h 00,S 0,S00 ) (60) Note that I 0 2 (S,S 0 ). But then S 0 snce g t, (h 00,S 0,S00 )= t (61) s not obvously domnant, a contradcton, u G (h 00,S 0,S00, 0 )= t apple + t (I 0 ) < 0 + t (I 0 )=u G (h, S,S, 0 ) (62) (The frst nequalty holds because of type s ncentve constrant.) Ths shows that part (1.d.v) of Defnton 15 holds. Its analog n Clause 2 s proved symmetrcally. Proposton 2. If G s a personal-clock aucton, then there exst S N and f y such that (G, S N ) OSP-mplements f y. Proof. Take any personal-clock aucton G. For any and any, the followng strategy S s obvously domnant: 1. If encounters an nformaton set consstent wth Clause 1 of Defnton 15, then, from that pont forward: (a) If + t (I ) > t and there exsts a 2 A(I ) \ A q,playa 2 A(I ) \ A q.. If A(I ) \ A q > 1, then play a 2 A(I 0 ) such that: For all z such that a 2 (z): 2 g y (z). (b) Else play some a 2 A q If encounters an nformaton set consstent wth Clause 2 of Defnton 15, then, from that pont forward: (a) If + t < t (I ) and there exsts a 2 A(I ) \ A q,playa 2 A(I ) \ A q. 3 If A q \ A(I ) > 1, the agent chooses determnstcally but arbtrarly. 17

18 . If A(I ) \ A q > 1, then play a 2 A(I 0 ) such that: For all z such that a 2 (z): /2 g y (z). (b) Else play some a 2 A q. The above strategy s well-defned for any agent n any personal-clock aucton, by nspecton of Defnton 15. Consder any devatng strategy S 0. At any earlest pont of departure, the agent wll have encountered an nformaton set consstent wth ether Clause 1 or Clause 2 of Defnton 15. Suppose that the agent has encountered an nformaton set covered by Clause 1. Take some earlest pont of departure I 2 (S,S 0 ). Notce that, by (1.a) of Defnton 15, no matter what strategy plays, condtonal on reachng I, ether agent s not n the allocaton and receves t, or agent s n the allocaton and receves a transfer ˆt apple t (I ). Suppose + t (I ) > t. Note that under S, condtonal on reachng I, the agent ether s not n the allocaton and receves t, or s n the allocaton and receves a transfer strctly above t.ifs 0(I ) 2 A q (.e. f agent quts), then the best outcome under S 0 s no better than the worst outcome under S.IfS 0(I ) /2 A q,then,snces 0(I ) 6= S (I ), A(I ) \ A q > 1. Then, by (1.d.) of Defnton 15, t wll fall no further. So S (I ) guarantees that wll be n the allocaton and receve transfer t (I ). But, by (1.a) of Defnton 15, the best possble outcome under S 0 condtonal on reachng I s no better, so the obvous domnance nequalty holds. Suppose + t (I ) apple t.then,unders, condtonal on reachng I, agent s not n the allocaton and has transfer t. 4 However, under S 0, ether the outcome s the same, or agent s n the allocaton for some transfer ˆt apple t (I ) apple t. Thus, the best possble outcome under S 0 s no better than the worst possble outcome under S, and the obvous domnance nequalty holds. The argument proceeds symmetrcally for Clause 2. Notce that the above strateges result n some allocaton and some payments, as a functon of the type profle. We defne these to be (f y,f t ), such that G OSP-mplements (f y,f t ). 2 Alternatve Emprcal Specfcatons Here we report alternatve emprcal specfcatons for the experment. 4 By (1.d.) of Defnton 15, ether wll have qut n the past, or wll have an opportunty to qut now, whch he exercses. 18

19 A natural measure of errors would be to take the sum, for k =1, 2, 3, 4, of the absolute d erence between the kth hghest bd and the kth hghest value. However we do not observe the hghest bd under AC, and we often do not observe the hghest bd under AC+X. We could nstead take the sum for k =2, 3, 4 of the absolute d erence between the kth hghest bd and the kth hghest value, averaged as before n fve-round blocks. Table 2 reports the results. Table 2: mean(sum(abs(kth bd - kth value))), for k = 2, 3, 4 Format Rounds SP OSP p-value <.001 Aucton (4.64) (1.89) (2.73) (0.91) X Aucton (3.70) (1.06) (3.70) (0.75) For each aucton, we sum the absolute d erences between the kth bd and the kth value, for k =2, 3, 4. We then take the mean of ths over each 5-round block. We then compute standard errors countng each group s 5-round mean as a sngle observaton. (18 observatons per cell.) p-values are computed usng a two-sample t-test, allowng for unequal varances. Another measure of errors would be to take the sum of the absolute d erence between each bdder s bd and that bdder s value, droppng all hghest bdders for symmetry. Table 3 reports the results. Table 4 reports the results of Table 4 n the man text, except that the p-values are calculated usng the Wlcoxon rank-sum test. 29.0% of rank-order lsts are ncorrect under SP-RSD. 2.6% of choces are ncorrect under OSP-RSD. However, ths s not a far comparson; rank-order lsts mechancally allow us to spot more errors than sngle choces. To compare lke wth lke, we compute the proporton of ncorrect choces we would have observed, f subjects played OSP-RSD as though they were mplementng the submtted rank-order lsts for SP-RSD. Ths s a cautous measure; t counts errors under SP-RSD only f they alter the outcome. Table 5 reports the results. In Table 6, we compute the mean d erence between the second-hghest bd and the second-hghest value, by aucton format and by fve-round blocks. Ths summarzes the average drecton of devatons (postve or negatve) from equlbrum play. I had no pror hypothess about these outcome varables, but report ths analyss for completeness. Table 7 breaks down the rate that subjects submt ncorrect rank-order lsts n one- 19

20 Table 3: mean(sum(abs( s bd - s value))), droppng hghest bdders Format Rounds SP OSP p-value <.001 Aucton (5.20) (1.88) (2.85) (0.72) X Aucton (4.10) (1.01) (3.72) (0.83) For each aucton, we sum the absolute d erences between each bdder s bd and ther value, droppng the hghest bdder. We then take the mean of ths over each 5-round block. We then compute standard errors countng each group s 5-round mean as a sngle observaton. (18 observatons per cell.) p-values are computed usng a two-sample t-test, allowng for unequal varances. Table 4: Proporton of seral dctatorshps not endng n domnant strategy outcome, p-values calculated usng Wlcoxon rank-sum test SP OSP p-value Rounds % 7.8%.0001 Rounds % 6.7%.0010 Ths s the same as Table 4 n the man text, except that the p-values are calculated usng the Wlcoxon rank-sum test. Table 5: Proporton of ncorrect choces under seral dctatorshp: SP (mputed) vs OSP (actual) SP OSP p-value Rounds % 2.6% <.001 (3.5%) (1.1%) Rounds % 2.6%.002 (2.0%) (1.3%) p-value For each group n SP-RSD, for each perod, we smulate the three choces that we would have observed under OSP-RSD. For each group, for each 5-round block, we record the proporton of choces that are ncorrect. We then compute standard errors countng each group-block par as a sngle observaton. (18 observatons per cell.) When comparng SP to OSP, we compute p-values usng a two-sample t-test. When comparng early to late rounds of the same game, we compute p-values usng a pared t-test. In the sample for OSP-RSD, there are 7 ncorrect choces n the frst fve rounds and 7 ncorrect choces n the last fve rounds. 20

21 shot SP-RSD by demographc category. Table 6: mean(2nd bd - 2nd value) Format Rounds SP OSP Aucton +X Aucton (2.07) (0.77) (1.28) (0.40) (0.70) (0.47) (0.97) (0.31) For each group, we take the mean d erence between the second-hghest bd and the second-hghest value over each 5-round block. We then compute standard errors countng each group s 5-round mean as a sngle observaton. (18 observatons per cell, standard errors n parentheses.) 3 Quantal Response Equlbrum Quantal response equlbrum (QRE) s defned for normal form games (McKelvey and Palfrey, 1995). Agent quantal response equlbrum (AQRE) adapts QRE to extensve form games (McKelvey and Palfrey, 1998); agents play a perturbed best response at each nformaton set, gven correct belefs about the (perturbed) play of other agents and ther future selves. Standard specfcatons of AQRE predct extreme under-bddng n ascendng auctons wth fne bd ncrements. I here provde a smple proof that, for any prvate-value ascendng aucton, there exst bd ncrements fne enough that logt-aqre predcts that all bds wll be close to 0 wth probablty close to 1. The proof also works for a class of related models, that allow agents to have arbtrary belefs about the strateges of other Table 7: Incorrect lsts n one-shot SP-RSD, by demographc categores Yes No p-value Economcs major 46% 31%.131 STEM major 32% 34%.876 Took a course n game theory 34% 33%.876 Male 37% 28%.214 Ths table dsplays the proporton of subjects who submtted ncorrect rank-order lsts n one-shot SP- RSD, by self-reported demographc categores. There are 192 subjects n ths treatment. p-values are computed wth Fsher s exact test. 21

22 agents and ther future selves. Defnton 1. An equlbrum wth -logt errors specfes, for every nformaton set, some (arbtrary) belefs about opponent strateges and the agent s own contnuaton strategy. Gven those belefs, let {v 1,...,v J } denote the mpled contnuaton values for actons {a 1,...,a J } at nformaton set I. The probablty that agent chooses a j at I s equal to e v j P J l=1 e v l for some nformaton sets and agents. 0, where ths randomzaton s ndependent across Ths contans logt-aqre (McKelvey and Palfrey, 1998) as a specal case. It also ncludes naïve logt models, such that agents beleve that ther opponents and ther future selves play domnant strateges. We now consder the followng envronment: There s a sngle object and some fnte set of bdders, wth types (prvate values) n the nterval [0, 1] and quas-lnear utlty. Types are drawn accordng to some jont cdf F :[0, 1] N! [0, 1]. Defnton 2. In a k-ncrement ascendng aucton, the prce rses step by step through the set { 0 k, 1 k, 2 k,...}. Agents observe only the clock prce, and whether the aucton has ended. At each prce, actve agents (n some arbtrary order) choose In or Out. Agents who choose Out become nactve; they make no payments and do not receve the object. At any pont when only one agent s actve, the aucton ends, and the actve agent wns the object at the current clock prce. Proposton 3. For any > 0 and any, there exsts k such that, for all k k, nany equlbrum wth -logt errors of the k-ncrement ascendng aucton, the probablty that the closng prce s above s less than. Proof. Let Mk := max{j 2 Z such that j k apple }; ths s the ndex of the hghest bd ncrement below. Take any agent, any type, any k-ncrement aucton, and any equlbrum wth -logt errors. Let v,j,k, In denote the contnuaton value (gven s arbtrary belefs) of playng In at clock prce j k. P ([ does not qut before ] \ [ some j 2 N \ does not qut before ]) apple P ( does not qut before some j 2 N \ does not qut before ) = M k Y j=0 e vin,j,k, e vin,j,k, + e 0 apple M k Y j=0 e e + e 0 (63) 22

23 The last nequalty holds because, for any belefs, s payo from the k-ncrement ascendng aucton s bounded above by 1. Moreover, P (closng prce above ) = P ( [ [[ does not qut before ] \ [ some j 2 N \ does not qut before ]]) 2N apple X 2N P ([ does not qut before ] \ [ some j 2 N \ does not qut before ]) lm k!1 Mk = 1, solm Q M k k!1 j=0 enough that, for all k completes the proof. e e +e 0 M k Y e apple N e + e 0 (64) j=0 = 0. Consequently, we can pck k large k, the rght-hand sde of the above equaton s less than, whch As Proposton 3 ndcates, n any ascendng aucton wth fne enough bd ncrements, models such as logt-aqre or naïve logt predct extreme underbddng. In the auctons n my experment, the range of types s $120 and bd ncrements are 25 cents. It s not feasble to compute AQRE for these auctons. Players have 481 possble types and there are 601 possble bds. Consequently, the pure strategy space n 2P and 2P+X s of the order of , and larger yet n AC and AC+X, snce there are multple payo -equvalent strateges of each knd (specfyng d erent actons at nformaton sets that are never reached). However, I compute logt-aqre for the followng smpler games: There are 4 agents, wth types drawn ndependently and unformly at random from 0.25 to 20 (n 0.25 ncrements). Agents ether play a second-prce aucton wth 80 possble bds (equal to the types), or an ascendng clock aucton wth 80 clock prces. I compute symmetrc logt-aqre for 2 {0, 2 4, 2 2, 1, 2 2, 2 4 }, and sample 10 6 plays for each game-parameter par. Table 8 dsplays the average absolute d erence between the second-hghest bd and the second-hghest value, by and aucton format. For every computed parameter value, AQRE predcts larger errors n the ascendng aucton than n the second-prce aucton, whch s the opposte of the e ect n the expermental data. For the smplfed auctons that I computed, Table 9 dsplays the average d erence between the second-hghest bd and the second-hghest value predcted by AQRE. Comparng Table 8 and Table 9 ndcates that the large errors n ascendng auctons under 23

24 Table 8: AQRE computed for smplfed auctons mean(abs(2nd bd - 2nd value)) 2nd-Prce Ascendng Table 9: Drectonal Errors under AQRE mean(2nd bd - 2nd value) 2nd-Prce Ascendng AQRE are drven almost entrely by under-bddng. As Fgure 2 and Fgure 3 (n the man text) llustrate, there s no evdence of systematc under-bddng n AC and AC+X. Table 6 shows ths rgorously. For further robustness, I compute a naïve alternatve to AQRE: Agents make logstc errors at each nformaton set, whle belevng that ther opponents (and ther future selves) play ther domnant strateges 5. Table 10 dsplays the average absolute d erence 5 In ascendng auctons, there are multple payo -equvalent domnant strateges, whch prescrbe d erent actons at prces above the agents values. For the purpose of these computatons, I assume that agents beleve that they wll qut at all such nformaton sets. Table 10: Naïve logstc errors computed for smplfed auctons mean(abs(2nd bd - 2nd value)) 2nd-Prce Ascendng

25 between the second-hghest bd and the second-hghest value, by and aucton format. Ths model also predcts larger mstakes n ascendng auctons than n second-prce auctons. The precedng results suggest that models n whch agents make payo -senstve mstakes ndependently at every nformaton set are substantally at odds wth the data. They predct that ascendng auctons wth fne ncrements wll nduce extreme underbddng, and that errors should be larger n ascendng auctons than n second-prce auctons. One could nvent new theores that permt non-ndependent errors across nformaton sets or d erent accuracy parameters for d erent games, but ths ntroduces many new degrees of freedom, and may have lttle testable content. References McKelvey, R. D. and Palfrey, T. R. (1995). Quantal response equlbra for normal form games. Games and economc behavor, 10(1):6 38. McKelvey, R. D. and Palfrey, T. R. (1998). Quantal response equlbra for extensve form games. Expermental economcs, 1(1):

26 Obvously Strategy-Proof Mechansms Onlne Appendx: Experment Instructons Shengwu L

27 WELCOME Ths s a study about decson-makng. Money earned wll be pad to you n cash at the end of the experment. Ths study s about 90 mnutes long. We wll pay you $5 for showng up, and $15 for completng the experment. Addtonally, you wll be pad n cash your earnngs from the experment. If you make choces n ths experment that lose money, we wll deduct ths from your total payment. However, your total payment (ncludng your show-up payment and completon payment) wll always be at least $20. You have been randomly assgned nto groups of 4. Ths experment nvolves 3 games played for real money. You wll play each game 10 tmes wth the other people n your group. We wll gve you nstructons about each game just before you begn to play t. Your choces n one game wll not affect what happens n other games. There s no decepton n ths experment. Every game wll be exactly as specfed n the nstructons. Anythng else would volate the IRB protocol under whch we run ths study. (IRB Protocol 34876) Please do not use electronc devces or talk wth other volunteers durng ths study. If we do fnd you usng electronc devces or talkng wth other volunteers, the rules of the study requre us to deduct $20 from your earnngs. If you have questons at any pont, please rase your hand and we wll answer your questons prvately. 1 of 5 AA

28 GAME 1 In ths game, you wll bd n an aucton for a money prze. The prze may have a dfferent dollar value for each person n your group. You wll play ths game for 10 rounds. All dollar amounts n ths game are n 25 cent ncrements. At the start of each round, we dsplay your value for ths round s prze. If you wn the prze, you wll earn the value of the prze, mnus any payments from the aucton. Your value for the prze wll be calculated as follows: 1. For each group we wll draw a common value, whch wll be between $10.00 and $ Every number between $10.00 and $ s equally lkely to be drawn. 2. For each person, we wll also draw a prvate adjustment, whch wll be between $0.00 and $ Every number between $0.00 and $20.00 s equally lkely to be drawn. In each round, your value for the prze s equal to the common value plus your prvate adjustment. At the start of each round, you wll learn your total value for the prze, but not the common value or the prvate adjustment. Ths means that each person n your group may have a dfferent value for the prze. However, when you have a hgh value, t s more lkely that other people n your group have a hgh value. The aucton proceeds as follows: Frst, you wll learn your value for the prze. Then, the aucton wll start. We wll dsplay a prce to everyone n your group, that starts low and counts upwards n 25 cent ncrements, up to a maxmum of $ At any pont, you can choose to leave the aucton, by clckng the button that says Stop Bddng. 2 of 5 AA

29 When there s only one bdder left n the aucton, that bdder wll wn the prze at the current prce. Ths means that we wll add to her earnngs her value for the prze, and subtract from her earnngs the current prce. All other bdders earnngs wll not change. At the end of each aucton, we wll show you the prces where bdders stopped, and the wnnng bdder s profts. If there s a te for the hghest bdder, no bdder wll wn the object. 3 of 5 AA

30 GAME 2 In ths game, you wll bd n an aucton for a money prze. You wll play ths game for 10 rounds. Your value for the prze wll be generated as before. However, each round, we wll also draw a new number, X, for each group. The rules of the aucton are dfferent, as follows: The prce wll count up from a low value, and you can choose to leave the aucton at any pont, by clckng the button that says Stop Bddng. When there s only one bdder left n the aucton, the prce wll contnue to rse for another X dollars, and then freeze. If the last bdder stays n the aucton untl the prce freezes, then she wll wn the prze at the fnal prce. Ths means that we wll add to her earnngs her value for the prze, and subtract from her earnngs the fnal prce. All other bdders earnngs wll not change. If the last bdder stops bddng before the prce freezes, then no bdder wll wn the prze. In that case, no bdder s earnngs wll change. X wll be between $0.00 and $3.00, wth every 25 cent ncrement equally lkely to be drawn. You wll be told your value for the prze at the start of each round, but wll not be told X. At the end of each round, we wll tell you the value of X. 4 of 5 AA

31 GAME 3 You wll play ths game for 10 rounds. In each round of ths game, there are four przes, labeled A, B, C, and D. Przes wll be worth between $0.00 and $1.25. For each prze, ts value wll be the same for all the players n your group. At the start of each round, you wll learn the value of each prze. You wll also learn your prorty score, whch s a random number between 1 and 10. Every whole number between 1 and 10 s equally lkely to be chosen. The game proceeds as follows: 1. The player wth the hghest prorty score wll pck one prze. 2. The player wth the second-hghest prorty score wll pck one of the przes that remans. 3. The player wth the thrd-hghest prorty score wll pck one of the przes that remans. 4. The player wth the lowest prorty score wll be assgned whatever prze remans. If two players have the same prorty score, we wll break the te randomly. When t s your turn to pck, you wll have 30 seconds to make your choce. You do ths by selectng a prze and then clckng the button that says Confrm Choce. Your choces wll not count unless you clck the button that says Confrm Choce. If you do not make a choce by the end of 30 seconds, we wll assgn przes as though you pcked whchever prze s earlest n the alphabet. At the end of each round, we wll add to your earnngs the value of the prze you pcked. 5 of 5 AA

32 WELCOME Ths s a study about decson-makng. Money earned wll be pad to you n cash at the end of the experment. Ths study s about 90 mnutes long. We wll pay you $5 for showng up, and $15 for completng the experment. Addtonally, you wll be pad n cash your earnngs from the experment. If you make choces n ths experment that lose money, we wll deduct ths from your total payment. However, your total payment (ncludng your show-up payment and completon payment) wll always be at least $20. You have been randomly assgned nto groups of 4. Ths experment nvolves 3 games played for real money. You wll play each game 10 tmes wth the other people n your group. We wll gve you nstructons about each game just before you begn to play t. Your choces n one game wll not affect what happens n other games. There s no decepton n ths experment. Every game wll be exactly as specfed n the nstructons. Anythng else would volate the IRB protocol under whch we run ths study. (IRB Protocol 34876) Please do not use electronc devces or talk wth other volunteers durng ths study. If we do fnd you usng electronc devces or talkng wth other volunteers, the rules of the study requre us to deduct $20 from your earnngs. If you have questons at any pont, please rase your hand and we wll answer your questons prvately. 1 of 6 BB

33 GAME 1 In ths game, you wll bd n an aucton for a money prze. The prze may have a dfferent dollar value for each person n your group. You wll play ths game for 10 rounds. All dollar amounts n ths game are n 25 cent ncrements. At the start of each round, we dsplay your value for ths round s prze. If you wn the prze, you wll earn the value of the prze, mnus any payments from the aucton. Your value for the prze wll be calculated as follows: 1. For each group we wll draw a common value, whch wll be between $10.00 and $ Every number between $10.00 and $ s equally lkely to be drawn. 2. For each person, we wll also draw a prvate adjustment, whch wll be between $0.00 and $ Every number between $0.00 and $20.00 s equally lkely to be drawn. In each round, your value for the prze s equal to the common value plus your prvate adjustment. At the start of each round, you wll learn your total value for the prze, but not the common value or the prvate adjustment. Ths means that each person n your group may have a dfferent value for the prze. However, when you have a hgh value, t s more lkely that other people n your group have a hgh value. The aucton proceeds as follows: Frst, you wll learn your value for the prze. Then you can choose a bd n the aucton. Each person n your group wll submt ther bds prvately and at the same tme. You do ths by typng your bd nto a text box and clckng confrm bd. You wll have 90 seconds to make your decson, and can revse your bd as many tmes as you lke. At the end of 90 seconds, your fnal bd wll be the one that counts. 2 of 6 BB

34 All bds must be between $0.00 and $150.00, and n 25 cent ncrements. The hghest bdder wll wn the prze, and make a payment equal to the secondhghest bd. Ths means that we wll add to her earnngs her value for the prze, and subtract from her earnngs the second-hghest bd. All other bdders earnngs wll not change. At the end of each aucton, we wll show you the bds, ranked from hghest to lowest, and the wnnng bdder s profts. If there s a te for the hghest bdder, no bdder wll wn the object. 3 of 6 BB

35 GAME 2 In ths game, you wll bd n an aucton for a money prze. You wll play ths game for 10 rounds. Your value for the prze wll be generated as before. However, each round, we wll also draw a new number, X, for each group. The rules of the aucton are dfferent, as follows: All bdders wll submt ther bds prvately and at once. However, the hghest bdder wll wn the prze f and only f ther bd exceeds the second-hghest bd by more than X. If the hghest bdder wns the prze, she wll make a payment equal to the secondhghest bd plus X. Ths means that we wll add to her earnngs her value for the prze, and subtract from her earnngs the second-hghest bd plus X. All other bdders earnngs wll not change. If the hghest bd does not exceed the second-hghest bd by more than X, then no bdder wll wn the prze. In that case, no bdder s earnngs wll change. X wll be between $0.00 and $3.00, wth every 25 cent ncrement equally lkely to be drawn. You wll be told your value for the prze at the start of each round, but wll not be told X. At the end of each round, we wll tell you the value of X. 4 of 6 BB

36 GAME 3 You wll play ths game for 10 rounds. In each round of ths game, there are four przes, labeled A, B, C, and D. Przes wll be worth between $0.00 and $1.25. For each prze, ts value wll be the same for all the players n your group. At the start of each round, you wll learn the value of each prze. You wll also learn your prorty score, whch s a random number between 1 and 10. Every whole number between 1 and 10 s equally lkely to be chosen. The game proceeds as follows: We wll ask you to lst the przes, n any order of your choce. All players wll submt ther lsts prvately and at the same tme. After all the lsts have been submtted, we wll assgn przes usng the followng rule: 1. The player wth the hghest prorty score wll be assgned the top prze on hs lst. 2. The player wth the second-hghest prorty score wll be assgned the top prze on hs lst, among the przes that reman. 3. The player wth the thrd-hghest prorty score wll be assgned the top prze on hs lst, among the przes that reman. 4. The player wth the lowest prorty score wll be assgned whatever prze remans. If two players have the same prorty score, we wll break the te randomly. You wll have 90 seconds to form your lst. You do ths by typng a number, from 1 to 4, next to each prze, and then clckng the button that says Confrm Choces. Each prze must be assgned a dfferent number, from 1 (top) to 4 (bottom). Your choces wll not count unless you clck the button that says Confrm Choces. 5 of 6 BB