6 Appendix 1: Proofs

We will intensively use the following notation. Given 2 G, h 2 H and i 2 N we will denote bybh the history at which i has played for the last time before h; namely,bh is such that (i) bh h; (ii) i^bh =i and (iii) there is no eh 6=bh such that bh eh h and i^eh =i^bh. Of course,bh depends oni but sincei will be clear from the context, we will omit its reference when denoting this earlier history.

6.1 Theorem 1

Let be a sequential allotment rule satisfying individually rational with respect to q 2X and let be the extensive game form obtained from the speci…c phase of the algorithm de…ned in Subsection 4.1.1.

We …rst present preliminary results that we will use in the proof of Theorem 1.

Lemma 1 Let h = (h⁰⁰; a_ih00; a_ih0) be the output of Step 2.t.b (for t 1), with h⁰ = The inequality in the …rst raw of (11) follows from strategy-proofness, single-peakedness, (10) and the fact thati^h00 2N_u^h00. The inequality in the second raw follows from the de…nition of i^h0 and q^h0 =q^h00. The inequality in the third raw follows from replacement monotonicity and (10). Since q^h00 is a feasible allotment, the inequality in (11) can be replaced by an equality, and since

To prove that the statement in (1.1) holds, we proceed by distinguishing between two cases, depending on the action a_ih0 2A^h0

i^h0 =fd; q^h0

i^h0 1g:

Case 1: Suppose a_ih0 =d: Then, N_d^h =N_d^h00 and, by (12), (0_N^h; q^h_NnNh) =q^h:

Case 2: Suppose a_ih0 = q_i^h_h0⁰ 1: Then, N_d^h = N_d^h0nfi^h0g = N_d^h00nfi^h0g; where the second equality follows from N_d^h0 =N_d^h00: Hence, by (12), strategy-proofness of implies

i^h0(0_N^h This …nishes the proof of (1.1) for the case t= 1:

Induction hypothesis: for t> 1; if h⁰⁰ = (h⁰⁰⁰⁰; a_ih0000; a_ih000) is the input of Step 2.t-1.a,

Observe that in the proof of (1.1) for the case t= 1, (10) can be replaced by (14) and using the same argument, it follows that

(0_N^h

d; q^h_NnNh

d) =q^h:

(1.2) Assume N_u^h 6=;: By condition (1) in the characterization of e¢ cient rules,

j(k_Nh

In order to show that the other inequality holds as well, we proceed by induction on t.

Suppose t= 1: Hence, h⁰⁰ is the output of Stage 1 and N^h00 =N, N_u^h00 6=;, N_d^h00 6=;;

Let i₁ 2 N_d^h00: By strategy-proofness (for i₁), single-peakedness, replacement monotonicity and (16), Let i₂ 2 N_d^h00nfi₁g; if any. By strategy-proofness (for i₂), single-peakedness, replacement monotonicity and (17),

Furthermore, by de…nition of i^h00; strategy-proofness (for i^h00) and replacement monotonicity,

j(k_N^h

This, together with (15), …nishes the proof of (1.2) for the case t= 1:

Induction hypothesis: fort>1;ifh⁰⁰= (h⁰⁰⁰⁰; a_ih0000; a_ih000)is the input of theStep 2.t-1.a, that (23) implies (18). Then, to obtain (22), the proof follows the same argument used in the caset= 1.

If i^h000 2=N_d^h00; then q_j^h0000 =q_j^h00 for all j 2N_d^h00: Therefore (23) implies (18) and the proof follows as in the case t= 1:

If i^h000 2 N_d^h00; then q_j^h0000 1 = q_j^h00 1 if j 2 N_d^h00nfi^h000g and q_j^h0000 1 = q^h_j⁰⁰ if j = i^h000. Then, by strategy-proofness (for i^h000)and (23),

i^h000(k_Nh00 Therefore, (24) and (25) imply (18) and the proof of (1.2) follows as in the case t= 1:

Lemma 2 Let h2H_T be a terminal history (and hence, N_u^h =; orN_d^h =;). Then,

Case 1: Suppose j has not played along Stage 2. Then, by replacement monotonicity and the de…nition of q_j^h,

Thus, (28) and (29) imply that

Proof of Theorem 1 It follows from the three statements that we will present and prove successively.

Statement 1.1 Let be an individually rational sequential allotment rule. Then, the extensive game form is well de…ned and …nite.

Proof of Statement 1.1 We …rst argue that the agents that are called to play along the speci…c phase of the algorithm are uniquely identi…ed and well de…ned. This is clear in any Step 1.t fort 1.

Consider now Step 2.t.a and Step 2.t.b, for t 1; with corresponding inputs h⁰⁰ and h⁰, where h⁰⁰ is not a terminal history. Since i2N_d^h00 implies 0< q_i^h00;

j(k_Nh00

u ;(q^h00 1)_Nh00 d ; q_N^h00h00

s ) (31)

is well de…ned for all j 2N. Moreover, the sum of the components of the vector where j

is applied in (31) is larger or equal than k. Hence, by e¢ ciency, P

is well de…ned for all j 2N. Moreover, the sum of the components of the vector where j

is applied in (32) is smaller or equal than k. Hence, by e¢ ciency, P

Hence, min_<fj 2N_d^h0 j ^j(q^h_ih00⁰ + 1;0_Nh0 d ; q^h0

N_d^h0[fi^h00g) q_j^h0 1gis well de…ned and so is i^h0. Second, the sets of agents de…ned along the speci…c phase of the algorithm are well de…ned because they evolve, for anyh h⁰, as follows.

(a) If i2N_u^h0 for someh⁰, theni =2N_d^h and i2N_u^h[N_s^h: (b) If i2N_d^h0 for someh⁰, theni =2N_u^h and i2N_d^h[N_s^h: (c) If i2N_s^h0 for someh⁰, theni2N_s^h:

(d) If N_u^h0; N_s^h0; N_d^h0 is a partition of N, then N_u^h; N_s^h; N_d^h is also a partition of N:

To see that the statements (a) to (d) hold, let h h⁰ be arbitrary. Assume i 2 N_u^h0. If i is not called to play anymore, i 2 N_u^h: Suppose i is called to play at h (i.e., i^h = i).

Then, d =2A^h_i. Hence, either i2 N_u^h or i2 N_s^h: Thus, (a) holds. Symmetrically ifi 2N_d^h0: Suppose now that i 2 N_s^h0: Then i will not be called to play anymore. Hence, N_s^h N_s^h0 which implies that i2 N_s^h: Thus, (c) holds. The proof of (d) follows immediately from the de…nitions of N_u^h; N_s^h and N_d^h:

Now, we argue that the extensive game form de…ned by the algorithm is …nite.

Stage 1 ends because NnN^h0 ) NnN^h holds, where h = (h⁰; a_ih0); and so, for some history h, N^h = N: In Stage 2, if h is an output history of some Step 2.t.b and N_u^h =; or N_d^h = ; then h is a terminal history and the game ends. If N_u^h 6= ; and N_d^h 6= ;, then h = (h⁰⁰; a_ih00; a_ih0) with h⁰ = (h⁰⁰; a_ih00), N_s^h00 N_s^h N_s^h00 [ fi^h00; i^h0g and q^h

i^h00 = q^h00

i^h00 + 1, q^h

i^h0 = q^h00

i^h0 1 and q^h_i = q_i^h00 for all i =2 fi^h00; i^h0g (since by de…nition, q_i^h0 = q_i^h00). Then, the algorithm stops at some Step 2.t.b with output h because N_u^h =; or N_d^h =; (recall that for any history h of Stage 2, with h being its immediate successor, q^h

i^h = k implies i^h 2N_s^h and q^h

i^h = 0 impliesi^h 2N_s^h ).

Let H¹ and H² be the sets of histories that are outputs of some steps in Stage 1 and Stage 2 of , respectively. In addition, let H be the set of histories that are output of Stage 1orStep 2.t.b, for somet 1:Note thatH¹ H :By the de…nition of the speci…c phase of the algorithm, the set of terminal historiesH_T of can be written as

H_T =fh2H jN_u^h =;or N_d^h =;g:

For each terminal historyh2H_T the output of the game is o(h) = q^h:

Statement 2.1 Let be an individually rational sequential allotment rule. Then, induces ; namely, for all R 2 Rⁿ;

(R) =o(h (?; ^R)):

Proof of Statement 2.1 Let R 2 Rⁿ be arbitrary, let = ( (R₁); : : : ; (R_n)) 2 f0; : : : ; kgⁿ be the pro…le of tops at R and set h := h (?; ^R): We …rst state and prove three claims.

Claim 1: If i2N_s^h ; then i =q_i^h .

Proof of Claim 1: Let i 2N_s^h : Then, by the de…nitions of the set N_s^h at any h and of the historybh, i=i^bh and ^R_i ⁱ(bh)2 f= u; dg: Ifbh2H¹; it follows that i = ^R_i ⁱ(bh) = q^bh_i =q^h_i . If bh 2 H²; and since bh h and h is the history induced by the pro…le of truth-telling strategies, i q^b_i^h+ 1 if i2N_u^bh and i q_i^bh 1 if i2N_d^bh:Then, since _iⁱ(bh)2 f= u; dg;

i =

( q^b_i^h+ 1 if i2N_u^bh

q^b_i^h 1 if i2N_d^bh: (33)

Then,

i =q_i^h ;

where the equality follows from (33) and the de…nition ofq_i^h =q_i^h; wherebh ^im h.

Claim 2: If i2N_d^h ; then i < q_i^h .

Proof of Claim 2: Let i 2 N_d^h : Then, by the de…nitions of the set N_d^h at any h and the history bh; i= i^bh and ^R_i ⁱ(bh) = d: If bh 2 H¹; it follows that i < q^b_i^h = q^h_i : If bh 2 H²; and sincebh h and i2 N_d^bh, by the de…nition of ^R_i ⁱ; we have that i < q^bh_i 1: But since q^bh_i 1 = q^h_i and h is a terminal history of Stage 2, it follows that i < q_i^h .

Claim 3: If i2N_u^h ; then i > q_i^h .

Proof of Claim 3: Let i 2 N_u^h : Then, by the de…nitions of the set N_u^h at any h and the history bh, i =i^bh and ^R_i ⁱ(bh) = u: Ifbh 2 H¹; it follows that i > q_i^bh = q^h_i : If bh 2 H²; and since bh h and i 2 N_u^bh, by the de…nition of ^R_i ⁱ; we have i > q^bh_i + 1. But since q^bh_i + 1 = q_i^h and h is a terminal history of Stage 2, it follows that i > q_i^h .

We proceed with the proof of Statement 2.1 by distinguishing between two cases.

Case 1: Assume N_u^h =;: By (2.1) in Lemma 2 and Claim 1, (0_N^h

d ; _N^h

s ) = q^h :

By Claim 2, 0 i < q_i^h for every i 2N_d^h : Therefore, by strategy-proofness and replace-ment monotonicity,

( _N^h

d ; _N^h

s ) =q^h : Hence, (R) = o(h (?; ^R)):

Case 2: Assume N_u^h 6=;: Then, N_d^h =; and, by (2.2) in Lemma 2 and Claim 1, k_Nu^h ; _Nh

s =q^h :

By Claim 3, i > q^h_i for every i2 N_u^h : Therefore, by strategy-proofness and replacement monotonicity,

N_u^h ; _Nh

s =q^h : Hence, (R) = o(h (?; ^R)):

Statement 3.1 Let be an individually rational sequential allotment rule and let R 2 Rⁿ be a pro…le. Then, for all i2N, the truth-telling strategy ^R_iⁱ is weakly dominant in .

Proof of Statement 3.1 Consider agenti with preferencesR_i and top allotment i. Let

Ri

i be i’s truth-telling strategy relative to R_i and let ⁰_i be any other strategy. We want to show that, for all i;

x_i =o_i(h (;;( ^R_i ⁱ; _i)))R_io_i(h (;;( ⁰_i; _i))) =x⁰_i (34) holds. Let i be arbitrary. Condition (34) holds trivially if x_i = x⁰_i: Assume x_i 6= x⁰_i: Let h⁰ be the earliest history at which ^R_iⁱ(h⁰) 6= ⁰_i(h⁰) along the equal play induced by both ( ^R_i ⁱ; _i) and ( ⁰_i; _i) up to h⁰: We proceed by distinguishing among several cases, depending on the step of the speci…c phase of the algorithm for which the history h⁰ is an input of, and agent i is called to play ath⁰.

Case 1: The history h⁰ is an input of some Step 1.t (i.e., i^h0 = i). Then, i2 NnN^h0. Let q be the allotment that is part of the input information at this step. We distinguish among three cases.

Subcase 1.1: i = q_i: Then, ^R_iⁱ(h⁰) = _i. Hence, i2 N_s^h for all h (h⁰; ⁰_i(h⁰)); agent i is not called to play anymore and x_i = _i: Thus, (34) holds.

Subcase 1.2: i > q_i. Then, ^R_iⁱ(h⁰) = u. Setting h = (h⁰; ^R_iⁱ(h⁰)), we have that i 2 N_u^h: Then, for all h h; by de…nition ofq_i^h;

q_i q_i^h: (35)

Denote by h (h⁰;( ^R_i ⁱ; i)) the terminal history obtained when agents play starting at h⁰ according to( ^R_iⁱ; i): Because h h (h⁰;( ^R_i ⁱ; i));

q_i^h0 x_i: (36)

Since h = (h⁰; ^R_i ⁱ(h⁰)) and h⁰ is an input of some Step 1.t, h is a history of Stage 1 or Stage 2. Hence, i2 N_u^h implies that i2Nu^{h (h0;( Rii ; i))}[Ns^{h (h0;( Rii ; i))}: Therefore, by the de…nition of ^R_iⁱ,

x_{i i}: (37)

By (35), (36) and (37),

q_i x_{i i}: (38)

On the other hand, since ⁰_i(h⁰) 6= ^R_iⁱ(h⁰) = u, we have that ⁰_i(h⁰) 2 fq_i; dg: Then, by setting h = (h⁰; ⁰_i(h⁰)); i 2 N_d^h [N_s^h which means that q_i^h q_i for all h h: Therefore, since h h (h⁰;( ⁰_i; _i));

x⁰_i q_i: (39)

By (38) and (39), single-peakedness of R_i implies that x_iR_ix⁰_i: Thus, (34) holds.

Subcase 1.3: i < q_i: Then, ^R_iⁱ(h⁰) = d. With a symmetric argument to the one used in Subcase 1.2, we obtain that i x_i q_i x⁰_i, and single-peakedness of R_i implies that x_iR_ix⁰_i. Thus, (34) holds.

Case 2: The historyh⁰ is an input of someStep 2.t.a, and to make the notation consistent with the one used to de…ne such step, set h⁰⁰ := h⁰ and so i^h00 = i. Then, i 2 N_u^h00. Let q^h_i⁰⁰ be i’s allotment that is part of the input information at this step. Since i 2 N_u^h00 and h⁰⁰ is a history in the path starting at the empty history ; when agents play according to ( ^R_i ⁱ; _i); agenti has chosenu before h⁰⁰. Hence, by de…nition of ^R_i ⁱ;

q^h_i⁰⁰+ 1 _i: (40)

We distinguish between two subcases.

Subcase 2.1: i = q_i^h00 + 1. Then, ^R_i ⁱ(h⁰⁰) = q_i^h00 + 1 = _i. Hence, i 2 N_s^h for all h (h⁰⁰; ^R_i ⁱ(h⁰⁰)); agenti is not called to play anymore, and so x_i = _i: Thus, (34) holds.

Subcase 2.2: i > q_i^h00+ 1. Then, ^R_i ⁱ(h⁰⁰) = u and ⁰_i(h⁰⁰) =q_i^h00+ 1. Hence, i 2N_s^h for all h (h⁰⁰; ⁰_i(h⁰⁰)); agent i is not called to play anymore, and so x⁰_i = q_i^h00+ 1: Furthermore, using a similar argument to the one used in Subcase 1.2, it is easy to see that q^h_i⁰⁰+ 1 x_{i i}. Sinceq_i^h00+ 1 =x⁰_i; single-peakedness ofR_i implies that x_iR_ix⁰_i: Thus, (34) holds.

Case 3: The history h⁰ is an input of some Step 2.t.b (i.e., i^h0 =i). Then, i 2 N_d^h0. Let q^h_i⁰ be i’s allotment that is part of the input information at this step. Since i 2 N_d^h0 and h⁰ is a history in the path starting at the empty history ; when agents play according to ( ^R_i ⁱ; _i), agent i has chosend before h⁰. Hence, by the de…nition of ^R_i ⁱ,

i q^h_i⁰ 1: (41)

We distinguish between two cases.

Subcase 3.1: i = q^h_i⁰ 1. Then, ^R_i ⁱ(h⁰) = q_i^h0 1 = _i. Hence, i 2 N_s^h for all h (h⁰; ^R_iⁱ(h⁰)); agenti is not called to play anymore, and so x_i = _i:Thus, (34) holds.

Subcase 3.2: i < q_i^h0 1. Then, ^R_i ⁱ(h⁰) = d and ⁰_i(h⁰) = q_i^h0 1. Hence, i 2 N_s^h for all h (h⁰; ⁰_i(h⁰));agentiis not called to play anymore, and sox⁰_i =q^h_i⁰ 1:Furthermore, using a similar argument to the one used in Subcase 2.2, it is easy to see that i x_i q^h_i⁰ 1.

Since q^h_i⁰ 1 = x⁰_i;single-peakedness of R_i implies that x_iR_ix⁰_i: Thus, (34) holds.

6.2 Theorem 2

Before starting with the proof of Theorem 2, we present an alternative (and equivalent) way of de…ning, along Stage A, the vectors q^h and q^h as images of evaluated respectively at m^h and m^h, according to either (6) or (7). Remember that the de…nitions of m^h and m^h use respectively the vectors q^h0 and q^h0, where h⁰ is the immediate predecessor of h:

It will be convenient to de…ne instead two vectors p^h and p^h as images of evaluated respectively at l^h and l^h which depend not only on h⁰ (and its embedded information) but also on all subhistories up to h (and their embedded information). In Lemma 6 we will show that p^h = q^h, p^h = q^h; l^h = m^h and l^h = m^h hold, and so both descriptions

are equivalent. Therefore, all histories, sets and vectors de…ned along Stage A using m^h and m^h (according to the original de…nition of the algorithm in Subsection 4.2.1) coincide with those that would have been obtained from having used instead the vectors l^h and l^h. However, the more involved description using all subhistories up tohis more helpful for the proofs of some of the statements required to prove Theorem 2.

For the empty history ;, de…ne

p^; = (k) and p^; = (0);

and, for any history h ;, de…ne

l^h_j = 8>

<

>:

k if j 2N_u^h[(NnN^h) p^bh

i^bh if j 2N_d^h and j =i^bh a_ibh if j 2N_s^h and j =i^bh and

l^h_j = 8>

<

>:

0 if j 2N_d^h[(NnN^h) p^bh

i^bh if j 2N_u^h and j =i^bh a_ibh if j 2N_s^h and j =i^bh;

where, given j, bh is the longest subhistory of h at which j has played at (i.e., j = i^bh and, for all bh eh h, j 6=i^eh). Then, de…ne

p^h = (l^h) and p^h = (l^h):

Let Hm be the set of histories obtained by evaluating at vectors m^h and m^h for histories h in Stage A (as de…ned in Subsection 4.2.1) and let Hl be the set of histories that would be obtained by evaluating at vectors l^h and l^h for histories h inStage A (as in Subsection 4.2.1, using l^h and l^h instead of m^h and m^h, respectively).

Remark 2 We want to emphasize that any history h contains all information needed to recover the sets N_u^h; N_d^h and N_s^h as well as all choices made by agents along h (speci…cally, a_ih0 and a_ibh) regardless of the vectors used to evaluate ; for instance, from the history h⁰ = (;; a₄ = 2; a₁ = u; a₂ = d; a₃ = u; a₁ = 4; a₂ = d; a₃ = u); input of Step 2.2.b in the example of Subsection 4.3, we can obtain in an unambiguous way the sets N_u^h0 = f3g; N_d^h0 = f2g; and N_s^h0 = f1;4g and all choices made by agents along h (for instance, a⁽₂^;;a4=2;a1=u) =d). In particular, for h2Hm[Hl, the corresponding setsN_u^h; N_d^h and N_s^h as well as all choices made by agents along h; can be obtained from h itself, independently of whether the vectors m^h and m^h orl^h and l^h have been used to generate h:

Claim 1 Let h 2 Hl be the output of Step A.t and let h^{0 im} h (i.e., h = (h⁰; a_ih0)).

Then,

(C1.1) if i^h0 2N_u^h; l^h =l^h0 and l^h = (l^h0_ih0; p^h_ih0⁰ ),

(C1.2) if i^h0 2N_d^h; l^h = (l^h0_i^h0; p^h0

holds, by replacement monotonicity and the de…nitions of p^h and p^h0; p^h_i⁰ p^h_i for all i6=i^h0:

Case 2: Assume i^h0 2N_d^h0: Then, i^h0 2N_d^h [N_s^h and by de…nition of l^h_i^h0; l^h_ih0 =

( p^h0

i^h0 if i^h0 2N_d^h a_ih0 if i^h0 2N_s^h:

Hence, by e¢ ciency and (3.1), l^h_i^h0 p^h_ih0⁰ = _ih0(l^h0) l^h_i^h00 . Since, by Claim 1, l^h_i^h0 =l^h0_i^h0 holds, by replacement monotonicity and the de…nitions of p^h and p^h0;

p^h_i⁰ p^h_i for all i6=i^h0:

(3.3) By the de…nition of l^h; i^h0 2N_s^h implies l^h_ih0 =a_ih0. By e¢ ciency of and (3.1),

p^h_ih0 = _ih0(l^h) a_ih0: (42)

By Claim 1 and since a_ih0 belongs to fp^h0

i^h0; : : : ; p^h_ih0⁰ g, l^h_ih0 = l^h0_ih0, l^h_ih0 = a_ih0 p^h_ih0⁰ =

i^h0(l^h0) l^h_i^h00 , where the last inequality holds by e¢ ciency and (3.1). By strategy-proofness,

p^h_ih0 = _ih0(l^h) a_ih0: (43)

By (42) and (43),

a_ih0 =p^h_ih0:

(3.4) Assume i^h0 2N_u^h: By Claim 1, l^h0 =l^h and sop^h0 =p^h.

(3.5) Let i 2 N_s^h and i 6= i^h0: Let bh be the history at which i has played for the last time before h and let h be the immediate successor of bh on the path towards h (i.e., bh ^imh ^{im im}h^{0 im}h): Then, i=i^bh 2N_s^h. Therefore, by (3.3),

a_ibh =p^h

i^bh: (44)

By (3.2), and since for any history h such thath h h; i^bh 2N_s^h and i^bh 6=i^h , l^h

i^bh =l^h_ibh =a_ibh =p^h

i^bh p^h0

i^bh p^h

i^bh; (45)

where the …rst equality follows from the de…nitions of l^h and l^h, the second equality from (44), and the two inequalities follow from (3.2), successively applied in the case of the …rst inequality. Furthermore, by e¢ ciency of and (3.1),

l^h

i^bh =l^h_ih^b i^bh(l^h) =p^h

i^hb: Therefore, by (45),

a_ibh =p^h0

i^bh =p^h

i^bh:

(3.6) By de…nition of l^h;we have that, for all i2N_d^h[(NnN^h),

l^h_i = 0: (46)

Let i2N_u^h: We distinguish between two cases. First, assume there exists bh h⁰ such that i2N_u^bh, i2N_u^h for all bh h h and i does not play ath : Then,

l^h_i =p^bh_i p^h_i; (47)

where the equality follows from the de…nition ofl^h and the inequality from (3.4) and (3.2).

Assume now thatbh=h⁰ (becausei=i^h0). Then, by the de…nition of l^h and (3.4),

l^h_i =p^h_i⁰ =p^h_i: (48)

Leti2N_s^h. We distinguish between two cases. First, assume i6=i^h0 and letbh be such that i^bh =i:Then,

l^h_i =a_ibh =p^h_i; (49)

where the …rst equality follows from the de…nition of l^h and the second one from (3.5).

Assume now that i=i^h0: Then, sincebh=h⁰; the de…nition of l^h and (3.3) imply that

l^h_i =a_ih0 =p^h_i: (50)

Therefore, by (46) to (50), for alli2N;

l^h_i p^h_i. Hence, by feasibility of p^h;

X

i2N

l^h_i k:

The proofs of (3.7), (3.8), (3.9) and (3.10) are symmetric to the proofs of (3.2), (3.3), (3.4) and (3.5), and so they are omitted.

Lemma 4 Let h= (h⁰; a_ih0)2Hlbe the output of Step A.tand assume N^h 6=N. Then, (4.1) if i^h0 2N_u^h; p^h

i^h0 =p^h

i^h0 =p^h0

i^h0, (4.2) if i^h0 2N_d^h,p^h

i^h0 =p^h_ih0 =p^h0

i^h0, (4.3) if i2N_s^h,p^h

i =p^h_i =a_ibh where bh is such that i=i^bh, (4.4) if i2N_d^h0nfi^h0g; then p^h_i =p^h_i⁰,

(4.5) if i2N_u^h0nfi^h0g; then p^h

i =p^h0

i . Proof of Lemma 4

(4.1) Let i^h0 2N_u^h. Then, by (C1.1) in Claim 1,

l^h =l^h0 and l^h = (l^h0_ih0; p^h_ih0⁰ ): (51) By de…nition of p^h and p^h0;

p^h =p^h0: (52)

By (3.6) in Lemma 3, (51), and e¢ ciency of ,

i^h0(l^h) p^h_ih0⁰ : If _i^h0(l^h)> p^h_ih0⁰ ;then

i^h0(l^h0_ih0; p^h_ih0⁰ ) = _ih0(l^h)> p^h_ih0⁰ > p^h0

i^h0 = _ih0(l^h0);

where the second strict inequality follows from the de…nition ofi^h0:But, by single-peakedness, it contradicts strategy-proofness of . Then, by (52) and the de…nition of p^h;

p^h

i^h0 = _ih0(l^h) = p^h_ih0⁰ =p^h_ih0; which is the statement in (4.1).

(4.2) The proof proceeds as in (4.1), using symmetric arguments.

(4.3) Let i2N_s^h and letbhbe such thati=i^bh:Ifbh6=h⁰ the proof follows from (3.5) and (3.10) in Lemma 3. Ifbh=h⁰ the proof follows from (3.3) and (3.8) in Lemma 3.

(4.4) The proof is by induction on the length of the histories.

Assume that h= (;; a_i;) is a history of length 1:Then, h⁰ =; and N_d^h0 =;: Therefore, (4.4) holds trivially.

Assume that (4.4) holds for all h h. We prove that it holds forh:

Let i 2N_d^h0nfi^h0g: Then, by the de…nition of l^h0; l^h_i⁰ =p^hb0

i where hb⁰ is such that i =i^hb0: Leth be such that h⁰ =hb⁰ and h h (i.e., hb^{0 im}h h). By (4.2),

p^hb0

i =p^h0

i =p^h

i =p^h_i: (53)

By the induction hypothesis and the de…nition of h, and since i 2 N_d^h and i 6= i^h for all h h h⁰,

p^h_i =p^h_i⁰: (54)

Hence,

l^h_i =l^h_i⁰ =p^h_i =p^h_i⁰; (55) where the …rst equality follows fromi6=i^h0;the second follows from (53) andl^h_i⁰ =p^hb0

i ; and the third from (54). Therefore, by (3.2) in Lemma 3, e¢ ciency of and (3.1) in Lemma 3, and (55),

p^h_i⁰ p^h_i l^h_i =p^h_i⁰: Thus,

p^h_i =p^h_i⁰:

(4.5) The proof proceeds as in (4.4), using symmetric arguments.

Lemma 5 Let h2Hl be the output of Step A.t. Then, N^h 6=N:

Proof of Lemma 5 If h = ; the statement follows trivially. Assume that h is a non-empty history and, to obtain a contradiction, that N^h = N: Then, by the de…nition of Stage A, there exists h h such that N^h =N_u^h[N_s^h[N_d^h =Nnfi^hg. By de…nition ofi^h

p^h

i^h < p^h_ih:

Without loss of generality, we can assume that agent i^h is not called to play betweenh and h. Then, since p^h is feasible, there exists i6=i^h such that

p^h_i < p^h

i:

Since N_u^h[N_s^h[N_d^h =N=fi^hg; i 2N_u^h[N_s^h[N_d^h. We can apply now Lemmata 3 and 4 to h; because N^h 6=N:

Case 1: If i2N_s^h we obtain a contradiction with (4.3).

Case 2: If i2N_u^h we obtain a contradiction with either (4.1) or (4.1), (4.5) and (3.2).

Case 3: If i2N_d^h we obtain a contradiction with either (4.2) or (4.2), (4.4) and (3.7).

Observe that by Lemma 5 the hypothesis that N^h 6=N in Lemmata 3 and 4 is without loss of generality. Hence, those two lemmata apply to any history inStage A, or output of Stage A. Note that in the proof of Lemma 5 we have used Lemmata 3 and 4, applied to history h; where N^h 6=N:

Lemma 6 Let h be a history such that h2Hm\Hl. Then, l^h =m^h and l^h =m^h.

Moreover, Hm=Hl:

Proof of Lemma 6 The proof is by induction on the length of the histories. The induction hypothesis is that for all t 1; the set of histories of length t in Hm and Hl coincide and if h2Hm has length t, then

l^h =m^h and l^h =m^h:

If t= 1;then the induction hypothesis holds trivially from the de…nitions.

Assume that the induction hypothesis holds for all t < t: We will prove that it holds for t:

Let h = (h⁰; a_ih0) be a history of length t in Hm (or Hl): Since h⁰ has length t 1; by the induction hypothesis, h⁰ is a history in Hm and Hl and

l^h0 =m^h0 and l^h0 =m^h0; which means that

q^h0 =p^h0 and q^h0 =p^h0: Then, h= (h⁰; a_ih0) is a history of length t inHl (orHm).

Now, we prove that l^h =m^h holds.

If i2N_u^h[(NnN^h); l^h =m^h holds by their de…nitions.

If i2N_s^hnfi^h0g; by (3.5) in Lemma 3 and Lemma 5,a_ibh =p^h0

i^bh where i=i^bh: Therefore, m^h_i =q^h_i⁰ =p^h_i⁰ =a_ibh =l^h_i:

If i 2N_d^hnfi^h0g; q^h_i⁰ =p^h_i⁰ by the induction hypothesis. Moreover, i 2N_d^h0nfi^h0g and by the de…nition of bh; i =i^bh and i 2N_d^bhnfi^h0g: By (4.4) in Lemma 4 and Lemma 5, p^h_i = p^h_i⁰: By (4.2) in Lemma 4 applied to i=i^bh; p^h0

i^bh =p^bh

i^hb: Therefore, m^h_i =q^h_i⁰ =p^h_i⁰ =p^bh

i =l^h_i: If i=i^h0 and i2N_s^h; then bh=h⁰ and, by their de…nitions,

m^h_i =a_ih0 =a_ihb =l^h_i: If i=i^h0 and i2N_d^h; then bh=h⁰ and, by their de…nitions,

m^h_i =q^h0

i =p^h0

i =p^bh

i^bh =l^h_i: Therefore,

l^h =m^h:

The proof thatl^h =m^h holds as well proceeds as the proof of l^h =m^h, using symmetric arguments.

Therefore, the sets of actions available to agenti^h0 if we usel andlormand mcoincide.

Hence, h= (h⁰; a_ih0) is the same history and belongs to Hl and Hm:

Lemma 7 Let h h be two histories in Stage A (h may be itself the output of Stage A): Then,

(7.1) if i2N_u^h and i=i^h0 =i^h0,q^h_i < q^h_i; (7.2) if i2N_d^h and i=i^h0 =i^h0 q^h

i > q^h

i, (7.3) if i2N_s^h,q^h

i =q^h_i =q^h

i =q^h_i. Proof of Lemma 7

(7.1) Assume i 2 N_u^h and i = i^h0 = i^h0: Without loss of generality we can assume that i=i^h0 does not play betweenh⁰ and h⁰:By Lemma 6,

q^h

i =q^h_i =q^h0

i < q^h_i⁰ =q^h_i,

where the …rst equality follows from (4.1) in Lemma 4 and i=i^h0, the second follows from iterate application of (4.5) in Lemma 4 (if needed) andi6=i^eh for allh eh h⁰ (if any), the strict inequality follows from the de…nition of i^h0 and i = i^h0 and the last equality follows from (4.1) in Lemma 4 andi=i^h0:

(7.2) The proof is symmetric to the proof of (7.1).

(7.3) The proof follows from (4.3) in Lemma 4.

Lemma 8 Let h be the output of Stage A and let h h: Then, (8.1) q^h_i =a_ibh for all i2N_s^h where bh is such that i=i^bh,

(8.2) q^h_i =q^bh_i for all i2N_u^h where bh is such that i=i^bh, (8.3) q^h_i =q^bh

i for all i2N_d^h where bh is such that i=i^bh, (8.4) q^h_i q^h_i for all i2N_u^h[N_s^h where i=i^h and a_ih =u, (8.5) q^h_i q^h

i for all i2N_d^h[N_s^h where i=i^h and a_ih =d:

Proof of Lemma 8

(8.1) The proof follows from Lemma 6 and (4.3) in Lemma 4.

(8.2) Let i=i^bh 2N_u^h and let h be such that bh ^im h h: Since i^bh 2N_u^h and i^bh 2N_u^h; by (4.1) and (4.5) in Lemma 4, and by Lemma 6,

q^bh_i =q^h_i =q^h

i =q^h

i: Now, since h is the output of Stage A,q^h

i =q_i^h. Hence, q^h_i =q^bh_i:

(8.3) The proof proceeds as the proof of (8.2), using symmetric arguments.

(8.4) Let i2N_u^h[N_s^h;where i=i^h anda_i^h =u. We distinguish between the two sets to which i can belong to.

Case 1: i 2 N_u^h. By (8.1), we can assume without loss of generality that h is the last history at which agent i has played; namely, h = bh. Then, q^h_i = q^h_i follows from (8.2) in Lemma 8.

Case 2: i 2 N_s^h. Then, since i = i^h and a_ih = u, h bh h. Consider h such that h ^imh bh: Then,

q^h_i =q^h_i =q^h

i q^bh

i;

where the …rst two equalities follow from (4.1) in Lemma 4, i=i^h and Lemma 6, and the inequality follows directly from Lemma 6 and (4.5) in Lemma 4 ifidoes not play betweenh andbh, and from (4.5) in Lemma 4 and (8.1) in Lemma 8 otherwise (perhaps, after applying them iteratively). Since i2N_s^h;by Lemma 6 and (4.3) in Lemma 4,

q_i^h =a^bh_i 2 fq^bh

i; : : : ; q^bh_ig: Thus,

q_i^h q_i^h:

(8.5) The proof proceeds as the proof of (8.4), using symmetric arguments.

Proof of Theorem 2 It follows from the three statements that we will present and prove successively. Along the proof we will use the statement of Lemma 6.

Statement 1.2 Let be a sequential allotment rule. Then, the extensive game form is well de…ned and …nite.

Proof of Statement 1.2 By Statement 1.1 it will be su¢ cient to show that Stage Aof the game is well de…ned and …nite. We …rst argue that the agents that are called to play along Stage A are uniquely identi…ed and well de…ned.

Consider Step A.t, with t 0, with input h⁰ and q^h0 6= q^h0. If t=0 and so h⁰ = ;; i^; = min_<fi 2 N j q^;

i < q^;_ig is well de…ned. If t 1 and so h⁰ 6= ;, m^h0 6= m^h0. Hence, NnN_s^h0 6= ;: Since q^h0 and q^h0 are feasible, P

i2N q^h0

i = P

i2N q^h_i⁰ = k: Hence, by (7.3) in Lemma 7, there exists at least one i 2 NnN_s^h0 such that q^h0

i < q^h_i⁰, and so i^h0 = min_<fi 2 NnN_s^h0 jq^h0

i < q^h_i⁰gis well de…ned.

As in the proof of Statement 1.1, it is easy to see that the sets of agentsN_u^h; N_d^h andN_s^h are well de…ned: The proof that Stage A is …nite follows from Lemma 8 and the following two facts: (i) if q^h_i⁰ =k and i=i^h0 2N_u^h0;then i2N_s^h and (ii) if q^h0

i = 0 and i=i^h0 2N_d^h0; then i2N_s^h.

We now proceed to state and prove that OSP-implements :

We now proceed to state and prove that OSP-implements :

Dans le document All Sequential Allotment Rules Are Obviously Strategy-proof (Page 26-46)