Proof of Lemma 6. Define Ez by letting the players move as in Ex; but only at dates that are multiples of m: That is, let zt D 0 for t D 0; : : : ; m 1; and for t m let zt D xnkCi; where k and i are the unique integers satisfying k 0; i 2 N; and
.nk C i/m t < .nk C i C 1/m:
In Ez player i moves only at dates .nk C i/m; since in Ex she moves only at dates nk C i: The path Ez is feasible for EN by (CY), since i 2 N.nkCi/m: Since Ex is an equilibrium path of 0. ; ENR/, it achieves some profile x by Lemma 1). Thus, Ez also converges to x. We use Lemma 5 to prove that Ez is an equilibrium path of 0. 1=m; EN/:
which we now show. If zsi D zt 1i for all s t; then (PS) implies (22). So suppose a date t
which we now show. The definitions of Ez and O imply .1 O/P
Because.Ex; / satisfies (8) at date p; we have .1 /P1
Putting the two previous displays together yields .1 O/P
s
Os ui.zs/ ui.zt 1i ; z i/ ui.zt 1i ; zt i/;
where the second inequality follows from (PS) and z i zt i: This proves (23).
Proof of Lemma 7. Let x 2 D: Then X 2 [ NY; YN]; by Corollary 1 .i/ and the fact that x is a satiation profile. Our definition of a neoclassical setting implies NY < YN: Choose a number OX as follows:
(a) If X D YN; choose OX 2 NY; YN so that for all coalitions S 6D N; OX > YS:
(b) If X < YN; choose OX 2 .X ; YN/ so that for all coalitions S; if OX > YSthen X YS:
Define Ox by Oxi :D vi0.YN/ OX: (Since P
Assume x is not strongly minimal. Then a coalition S and profile z < x exist such that z S D 0 and uS.x/ uS.z/: Since x is a satiation profile, the argument used to prove Lemma B2 increasing on [0; YS]; implies
X > YS: (26)
This proves S 6D N; since X < YN: The remainder of the proof depends on the case.
Case (a). In this case X D YN > YS: Furthermore, since S 6D N; the way OX was chosen
So (27) again holds, and the remaining proof is the same as in case (a).
The following lemma will be used to prove Proposition 4.
Lemma C1. In a neoclassical setting satisfying (PD), for any x 2 D0; a neighborhood of x exists such that every Ox in it that satisfies Ox < x is also in D0:
Proof. Assume the lemma is false. Then an infinite sequence fxkg exists such that xk ! x;
xk < x; and xk 2 D= 0: Since (PD) implies each xk is a satiation profile, each xk must not be strongly minimal. Thus, for each k a coalition Sk and a profile zk < xk exist such that zkSk D 0 and uSk.zk/ uS.xk/: By taking a subsequence we may assume N .xk/ D N .x/
and Sk D S for all k, and that fzkg converges to a profile z (as each zk is in the compact set [0; x]n/: Taking k ! 1 in the inequalities zk < x and uS.zk/ uS.xk/ yields z x and uS.z/ uS.x/: Since zkS D 0 for all k; z S D 0: Therefore, since x is strongly minimal, it must not be true that z < x: Hence, z D x: This implies N .x/ S: Since N .xk/ D N .x/;
we have XkS D Xk: Because uSk.zk/ uS.xk/; (PD) implies uS.zk/ uS.xk/: Summing these inequalities over S yields fS.Zk/ fS.Xk/: Thus, since fSis strictly increasing on [0; YS] and Xk < X YN.x/ YS; we conclude that Zk Xk: This contradicts zk < xk:
Proof of Proposition 4. Observe first that once we find 2 .0; 1/ and an equilibrium path of 0. ; EN/ that converges to x; then Ex is also an equilibrium path of 0.1; EN/: For, by Lemma 2, the pair.Ex; / must satisfy (11). Hence, using (PD) and (9), we see that for all t 1 and i 2 N;
ui.xit 1; xti/ D ui.xit 1; xti/ ui.x/ D Ut.Ex; 1/:
Thus, Ex is an equilibrium path of 0.1; EN/ by Lemma 5.
Accordingly, we only need to find a path Ex that converges to x and a number < 1 such that Ex is an equilibrium path of 0. ; EN/ for all 2 [ ; 1/: By Lemma 6, it suffices to prove this for EN D ENR: If x D 0 we are done, since (PD) implies that the passive strategy profile is an equilibrium that achieves the origin. So we can assume x > 0: Define d 2 RnCby di :D 0 if i 2 N .x/; and=
di :D vi0.X / P
j 2N .x/v0j.X / for i 2 N .x/:
Since X < YN.x/; we haveP
j 2N .x/v0j.X / > 1: Hence, 0 < di < v0i.X / for i 2 N .x/: Choose N > 0 small enough that Nx :D x N d 0: Since x is strongly minimal, Lemma C1 implies the existence of O 2 .0; N/ such that Ox :D x O d is strongly minimal. We have 0 Nx < Ox < x:
We also have u. Nx/ u. Ox/ u.x/; since the concavity of each vi implies that for any 0;
@ui.x d/=@ D di vi0.X / di v0i.X / < 0:
Define fxtg1kD0 to be a round-robin sequence if for each t > 0 and i D t .mod n/; xti D xt 1i : The rest of the proof consists of three steps.
Step 1. A discount factor 0 < 1 and a nondecreasing round-robin sequence fxtg1t D0exist such that x0D Nx; xt ! x; and positive numbers a and" such that
.1 C "/di
This fxtg1t D0 is a round-robin sequence that starts at Nx and converges to x: Fix t > 0; and let i D t .mod n/: Let q 0 be the integer for which t D i C qn: At the end of period t 1;
players j D 1; : : : ; i 1 have raised their actions q C 1 times, and players j D i; : : : ; n have raised theirs just q times. Hence, since x Nx D N d;
xt 1j D
Turning to the desired inequality (28), note that it is equivalent to A :DP
s t
s t ui.xs/ ui.xit 1; xti/ 0:
Observe that A DP1
Each Akis a sum over n consecutive dates, and player i moves only at the first one, t C .k 1/n.
Hence, for each of these dates s; xis D xitC.k 1/n. This implies that
where the inequality follows from Xs XtC.k 1/nfor s t C.k 1/n: Using now the concavity ofvi and Xt 1 Xt C.k 1/n X; we obtain This expression can be bounded from below. From (32) and (34) we have
XtC.k 1/n Xt 1 D N aq
Therefore,
Step 2. A finite, nonincreasing round-robin sequence fxkgkD0K exists such that x0D Nx; xK D 0;
and u.xk/ u. Ox/ for each k D 0; : : : ; K:
This defines a nonincreasing and bounded round-robin sequence fxkg1kD0: Let z be its limit.
We have z xk for all k> 0; and u.z/ u. Ox/: minimal. This contradiction proves that in fact, z D 0:
If ui.0/ ui. Ox/; then Ox would not be strongly minimal (let S D fig and z D 0 in the definition). Hence, u.0/ u. Ox/: Since xk ! 0; this implies that K0 exists such that ui.0; xki/ < u. Ox/ for all k K0and i 2 N: The construction of the sequence thus implies the existence of K K0C n such that xK D 0:
Step 3. A discount factor < 1 and a path Ex ! x exist such that Ex is an equilibrium path of 0. ; ENR/ for 2 [ ; 1/:
Proof of Step 3. Reverse the round-robin sequence obtained in Step 2, and add enough copies of 0 to its beginning and Nx to its end to obtain a finite, nondecreasing round-robin path. This
yields a path, fztgTt D0, from z0D 0 to zT D Nx; that has player 1 moving first and player n moving last.zTn1 D Nx n/: To the end of of this path add the round-robin sequence obtained in Step 1:
zT Cs D xs for all integers s 0: This yields a path Ez D fztg1tD0 that is feasible for ENR and converges to x: To be notationally consistent, relabel it as Ex :D Ez:
Let t 1 and i 2 NtR; so that i D t .mod n/: If t > T and > 0Step 1 implies
ui.xit 1; xti/ Uit.Ex; /: (35) If t T; then since xti D xt 1i ; Step 2 implies
ui.xit 1; xti/ D ui.xt 1/ ui. Ox/ < ui.x/:
Therefore, since Uit.Ex; / ! ui.x/ as ! 1; t < 1 exists such that (35) holds for > t: We conclude that (35) holds for all t 1; i 2 NtR; and > :D max. 0; 1; : : : ; T/. Lemma 5 now implies that Ex is an equilibrium path of 0. ; ENR/ for all 2 . ; 1/: