Imperfect durability and the Coase conjecture.

To prove (ii), suppose that [??] is a steady state so that P([??]) = [bar.v]. Then because P(*) is decreasing, and because (as shown in the proof of Lemma 1) P(y) [less than or equal to] [bar.v] for y [less than or equal to] [??], we have P(y) = [bar.v] for all y [member of] [0, [??]]. Hence Lemma 4 shows that y < [??] cannot be a steady state. We conclude that the only possible steady state other than [??] is 1.

To prove (iii), suppose that 0 < [y.sup.*] < [??] is a steady state. Then we must have P(y) = [v.bar] for all y [member of] [0, [y.sup.*]], so Lemma 4 implies that no y < [y.sup.*] can be a steady state, and that P(q) < [bar.v] for q [member of] ([y.sup.*], [??]] (otherwise [y.sup.*] could not be a steady state). Thus, by Lemma 1, no y [member of] ([y.sup.*], [??]] can be a steady state, leaving y = 1 as the only other possible steady state. Suppose y = 1 were not a steady state, so P(1) > [v.bar]. This would imply P(q) > [v.bar] for all q [member of] ([??], 1]. We can then use the argument at the end of the proof of Lemma 1 to obtain a contradiction. To prove (i), note that if there is no steady state in [0, [??]], then by the first paragraph, y = 1 is the only possible steady state. It then follows from Lemma 1 that y = 1 must be a steady state. Q.E.D

Proof of Theorem 4. For all i = 1, ..., N - 1, analogously to the two-step case, given [q.sub.i] as a steady state, construct a sequence [{[[??].sup.i.sub.k]}.sup.[infinity].sub.k=0] to the right of [q.sub.i] as follows. Let [[??].sup.i.sub.0] = (I - [mu])[q.sub.i] and let [[??].sup.i](q) = (l - [rho])f(q) + [rho]f([q.sub.i]) for q [greater than or equal to] [q.sub.i]. Given [[??].sup.i](*), let [[??].sup.i.sub.1] be the highest state such that the seller prefers selecting y = [q.sub.i] to selecting y [member of] ([q.sub.i], [[??].sup.i.sub.1]. For k [greater than or equal to] 2, let [[??].sup.i](q) be the seller's optimal choice over ([[??].sup.i.sub.k-2], [[??].sup.i.sub.k-1]] and redefine [[??].sup.i](q) = (1 - [rho])f(q) + [rho]f([[??].sup.i](q)) for q [greater than or equal to] [[??].sup.i.sub.k-1]. Finally, let [[??].sup.i.sub.k] be the highest state such that the seller prefers selecting y [member of] ([[??].sup.i.sub.k-2], [[??].sup.i.sub.k-1]] to selecting y [member of] ([[??].sup.i.sub.k-1], [[??].sup.i.sub.k]]. Let [[??].sup.i] (q) be the seller's payoff function associated with [[??].sup.i](q).

Also analogously to the two-step case, given [q.sub.i] as a steady state, construct a sequence {[[??].sup.i.sub.k]} to the left of [q.sub.i] as follows. Let [[bar.x].sup.i.sub.0] = (1 - [mu])[q.sub.i] and let [[bar.P].sup.i](q) = (1 - [rho])f(q) + [rho] f ([q.sub.i] for q [less than or equal to] [q.sub.i]. Given [[bar.P].sup.i](*), let [[bar.x].sup.i.sub.1] be the smallest state such that the seller prefers selecting y = [q.sub.i] to selecting y [member of] ([[bar.y].sup.i.sub.1], [q.sub.i-1]]. For k [greater than or equal to] 2, let [[bar.t].sup.i](q) be the seller's optimal choice over ([[bar.y].sup.i.sub.k-1], [[bar.y].sup.i.sub.k-2]] and redefine [[bar.P].sup.i](q) = (1 - [rho])f(q) + [rho] f([[bar.t].sup.i](q)) for q < [[bar.y].sup.i.sub.k-1]. Finally, let [[bar.x].sup.i.sub.k] be the smallest state such that the seller prefers selecting y [member of] ([[bar.y].sup.i.sub.k-1], [[bar.y].sup.i.sub.k-2]] to selecting y [member of] ([[bar.y].sup.i.sub.k], [[bar.y].sup.i.sub.k-1]. Let [m.sup.i] be the largest value of k such that 0 [less than or equal to] [[bar.x].sup.i.sub.j] < [[bar.x].sup.i.sub.j-1] for all j [less than or equal to] k, and let [[bar.R].sup.i](q) be the seller's payoff function associated with [[bar.P].sup.i](q).

Let [A.sub.0] = {i|[lim.sub.k[right arrow][infinity]] [[??].sup.i.sub.k] [greater than or equal to] 1 - [mu]}. If [A.sub.0] = [empty set], let [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Otherwise, let [i.sub.0] = min [A.sub.0]. Then define P(q) = [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Also let R(q) and t(q) be the equilibrium value function and policy function associated with P(q).

(P) Now we will construct P(q) for [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Define [i.sub.1] such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for all i < [i.sub.0]. Note that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] because [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] be such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and set [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Now return to (P) with [i.sub.1] taking the role of [i.sub.0]. (P') If [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] a proof similar to the one showing [mu] < [bar.[mu]]in the two-step case yields[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] . Hence, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] exists for[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Define [i.sub.2] such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. for all i < [i.sub.0] - 1. If [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] q' be such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and set P(q) = [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and P(q) = [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Now return to (P) with [i.sub.2] taking the role of [i.sub.0]. If[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] , then return to (P') with [i.sub.2] taking the role of [i.sub.1] and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] taking the role of[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Q.E.D

We would like to thank the editor, Chaim Fershtman, and two referees for helpful comments and suggestions. Liang acknowledges financial support from the Taiwan NSC, grant 95-2415-H-001-009.

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