Market forces meet behavioral biases: cost misallocation and irrational pricing.

To prove the lemma, suppose, by way of contradiction, that {[z.sup.k]} has as an accumulation point the unit vector z with [z'.sub.n] = 1 and all other entries equal to zero. Passing to subsequences if necessary, we may assume that [z.sup.k] [right arrow] z'. This implies that [[??].sup.k.sub.m] - [[bar.p].sup.k.sub.m]/[parallel][[??].sup.k] - [[bar.p].sup.k][parallel] [right arrow] 0 for every m [not equal to] n, and [[??].sup.k.sub.n] - [[bar.p].sup.k.sub.n]/[parallel][[??].sup.k] - [[bar.p].sup.k][parallel] [right arrow] 1. Thus, given any [tau] > 0, for [epsilon] small and for all k large enough, [[??].sup.k.sub.n] - [[bar.p].sup.k.sub.n]/[parallel][[??].sup.k] - [[bar.p].sup.k][parallel] [right arrow] [tau], hence [parallel][[??].sup.k] - [[bar.p].sup.k][parallel] < [[??].sup.k.sub.n] - [[bar.p].sup.k.sub.n]/1 - [tau]. Then for m [not equal to] n,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

contradicting the assumption that [z.sup.k] [right arrow] z'.

Lemma 2. Suppose that [s.sup.k] [right arrow] s, and [p.sup.k] [right arrow] p where for each k, [p.sup.k] is an equilibrium of [GAMMA]([s.sup.k]). Thenp is an equilibrium of [GAMMA](s).

Proof. If p were not an equilibrium of [GAMMA](s), then there would be a firm m and a strategy [p'.sub.m] such that

[[pi].sub.m] ([p.sub.-m], [p'.sub.m], [s.sub.m]) > [[pi].sub.m] ([p.sub.-m], [p.sub.m], [s.sub.m]).

Because [[pi].sub.m] is continuous p and [s.sub.m], for all-large enough k,

This contradicts the fact that [p.sup.k] is an equilibrium of [GAMMA]([s.sup.k]) for all k.

Lemma 3. For any [s.sub.-n], [s.sub.n] [greater than or equal to] [DELTA], and p [member of] [epsilon](s),

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Proof. First, note that, because best responses are strictly increasing in opponents' prices and cost, [p.sub.0] << p. Define a sequence of price vectors [{[p.sub.t]}.sup.[infinity].sub.t=0] such that [p.sub.0] = sup{z [member of] [epsilon]([s.sub.-n], 0), z [less than or equal to] p}, and for t > 0,

[p.sub.t] = ([r.sub.m]([p.sub.t-1], [s.sub.m]), m [not equal to] n, [r.sub.n]([ps.bu.t-1], [DELTA])).

We show by induction that [p.sub.t] << p for all t. The claim holds by definition for t = 0. Assume that it holds for t [greater than or equal to] 0, then because [p.sub.t,-m] << [p.sub.-m] for all m, and best responses are strictly increasing in opponents' prices, [r.sub.]([p.sub.t], [s.sub.m]) < [r.sub.m](p, [s.sub.m]). A similar argument shows that [p.sub.t] >> [P.sub.0] for all t > 0. Because [p.sub.0], p] is compact, passing to subsequences if necessary, we may assume that [p.sub.t] [right arrow] [P.sub.[DELTA]] [member of] [[P.sub.0], P]. Because {[p.sub.t]} is an adaptive pricing adjustment starting at [p.sub.0], by Proposition 1, p = inf{z [member of] [epsilon]([s.sub.-n], [DELTA]), z [greater than or equal to] [p.sub.0]}. Clearly, [p.sub.[DELTA]] [less than or equal to] < sup{z [member of] [epsilon]([s.sub.-n], [DELTA]), z [less than or equal to] p}. Lemma 4. There is [phi] > 0 such that for any s and p [member of] [epsilon](s),

[[PI].sub.n]([s.sub.-n], [DELTA], p) - [[PI].sub.n]([s.sub.-n], 0, p) > [phi].

Proof. Consider any s', p' [member of] [epsilon](s'). Assume first that [[??].sub.n] > [DELTA]. By Proposition 1,

[[PI].sub.n]([s'.sub.-n], [DELTA], p') - [[PI].sub.n]([s'.sub.-n], 0, p') = [[pi].sup.e.sub.n](sup{z [member of [epsilion](s'.sub.-n], [DELTA]), z [less than or equal to] p'}) -[[pi].sup.e.sub.n(sup{z [member of] [epsilon]([s'.sub.-n], 0), z [less than or equal to] p'}).

Combining Lemma 3 and Proposition 9, this expression is strictly positive for every s', p' [member of] [epsilon](s'). The conclusion now follows from the facts that s' ranges over a finite set, [epsilon](s') is compact for every s', and [[pi].sup.e.sub.n] is continuous. The remaining cases [s'.sub.n] = 0, [DELTA] can be dealt with similarly.

Proof of Theorem 2. Choose [DELTA] as in Proposition 9. Fix an arbitrary firm n and define [L.sub.[tau](s) to be the sum of payoffs in all periods [tau]' < [tau] such that [[tau]' was an experimental period and the distortion was chosen. For any 0 < [delta] < [DELTA] define

[X.sub.[tau]] = [L.sub.[tau]]([delta]) - [L.sub.[tau]](0).

Because experiments are independent across periods and all distortions are picked with equal probability, the strong law of large numbers implies that, almost surely, for all [tau] large enough, all distortions are played with equal frequencies. Thus, for all large nonexperimental periods [tau], [X.sub.[tau]], > 0 implies that firm n does not choose [s.sub.n] = 0. Thus, to prove the theorem, it suffices to show that [X.sub.[tau]] [right arrow] [infinity] almost surely.

Define

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Clearly,

[X.sub.[tau]] = [Y.sub.[tau]] + [Z.aub.[tau]], (29) (A3)

and {[Y.sub.[tau]]} is a martingale. (30) Because each [Y.sub.[tau]] has bounded support, by Theorem 6.8.5 in Ash and Doleans-Dade (2000), the random variables [Y.sub.[tau]] - [Y.sub.[tau] - 1]] = [y.sub.[tau] are orthogonal, and by Theorem 5.1.2 Chung (1974) [Y.sub.[tau]]/[tau] = [[summation].sup.[infinity].sub.s=1] [y.sub.s]/[tau], [right arrow] 0 almost surely (i.e. the conclusion of the law of large numbers holds). From this and equation (A3), it follows that

[X.sub.[tau]] - [Z.aub.[tau]]/[tau] [right arrow] 0 a.s. (A4)

On the other hand,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

where the last two inequalities follow from Lemma 4, and K is the number of distortions in the grid. This implies that [Z.sub.[tau]]/[tau] [greater than or equal to] 1/N [epsilon]/K [phi]. This fact and equation (A4) imply [X.sub.[tau] [right arrow] [infinity] almost surely.

Proof of Proposition 2. In any equilibrium of the pricing game, the lowest price offered must be at or just below the second-highest cost including any distortion. Suppose this is not the case and let [p.sup.*] be the lowest-price. Then, of the two lowest cost firms, at least one is not capturing the entire market. But that firm can price just under [p.sup.*] and capture the entire market and increase its accounting profit, a contradiction. Also, in any equilibrium, the firm with the lowest cost including distortions must be setting the lowest price. Otherwise, it is making zero profits and can deviate and make positive accounting profit by matching the lowest price. If its cost including distortion is lower than the second- highest cost, it will price slightly below the second-highest cost to capture the entire market. This observation implies that in any equilibrium with no distortions, the lowest price must be at or just below [c.sub.2]. Also, firm 1 must be setting the lowest price and either capturing the entire market, if [c.sub.2] > [c.sub.1], or sharing it, if [c.sub.1] = [c.sub.2]. In either case, firms 2 and above make zero profits and firm 1 makes positive profits only if [c.sub.2] > [c.sub.1].

Suppose firm 2 or higher distorts its relevant costs upward by s. Then, let [c.sup.*] he the lowest cost including the distortion among all the firms apart from firm 1. By our argument above, any equilibrium of the pricing game must have the price set at or just below [c.sup.*]. Therefore, firms 2 and higher make zero profits in equilibrium and hence have no incentive to distort their relevant costs upward. Now, suppose firm 1 distorts upward by s. If [c.sub.1] + s < [c.sub.2], then there is no change in the equilibrium in the pricing game and hence no incentive to distort. If [c.sub.1] + s > [c.sub.2], in any equilibrium in the pricing game, firm 1 must make zero profits as firm 2 has the lowest cost and will always undercut the price set by the firm with the second-highest cost including any distortion. Hence, firm 1 has no incentive to distort relevant costs upward. Moreover, there is no incentive for firm 1 to distort downward as it does not alter the pricing equilibrium. Similarly, if firm n > 1 distorts downward by s it still makes zero profits, when [c.sub.n] - s [greater than or equal to] > [c.sub.1], or makes a loss, when [c.sub.n] - s < [c.sub.1], and it sets a price just under c j which is less than its true cost [c.sub.n]. []

Proof of Proposition 3. We establish existence by showing that [[bar.[pi]].sup.e.sub.n](s) is supermodular in s (Milgrom and Roberts, 1990). Differentiating [[bar.[pi]].sup.e.sub.n](s) with respect to [s.sub.n] we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Using (3), we can write this as

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (A5)

where [partial derivative][[bar.p].sub.n][partial derivative][s.sub.n] and [partial derivative][[bar.p].sub.n][partial derivative][s.sub.m] are given by (5) and (6), respectively. This implies

[[partial derivative].sup.2][[bar.[pi]].sup.e.sub.n](s)/[partial derivative][s.sub.n][partial derivative][s.sub.m] = [(N - 1).sub.[gamma]] ([partial derivative][[bar.p].sub.n]/[partial derivative][s.sub.n]) ([partial derivative][[bar.p].sub.m]/[partial derivative][s.sub.n]) > 0.

Thus, the first-stage game is supermodular in s. To establish uniqueness we establish the contraction condition [LAMBDA] [equivalent to] [[partial derivative].sup.2][[bar.[pi]].sup.e.sub.n](s)/[partial derivative][s.sup.2.sub.n] + [[summation].sub.n[not equal to]m] [[partial derivative].sup.2][[bar.[pi]].sup.e.sub.n](s)/[partial derivative][s.sub.n] [partial derivative][s.sub.m] < 0 for all s [greater than or equal to] 0. We first note that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Thus, we can write