Cournot Oligopoly and the Theory of Supermodular Games

118 RABAH AMIRprominent setting of noncooperative common-property resource extraction, AssumptionA.3 is undoubtedly appropriate and is easily interpreted in terms ofcapital and labor (bounded) capacities.In the following, it is convenient to refer to the (n + 1)-period horizon game(including the 0th period) as G n , with G ∞ being the infinite horizon game.Another related (auxiliary game) is defined in Section 3.We are now in a position to state theMAIN THEOREM. The infinite-horizon game G ∞ has a stationary equilibrium.Furthermore, every finite-horizon game G n has a unique Markovian equilibrium.In both cases, equilibrium strategies consist of (consumption) decisionfunctions which are continuous and nondecreasing and have all slopes below 1,and corresponding value functions are continuous and nondecreasing.3. PROOFSThe proof of the Main Theorem proceeds via two intermediate propositions,each of which requires, in turn, a few lemmas. Many of these results are ofindependent interest. Note that we assume the reader is familiar with Topkis’theorem (Topkis, 1978).We start with two lemmas which establish useful facts invoked repeatedlythroughout our analysis.LEMMA 0.1. Let F 1 and F 2 be probability distributions over [0, +∞), withits Borel subsets. Then F 2 first-order stochastically dominates F 1 , or∫ ∫F 1 (x) ≥ F 2 (x), ∀x if and only if v dF 1 ≤ vdF 2 ,for all real-valued nondecreasing functions v on [0, +∞).Proof. This result is well known. See, e.g., Stoyan (1983).LEMMA 0.2. Let U: [0,+∞) → [0, +∞) be a bounded concave function.Let h: [0,+∞) → [0, +∞) satisfy h(0) = 0 and [h(x 1 ) − h(x 2 )]/(x 1 −x 2 ) ≤ 1, for all distinct x 1 , x 2 ∈ [0, +∞). Then (i) the set ϕ ={(x,y):x∈[0, +∞) and y ∈ [0, x − h(x)]} is a lattice, and (ii) U[x − y − h(x)] is supermodularon ϕ.Proof. (i) This follows directly from the fact that x − h(x) is nondecreasingin x. (ii) We show the equivalent fact that U[x − y − h(x)] has nondecreasingdifferences in (x, y) ∈ ϕ. Let x 2 ≥ x 1 , and y 2 ≥ y 1 be fixed with 0 ≤ y i ≤x i − h(x i ), i = 1, 2. Then, clearly for i = 1, 2, we havex 2 − y 1 − h(x 2 ) ≥ x i − y i − h(x i ) ≥ x 1 − y 2 − h(x 1 ),