Endogenous Heterogeneity in Strategic Models: Symmetry-breaking via Strategic Substitutes and Nonconcavities



set and hence we may use Tarski’s Fixed Point Theorem to conclude that there
exists a pair
(x,y) such that B(x,y) = (rι(y), r2(x)). (x, y) is a pure strategy
Nash equilibrium.

Finally, by Lemma 4.1 we know that no equilibrium can be symmetric. ■
Proof. (Theorem 4.2) Define R, rι(y)U and r2(x)u as before. From
assumption B
5, rι(y)U R and r2(x)u R and from assumption B10
they are continuous. Consider the mapping B : R R such that B(x, y)=
(
r1(y ), r2(x)). B is a continuous correspondence, given that both its compo-
nents are continuous, R is a compact set and hence we may use Brouwer’s
Fixed Point Theorem to conclude that there exists a pair
(x, y∕) such that
B
(x, y∕) = (r1 (y∕), r2 (x∕)). (x, y∕) is a pure strategy Nash equilibrium.

By Lemma 4.2 we know that no equilibrium can be symmetric. ■

8.4 Proofs of Section 6

To prove the theorem we first formulate a useful lemma.

Lemma 8.5 If A10, A2 and A4 hold, then there exist exactly one point d [0, c]
such that
r1(d - ε) >d>r1(d + ε),ε > 0.

Proof. (Lemma 8.5) Consider U . From A10 and Topkis’s Monotonicity The-
orem we know that reaction curve is decreasing in this area. The generalized
first order condition for a maximum to occur in
(x, x) is (in the absence of dif-
ferentiability in this point) U
1 (x, x) ≤ L1(x, x). Assumption A2 rules out this
possibility, thus no x
(0, c) can be a best response to itself. Moreover, neither
(0, 0) nor (c, c) can be an equilibrium, since from A3 follows, that for player 1
it is always profitable to deviate from any of these points. Hence, the reaction
curve does not cross the
45o line, and there must exist a point, call it d [0, c]
such that r1(d - ε) >d>r1(d + ε).

Now, we prove uniqueness of this point. Consider r1(y) and r1(y) defined
as in Section 4. Denote W
(y) = L(rι(y),y) and V(y) = U(rι(y),y). From the
Envelope Theorem,
dwy(y) = L2(r1(y), y), and dVy(y) = U2(r1(y),y). Hence, if

41



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