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

Proof. (Lemma 4.1) We consider the area ∆_U .FromB1 and Topkis’s Monotonic-
ity Theorem, we know that the reaction curves are increasing. The generalized
first order condition for a maximum to occur in the point (x, x) in the absence
of differentiability of F (and G, by symmetry) is that U1 (x, x) ≤ L1 (x, x) . As-
sumption B2 rules out this possibility so we have that no y (nor x), belonging
to (0, c), can be best reply to itself, meaning that the reaction curves do not
cross the 45^o line. Assumption B4 excludes that 0 can be a best reply to 0 and
that c can be a best reply to c. Hence, there must exist a d ∈ [0, c], such that
r1 (d - ε) >d >r1 (d + ε).

To exclude the possibility of another jump we use assumption B3. Consider
r₁(y) and r₁(y) defined as in Section 4. Denote W(y) = L(r₁(y),y) and V(y) =
U(r₁(y),y). From the Envelope Theorem, ^dvζ(^y) = L2(r₁(y),y), and ^dV_y(^y) =
U2(r1(y),y). Hence, when B3 holds, we know that W increases in y quicker
then V does. It means that when the overall reaction curve jumps down along
the diagonal, it never jumps up again. ■

Proof. (Lemma 4.2 ) If B 1⁰ holds reaction curves are continuous in ∆U and
in ∆L . B2 and B4 rules out the possibility that F (and G, by symmetry) has
a maximum in a point [x, x] , that secures that the reaction curve must have a
jump down in a point d ∈ [0, c]. As in the proof of Lemma 4.1, B3 rules out
other possible upward jumps between Δl and ∆u. ■

We may now show that only asymmetric pure strategy Nash equilibria exist.
Proof. (Theorem 4.1) Consider R ⊂ ∆U as defined in Section 4. Define as
before the restricted reaction curves as rι(y)∣∆U and r2(x)∣∆_u. From assumption
B5, the best reply of player 1 to y = 0 cannot be less than d as U is decreasing in
x for y =0when x<d.For y>0, and given assumption B 1 (supermodularity),
Topkis’s Monotonicity Theorem allows us to conclude that the best reply of 1
is increasing. Hence r1 (y)∣∆_U ∈ R. Seemingly the best reply of player 2 for
x = c cannot exceed d as L is decreasing in y when x = c. Also by Topkis’s,
the reaction curve is an increasing map. Therefore r2(x)∣∆_U ∈ R. Consider the
mapping, B : R → R such that B(x, y) = (r1(y), r2(x)). B(x, y) is an increasing
correspondence, given that both its components are increasing. R is a compact

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