that, if the stated parameter restrictions are satisfied, there exists always one and only one real and
positive solution θ0 ∈ (γ, 1). The proof follows from the fact that the specified parameter restriction
allows the intercept (the value of the polynomial at θ0 = 0) of the LHS polynomial to be bigger than
in intercept of the RHS polynomial. Specifically LHS (0) > RHS(0) implies:
n n n nγ Ω - ΓG4
(1 - 2γ) φ∕2 > , (----v(-—),
μσ(ρ + n∕μ — n) Γ
which rearranged leads to the parameter restriction. It is easy to see that this condition allows for a
unique solution36. Moreover for Minkowski’s inequality Ω — ΓG < 0, therefore when 1 — 2γ > 0 no
restriction on parameters is needed for a unique solution.
Proof of Proposition 1.a. Solving (18) for x (ω) we get:
and deriving w.r.t. Gωwe obtain
+ Gω
θo — γ
bσ(ρ + n∕μ — n)
= x (ω) ,
∂x (ω) / λ (ω) — 1 ∖ θo — γ
∂Gω y λ (ω) j bσ(ρ + n∕μ — n),
which is always positive since λ (ω) > 1, θ0 > γ and ρ > n. From this derivative we can also see
that ∂x (ω) ∕∂Gω > ∂x (ω) ∕∂Gω∣ when (λ (ω) — 1) ∕λ (ω) > (λ (ω) — 1) ∕λ (ω) which is always true if
λ (ω) > λ (ω).
Proof of Proposition 1.b Rearranging (A11) we get a single polynomial in θo and Ω:
F (θo; Ω) = ---j-θ----г [(θo — ω) (γ 1 — 1) + (G — ω)] — (θ0 + 1 — 2Y)(1 — θθ) φ^. (A.1.2)
μσ(ρ + n∕μ — n) l
Using the Implicit Function Theorem we get:
dθo —∂F∕∂Ω
'dΩ = ∂F∕∂θ0 =
n(θθ-γ)
μσ⅛l<√μ-nl∣' ∩
=>> 0
, .n.---τ [(θo — Ω) (Γ-1 — 1) + (G — Ω)] + n(θ0-γ) ʌ(Γ-1 — 1) + φ(θo — γ)
μσ(ρ+n∕μ-n) o μσ(ρ+n∕μ-n) 0 o 0 o
This results follows from the fact that θ0 > γ, ρ > n, Γ-1 > 1 and finally, from (A1) we know
that the expression inside the square brackets is greater than zero.
10 Appendix II: transitional dynamics
The schooling choice leading to eq. (4) in the paper implies, off steady states, the following ability
threshold θo(t):
36It is easy to check that all parameters restriction are satisfied by the number we use in the calibration excercise.
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