Income Taxation when Markets are Incomplete

show that this property holds we can proceed by appending

(b 1 ,...,bH)KZ = 0,

bb^T - 1 = 0

to the equilibrium system, F = 0, and show that the new system as no
solution at t^I = 0 (for this we need that H ≤ S - J). This property is
implied by the well-known property of local controllability of individual
state prices, λ. Thus, restrict attention to the set Ψ* for which Theorem 2
in Geanakoplos et al. (1990) holds, and (i) through (iv) are satisfied; these
properties can be shown to hold generically in consumers’ endowments, e.
Observe that, at t¹ = 0, F and the system of first-order GEI equilibrium
conditions, F, are identical, and that so are their Jacobians with respect to
ξ '. Therefore,

rank (Dξ,F(^{ξ,, t}I^{, ψ}) |f(ξ,,ti,ψ)=o)_t, =0 = n^,,

and the set of regular equilibria, F ^-1 (0), is a manifold of dimension k =
dim(Ψ). Next, we add the government budget constraint to F to get F,
the complete system of first-order conditions of the E - t^I equilibrium.
We want to show that the Jacobian of F , evaluated at a no-tax equilibrium,
has full rank. Now (Dξ,,_tι F F(ξ,,_tι,ψ*=₀)_tI=0 has the following structure (its
complete representation is in (5) below, with the last column and row blocks
deleted):

var.                                                                         ξ^, t¹

equ.                                                         ---------

F                                                   | Dξ, F ..........

n ^,× n^,

∑ h ∈ H ( tS + tS (∑ j ∈ J, ^θhR^j + ∑ ∈ J2 ^θj )) = 0 | 0     [∑j ∈ J2 y^j ] ( * *

S×S

(*) : [ξ_s]denotes a block-diagonal matrix with typical element ξ_s.

Since we take ψ ∈ Ψ*, we have that ∑j-ys^j = 0 for all s. Thus, if0is a
regular value of F, it is also a regular value of F. □

By straightforward application of the implicit function theorem, we ob-
tain the following.

Corollary 2. In the context of Theorem 2, for every ψ ∈ Ψ *, there exists an
open neighborhood N (0) of tⁱ = 0 such that, for all tⁱ ∈ N (0),

rank (Dξ,,tiF(ξ^,, t¹, ψ) |F(ξ,,ti,ψ*=o) = n,

and F(ξ^,, t¹, ψ) is transverse to zero.

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