The trace of J is :
trJ = α + 1+ρ - (1 ~ α)(σ - 1)
ασ
C-" O
The determinant of J is:
DetJ = 1 + ρ > 1
Rearranging the terms of the trace of J yields:
trJ=2+ρ+1 μ (ι - α )(1+p - α ) )
σα
Since the term in brackets is positive, it is straightforward to conclude that |1 + DetJ | <
|trJ |. As a result, the steady state is a saddle point for any σ and ρ.
B Proof of proposition 4
By eliminating the Lagrange multipliers from the equations (16)-(18), we obtain the
following system:
kt+1 = i + β(1 - τt)(1 - α)Akα (35)
τt+1 = 1 - Tl ʌ Λb,a (36)
(1 - qt+1)Akt+1
[ _ [(1 + ρ)(1 + β)]-τtkα(1 - qt)- /O7X
qt+1 = --------1—≡----α ------ (37)
α 1 -σ qt σ τt+1 kt+11 σ
40
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