four eigenvalues are determined by the upper left 4 × 4 submatrix of ACRP I , denoted by
rR , , . . . . rR .
ACPI. Then the characteristic equation of ACPI is
r r3 + a3r2 + a2r + a1 = 0
where
a3 = -1
1 Λ1
ββ
1 α(1 - Λ2)(2a - 1)∙
. Λ2 .
— 2(1 — a)μ
1 λ-1Λ 1 α(1 - Л2)(2а - 1)
a2 = β + — 1 +-----Λ2-----,
ι αΛ1(2a — 1) ι 2(1 — a)μ(1 + β)
+ Λ2β + β
μΓ Λ'α(2a - 1)
a1 = - β 2(1 - a) +----Λ----
Hence one eigenvalue is zero and the three remaining eigenvalues are the solutions to the
cubic equation r3 + a3r2 + a2r + a1 = 0. Determinacy requires two eigenvalues to lie inside
the unit circle and the other three eigenvalues to lie outside the unit circle. First suppose
the eigenvalue eK is outside the unit circle |eK | > 1, which requires either a > 0.5 or
0.5 > a > 2-1δ [1 — δ — 1 =]. Then using Proposition C.2 of Woodford (2003), two of the
remaining three eigenvalues are outside the unit circle and one eigenvalue is inside the
unit circle if and only if: (Case I)
1 + a3 + a2 + a1 > 0 ⇔
(μ ɪ^1 [1 — (2a — 1)α] < 0,
β
(B1)
— 1+a3—a2+a1 > 0 -⇔ —2(1+β)—4μ(1—a)(1+β) — (μ+1)Λι
1 I (2a — 1)α(2 — ^) 1 0
1+ Λ2 J >0,
(B2)
or (Case II)
1 + a3 + a2 + ɑɪ > 0 ⇔ ———1 [1 — (2a — 1)α] > 0, (B3)
— 1+a3—a2+a1 > 0 ⇔ —2(1+β)—4μ(1—a)(1+β) — (μ+1)Λι
and either
1 (2a — 1)α(2 — Λ2)'
. + Λ2 .
< 0,
(B4)
a1 — a1a3 + a2 — 1 > 0, (B5a)
31
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