σ -αq 1 1+a (1 -σ )
1—σ — q+τ I ασ
(1 -τ)( σ+a (1 -σ ))
aσ
)’
and after simplifications, the second term of the trace becomes
—q
—ql τ
(τ( σ+α-ασ )
—
+ 1)∙
As for the first additive term of the trace, Ω, it can be rewritten as follows:
σ Λ _ l(1+β)α (1—q)α-2
1—σ y1 A[lβ(1—a)]α
1 l (1+β )α (1—q)α-2
1 A [ lβ (1—a )] α
(45)
The big term that is common to the nominator and the denominator can be identified,
by replacing 1 — q by its steady state expression, as (1 — τ)/(1 — q).
After simplifications, Ω becomes:
— [ τ(1 — <)] + α<
—q + τ
As a result, we obtain the final expression for the trace J:
trJ =
— [1 — q(2 — α + a )1 — <7(1 — α )
—q + τ
(46)
The final part of this proof consists in comparing ∖trJ∣ and ∣1 + DetJ∣, i.e.,
ισ [1—q(2—a+α )] —9(1—a)
—q+τ
I and I I.
l l —q+τ 1
This depends on σ, the magnitude of q, which itself depends on the value of the parameters
ρ and l, and on whether ∖DetJ∣ ^ 1 (conditions 1 and 2, established above).
When σ > 1, it can be shown that:
• If q ∈
2 —1
______σ
2 a» (1—σ )
, 1 , then either ɪ [1 — q(2
— α + a )1 < 0 or
■—:[1 — <(2 — α + a)1 > 1 and, therefore, ∖trJ∣ > 11 + DetJ∣. The steady state is a
45