Using the expression of Υ, the determinant becomes:
DetJ
τ⅛(1 — q)1 σ+1+a [(1 + ρ )β (1 — α )l ] 1 -σ
( αq)1-σ
(1 -q)2 A [ lβ (1 -a )]α
(1+β )α
— l (1 — [)a)
_-_ (1 — [) AA+1+a ( ɪ ) -
α-101⅛ l (1 — [a (--q)2 A[lβ(1 -a)]α
α q 1(1 q) l (1+ β) αl (1 -q)α
-1
Simplifying and replacing (1 — q)a by its steady state expression yields
DetJ = -
αq
(1 — [)
(1 -q)2 Aa-α [lβ(1 -a)]α [β(1+ ρ)(1 -a)l+a]α
l (1+β )α
-1
Replacing (1 — q) by its steady state expression, the large term in the denominator can
be identified as (1 — q)/(1 — τ). Finally, after simplification, we obtain:
DetJ =
-αq[
(44)
Condition 1: if q ∈ ] 1 +a , ^-α[, then ∖DetJ| > 1;
Condition 2: if q ∈ ] 1 ∖ι, 1[, then ∖DetJ| < 1;
where q is a function of the parameters ρ and l ; τ is a function of the parameters l, ρ and
A (see equation (29)); and 0 < τ,q < 1.
As for the trace, rearranging (42) yields
where 1 — τ = a(1 -l-)^α (see equation (36)).
trJ = Ω + J
_ μ α ∖ a (1 — q)- + a[a
σ y(1 + ρ ) β (1 — α ) l + α J ∣
1+a (1 - σ ) (1 - ∙
aσ
(1 + ρ ) -1σ Υ
τ)(σ+a(1 -σ)) ´
aσ I
The second additive term of the trace can be simplified in the same way as the determinant.
After some computations, we obtain:
44