lβ (1 -α )
(1+β )(1 -q)2
A (1 -q) 1-σ+α+ 2 g —1
(1+ ρ ) - Υ
A (1 - g) 1-σ + α+ 2 ( ( + + α )k α (1 σ + )
a (1 q) l( 1 σ + α) - A(1 — q)
Ω H--_i---------
α (1+ ρ )- Υ
where,
Υ = (αq)1-σ [lβ(1 - α)]- 1-σ ((1 -^( 1 -α)]α - l(1 - q)α´
- (1 - g) _—1( σ____l ) l (1+β )α
(1 q) ( (1 — σ) 1 — q ) A [lβ(1 — )]α
1__l (1+β )α
A [lβ(1 — a)]α (1 — q)2 __
The trace of J is:
A (1 - ɑ) 1-σ+α+2 ( ( _i—l α ʌ kα - l ( 1—σ + α ) ʌ
-l q ) lkɪ-σ ' α - A (1 -g)
ω l :-----r—ï------------------
α (1 + ρ )1—σ Υ
(42)
and the determinant of J is:
DetJ = -
μ lβ(1 - α) ∖ Aa(1 - g) 1—-+αkα-1
∖ 1+ β J ∖ (1+ ρ ) 1—- Υ
(43)
Then, we rearrange the determinant and the trace by removing the k0s and express them
in terms of q and k to analyze them more easily.
DetJ = - (1 - q)
11~k(1 - q) 1—σ + α
∖ (1 + ρ ) γ
43