hold. Using the first inequality, we may further simplify the second and finally arrive at
the following condition:10
t-1
ln πt2-ι(zt-1) < ∑ αi ln At-ι-i + (1-⅛ + (α-α)2) (1- α + γ)ln B
i=0
+ ι-γ-(1 — αt) ln ha,o + (1 — αt) ln ho + αt ln ko.
Applying the expectations operator with respect to the information set available in period
0 to all At-1-i gives:
E0 [ln πt-1(zt1)] < ρρ-- ln A0 + ɪ (1 - αt)ln h0 + (1 - αt)ln ho
+ ( 1—— + (--α)2 ) (1 - α + γ)ln B + αt ln ko.
Since F(ht, kt, ht+1, kt+1,At; ha,t) ≤ F(ht, kt,0, 0,At; ha,t) holds for the per-period re-
turn function we know that:
F πt1-1(zt-1), πt2-1(zt-1), πt1(zt), πt2(zt), zt
≤ lnAt + γ ln ha,t + (1 — α) lnπt1-1(zt-1) + α ln πt2-1(zt-1) (46)
must also hold. Hence for any pair (ho , ko , Ao) and for any feasible plan π, the sequence
of expected one period returns satisfies:
Eo [F(∙,t)] < PtlnAo + α (ρρ - ) lnAo + (1-- + -1-α)- ) (1 - α + γ)lnB
(1-α+γ-(α-αt+1 ) ln ho + αt+1 ln ko,
where F(∙,t) := F(kt, ht, kt+1, ht+1, At). Then for any feasible plan, the expected total
returns are bounded from above:
n
lim Eo
n→∞
[∑βtF (∙,t)]
t=o
— α ln ko 1 (1-α) ln ho ∣ γ ln ha,o .
β(1-α+γ) ln B +
(1-β)2(1-αβ) +
ln Ao
(1-ρβ)(1-—e)
≤ 1-αβ + (1-αβ)(1-β) + (1-αβ)(1-β) +
This concludes our search for an upper bound of the value function, i.e. we have shown
that the limit in Assumption 2 exists although it may be minus infinity.
We know from Section 4 that:
v(h, k, A) = φ + φk ln k + ψh ln h + ; h ln ha + ψA ln A + ψB ln B (47)
solves the functional equation. The coefficients φi, with i ∈ {k, h, ha, A, B}, were defined
as follows11:
- 1-- γ
^k := 1-αβ, Ψh := (1-β)(1-αβ) , Ψha := (1-β)(1-αβ) ,
_ (1-α+γ)β _ 1
ψB := (1-β)2(1-αβ) , 7A := (1-ρβ)(1-αβ) .
10Note that Y't ∩ sαs = α 1-—α,, — α1 + t holds.
s=0 (1-α)2 1-α .
i αβ ln α
+ (1—β)(1-αβ)
I (1 —αβ+γ)β ln β
+ (1 —β)2(1-αβ) ∙
11The constant in is OTVen bv- in — ln[1 —αβ] + (1 —α) ln[1—β]
The constant ψ Is glven by: ψ :— 1-β + (1-β)(1-αβ)
20
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