Appendix
Appendix: solution of the model with contemporaneous feedback
The Lagrangian of the problem is the following:
t = ^ β t Il - - [(1 - q ) Dt - Ft I NW ] I--ɪtɪ I -rB - rD I EL }[(1 - q ) Dt - Ft + nw ] +
t=0
- 2v [(1 - q)Dt - Ft + NW]2 - uDt - z [(1 - q)Dt - Ft + NW] + VtFt + PtqDt +
-μt{Dt - γγDt-1 - к [(1 - q)Dt - Ft + NW] - g3rD + g4rB }}. (55)
The first order conditions are:
——— = βt+jIlV+~)∣^(1-q)Dt+j-Ft+j + nw 1--( t+j -r-rj j + rD+ j+z-eL+ j + Vt+j-κμt+j} = 0,
UFt + j к ∖ O b L J O O J
(56)
∂t
U
= βt+jEt+j [(1 - q) [ ( O+' + brB+ j - rD+ j + eL+ j 1 - ∣v + O)[(1 - q)2Dt+j +
- (1 - q) Ft+j + (1 - q) NW] - u - (1 - q) z + ρt+j q +
- [l - к(1 - qH μt+j + βγγe[μt+j+1]j = 0,
∂t
∂μt+j
-Dt+j + YYDt+j-1 + kE [(1 - q)Dt+j - Ft+j + nw 1 + g3rD+ j - g4rB+ j = 0.
(57)
(58)
For j = T, Condition (57) implies:
βt+τe|(1 - q) [ ( t+ t') + brB+ τ - rD+ τ + eL+ T1 - ∣v + O)[(1 - q)2Dt+T +
-(1 - q)Ft+T + (1 - q)NW - u - (1 - q)z + ρt+τq - μt+τ f = 0.
(59)
The transversality condition of the problem is:
τlim βt+τEt ∣(1 - q)ωt+τ - a [(1 - q)2Dt+τ - (1 - q)Ft+τ + (1 - q)NW] - u - (1 - q)z +
+Pt+τq} - κ I - ωt+τ + v [(1 - q)Dt+τ - Ft+τ + NW1 + Vt+τ }} = 0.
(60)
To keep the notation simpler, we omit from now on the index j , to show it only when it will be
necessary, in the final solutions.
Besides we have used the definitions
bβ+b1') = a and a b' +
drB - rD + EL = ωt. From Equation (56) we can obtain an equation that can be solved in order
to get a solution for the multiplier μt.
μt = 11 a [(1 - q)Dt - Ft + NW1 - ωt + z + Vt}.
(61)
This value can be substituted in the other first order condition, shown in equation (57), that
provides the Euler equation of the system:
βt IPtq +(1 - q)(ωt - z) - u - a [(1 - q)2Dt - (1 - q)Ft + (1 - q)NW] +
----κ(K—q) I a [(1 - q ) Dt - Ft + NWt 1 - ( ωt - z ) + Vt} +
+ γγβEt I a [(1 - q)Dt +1
κ I L
- Ft +1 + NW^ - ( ωt+1 - z ) + Vt+ι∣∣
= 0.
(62)
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