Unemployment cycles

λ(1) − λ(0) = λ 1 :
(y − b)λ 1 m(θ(0)) − k(r + δ) + λ 1 m(θ(0)) [ π(y − y) ] < 0
or
(
)
θ(0) < m −1 k(r + δ)
[ ] .
λ 1 πy + (1 − π)y − b
2.1. No deviation in y job when all search: E 1 (1|1) > E 1 (0|1).
In this case, when all actively search on-the-job (Ω = 1), a worker in a low productivity job deviating
for an interval dt chooses ω = 0 and gets a payoff
E 1 (0|1) =
1 [
dtw1 (1) + dtλ(0)(1 − δdt)m(θ(1)) [ ]
(1 − π)E
1 + rdt
2 + πE 2 ′ + (1 − δdt)(1 − dtλ(0)m(θ(1)))E1 (1|1) + δdtU ] .
There is no deviation provided E 1 (1|1) > E 1 (0|1):
E 1 (1|1)(1 + rdt) > dtw 1 (1) + dtλ(0)(1 − δdt)m(θ(1)) [ ]
(1 − π)E 2 + πE 2 ′
+(1 − δdt − dtλ(0)m(θ(1)) + dt 2 δλ(0)m(θ(1)))E 1 (1|1) + δdtU.
After subtracting E 1 (1|1) from both sides and dividing by dt and taking the limit dt → 0, we obtain:
rE 1 (1|1) > w 1 (1) + λ(0)m(θ(1)) [ (1 − π)E 2 + πE 2 ′]
+ (−δ − λ(0)m(θ(1)))E1 (1|1) + δU.
Substituting the equilibrium values for E 1 (1|1), E 2 , E 2 ′, U and w 1 (1) we get:
(y − b)[λ(1) − λ(0)]m(θ(1)) − k(r + δ) > 0.
So there is no deviation provided that:
( ) k(r + δ)
θ(1) > m −1 λ 1 (y − b))
2.2. No deviation in y job when all search: E 1 (1|1) > E 1 (0|1).
In this case, when all actively search on-the-job (Ω = 1), a worker in a high productivity job
deviating for an interval dt chooses ω = 0 and gets a payoff
E 1 (0|1) =
1 [
dtw1 (1) + dtλ(0)(1 − δdt)m(θ(1)) [ ]
(1 − π)E 2 + πE 2 ′ + (1 − δdt)(1 − dtλ(0)m(θ(1)))E1 (1|1) + δdtU ] .
1 + rdt