Unemployment cycles

Now using the equilibrium wages and substituting all explicit solutions for the values, we obtain the
following complete set of equilibrium Bellman equations:
U = E 1 = E 1 = pb
r
E 2 =
pb
py + δ
r
E 2 ′ = E 2 =
r + δ
E 2 ′ =
pb
py + δ
r
r + δ
⎡ ⎛
V = −c + q(θ(Ω)) ⎣
u(1 − π) p(y − b)
⎝
s r + δ
−
+ uπ ( p(y − b)
pkΩ
−
s r + δ r + δ + λ(Ω)m(θ(Ω))
J 1 =
p(y−y)
p(y − b) pkΩ − λ(Ω)m(θ(Ω))(1 − π)
r+δ
−
r + δ r + δ + λ(Ω)m(θ(Ω))
J 1 =
p(y − b)
pkΩ
−
r + δ r + δ + λ(Ω)m(θ(Ω))
J 2 =
p(y − y)
J 2 ′ = J 2 =
r + δ
J 2 ′ = 0
p(y−y)
pkΩ − λ(Ω)m(θ(Ω))(1 − π)
r+δ
r + δ + λ(Ω)m(θ(Ω))
)
+ λ(Ω)γπ
s
]
p(y − y)
= 0
r + δ
⎞
⎠
Multiple Equilibria. Now we need to verify the two no-deviation conditions for those workers matched
in two types of matches, a match y and y. This implies that we need to check the conditions:
1. No deviation when non one searches: E 1 (0|0) > E 1 (1|0) and E 1 (0|0) > E 1 (1|0)
2. No deviation when all search: E 1 (1|1) > E 1 (0|1) and E 1 (1|1) > E 1 (0|1)
The next proof, adapted from the proof of the baseline model, shows that the condition for multiplicity
is very similar to (but stronger than) the condition from the baseline model, i.e.,
Proof.
(
)
θ(0) < m −1 k(r + δ)
[ ]
λ 1 πy + (1 − π)y − b
( ) k(r + δ)
< m −1 < θ(1).
λ 1 (y − b))
1.1. No deviation in y jobs when no one searches: E 1 (0|0) > E 1 (1|0).
In this case, when no one actively searches on-the-job (Ω = 0), a worker in a low productivity job
deviating during an interval dt chooses ω = 1 and gets a payoff
E 1 (1|0) =
1
1 + rdt [dt(w 1(0) − pk) + (1 − δdt)dtλ(1)m(θ(0)) [(1 − π)E 2 + πE 2 ′] + (1 − δdt)(1 − dtλ(1)m(θ(0)))E 1 (0|0) + δdtU