Notes on Unemployment Dynamics
Jorge F. Chavez
November 20, 2014
Let Lt denote the size of the labor force at time t. This number of workers can be
divided between two mutually exclusive subsets: the number of employed workers at time
t (denoted Et and the number of unemployed workers also at time t denoted Ut :
Lt = Ut + Et
Let s and f denote the job separation rate and the job finding rate, correspondingly.
The assumption now is that each period, the new members of the labor force will find a
job with probability f , or which is equivalent, a proportion f of the additional workers will
become part of Et .
Our objective is to describe how unemployment evolves in an economy, given certain
assumptions, and in particular we are interested in knowing the behavior of this variable
in the long-run. It is assumed (later we can prove this) that in the long-run the economy
will reach a steady-state (an equilibrium) and as such the unemployment rate will be equal
to its natural level, which is constant. To start the analysis we first need to make use
of an assumption regarding the behaviour of Lt . To simplify the discussion it is usually
assumed that Lt is fixed, so that the size of the labor force does not change (e.g. there is
no population growth). Later I’ll relax this assumption to see how the analysis change.
1
Unemployment dynamics without population growth
For now suppose that the size of the labor force t is constant and equal to L for all t.
This assumption simplifies the analysis greatly as will become clear shortly. When L is
constant, it is indifferent whether we look at the evolution of Ut or at the evolution of µt : if
µt becomes constant in the long-run, given that L is fixed all the time, Ut must necessarily
be also constant (see why?).
We can start writing an equation that describes how Ut evolves over time as follows:
Ut = Ut−1 + sEt−1 − f Ut−1
This equation can be read as follows. The number (stock) of unemployed individuals today
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(at time t) will be equal to the number of individuals that yesterday (at time t − 1) were
unemployed plus those that were employed at t − 1 but lose their jobs (sEt−1 ) minus those
that being unemployed at time t − 1 were able to find a job (and so today are part of Et ).
In principle we are interested in describing the evolution of the unemployment rate and
not really the number of unemployed individuals. However when there is no population
growth, it is irrelevant if we analyze the evolution of Ut or the evolution of µt . The
reason relies precisely on the assumption that L is constant: because the natural rate
of unemployment is constant in steady-state (i.e. in the long-run), then the number of
unemployed individuals will also be constant.
Replacing Et−1 with L − Ut−1 , and rearranging terms a little bit we get a difference
equation or law of motion for Ut :
Ut = sL + (1 − s − f ) Ut−1
(1)
Dividing the RHS and the LHS of (1) of the above equation by L we can get he law of
motion for µt :
µt = s + (1 − s − f ) µt−1
Both equations govern how Ut and µt evolve over time. Recall that a law of motion must
be satisfied all the time, including in steady state. Then in steady-state:
µ = s + (1 − s − f ) µ
s
µ =
s+f
1.1
Why this works?
In general, the way to solve a difference equation such as (1) is to find the path over time
of the variable of interest, given parameters and a known initial value.1 In this case, our
variable of interest is Ut and so our objective in solving the difference equation (1) is to
find Ut as function of t and known parameters including the known initial value which we
will denote U0 .
There are several methods to solve difference equations. One way is called brute-force
method and involves replacing the lagged difference equation several times until we find a
pattern. That is, we know the difference equation (1) for Ut as a function of Ut−1 , and so
we also know the equation of Ut−1 as a function of Ut−2 . Then, first :
Ut = sL + (1 − s − f ) [sL + (1 − s − f ) Ut−2 ]
= sL + (1 − s − f ) sL + (1 − s − f )2 Ut−2
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This is not possible in all cases, specially when the equation is non-linear. Different techniques are
applied in those situations.
2
We can then also replace Ut−2 as a function of Ut−1 :
Ut = sL + (1 − s − f ) sL + (1 − s − f )2 [sL + (1 − s − f ) Ut−3 ]
= sL + (1 − s − f ) sL + (1 − s − f )2 sL + (1 − s − f )3 Ut−3
Now you should see a pattern (if you don’t do this once more, so replace Ut−3 as a function
of Ut−2 ). Now think what would happend if we do this t times:
Ut = sL + (1 − s − f ) sL + (1 − s − f )2 sL + . . . (1 − s − f )t sL + (1 − s − f )t+1 U0
Using the formulas for the sum of a geometric progression, we can write:2
(
)
1 − (1 − s − f )t+1
Ut = sL
+ (1 − s − f )t+1 U0
1 − (1 − s − f )
Now we are ready to see what happens in the long-run:
(
)
1 − (1 − s − f )t+1
lim Ut = sL lim
+ lim (1 − s − f )t+1 U0
t→∞
t→∞
t→∞
1 − (1 − s − f )
|
{z
}
{z
}
|
=0
(
=
s
s+f
)
(2)
1
= s+f
(3)
L=U
As we can see from the last equality, both the number of unemployed individuals and the
unemployment rate in the long-run will both become constant.
2
Unemployment dynamics with population growth
Now we will allow entrants to this labor force (i.e. people that become the right age to look
for a job, or people that just finished their studies and therefore will start looking for a
job). Suppose that Lt grows at a constant positive rate n ∈ (0, 1), that is Lt+1 = (1 + n)Lt
for all t. As before Lt is composed by those that have a job at time t, denoted Et , plus
those that are unemployed Ut in that period.
The assumption now is that each period, the new members of the labor force will find
a job with probability f , or which is equivalent, a proportion f of the additional workers
2
Suppose we have the following sum of a geometric progression
1 + a + a2 + a3 + . . . + ak = M
Then
(
)
1 + a 1 + a + a2 + a3 + . . . + ak−1 = M
(
)
and hence 1 + a M − ak = M , which finally gives us: M =
3
1−ak+1
1−a
will become part of Et :
Et+1 = Et + f Ut − sEt + f ∆Lt+1
(4)
We can get an analogous law of motion for Ut :
Ut+1 = Ut + sEt − f Ut + (1 − f ) ∆Lt+1
(5)
where we are taking into account that a proportion (1 − f ) of the new workers are unable
to find a job.
Take (5) as base, divide it by Lt and define the unemployment rate as µt ≡ Ut /Lt :
Ut+1
= µt + s (1 − µt ) − f µt + (1 − f ) n
Lt
(6)
Then divide and multiply by both sides of (6) by Lt+1 to get µt+1 :
Ut+1 Lt+1
= µt+1 (1 + n) = µt + s (1 − µt ) − f µt + (1 − f ) n
Lt+1 Lt
If we had chose to work with (4) instead, the equation we would reach would be:
(1 − µt+1 ) (1 + n) = 1 − µt + f µt − s (1 − µt ) + f n
Finally, in steady-state µt = µt+1 = µ, therefore applying this condition we get:
µ=
s + (1 − f ) n
s+f +n
Note than in the special case in which n = 0 we are back in the case in which there was
not population growth and therefore the expression for the unemployment rate in steady
s
state is again µ = s+f
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A very general exercise
Consider the simplest model of unemployment dynamics. Let Ut denote the number of
unemployed workers at time t, and let s and f denote the job-separation rate and the job
finding rate in this economy. Assume initially that the total size of labor force is constant
(Lt−1 = Lt = Lt+1 = L, all t).
(a) Interpret s and f in terms of probabilities.
Answer. The job separation rate s is the probability that a worker will loose its job. The job
finding rate f is the probability that a worker that is actively seeking for a job finds one.
(b) Let µt denote the unemployment rate at time t. Show that the unemployment rate in
steady-state will be a function of s and f only.
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Answer. Start from:
Ut+1 = Ut + sEt − f Ut
Replace Et = L − Ut and divide both sides by L to put everything in terms of unemployment
rates. Rearranging we get:
µt+1 = s + (1 − s − f ) µt
In steady-state µ̄ = µt+1 = µt . Hence, solving for µ̄:
µ̄ =
s
s+f
(c) Unemployment tends to evolve over time before it will reach a steady state. Express
the change in the number of unemployed persons as a function of s, f , and U , and
show that if unemployment is above its steady state level, unemployment falls (and
conversely). Answer. First express the change in the number of employed persons as a
function of known parameters. Start from:
Ut+1 = Ut + sEt − f Ut
replacing Et = L − Ut we get:
∆Ut+1 ≡ Ut+1 − Ut = s (L − Ut ) − f Ut = sL − (s + f ) Ut
Suppose that we start from a situation in which Ut > Ū . Without loss of generality we can
assume that [Ut = Ū + m. Then, plugging this value in the expression for ∆Ut+1 :
∆Ut+1
(
)
= sL − (s + f ) Ū + m
= sL − (s + f ) Ū − (s + f ) m < 0
{z
}
|
=0
where the term that vanishes comes from the steady-state condition.
(d) Suppose s = 0.03 (or 3 percent) and f = 0.7 (or 70 percent). Initially (at time t = 0),
the unemployment rate is equal to µ0 = 0.08. Further assume that at time t = 200
(in the long run) this economy will experience a permanent shock on f such that from
period t = 200 onwards, f = 0.4. Sketch the evolution of the unemployment rate over
time in a simple graph (Hint: consider how would the path of µt would look in a graph
with time in the horizontal axis, where t goes from 0 to 1000).
Answer. With s = 0.03 and f = 0.7 the unemployment rate in steady-state will be 0.041.
Hence, at time t = 0 we are above the steady-state and from our knowledge of how µt evolves
over time we know that we will progressively tend towards the steady-state from above. This
might not take too long, so by the time we are at t = 199 the economy is already in steadystate.
Then, the permanent shock on f translates into an increase of the steady-state rate of unemployment to µ = 0.697 ≈ 0.07. This means that right after the shock, the economy is below the
steady-state, and therefore the evolution of µt will be such that we will progressively converge
to the new steady-state from below. See figure 1.
(e) In macroeconomics it is always useful to express a dynamic variable (say xt ) in terms
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Figure 1: Path of µt
µt
0.08
0.07
0.041
t=0
t = 200
t
of deviations from its value in steady state: x̃t ≡ xt − x̄. Show that the deviation of the
unemployment rate at t + 1 with respect to its steady-state value can be expressed as a
linear function of the deviation of the unemployment rate at time t, that is µ̃t+1 = γ µ̃t .
Hint: Try to express the γ parameter in terms of s and f .
Answer. It is just a matter of playing with the expression we got above as follows:
µt+1 − µ̄ = s + (1 − s − f ) (µt − µ̄) + (1 − s − f ) µt − µ̄
Note that besides putting µ̄ on both sides of the equality, I am adding and subtracting (1 −
s − f )µ̄ from the RHS. Rearranging appropriately we get:
µt+1 − µ̄ = (1 − s − f ) (µt − µ̄)
Note that s − (s + f )µ̄ = 0 because of the formula for the natural rate of unemployment.
Hence γ ≡ (1 − s − f ) < 1. Making sure that γ < 1 is very important for the stability of the
path followed by µt , can you see why?.
(Optional) Now suppose that the labor force grows at a rate n: Lt+1 = (1 + n)Lt all t.
(f) Show that the unemployment rate in steady-state will be a function of s, f and n only.
Answer. This is what section 2 of this handout was about!
(g) (Harder) Show that we can also write µ̃t+1 = ξ µ̃t . What is the relationship between ξ
and γ?
Answer. Follow similar steps as for the case of no population growth. At the end you will find
that both parameters are exactly the same.
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