Notes on Unemployment Dynamics Jorge F. Chavez November 20, 2014 Let L t 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 E t and the number of unemployed workers also at time t denoted U t : L t = U t + E t 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 E t. 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 L t. To simplify the discussion it is usually assumed that L t 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 U t or at the evolution of µ t : if µ t becomes constant in the long-run, given that L is fixed all the time, U t must necessarily be also constant (see why?). We can start writing an equation that describes how U t evolves over time as follows: U t = U t 1 + se t 1 fu t 1 This equation can be read as follows. The number (stock) of unemployed individuals today 1
(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 (se t 1 ) minus those that being unemployed at time t 1 were able to find a job (and so today are part of E t ). 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 U t or the evolution of µ t. 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 E t 1 with L U t 1, and rearranging terms a little bit we get a difference equation or law of motion for U t : U t = sl + (1 s f) U t 1 (1) Dividing the RHS and the LHS of (1) of the above equation by L we can get he law of motion for µ t : µ t = s + (1 s f) µ t 1 Both equations govern how U t and µ t 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 U t and so our objective in solving the difference equation (1) is to find U t as function of t and known parameters including the known initial value which we will denote U 0. 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 U t as a function of U t 1, and so we also know the equation of U t 1 as a function of U t 2. Then, first : U t = sl + (1 s f) [sl + (1 s f) U t 2 ] = sl + (1 s f) sl + (1 s f) 2 U t 2 1 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 U t 2 as a function of U t 1 : U t = sl + (1 s f) sl + (1 s f) 2 [sl + (1 s f) U t 3 ] = sl + (1 s f) sl + (1 s f) 2 sl + (1 s f) 3 U t 3 Now you should see a pattern (if you don t do this once more, so replace U t 3 as a function of U t 2 ). Now think what would happend if we do this t times: U t = sl + (1 s f) sl + (1 s f) 2 sl +... (1 s f) t sl + (1 s f) t+1 U 0 Using the formulas for the sum of a geometric progression, we can write: 2 U t = sl ( ) 1 (1 s f) t+1 + (1 s f) t+1 U 0 1 (1 s f) Now we are ready to see what happens in the long-run: ( ) lim U 1 (1 s f) t+1 t = sl lim t t 1 (1 s f) = }{{} = 1 s+f + lim t (1 s f) t+1 U 0 }{{} =0 ( ) s L = U (3) s + f 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) 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 L t grows at a constant positive rate n (0, 1), that is L t+1 = (1 + n)l t for all t. As before L t is composed by those that have a job at time t, denoted E t, plus those that are unemployed U t 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 + a 2 + a 3 +... + a k = M Then ( 1 + a 1 + a + a 2 + a 3 +... + a k 1) = M and hence 1 + a ( M a k) = M, which finally gives us: M = 1 ak+1 1 a 3
will become part of E t : E t+1 = E t + fu t se t + f L t+1 (4) We can get an analogous law of motion for U t : U t+1 = U t + se t fu t + (1 f) L t+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 L t and define the unemployment rate as µ t U t /L t : U t+1 L t = µ t + s (1 µ t ) fµ t + (1 f) n (6) Then divide and multiply by both sides of (6) by L t+1 to get µ t+1 : U t+1 L t+1 L t+1 L t = µ t+1 (1 + n) = µ t + s (1 µ t ) fµ t + (1 f) n If we had chose to work with (4) instead, the equation we would reach would be: (1 µ t+1 ) (1 + n) = 1 µ t + fµ t s (1 µ t ) + fn Finally, in steady-state µ t = µ t+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 state is again µ = s s+f 3 A very general exercise Consider the simplest model of unemployment dynamics. Let U t 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 (L t 1 = L t = L t+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 µ t denote the unemployment rate at time t. Show that the unemployment rate in steady-state will be a function of s and f only. 4
Answer. Start from: U t+1 = U t + se t fu t Replace E t = L U t and divide both sides by L to put everything in terms of unemployment rates. Rearranging we get: µ t+1 = s + (1 s f) µ t In steady-state µ = µ t+1 = µ t. Hence, solving for µ: µ = 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: replacing E t = L U t we get: U t+1 = U t + se t fu t U t+1 U t+1 U t = s (L U t ) fu t = sl (s + f) U t Suppose that we start from a situation in which U t > Ū. Without loss of generality we can assume that [U t = Ū + m. Then, plugging this value in the expression for U t+1: U t+1 = sl (s + f) ( Ū + m ) = sl (s + f) }{{ Ū (s + f) m < 0 } =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 µ 0 = 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 µ t 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 µ t 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 µ t 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 x t ) in terms 5
Figure 1: Path of µ t µ t 0.08 0.07 0.041 t = 0 t = 200 t of deviations from its value in steady state: x t x t x. 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 µ t+1 = γ µ t. 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: µ t+1 µ = s + (1 s f) (µ t µ) + (1 s f) µ t µ Note that besides putting µ on both sides of the equality, I am adding and subtracting (1 s f) µ from the RHS. Rearranging appropriately we get: µ t+1 µ = (1 s f) (µ t µ) Note that s (s + f) µ = 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 µ t, can you see why?. (Optional) Now suppose that the labor force grows at a rate n: L t+1 = (1 + n)l t 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 µ t+1 = ξ µ t. 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. 6