# DMP: Two Labor Markets

Consider the following extension of the standard DMP model, in which we have two labor markets with different returns, indexed $i$. The markets are completely separated and firms and workers can only go to one market. It might help to think of them as different countries, for example.

I am ultimately interested in the different waiting time of unemployed to find a job.

As usual, we match using a CRS function $M(u,v)$, such that the likelihoods of finding a job and finding a worker are given by $f(\theta) = M(u,v)/u = M(1, \theta)$, and $q(\theta) = M(u, v)/v = M(\theta^{-1}, 1)$.

The vacancy-value equation is

$$\rho V_i = -c q(\theta_i) (J_i - V_i)$$

Free entry implies $V_i = 0$. Manipulating the equations we get

$$\frac{q(\theta_1)}{q(\theta_0)} = \frac{J_0}{J_1}$$

which makes sense - the job-filling-rates are inverse to the value of the jobs. Say $J_1 > J_0$. Then, market $1$ is tighter, leading to a lower job-filling-rate for an individual firm.

Using the earlier definitions, I can use that $f(\theta) = q(\theta)\theta$:

$$\frac{f(\theta_1)}{f(\theta_0)} = \frac{J_0}{J_1}\frac{\theta_1}{\theta_0}$$

Let $W_i$ denote the average waiting time of an unemployed in market $i$. As we're using Poisson rates, $W_i = 1/f(\theta_i)$:

$$\frac{W_0}{W_1} = \frac{J_0}{J_1}\frac{\theta_1}{\theta_0}$$

Which doesn't make sense anymore. Consider again $J_1 > J_0$: A worker in market $1$ is much more valuable to firms than a worker in market $0$. However, the relative waiting time for workers in market $1$ increases in the relative value of workers in market $1$.

I don't understand this. Was my derivation wrong? Or am I not allowed to to the ceteris paribus argument while ignoring $\theta_1/\theta_0$ on the right-hand-side?

• On the fly, maybe this explains why middle-level executives find it hard to find a job after leaving one (i.e. they have to wait longer compared to non-managerial employees)? (I am only half-joking, I will return here) Apr 21, 2015 at 16:00

Define $q^{-1}$ be the inverse of $q$. Then,
$$\theta = q^{-1} \left( \frac{c}{J} \right)$$
$$\frac{W_1}{W_0} = \frac{f(\theta_0)}{f(\theta_1)} = \frac{f \left( q^{-1} \left(\frac{c}{J_1}\right)\right)}{f \left( q^{-1} \left(\frac{c}{J_0}\right)\right)}$$
$J_1 > J_0$, hence $\theta_1 > \theta_0$ and $W_1 < W_0$.