Consider the following one-shot version of a labour market matching model. Let the labour force be normalized at 1, who, because there is only one period, all start out as unemployed. There is a very large number of firms who can enter the market and search for a worker. Firms who engage in search first have to pay a fixed cost, $k$. If a measure $v$ of firms enters the labour market, a constant returns to scale matching function $m(1,v)$ gives us the total measure of matches in the economy.

Within each match, the firm and the worker bargain for the wage, $w$, so that the workers get a constant proportion of $y$. Denote this proportion by $\beta$, which is interpreted as the bargaining power of the worker. Assume $\frac{k}{y} < 1 - \beta$ for the firm.

Define market tightness as $ b \equiv \frac{1}{v}$ and assume that the arrival rate for a firm is given by: $a_{F} = 1 - e^{-b}$.

Consider the firms can enter the labour market freely if they pay the entry cost, then what is the equilibrium value of $b$? Describe it graphically? Does it always exist? Is it unique?

My solution: The value of a vacancy: $$V = -k + a_{F}(b)(J-V),$$ and the value of a filled job: $$J = y-w.$$ If the firms enter the labour market freely then $V=0$. Then from these two equations I am left with this equation $$ 1 - e^{-b} = a_{F}(b) = \frac{k}{y (1-\beta)}.$$

Graphically, the function looks something like this:

enter image description here

From the graph, it looks like that $b^*$ is unique, but how do I know if it always exists or not?



$$ 1 - e^{-b} = a_{F}(b) = \frac{k}{y (1-\beta)}$$

we get

$$b = -\ln\left(1-\frac{k}{y (1-\beta)}\right) \tag{1}$$

For this to exist (be a real number) it must be the case that

$$1-\frac{k}{y (1-\beta)} > 0 \implies \frac k y < (1-\beta) \tag{2}$$

which is already assumed.

Then, uniqueness of the solution of $(1)$ is guaranteed because all arguments/parameters in the right-hand side enter only once. So for any vector $(y,k,\beta)$ satisfying what it needs to satisfy, we get a unique $b$.

  • $\begingroup$ Wow. I missed that simple assumption for the existence part. $\endgroup$
    – OGC
    Feb 29 '16 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.