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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?

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From

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

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 $b$.

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

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