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Let $L=E+U$ where $L$ is labour force, $E$ is number of employed, and $U$ is unemployed people.

Let $u = \frac{U}{L}$ and $v = \frac{V}{L}$.

Given $m(u,v)$ as a matching function that determines the flow of matches between workers and firms, and assume $m$ has constant return to scale (CRS) in the two arguments $u$ and $v$.

The elasticity of the arrival rate for workers is given by $$\theta q(\theta),$$ where $\theta = \frac{v}{u}$ represents "tightness" in the labour market.

Elasticity ($\epsilon$) = $$\frac{\mathrm{d} f(x)}{\mathrm{d} x} \frac{x}{f(x)} = \frac{\partial m}{\partial \theta} \frac{\theta}{m(1,\theta)},$$

where $\theta q(\theta) = m(\theta^{-1},1) = m(1,\theta)$ (by using CRS property of $m$ in the two arguments $u$,$v$).

The rate at which unemployed workers find job: $$\theta q(\theta) = m(\theta^{-1},1)=m(1,\theta),$$ since the matching function, $m$, is constant returns to scale in the each argument.

Since $\theta q(\theta) = m(1,\theta)$, $$\frac{\partial m(1,\theta)}{\partial \theta} = q(\theta) + \theta q'(\theta).$$

So the elasticity of the arrival rate for workers, $$\epsilon = \frac{\partial m}{\partial \theta} \frac{\theta}{m} = \left [q(\theta) + \theta q'(\theta) \right ] \frac{\theta}{\theta q(\theta)} = 1 + \frac{\theta q'(\theta)}{q(\theta)}$$.

Since $q'(\theta)<0$, $\epsilon<1$, but how can I formally show that it is between $0$ and $1$?

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  • $\begingroup$ What does the function $q(\cdot)$ represent? $\endgroup$ – Giskard Feb 28 '16 at 18:48
  • $\begingroup$ $q(\theta)$ is the rate at which jobs get filled. It is positive and decreasing in $\theta$. $\endgroup$ – OGC Feb 28 '16 at 18:59
  • $\begingroup$ @denesp So I'm not sure if $\theta>0$ I thought it can be $0$, but that's the extreme case. $\endgroup$ – OGC Feb 28 '16 at 19:02
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    $\begingroup$ That is not a problem. Given $\theta \geq 0$ and $q'(\theta) < 0$ even if $\theta = 0$ you have $$\epsilon = 1 + \frac{\theta q'(\theta)}{q(\theta)} \leq 1.$$ Basically you want to show $$\epsilon = 1 + \frac{\theta q'(\theta)}{q(\theta)} \geq 0.$$ For this I think you will need more information about $q$ as it does not hold for all functions. $\endgroup$ – Giskard Feb 28 '16 at 19:05
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    $\begingroup$ Afraid not. You need the specific form of either $q$ or $m$ or some additional properties. $\endgroup$ – Giskard Feb 28 '16 at 21:36
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Let $m$ cobb-douglas. This means $m(u,v) = A u^\alpha v^{1-\alpha}$ for $\alpha < 1$ and $q(\theta) = A \theta^{-\alpha}$.

Then

$$\frac{\theta q'(\theta)}{q(\theta)} = -\alpha > -1$$

To what extent this holds for other functional forms I'm not sure, but I haven't seen anything but Cobb-Douglas being used as the matching function.

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    $\begingroup$ Got it! Works perfectly for this special case, i.e, $\epsilon > 0$. However, not so clear if $q$ is general. $\endgroup$ – OGC Feb 28 '16 at 20:00

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