<p>Let $m$ cobb-douglas. This means $m(u,v) = A u^\alpha v^{1-\alpha}$ for $\alpha &lt; 1$ and $q(\theta) = A \theta^{-\alpha}$.</p>

<p>Then </p>

<p>$$\frac{\theta q'(\theta)}{q(\theta)} = -\alpha &gt; -1$$</p>

<p>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.</p>

CC BY-SA 3.0
2016-02-28T20:00:00.903Z
Got it! Works perfectly for this special case, i.e, $\epsilon > 0$. However, not so clear if $q$ is general.
OGC
CC BY-SA 3.0
2016-02-28T19:44:37.920Z
2016-02-28T19:44:37.920Z
FooBar
<p>Let $L=E+U$ where $L$ is labour force, $E$ is number of employed, and $U$ is unemployed people. </p>

<p>Let $u = \frac{U}{L}$ and $v = \frac{V}{L}$.</p>

<p>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 $v$. </p>

<p>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. </p>

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

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

<p>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. </p>

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

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

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

CC BY-SA 3.0
2016-02-28T18:48:04.273Z
What does the function $q(\cdot)$ represent?
Giskard
CC BY-SA 3.0
2016-02-28T18:59:52.653Z
$q(\theta)$ is the rate at which jobs get filled. It is positive and decreasing in $\theta$.
OGC
CC BY-SA 3.0
2016-02-28T19:02:50.500Z
@denesp So I'm not sure if $\theta>0$ I thought it can be $0$, but that's the extreme case.
OGC
CC BY-SA 3.0
2016-02-28T19:05:34.580Z
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.
Giskard
CC BY-SA 3.0
2016-02-28T19:21:18.097Z
@denesp So $q(\theta) = \frac{m(u,v)}{v}=m(\frac{u}{v},1)=m(\theta^{-1},1)$. Is this enough to show that $\epsilon \geq 0$?
OGC
CC BY-SA 3.0
2016-02-28T21:36:41.137Z
Afraid not. You need the specific form of either $q$ or $m$ or some additional properties.
Giskard
<macroeconomics><labor-economics><search-and-matching>
Showing that the elasticity of arrival rate for workers wrt $\theta$ is between $0$ and $1$
OGC
# Showing that the elasticity of arrival rate for workers wrt $\theta$ is between $0$ and $1$

(Asked by: username_0 on 2016-02-28)

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 $1$?

**Tags:** macroeconomics, labor-economics, search-and-matching

> **Comments:**
> **username_1**: What does the function $q(\cdot)$ represent?  
> **username_0**: $q(\theta)$ is the rate at which jobs get filled. It is positive and decreasing in $\theta$.  
> **username_0**: @denesp So I'm not sure if $\theta>0$ I thought it can be $0$, but that's the extreme case.  
> **username_1**: 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.  
> **username_0**: @denesp So $q(\theta) = \frac{m(u,v)}{v}=m(\frac{u}{v},1)=m(\theta^{-1},1)$. Is this enough to show that $\epsilon \geq 0$?  
> **username_1**: Afraid not. You need the specific form of either $q$ or $m$ or some additional properties.  

---

## 1 answers

### Answer by username_2 on 2016-02-28. Score: 3 | Accepted answer

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.

> **Comments:**
> **username_0**: Got it! Works perfectly for this special case, i.e, $\epsilon > 0$. However, not so clear if $q$ is general.  

---

