<p>Consider the following equations:</p>
<p><span class="math-container">$$ \tag{1}
E_t - E_t^* = \gamma \cdot (Y_t - Y_t^*) + \eta_t, \gamma &gt; 0,
$$</span>
<span class="math-container">$$ \tag{2}
U_t - U_t^* = \delta \cdot (E_t - E_t^*) + \mu_t, \delta &lt; 0
$$</span>
where <span class="math-container">$E_t$</span> is the log of employment, <span class="math-container">$Y_t$</span> is the log of output, <span class="math-container">$U_t$</span> is the unemployment rate ,and <span class="math-container">$^*$</span> indicates a long-run level.</p>
<p>We can then derive Okun's Law by substituting <span class="math-container">$(1)$</span> in <span class="math-container">$(2)$</span>:
<span class="math-container">$$ \tag{3}
U_t -U_t^* = \beta \cdot (Y_t - Y_t^*) + \epsilon_t, \beta &lt; 0
$$</span>
where <span class="math-container">$\beta = \gamma \cdot \delta$</span> and <span class="math-container">$\epsilon_t = \mu_t + \delta \cdot \eta_t$</span>.</p>
<p>I now quote the authors:</p>
<blockquote>
<p>Past research provides guidance about the values of the parameters in equations (1)-
(3). To see this, suppose first that changes in output and employment are movements along a
neoclassical production function: more labor produces more output. For the United States,
economists believe that the elasticity of output with respect to labor is about 2/3, based on
factor shares of income. If we invert the production function, we get equation (1) with γ = 3/2
= 1.5.</p>
</blockquote>
<p>How does one go about 'inverting the production function' and getting equation <span class="math-container">$(1)$</span>? I am quite lost and would appreciate any help.</p>

CC BY-SA 4.0
2020-12-14T07:36:08.297Z
Could it be that $y = F(e)$ with functional form assumption $y = e^\alpha$ so inverse is $y^{1/\alpha} = e$ then around some initial point $(y^*,e^*)$ in log form you get (1).
Jesper Hybel
CC BY-SA 4.0
2020-12-14T13:36:58.363Z
Hi. Could you expand a little bit more on the Taylor expansion bit (I'm assuming that's what you meant by 'then around...' ? I've tried to get equation (1) to no avail. Thanks.
Pedro Cunha
CC BY-SA 4.0
2020-12-14T20:38:16.673Z
$\log y = \alpha \log e$ and evaluated at $e^*$ you have $\log y^* = \alpha \log e^*$ so deviation from some equilibrium or steady state $\log y - \log y^* = \alpha (\log e - \log e^*)$.
Jesper Hybel
CC BY-SA 4.0
2020-12-14T21:42:30.473Z
Perfect. Thank you, @JesperHybel
Pedro Cunha
<macroeconomics><production-function><unemployment>
Inverting the production function - Okun's Law: Fit at 50? (BALL; LEIGH; LOUNGANI, 2012)
Pedro Cunha
# Inverting the production function - Okun's Law: Fit at 50? (BALL; LEIGH; LOUNGANI, 2012)

(Asked by: username_0 on 2020-12-14)

Consider the following equations:

$$ \tag{1}
E\_t - E\_t^\* = \gamma \cdot (Y\_t - Y\_t^\*) + \eta\_t, \gamma > 0,
$$
$$ \tag{2}
U\_t - U\_t^\* = \delta \cdot (E\_t - E\_t^\*) + \mu\_t, \delta < 0
$$
where $E\_t$ is the log of employment, $Y\_t$ is the log of output, $U\_t$ is the unemployment rate ,and $^\*$ indicates a long-run level.

We can then derive Okun's Law by substituting $(1)$ in $(2)$:
$$ \tag{3}
U\_t -U\_t^\* = \beta \cdot (Y\_t - Y\_t^\*) + \epsilon\_t, \beta < 0
$$
where $\beta = \gamma \cdot \delta$ and $\epsilon\_t = \mu\_t + \delta \cdot \eta\_t$.

I now quote the authors:

> Past research provides guidance about the values of the parameters in equations (1)-
> (3). To see this, suppose first that changes in output and employment are movements along a
> neoclassical production function: more labor produces more output. For the United States,
> economists believe that the elasticity of output with respect to labor is about 2/3, based on
> factor shares of income. If we invert the production function, we get equation (1) with γ = 3/2
> = 1.5.

How does one go about 'inverting the production function' and getting equation $(1)$? I am quite lost and would appreciate any help.

**Tags:** macroeconomics, production-function, unemployment

> **Comments:**
> **username_1**: Could it be that $y = F(e)$ with functional form assumption $y = e^\alpha$ so inverse is $y^{1/\alpha} = e$ then around some initial point $(y^*,e^*)$ in log form you get (1).  
> **username_0**: Hi. Could you expand a little bit more on the Taylor expansion bit (I'm assuming that's what you meant by 'then around...' ? I've tried to get equation (1) to no avail. Thanks.  
> **username_1**: $\log y = \alpha \log e$ and evaluated at $e^*$ you have $\log y^* = \alpha \log e^*$ so deviation from some equilibrium or steady state $\log y - \log y^* = \alpha (\log e - \log e^*)$.  
> **username_0**: Perfect. Thank you, @username_1  

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## 0 answers

