<p>The production function is:
<span class="math-container">$$q = (l^\rho + k^\rho)^\frac{1}{\rho}$$</span>
The MPL and MPK are respectively:
<span class="math-container">$$q_l = \frac{\partial q}{\partial l} = \frac{1}{\rho} \cdot (l^\rho + k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot l^{\rho-1}$$</span>
<span class="math-container">$$q_k = \frac{\partial q}{\partial k} = \frac{1}{\rho} \cdot (l^\rho 
+ k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot k^{\rho-1}$$</span>
What is the rate that <span class="math-container">$l$</span> can be substituted for <span class="math-container">$k$</span>?</p>
<p>Where <span class="math-container">$f$</span> is a differentiable real-valued function of a single variable, we define the elasticity of <span class="math-container">$f(x)$</span> with respect to <span class="math-container">$x$</span> (at the point <span class="math-container">$x$</span>) to be
<span class="math-container">$$\sigma(x) =  \frac{x f'(x)}{f(x)}\equiv \frac{\frac{df(x)}{f(x)}}{\frac{dx}{x}}$$</span></p>
<ol>
<li>Do a change of variables such that <span class="math-container">$u = ln(x)$</span> (<span class="math-container">$\rightarrow x = e^u$</span>) and <span class="math-container">$v=ln(f(x))$</span> (<span class="math-container">$\rightarrow f(x) = e^v$</span>)</li>
<li>Note that <span class="math-container">$v' = f'(x) / f(x)$</span> and <span class="math-container">$u'=\frac{1}{x}$</span> so that
<span class="math-container">$$\frac{v'}{u'}=\frac{\frac{f'(x)}{f(x)}}{\frac{1}{x}} = \sigma(x)$$</span></li>
<li>Note that this is also the result you get by solving for <span class="math-container">$ \frac{d ln f(x)}{d ln(x)}$</span> because  <span class="math-container">$ \frac{d ln f(x)}{d ln(x)} = \frac{d v}{d u}$</span> which we solve via the chain rule:
<span class="math-container">$$ \frac{d v}{d u} = \frac{d v}{d x} \cdot \frac{d x}{d u} = \frac{f'(x)}{f(x)} \cdot x $$</span>
which happens to be exactly the definition of <span class="math-container">$\sigma(x)$</span>.</li>
</ol>
<p>Now let's tackle your elasticity problem.</p>
<p><span class="math-container">$$ ln(\frac{q_k}{q_l})= log(\frac{\frac{1}{\rho} \cdot (l^\rho + k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot l^{\rho-1}}{\frac{1}{\rho} \cdot (l^\rho 
+ k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot k^{\rho-1}}) = ln (\frac{l}{k})^{\rho-1} = (\rho-1) ln (l/k) = (1 - \rho) ln (k/l)$$</span>
<span class="math-container">$$ \Rightarrow ln (k/l) = \frac{1}{1-\rho} \cdot ln(\frac{q_k}{q_l})$$</span></p>
<p>So <span class="math-container">$\sigma = \frac{1}{1-\rho}$</span></p>

CC BY-SA 4.0
2018-05-11T15:28:44.787Z
$\frac{1}{\rho}$ and  $\rho$ can be reduced from derivatives MPL and MPK to simplify the exposition.
garej
CC BY-SA 4.0
2016-10-13T21:01:10.383Z
2024-07-23T21:46:39.730Z
2024-07-23T21:46:39.730Z
BKay
<p>I would like to add a little bit to the answer above. I wrote a comment earlier, but I thought it'd be helpful to flesh out the argument a little more.</p>

<p>We have a firm that uses two factors of production, labor <span class="math-container">$l$</span> and capital <span class="math-container">$k$</span>, to produce output. Quantity of output is written <span class="math-container">$q$</span>.</p>

<p>The elasticity of a function of a <em>single variable</em> measures the percentage response of a dependent variable to a percentage change in the independent variable.</p>

<p>On the other hand, the elasticity of substitution between two factor inputs measures the percentage response of the ratio of their quantities to a percentage change in the relative marginal products.</p>

<p>Pertaining to the above, we have that the elasticity is given by</p>

<p><span class="math-container">\begin{align}
\sigma \equiv \frac{d \ln{\left( k/l \right)}}{d\ln{ \left( MPL/MPK \right)}} 
\end{align}</span></p>

<p>where <span class="math-container">$MPL$</span> is the marginal product of labor and <span class="math-container">$MPK$</span> is the marginal product of capital.</p>

<p>The reason I am writing this out is that there is a small error in the above answer. In the equation right after "Now let's tackle your elasticity problem," <span class="math-container">$\ln{\frac{q_k}{q_l}}$</span> is immediately followed by an expression for <span class="math-container">$\ln{\frac{q_l}{q_k}}$</span> switching the numerator with the denominator. </p>

<p>If you correct this, you get that <span class="math-container">$\sigma = -\frac{1}{1-\rho}$</span>, which is close but not quite correct. To get the correct answer, you follow otherwise exactly the same calculations given by the above answer to get</p>

<p><span class="math-container">$$ \ln{k/l} = \frac{1}{1-\rho}\cdot \ln {\frac{q_l}{q_k}}$$</span></p>

<p>to get that <span class="math-container">$\sigma = \frac{1}{1-\rho}$</span>, where the corrections are for reasons outlined above. </p>

CC BY-SA 4.0
2020-05-20T14:48:51.517Z
2020-05-20T14:48:51.517Z
mathsquestions1
<p>I have to prove that $\sigma = 1/(1 + \rho)$ for the CES production function:</p>

<p>\begin{align}
q = (l^\rho + k^\rho)^\frac{1}{\rho}
\end{align}</p>

<p>I found out that I need to solve the following equation:
\begin{align}
\sigma = \frac{\frac{d(k/l)}{k/l}}{\frac{dRTS}{RTS}} = \frac{d(k/l)}{dRTS}\frac{RTS}{k/l} = \frac{d(k/l)}{d((k/l)^{1-\rho})}\frac{(k/l)^{1-\rho}}{k/l}
\end{align}</p>

<p>But i just don't know how to rewrite this expression to $\sigma = 1/(1 + \rho)$</p>

CC BY-SA 3.0
2016-04-11T13:56:14.493Z
Check the example for Cobb Douglas production and try to solve it for CES. https://en.wikipedia.org/wiki/Elasticity_of_substitution
clueless
<production-function>
CES: Production function: Elasticity of substitution $\sigma = 1/(1 + \rho)$
frankfranfrank
# CES: Production function: Elasticity of substitution $\sigma = 1/(1 + \rho)$

(Asked by: username_0 on 2016-04-05)

I have to prove that $\sigma = 1/(1 + \rho)$ for the CES production function:

\begin{align}
q = (l^\rho + k^\rho)^\frac{1}{\rho}
\end{align}

I found out that I need to solve the following equation:
\begin{align}
\sigma = \frac{\frac{d(k/l)}{k/l}}{\frac{dRTS}{RTS}} = \frac{d(k/l)}{dRTS}\frac{RTS}{k/l} = \frac{d(k/l)}{d((k/l)^{1-\rho})}\frac{(k/l)^{1-\rho}}{k/l}
\end{align}

But i just don't know how to rewrite this expression to $\sigma = 1/(1 + \rho)$

**Tags:** production-function

> **Comments:**
> **username_1**: Check the example for Cobb Douglas production and try to solve it for CES. https://en.wikipedia.org/wiki/Elasticity_of_substitution  

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## 2 answers

### Answer by username_2 on 2016-10-13. Score: 9

The production function is:
$$q = (l^\rho + k^\rho)^\frac{1}{\rho}$$
The MPL and MPK are respectively:
$$q\_l = \frac{\partial q}{\partial l} = \frac{1}{\rho} \cdot (l^\rho + k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot l^{\rho-1}$$
$$q\_k = \frac{\partial q}{\partial k} = \frac{1}{\rho} \cdot (l^\rho
+ k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot k^{\rho-1}$$
What is the rate that $l$ can be substituted for $k$?

Where $f$ is a differentiable real-valued function of a single variable, we define the elasticity of $f(x)$ with respect to $x$ (at the point $x$) to be
$$\sigma(x) = \frac{x f'(x)}{f(x)}\equiv \frac{\frac{df(x)}{f(x)}}{\frac{dx}{x}}$$

1. Do a change of variables such that $u = ln(x)$ ($\rightarrow x = e^u$) and $v=ln(f(x))$ ($\rightarrow f(x) = e^v$)
2. Note that $v' = f'(x) / f(x)$ and $u'=\frac{1}{x}$ so that
   $$\frac{v'}{u'}=\frac{\frac{f'(x)}{f(x)}}{\frac{1}{x}} = \sigma(x)$$
3. Note that this is also the result you get by solving for $ \frac{d ln f(x)}{d ln(x)}$ because $ \frac{d ln f(x)}{d ln(x)} = \frac{d v}{d u}$ which we solve via the chain rule:
   $$ \frac{d v}{d u} = \frac{d v}{d x} \cdot \frac{d x}{d u} = \frac{f'(x)}{f(x)} \cdot x $$
   which happens to be exactly the definition of $\sigma(x)$.

Now let's tackle your elasticity problem.

$$ ln(\frac{q\_k}{q\_l})= log(\frac{\frac{1}{\rho} \cdot (l^\rho + k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot l^{\rho-1}}{\frac{1}{\rho} \cdot (l^\rho
+ k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot k^{\rho-1}}) = ln (\frac{l}{k})^{\rho-1} = (\rho-1) ln (l/k) = (1 - \rho) ln (k/l)$$
$$ \Rightarrow ln (k/l) = \frac{1}{1-\rho} \cdot ln(\frac{q\_k}{q\_l})$$

So $\sigma = \frac{1}{1-\rho}$

> **Comments:**
> **username_3**: $\frac{1}{\rho}$ and  $\rho$ can be reduced from derivatives MPL and MPK to simplify the exposition.  

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### Answer by username_4 on 2020-05-20. Score: 0

I would like to add a little bit to the answer above. I wrote a comment earlier, but I thought it'd be helpful to flesh out the argument a little more.

We have a firm that uses two factors of production, labor $l$ and capital $k$, to produce output. Quantity of output is written $q$.

The elasticity of a function of a *single variable* measures the percentage response of a dependent variable to a percentage change in the independent variable.

On the other hand, the elasticity of substitution between two factor inputs measures the percentage response of the ratio of their quantities to a percentage change in the relative marginal products.

Pertaining to the above, we have that the elasticity is given by

\begin{align}
\sigma \equiv \frac{d \ln{\left( k/l \right)}}{d\ln{ \left( MPL/MPK \right)}}
\end{align}

where $MPL$ is the marginal product of labor and $MPK$ is the marginal product of capital.

The reason I am writing this out is that there is a small error in the above answer. In the equation right after "Now let's tackle your elasticity problem," $\ln{\frac{q\_k}{q\_l}}$ is immediately followed by an expression for $\ln{\frac{q\_l}{q\_k}}$ switching the numerator with the denominator.

If you correct this, you get that $\sigma = -\frac{1}{1-\rho}$, which is close but not quite correct. To get the correct answer, you follow otherwise exactly the same calculations given by the above answer to get

$$ \ln{k/l} = \frac{1}{1-\rho}\cdot \ln {\frac{q\_l}{q\_k}}$$

to get that $\sigma = \frac{1}{1-\rho}$, where the corrections are for reasons outlined above.


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