<p>As a follow up to Ben's answer. Here's the derivation to get to the solution of the SODE.</p>
<p>For notational convenience, let <span class="math-container">$y = f(k)$</span> and <span class="math-container">$y' = f'(k)$</span>. We have:
<span class="math-container">$$
\frac{y'(ky' - y)}{k y y''} = s.
$$</span>
This is equivalent to:
<span class="math-container">$$
\frac{y'}{y}-\frac{1}{k} = s\frac{y''}{y'}.
$$</span>
Integrating both sides gives:
<span class="math-container">$$
\begin{align*}
&amp;\int \frac{y'}{y} dk - \int \frac{1}{k} dk = s\int \frac{y''}{y'} dk,\\
\iff &amp;\ln(y) - \ln(k) = s \ln(y') + C
\end{align*}
$$</span>
For some constant <span class="math-container">$C$</span>.
Taking exponents on both sides, produces:
<span class="math-container">$$
D (y')^{s} = \frac{y}{k} \iff y^{-1/s} y' = Ek^{-1/s}
$$</span>
where <span class="math-container">$D$</span> and <span class="math-container">$E$</span> are positive constants.
Integrating once more the last expression gives:
<span class="math-container">$$
\begin{align*}
&amp;\int y^{-1/s} y' dk = E \int k^{-1/s} dk,\\
\iff&amp; \frac{s}{s-1} y^{\frac{s-1}{s}} = F + E \frac{s}{s-1} k^{\frac{s-1}{s}} 
\end{align*}
$$</span>
For some constant <span class="math-container">$F$</span>.
Rewriting gives:
<span class="math-container">$$
y = \left(A + E k^{\frac{s-1}{s}}\right)^{\frac{s}{s-1}}.
$$</span>
For some constants <span class="math-container">$A$</span> and <span class="math-container">$E$</span>.</p>

CC BY-SA 4.0
2024-06-14T11:23:16.560Z
2024-06-14T11:23:16.560Z
tdm
<p>The CES function can be derived directly from the condition of constant elasticity of substitution.  There are various ways to do this, but the simplest derivation occurs for a homothetic production function.  Suppose we start with a homothetic production function <span class="math-container">$Q = f^*(K, L)$</span> and we rewrite this in intensive form as:</p>
<p><span class="math-container">$$\begin{matrix} q = f(k) &amp; &amp; q \equiv Q/L &amp; &amp; k \equiv K/L. \end{matrix}$$</span></p>
<p>For this case the elasticity of substitution <span class="math-container">$s$</span> can be shown to be:</p>
<p><span class="math-container">$$s = - \frac{f'(k)(f(k) - kf'(k))}{kf(k)f''(k)}.$$</span></p>
<p>Letting <span class="math-container">$r \equiv (s-1)/s$</span> and re-arranging this equation gives the second-order differential equation:</p>
<p><span class="math-container">$$\frac{kf(k)f''(k)}{1-r} + f'(k)(f(k) - kf'(k)) = 0.$$</span></p>
<p>This equation has general solution <span class="math-container">$q = f(k) =  c_0 (1 + c_1 k^r)^{1/r}$</span> where <span class="math-container">$c_0$</span> and <span class="math-container">$c_1$</span> are constants.  (See Appendix below to show that this gives a solution to the differential equation.)  Parameterising with <span class="math-container">$a \equiv c_1/(1+c_1)$</span> and <span class="math-container">$F \equiv c_0 (1+c_1)^{1/r}$</span> and substituting to obtain the extensive form gives:</p>
<p><span class="math-container">$$\begin{equation} \begin{aligned}
Q = Lq = Lf(K/L) 
&amp;= c_0 L \left( 1 + c_1 \left( \frac{K}{L} \right)^r \right)^{1/r} \\[6pt]
&amp;= c_0 \left( L^r + c_1 K^r \right)^{1/r} \\[6pt]
&amp;= c_0 (1+c_1)^{1/r} \left( \frac{1}{1+c_1} L^r + \frac{c_1}{1+c_1} K^r \right)^{1/r} \\[6pt]
&amp;= F \left(a K^r + (1-a) L^r \right)^{1/r}. \\[6pt]
\end{aligned} \end{equation}$$</span></p>
<p>The parameter <span class="math-container">$a$</span> can be interpreted as the capital intensity in production and the parameter <span class="math-container">$F$</span> can be interpreted as the overall efficiency of production.</p>
<hr />
<p><strong>Appendix --- Solution to differential equation:</strong> We can confirm this solution by noting the following derivatives:</p>
<p><span class="math-container">$$\begin{align}
f'(k)
&amp;= \frac{d}{dk} c_0 (1 + c_1 k^r)^{1/r} \\[10pt]
&amp;= \frac{c_0}{r} (r c_1 k^{r-1}) (1 + c_1 k^r)^{1/r-1} \\[6pt]
&amp;= c_0 c_1 k^{r-1} (1 + c_1 k^r)^{1/r-1}, \\[12pt]
f''(k)
&amp;= \frac{d}{dk} c_0 c_1 k^{r-1} (1 + c_1 k^r)^{1/r-1} \\[6pt]
&amp;= c_0 c_1 (r-1) k^{r-2} (1 + c_1 k^r)^{1/r-1} + c_0 c_1 k^{r-1} (1/r-1) (c_1 r k^{r-1}) (1 + c_1 k^r)^{1/r-2} \\[6pt]
&amp;= c_0 c_1 k^{r-2} (1 + c_1 k^r)^{1/r-2} \bigg[ (r-1) (1 + c_1 k^r) + (1-r) c_1 k^r \bigg] \\[6pt]
&amp;= c_0 c_1 k^{r-2} (r-1) (1 + c_1 k^r)^{1/r-2}, \\[12pt]
\end{align}$$</span></p>
<p>which gives:</p>
<p><span class="math-container">$$\begin{align}
f(k) - kf'(k)
&amp;= c_0 (1 + c_1 k^r)^{1/r} - c_0 c_1 k^r (1 + c_1 k^r)^{1/r-1} \\[6pt]
&amp;= c_0 (1 + c_1 k^r)^{1/r-1} \bigg[ (1 + c_1 k^r) - c_1 k^r \bigg] \\[6pt]
&amp;= c_0 (1 + c_1 k^r)^{1/r-1}, \\[12pt]
f'(k)(f(k) - kf'(k))
&amp;= c_0 c_1 k^{r-1} (1 + c_1 k^r)^{1/r-1} c_0 (1 + c_1 k^r)^{1/r-1} \\[12pt]
&amp;= c_0^2 c_1 k^{r-1} (1 + c_1 k^r)^{2/r-2}, \\[12pt]
\frac{kf(k)f''(k)}{1-r}
&amp;= \frac{k c_0 (1 + c_1 k^r)^{1/r} c_0 c_1 k^{r-2} (r-1) (1 + c_1 k^r)^{1/r-2}}{1-r} \\[6pt]
&amp;= \frac{c_0^2 c_1 (r-1) k^{r-1} (1 + c_1 k^r)^{2/r-2}}{1-r} \\[6pt]
&amp;= - c_0^2 c_1 k^{r-1} (1 + c_1 k^r)^{2/r-2}. \\[6pt]
\end{align}$$</span></p>
<p>Substitution of these results confirms that the second-order differential equation is satisfied.</p>

CC BY-SA 3.0
2018-02-02T01:16:36.967Z
Would you mind showing how you derived the elasticity? Its a shame ODE's aren't a requirement for Statisitcs and Economics students because it seems quite essential as demonstrated in the answer to fully understand
Bensstats
CC BY-SA 4.0
2018-02-02T00:45:09.777Z
2024-06-14T07:25:26.500Z
2024-06-14T07:25:26.500Z
Ben
<p>The <a href="https://en.wikipedia.org/wiki/Constant_elasticity_of_substitution" rel="nofollow noreferrer">Constant Elasticity of Substitution</a> production function is defined as:</p>
<p>(<em>Taken from Wikipedia</em>)</p>
<p><span class="math-container">$$Q=F \boldsymbol{\cdot}\left(a\boldsymbol{\cdot}K^r+(1-a)\boldsymbol{\cdot}L^r \right)^{1\over{r}}$$</span></p>
<p>Where:</p>
<p><span class="math-container">$Q=$</span> the quantity of output</p>
<p><span class="math-container">$F=$</span> <a href="https://en.wikipedia.org/wiki/Total_factor_productivity" rel="nofollow noreferrer">factor productivity</a></p>
<p><span class="math-container">$a=$</span> Share parameter (i.e. <span class="math-container">$0 &lt; a &lt;1$</span>)</p>
<p><span class="math-container">$K, L=$</span> Quantities of production factors</p>
<p><span class="math-container">$r= {\left(s-1 \right)\over{s}}$</span></p>
<p><span class="math-container">$s= {1\over{1-r}}=$</span> Elasticity of subsitution</p>
<h1>My question:</h1>
<p>Though this is a quite elegant formula, how is it derived?</p>

CC BY-SA 3.0
2018-02-01T20:57:14.833Z
I am perplexed by your question. What do you mean by "how is it derived"?
Giskard
CC BY-SA 3.0
2018-02-01T21:39:43.890Z
@denesp: Perhaps the OP wanted to ask who first proposed this functional form?
Herr K.
CC BY-SA 3.0
2018-02-01T23:11:53.867Z
@HerrK.  Let us hope for clarification,  as that is a very different question.
Giskard
CC BY-SA 3.0
2018-02-02T00:46:06.053Z
His question makes sense - he is asking for the derivation of the functional form from its defining condition (i.e., constant elasticity of substitution).
Ben
<mathematical-economics><production-function>
How was the CES production function derived?
Bensstats
# How was the CES production function derived?

(Asked by: username_0 on 2018-02-01)

The [Constant Elasticity of Substitution](https://en.wikipedia.org/wiki/Constant_elasticity_of_substitution) production function is defined as:

(*Taken from Wikipedia*)

$$Q=F \boldsymbol{\cdot}\left(a\boldsymbol{\cdot}K^r+(1-a)\boldsymbol{\cdot}L^r \right)^{1\over{r}}$$

Where:

$Q=$ the quantity of output

$F=$ [factor productivity](https://en.wikipedia.org/wiki/Total_factor_productivity)

$a=$ Share parameter (i.e. $0 < a <1$)

$K, L=$ Quantities of production factors

$r= {\left(s-1 \right)\over{s}}$

$s= {1\over{1-r}}=$ Elasticity of subsitution

# My question:

Though this is a quite elegant formula, how is it derived?

**Tags:** mathematical-economics, production-function
> **Comments:**
> **username_1**: I am perplexed by your question. What do you mean by "how is it derived"?  
> **username_2**: @denesp: Perhaps the OP wanted to ask who first proposed this functional form?  
> **username_1**: @HerrK.  Let us hope for clarification,  as that is a very different question.  
> **username_3**: His question makes sense - he is asking for the derivation of the functional form from its defining condition (i.e., constant elasticity of substitution).  

---

## 2 answers

### Answer by username_3 on 2018-02-02. Score: 12 | Accepted answer

The CES function can be derived directly from the condition of constant elasticity of substitution. There are various ways to do this, but the simplest derivation occurs for a homothetic production function. Suppose we start with a homothetic production function $Q = f^\*(K, L)$ and we rewrite this in intensive form as:

$$\begin{matrix} q = f(k) & & q \equiv Q/L & & k \equiv K/L. \end{matrix}$$

For this case the elasticity of substitution $s$ can be shown to be:

$$s = - \frac{f'(k)(f(k) - kf'(k))}{kf(k)f''(k)}.$$

Letting $r \equiv (s-1)/s$ and re-arranging this equation gives the second-order differential equation:

$$\frac{kf(k)f''(k)}{1-r} + f'(k)(f(k) - kf'(k)) = 0.$$

This equation has general solution $q = f(k) = c\_0 (1 + c\_1 k^r)^{1/r}$ where $c\_0$ and $c\_1$ are constants. (See Appendix below to show that this gives a solution to the differential equation.) Parameterising with $a \equiv c\_1/(1+c\_1)$ and $F \equiv c\_0 (1+c\_1)^{1/r}$ and substituting to obtain the extensive form gives:

$$\begin{equation} \begin{aligned}
Q = Lq = Lf(K/L)
&= c\_0 L \left( 1 + c\_1 \left( \frac{K}{L} \right)^r \right)^{1/r} \\[6pt]
&= c\_0 \left( L^r + c\_1 K^r \right)^{1/r} \\[6pt]
&= c\_0 (1+c\_1)^{1/r} \left( \frac{1}{1+c\_1} L^r + \frac{c\_1}{1+c\_1} K^r \right)^{1/r} \\[6pt]
&= F \left(a K^r + (1-a) L^r \right)^{1/r}. \\[6pt]
\end{aligned} \end{equation}$$

The parameter $a$ can be interpreted as the capital intensity in production and the parameter $F$ can be interpreted as the overall efficiency of production.

---

**Appendix --- Solution to differential equation:** We can confirm this solution by noting the following derivatives:

$$\begin{align}
f'(k)
&= \frac{d}{dk} c\_0 (1 + c\_1 k^r)^{1/r} \\[10pt]
&= \frac{c\_0}{r} (r c\_1 k^{r-1}) (1 + c\_1 k^r)^{1/r-1} \\[6pt]
&= c\_0 c\_1 k^{r-1} (1 + c\_1 k^r)^{1/r-1}, \\[12pt]
f''(k)
&= \frac{d}{dk} c\_0 c\_1 k^{r-1} (1 + c\_1 k^r)^{1/r-1} \\[6pt]
&= c\_0 c\_1 (r-1) k^{r-2} (1 + c\_1 k^r)^{1/r-1} + c\_0 c\_1 k^{r-1} (1/r-1) (c\_1 r k^{r-1}) (1 + c\_1 k^r)^{1/r-2} \\[6pt]
&= c\_0 c\_1 k^{r-2} (1 + c\_1 k^r)^{1/r-2} \bigg[ (r-1) (1 + c\_1 k^r) + (1-r) c\_1 k^r \bigg] \\[6pt]
&= c\_0 c\_1 k^{r-2} (r-1) (1 + c\_1 k^r)^{1/r-2}, \\[12pt]
\end{align}$$

which gives:

$$\begin{align}
f(k) - kf'(k)
&= c\_0 (1 + c\_1 k^r)^{1/r} - c\_0 c\_1 k^r (1 + c\_1 k^r)^{1/r-1} \\[6pt]
&= c\_0 (1 + c\_1 k^r)^{1/r-1} \bigg[ (1 + c\_1 k^r) - c\_1 k^r \bigg] \\[6pt]
&= c\_0 (1 + c\_1 k^r)^{1/r-1}, \\[12pt]
f'(k)(f(k) - kf'(k))
&= c\_0 c\_1 k^{r-1} (1 + c\_1 k^r)^{1/r-1} c\_0 (1 + c\_1 k^r)^{1/r-1} \\[12pt]
&= c\_0^2 c\_1 k^{r-1} (1 + c\_1 k^r)^{2/r-2}, \\[12pt]
\frac{kf(k)f''(k)}{1-r}
&= \frac{k c\_0 (1 + c\_1 k^r)^{1/r} c\_0 c\_1 k^{r-2} (r-1) (1 + c\_1 k^r)^{1/r-2}}{1-r} \\[6pt]
&= \frac{c\_0^2 c\_1 (r-1) k^{r-1} (1 + c\_1 k^r)^{2/r-2}}{1-r} \\[6pt]
&= - c\_0^2 c\_1 k^{r-1} (1 + c\_1 k^r)^{2/r-2}. \\[6pt]
\end{align}$$

Substitution of these results confirms that the second-order differential equation is satisfied.

> **Comments:**
> **username_0**: Would you mind showing how you derived the elasticity? Its a shame ODE's aren't a requirement for Statisitcs and Economics students because it seems quite essential as demonstrated in the answer to fully understand  

---

### Answer by username_4 on 2024-06-14. Score: 4

As a follow up to Ben's answer. Here's the derivation to get to the solution of the SODE.

For notational convenience, let $y = f(k)$ and $y' = f'(k)$. We have:
$$
\frac{y'(ky' - y)}{k y y''} = s.
$$
This is equivalent to:
$$
\frac{y'}{y}-\frac{1}{k} = s\frac{y''}{y'}.
$$
Integrating both sides gives:
$$
\begin{align\*}
&\int \frac{y'}{y} dk - \int \frac{1}{k} dk = s\int \frac{y''}{y'} dk,\\
\iff &\ln(y) - \ln(k) = s \ln(y') + C
\end{align\*}
$$
For some constant $C$.
Taking exponents on both sides, produces:
$$
D (y')^{s} = \frac{y}{k} \iff y^{-1/s} y' = Ek^{-1/s}
$$
where $D$ and $E$ are positive constants.
Integrating once more the last expression gives:
$$
\begin{align\*}
&\int y^{-1/s} y' dk = E \int k^{-1/s} dk,\\
\iff& \frac{s}{s-1} y^{\frac{s-1}{s}} = F + E \frac{s}{s-1} k^{\frac{s-1}{s}}
\end{align\*}
$$
For some constant $F$.
Rewriting gives:
$$
y = \left(A + E k^{\frac{s-1}{s}}\right)^{\frac{s}{s-1}}.
$$
For some constants $A$ and $E$.


---

