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The 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

$a=$ 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?

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  • $\begingroup$ I am perplexed by your question. What do you mean by "how is it derived"? $\endgroup$
    – Giskard
    Commented Feb 1, 2018 at 20:57
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    $\begingroup$ @denesp: Perhaps the OP wanted to ask who first proposed this functional form? $\endgroup$
    – Herr K.
    Commented Feb 1, 2018 at 21:39
  • $\begingroup$ @HerrK. Let us hope for clarification, as that is a very different question. $\endgroup$
    – Giskard
    Commented Feb 1, 2018 at 23:11
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    $\begingroup$ His question makes sense - he is asking for the derivation of the functional form from its defining condition (i.e., constant elasticity of substitution). $\endgroup$
    – Ben
    Commented Feb 2, 2018 at 0:46

2 Answers 2

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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.

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  • $\begingroup$ 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 $\endgroup$
    – Bensstats
    Commented Feb 2, 2018 at 1:16
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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 $E$.

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