The Constant Elasticity of Substitution production function is defined as:

(Taken from Wikipedia)

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


$Q=$ 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?

  • $\begingroup$ I am perplexed by your question. What do you mean by "how is it derived"? $\endgroup$
    – Giskard
    Commented Feb 1, 2018 at 20:57
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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


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.

  • $\begingroup$ 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

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 $E$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.