# Constant elasticity of substitution function and its properties

Is constant elasticity of substitution (CES) function famous? Maybe yes to you but maybe not for others. However, if you are major in economics or econ-related field, you must know something about Cobb-Douglas function. Then I would like to tell you that the Cobb-Douglas function is a special form of CES function.

Why is it a special form of CES function? To understand this, we should be clear in mind what does elasticity of substitution mean. Here’s one paragraph quoted from Wikipedia (click [ww wiki=”elasticity of substitution|here”]):

Elasticity of substitution is the elasticity of the ratio of two inputs to a production (or utility) function with respect to the ratio of their marginal products (or utilities). It measures the curvature of an isoquant and thus, the substitutability between inputs (or goods), i.e. how easy it is to substitute one input (or good) for the other.

According to this definition, we can easily write the definition formulae for the elasticity of substitution of a function $$f (x_1, x_2)$$:
$$\sigma_{1,2} = \frac{d \ln (x_2 / x_1)}{d \ln ({MRS}_{1,2})}$$

where $${MRS}_{1,2}$$ means the ratio of each input’s marginal rate of substitution which is defined as:
$${MRS}_{1,2} = \frac{\partial{f} / \partial{x_1}}{\partial{f} / \partial{x_2}}$$

Now let’s turn to CES function, the general form of CES production function (the typical function we use in economics and we doesn’t impose constant returns to scale assumption here) is (right click the formulae to change zoom options):
$$f(K, L) = A {\left[ \alpha K^{-\rho} + (1-\alpha) L^{-\rho} \right]}^{ -\frac{\nu}{\rho} }$$

in which $$\rho \in [-1,\infty)$$.

Then we can get the marginal rate of substitution of CES function as:
\begin{align} {MRS}_{1,2} &= \frac{\partial{f} / \partial{K}}{\partial{f} / \partial{L}} \\ &= \frac{A \left( -\frac{\nu}{\rho} \right) {\left[ \alpha K^{-\rho} + (1-\alpha) L^{-\rho} \right]}^{-\frac{\nu}{\rho}-1} \left[ \alpha (-\rho) K^{-\rho-1} \right]}{A \left( -\frac{\nu}{\rho} \right) {\left[ \alpha K^{-\rho} + (1-\alpha) L^{-\rho} \right]}^{-\frac{\nu}{\rho}-1} \left[ (1-\alpha) (-\rho) L^{-\rho-1} \right]} \\ &= \frac{\alpha}{1-\alpha} {\left( \frac{L}{K} \right)}^{\rho+1} \end{align}

From above definitions, we can calculate the elasticity of substitution of CES function as:
\begin{align} \sigma &= \frac{d \ln (L/K)}{d \ln ({MRS}_{1,2})} \\ &= \frac{d \ln (L/K)}{d \ln \left( \frac{\alpha}{1-\alpha} {(L/K)}^{\rho+1} \right)} \\ &= \frac{d \ln (L/K)}{\frac{1-\alpha}{\alpha} \frac{1}{{(L/K)}^{\rho+1}} \left( \frac{\alpha}{1-\alpha} \right) (\rho+1) {(L/K)}^{\rho} d(L/K)} \\ &= \frac{1}{\rho+1} \end{align}

Since $$\rho$$ is a exogenous parameter, CES function has a constant elasticity of substitution $$\frac{1}{\rho+1}$$.

Next, we can calculate the elasticity of substitution for Cobb-Douglas function using the same method:
$$f(K,L) = A K^{\alpha} L^{\beta}$$

and
$$\sigma = \frac{d \ln (L/K)}{d \ln ({MRS}_{1,2})} = \frac{d \ln \left( \frac{\beta}{\alpha} (L/K) \right)}{d \ln (L/K)} = 1$$

We can see that the elasticity of substitution of Cobb-Douglas function is a constant “1”, this happens to the CES function when $$\rho = 0$$.

You can understand now why I said that the Cobb-Douglas function is a special form of CES function. However, this is just intuition to some extent, can we prove that mathematically? The answer is absolutely YES!

## Converging CES to Cobb-Douglas function

As what has mentioned above, we are guessing that when $$\rho \to 0$$, the CES function may converging to Cobb-Douglas function. Now let’s prove it!

First, let’s take log transformation of both sides of CES function:
$$\ln f(K,L) = \ln A – \frac{\nu}{\rho} \ln \left[ \alpha K^{-\rho} + (1-\alpha) L^{-\rho} \right]$$

The result is believable for the reason that monotonic transformation (log transformation) will not change properties.

Next, we take the limit of both sides when $$\rho$$ converges to zero:
$$\lim\limits_{\rho \to 0} \ln f(K,L) = \ln A -\nu \lim\limits_{\rho \to 0} \frac{\ln \left[ \alpha K^{-\rho} + (1-\alpha) L^{-\rho} \right]}{\rho}$$

Using L’Hôpital’s rule on the right side, we can get:
\begin{align} \lim\limits_{\rho \to 0} \ln f(K,L) &= \ln A – \nu \lim\limits_{\rho \to 0} \frac{-\alpha \ln K K^{-\rho} + (\alpha-1) \ln L L^{-\rho}}{\alpha K^{-\rho} + (1-\alpha) L^{-\rho}} \\ &= \ln A + \nu \frac{\alpha \ln K + (1-\alpha) \ln L}{\alpha + 1 – \alpha} \\ &= \ln A + \alpha \nu \ln K + (1-\alpha) \nu \ln L \end{align}

Finally we get:
$$\lim\limits_{\rho \to 0} f(K,L) = A K^{\alpha\nu} L^{(1-\alpha)\nu}$$

This is exactly the form of Cobb-Douglas production function (parameter $$\nu$$ is to measure the level of returns to scale).

At this time, the elasticity of substitution $$\sigma = 1$$.

## Converging CES to Linear function

When $$\rho$$ converges to $$-1$$, the CES function converges to the linear combination function. This can be easily seen from the following calculation:
\begin{align} \lim\limits_{\rho \to -1} \ln f(K,L) &= \ln A – \nu \lim\limits_{\rho \to 0} \frac{\ln \left[ \alpha K^{-\rho} + (1-\alpha) L^{-\rho} \right]}{\rho} \\ &= \ln A – \nu \frac{\ln \left[ \alpha K + (1-\alpha) L \right]}{-1} \\ &= \ln A + \ln {\left[ \alpha K + (1-\alpha) L \right]}^{\nu} \end{align}

Then we get:
$$\lim\limits_{\rho \to -1} f(K,L) = A {\left[ \alpha K + (1-\alpha) L \right]}^{\nu}$$

This is exactly linear combination of $$K$$ and $$L$$ without $$\rho$$ appearing on the superscript.

At this time, the elasticity of substitution $$\sigma = \infty$$.

## Converging CES to Leontif function

Leontif function is a quite important function used in everywhere in economics fields. In CGE modeling, it is much easier to use Leontif function to simulate some kind of economic behavior than using other tricks. However, it is also a special form of CES function when $$\rho \to \infty$$ and we can see this from following prove.

\begin{align} \lim\limits_{\rho \to \infty} \ln f(K,L) &= \ln A -\nu \lim\limits_{\rho \to \infty} \frac{\ln \left[ \alpha K^{-\rho} + (1-\alpha) L^{-\rho} \right]}{\rho} \\ &= \ln A + \nu \lim\limits_{\rho \to \infty} \frac{\ln K \left[ \alpha K^{-\rho} + (1-\alpha) L^{-\rho} \right] – (1-\alpha)\ln K L^{-\rho} + (1-\alpha)\ln L L^{-\rho}}{\alpha K^{-\rho} + (1-\alpha) L^{-\rho}} \\ &= \ln A + \nu \nu \lim\limits_{\rho \to \infty} \left[ \ln K + (1-\alpha) \frac{\ln L L^{-\rho} – \ln K L^{-\rho}}{\alpha K^{-\rho} + (1-\alpha) L^{-\rho}} \right] \\ &= \ln A + \nu \lim\limits_{\rho \to \infty} \left[ \ln K + (1-\alpha) \frac{\ln L – \ln K}{\alpha {\left( \frac{L}{K} \right)}^{\rho} + (1-\alpha)} \right] \end{align}

If L < K[/latex], then $\lim\limits_{\rho \to \infty} {\left( \frac{L}{K} \right)}^{\rho} \to 0$, then we get: \begin{align} \lim\limits_{\rho \to \infty} \ln f(K,L) &= \ln A + \nu [\ln K + \ln L - \ln K] \\ &= \ln A + \nu \ln L \\ \end{align} Then we get $\lim\limits_{\rho \to \infty} f(K,L) = A L^{\nu}$. The situation is similar when $K < L$, if this happen, we will get $\lim\limits_{\rho \to \infty} f(K,L) = A K^{\nu}$. Finally, combine both situations, we get: $$\lim\limits_{\rho \to \infty} f(K,L) = A { \min \left\{ L, K \right\}}^{\nu}$$ At this time, the elasticity of substitution $\sigma = 0$.