# CES Production Function

To answer how a CES production function becomes Cobb-Douglas or Leontief.

### CES Form

Constant Elasticity of Substitution neoclassical production. A two-factor production takes the following general form:

$y = k \left(\alpha x_1^{\frac{\sigma-1}{\sigma}} + (1-\alpha) x_2^{\frac{\sigma-1}{\sigma}}\right)^{\frac{\sigma}{\sigma-1}},$

where $\sigma$ is the Elasticity of Substitution, $k$ is a shift parameter, or efficiency. The production employs two inputs $x_1$ and $x_2$, with substitutability between them governed by $\sigma$, parameter $\alpha \in [0,1]$ shows the weight/importance of each input factor. The production function is homogenous of degree 1.

For notational ease, let:

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

which is called Substitution parameter. Interchangably:

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

We can rewrite the CES as follows.

$y = k \left( \alpha x_1^\rho + (1-\alpha)x_2^\rho \right)^{1/\rho}.$

The elasticity of substitution $\sigma$ varies between 0 and $+\infty$.

(1) When $\sigma \to 0, \rho \to -\infty$: Leontief Production function (inputs are perfect complements).

(2) When $\sigma \to 1, \rho \to 0$: Cobb-Douglas Production function.

(3) When $\sigma \to +\infty, \rho \to 1$: Linear Production (inputs are perfect substitutes).

So the range of $\rho$ is $-\infty < \rho < 1$.

### (1) Leontief

When $\sigma \to 0, \rho \to -\infty$, inputs are perfect complements.

The idea is to show that $\lim_{\rho\to -\infty} y$ equals the smallest factor.

Since $\rho \to -\infty$, we treat $\rho$ as strictly negative.

Suppose that $x_1, x_2 > 0$ and $x_1 \geq x_2$, then $x_1^\rho \leq x_2^\rho$ (i).

Multiplying both sides of (i) with $\alpha$, we have:

$x_1^\rho \leq x_2^\rho \\ \Rightarrow \alpha x_1^\rho \leq \alpha x_2^\rho \\ \Rightarrow \alpha x_1^\rho + (1-\alpha) x_2^\rho \leq \alpha x_2^\rho + (1-\alpha)x_2^\rho \\ \Rightarrow (\alpha x_1^\rho + (1-\alpha)x_2^\rho)^{1/\rho} \leq (x_2^\rho)^{1/\rho} = x_2 \\ \Rightarrow k(\alpha x_1^\rho + (1-\alpha)x_2^\rho)^{1/\rho} \leq k x_2. \\$

The LHS is precisely $y$.

It is obvious that:

$(1-\alpha)x_2^\rho < \alpha x_1^\rho + (1-\alpha)x_2^\rho \\ \Rightarrow (1-\alpha)^{1/\rho} x_2 \leq (\alpha x_1^\rho + (1-\alpha)x_2^\rho)^{1/\rho} \\ \Rightarrow k (1-\alpha)^{1/\rho} x_2 \leq k (\alpha x_1^\rho + (1-\alpha)x_2^\rho)^{1/\rho}.$

The RHS is, again, $y$. So we have:

$k (1-\alpha)^{1/\rho} x_2 \leq y \leq k x_2.$

Furthermore, take lim of the LHS:

$\lim_{\rho\to-\infty} k (1-\alpha)^{1/\rho} x_2 = k x_2.$

Therefore:

$\lim_{\rho\to-\infty} y = k x_2.$

Thus far, we suppose that $x_1 \geq x_2$. By the same process, we can show that if $x_1 \leq x_2$, then $\lim_{\rho\to-\infty}y = k x_1$. Therefore, in general, for $\rho\to-\infty$:

$\lim_{\rho\to-\infty} y = k \ \min (x_1, x_2),$

which is the Leontief pf, the isoquants become L-shaped. Production takes a fixed proportions of inputs because there is no substitutability between factors.

### (2) Cobb-Douglas

Consider when $\sigma \to 1, \rho \to 0$, to show case (2), we need l’Hospital’s rule. The idea is to take logs and use calculus to directly derive.

L’Hospital’s Rule: Suppose that $f(x)$ and $g(x)$ both tend to 0 as $x\to 0$. If the ratio of $f’(x)/g’(x)$ exists: $\lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x\to 0}\frac{f'(x)}{g'(x)}.$ In words: the limit of the ratio of functions, if it exists, equals the ratio of their respective derivatives.

Take logs of the CES function:

$\ln(y) = \ln(k) + \frac{\ln \left( \alpha x_1^\rho + (1-\alpha) x_2^\rho \right)}{\rho} = \ln (k) + \frac{f(\rho)}{\rho} \ \ (2)$

Note that:

$y = \ln(u) \Rightarrow y' = \frac{dy}{du} = \frac{u'}{u}, \\ y = a^{u} \Rightarrow y' = \frac{dy}{du} = u' a^u \ln(a).$

Take the derivative of $f(\rho)$:

$f'(\rho) = \frac{1}{\alpha x_1^\rho + (1-\alpha)x_2^\rho} \frac{d \left[\alpha x_1^\rho + (1-\alpha)x_2^\rho\right]}{d\rho} \\ = \frac{1}{\alpha x_1^\rho + (1-\alpha)x_2^\rho} \left[ \alpha x_1^\rho \ln(x_1) + (1-\alpha)x_2^\rho \ln(x_2) \right].$

Take lim as $\rho \to 0$:

$\lim_{\rho\to 0}\frac{f(\rho)}{\rho} = \lim_{\rho\to 0} \frac{f'(\rho)}{1} = \alpha \ln(x_1) + (1-\alpha) \ln(x_2).$

By plugging back to Eq. (2), we can say that, when $\rho \to 0$:

$\ln(y) = \ln(k) + [\alpha \ln(x_1) + (1-\alpha)\ln(x_2)]$

Take the exponential of $y$:

$y = k x_1^\alpha x_2^{1-\alpha},$

which is the Cobb-Douglas pf. The isoquants become curved (rectangular hyperbola).

### (3) Linear function

Just plugging $\rho = 1$ to the function, we obtain:

$y = k (\alpha x_1 + (1-\alpha) x_2),$

which is linear. $x_1$ and $x_2$ are perfect substitutes. The isoquants become a straight line.