# CES: Production function: Elasticity of substitution $\sigma = 1/(1 + \rho)$

I have to prove that $\sigma = 1/(1 + \rho)$ for the CES production function:

\begin{align} q = (l^\rho + k^\rho)^\frac{1}{\rho} \end{align}

I found out that I need to solve the following equation: \begin{align} \sigma = \frac{\frac{d(k/l)}{k/l}}{\frac{dRTS}{RTS}} = \frac{d(k/l)}{dRTS}\frac{RTS}{k/l} = \frac{d(k/l)}{d((k/l)^{1-\rho})}\frac{(k/l)^{1-\rho}}{k/l} \end{align}

But i just don't know how to rewrite this expression to $\sigma = 1/(1 + \rho)$

The production function is: $$q = (l^\rho + k^\rho)^\frac{1}{\rho}$$ The MPL and MPK are respectively: $$q_l = \frac{\partial q}{\partial l} = \frac{1}{\rho} \cdot (l^\rho + k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot l^{\rho-1}$$ $$q_k = \frac{\partial q}{\partial k} = \frac{1}{\rho} \cdot (l^\rho + k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot k^{\rho-1}$$ What is the rate that l can be substituted for k?

Where $f$ is a differentiable real-valued function of a single variable, we define the elasticity of f(x) with respect to x (at the point x) to be $$\sigma(x) = \frac{x f'(x)}{f(x)}\equiv \frac{\frac{df(x)}{f(x)}}{\frac{dx}{x}}$$

1. Do a change of variables such that $u = ln(x)$ ($\rightarrow x = e^u$) and $v=ln(f(x))$ ($\rightarrow f(x) = e^v$)
2. Note that $v' = f'(x) / f(x)$ and $u'=\frac{1}{x}$ so that $$\frac{v'}{u'}=\frac{\frac{f'(x)}{f(x)}}{\frac{1}{x}} = \sigma(x)$$
3. Note that this is also the result you get by solving for $\frac{d ln f(x)}{d ln(x)}$ because $\frac{d ln f(x)}{d ln(x)} = \frac{d v}{d u}$ which we solve via the chain rule: $$\frac{d v}{d u} = \frac{d v}{d x} \cdot \frac{d x}{d u} = \frac{f'(x)}{f(x)} \cdot x$$ which happens to be exactly the definition of $\sigma(x)$.

Now let's tackle your elasticity problem.

$$ln(\frac{q_k}{q_l})= log(\frac{\frac{1}{\rho} \cdot (l^\rho + k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot l^{\rho-1}}{\frac{1}{\rho} \cdot (l^\rho + k^\rho)^{\frac{1}{\rho}-1} \cdot \rho\cdot k^{\rho-1}}) = ln (\frac{l}{k})^{\rho-1} = (\rho-1) ln (l/k) = (1 - \rho) ln (k/l)$$ $$\Rightarrow ln (k/l) = \frac{1}{1-\rho} \cdot ln(\frac{q_k}{q_l})$$

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

• $\frac{1}{\rho}$ and $\rho$ can be reduced from derivatives MPL and MPK to simplify the exposition. – garej May 11 '18 at 15:28