How can I obtain Leontief and Cobb-Douglas production function from CES function?

Question

In most Microeconomics textbooks it is mentioned that the Constant Elasticity of Substitution (CES) production function, $$Q=\gamma[a K^{-\rho} +(1-a) L^{-\rho} ]^{-\frac{1}{\rho}}$$

(where the elasticity of substitution is $\sigma = \frac 1{1+\rho},\rho > -1$), has as its limits both the Leontief production function and the Cobb-Douglas one. Specifically,

$$\lim_{\rho\to \infty}Q= \gamma \min \left \{K , L\right\}$$

and

$$\lim_{\rho\to 0}Q= \gamma K^aL^{1-a}$$

But they never provide the mathematical proof for these results.

Can somebody please provide these proofs?

Moreover, the above CES function incorporates constant-returns-to-scale (homogeneity of degree one), due to the outside exponent being $-1/\rho$. If it was, say $-k/\rho$, then the degree of homogeneity would be $k$.

How are the limiting results affected if $k\neq 1$?

Answer 1

The proofs I will present are based on techniques relevant to the fact that the CES production function has the form of a generalized weighted mean.
This was used in the original paper where the CES function was introduced, Arrow, K. J., Chenery, H. B., Minhas, B. S., & Solow, R. M. (1961). Capital-labor substitution and economic efficiency. The Review of Economics and Statistics, 225-250.
The authors there referred their readers to the book Hardy, G. H., Littlewood, J. E., & Pólya, G. (1952). Inequalities , chapter $2 $.

We consider the general case $$Q_k=\gamma[a K^{-\rho} +(1-a) L^{-\rho} ]^{-\frac{k}{\rho}},\;\; k>0$$

$$\Rightarrow \gamma^{-1}Q_k = \frac 1{[a (1/K^{\rho}) +(1-a) (1/L^{\rho}) ]^{\frac{k}{\rho}}}$$

1) Limit when $\rho \rightarrow \infty$
Since we are interested in the limit when $\rho\rightarrow \infty$ we can ignore the interval for which $\rho \leq0$, and treat $\rho$ as strictly positive.

Without loss of generality, assume $K\geq L \Rightarrow (1/K^{\rho})\leq (1/L^{\rho})$. We also have $K, L >0$. Then we verify that the following inequality holds:

$$(1-a)^{k/\rho}(1/L^{k})\leq \gamma Q_k^{-1} \leq (1/L^{k}) $$

$$\implies (1-a)^{k/\rho}(1/L^{k})\leq [a (1/K^{\rho}) +(1-a) (1/L^{\rho}) ]^{\frac{k}{\rho}} \leq (1/L^{k}) \tag{1}$$

by raising throughout to the $\rho/k$ power to get

$$(1-a)(1/L^{\rho}) \leq a (1/K^{\rho}) +(1-a) (1/L^{\rho}) \leq (1/L^{\rho}) \tag {2}$$ which indeed holds, obviously, given the assumptions. Then go back to the first element of $(1)$ and

$$\lim_{\rho\rightarrow \infty} (1-a)^{k/\rho}(1/L^{k}) =(1/L^{k})$$

which sandwiches the middle term in $(1)$ to $(1/L^{k})$ , so

$$\lim_{\rho\rightarrow \infty}Q_k = \frac {\gamma }{1/L^k} = \gamma L^k = {\gamma }\big[\min\{K,L\}\big]^{k} \tag{3}$$

So for $k=1$ we obtain the basic Leontief production function.

2) Limit when $\rho \rightarrow 0$
Write the function using exponential as

$$\gamma^{-1}Q_k=\exp\left\{-\frac k{\rho}\cdot \ln\big[a (K^{\rho})^{-1} +(1-a) (L^{\rho})^{-1}\big]\right\} \tag {4}$$

Consider the first-order Maclaurin expansion (Taylor expansion centered at zero) of the term inside the logarithm, with respect to $\rho$:

$$a (K^{\rho})^{-1} +(1-a) (L^{\rho})^{-1} \\= a (K^{0})^{-1} +(1-a) (L^{0})^{-1} -a (K^{0})^{-2}K^{0}\rho\ln K- (1-a) (L^{0})^{-2}L^{0}\rho\ln L + O(\rho^2) \\$$

$$=1 - \rho a\ln K - \rho(1-a)\ln L+ O(\rho^2) = 1 +\rho \big[\ln K^{-a}L^{-(1-a)}\big]+ O(\rho^2)$$

Insert this back into $(4)$ and get rid of the outer exponential,

$$\gamma^{-1}Q_k = \left(1 +\rho \big[\ln K^{-a}L^{-(1-a)}\big]+ O(\rho^{2})\right)^{-k/\rho}$$

In case it is opaque, define $r\equiv 1/\rho$ and re-write

$$\gamma^{-1}Q_k = \left(1 +\frac{\big[\ln K^{-a}L^{-(1-a)}\big]}{r}+ O(r^{-2})\right)^{-kr}$$

Now it does look like an expression whose limit at infinity will give us something exponential:

$$\lim_{\rho\rightarrow 0}\gamma^{-1}Q_k = \lim_{r\rightarrow \infty}\gamma^{-1}Q_k = \left(\exp\left\{ \ln K^{-a}L^{-(1-a)}\right\} \right)^{-k}$$

$$\Rightarrow \lim_{\rho\rightarrow 0}Q_k =\gamma\left(K^{a}L^{1-a}\right)^k$$

The degree of homogeneity $k$ of the function is preserved, and if $k=1$ we obtain the Cobb-Douglas function.

It was this last result that made Arrow and Co to call $a$ the "distribution" parameter of the CES function.

Answer 2

The regular method of obtaining Cobb-Douglas and Leotief is L'Hôpital's rule.

Another methods should be used too. Setting $ \gamma=1$ will be return $Q=[a K^{-\rho} +(1-a) L^{-\rho} ]^{-\frac{1}{\rho}}$ and $$Q^{-\rho}=[a K^{-\rho} +(1-a) L^{-\rho} ]$$ By The total derivative via differentials we will have $$-\rho Q^{-\rho-1}dQ=- a\rho K^{-\rho-1}dK -(1-a)\rho L^{-\rho-1}dL $$ With some manupulations our main equation will be obtained.

$$ dQ= a {(\frac{Q}{ K})}^{1+\rho}dK +(1-a){(\frac{Q}{ L})}^{1+\rho}dL $$

Linear Function : $\lim_{\rho\to -1}{dQ}\Rightarrow Q=aK+(1-a)L$

Cobb-Douglas Function : $$\lim_{\rho\to 0}{dQ}\Rightarrow \frac{1}{Q}dQ= a {(\frac{1}{ K})} dK +(1-a){(\frac{1}{ L})} dL$$ Taking the Integral from both side would produce

$$ \int\frac{1}{Q}dQ= a \int {(\frac{1}{ K})} dK +(1-a)\int{(\frac{1}{ L})} dL$$

$$Q=K^a L^{(1-a)}e^{C}=AK^a L^{(1-a)}$$

Leontief Function: $\lim_{\rho\to \infty}{dQ}\Rightarrow min(aK,(1-a)L)$