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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 $k\neq 1$?

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    $\begingroup$ This seems to be a homework question with no prior effort of solving it, see: meta.economics.stackexchange.com/questions/24/… $\endgroup$ – FooBar Nov 28 '14 at 22:27
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    $\begingroup$ It is certainly an on-topic subject, but a low-quality question. Even if it is not homework Huseyin, we expect from you to a) Be careful with your notation (you used $\rho$ and $p$) and b) Contribute some thoughts and ways you have tried to solve the problem. We are here to help people who help themselves, and not to offer professional services pro bono. $\endgroup$ – Alecos Papadopoulos Nov 28 '14 at 23:04
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    $\begingroup$ Mathematics do things differently to pretty much the entire rest of the stackexchange network. Only on math.se can you submit problems for other people to solve without showing effort. Please save that sort of question for math.se, not here. $\endgroup$ – EnergyNumbers Nov 29 '14 at 4:31
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    $\begingroup$ When you say "I need to prove" without any indication of why you need to prove it, people are going to assume this is homework. $\endgroup$ – Steven Landsburg Nov 30 '14 at 4:16
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    $\begingroup$ @Huseyin Now that the question has been re-opened and an answer has been provided, won't you post your answer for the Cobb-Douglas limit? $\endgroup$ – Alecos Papadopoulos Dec 2 '14 at 20:18
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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.

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

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    $\begingroup$ (+1) I like especially how the Cobb-Douglas function is obtained. $\endgroup$ – Alecos Papadopoulos Dec 7 '14 at 23:45
  • $\begingroup$ Thanks @AlecosPapadopoulos. but I don't know why somebodies make dislike this post yet? I think this type of questions may provide brain storm at least to me. $\endgroup$ – Huseyin Dec 8 '14 at 20:41
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    $\begingroup$ Strictly speaking Huseyin, they are right: you should have included at least part of your answer in your question: "here is my way of doing things, is there some other way?" $\endgroup$ – Alecos Papadopoulos Dec 8 '14 at 20:56
  • $\begingroup$ Is taking a differential and integrating "equivalent" to taking a limit? In general, can we take differential and integrate to find a limit? Or is this a special application? $\endgroup$ – PGupta Nov 16 '18 at 5:15

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