Constant Elasticity of Substitution: Special Cases

Take an $n$-commodity constant elasticity of substitution utility function,

$$U = \left[\sum^n_{i=1} \alpha_i x^\rho_i \right]^\frac{1}{\rho}$$

How do we show the following:

• Show that as $\rho \rightarrow 0$ that this utility function represents preferences for Cobb-Douglass utility. $U(x) = \prod^n_{i=1} x_i^{\alpha_i}$
• Show that as $\rho \rightarrow - \infty$ that this utility function has the indifference curves as in Leontief utility. $U(x) = \min\left\{x_1,...,x_n\right\}$

We know that if $u$ represents $\succeq$ on $X$, then for any strictly increasing function $f: \mathbb{R} \rightarrow \mathbb{R}$, then $v(x) = f(u(x))$ represents $\succeq$ on $X$

($X$ in this case is $\mathbb{R^n}$)

Consider $v(x, \rho) = \ln(u(x, \rho)) - \frac{\ln\left[\sum^n_{i=1}\alpha_i \right]}{\rho}$, which is strictly increasing.

$$v(x, \rho) = \frac{\ln\left[\sum^n_{i=1} \alpha_i x^\rho_i \right]}{\rho} - \frac{\ln\left[\sum^n_{i=1}\alpha_i \right]}{\rho} = \frac{\ln\left[\sum^n_{i=1} \alpha_i x^\rho_i \right] - \left [\ln\sum^n_{i=1}\alpha_i \right]}{\rho}$$

The limit of this as $\rho \rightarrow 0$ is indeterminate, $\frac{0}{0}$. So we can use L'Hopital's Rule and take the derivative with respect to $\rho$ of the numerator and denominator.

$$\lim_{\rho \rightarrow 0} \frac{\ln\left[\sum^n_{i=1} \alpha_i x^\rho_i \right] - \left [\ln\sum^n_{i=1}\alpha_i \right]}{\rho} = \lim_{\rho \rightarrow 0} \frac{1}{\sum^n_{i=1} \alpha_i x_i^\rho} \cdot \left(\sum^n_{i=1} \alpha_i x_i^\rho \ln x_i\right)$$

by the Chain Rule.

$$= \lim_{\rho \rightarrow 0} \frac{\sum^n_{i=1} \alpha_i x_i^\rho \ln x_i}{\sum^n_{i=1} \alpha_i x_i^\rho} = \frac{\sum^n_{i=1} \alpha_i \ln x_i}{\sum^n_{i=1} \alpha_i} = \frac{1}{\sum^n_{i=1} \alpha_i} \cdot \ln\left(\prod^n_{i=1} x_i^{\alpha_i}\right)$$

Consider $w(x, \rho) = \mathrm{e}^{(\sum^n_{i=1} \alpha_i) \cdot v(x, \rho)}$, which is another monotonic transformation, strictly increasing. So $w$ still represents the same preference as $u$.

$$\lim_{\rho \rightarrow 0} w(x, \rho) = \mathrm{e}^{(\sum^n_{i=1} \alpha_i) \cdot \lim_{\rho \rightarrow 0} v(x, \rho)} = \prod^n_{i=1} x_i^{\alpha_i}$$

which is a Cobb-Douglas function.

$\square$

To show the second point, it is sufficient to show that

$$\lim_{\rho \rightarrow -\infty} u(x) = \left\{x_k \ \forall j \neq k \mid x_j \geq x_k \right\}$$

$$u(x) = \left[\sum^n_{i=1} \alpha_i x^\rho_i \right]^\frac{1}{\rho} = x_k \left[(\sum^n_{i=1, i \neq k} \alpha_i x^\rho_i) + \alpha_k \right]^\frac{1}{\rho}$$

$(\frac{x_j}{x_k})^\rho \rightarrow 0$ as $\rho \rightarrow -\infty$ if $x_j > x_k$

$(\frac{x_j}{x_k})^\rho \rightarrow 1$ as $\rho \rightarrow -\infty$ if $x_j = x_k$

So

$$\lim_{\rho \rightarrow -\infty} x_k \left[(\sum^n_{i=1, i \neq k} \alpha_i x^\rho_i) + \alpha_k \right]^\frac{1}{\rho} = x_k$$

since $1/\rho \rightarrow 0$ and a constant to the zeroth power is 1.

Construct a similar argument for any $k$. Thus $\lim_{\rho \rightarrow -\infty} u(x) = \min \left\{x_1,...,x_n \right\}$

$\square$

• Is this a request for confirmation? You posted the question and the answer within 1 minute. Oct 18 '15 at 19:44
• I was under the impression that you could make Q+A style posts by asking a question and an answer at the same time. If it is more appropriate for this to be a community wiki post, then my bad. It's also a request for a better answer if one exists. Oct 18 '15 at 19:50
• @KitsuneCavalry There is nothing wrong with posting an answer to your own question---even if it is immediately. In fact, we should encourage this! The goal is to increase the amount of good content available to the site so that it can grow. Please keep it up! :) Oct 24 '15 at 12:36

These are standard mathematical results for generalized means. For example,for the $\rho \rightarrow 0$ result, write (setting without loss of generality $\sum_{i=1}^na_i =1$),

$$U = \left[\sum^n_{i=1} \alpha_i x^\rho_i \right]^\frac{1}{\rho} = \exp\left\{\frac 1\rho\ln \left(\sum^n_{i=1} \alpha_i x^\rho_i\right)\right\}$$

Apply L'Hopital's rule on

$$\frac 1\rho\ln \left(\sum^n_{i=1} \alpha_i x^\rho_i\right)$$

to get

$$\frac {\sum^n_{i=1} \alpha_ix_i^{\rho}\ln x_i}{\sum^n_{i=1} \alpha_i x^\rho_i} \rightarrow \sum^n_{i=1} \alpha_i\ln x_i,\;\;\; \rho\rightarrow 0$$

So (by the uniform continuity of the exponential function)

$$\rho\rightarrow 0,\;\;\; U = \exp\left\{\frac 1\rho\ln \left(\sum^n_{i=1} \alpha_i x^\rho_i\right)\right\} \rightarrow \exp\left\{\sum^n_{i=1} \alpha_i\ln x_i\right\} = \prod_{i=1}^nx_i^{a_i}$$

• Looks good! A very cool streamlined way of doing this problem. Oct 19 '15 at 1:24
• @KitsuneCavalry Well, I try not to forget that in Economics we use a tiny fraction of Mathematics, and mathematicians are likely to have proven a ton of properties before us and in a more compact and abstract level. The above is a simpler version of the proof to be found in Hardy, Littlewood & Polya "Inequalities" book, ch. 2 (they use a Taylor expansion). As for the second result, $\rho \rightarrow -\infty$, they prove it in five lines (of text mostly), together with the result for $\rho \rightarrow \infty$. Oct 19 '15 at 10:22
• I'm trying to prove this for a two input case: $f(k,l)=(k^\rho+l^\rho)^(1/\rho)$. (Sorry for the bad notation, I am not being able to put the fraction $1/\rho$ as an exponent. I have a natural logarithm of the function but I am not getting 0/0 form so I can't see how to apply L'opital rule. Aug 8 '18 at 15:37
• @PGupta You should. Just follow my post exactly setting $n=2$ and you will get the $0/0$ form. Aug 8 '18 at 15:43
• After taking logs I get $(1/\rho)ln(k^\rho+l^\rho)$. The numerator tends to $ln 2$ when $\rho$ tends to 0. So, how do we get 0/0 form? Aug 8 '18 at 15:51