Constant Elasticity of Substitution: Special Cases

Question

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. $U(x) = \min\left\{x_1,...,x_n\right\}$

Answer 1

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.

$\square$

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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\} $

$\square$

Answer 2

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)

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