Tastes in Cobb-Douglas vs CES: Different degree of homogeneity

Question

The question has been reworded based on Giskard's comment

---

If we have a Cobb-Douglas function:

$$U_1 = x_1^{\alpha_1} x_2^{\alpha_2}$$

The degree of homogenity depends on tastes $\alpha_1$ and $\alpha_2$. This is associated with returns to scale, which also depend on tastes $\boldsymbol{\alpha}$. We can change returns to scale by changing tastes. For example, if $\alpha_1 + \alpha_2 < 1$ then the function has decreasing returns to scale.

Simultaneously, we know, that Cobb-Douglas function is a special case of CES function:

$$U_2 = \left( \alpha_1 x_1^\rho + \alpha_2 x_2^\rho \right)^{\frac{1}{\rho}}$$

If $\rho = 0$, then the limit case of CES function provides results for Cobb-Douglas.

So, the thing I do not understand is, why, in general, returns to scale of CES are constant... And do not depend on tastes $\boldsymbol{\alpha}$ like in the case of CD?

In other words, why when CES is homogenous of degree 1 for all elements, the same does not hold for its special limit case?

What kind of Shenanigans are these?

The question

• Why does degree of homogeneity differ between CES and CD?
• Should limit operations conserve degree of homogeneity, or under what conditions do they do so, etc.
• Does this mean that, in principle, we cannot consider CD as a full-fledged special case of CES?

---

Proof:

$$U_2 = \left( \alpha_1 \left( s \cdot x_1 \right)^\rho + \alpha_2 \left( s \cdot x_2 \right)^\rho \right)^{\frac{1}{\rho}}$$

$$U_2 = \left( s^{\rho} \left( \alpha_1 x_1^\rho + \alpha_2 x_2^\rho \right) \right)^{\frac{1}{\rho}}$$

$$ U_2 = s \cdot \left( \alpha_1 x_1^\rho + \alpha_2 x_2^\rho \right)^{\frac{1}{\rho}}$$

Answer 1

(I am adding another answer since the question has been reworded.)

I think your question contains a misunderstanding which stems from your usage of the same parameter names $\alpha_1$ and $\alpha_2$ in both the CD and the CES function. As @tdm's answer shows, if you call the CES "taste parameters" $\alpha_1$ and $\alpha_2$ and take the limit of $\rho\to 0$, then the resulting CD function has "taste parameters" $\frac{\alpha_1}{\alpha_1+\alpha_2}$ and $\frac{\alpha_2}{\alpha_1+\alpha_2}$, which sum to $1$.

As a consequence, the resulting CD function always has constant RTS. Or, put the other way round, any CD function with non-constant RTS is simply not a special case of the class of CES functions you presented.

Answer 2

Consider the CES production function: $$ f(x_1, x_2) = (\alpha x_1^\rho + \beta x_2^\rho)^{1/\rho}. $$ The limit for $\rho$ going to zero of this function gives a Cobb-Douglas function: $$ g(x_1, x_2) = x_1^a x_2^b, $$ where $a = \frac{\alpha}{\alpha + \beta}$ and $b = \frac{\beta}{\alpha + \beta}$, so $a + b = 1$ which means that the limiting Cobb-Douglas will indeed have CRS.

Proof: Take the log of the CES: $$ \frac{\ln(\alpha x_1^\rho + \beta x_2^\rho)}{\rho}. $$ Take the limit for $\rho$ going to zero (using l'Hôpital): $$ \frac{\alpha}{\alpha + \beta} \ln(x_1) + \frac{\beta}{\alpha + \beta} \ln(x_2), $$ So we get the log of a Cobb-Douglas with parameters $a = \frac{\alpha}{\alpha + \beta}$ and $b = \frac{\beta}{\alpha + \beta}$.

No shenanigans!

Answer 3

In consumer theory, utility functions are ordinal and can thus be rescaled by any strictly increasing transformation. Therefore, "returns to scale" are meaningless for such a utility function. E.g. $u(x_1,x_2)=x_1^{1/3}x_2^{1/3}$ (with decreasing returns to scale) and $v(x_1,x_2)=(u(x_1,x_2))^2=x_1^{2/3}x_2^{2/3}$ (with increasing returns to scale) represent the same preferences.