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?


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

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

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

  • $\begingroup$ A possible rewording of the question is, that for a specific series of CES functions $U_{\rho}(x_1,x_2) = \left( \alpha_1 x_1^\rho + \alpha_2 x_2^\rho \right)^{\frac{1}{\rho}}$ you have degree of homogeneity 1 for all elements. But for the limit case $\lim \rho \to 0$ you have a degree of homogeneity $\alpha_1+\alpha_2$. Should limit operations conserve degree of homogeneity, or under what conditions do they do so, etc. $\endgroup$
    – Giskard
    Commented Mar 6 at 16:39
  • $\begingroup$ @Giskard Yeah, that's it. I will try to reword it $\endgroup$
    – Athaeneus
    Commented Mar 6 at 19:25

3 Answers 3


(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.

  • 1
    $\begingroup$ Thank you. Now, I understand completely what I have missed. I can also see and finally appreciate how great @tdm's answer really was, while I did not understand it at the time. The usage of the same parameter names is what really caused my confusion. So, I would like to thank you both for your great answers, clarifying this. $\endgroup$
    – Athaeneus
    Commented Mar 8 at 7:45

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!

  • $\begingroup$ The shenanigans, as I see it, come from the fact that we could introduce different returns to scale to cobb-douglas by shifting the taste parameters, but if we change taste parameters within CES, the returns to scale stay the same no matter what. $\endgroup$
    – Athaeneus
    Commented Mar 6 at 15:19
  • $\begingroup$ @Athaeneus I do not see the issue, you can easily modify the RTS of the CES by adding a parameter $\delta$ to get $(\alpha x_1^\rho + \beta x_2^\rho)^{\beta/\rho}$ $\endgroup$
    – tdm
    Commented Mar 6 at 15:32
  • $\begingroup$ But that is not the same as those tastes, right? By an additional parameter, I could change RTS, but the interesting thing is that RTS within CD depend on tastes. Say, we have $\alpha_1 = 0.5$ and $\alpha_2 = 0.5$, we would have constant RTS both within CD and CES... BUT... If we change $\alpha_2 = 0.1$, CES still has constant RTS, but CD has decreasing. Am I correct and are these two different things (than adding additional parameter in exponent of CES) or am I missing something? $\endgroup$
    – Athaeneus
    Commented Mar 6 at 15:39

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.

  • $\begingroup$ Maybe I did not express myself the best. The thing I do not understand is why cobb-douglass can allow us to control for returns to scale by adjusting the taste parameters, while CES function does not have this property. If we adjust the taste parameters within CES, the original funtion will still have constant returns, which is contradictory to CD. $\endgroup$
    – Athaeneus
    Commented Mar 6 at 15:23
  • $\begingroup$ @Athaeneus, O.k., I understand. The reworded question is much clearer now. $\endgroup$
    – VARulle
    Commented Mar 7 at 12:12

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