<p>Consider the CES production function:
<span class="math-container">$$
f(x_1, x_2) = (\alpha x_1^\rho + \beta x_2^\rho)^{1/\rho}.
$$</span>
The limit for <span class="math-container">$\rho$</span> going to zero of this function gives a Cobb-Douglas function:
<span class="math-container">$$
g(x_1, x_2) = x_1^a x_2^b,
$$</span>
where <span class="math-container">$a = \frac{\alpha}{\alpha + \beta}$</span> and <span class="math-container">$b = \frac{\beta}{\alpha + \beta}$</span>, so <span class="math-container">$a + b = 1$</span> which means that the limiting Cobb-Douglas will indeed have CRS.</p>
<p>Proof: Take the log of the CES:
<span class="math-container">$$
\frac{\ln(\alpha x_1^\rho + \beta x_2^\rho)}{\rho}.
$$</span>
Take the limit for <span class="math-container">$\rho$</span> going to zero (using l'Hôpital):
<span class="math-container">$$
\frac{\alpha}{\alpha + \beta} \ln(x_1) + \frac{\beta}{\alpha + \beta} \ln(x_2),
$$</span>
So we get the log of a Cobb-Douglas with parameters <span class="math-container">$a = \frac{\alpha}{\alpha + \beta}$</span> and <span class="math-container">$b = \frac{\beta}{\alpha + \beta}$</span>.</p>
<p>No shenanigans!</p>

CC BY-SA 4.0
2024-03-06T15:19:28.143Z
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.
Athaeneus
CC BY-SA 4.0
2024-03-06T15:32:43.237Z
@Athaeneus I do not see the issue, you can easily modify the RTS of the CES by adding a parameter $\delta$ to get $(\alpha x_1^\rho + \beta x_2^\rho)^{\beta/\rho}$
tdm
CC BY-SA 4.0
2024-03-06T15:39:26.500Z
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?
Athaeneus
CC BY-SA 4.0
2024-03-06T12:57:03.203Z
2024-03-07T12:13:05.187Z
2024-03-07T12:13:05.187Z
tdm
<p>In consumer theory, utility functions are ordinal and can thus be rescaled by any strictly increasing transformation. Therefore, &quot;returns to scale&quot; are meaningless for such a utility function. E.g. <span class="math-container">$u(x_1,x_2)=x_1^{1/3}x_2^{1/3}$</span> (with decreasing returns to scale) and <span class="math-container">$v(x_1,x_2)=(u(x_1,x_2))^2=x_1^{2/3}x_2^{2/3}$</span> (with increasing returns to scale) represent the same preferences.</p>

CC BY-SA 4.0
2024-03-06T15:23:47.030Z
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.
Athaeneus
CC BY-SA 4.0
2024-03-07T12:12:16.690Z
@Athaeneus, O.k., I understand. The reworded question is much clearer now.
VARulle
CC BY-SA 4.0
2024-03-06T13:04:16.553Z
2024-03-06T13:04:16.553Z
VARulle
<p>(I am adding another answer since the question has been reworded.)</p>
<p>I think your question contains a misunderstanding which stems from your usage of the same parameter names <span class="math-container">$\alpha_1$</span> and <span class="math-container">$\alpha_2$</span> in both the CD and the CES function. As @tdm's answer shows, if you call the CES &quot;taste parameters&quot; <span class="math-container">$\alpha_1$</span> and <span class="math-container">$\alpha_2$</span> and take the limit of <span class="math-container">$\rho\to 0$</span>, then the resulting CD function has &quot;taste parameters&quot; <span class="math-container">$\frac{\alpha_1}{\alpha_1+\alpha_2}$</span> and <span class="math-container">$\frac{\alpha_2}{\alpha_1+\alpha_2}$</span>, which sum to <span class="math-container">$1$</span>.</p>
<p>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 <em>not</em> a special case of the class of CES functions you presented.</p>

CC BY-SA 4.0
2024-03-08T07:45:33.313Z
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.
Athaeneus
CC BY-SA 4.0
2024-03-07T12:42:56.423Z
2024-03-07T12:42:56.423Z
VARulle
<p><strong>The question has been reworded based on Giskard's comment</strong></p>
<hr />
<p>If we have a Cobb-Douglas function:</p>
<p><span class="math-container">$$U_1 = x_1^{\alpha_1} x_2^{\alpha_2}$$</span></p>
<p>The degree of homogenity depends on tastes <span class="math-container">$\alpha_1$</span> and <span class="math-container">$\alpha_2$</span>. This is associated with returns to scale, which also <strong>depend</strong> on tastes <span class="math-container">$\boldsymbol{\alpha}$</span>. We can change returns to scale by changing tastes. For example, if <span class="math-container">$\alpha_1 + \alpha_2 &lt; 1$</span> then the function has decreasing returns to scale.</p>
<p>Simultaneously, we know, that Cobb-Douglas function is a special case of CES function:</p>
<p><span class="math-container">$$U_2 = \left( \alpha_1 x_1^\rho + \alpha_2 x_2^\rho \right)^{\frac{1}{\rho}}$$</span></p>
<p>If <span class="math-container">$\rho = 0$</span>, then the limit case of CES function provides results for Cobb-Douglas.</p>
<p>So, the thing I do not understand is, why, in general, <strong>returns to scale of CES are constant... And do not depend on tastes <span class="math-container">$\boldsymbol{\alpha}$</span> like in the case of CD?</strong></p>
<p>In other words, why when <strong>CES is homogenous of degree 1 for all elements, the same does not hold for its special limit case?</strong></p>
<p>What kind of Shenanigans are these?</p>
<h2>The question</h2>
<ul>
<li>Why does degree of homogeneity differ between CES and CD?</li>
<li>Should limit operations conserve degree of homogeneity, or under what conditions do they do so, etc.</li>
<li>Does this mean that, in principle, we cannot consider CD as a full-fledged special case of CES?</li>
</ul>
<hr />
<p><em>Proof:</em></p>
<p><span class="math-container">$$U_2 = \left( \alpha_1 \left( s \cdot x_1 \right)^\rho + \alpha_2 \left( s \cdot x_2 \right)^\rho \right)^{\frac{1}{\rho}}$$</span></p>
<p><span class="math-container">$$U_2 = \left( s^{\rho} \left( \alpha_1 x_1^\rho + \alpha_2 x_2^\rho \right) \right)^{\frac{1}{\rho}}$$</span></p>
<p><span class="math-container">$$ U_2 = s \cdot \left(   \alpha_1 x_1^\rho + \alpha_2 x_2^\rho \right)^{\frac{1}{\rho}}$$</span></p>

CC BY-SA 4.0
2024-03-06T16:39:06.263Z
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.
Giskard
CC BY-SA 4.0
2024-03-06T19:25:37.223Z
@Giskard Yeah, that's it. I will try to reword it
Athaeneus
<microeconomics><cobb-douglas><ces-function><returns-to-scale>
Tastes in Cobb-Douglas vs CES: Different degree of homogeneity
Athaeneus
# Tastes in Cobb-Douglas vs CES: Different degree of homogeneity

(Asked by: username_0 on 2024-03-06)

**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:*

$$U\_2 = \left( \alpha\_1 \left( s \cdot x\_1 \right)^\rho + \alpha\_2 \left( s \cdot x\_2 \right)^\rho \right)^{\frac{1}{\rho}}$$

$$U\_2 = \left( s^{\rho} \left( \alpha\_1 x\_1^\rho + \alpha\_2 x\_2^\rho \right) \right)^{\frac{1}{\rho}}$$

$$ U\_2 = s \cdot \left( \alpha\_1 x\_1^\rho + \alpha\_2 x\_2^\rho \right)^{\frac{1}{\rho}}$$

**Tags:** microeconomics, cobb-douglas, ces-function, returns-to-scale
> **Comments:**
> **username_1**: 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.  
> **username_0**: @Giskard Yeah, that's it. I will try to reword it  

---

## 3 answers

### Answer by username_3 on 2024-03-07. Score: 2 | Accepted answer

(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.

> **Comments:**
> **username_0**: 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.  

---

### Answer by username_2 on 2024-03-06. Score: 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!

> **Comments:**
> **username_0**: 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.  
> **username_2**: @Athaeneus I do not see the issue, you can easily modify the RTS of the CES by adding a parameter $\delta$ to get $(\alpha x_1^\rho + \beta x_2^\rho)^{\beta/\rho}$  
> **username_0**: 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?  

---

### Answer by username_3 on 2024-03-06. Score: 1

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.

> **Comments:**
> **username_0**: 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.  
> **username_3**: @Athaeneus, O.k., I understand. The reworded question is much clearer now.  

---

