<p>Note that the <a href="https://en.wikipedia.org/wiki/Homothetic_preferences" rel="nofollow noreferrer">wikipedia article</a>  is very specific:</p>

<blockquote>
  <p>[...] defined a preference to be <em>homothetic</em>, if they <strong>CAN</strong> be represented by <strong>A</strong> utility function [...]</p>
</blockquote>

<p>You chose a specific utility function to represent your Cobb-Douglas preferences. However there are infinitely many others. All monotonic transformations of your utility function represent the same preference. Take
<span class="math-container">$$
\hat{u}(x_1,x_2) = \left(u(x_1,x_2)\right)^{\frac{1}{a+b}} = x_1^{\frac{a}{a+b}} \cdot x_2^{\frac{b}{a+b}}.
$$</span>
As <span class="math-container">$\hat{u}$</span> is a monotonic transformation of <span class="math-container">$u$</span>, it represents the same preference. It is straightforward to check that <span class="math-container">$\hat{u}$</span> fullfils the condition set forth in the wiki article. So there is indeed such a utility function, that also represents the preference, hence the preference is homothetic. </p>

CC BY-SA 3.0
2017-07-30T03:07:47.223Z
Thank you. This definitely answers my question. Do you by any chance, also have a general proof, how the two definitions are equivalent?
user7802048
CC BY-SA 3.0
2017-07-30T09:40:28.617Z
It's easy to show that they represent the same preferences. It's a general property of u functions that they are ordinal, not cardinal, and that any monotone transformation of u represents the same preference relation. You can just show that u(a1, b1) > u(a2, b2) always implies that uhat(a1, b1) > uhat(a2, b2) and you are done imo.
Tobias
CC BY-SA 4.0
2019-07-31T07:12:14.057Z
Also note that in the Cobb-Douglas case with Constant Returns to Scale (CRS) your final equation holds true. CRS would imply any case such that $a+b=1$ and hence $k^{a+b}=k\forall a+b=1$
Brennan
CC BY-SA 4.0
2017-07-29T19:06:40.163Z
2019-07-31T06:25:35.317Z
2019-07-31T06:25:35.317Z
Giskard
<p>Our lecture defined a preference to be <em>homothetic</em>, if the following is true:</p>

<p>$$(x_1, x_2)  \thicksim (y_1, y_2) \Leftrightarrow (kx_1, kx_2) \thicksim (ky_1, ky_2)$$</p>

<p><em>Cobb-Douglas</em> preferences can be displayed as some utility function of the following form: </p>

<p>$$u(x_1, x_2) = x_1^a \cdot x_2^b$$
Therefore:
$$(x_1, x_2)  \thicksim (y_1, y_2) \\ 
\Leftrightarrow x_1^a \cdot x_2^b = y_1^a \cdot y_2^b \\
\Leftrightarrow k^ax_1^a \cdot k^bx_2^b = k^ay_1^a \cdot k^by_2^b \\
 \Leftrightarrow (kx_1, kx_2)  \thicksim (ky_1, ky_2)$$</p>

<p>With this argumentation the <em>Cobb-Douglas</em> preferences <strong>should be homothetic</strong>. </p>

<p>The <a href="https://en.wikipedia.org/wiki/Homothetic_preferences" rel="nofollow noreferrer">wikipedia article</a> about <em>Homothetic preferences</em> however defined a preference to be <em>homothetic</em>, if they can be represented by a utility function  and the following is true:</p>

<p>$$ u(kx_1, kx_2) = k \cdot u(x_1, x_2)$$
And I am pretty sure, that this is <strong>not true</strong> for <em>Cobb Douglas</em> preferences:</p>

<p>$$ u(kx_1, kx_2) = (kx_1)^a (kx_2)^b = k^{a+b} x_1^a x_2^b \neq k \cdot u(x_1, x_2)$$</p>

<p>So what am I missing here? Are the definitions not equivalent? Did I calculate something wrong?</p>

<utility><consumer-theory><preferences><cobb-douglas>
Are Cobb-Douglas preferences homothetic?
user7802048
# Are Cobb-Douglas preferences homothetic?

(Asked by: username_0 on 2017-07-29)

Our lecture defined a preference to be *homothetic*, if the following is true:

$$(x\_1, x\_2) \thicksim (y\_1, y\_2) \Leftrightarrow (kx\_1, kx\_2) \thicksim (ky\_1, ky\_2)$$

*Cobb-Douglas* preferences can be displayed as some utility function of the following form:

$$u(x\_1, x\_2) = x\_1^a \cdot x\_2^b$$
Therefore:
$$(x\_1, x\_2) \thicksim (y\_1, y\_2) \\
\Leftrightarrow x\_1^a \cdot x\_2^b = y\_1^a \cdot y\_2^b \\
\Leftrightarrow k^ax\_1^a \cdot k^bx\_2^b = k^ay\_1^a \cdot k^by\_2^b \\
\Leftrightarrow (kx\_1, kx\_2) \thicksim (ky\_1, ky\_2)$$

With this argumentation the *Cobb-Douglas* preferences **should be homothetic**.

The [wikipedia article](https://en.wikipedia.org/wiki/Homothetic_preferences) about *Homothetic preferences* however defined a preference to be *homothetic*, if they can be represented by a utility function and the following is true:

$$ u(kx\_1, kx\_2) = k \cdot u(x\_1, x\_2)$$
And I am pretty sure, that this is **not true** for *Cobb Douglas* preferences:

$$ u(kx\_1, kx\_2) = (kx\_1)^a (kx\_2)^b = k^{a+b} x\_1^a x\_2^b \neq k \cdot u(x\_1, x\_2)$$

So what am I missing here? Are the definitions not equivalent? Did I calculate something wrong?

**Tags:** utility, consumer-theory, preferences, cobb-douglas

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## 1 answers

### Answer by username_1 on 2017-07-29. Score: 5 | Accepted answer

Note that the [wikipedia article](https://en.wikipedia.org/wiki/Homothetic_preferences) is very specific:

> [...] defined a preference to be *homothetic*, if they **CAN** be represented by **A** utility function [...]

You chose a specific utility function to represent your Cobb-Douglas preferences. However there are infinitely many others. All monotonic transformations of your utility function represent the same preference. Take
$$
\hat{u}(x\_1,x\_2) = \left(u(x\_1,x\_2)\right)^{\frac{1}{a+b}} = x\_1^{\frac{a}{a+b}} \cdot x\_2^{\frac{b}{a+b}}.
$$
As $\hat{u}$ is a monotonic transformation of $u$, it represents the same preference. It is straightforward to check that $\hat{u}$ fullfils the condition set forth in the wiki article. So there is indeed such a utility function, that also represents the preference, hence the preference is homothetic.

> **Comments:**
> **username_0**: Thank you. This definitely answers my question. Do you by any chance, also have a general proof, how the two definitions are equivalent?  
> **username_2**: It's easy to show that they represent the same preferences. It's a general property of u functions that they are ordinal, not cardinal, and that any monotone transformation of u represents the same preference relation. You can just show that u(a1, b1) > u(a2, b2) always implies that uhat(a1, b1) > uhat(a2, b2) and you are done imo.  
> **username_3**: Also note that in the Cobb-Douglas case with Constant Returns to Scale (CRS) your final equation holds true. CRS would imply any case such that $a+b=1$ and hence $k^{a+b}=k\forall a+b=1$  

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