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 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?


1 Answer 1


Note that the wikipedia article 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.

  • $\begingroup$ Thank you. This definitely answers my question. Do you by any chance, also have a general proof, how the two definitions are equivalent? $\endgroup$ Commented Jul 30, 2017 at 3:07
  • 2
    $\begingroup$ 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. $\endgroup$
    – Tobias
    Commented Jul 30, 2017 at 9:40
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    $\begingroup$ 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 $k^{a+b}=k\forall a+b=1$ $\endgroup$
    – Brennan
    Commented Jul 31, 2019 at 7:12

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