<p>Yes.
We know that a monotonic transformation of a utility function still represents the same preferences and as the old utility function represented homothetic preferences the new one does, too.</p>

<p>As an easy example you could look at Cobb-Douglas utility functions of the form $u(x,y) = a\left(x y\right)^\alpha$. For $\alpha = \frac12$ the utility function is homogeneous of degree 1, but for every $\alpha,a&gt;0,$ the preference relation is homogeneous of degree 1.</p>

<p>We never used homothetic as a property of the utility function to answer your question (as it works for every utility function).
A homothetic utility function is, to the best of my knowledge, not very clearly defined, I have see two different versions:</p>

<ol>
<li>Homothetic as different name for homogeneous of degree 1</li>
<li>A homothetic utility function is a utility function that represents
a homothetic preference relation.</li>
</ol>

<p>Which means, with definition 2 Cobb-Douglas utility functions with equal weights are always homothetic, with definition 2 only some of them are:
$$u(tx,ty) = (tx)^\frac12 (ty)^\frac12 = t xy = t u(x,y)$$
and some are not
$$u(tx,ty) = (tx) (ty) = t^2 xy \neq t u(x,y)$$</p>

CC BY-SA 3.0
2015-04-05T14:14:41.893Z
Wait, I thought Cobb-Douglas was homothetic for all the utility functions because $$u(tx,ty) = (tx)^\alpha (ty)^{1-\alpha}= t x^\alpha y^{1-\alpha} $$ is that wrong? Is only the preference relation homothetic for all $\alpha$? If so, why?
Stan Shunpike
CC BY-SA 3.0
2015-04-05T14:58:01.707Z
@StanShunpike Your CD function and mine are not the same, they are both special cases. In mine both exponents are the same, in yours they add up to 1. I actually don't use homothetic for utility functions, as people seem to use different definitions for homothetic utility functions: One being "its preference relation is homothetic" the other one is "it is homogeneous of degree 1", I think I will try to make that clear in the answer. Just do the same calculations you have done with your utility function with mine (and $\alpha =1$), then you will see it is not homogeneous of degree 1.
The Almighty Bob
CC BY-SA 3.0
2015-04-05T15:10:10.893Z
So, I hope this helps.
The Almighty Bob
CC BY-SA 3.0
2015-04-05T17:28:43.637Z
So if I understand you correctly, in your answer you showed the special cases you and I mentioned are homothetic utility functions. They therefore have a homothetic preference relation. Is that the basic idea?
Stan Shunpike
CC BY-SA 3.0
2015-04-05T18:16:39.323Z
Mine is only homogeneous of degree 1 for the case $\alpha = 1/2$, yours always is. But having one is enough, as every other of mine is just a monotonic transformation of the one that is. But yes, it is the basic idea.
The Almighty Bob
CC BY-SA 3.0
2015-05-03T01:28:33.927Z
I don't think that I've ever seen your definition 1. Instead, I've seen homothetic functions defined as monotonic transformations of homogenous of degree 1 function. This is still not quite the same as definition 2, since it is possible for homothetic preferences (defined in the sense of "demanding constant proportions of goods as income changes") not to be representable with homogenous of degree 1 functions.
nominally rigid
CC BY-SA 3.0
2015-04-05T08:21:34.320Z
2015-04-05T15:09:37.853Z
2015-04-05T15:09:37.853Z
The Almighty Bob
<p>Does a monotonic transformation of a homothetic utility function imply the preference relation on the set of consumption bundles is still homothetic?   </p>

<p>Obviously, if a utility function on a set of  consumption bundles is homothetic, then the preference relation is homothetic. </p>

<consumer-theory>
Does a monotonic transformation of a homothetic utility function imply the preference relation on the set of consumption bundles is still homothetic?
Stan Shunpike
# Does a monotonic transformation of a homothetic utility function imply the preference relation on the set of consumption bundles is still homothetic?

(Asked by: username_0 on 2015-04-05)

Does a monotonic transformation of a homothetic utility function imply the preference relation on the set of consumption bundles is still homothetic?

Obviously, if a utility function on a set of consumption bundles is homothetic, then the preference relation is homothetic.

**Tags:** consumer-theory


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## 1 answers

### Answer by username_1 on 2015-04-05. Score: 8 | Accepted answer

Yes.
We know that a monotonic transformation of a utility function still represents the same preferences and as the old utility function represented homothetic preferences the new one does, too.

As an easy example you could look at Cobb-Douglas utility functions of the form $u(x,y) = a\left(x y\right)^\alpha$. For $\alpha = \frac12$ the utility function is homogeneous of degree 1, but for every $\alpha,a>0,$ the preference relation is homogeneous of degree 1.

We never used homothetic as a property of the utility function to answer your question (as it works for every utility function).
A homothetic utility function is, to the best of my knowledge, not very clearly defined, I have see two different versions:

1. Homothetic as different name for homogeneous of degree 1
2. A homothetic utility function is a utility function that represents
   a homothetic preference relation.

Which means, with definition 2 Cobb-Douglas utility functions with equal weights are always homothetic, with definition 2 only some of them are:
$$u(tx,ty) = (tx)^\frac12 (ty)^\frac12 = t xy = t u(x,y)$$
and some are not
$$u(tx,ty) = (tx) (ty) = t^2 xy \neq t u(x,y)$$

> **Comments:**
> **username_0**: Wait, I thought Cobb-Douglas was homothetic for all the utility functions because $$u(tx,ty) = (tx)^\alpha (ty)^{1-\alpha}= t x^\alpha y^{1-\alpha} $$ is that wrong? Is only the preference relation homothetic for all $\alpha$? If so, why?  
> **username_1**: @username_0 Your CD function and mine are not the same, they are both special cases. In mine both exponents are the same, in yours they add up to 1. I actually don't use homothetic for utility functions, as people seem to use different definitions for homothetic utility functions: One being "its preference relation is homothetic" the other one is "it is homogeneous of degree 1", I think I will try to make that clear in the answer. Just do the same calculations you have done with your utility function with mine (and $\alpha =1$), then you will see it is not homogeneous of degree 1.  
> **username_1**: So, I hope this helps.  
> **username_0**: So if I understand you correctly, in your answer you showed the special cases you and I mentioned are homothetic utility functions. They therefore have a homothetic preference relation. Is that the basic idea?  
> **username_1**: Mine is only homogeneous of degree 1 for the case $\alpha = 1/2$, yours always is. But having one is enough, as every other of mine is just a monotonic transformation of the one that is. But yes, it is the basic idea.  
> **username_2**: I don't think that I've ever seen your definition 1. Instead, I've seen homothetic functions defined as monotonic transformations of homogenous of degree 1 function. This is still not quite the same as definition 2, since it is possible for homothetic preferences (defined in the sense of "demanding constant proportions of goods as income changes") not to be representable with homogenous of degree 1 functions.  

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