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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.

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up vote 6 down vote accepted

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 $$u(tx,ty) = (tx) (ty) = t^2 xy \neq t u(x,y)$$

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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 Apr 5 '15 at 14:14
@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 Apr 5 '15 at 14:58
So, I hope this helps. – The Almighty Bob Apr 5 '15 at 15:10
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 Apr 5 '15 at 17:28
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 Apr 5 '15 at 18:16

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