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The understanding that I am not clear is in when do homothetic preferences represent a utility function and vice-versa. My solution to the problem is posted below the problem:

A consumer’s preferences are described by a utility function that is homogeneous of degree two: For all $\alpha > 0$ and $x \in R^{L}_{+} $ ,

$u(\alpha x) = \alpha^2 u(x)$

The problem that I am not getting clear is: Q) "Are this consumer’s preferences homothetic? Show that they are or give a counterexample."

My solution:

According to Mas Colell et al. "Microeconomic Theory" (chapter 3, page 50) enter image description here

Therefore, this given consumer's preferences are not homothetic as it doesn't generate a utility function that is homogeneous of degree 1 (HOD(1)). A counter example would be a utility function that is HOD(1) like the Cobb Douglas Utility Function

$ U(x_1, x_2) = x_{1}^{\alpha} x_{2}^{1-\alpha} $

To conclude, this consumer's preferences are not homothetic as it represents a utility function of HOD(2). While , according to Mas Colell et al. preference $\pmb{\succsim}$ is homothetic $\textbf{if and only if}$ it admits a utility function that is HOD(1).

Could you please help me in understanding where I am going wrong with what Mas-Colell mentioned above "necessary and sufficient condition" and how a utility function that is HOD(2) implies that $\pmb{\succsim}$ is homothetic.

Thanks.

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First of all, in order to provide a counterexample, you need to construct a utility function that is homogeneous of degree two, but is not homothetic. Therefore, the counterexample you gave in your solution doesn't work.

To prove the statement directly, let $u(x)$ be a utility representation that is homogeneous of degree two. That is, $u(\alpha x)=\alpha^2 u(x)$. Therefore, if $x\sim y$, which means $u(x)=u(y)$, we have $$u(\alpha x)= \alpha^2 u(x)=\alpha^2 u(y)=u(\alpha y).$$ This means $\alpha x\sim \alpha y$, and hence the preferences are homothetic.

We can also use the proposition in MWG: A continuous $\succeq$ is homothetic if and only if it admits a utility function $u(x)$ that is homogeneous of degree one. One caveat is that the utility representation is unique up to monotone transformations, so even if one representation $u(x)$ is not homogeneous of degree one, the preferences could still be homothetic if a monotone transformation of the representation, $\phi(u(x))$, is.

In this question, if we consider a monotone transformation $\hat{u} (x)=(u(x))^\frac{1}{2}$, this $\hat{u}(x)$ still represents the preferences $\succeq$. Notice that $$\hat{u} (\alpha x)=(u(\alpha x))^\frac{1}{2}=(\alpha^2 u(x))^\frac{1}{2}=\alpha (u(x))^\frac{1}{2}=\alpha\hat{u} (x),$$ meaning that this new representation is homogeneous of degree one. Therefore, by the proposition above, the preferences are homothetic.

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actually if a utility function is HOD(2) then it is not HOD(1), therefore, you can conclude (as Mas Colell states), that it does not represent an homotetic preference relation.

As a supportive example, consider an economy with just 1 good and a consumer whose preferences are HOD(2) and can be represented by the utiliy function $u(x)=x^2$.

It is easy to see that $u(\cdot)$ is homogeneous of degree 2 (for $\alpha>0$): $$u(\alpha x)=(\alpha x)^2=\alpha^2x^2=\alpha^2 u(x)$$ however you also know that $\forall \alpha\neq1$ you have $\alpha^2u(x)\neq \alpha u(x)$ therefore you conclude that $u(\cdot)$ is not HOD(1). At this point you can use Mas Colell's proposition and conclude that this kind of preferences is not necessarily homotetic.

What Mas Colell means with "necessary and sufficient" is that as long as $\succeq$ is continuous, homotetic and rational you can always rationalize it as with a HOD(1) utility function; furthermore, as long as the utility representation of a preference relation is HOD(1), you can always prove that $U(a)>U(b)\Rightarrow a\succeq b$ where $\succeq$ is rational, continuous and homotetic (cfr. Mas Colell pag.96, excercise 3.C.5).

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    $\begingroup$ The symmetric Cobb-Douglas preference is homothetic. Here is a HOD(1) representation: $$ U(x,y) = \left(xy\right)^{1/2} $$ Here is a HOD(2) representation: $$ U(x,y) = xy. $$ $\endgroup$ – Giskard Sep 30 '17 at 20:06

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