Someone gave me a proof of this, but I am not sure if it is correct.

Let $B(p,w) = \{x: p\cdot x \leq w\}$ (the budget set). Then:

\begin{align} x(p,w) &= \arg \max_{x\in B(p,w)} u(x)\\ &=\arg \max_{\alpha x\in \alpha B(p,w)} u(\alpha x) \\ &=\arg \max_{y\in B(p,\alpha w)} u(y) \\ &=\frac{1}{\alpha} \arg \max_{y\in B(p,\alpha w)} u(y) \\ &=\frac{1}{\alpha} x(p, \alpha w) \end{align} Where the result follows from taking $\alpha=\frac{1}{w}$.

Is this proof correct (I am not sure of the middle three equalities)? Where is homotheticity used?

EDIT: A monotone preference relation $\succsim$ on $X= \mathbb{R}^{L}_{+}$ is homothetic if all indifference sets are related by proportional expansion along rays; that is, if $x \sim y$, then $\alpha x \sim \alpha y$ for any $\alpha \geq 0$.

Also, recall a continuous $\succsim$ on $X = \mathbb{R}_{+}^{L}$ is homothetic iff it admits a utility function that is homogenous of degree one; $u(\alpha x) = \alpha u(x)$.

  • $\begingroup$ It is used in the step between equations 2 and 3. But if you don't believe the guy that his proof is correct why would you believe me...? $\endgroup$
    – Giskard
    Jan 19 '16 at 12:56
  • $\begingroup$ @denesp Like I said I am unsure of the middle three steps. I am looking for someone to rationalize these steps so I can confirm for myself that the proof is correct. (In addition, you specifically have provided some useful answers for me before! I already trust you more than the other guy...) $\endgroup$
    – möbius
    Jan 19 '16 at 13:02
  • $\begingroup$ Upon rereading the proof I now think homotheticity is used in several places. Can you edit your question to show the exact definition of homothetic functions you use? $\endgroup$
    – Giskard
    Jan 19 '16 at 13:42
  • $\begingroup$ @denesp I have edited as requested using the only definitions I have studied for homotheticity. Hopefully it is sufficient. $\endgroup$
    – möbius
    Jan 19 '16 at 16:25

An indirect proof. Suppose $$ x(p,w) = w\cdot x(p,1) $$ does not hold. This is equivalent with stating $$ U(x(p,w)) \neq U(w\cdot x(p,1)). $$ (To be precise: $x(p,w)$ and $x(p,1)$ may be set valued. In this case we are talking about two elements at least one of which is not included in both sets.)

Case 1.
$$ U(x(p,w)) > U(w\cdot x(p,1)) $$ As $U$ is homothetic $$ U(x(p,w)) = U(w \cdot \frac{1}{w}\cdot x(p,w)) = w \cdot U(\frac{1}{w}\cdot x(p,w)). $$ Using this we have $$ w \cdot U(\frac{1}{w}\cdot x(p,w)) = U(x(p,w)) > U(w\cdot x(p,1)) = w\cdot U(x(p,1)) $$ and thus $$ U(\frac{1}{w}\cdot x(p,w)) > U(x(p,1)) $$ However as $\frac{1}{w} \cdot x(p,w)$ is clearly an element of $B(p,1)$ this is impossible as $x(p,1)$ gives maximal utility in that budget set.

Case 2.
$$ U(x(p,w)) < U(w\cdot x(p,1)) $$ As $w \cdot x(p,1)$ is clearly an element of $B(p,w)$ this is impossible as $x(p,w)$ gives maximal utility in that budget set.

Thus we have proven that $$ U(x(p,w)) = U(w\cdot x(p,1)) $$ which is equivalent with $$ x(p,w) = w\cdot x(p,1). $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.