<p>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.)</p>

<p>Case 1. <br>
$$
U(x(p,w)) &gt; U(w\cdot x(p,1))
$$
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
$$
w \cdot U(\frac{1}{w}\cdot x(p,w)) = U(x(p,w)) &gt; U(w\cdot x(p,1)) = w\cdot U(x(p,1))
$$
and thus
$$
U(\frac{1}{w}\cdot x(p,w)) &gt; U(x(p,1))
$$
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.</p>

<p>Case 2. <br>
$$
U(x(p,w)) &lt; U(w\cdot x(p,1))
$$
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.</p>

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

CC BY-SA 3.0
2016-02-09T19:09:25.593Z
2016-02-09T19:09:25.593Z
Giskard
<p>Someone gave me a proof of this, but I am not sure if it is correct. </p>

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

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

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

<p>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 $\alpha \geq 0$. </p>

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

CC BY-SA 3.0
2016-01-19T12:56:47.170Z
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...?
Giskard
CC BY-SA 3.0
2016-01-19T13:02:46.977Z
@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...)
möbius
CC BY-SA 3.0
2016-01-19T13:42:38.640Z
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?
Giskard
CC BY-SA 3.0
2016-01-19T16:25:32.720Z
@denesp I have edited as requested using the only definitions I have studied for homotheticity. Hopefully it is sufficient.
möbius
<utility><consumer-theory><walrasian>
Show that $x(p,w)=w\cdot x(p,1)$ with homothetic preferences
möbius
# Show that $x(p,w)=w\cdot x(p,1)$ with homothetic preferences

(Asked by: username_0 on 2016-01-19)

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; $u(\alpha x) = \alpha u(x)$.

**Tags:** utility, consumer-theory, walrasian
> **Comments:**
> **username_1**: 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...?  
> **username_0**: @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...)  
> **username_1**: 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?  
> **username_0**: @denesp I have edited as requested using the only definitions I have studied for homotheticity. Hopefully it is sufficient.  

---

## 1 answers

### Answer by username_1 on 2016-02-09. Score: 2

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
$$
x(p,w) = w\cdot x(p,1).
$$


---

