<p>You have exactly the right idea. For a clearer and more efficient proof, I would suggest the following:</p>
<ul>
<li>I am not entirely sure what the index <span class="math-container">$i$</span> is referring to. A specific good of the bundle? I don't think you need this.</li>
<li><span class="math-container">$\hat x (p) $</span> denotes an optimal bundle for an income equal to <span class="math-container">$1$</span>, right? It is clearer to write <span class="math-container">$\hat x (p, 1)$</span>  in order to be consistent with your notation.</li>
<li>The claim which you want to prove is &quot;If <span class="math-container">$\hat x (p, 1)$</span>  is optimal at an income of 1, then <span class="math-container">$I \hat x (p, 1)$</span> is optimal at an income of <span class="math-container">$I$</span>&quot; . By the way, it is not generally true that every optimal bundle at <span class="math-container">$I$</span> is given by <span class="math-container">$I \hat x (p, 1)$</span>.</li>
<li>To prove the above claim, you only need to note that <span class="math-container">$I \hat x (p, 1)$</span> is feasible at <span class="math-container">$I$</span> and then argue that it is weakly preferred to every other <span class="math-container">$y$</span> that's feasible at <span class="math-container">$I$</span>. You basically did this in your answer when you noted that the bundle <span class="math-container">$y / I$</span> is feasible at an income of 1, and then used optimality of <span class="math-container">$\hat x (p, 1)$</span>.</li>
<li>As a final remark: Homotheticity is really a property of preferences, not derivatives of utility representations. Indeed, all you need for your argument is that scaling any two bundles by the same positive scalar does not change the preference between then.</li>
</ul>

CC BY-SA 4.0
2020-09-21T09:16:01.857Z
2020-09-21T09:16:01.857Z
induction_is_a_laddah
<p>Can I please have some feedback/help proving the following. My proof is below but I am quite uncertain as to whether my solution is efficient. Thank you.</p>
<p>If <span class="math-container">$u(x)$</span> is a homothetic utility, then show that Marshallian demand is of the form: <span class="math-container">$x_i^*(p,I) = \hat{x}_i^*(p)I.$</span></p>
<p><span class="math-container">$\textit{Proof.}$</span> If <span class="math-container">$u(x)$</span> is homothetic then
<span class="math-container">$$
\forall \alpha \in \mathbb{R}_{+}, \forall x : \hskip 6pt
\frac{\partial u(x)}{\partial x} = 
\frac{\partial u(\alpha \cdot x)}{\partial x}.
$$</span>
Now suppose <span class="math-container">$$
x_i^*(p,I) = \hat{x}_i^*(p)I
$$</span>
does not hold, which is equivalent to
<span class="math-container">$$
u(x_i^*(p,I)) \ne u(\hat{x}_i^*(p)I).
$$</span>
Precisely, <span class="math-container">$x_i^*(p,I)$</span> and <span class="math-container">$\hat{x}_i^*(p)I$</span> may be set valued. In this case we refer to two elements, with at least one of which is not included in both sets.</p>
<p>Case 1.
<span class="math-container">$$
u(x_i^*(p,I)) &gt; u(\hat{x}_i^*(p)I)
$$</span>
As <span class="math-container">$u$</span> is homothetic
<span class="math-container">$$
u(x_i^*(p,I)) =  u(I \cdot \frac{1}{I}\cdot x_i^*(p,I)) = I \cdot u(\frac{1}{I}\cdot \hat{x}_i^*(p)I).
$$</span>
Using this we have
<span class="math-container">$$
I \cdot u(\frac{1}{I}\cdot x_i^*(p,I)) = u(x_i^*(p,I)) &gt; u(I\cdot \hat{x}_i^*(p)I) = I\cdot u(x_i^*(p,I))
$$</span>
thus we have
<span class="math-container">$$
u(\frac{1}{I}\cdot x_i^*(p,I)) &gt; \hat{x}_i^*(p)I)
$$</span>
However as <span class="math-container">$\frac{1}{I} \cdot x_i^*(p,I)$</span> is clearly an element of <span class="math-container">$B(p)I$</span> this is impossible as <span class="math-container">$\hat{x}_i^*(p)I$</span> gives maximal utility in that budget set.</p>
<p>Case 2.
<span class="math-container">$$
u(x_i^*(p,I)) &lt; u(\hat{x}_i^*(p)I)
$$</span>
As <span class="math-container">$\hat{x}_i^*(p)I$</span> is clearly an element of <span class="math-container">$B(p,I)$</span> this is impossible as <span class="math-container">$x_i^*(p,I)$</span> gives maximal utility in that budget set.</p>
<p>Thus we have proven that
<span class="math-container">$$
u(x_i^*(p,I)) = u(\hat{x}_i^*(p)I)
$$</span>
which is equivalent with
<span class="math-container">$$
x_i^*(p,I) = \hat{x}_i^*(p)I.
$$</span></p>

<microeconomics><mathematical-economics><utility><consumer-theory>
Proving that Marshallian demand is of the form: $x_i^*(p,I) = \hat{x}_i^*(p)I$ with certain conditions
rudinsimons12
# Proving that Marshallian demand is of the form: $x_i^*(p,I) = \hat{x}_i^*(p)I$ with certain conditions

(Asked by: username_0 on 2020-09-17)

Can I please have some feedback/help proving the following. My proof is below but I am quite uncertain as to whether my solution is efficient. Thank you.

If $u(x)$ is a homothetic utility, then show that Marshallian demand is of the form: $x\_i^\*(p,I) = \hat{x}\_i^\*(p)I.$

$\textit{Proof.}$ If $u(x)$ is homothetic then
$$
\forall \alpha \in \mathbb{R}\_{+}, \forall x : \hskip 6pt
\frac{\partial u(x)}{\partial x} =
\frac{\partial u(\alpha \cdot x)}{\partial x}.
$$
Now suppose $$
x\_i^\*(p,I) = \hat{x}\_i^\*(p)I
$$
does not hold, which is equivalent to
$$
u(x\_i^\*(p,I)) \ne u(\hat{x}\_i^\*(p)I).
$$
Precisely, $x\_i^\*(p,I)$ and $\hat{x}\_i^\*(p)I$ may be set valued. In this case we refer to two elements, with at least one of which is not included in both sets.

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

Case 2.
$$
u(x\_i^\*(p,I)) < u(\hat{x}\_i^\*(p)I)
$$
As $\hat{x}\_i^\*(p)I$ is clearly an element of $B(p,I)$ this is impossible as $x\_i^\*(p,I)$ gives maximal utility in that budget set.

Thus we have proven that
$$
u(x\_i^\*(p,I)) = u(\hat{x}\_i^\*(p)I)
$$
which is equivalent with
$$
x\_i^\*(p,I) = \hat{x}\_i^\*(p)I.
$$

**Tags:** microeconomics, mathematical-economics, utility, consumer-theory

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## 1 answers

### Answer by username_1 on 2020-09-21. Score: 1

You have exactly the right idea. For a clearer and more efficient proof, I would suggest the following:

* I am not entirely sure what the index $i$ is referring to. A specific good of the bundle? I don't think you need this.
* $\hat x (p) $ denotes an optimal bundle for an income equal to $1$, right? It is clearer to write $\hat x (p, 1)$ in order to be consistent with your notation.
* The claim which you want to prove is "If $\hat x (p, 1)$ is optimal at an income of 1, then $I \hat x (p, 1)$ is optimal at an income of $I$" . By the way, it is not generally true that every optimal bundle at $I$ is given by $I \hat x (p, 1)$.
* To prove the above claim, you only need to note that $I \hat x (p, 1)$ is feasible at $I$ and then argue that it is weakly preferred to every other $y$ that's feasible at $I$. You basically did this in your answer when you noted that the bundle $y / I$ is feasible at an income of 1, and then used optimality of $\hat x (p, 1)$.
* As a final remark: Homotheticity is really a property of preferences, not derivatives of utility representations. Indeed, all you need for your argument is that scaling any two bundles by the same positive scalar does not change the preference between then.


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