<p>Proposition 1. For <span class="math-container">$a&gt;0,b&gt;0,c&gt;0,d&gt;0$</span>, if <span class="math-container">$\dfrac{a}{b}=\dfrac{c}{d}$</span> then <span class="math-container">$\dfrac{a+c}{b+d}=\dfrac{a}{b}$</span>.</p>
<p>Proof is immediate.</p>
<p>Proposition 2. For <span class="math-container">$a_i&gt;0,b_i&gt;0$</span>, where <span class="math-container">$i\in\{1,2,\ldots,n\}$</span>, if <span class="math-container">$\dfrac{a_i}{b_i}=\dfrac{a_1}{b_1}$</span> (for all <span class="math-container">$i\in\{1,2,\ldots,n\}$</span>), then <span class="math-container">$\dfrac{\sum_{i=1}^{n}a_i}{\sum_{i=1}^{n}b_i}=\dfrac{a_1}{b_1}$</span>.</p>
<p>Proof. Use induction on proposition 1.</p>
<p>Using this proposition, it follows that <span class="math-container">$\dfrac{3(tx)^2(ty)^6}{6(tx)^3(ty)^5} = \dfrac{6(tx)(ty)^4}{12(tx)^2(ty)^3} = \dfrac{6(ty)^2}{12(tx)(ty)}=\dfrac{3x^2y^6}{6x^3y^5}=\dfrac{6xy^4}{12x^2y^3}  = \dfrac{6y^2}{12xy}$</span> implies <span class="math-container">$\dfrac{3(tx)^2(ty)^6 + 6(tx)(ty)^4 + 6(ty)^2}{6(tx)^3(ty)^5 + 12(tx)^2(ty)^3 + 12(tx)(ty)}=\dfrac{3x^2y^6 + 6xy^4 + 6y^2}{6x^3y^5 + 12x^2y^3 + 12xy}$</span> for all <span class="math-container">$t&gt;0$</span>.</p>

CC BY-SA 4.0
2023-10-05T05:34:48.550Z
2023-10-05T06:24:54.447Z
2023-10-05T06:24:54.447Z
Amit
<p>I am studying homothetic function and got the following problem:</p>
<h2>Problem</h2>
<blockquote>
<p>Determine whether the function <span class="math-container">$x^3y^6 + 3x^2y^4 + 6xy^2 + 9$</span> is homothetic or not.</p>
</blockquote>
<p>Here is my attempt.</p>
<h2><strong>My Attempt</strong></h2>
<p>I did it in two different ways, but I got two opposite answers. First by using the following definition of a homothetic function, I showed that the above function is homothetic.</p>
<blockquote>
<p><strong>Definition</strong><span class="math-container">$\quad$</span> A function <span class="math-container">$v:\mathbb{R}_{+}^{\mathbf{n}} \to \mathbb{R}$</span> is called <strong>homothetic</strong> if it is a monotonic transformation of a homogeneous function, that is, if there is a monotonic transformation <span class="math-container">$z \mapsto g(z)$</span>, where <span class="math-container">$g:\mathbb{R}_{+} \to \mathbb{R}$</span>, and a homogeneous function <span class="math-container">$u:\mathbb{R}_{+}^{\mathbf{n}} \to \mathbb{R}_{+}$</span> such that <span class="math-container">$v(\mathbf{x}) = g(u(\mathbf{x}))$</span> for all <span class="math-container">$\mathbf{x}$</span> in <span class="math-container">$\mathbb{R}_{+}^{\mathbf{n}}$</span>.</p>
</blockquote>
<blockquote>
<p><strong>Answer 1</strong><span class="math-container">$\quad$</span> <span class="math-container">$v(x,y) = x^3y^6 + 3x^2y^4 + 6xy^2 + 9 = (xy^2)^3 + 3(xy^2)^2 + 6xy^2 + 9$</span> is a monotonic transformation of a homogeneous function. Define <span class="math-container">$g:\mathbb{R}_{+} \to \mathbb{R}$</span> by <span class="math-container">$g(z) = z^3 + 3z^2 + 6z + 9$</span>. Since <span class="math-container">$g'(z) = 3z^2 + 6z + 6 = 3(z+1)^2 + 3 &gt; 0$</span>, <span class="math-container">$g$</span> is strictly increasing. Define <span class="math-container">$u:\mathbb{R}_{+}^{2} \to \mathbb{R}_{+}$</span> by <span class="math-container">$u(x,y) = xy^2$</span>. Then <span class="math-container">$u$</span> is homogeneous of degree <span class="math-container">$3$</span>. Thus, <span class="math-container">$v = g \circ u$</span> is homothetic.</p>
</blockquote>
<p>However, if I use the following theorem, I would get that <span class="math-container">$v$</span> is not homothetic.</p>
<blockquote>
<p><strong>Theorem</strong><span class="math-container">$\quad$</span> Let <span class="math-container">$u$</span> be a <span class="math-container">$C^1$</span> function on <span class="math-container">$\mathbb{R}_{+}^{\mathbf{n}}$</span>. <span class="math-container">$u$</span> is homothetic if and only if, for all <span class="math-container">$\mathbf{x}$</span> in <span class="math-container">$\mathbb{R}_{+}^{\mathbf{n}}$</span>, all <span class="math-container">$t &gt; 0$</span>, and all <span class="math-container">$i$</span>, <span class="math-container">$j$</span>,
<span class="math-container">\begin{align*}
\frac{\frac{\partial u}{\partial x_i}(t\mathbf{x})}{\frac{\partial u}{\partial x_j}(t\mathbf{x})} = \frac{\frac{\partial u}{\partial x_i}(\mathbf{x})}{\frac{\partial u}{\partial x_j}(\mathbf{x})}.
\end{align*}</span></p>
</blockquote>
<blockquote>
<p><strong>Answer 2</strong><span class="math-container">$\quad$</span> Since for all <span class="math-container">$x$</span> in <span class="math-container">$\mathbb{R}_{+}^{\mathbf{2}}$</span>,
<span class="math-container">\begin{align*}
\frac{\frac{\partial v}{\partial x}(tx,ty)}{\frac{\partial v}{\partial y}(tx,ty)} = \frac{3(tx)^2(ty)^6 + 6(tx)(ty)^4 + 6(ty)^2}{6(tx)^3(ty)^5 + 12(tx)^2(ty)^3 + 12(tx)(ty)}
\end{align*}</span>
which does not equal to
<span class="math-container">\begin{align*}
\frac{\frac{\partial v}{\partial x}(x,y)}{\frac{\partial v}{\partial y}(x,y)} = \frac{3x^2y^6 + 6xy^4 + 6y^2}{6x^3y^5 + 12x^2y^3 + 12xy},
\end{align*}</span>
for all <span class="math-container">$t &gt; 0$</span>, we conclude that <span class="math-container">$v$</span> is not homothetic.</p>
</blockquote>
<h2>My Question</h2>
<p>Could someone please help me explain why would I get opposite conclusion in the above problem? What am I missing? I really appreciate it!</p>

<microeconomics><mathematical-economics>
About Characterization of Homothetic Function - Mathematics for Economists by Simon and Blume Chapter 20 Exercise 18(3)
Beerus
# About Characterization of Homothetic Function - Mathematics for Economists by Simon and Blume Chapter 20 Exercise 18(3)

(Asked by: username_0 on 2023-10-04)

I am studying homothetic function and got the following problem:

## Problem

> Determine whether the function $x^3y^6 + 3x^2y^4 + 6xy^2 + 9$ is homothetic or not.

Here is my attempt.

## **My Attempt**

I did it in two different ways, but I got two opposite answers. First by using the following definition of a homothetic function, I showed that the above function is homothetic.

> **Definition**$\quad$ A function $v:\mathbb{R}\_{+}^{\mathbf{n}} \to \mathbb{R}$ is called **homothetic** if it is a monotonic transformation of a homogeneous function, that is, if there is a monotonic transformation $z \mapsto g(z)$, where $g:\mathbb{R}\_{+} \to \mathbb{R}$, and a homogeneous function $u:\mathbb{R}\_{+}^{\mathbf{n}} \to \mathbb{R}\_{+}$ such that $v(\mathbf{x}) = g(u(\mathbf{x}))$ for all $\mathbf{x}$ in $\mathbb{R}\_{+}^{\mathbf{n}}$.

> **Answer 1**$\quad$ $v(x,y) = x^3y^6 + 3x^2y^4 + 6xy^2 + 9 = (xy^2)^3 + 3(xy^2)^2 + 6xy^2 + 9$ is a monotonic transformation of a homogeneous function. Define $g:\mathbb{R}\_{+} \to \mathbb{R}$ by $g(z) = z^3 + 3z^2 + 6z + 9$. Since $g'(z) = 3z^2 + 6z + 6 = 3(z+1)^2 + 3 > 0$, $g$ is strictly increasing. Define $u:\mathbb{R}\_{+}^{2} \to \mathbb{R}\_{+}$ by $u(x,y) = xy^2$. Then $u$ is homogeneous of degree $3$. Thus, $v = g \circ u$ is homothetic.

However, if I use the following theorem, I would get that $v$ is not homothetic.

> **Theorem**$\quad$ Let $u$ be a $C^1$ function on $\mathbb{R}\_{+}^{\mathbf{n}}$. $u$ is homothetic if and only if, for all $\mathbf{x}$ in $\mathbb{R}\_{+}^{\mathbf{n}}$, all $t > 0$, and all $i$, $j$,
> \begin{align\*}
> \frac{\frac{\partial u}{\partial x\_i}(t\mathbf{x})}{\frac{\partial u}{\partial x\_j}(t\mathbf{x})} = \frac{\frac{\partial u}{\partial x\_i}(\mathbf{x})}{\frac{\partial u}{\partial x\_j}(\mathbf{x})}.
> \end{align\*}

> **Answer 2**$\quad$ Since for all $x$ in $\mathbb{R}\_{+}^{\mathbf{2}}$,
> \begin{align\*}
> \frac{\frac{\partial v}{\partial x}(tx,ty)}{\frac{\partial v}{\partial y}(tx,ty)} = \frac{3(tx)^2(ty)^6 + 6(tx)(ty)^4 + 6(ty)^2}{6(tx)^3(ty)^5 + 12(tx)^2(ty)^3 + 12(tx)(ty)}
> \end{align\*}
> which does not equal to
> \begin{align\*}
> \frac{\frac{\partial v}{\partial x}(x,y)}{\frac{\partial v}{\partial y}(x,y)} = \frac{3x^2y^6 + 6xy^4 + 6y^2}{6x^3y^5 + 12x^2y^3 + 12xy},
> \end{align\*}
> for all $t > 0$, we conclude that $v$ is not homothetic.

## My Question

Could someone please help me explain why would I get opposite conclusion in the above problem? What am I missing? I really appreciate it!

**Tags:** microeconomics, mathematical-economics


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## 1 answers

### Answer by username_1 on 2023-10-05. Score: 3 | Accepted answer

Proposition 1. For $a>0,b>0,c>0,d>0$, if $\dfrac{a}{b}=\dfrac{c}{d}$ then $\dfrac{a+c}{b+d}=\dfrac{a}{b}$.

Proof is immediate.

Proposition 2. For $a\_i>0,b\_i>0$, where $i\in\{1,2,\ldots,n\}$, if $\dfrac{a\_i}{b\_i}=\dfrac{a\_1}{b\_1}$ (for all $i\in\{1,2,\ldots,n\}$), then $\dfrac{\sum\_{i=1}^{n}a\_i}{\sum\_{i=1}^{n}b\_i}=\dfrac{a\_1}{b\_1}$.

Proof. Use induction on proposition 1.

Using this proposition, it follows that $\dfrac{3(tx)^2(ty)^6}{6(tx)^3(ty)^5} = \dfrac{6(tx)(ty)^4}{12(tx)^2(ty)^3} = \dfrac{6(ty)^2}{12(tx)(ty)}=\dfrac{3x^2y^6}{6x^3y^5}=\dfrac{6xy^4}{12x^2y^3} = \dfrac{6y^2}{12xy}$ implies $\dfrac{3(tx)^2(ty)^6 + 6(tx)(ty)^4 + 6(ty)^2}{6(tx)^3(ty)^5 + 12(tx)^2(ty)^3 + 12(tx)(ty)}=\dfrac{3x^2y^6 + 6xy^4 + 6y^2}{6x^3y^5 + 12x^2y^3 + 12xy}$ for all $t>0$.


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