<p>Let's look at the use of monotonic transformations of utility functions (which I guess is the most frequent occurrence of this concept in econ).</p>
<p>Let <span class="math-container">$u: \mathbb{R}^n_+ \to \mathbb{R}$</span> be a utility function. We say that <span class="math-container">$g: \mathbb{R}^n_+ \to \mathbb{R}$</span> is a monotonic transformation of <span class="math-container">$u$</span> if for all <span class="math-container">$x, y \in \mathbb{R}^n_+$</span>:
<span class="math-container">$$
u(x) \ge u(y) \iff g(x) \ge g(y).
$$</span></p>
<p>Let <span class="math-container">$D \subseteq \mathbb{R}$</span>. Then <span class="math-container">$h: D \to \mathbb{R}$</span> is called strictly increasing if for all <span class="math-container">$x, y \in D$</span>:
<span class="math-container">$$
x \ge y \iff h(x) \ge h(y).
$$</span></p>
<p>There's the following result:</p>
<p><strong>Th</strong> Let <span class="math-container">$D$</span> be the range of <span class="math-container">$u$</span>, i.e. <span class="math-container">$D = u(\mathbb{R}^n_+)$</span>. Then <span class="math-container">$g$</span> is a monotonic  transformation of <span class="math-container">$u$</span> iff there exists a strictly increasing function <span class="math-container">$h: D \to \mathbb{R}$</span> such that <span class="math-container">$g(x) = h(u(x))$</span>.</p>
<p><em>proof</em>: (<span class="math-container">$\leftarrow$</span>) If <span class="math-container">$g(x) = h(u(x))$</span> and <span class="math-container">$u(x) \ge (&gt;) u(y)$</span> then <span class="math-container">$h(u(x)) \ge (&gt;) h(u(y))$</span> so <span class="math-container">$g$</span> is indeed a monotone transformation of <span class="math-container">$u$</span>.</p>
<p>(<span class="math-container">$\rightarrow$</span>) For the reverse, define
<span class="math-container">$$
h(z) = g(x) \text{ whenever } u(x) = z.
$$</span>
First we need to check that <span class="math-container">$h$</span> is indeed a function. As such, let <span class="math-container">$x, y$</span> be such that <span class="math-container">$u(x) = u(y) = z$</span>. We need to show that <span class="math-container">$g(x) = g(y)$</span>. Indeed as <span class="math-container">$g$</span> is a monotone transformation, we have that <span class="math-container">$g(x) \ge g(y)$</span> and <span class="math-container">$g(y) \ge g(x)$</span> so <span class="math-container">$g(x) = g(y)$</span>. To see that <span class="math-container">$h$</span> is strictly increasing, let <span class="math-container">$z \ge (&gt;) w$</span> and let <span class="math-container">$x$</span> and <span class="math-container">$y$</span> be such that <span class="math-container">$u(x) = z \ge (&gt;) w = u(y)$</span>. Then as <span class="math-container">$g$</span> is a monotone transformation <span class="math-container">$h(z) = g(x) \ge (&gt;) g(y) = h(w)$</span>. <span class="math-container">$\blacksquare$</span></p>
<p>To see the connection with the derivatives we have the following:</p>
<p><strong>Th</strong> if <span class="math-container">$g: D \to \mathbb{R}$</span> is differentiable, <span class="math-container">$D$</span> is convex and if <span class="math-container">$g'(x) &gt; 0$</span> for all <span class="math-container">$x$</span> in <span class="math-container">$D$</span>, then <span class="math-container">$g$</span> is strictly monotone.</p>
<p><em>proof</em>: Let <span class="math-container">$x \ge (&gt;) y$</span> then by the mean value theorem there is a <span class="math-container">$c \in [x,y]$</span> such that:
<span class="math-container">$$
g(y) - g(x) = g'(c)(y - x) \ge (&gt;) 0.
$$</span></p>
<p><strong>Remark</strong>: as @Michael Greinecker said, the reverse is not true, there are strictly increasing functions whose derivative is zero at some points, like <span class="math-container">$g(x) = x^3$</span>.</p>
<blockquote>
<p>But how is that true for a negative monotonic transformation such as g(x)=−x where g′(x)=−1?</p>
</blockquote>
<p>In principle we could define the reverse notions. Say that <span class="math-container">$\tilde g$</span> is a &quot;negative&quot; monotonic transformation of <span class="math-container">$u$</span> if for all <span class="math-container">$x, y \in \mathbb{R}^n_+$</span>:
<span class="math-container">$$
u(x) \ge u(y) \iff \tilde g(x) \le \tilde g(y).
$$</span></p>
<p>Notice that <span class="math-container">$\tilde g$</span> reverses the ranking given by <span class="math-container">$u$</span>.</p>
<p>Next, we can look at functions <span class="math-container">$\tilde h: D \to \mathbb{R}$</span> that are strictly decreasing:
<span class="math-container">$$
x \ge y \iff \tilde h(x) \le \tilde h(y).
$$</span></p>
<p>We have the following:</p>
<p><strong>Th</strong> The function <span class="math-container">$\tilde g$</span> is a negative monotone transformation of <span class="math-container">$u$</span> iff there exists a strictly decreasing function <span class="math-container">$\tilde h: D \to \mathbb{R}$</span> such that <span class="math-container">$\tilde g(x) = \tilde h(u(x))$</span>.</p>
<p>The proof is similar to the proof for the monotone/strictly increasing case.</p>
<p><strong>Remark:</strong> In economics, &quot;negative&quot; monotonic transformations are not really used. We would like the functions <span class="math-container">$u$</span> and <span class="math-container">$g$</span> to give the same ranking over all bundles <span class="math-container">$x \in \mathbb{R}^n_+$</span> then they should be monotone transformations of each other. If you use a negative monotonic transformation, you are reversing the order.</p>

CC BY-SA 4.0
2021-04-30T08:10:24.417Z
2021-04-30T08:10:24.417Z
tdm
<p>I'm looking into utility functions and their relation to indifference curves.</p>
<p>Now, I understand a positive monotonic transformation does not change the order (it's a rank-preserving transformation).</p>
<p>I thought the same is true for a negative monotonic transformation (I'm reading H. Varian, he distinguishes between negative and positive monotonic transformations) except it reverses the order.</p>
<p>However in a paper I've read just recently: <a href="https://ocw.mit.edu/courses/economics/14-03-microeconomic-theory-and-public-policy-fall-2016/lecture-notes/MIT14_03F16_lec3.pdf" rel="nofollow noreferrer">https://ocw.mit.edu/courses/economics/14-03-microeconomic-theory-and-public-policy-fall-2016/lecture-notes/MIT14_03F16_lec3.pdf</a></p>
<p>It says:</p>
<blockquote>
<p>Definition: Monotonic Transformation
Let I be an interval on the real line (<span class="math-container">$R1$</span>) then: <span class="math-container">$g : I → R1$</span> is a monotonic transformation if
g is a strictly increasing function on I.
If <span class="math-container">$g(x)$</span> is differentiable then <span class="math-container">$g'(x) \gt 0\ \forall x$</span>
Informally: A monotone transformation of a variable is a rank-preserving transformation. [Note: not all rank-preserving transformations are differentiable.]</p>
</blockquote>
<p>But how is that true for a negative monotonic transformation such as <span class="math-container">$g(x) = -x$</span> where <span class="math-container">$g'(x)=-1$</span>?</p>
<p>I don't understand. Does the paper not take into account that there are negative monotonic transformations?</p>

CC BY-SA 4.0
2021-04-29T19:40:22.777Z
It is not true that every strictly increasing differentiable function must have a strictly positive derivative, the increasing function $x\mapsto x^3$ has derivative $0$ at $0$.
Michael Greinecker
CC BY-SA 4.0
2021-04-29T20:10:14.750Z
I frequently used "monotonic transformation" as a shorthand for "positive monotonic transformation", as in microeconomics negative monotonic transformations are seldom used.
Giskard
CC BY-SA 4.0
2021-04-29T21:46:03.420Z
Alright cool thank you both! Btw that's sort of what I was assuming because the term "negative monotonic transform" seems to be rarely used anywhere. @Giskard , Cheers. Feel free to sum that up in an answer btw.
j3141592653589793238
CC BY-SA 4.0
2021-04-29T22:58:13.397Z
This is not a paper, it's just a lecture note. It contains mistakes/sloppiness, as @MichaelGreinecker pointed out. The author defines a monotonic transformation meaning a positive monotonic transformation. He doesn't consider negative monotonic transformations, as these are not used in consumer theory.
VARulle
<microeconomics><utility>
Why is the derivative of a monotonic transformation of a utility function assumed to always be greater than 0?
j3141592653589793238
# Why is the derivative of a monotonic transformation of a utility function assumed to always be greater than 0?

(Asked by: username_0 on 2021-04-29)

I'm looking into utility functions and their relation to indifference curves.

Now, I understand a positive monotonic transformation does not change the order (it's a rank-preserving transformation).

I thought the same is true for a negative monotonic transformation (I'm reading H. Varian, he distinguishes between negative and positive monotonic transformations) except it reverses the order.

However in a paper I've read just recently: <https://ocw.mit.edu/courses/economics/14-03-microeconomic-theory-and-public-policy-fall-2016/lecture-notes/MIT14_03F16_lec3.pdf>

It says:

> Definition: Monotonic Transformation
> Let I be an interval on the real line ($R1$) then: $g : I → R1$ is a monotonic transformation if
> g is a strictly increasing function on I.
> If $g(x)$ is differentiable then $g'(x) \gt 0\ \forall x$
> Informally: A monotone transformation of a variable is a rank-preserving transformation. [Note: not all rank-preserving transformations are differentiable.]

But how is that true for a negative monotonic transformation such as $g(x) = -x$ where $g'(x)=-1$?

I don't understand. Does the paper not take into account that there are negative monotonic transformations?

**Tags:** microeconomics, utility
> **Comments:**
> **username_1**: It is not true that every strictly increasing differentiable function must have a strictly positive derivative, the increasing function $x\mapsto x^3$ has derivative $0$ at $0$.  
> **username_2**: I frequently used "monotonic transformation" as a shorthand for "positive monotonic transformation", as in microeconomics negative monotonic transformations are seldom used.  
> **username_0**: Alright cool thank you both! Btw that's sort of what I was assuming because the term "negative monotonic transform" seems to be rarely used anywhere. @Giskard , Cheers. Feel free to sum that up in an answer btw.  
> **username_3**: This is not a paper, it's just a lecture note. It contains mistakes/sloppiness, as @MichaelGreinecker pointed out. The author defines a monotonic transformation meaning a positive monotonic transformation. He doesn't consider negative monotonic transformations, as these are not used in consumer theory.  

---

## 1 answers

### Answer by username_4 on 2021-04-30. Score: 5 | Accepted answer

Let's look at the use of monotonic transformations of utility functions (which I guess is the most frequent occurrence of this concept in econ).

Let $u: \mathbb{R}^n\_+ \to \mathbb{R}$ be a utility function. We say that $g: \mathbb{R}^n\_+ \to \mathbb{R}$ is a monotonic transformation of $u$ if for all $x, y \in \mathbb{R}^n\_+$:
$$
u(x) \ge u(y) \iff g(x) \ge g(y).
$$

Let $D \subseteq \mathbb{R}$. Then $h: D \to \mathbb{R}$ is called strictly increasing if for all $x, y \in D$:
$$
x \ge y \iff h(x) \ge h(y).
$$

There's the following result:

**Th** Let $D$ be the range of $u$, i.e. $D = u(\mathbb{R}^n\_+)$. Then $g$ is a monotonic transformation of $u$ iff there exists a strictly increasing function $h: D \to \mathbb{R}$ such that $g(x) = h(u(x))$.

*proof*: ($\leftarrow$) If $g(x) = h(u(x))$ and $u(x) \ge (>) u(y)$ then $h(u(x)) \ge (>) h(u(y))$ so $g$ is indeed a monotone transformation of $u$.

($\rightarrow$) For the reverse, define
$$
h(z) = g(x) \text{ whenever } u(x) = z.
$$
First we need to check that $h$ is indeed a function. As such, let $x, y$ be such that $u(x) = u(y) = z$. We need to show that $g(x) = g(y)$. Indeed as $g$ is a monotone transformation, we have that $g(x) \ge g(y)$ and $g(y) \ge g(x)$ so $g(x) = g(y)$. To see that $h$ is strictly increasing, let $z \ge (>) w$ and let $x$ and $y$ be such that $u(x) = z \ge (>) w = u(y)$. Then as $g$ is a monotone transformation $h(z) = g(x) \ge (>) g(y) = h(w)$. $\blacksquare$

To see the connection with the derivatives we have the following:

**Th** if $g: D \to \mathbb{R}$ is differentiable, $D$ is convex and if $g'(x) > 0$ for all $x$ in $D$, then $g$ is strictly monotone.

*proof*: Let $x \ge (>) y$ then by the mean value theorem there is a $c \in [x,y]$ such that:
$$
g(y) - g(x) = g'(c)(y - x) \ge (>) 0.
$$

**Remark**: as @Michael Greinecker said, the reverse is not true, there are strictly increasing functions whose derivative is zero at some points, like $g(x) = x^3$.

> But how is that true for a negative monotonic transformation such as g(x)=−x where g′(x)=−1?

In principle we could define the reverse notions. Say that $\tilde g$ is a "negative" monotonic transformation of $u$ if for all $x, y \in \mathbb{R}^n\_+$:
$$
u(x) \ge u(y) \iff \tilde g(x) \le \tilde g(y).
$$

Notice that $\tilde g$ reverses the ranking given by $u$.

Next, we can look at functions $\tilde h: D \to \mathbb{R}$ that are strictly decreasing:
$$
x \ge y \iff \tilde h(x) \le \tilde h(y).
$$

We have the following:

**Th** The function $\tilde g$ is a negative monotone transformation of $u$ iff there exists a strictly decreasing function $\tilde h: D \to \mathbb{R}$ such that $\tilde g(x) = \tilde h(u(x))$.

The proof is similar to the proof for the monotone/strictly increasing case.

**Remark:** In economics, "negative" monotonic transformations are not really used. We would like the functions $u$ and $g$ to give the same ranking over all bundles $x \in \mathbb{R}^n\_+$ then they should be monotone transformations of each other. If you use a negative monotonic transformation, you are reversing the order.


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