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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?

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    $\begingroup$ 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 $0$. $\endgroup$ – Michael Greinecker Apr 29 at 19:40
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    $\begingroup$ I frequently used "monotonic transformation" as a shorthand for "positive monotonic transformation", as in microeconomics negative monotonic transformations are seldom used. $\endgroup$ – Giskard Apr 29 at 20:10
  • $\begingroup$ 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. $\endgroup$ – j3141592653589793238 Apr 29 at 21:46
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    $\begingroup$ 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. $\endgroup$ – VARulle Apr 29 at 22:58
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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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