# Why is the derivative of a monotonic transformation of a utility function assumed to always be greater than 0?

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?

• 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 $0$. Apr 29 '21 at 19:40
• I frequently used "monotonic transformation" as a shorthand for "positive monotonic transformation", as in microeconomics negative monotonic transformations are seldom used. Apr 29 '21 at 20:10
• 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. Apr 29 '21 at 21:46
• 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. Apr 29 '21 at 22:58

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.