# Ordinal utility and monotonic transformations

If u(x) is an ordinal utility function that represents the (weak) preference relation R, then

(a) any strictly monotonic transformation of u(x) also represents $$R$$, or

(b) any monotonic transformation of u(x) also represents $$R$$.

Which is the right proposition, (a) or (b)?

I thought that (b) is the right answer, but when I looked up various online sources I found both definitions, so I'm no longer sure.

I thought (a) cannot be right, because the condition for a monotonic transformation is usually formulated as a conditional: F is a strictly monotonic transformation of u if the following holds: (1) if $$u(x)>u(y)$$, then $$F(u(x))>F(u(y))$$. But that doesn't deal with the case (2) $$u(x)=u(y)$$, which represents xIy. Wouldn't $$F(u(x))>F(u(y))$$ be compatible with (1) and (2), but represent xPy? Thinking about it, however, it seems that the same as (1) with "greater than or equal" wouldn't do it either. Are the monotonicity conditions formulated as biconditionals? I'm confused.

• I would be very surprised if this is not a duplicate. Maybe this one, this one, this one, this one,... – luchonacho Sep 25 '17 at 13:18
• None of these answer my question. – Eric '3ToedSloth' Sep 26 '17 at 15:31
• What is your definition of a monotonic transformation? – user18 Sep 30 '17 at 4:17

You need to be clear about the definitions. Lets take:

(1) $u: X \to \mathbb{R}$ represents $\succsim$ if $u(x) \geq u(y) \iff x \succsim y$.

(2) A function $h: \mathbb{R} \to \mathbb{R}$ is monotone if $z \geq w \implies h(z) \geq h(w)$. $h$ is strictly monotone if $z > w \implies h(z) > h(w)$.

First, note that every strictly monotone function is monotone. Why? Well let $z \geq w$, and there are two cases to check: (i) $z > w$: apply the definition; (ii) $z = w$: then $h(z) = h(w)$ by the definition of equality (this is because $\mathbb{R}$ is a totally ordered set).

So immediately, we see that your condition (b) implies your condition (a). However, (b) is false (I do not understand Kanak's line of reasoning, but it is certainly wrong, although perhaps can be rationalized by non-standard definitions). To show that it is wrong, we need a counter example. Lets let $X = \{x,y\}$ and let $x \succ y$. Then $u(x) = 1$ and $u(y) = 0$ represents $\succsim$. Moreover, $h: z \mapsto 0$ is a monotone transformation. But, $h(u(x)) = h(u(y)) = 0$ does not.

Indeed, this example shows that it is weakly monotone transformations that destroy information (they need not be invertible), which in terms of the representation, indicates that strict preference gets collapsed into weak inequality.

Showing that (a) holds is a straightforward application of the definitions. (Hint: show that strict monotonicity can be defined via a bi-conditional statement).

• It is kind (and maybe not that false) to say that my answer can be rationalized by non-standard definitions. I am afraid I am simply wrong ! Actually what I was saying was briefly that, e.g. a staircase function can be smoothed by a strictly monotonic transformation. If the opposite is true, this is flat wrong. This will still be a staircase function. – keepAlive Sep 30 '17 at 17:03
• This answer is correct and Kanak's was wrong. I suspect OP's confusion arose because in some texts, "monotone" is defined using the definition of "strictly monotone" given in this answer. – user18 Sep 30 '17 at 23:26