# If an ordinal-scaled utility function is defined via strictly increasing transformation, how can it represent a case of indifference?

Problem: According to Wulf Gaertner’s (2009, p. 13) A Primer in Social Choice Theory, any strictly increasing transformation of an individual’s ordinal utility function is informationally equivalent. He says that:

“The only thing which matters in terms of available information is whether, say, commodity bundle $$a$$ is preferred to or indifferent to or dispreferred to commodity bundle $$b$$ or, expressed in terms of utility indices, whether $$a$$ has a higher or equally high or lower utility index than $$b$$.”

This makes sense, however, when I looked up the definition of a `strictly increasing function/transformation’ I became confused regarding how this would allow one to represent cases of indifference as well as cases of strict preference.

Ken Binmore’s (1983, p. 109) Mathematical Analysis: A Straightforward Approach formally defines an increasing and strictly increasing function as:

Increasing function: $$\forall x,y \in X: (x > y) \Longrightarrow ( \ f(x) \geq f(y) \ )$$

Strictly increasing function: $$\forall x,y \in X: (x > y) \Longrightarrow ( \ f(x) > f(y) \ )$$

So a strictly increasing transformation in utility terms–––for individual $$i$$–––would (I think) be defined formally as:

Strictly increasing transformation: $$\forall x,y \in X: ( \ u_i(x) > u_i(y) \ ) \Longrightarrow ( \ f(u_i(x)) > f(u_i(y)) \ )$$

But, since this is defined solely in terms of strict inequality ($$\ u_i(x) > u_i(y) \$$) I’m confused as to how this can allow for cases where an individual is indifferent between two alternatives: $$xI_iy = ( \ u_i(x) = u_i(y) \ )$$. Or cases of three or more alternatives where $$i$$ strictly prefers some and is indifference regarding others: e.g. $$( \ u_i(x) = u_i(y) \ ) > u_i(z)$$. Wouldn’t these be ordering cases, where the orderings are not characterised by strict inequality of utilities?

Duplication information: I am aware that a question very similar to this has been asked before by Eric '3ToedSloth' (Ordinal utility and monotonic transformations). Eric '3ToedSloth'’s question was focused on the definitions of both ordinal-scaled utility functions and monotone transformations. He also mentioned the difficulty of representing indifference. The reason that I have asked this question is that the accepted answer (by 201p) was focused on answering the question by showing that increasing (/weakly monotone) transformations destroy ordering information, rather than the issue which I am most confused about: how ordinal utility functions (defined in terms of strictly increasing transformations) can represent indifference. (There is a chance that 201p’s answer did cover this, but I am too stupid to be able to see it). Ideally I would have posed this as a clarificatory comment, but I lack the reputation level to post one. I am happy for my question to be closed if the community thinks it is too similar :).

You should think it in the following way, a utility function represents someone’s preferences if the function gives a higher/lower/equal value from bundle a than from bundle b if that person prefers more/less/the same bundle a than bundle b. Suppose that such function is $$u(x)$$, then you can ask, is there a unique function with such properties (I.e. that represents the person’s preferences) and the answer is NO. Any strictly increasing monotonic transformation of $$u(x)$$, say $$f(u(x))$$ is also a function describing the preferences of the person. Hence we say that these two are valid representations, and are equivalent (or informationally equivalent).
• Thank you for your help; I really appreciate it. Clearly I’m still missing something, as I feel comfortable with the idea of informational equivalence (as outlined by you), but am still unsure how a strictly increasing function applies to equal utilities. A strictly increasing transformation is a conditional statement: if we have a case in which $u_i(x) > u_i(y)$, then $f(u_i(x)) > f(u_i(y))$. Since the antecedent only refers to cases of strict inequality, I can’t see how the consequent has any bearing on cases which are not characterised by strict inequality. – Nikelmouse Dylar May 29 '19 at 6:19
• Well, but for any function it is true that $u_i(x)=u_i(y)$ implies $f(u_i(x))=f(u_i(y))$. Otherwise, $f$ will not be a well-defined function. Perhaps this is why that case is not addressed directly. I would appreciate a lot if you could mark mine as the accepted answer. – Regio May 29 '19 at 16:37