Call a utility function $u(x,y)$ additive if there exist functions $v_x,v_y$ such that: $$u(x,y)=v_x(x)+v_y(y)$$

Consider the function $u(x,y)=xy$. It is not additive, but, it can transformed using a positive-monotonic-transformation (PMT) to the function: $u'(x,y)=\log u(x,y) = \log{x}+\log{y}$, and the function $u'$ is additive.

My question is: what conditions on a function $u(x,y)$ guarantee that it can be transformed using a PMT to an additive function?

I.e, if I see a function $u(x,y)$, how can I know whether it represents a preference relation which can also be represented by an additive utility function?

  • 1
    $\begingroup$ I don't think there is a general rule which applies for the use of PMT but for example, you can not use this one if you have Constant Elasticity of Substitution (CES) utility function. By the way, the specification of utility functions is generally made for analytical reasons. People use them wheter it makes a tractable model or not. $\endgroup$ Commented Aug 16, 2015 at 10:42
  • $\begingroup$ This is why we have axioms on preference when definition a utility function. Without axioms we don't know if a utility function is equivalent to another one. $\endgroup$
    – High GPA
    Commented Aug 17, 2023 at 12:34

2 Answers 2


Ted Bergstrom has Lecture Notes on Separable Preferences that seem to have what you are looking for. For example:

When are preferences additively separable?

The most useful necessary and sufficient condition for preferences to be additively separable is that every subset of the set of all commodities is separable. The proofs that I know of for this proposition are a bit more elaborate than seems appropriate here. A somewhat more general version of this theorem can be found in a paper by Gerard Debreu (Topological methods in cardinal utility). Debreu’s paper seems to be the first satisfactorily general solution to this problem. Other proofs can be found in (Foundations of Measurement) and (Utility theory for decision making).


If and only if (If $(x1, x2)R(y1, y2)$ and $(y1, z2)R(z1, x2)$, then $(x1, z2)R(z1, y2)$) then with a two good functions, then the variables are separable. Hence additively separable. This is the Debreu theorem.

  • $\begingroup$ Sorry for the poor formating, writting from a phone $\endgroup$
    – VicAche
    Commented Aug 15, 2015 at 22:58
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    $\begingroup$ Added latex, but I think a few more details (or clearer editing/phrasing) could make this more helpful (I definitely understand typing is hard from a phone though!). $\endgroup$
    – cc7768
    Commented Aug 16, 2015 at 2:35

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