<p>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.</p>

CC BY-SA 3.0
2015-08-15T22:58:43.090Z
Sorry for the poor formating, writting from a phone
VicAche
CC BY-SA 3.0
2015-08-16T02:35:25.113Z
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!).
cc7768
CC BY-SA 3.0
2015-08-15T22:57:46.107Z
2015-08-16T07:23:09.873Z
2015-08-16T07:23:09.873Z
VicAche
<p>Ted Bergstrom has <a href="http://www.econ.ucsb.edu/~tedb/Courses/GraduateTheoryUCSB/separabilitynotes.pdf" rel="nofollow">Lecture Notes on Separable Preferences</a> that seem to have what you are looking for. For example:</p>

<blockquote>
  <p><strong>When are preferences additively separable?</strong> </p>
  
  <p>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).</p>
</blockquote>

CC BY-SA 3.0
2015-08-17T13:33:11.653Z
2015-08-17T13:33:11.653Z
BKay
<p>Call a utility function $u(x,y)$ <em>additive</em> if there exist functions $v_x,v_y$ such that:
$$u(x,y)=v_x(x)+v_y(y)$$</p>

<p>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.</p>

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

<p>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?</p>

CC BY-SA 3.0
2015-08-16T10:42:09.567Z
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.
optimal control
CC BY-SA 4.0
2023-08-17T12:34:35.857Z
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.
High GPA
<utility><preferences>
What utility functions are equivalent to additive functions?
Erel Segal-Halevi
# What utility functions are equivalent to additive functions?

(Asked by: username_0 on 2015-08-15)

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?

**Tags:** utility, preferences
> **Comments:**
> **username_1**: 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.  
> **username_2**: 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.  

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## 2 answers

### Answer by username_5 on 2015-08-17. Score: 4 | Accepted answer

Ted Bergstrom has [Lecture Notes on Separable Preferences](http://www.econ.ucsb.edu/~tedb/Courses/GraduateTheoryUCSB/separabilitynotes.pdf) 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).


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### Answer by username_3 on 2015-08-15. Score: 1

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.

> **Comments:**
> **username_3**: Sorry for the poor formating, writting from a phone  
> **username_4**: 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!).  

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