<p>The other answer explained why the result is trivial; here is why it is not true without the modifier &quot;on the range&quot; and under the most literal reading.</p>
<p>Consider <span class="math-container">$[0,1]\cup(2,3]$</span> with the usual order. Consider the utility representations <span class="math-container">$u_1$</span> and <span class="math-container">$u_2$</span> with <span class="math-container">$u_1$</span> given by <span class="math-container">$u_1(x)=x$</span> for all <span class="math-container">$x$</span> and <span class="math-container">$u_2(x)=x$</span> for <span class="math-container">$x\leq 2$</span> and <span class="math-container">$u_2(x)=x-1$</span> for <span class="math-container">$x&gt;2$</span>. There is no strictly increasing function <span class="math-container">$f:\mathbb{R}\to\mathbb{R}$</span> such that <span class="math-container">$u_2=f\circ u_1$</span>. Indeed, that would require <span class="math-container">$f(3/2)&gt;1$</span> and <span class="math-container">$f(3/2)&lt;1+\epsilon$</span> for all <span class="math-container">$\epsilon&gt;0$</span>, which is impossible.</p>

CC BY-SA 4.0
2023-11-03T17:48:53.307Z
Great example!! I think we need the "connectness" assumption in Debreu 1954. Is you $[0,1][2,3]$ connected?
dodo
CC BY-SA 4.0
2023-11-03T18:20:43.013Z
No, it is not. But Debreu did not use connectedness either. Eilenberg did; that's the proof in Debreu's book.
Michael Greinecker
CC BY-SA 4.0
2023-11-05T08:50:17.760Z
I think I missed something but in Debreu (1954) paper "Representation of a Preference Ordering by a Numerical Function" in: Thrall, M., Davis, R.C. and Coombs, C.H., Eds., Decision Processes, his Theorem I requires connected and separable.
dodo
CC BY-SA 4.0
2023-11-05T10:04:17.050Z
@dodo As Debreu writes there about Theorem 1: "This theorem can easily be derived from the [results of S. Eilenberg.](https://www.jstor.org/stable/2371274)" Indeed, for the linearly ordered case (obtainable by taking a quotient), this is simply Eilenberg's Theorem I and 6.1. The novelty of Debreu's paper is Theorem 2, though the proof had a huge gap; one that concerns gaps.
Michael Greinecker
CC BY-SA 4.0
2023-11-05T11:55:18.797Z
Oh, I see. So in the "second countable" case, there is no uniqueness result. I think I get your point? Could you please explain a bit more about the term "on the range"?
dodo
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2023-11-05T12:30:04.030Z
If $u_1:X\to\mathbb{R}$ and $u_2:X\to\mathbb{R}$ represent the same preferences, then there exists a strictly increasing invertible function $f:u_1(X)\to u_2(X)$ such that $u_2=f\circ u_1$ and $u_2=f^{-1}\circ u_1$. The issue is that $f$ cannot in general be extended to a strictly increasing function defined on all of $\mathbb{R}$.
Michael Greinecker
CC BY-SA 4.0
2023-11-02T23:58:29.107Z
2023-11-02T23:58:29.107Z
Michael Greinecker
<p>It is not a 'result' because it is trivial to show that a positive strictly monotonic transformation function applied to the function that represents the preferences leaves the ordering unchanged. Say <span class="math-container">$f(x)$</span> represents a set of preferences on <span class="math-container">$X$</span>, in the sense that <span class="math-container">$x \succeq y$</span> if and only if <span class="math-container">$f(x) \geq f(y)$</span>. Then let <span class="math-container">$g(\cdot)$</span> represent a strictly positive monotonic transformation on the range of <span class="math-container">$f$</span>. That also means that <span class="math-container">$g(a) \geq g(b)$</span> if and only if <span class="math-container">$a \geq b$</span>. Then it must be true that <span class="math-container">$x \succeq y$</span> if and only if <span class="math-container">$g(f(x)) \geq g(f(y))$</span>. I'll let you fill in the rest of the details.</p>
<p>What that means is that you can create equivalence classes of functions that represent the same ordering. This is what is meant by 'unique up to a positive monotonic transformation'.</p>

CC BY-SA 4.0
2023-11-02T20:34:20.770Z
It is nice to see that people believe it trivial. So you think I don't have to cite sources for this observation?
dodo
CC BY-SA 4.0
2023-11-03T00:20:01.403Z
There's no need to cite a source for this.
VARulle
CC BY-SA 4.0
2023-11-03T01:08:14.350Z
Indeed. In general, if a result can be shown as a one-liner, it's not worth citing. You can prove it yourself with fewer words, right? So why cite it?
Surge
CC BY-SA 4.0
2023-11-02T17:08:12.537Z
2023-11-02T17:08:12.537Z
Surge
<p>In <a href="https://en.wikipedia.org/wiki/Debreu%27s_representation_theorems" rel="nofollow noreferrer">Debreu's 1954</a> ordinal utility representation theorem, the utility is unique up to <em>a positive monotonic</em> transformation.</p>
<p>While the uniqueness result is well-known, I fail to find a proper reference. Debreu's original paper does not contain that uniqueness result.</p>
<p>Is there a good reference? Who was the first researcher find this uniqueness result?</p>

<microeconomics><reference-request><theory><decision-theory>
Debreu's ordinal representation theorem is unique up to a positive monotonic transformation, what is the source?
dodo
# Debreu's ordinal representation theorem is unique up to a positive monotonic transformation, what is the source?

(Asked by: username_0 on 2023-11-02)

In [Debreu's 1954](https://en.wikipedia.org/wiki/Debreu%27s_representation_theorems) ordinal utility representation theorem, the utility is unique up to *a positive monotonic* transformation.

While the uniqueness result is well-known, I fail to find a proper reference. Debreu's original paper does not contain that uniqueness result.

Is there a good reference? Who was the first researcher find this uniqueness result?

**Tags:** microeconomics, reference-request, theory, decision-theory

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## 2 answers

### Answer by username_1 on 2023-11-02. Score: 3

The other answer explained why the result is trivial; here is why it is not true without the modifier "on the range" and under the most literal reading.

Consider $[0,1]\cup(2,3]$ with the usual order. Consider the utility representations $u\_1$ and $u\_2$ with $u\_1$ given by $u\_1(x)=x$ for all $x$ and $u\_2(x)=x$ for $x\leq 2$ and $u\_2(x)=x-1$ for $x>2$. There is no strictly increasing function $f:\mathbb{R}\to\mathbb{R}$ such that $u\_2=f\circ u\_1$. Indeed, that would require $f(3/2)>1$ and $f(3/2)<1+\epsilon$ for all $\epsilon>0$, which is impossible.

> **Comments:**
> **username_0**: Great example!! I think we need the "connectness" assumption in Debreu 1954. Is you $[0,1][2,3]$ connected?  
> **username_1**: No, it is not. But Debreu did not use connectedness either. Eilenberg did; that's the proof in Debreu's book.  
> **username_0**: I think I missed something but in Debreu (1954) paper "Representation of a Preference Ordering by a Numerical Function" in: Thrall, M., Davis, R.C. and Coombs, C.H., Eds., Decision Processes, his Theorem I requires connected and separable.  
> **username_1**: @dodo As Debreu writes there about Theorem 1: "This theorem can easily be derived from the [results of S. Eilenberg.](https://www.jstor.org/stable/2371274)" Indeed, for the linearly ordered case (obtainable by taking a quotient), this is simply Eilenberg's Theorem I and 6.1. The novelty of Debreu's paper is Theorem 2, though the proof had a huge gap; one that concerns gaps.  
> **username_0**: Oh, I see. So in the "second countable" case, there is no uniqueness result. I think I get your point? Could you please explain a bit more about the term "on the range"?  
> **username_1**: If $u_1:X\to\mathbb{R}$ and $u_2:X\to\mathbb{R}$ represent the same preferences, then there exists a strictly increasing invertible function $f:u_1(X)\to u_2(X)$ such that $u_2=f\circ u_1$ and $u_2=f^{-1}\circ u_1$. The issue is that $f$ cannot in general be extended to a strictly increasing function defined on all of $\mathbb{R}$.  

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### Answer by username_2 on 2023-11-02. Score: 2

It is not a 'result' because it is trivial to show that a positive strictly monotonic transformation function applied to the function that represents the preferences leaves the ordering unchanged. Say $f(x)$ represents a set of preferences on $X$, in the sense that $x \succeq y$ if and only if $f(x) \geq f(y)$. Then let $g(\cdot)$ represent a strictly positive monotonic transformation on the range of $f$. That also means that $g(a) \geq g(b)$ if and only if $a \geq b$. Then it must be true that $x \succeq y$ if and only if $g(f(x)) \geq g(f(y))$. I'll let you fill in the rest of the details.

What that means is that you can create equivalence classes of functions that represent the same ordering. This is what is meant by 'unique up to a positive monotonic transformation'.

> **Comments:**
> **username_0**: It is nice to see that people believe it trivial. So you think I don't have to cite sources for this observation?  
> **username_3**: There's no need to cite a source for this.  
> **username_2**: Indeed. In general, if a result can be shown as a one-liner, it's not worth citing. You can prove it yourself with fewer words, right? So why cite it?  

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