<p>For the sake of simplicity, let us consider the domain $X$ to be $\mathbb{R_+^n}$, and let $\succsim$be any preference relation(satisfying all the required choice axioms) on this domain.</p>

<p>Let $\textbf{x}$ and $\textbf{y}$ belong to the preference relation $\succsim$ (both $\textbf{x}$ and $\textbf{y}$ are points in $\mathbb{R_+^n}$), such that $\textbf{x}$$\succsim$$\textbf{y}$. This implies that there exists some utility representation $U$ of this relation, such that $\forall$ $\textbf{x}$,$\textbf{y}$$\in$$\mathbb{R_+^n}$, we have $U(\textbf{x})$$\geq$$U(\textbf{y})$. Now, if we consider any monotonically increasing function $f(.)$, then the former inequality can very well be written as $f(U(\textbf{x}))$$\geq$$f(U(\textbf{y}))$, $\forall$ $\textbf{x}$,$\textbf{y}$$\in$$\mathbb{R_+^n}$.</p>

<p>Coming back to your question, there can be two cases. If it is already known that $U$ and $U'$ induce the same preference relation on $X$, then all the assumptions are covered under the standard choice axioms. However, to prove that one is a monotonic transformation of the other(let's say $U'$ is a monotonic transformation of $U$) you can simply find out the <em>demand correspondences</em> corresponding to the two utility functions, and show that they are essentially the same. To see that this is true, consider the lagrangian $L$$=$$f(U(\textbf{x}))$ + $\lambda$$(w -\textbf{p.x}) $, where $w$$\in$$\mathbb{R_+}$ and $\textbf{p}$$\in$$\mathbb{R_+^n}$, and $f(.)$ is a monotonically increasing function as assumed earlier(you can also replace $f(.)$ with $U'(.)$, it's essentially the same thing). Taking the first order conditions, we get $f_U(U(\textbf{x})).U_{x_i}(\textbf{x}) -\lambda.p_{x_i}= 0$. Now, for any $i,j$$\in$ $\{1,2,..,n\}$, we have $f_U(U(\textbf{x})).U_{x_i}(\textbf{x})/p_{x_i}$$=$$f_U(U(\textbf{x})).U_{x_j}(\textbf{x})/p_{x_j}$,  or $U_{x_i}(\textbf{x})/p_{x_i}=U_{x_j}(\textbf{x})/p_{x_j}$, which is the same condition that we would've obtained if the utility function was only $U(.)$. Thus, in layman language, the demand correspondences are exactly the same 'algebraic expressions' for $f(U(.))$ and $U(.)$.</p>

<p>The second case can be when you are not aware of any preference relation, you're just given two functions, $U \&amp; U'$. In that case, you can similarly proceed to solve the Lagrangian, and check whether the expressions(read as solutions for all $x_i$) obtained are same or not. If they are not same, then neither of the two functions can be represented as a monotonic transformation of the other. If they are same, you'll still need to prove that the 'expression' obtained represents a demand correspondence, and hence a preference relation exists for the given utility functions. To prove that, you simply need to check whether the demand correspondences obtained satisfies all the properties pertaining to the Slutsky's matrix(you can refer to MWG for this section).</p>

CC BY-SA 4.0
2018-09-08T11:57:47.880Z
2018-09-09T06:35:38.300Z
2018-09-09T06:35:38.300Z
superhulk
<p>Depending onhow one reads the question, the converse holds either always or basically never. Let $u:X\to\mathbb{R}$ and $v:X\to\mathbb{R}$ be arbitrary functions with ranges $u(X)$ and $v(X)$, respectively. It is not hard to show that $u$ and $v$ represent the same preferences if and only if there exists a strictly increasing function $h:v(X)\to\mathbb{R}$ such that $u=h\circ v$. So this is the "always"-answer.</p>

<p>Now one might require $h$ to be defined and strictly increasing on all of $\mathbb{R}$. In that case, one can basically never guarantee the existence of such an $h$, at least for standard economic applications. Let $X=[0,1]$, $u$ be given by $u(x)=x$ and $v$ be given by $v(x)=x$ for $x\leq 1/2$ and $v(x)=x+1$ for $x&gt;1/2$. Then $v(X)=[0, 1/2]\cup (3/2,2]$. If $h:\mathbb{R}\to\mathbb{R}$ is a nondecreasing function such that $u=h\circ v$, all $r\in(1/2,3/2]$ must be mapped to the same point by $h$, so $h$ cannot be strictly increasing. </p>

<p>Such a problem occurs whenever $X$ is a connected topological space (say, a convex subset of some $\mathbb{R}^n$) and some continuous utility representation exists that is not constant.</p>

<p>The problem can even occur if both $v$ and $u$ are continuous. Let $X=\mathbb{R}$, $u$ be given by $u(x)=x$ and $v$ be given by $v(x)=e^x$. Then $v(X)=\mathbb{R}_{++}$, the set of strictly positive numbers. Suppose there were some strictly increasing function $h:\mathbb{R}\to\mathbb{R}$ such that $u=h\circ v$. Then $h$ would be a strictly increasing extension of the logarithm function to all of $\mathbb{R}$, which is impossible.</p>

<p>Finally, some positive result: If $X$ is a connected topological space, and $u:X\to\mathbb{R}$ and $v:X\to\mathbb{R}$ are continuous functions that represent the same preferences and are either both bounded above or unbounded above and are either both bounded below or unbounded below, then there exists a strictly increasing continuous function $h:\mathbb{R}\to\mathbb{R}$ such that $u=h\circ v$.</p>

CC BY-SA 4.0
2018-09-08T15:24:40.863Z
2018-09-09T11:05:44.013Z
2018-09-09T11:05:44.013Z
Michael Greinecker
<p>Given a choice set $X$ (NOT assumed to be a commodity set...), and utility functions $u,u'$ on $X$, it is clear that if $u'$ is a strictly monotonic transformation of $u$ then they induce the same preference relation on $X$.</p>

<p>My question is, under what conditions does the opposite hold? <strong>What are the least strict conditions we can think of that imply that if $u,u'$ induce the same preference relation on $X$, that $u'$ must then necessarily be a strictly increasing transformation of $u$?</strong></p>

<preferences>
preference relations and monotonic transformations of utillity functions
user56834
# preference relations and monotonic transformations of utillity functions

(Asked by: username_0 on 2018-09-08)

Given a choice set $X$ (NOT assumed to be a commodity set...), and utility functions $u,u'$ on $X$, it is clear that if $u'$ is a strictly monotonic transformation of $u$ then they induce the same preference relation on $X$.

My question is, under what conditions does the opposite hold? **What are the least strict conditions we can think of that imply that if $u,u'$ induce the same preference relation on $X$, that $u'$ must then necessarily be a strictly increasing transformation of $u$?**

**Tags:** preferences

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## 2 answers

### Answer by username_2 on 2018-09-08. Score: 2 | Accepted answer

Depending onhow one reads the question, the converse holds either always or basically never. Let $u:X\to\mathbb{R}$ and $v:X\to\mathbb{R}$ be arbitrary functions with ranges $u(X)$ and $v(X)$, respectively. It is not hard to show that $u$ and $v$ represent the same preferences if and only if there exists a strictly increasing function $h:v(X)\to\mathbb{R}$ such that $u=h\circ v$. So this is the "always"-answer.

Now one might require $h$ to be defined and strictly increasing on all of $\mathbb{R}$. In that case, one can basically never guarantee the existence of such an $h$, at least for standard economic applications. Let $X=[0,1]$, $u$ be given by $u(x)=x$ and $v$ be given by $v(x)=x$ for $x\leq 1/2$ and $v(x)=x+1$ for $x>1/2$. Then $v(X)=[0, 1/2]\cup (3/2,2]$. If $h:\mathbb{R}\to\mathbb{R}$ is a nondecreasing function such that $u=h\circ v$, all $r\in(1/2,3/2]$ must be mapped to the same point by $h$, so $h$ cannot be strictly increasing.

Such a problem occurs whenever $X$ is a connected topological space (say, a convex subset of some $\mathbb{R}^n$) and some continuous utility representation exists that is not constant.

The problem can even occur if both $v$ and $u$ are continuous. Let $X=\mathbb{R}$, $u$ be given by $u(x)=x$ and $v$ be given by $v(x)=e^x$. Then $v(X)=\mathbb{R}\_{++}$, the set of strictly positive numbers. Suppose there were some strictly increasing function $h:\mathbb{R}\to\mathbb{R}$ such that $u=h\circ v$. Then $h$ would be a strictly increasing extension of the logarithm function to all of $\mathbb{R}$, which is impossible.

Finally, some positive result: If $X$ is a connected topological space, and $u:X\to\mathbb{R}$ and $v:X\to\mathbb{R}$ are continuous functions that represent the same preferences and are either both bounded above or unbounded above and are either both bounded below or unbounded below, then there exists a strictly increasing continuous function $h:\mathbb{R}\to\mathbb{R}$ such that $u=h\circ v$.


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### Answer by username_1 on 2018-09-08. Score: 0

For the sake of simplicity, let us consider the domain $X$ to be $\mathbb{R\_+^n}$, and let $\succsim$be any preference relation(satisfying all the required choice axioms) on this domain.

Let $\textbf{x}$ and $\textbf{y}$ belong to the preference relation $\succsim$ (both $\textbf{x}$ and $\textbf{y}$ are points in $\mathbb{R\_+^n}$), such that $\textbf{x}$$\succsim$$\textbf{y}$. This implies that there exists some utility representation $U$ of this relation, such that $\forall$ $\textbf{x}$,$\textbf{y}$$\in$$\mathbb{R\_+^n}$, we have $U(\textbf{x})$$\geq$$U(\textbf{y})$. Now, if we consider any monotonically increasing function $f(.)$, then the former inequality can very well be written as $f(U(\textbf{x}))$$\geq$$f(U(\textbf{y}))$, $\forall$ $\textbf{x}$,$\textbf{y}$$\in$$\mathbb{R\_+^n}$.

Coming back to your question, there can be two cases. If it is already known that $U$ and $U'$ induce the same preference relation on $X$, then all the assumptions are covered under the standard choice axioms. However, to prove that one is a monotonic transformation of the other(let's say $U'$ is a monotonic transformation of $U$) you can simply find out the *demand correspondences* corresponding to the two utility functions, and show that they are essentially the same. To see that this is true, consider the lagrangian $L$$=$$f(U(\textbf{x}))$ + $\lambda$$(w -\textbf{p.x}) $, where $w$$\in$$\mathbb{R\_+}$ and $\textbf{p}$$\in$$\mathbb{R\_+^n}$, and $f(.)$ is a monotonically increasing function as assumed earlier(you can also replace $f(.)$ with $U'(.)$, it's essentially the same thing). Taking the first order conditions, we get $f\_U(U(\textbf{x})).U\_{x\_i}(\textbf{x}) -\lambda.p\_{x\_i}= 0$. Now, for any $i,j$$\in$ $\{1,2,..,n\}$, we have $f\_U(U(\textbf{x})).U\_{x\_i}(\textbf{x})/p\_{x\_i}$$=$$f\_U(U(\textbf{x})).U\_{x\_j}(\textbf{x})/p\_{x\_j}$, or $U\_{x\_i}(\textbf{x})/p\_{x\_i}=U\_{x\_j}(\textbf{x})/p\_{x\_j}$, which is the same condition that we would've obtained if the utility function was only $U(.)$. Thus, in layman language, the demand correspondences are exactly the same 'algebraic expressions' for $f(U(.))$ and $U(.)$.

The second case can be when you are not aware of any preference relation, you're just given two functions, $U \& U'$. In that case, you can similarly proceed to solve the Lagrangian, and check whether the expressions(read as solutions for all $x\_i$) obtained are same or not. If they are not same, then neither of the two functions can be represented as a monotonic transformation of the other. If they are same, you'll still need to prove that the 'expression' obtained represents a demand correspondence, and hence a preference relation exists for the given utility functions. To prove that, you simply need to check whether the demand correspondences obtained satisfies all the properties pertaining to the Slutsky's matrix(you can refer to MWG for this section).


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