<p>In general, it will not represent the same preferences. There seems to be confusion on what &quot;monotonic transformation&quot; means in this context. It does not have much to do with monotonic preferences.</p>
<p>We say that the utility function <span class="math-container">$v:X\to\mathbb{R}$</span> is a <em>monotonic transformation</em> of the utility function <span class="math-container">$u:X\to\mathbb{R}$</span> if there exists a strictly increasing function <span class="math-container">$g:u(X)\to\mathbb{R}$</span> (the domain is the range of <span class="math-container">$u$</span>) such that <span class="math-container">$v(x)=g(u(x))$</span> for all <span class="math-container">$x\in X$</span>. A monotonic transformation of a utility function does always represent the same preferences, and this has nothing to do with whether these preferences are monotone, convex, or anything like that.</p>

CC BY-SA 4.0
2021-11-14T14:10:31.557Z
But by the book's solution manual, it is indeed a monotonic transformation, f' is increasing, isn't it?
Karan Kumar
CC BY-SA 4.0
2021-11-14T14:14:38.137Z
@KaranKumar Well, $f$ is increasing in the arguments and, therefore, represents monotone preferences. But it does not represent the same preferences as $u$. And $f$ is not a monotonic transformation of $u$ according to the definition given in the answer.
Michael Greinecker
CC BY-SA 4.0
2021-11-14T15:12:06.060Z
I disagree. Utility function in regards to representing the preferences are invariant to positive monotonic transformations. Since f' is greater than zero, therefore f and u represent the same preferences. Also, there does not seem to be any ambiguity as to how the function f is not a monotonic transformation as provided in the answer. Perhaps you could illustrate with an example or some argument as I am utterly missing your point.
Karan Kumar
CC BY-SA 4.0
2021-11-14T15:50:05.467Z
@KaranKumar If $f$ would be a positive monotonic transformation of $u$, it would represent the same preferences. It does not, as your own example shows. The other "proofs" given by you do not show that it is a monotonic transformation. Where is the function $g$?
Michael Greinecker
CC BY-SA 4.0
2021-11-14T16:00:28.760Z
I actually found the "solution manual" you quote from. It is not an official solution manual for the book; it is someone typing up their own solutions. And making a mistake, here.
Michael Greinecker
CC BY-SA 4.0
2021-11-14T22:27:39.883Z
@KaranKumar First of all, if $f$ is a function on a vector space, then the proper terminology is $\nabla f$ rather than $f'$. Second, "$f: X \rightarrow Y$ is increasing" means $a <_X b \rightarrow f(a) <_Y f(b)$, which requires some order $<_X$ to be defined. Saying that $\nabla f$ is "greater than zero" again requires some ordering of the vector space, and the precise statement would then be that it's greater than the zero **vector**.
Acccumulation
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2021-11-14T23:00:47.993Z
@KaranKumar: In particular, $v$ is a monotonic transformation of $u$ if (and only if) they both define the same ordering between preferences, i.e. if $v(x) ≥ v(y) \iff u(x) ≥ u(y)$ for all $x, y \in X$. When $X = \mathbb R$, it's easy to show that any two strictly increasing functions are monotonic transformations of each other, but this does not generalize easily to $X = \mathbb R^n$ for $n > 1$.
Ilmari Karonen
CC BY-SA 4.0
2021-11-14T23:00:55.210Z
… In particular, $u$ and $v$ being monotonic transformations implies that they must have the same level sets (i.e. indifference curves); this is trivial for strictly increasing functions in one dimension (where the level sets are all just single points), but not in more than one dimension. And this (that $f$ and $u$ have different level sets) is exactly what your counterexample shows.
Ilmari Karonen
CC BY-SA 4.0
2021-11-19T15:09:02.103Z
@MichaelGreinecker The $g$ that you are asking is g(x) = $u(\textbf x)$ + $<(u^-1(\textbf x))^T,e>$
Clearly g is defined on reals as $u(\textbf x)$ is a scalar and the dot product also gives a scalar. But I don't know how to differentiate a dot product
Karan Kumar
CC BY-SA 4.0
2021-11-19T15:10:39.200Z
@Accumulation the problem is not to prove whether f or u defined on vectors is having a positive gradient but whether g defined on reals (scalars) has a positive differentiation for all the domain values
Karan Kumar
CC BY-SA 4.0
2021-11-19T15:14:30.460Z
@Accumulation your logical example is extremely helpful by the way. But shouldn't it apply on say more things. Like if I consume 2 half litre cans of coke and 1 litre can of coke we get the same utility - also assume they are  priced so that half a can of coke costs half the price of coke. but by your logic if we add the two cans we will get  utility of say 3 and utility of 1 can will be 2! What then? Shouldn't we be indifferent?
Karan Kumar
CC BY-SA 4.0
2021-11-19T17:09:23.700Z
@KaranKumar $u^{-1}(x)$ is a set of points (an indifference curve), not a single point.
Michael Greinecker
CC BY-SA 4.0
2021-11-14T10:24:49.007Z
2021-11-14T10:24:49.007Z
Michael Greinecker
<p>We can simplify by assuming that <span class="math-container">$u$</span> is linear, in which we can treat <span class="math-container">$u$</span> as the total dollar value of the set of items, and <span class="math-container">$f$</span> to be the total dollar value plus the count of the items. Or, in other words, <span class="math-container">$f$</span> adds <span class="math-container">$1$</span> to the dollar value of each item. Would you prefer <span class="math-container">$1000$</span> pennies to one diamond worth <span class="math-container">$\\\$100$</span>? <span class="math-container">$u(1000 \text 
 { pennies}) = 10$</span>, <span class="math-container">$u(\text{diamond})=100$</span>, so <span class="math-container">$u$</span> says the diamond is better. <span class="math-container">$f(1000 \text 
 { pennies}) = 1010$</span>, <span class="math-container">$f(\text{diamond})=101$</span>, so <span class="math-container">$f$</span> says the pennies are better.</p>
<p>The fact that <span class="math-container">$f$</span> is &quot;monotonic&quot; (a more precise phrasing would be that it's monotonic with respect to each of the components) just means that <span class="math-container">$f$</span> gives a higher number than <span class="math-container">$u$</span> for the same basket of goods. It doesn't say anything about what happens when comparing different baskets of goods.</p>

CC BY-SA 4.0
2021-11-19T15:14:43.837Z
@Accumulation your logical example is extremely helpful by the way. But shouldn't it apply on say more things. Like if I consume 2 half litre cans of coke and 1 litre can of coke we get the same utility - also assume they are priced so that half a can of coke costs half the price of coke. but by your logic if we add the two cans we will get utility of say 3 and utility of 1 can will be 2! What then? Shouldn't we be indifferent?
Karan Kumar
CC BY-SA 4.0
2021-11-20T00:55:11.550Z
@KaranKumar $f$ is given in terms of a particular meaning of $x_i$. Changing what $x_i$ means changes what $f$ is in terms of $x_i$.
Acccumulation
CC BY-SA 4.0
2021-11-27T10:19:30.897Z
Thank you all, we can close this question
Karan Kumar
CC BY-SA 4.0
2021-11-14T22:52:51.190Z
2021-11-14T22:52:51.190Z
Acccumulation
<p>Hi I am reading Jehle and Reny in my master's course and I have come across a problem in one of the exercises. My instructor herself was a bit confused when a student gave her a counter-example and then said that the example illustrates that our logic is faulty. I want to reconcile the example and math logic.
Here is the question with the solutions discussed.</p>
<p>Q1.24 Let u(<span class="math-container">$\textbf x$</span>) represent some consumer’s monotonic preferences over <span class="math-container">$\textbf x ∈ \mathbb{R}^n_+$</span>. For each of the functions <span class="math-container">$f (x)$</span> that follow, state whether or not f also represents the preferences of this consumer. In each case, be sure to justify your answer with either an argument or a counterexample.</p>
<p>Part (c) <span class="math-container">$f (x) = u(\textbf x) +\Sigma_{i=1}^{n} x_i$</span></p>
<p>Now I believe that this function is a monotonic transformation of <span class="math-container">$u(\textbf x)$</span> and thus is a representation of the inherent &quot;monotonic&quot; preferences. I provide two proofs stating this fact and one counter-example which is creating difficulty in understanding</p>
<p>Proof 1: (OP's proof)</p>
<p>Let <span class="math-container">$\textbf x^1 \ge \textbf x^2.$</span> (1)</p>
<p>Clearly, <span class="math-container">$\textbf x^1 \succsim  \textbf x^2 $</span> (Since preferences are monotonic) (2)</p>
<p><span class="math-container">$\because \textbf x^1 \ge \textbf x^2 \implies \Sigma_{i=1}^{n} x_i^1 \ge \Sigma_{i=1}^{n}x_i^2$</span>...(3)</p>
<p><span class="math-container">$\therefore u(\textbf x^1)\ge u(\textbf x^2)$</span> (by (2))....(4)</p>
<p><span class="math-container">$\therefore f(\textbf x^1)\ge f(\textbf x^2)$</span> (by 3 and 4)</p>
<p>Proof (2):Book's solution manual
<a href="https://i.sstatic.net/pKtGg.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/pKtGg.png" alt="enter image description here" /></a></p>
<p>Counter example: <span class="math-container">$u(x_1,x_2)=x_1x_2$</span>
<span class="math-container">$u(1,4)=4$</span> and <span class="math-container">$u(2,2)=4$</span> and <span class="math-container">$(2,2)~(1,4)$</span> but <span class="math-container">$f(1,4)=9$</span> and <span class="math-container">$f(2,2)=8$</span></p>
<p>I believe that we are getting this counter example because we are inherently assuming strict convexity / convexity of preferences. We are not given that preferences are convex/strictly convex.</p>
<p>any thoughts?</p>

<mathematical-economics><consumer-theory><preferences>
Violation of Monotonicity of preferences
Karan Kumar
# Violation of Monotonicity of preferences

(Asked by: username_0 on 2021-11-14)

Hi I am reading Jehle and Reny in my master's course and I have come across a problem in one of the exercises. My instructor herself was a bit confused when a student gave her a counter-example and then said that the example illustrates that our logic is faulty. I want to reconcile the example and math logic.
Here is the question with the solutions discussed.

Q1.24 Let u($\textbf x$) represent some consumer’s monotonic preferences over $\textbf x ∈ \mathbb{R}^n\_+$. For each of the functions $f (x)$ that follow, state whether or not f also represents the preferences of this consumer. In each case, be sure to justify your answer with either an argument or a counterexample.

Part (c) $f (x) = u(\textbf x) +\Sigma\_{i=1}^{n} x\_i$

Now I believe that this function is a monotonic transformation of $u(\textbf x)$ and thus is a representation of the inherent "monotonic" preferences. I provide two proofs stating this fact and one counter-example which is creating difficulty in understanding

Proof 1: (OP's proof)

Let $\textbf x^1 \ge \textbf x^2.$ (1)

Clearly, $\textbf x^1 \succsim \textbf x^2 $ (Since preferences are monotonic) (2)

$\because \textbf x^1 \ge \textbf x^2 \implies \Sigma\_{i=1}^{n} x\_i^1 \ge \Sigma\_{i=1}^{n}x\_i^2$...(3)

$\therefore u(\textbf x^1)\ge u(\textbf x^2)$ (by (2))....(4)

$\therefore f(\textbf x^1)\ge f(\textbf x^2)$ (by 3 and 4)

Proof (2):Book's solution manual
[![enter image description here](https://i.sstatic.net/pKtGg.png)](https://i.sstatic.net/pKtGg.png)

Counter example: $u(x\_1,x\_2)=x\_1x\_2$
$u(1,4)=4$ and $u(2,2)=4$ and $(2,2)~(1,4)$ but $f(1,4)=9$ and $f(2,2)=8$

I believe that we are getting this counter example because we are inherently assuming strict convexity / convexity of preferences. We are not given that preferences are convex/strictly convex.

any thoughts?

**Tags:** mathematical-economics, consumer-theory, preferences

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## 2 answers

### Answer by username_1 on 2021-11-14. Score: 6

In general, it will not represent the same preferences. There seems to be confusion on what "monotonic transformation" means in this context. It does not have much to do with monotonic preferences.

We say that the utility function $v:X\to\mathbb{R}$ is a *monotonic transformation* of the utility function $u:X\to\mathbb{R}$ if there exists a strictly increasing function $g:u(X)\to\mathbb{R}$ (the domain is the range of $u$) such that $v(x)=g(u(x))$ for all $x\in X$. A monotonic transformation of a utility function does always represent the same preferences, and this has nothing to do with whether these preferences are monotone, convex, or anything like that.

> **Comments:**
> **username_0**: But by the book's solution manual, it is indeed a monotonic transformation, f' is increasing, isn't it?  
> **username_1**: @KaranKumar Well, $f$ is increasing in the arguments and, therefore, represents monotone preferences. But it does not represent the same preferences as $u$. And $f$ is not a monotonic transformation of $u$ according to the definition given in the answer.  
> **username_0**: I disagree. Utility function in regards to representing the preferences are invariant to positive monotonic transformations. Since f' is greater than zero, therefore f and u represent the same preferences. Also, there does not seem to be any ambiguity as to how the function f is not a monotonic transformation as provided in the answer. Perhaps you could illustrate with an example or some argument as I am utterly missing your point.  
> **username_1**: @KaranKumar If $f$ would be a positive monotonic transformation of $u$, it would represent the same preferences. It does not, as your own example shows. The other "proofs" given by you do not show that it is a monotonic transformation. Where is the function $g$?  
> **username_1**: I actually found the "solution manual" you quote from. It is not an official solution manual for the book; it is someone typing up their own solutions. And making a mistake, here.  
> **username_2**: @KaranKumar First of all, if $f$ is a function on a vector space, then the proper terminology is $\nabla f$ rather than $f'$. Second, "$f: X \rightarrow Y$ is increasing" means $a <_X b \rightarrow f(a) <_Y f(b)$, which requires some order $<_X$ to be defined. Saying that $\nabla f$ is "greater than zero" again requires some ordering of the vector space, and the precise statement would then be that it's greater than the zero **vector**.  
> **username_3**: @KaranKumar: In particular, $v$ is a monotonic transformation of $u$ if (and only if) they both define the same ordering between preferences, i.e. if $v(x) ≥ v(y) \iff u(x) ≥ u(y)$ for all $x, y \in X$. When $X = \mathbb R$, it's easy to show that any two strictly increasing functions are monotonic transformations of each other, but this does not generalize easily to $X = \mathbb R^n$ for $n > 1$.  
> **username_3**: … In particular, $u$ and $v$ being monotonic transformations implies that they must have the same level sets (i.e. indifference curves); this is trivial for strictly increasing functions in one dimension (where the level sets are all just single points), but not in more than one dimension. And this (that $f$ and $u$ have different level sets) is exactly what your counterexample shows.  
> **username_0**: @MichaelGreinecker The $g$ that you are asking is g(x) = $u(\textbf x)$ + $<(u^-1(\textbf x))^T,e>$ Clearly g is defined on reals as $u(\textbf x)$ is a scalar and the dot product also gives a scalar. But I don't know how to differentiate a dot product  
> **username_0**: @Accumulation the problem is not to prove whether f or u defined on vectors is having a positive gradient but whether g defined on reals (scalars) has a positive differentiation for all the domain values  
> **username_0**: @Accumulation your logical example is extremely helpful by the way. But shouldn't it apply on say more things. Like if I consume 2 half litre cans of coke and 1 litre can of coke we get the same utility - also assume they are  priced so that half a can of coke costs half the price of coke. but by your logic if we add the two cans we will get  utility of say 3 and utility of 1 can will be 2! What then? Shouldn't we be indifferent?  
> **username_1**: @KaranKumar $u^{-1}(x)$ is a set of points (an indifference curve), not a single point.  

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### Answer by username_2 on 2021-11-14. Score: 0

We can simplify by assuming that $u$ is linear, in which we can treat $u$ as the total dollar value of the set of items, and $f$ to be the total dollar value plus the count of the items. Or, in other words, $f$ adds $1$ to the dollar value of each item. Would you prefer $1000$ pennies to one diamond worth $\\\$100$? $u(1000 \text
{ pennies}) = 10$, $u(\text{diamond})=100$, so $u$ says the diamond is better. $f(1000 \text
{ pennies}) = 1010$, $f(\text{diamond})=101$, so $f$ says the pennies are better.

The fact that $f$ is "monotonic" (a more precise phrasing would be that it's monotonic with respect to each of the components) just means that $f$ gives a higher number than $u$ for the same basket of goods. It doesn't say anything about what happens when comparing different baskets of goods.

> **Comments:**
> **username_0**: @Accumulation your logical example is extremely helpful by the way. But shouldn't it apply on say more things. Like if I consume 2 half litre cans of coke and 1 litre can of coke we get the same utility - also assume they are priced so that half a can of coke costs half the price of coke. but by your logic if we add the two cans we will get utility of say 3 and utility of 1 can will be 2! What then? Shouldn't we be indifferent?  
> **username_2**: @KaranKumar $f$ is given in terms of a particular meaning of $x_i$. Changing what $x_i$ means changes what $f$ is in terms of $x_i$.  
> **username_0**: Thank you all, we can close this question  

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