<p>Statement: Consider goods <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> (and we denote the quantities of  by the same notation) such that they are perfect substitutes with the substitution ratio <span class="math-container">$1:n$</span>.</p>
<p>Assume the basic axioms including monotonicity and continuity of preferences to prove or disprove the equivalence of the two equations (given as (1) and (2) below) with the statement above.</p>
<p>I asked two questions earlier and I learnt:</p>
<ol>
<li><p><a href="https://economics.stackexchange.com/a/50353/33107">The above &quot;statement&quot; is equivalent to <span class="math-container">$U(x+1, y) = U(x, y+n)$</span> <span class="math-container">$\forall$</span> <span class="math-container">$(x,y) \in \mathbb{R}^2$</span></a>.</p>
</li>
<li><p><a href="https://economics.stackexchange.com/a/50354/33107">The same &quot;statement&quot; is also equivalent to <span class="math-container">$\frac{\partial U}{\partial x} = n \cdot \frac{\partial U}{\partial y}$</span> <span class="math-container">$\forall$</span> <span class="math-container">$(x,y) \in \mathbb{R}^2$</span></a>.</p>
</li>
</ol>
<p><strong>The goal is find under which conditions are (1) and (2) equivalent to the statement. If not, we will disregard one of the two (or both) from being a definition of the given 'statement'.</strong></p>
<p><strong>My thoughts:</strong> We know that the partial differential eq. is equivalent to <span class="math-container">$U = g(nx+y)$</span> (for differentiable <span class="math-container">$g$</span>) <span class="math-container">$\implies U(x+1,y) = U(x,y+n)$</span>. But the converse (that is, 1 <span class="math-container">$\implies$</span> 2) does not necessarily hold in general.</p>
<p><strong>Some observations</strong> (which need not be answered)<strong>:</strong></p>
<ol>
<li>If we take an example like <span class="math-container">$U(x,y) = \sin(2 \pi y)$</span>, then we see that the two above are not equivalent. But again, <span class="math-container">$U$</span> here does not describe a perfect substitute. This shows that without added conditions, they don't act as definitions since <span class="math-container">$U$</span> satisfies these but not the statement.</li>
<li>If a function is monotone and satisfies <span class="math-container">$U(x+1,y) = U(x, y+n)$</span>, then by monotonicity of <span class="math-container">$U$</span>, it is immediate that every point <span class="math-container">$(x,y)$</span> on the line <span class="math-container">$nx + y = c$</span> (for <span class="math-container">$c \in \mathbb{R})$</span> maps to <span class="math-container">$f(c)$</span> which is constant. Does this say <span class="math-container">$U$</span> has to be constant in general? (Probably not, but I have to think.)</li>
</ol>

CC BY-SA 4.0
2022-02-09T12:49:30.090Z
Are $x,y,n$ integer numbers here?
Giskard
CC BY-SA 4.0
2022-02-09T13:14:06.640Z
@Giskard Not really. That's what you said - that the "definition is general enough" in the earlier post. You can take $n$ to be an integer, but the rest, no.
not tdm's twin
CC BY-SA 4.0
2022-02-09T13:20:28.457Z
tsk, tsk, that is not [the exact quote](https://economics.stackexchange.com/questions/50352/mathematical-definition-of-perfect-substitutes/50353#comment81164_50353), you left out a fairly important word! (And you should never use quotation marks when now exactly quoting, that is bad practice.)
Giskard
CC BY-SA 4.0
2022-02-09T13:37:49.087Z
And do you no longer wish to assume monotonicity, like you did in the last question?
Giskard
CC BY-SA 4.0
2022-02-09T13:45:27.360Z
So it seems that this can be resolved quickly: when I wrote 

"*your definition seems general enough.*"

in a comment under your earlier question I was wrong! (To be fair in your question you also wrote that we should assume monotonicity. ) The problem is that

$$ \forall (x,y) \in \mathbb{R}^2: U(x+1, y) = U(x, y+n) $$

only checks this indifference at certain points of a line: e.g., starting at $(x,y)$, the points $(x+1,y-n)$, $(x+2,y-2n)$, $(x+3,y-3n)$,... should be on the same indifference curve.
Giskard
CC BY-SA 4.0
2022-02-09T13:49:00.017Z
@Giskard Um, you said that if $x,y,n$ are reals and the equation holds for all reals, then it's true. (Or is it the "*seems* general enough" rather than "general enough" is what you're pointing out?)

As for the monotonicity, I said I am not sure if a function exists that is monotonous and satisfy both the equations (the recurrence and the PDE) simultaneously. So I decided to leave it to the answerer. Ideally, the desired utility function should be (at least weakly) monotonous.
not tdm's twin
CC BY-SA 4.0
2022-02-09T13:53:30.527Z
@Giskard Yes, I did say we should assume monotonicity. I still do for this question. I am looking for a proof of their equivalence this time (although I would like to see the case when monotonicity is not assumed). As an additional note, $x$ and $y$ are variables while $n$ is a given constant.
not tdm's twin
CC BY-SA 4.0
2022-02-09T13:59:10.430Z
Yes, I included the word "seems" to note that it seemed to be right, but I was not sure as I hadn't given this case much thought. My semi-educated guess was wrong.
Giskard
CC BY-SA 4.0
2022-02-09T15:37:29.937Z
@Giskard Was wrong for even for the monotonous case?
not tdm's twin
CC BY-SA 4.0
2022-02-09T15:40:39.603Z
I didn't think much about that before. Without also assuming continuity, yes, monoticity is not enough.
Giskard
CC BY-SA 4.0
2022-02-09T16:26:03.420Z
@Giskard Sure, let's assume continuity as well. Then? I am looking for a proof rather a yes/no answer.
not tdm's twin
CC BY-SA 4.0
2022-02-09T16:31:51.990Z
I am looking for a clear, well-defined question with all assumptions edited into it rather than back-and-forth comments. I suspect other users prefer the same.
Giskard
CC BY-SA 4.0
2022-02-09T16:33:53.177Z
@Giskard I tried to, but since you're adding assumptions, I am adding them as well. This is not a well-defined question; I am trying to work out the various possibilities. Let me fit these two criteria (assumptions) in the question. Edited.
not tdm's twin
CC BY-SA 4.0
2022-02-09T16:37:38.593Z
Hey, I am not adding assumptions to the question. You have to state what it is exactly that you want. You asked me about the monotonous case, I answered. I pointed out what feature you could use to produce a counterexample. This is not "adding assumptions" in any sense of the word, please do not attribute things to me that I did not do. Good day.
Giskard
CC BY-SA 4.0
2022-02-09T16:40:56.747Z
@Giskard Sorry if anything sounded offending. English is not my first language and I didn't have a better way to write that. But I appreciate your help (in case you feel offended and don't intend to discuss this anymore).
not tdm's twin
CC BY-SA 4.0
2022-02-09T16:41:54.427Z
It's allright. Please gather all details/assumption that you want to include and edit your question accordingly. Do not be afraid to remove unnecessary details to make it clear and concise. E.g., having "*statement*" instead of an actual definition, such as "preference represented by the utility function $U(x,y) = nx + y$" is quite unclear and probably what got the question its downvote.
Giskard
CC BY-SA 4.0
2022-02-09T17:03:17.593Z
@Giskard I didn't exactly remove the statement but I have tried to make the question more clear and added what each paragraph is for (like, goal, assumptions, thoughts and observations that might help the answerer).
not tdm's twin
<utility><preferences><marginal><choice-theory><perfect-substitutes>
Perfect substitutes mathematical definitions not equivalent
not tdm's twin
# Perfect substitutes mathematical definitions not equivalent

(Asked by: username_0 on 2022-02-09)

Statement: Consider goods $X$ and $Y$ (and we denote the quantities of by the same notation) such that they are perfect substitutes with the substitution ratio $1:n$.

Assume the basic axioms including monotonicity and continuity of preferences to prove or disprove the equivalence of the two equations (given as (1) and (2) below) with the statement above.

I asked two questions earlier and I learnt:

1. [The above "statement" is equivalent to $U(x+1, y) = U(x, y+n)$ $\forall$ $(x,y) \in \mathbb{R}^2$](https://economics.stackexchange.com/a/50353/33107).
2. [The same "statement" is also equivalent to $\frac{\partial U}{\partial x} = n \cdot \frac{\partial U}{\partial y}$ $\forall$ $(x,y) \in \mathbb{R}^2$](https://economics.stackexchange.com/a/50354/33107).

**The goal is find under which conditions are (1) and (2) equivalent to the statement. If not, we will disregard one of the two (or both) from being a definition of the given 'statement'.**

**My thoughts:** We know that the partial differential eq. is equivalent to $U = g(nx+y)$ (for differentiable $g$) $\implies U(x+1,y) = U(x,y+n)$. But the converse (that is, 1 $\implies$ 2) does not necessarily hold in general.

**Some observations** (which need not be answered)**:**

1. If we take an example like $U(x,y) = \sin(2 \pi y)$, then we see that the two above are not equivalent. But again, $U$ here does not describe a perfect substitute. This shows that without added conditions, they don't act as definitions since $U$ satisfies these but not the statement.
2. If a function is monotone and satisfies $U(x+1,y) = U(x, y+n)$, then by monotonicity of $U$, it is immediate that every point $(x,y)$ on the line $nx + y = c$ (for $c \in \mathbb{R})$ maps to $f(c)$ which is constant. Does this say $U$ has to be constant in general? (Probably not, but I have to think.)

**Tags:** utility, preferences, marginal, choice-theory, perfect-substitutes

> **Comments:**
> **username_1**: Are $x,y,n$ integer numbers here?  
> **username_0**: @username_1 Not really. That's what you said - that the "definition is general enough" in the earlier post. You can take $n$ to be an integer, but the rest, no.  
> **username_1**: tsk, tsk, that is not [the exact quote](https://economics.stackexchange.com/questions/50352/mathematical-definition-of-perfect-substitutes/50353#comment81164_50353), you left out a fairly important word! (And you should never use quotation marks when now exactly quoting, that is bad practice.)  
> **username_1**: And do you no longer wish to assume monotonicity, like you did in the last question?  
> **username_1**: So it seems that this can be resolved quickly: when I wrote   "*your definition seems general enough.*"  in a comment under your earlier question I was wrong! (To be fair in your question you also wrote that we should assume monotonicity. ) The problem is that  $$ \forall (x,y) \in \mathbb{R}^2: U(x+1, y) = U(x, y+n) $$  only checks this indifference at certain points of a line: e.g., starting at $(x,y)$, the points $(x+1,y-n)$, $(x+2,y-2n)$, $(x+3,y-3n)$,... should be on the same indifference curve.  
> **username_0**: @username_1 Um, you said that if $x,y,n$ are reals and the equation holds for all reals, then it's true. (Or is it the "*seems* general enough" rather than "general enough" is what you're pointing out?)  As for the monotonicity, I said I am not sure if a function exists that is monotonous and satisfy both the equations (the recurrence and the PDE) simultaneously. So I decided to leave it to the answerer. Ideally, the desired utility function should be (at least weakly) monotonous.  
> **username_0**: @username_1 Yes, I did say we should assume monotonicity. I still do for this question. I am looking for a proof of their equivalence this time (although I would like to see the case when monotonicity is not assumed). As an additional note, $x$ and $y$ are variables while $n$ is a given constant.  
> **username_1**: Yes, I included the word "seems" to note that it seemed to be right, but I was not sure as I hadn't given this case much thought. My semi-educated guess was wrong.  
> **username_0**: @username_1 Was wrong for even for the monotonous case?  
> **username_1**: I didn't think much about that before. Without also assuming continuity, yes, monoticity is not enough.  
> **username_0**: @username_1 Sure, let's assume continuity as well. Then? I am looking for a proof rather a yes/no answer.  
> **username_1**: I am looking for a clear, well-defined question with all assumptions edited into it rather than back-and-forth comments. I suspect other users prefer the same.  
> **username_0**: @username_1 I tried to, but since you're adding assumptions, I am adding them as well. This is not a well-defined question; I am trying to work out the various possibilities. Let me fit these two criteria (assumptions) in the question. Edited.  
> **username_1**: Hey, I am not adding assumptions to the question. You have to state what it is exactly that you want. You asked me about the monotonous case, I answered. I pointed out what feature you could use to produce a counterexample. This is not "adding assumptions" in any sense of the word, please do not attribute things to me that I did not do. Good day.  
> **username_0**: @username_1 Sorry if anything sounded offending. English is not my first language and I didn't have a better way to write that. But I appreciate your help (in case you feel offended and don't intend to discuss this anymore).  
> **username_1**: It's allright. Please gather all details/assumption that you want to include and edit your question accordingly. Do not be afraid to remove unnecessary details to make it clear and concise. E.g., having "*statement*" instead of an actual definition, such as "preference represented by the utility function $U(x,y) = nx + y$" is quite unclear and probably what got the question its downvote.  
> **username_0**: @username_1 I didn't exactly remove the statement but I have tried to make the question more clear and added what each paragraph is for (like, goal, assumptions, thoughts and observations that might help the answerer).  

---

## 0 answers

