<p>No. Let <span class="math-container">$X=[-1,1]$</span> and <span class="math-container">$u:X\to\mathbb{R}$</span> be given by <span class="math-container">$u(x)=-|x|$</span>. Then, <span class="math-container">$u$</span> represents strictly convex preferences. Moreover, <span class="math-container">$0.5\succ -1$</span>. Now, the only points that <span class="math-container">$0.5$</span> is revealed preferred to are larger points, since every convex set containing <span class="math-container">$0.5$</span> and smaller points must contain strictly better points. Now, every larger point in turn can only be revealed preferred to even larger points, and so on. It follows that <span class="math-container">$0.5$</span> is only indirectly revealed preferred to larger points but not <span class="math-container">$-1$</span>.</p>

CC BY-SA 4.0
2023-12-07T05:56:38.270Z
So it seems like that we need some path-continuity assumption, and one-dimensional case is not rich enough. Perhaps in higher dimension it works. Anyways I think if can make it work with normal economic assumptions, then its economic intuition would be nice.
High GPA
CC BY-SA 4.0
2023-12-07T14:41:55.733Z
If preferences are continuous and strictly monotone, something like the following might work: Pick a utility representation $u$. If $u(x)>u(y)$, find a sufficiently well-behaved path from $y$ to $x$ along which $u$ increases. Replace the path by a piecewise-linear path with the same property. The linear pieces are the budget sets used to verify that $x P' y$.
Michael Greinecker
CC BY-SA 4.0
2023-12-07T17:53:56.803Z
So now the key point is if this path exists
High GPA
CC BY-SA 4.0
2023-12-08T02:38:41.100Z
I think strict monotonicity might be insufficient. Consider a utility function $u(x,y)=x+y$. Let the revealed demand at wealth=1, price=1 be (x,y)=(1/2,1/2). It follows that (1/2,1/2) is revealed preferred to (1/3,1/3). However, the point (1,0) is not revealed preferred to (1/3,1/3), as (1,0) was never chosen in any budget set. Am I correct?
High GPA
CC BY-SA 4.0
2023-12-08T06:59:46.473Z
Theses preferences are not strictly convex. But more importantly, your question allows for all nonempty convex compact subsets as "budget sets," not just the traditional closed halfspaces. So you can just take the line segment between (1/3,1/3) and (1,0) as a budget set to show that the latter is revealed preferred to the former.
Michael Greinecker
CC BY-SA 4.0
2023-12-08T07:33:13.217Z
One can also find a traditional budget set here that does the job: $p_x=1$, $p_y=2$, and $w=1$.
Michael Greinecker
CC BY-SA 4.0
2023-12-06T23:25:22.670Z
2023-12-06T23:25:22.670Z
Michael Greinecker
<p>Let <span class="math-container">$X$</span> be the closed compact convex set of alternative and <span class="math-container">$B$</span> be a closed compact convex subset of <span class="math-container">$X$</span>. <span class="math-container">$C$</span> is defined on all closed compact convex set <span class="math-container">$B\subseteq X$</span>.</p>
<p><span class="math-container">$X$</span> is ordered by a strictly convex preference (not necessarily continuous).</p>
<p><span class="math-container">$C$</span> is a well-defined choice function choosing the maximal element from <span class="math-container">$B$</span>. Function <span class="math-container">$C:2^B\to B$</span>. If follows that <span class="math-container">$C(B)$</span> is unique for each <span class="math-container">$B$</span></p>
<p>Define <span class="math-container">$xPy$</span> if <span class="math-container">$x=C(B)$</span> and <span class="math-container">$y\in B$</span>. If <span class="math-container">$xPy$</span> and <span class="math-container">$yPz$</span> , we say <span class="math-container">$xP'z$</span>.</p>
<p><span class="math-container">$P$</span> is the direct revealed preference relation. <span class="math-container">$P'$</span> is the indirect one?</p>
<p><strong>Can we say <strong><span class="math-container">$P'$</span> is complete</strong>, in a sense that if <span class="math-container">$x\succ y$</span> then <span class="math-container">$xP'y$</span>?</strong>
This means any pair of <span class="math-container">$x,y$</span> can be ranked by the indirect revealed preference.</p>
<p>To prove that <span class="math-container">$xP'y$</span>, we need to find a finite sequence that <span class="math-container">$z_1,z_2...z_n$</span> such that <span class="math-container">$xPz_1P...Pz_nPy$</span>.</p>
<p>If you draw two indifference curve, and choose one point from each, you can see that it is usually not the case the <span class="math-container">$xPy$</span>, but usually you can draw a new IC in between, find a <span class="math-container">$z$</span>, to make <span class="math-container">$xPzPy$</span>.</p>
<hr />
<p>Side-Note that in classic RP theory, the data set is usually finite, and therefore <span class="math-container">$P'$</span> cannot be complete. Here <span class="math-container">$C$</span> is a function so completeness might be possible.</p>
<p>Strong Axiom of revealed preference: <span class="math-container">$xP' z$</span> and <span class="math-container">$x\neq z$</span> implies <span class="math-container">$z\not P' x$</span>. SARP implies the existence of <span class="math-container">$u$</span> such that <span class="math-container">$x=\arg\max_{x\in B}u(x)$</span>. It seems to me that SARP is not enough as it is possible that <span class="math-container">$x\sim y$</span> and then <span class="math-container">$x\not P'y$</span> and <span class="math-container">$y\not P'x$</span>.</p>

CC BY-SA 4.0
2023-12-06T15:25:52.380Z
You might want to consult the papers of [Mas-Collel(1977)](https://doi.org/10.2307/1912308) and [Mas-Collel(1978)](https://doi.org/10.2307/2297089).
tdm
CC BY-SA 4.0
2023-12-06T15:42:51.943Z
You probably need some substitute for continuity to guarantee $C(B)\neq\emptyset$.
Michael Greinecker
CC BY-SA 4.0
2023-12-06T18:39:46.487Z
@MichaelGreinecker Yes you are right. I think we define $C$ as well-defined function.
High GPA
<microeconomics><preferences><decision-theory><choice-theory>
In revealed preference (RP), is any two points $x,y$ related by the indirect revealed preference relation?
High GPA
# In revealed preference (RP), is any two points $x,y$ related by the indirect revealed preference relation?

(Asked by: username_0 on 2023-12-06)

Let $X$ be the closed compact convex set of alternative and $B$ be a closed compact convex subset of $X$. $C$ is defined on all closed compact convex set $B\subseteq X$.

$X$ is ordered by a strictly convex preference (not necessarily continuous).

$C$ is a well-defined choice function choosing the maximal element from $B$. Function $C:2^B\to B$. If follows that $C(B)$ is unique for each $B$

Define $xPy$ if $x=C(B)$ and $y\in B$. If $xPy$ and $yPz$ , we say $xP'z$.

$P$ is the direct revealed preference relation. $P'$ is the indirect one?

**Can we say **$P'$ is complete**, in a sense that if $x\succ y$ then $xP'y$?**
This means any pair of $x,y$ can be ranked by the indirect revealed preference.

To prove that $xP'y$, we need to find a finite sequence that $z\_1,z\_2...z\_n$ such that $xPz\_1P...Pz\_nPy$.

If you draw two indifference curve, and choose one point from each, you can see that it is usually not the case the $xPy$, but usually you can draw a new IC in between, find a $z$, to make $xPzPy$.

---

Side-Note that in classic RP theory, the data set is usually finite, and therefore $P'$ cannot be complete. Here $C$ is a function so completeness might be possible.

Strong Axiom of revealed preference: $xP' z$ and $x\neq z$ implies $z\not P' x$. SARP implies the existence of $u$ such that $x=\arg\max\_{x\in B}u(x)$. It seems to me that SARP is not enough as it is possible that $x\sim y$ and then $x\not P'y$ and $y\not P'x$.

**Tags:** microeconomics, preferences, decision-theory, choice-theory
> **Comments:**
> **username_1**: You might want to consult the papers of [Mas-Collel(1977)](https://doi.org/10.2307/1912308) and [Mas-Collel(1978)](https://doi.org/10.2307/2297089).  
> **username_2**: You probably need some substitute for continuity to guarantee $C(B)\neq\emptyset$.  
> **username_0**: @MichaelGreinecker Yes you are right. I think we define $C$ as well-defined function.  

---

## 1 answers

### Answer by username_2 on 2023-12-06. Score: 3

No. Let $X=[-1,1]$ and $u:X\to\mathbb{R}$ be given by $u(x)=-|x|$. Then, $u$ represents strictly convex preferences. Moreover, $0.5\succ -1$. Now, the only points that $0.5$ is revealed preferred to are larger points, since every convex set containing $0.5$ and smaller points must contain strictly better points. Now, every larger point in turn can only be revealed preferred to even larger points, and so on. It follows that $0.5$ is only indirectly revealed preferred to larger points but not $-1$.

> **Comments:**
> **username_0**: So it seems like that we need some path-continuity assumption, and one-dimensional case is not rich enough. Perhaps in higher dimension it works. Anyways I think if can make it work with normal economic assumptions, then its economic intuition would be nice.  
> **username_2**: If preferences are continuous and strictly monotone, something like the following might work: Pick a utility representation $u$. If $u(x)>u(y)$, find a sufficiently well-behaved path from $y$ to $x$ along which $u$ increases. Replace the path by a piecewise-linear path with the same property. The linear pieces are the budget sets used to verify that $x P' y$.  
> **username_0**: So now the key point is if this path exists  
> **username_0**: I think strict monotonicity might be insufficient. Consider a utility function $u(x,y)=x+y$. Let the revealed demand at wealth=1, price=1 be (x,y)=(1/2,1/2). It follows that (1/2,1/2) is revealed preferred to (1/3,1/3). However, the point (1,0) is not revealed preferred to (1/3,1/3), as (1,0) was never chosen in any budget set. Am I correct?  
> **username_2**: Theses preferences are not strictly convex. But more importantly, your question allows for all nonempty convex compact subsets as "budget sets," not just the traditional closed halfspaces. So you can just take the line segment between (1/3,1/3) and (1,0) as a budget set to show that the latter is revealed preferred to the former.  
> **username_2**: One can also find a traditional budget set here that does the job: $p_x=1$, $p_y=2$, and $w=1$.  

---

