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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 $y\not P'x$.

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    $\begingroup$ You might want to consult the papers of Mas-Collel(1977) and Mas-Collel(1978). $\endgroup$
    – tdm
    Commented Dec 6, 2023 at 15:25
  • $\begingroup$ You probably need some substitute for continuity to guarantee $C(B)\neq\emptyset$. $\endgroup$
    – Michael Greinecker
    Commented Dec 6, 2023 at 15:42
  • $\begingroup$ @MichaelGreinecker Yes you are right. I think we define $C$ as well-defined function. $\endgroup$
    – High GPA
    Commented Dec 6, 2023 at 18:39

1 Answer 1

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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 $-1$.

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  • $\begingroup$ 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. $\endgroup$
    – High GPA
    Commented Dec 7, 2023 at 5:56
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    $\begingroup$ 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 $x P' y$. $\endgroup$
    – Michael Greinecker
    Commented Dec 7, 2023 at 14:41
  • $\begingroup$ So now the key point is if this path exists $\endgroup$
    – High GPA
    Commented Dec 7, 2023 at 17:53
  • $\begingroup$ 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? $\endgroup$
    – High GPA
    Commented Dec 8, 2023 at 2:38
  • $\begingroup$ 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. $\endgroup$
    – Michael Greinecker
    Commented Dec 8, 2023 at 6:59

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