# [Economics] Convex rationalization when the budget sets are segments?

## Solution 1:

Take a dataset $D = (B^t, x^t)_{t \in T}$ such that for all $t$, $x^t \in B^t$. I'll say that $D$ is **rationalisable** by the utility function $u$ if for all $t$ and all $x \in B$: $u(x^t) \ge u(x)$.

If you only impose convexity on $u$, then any dataset $D$ is rationalisable by the constant (convex) utility function $u(x) = k$. So this has not testable implications.

Assume now that you impose **strict monotonicity** on $u$ in the sense that $x > y$ implies $u(x) > u(y)$.

Consider a budget set $\overline{y^t z^t}$ and take the comprehensive hull: $$ CH(\overline{y^t z^t}) = \left\{x \in \mathbb{R}^n: x \le w \text{ for some } w \in \overline{y^t z^t}\right\}. $$

Then we can show that the dataset $D = (\overline{y^t z^t}, x^t)_{t \in T}$ is rationalisable by a montone utility function $u$ **if and only if** the dataset $D' = (CH(\overline{y^t z^t}), x^t)_{t \in T}$ is rationalisable by the same monontone utility function $u$.

For the proof: $(\to)$ Let $D$ be rationalisable by the montone function $u$. We want to show that for all $t$ and all $x \in CH(\overline{y^t z^t})$ we have $u(x^t) \ge u(x)$. Indeed, if $x \in CH(\overline{y^t z^t})$, we know that $x \le w$ for some $w \in \overline{y^t z^t}$. As such: $$ u(x) \le u(w) \le u(x^t), $$ where the first inequality comes from the monotonicity conditionon $u$ and the last one from rationalisability of D.

$(\leftarrow)$ Let $D'$ be rationalisable by $u$. Let us first show that for all $t$, $x^t \in \overline{y^t z^t}$. If not, then as $x^t \in CH(\overline{y^t z^t})$ there must exist a $w \in \overline{y^t z^t}$ such that $x^t < w$; Then, by monotonicity of $u$ and rationalisability of $D'$: $$ u(x^t) < u(w) \le u(x^t), $$ a contradiction. Now for any $z \in \overline{y^t z^t}$ it then follows from rationalisability of $D'$ that $u(z) \le u(x^t)$, so $D$ is also rationalisable by $u$.

This shows the following equivalence:

- The dataset $D = (\overline{y^t, z^t}, x^t)_{t \in T}$ is rationalisable by a convex and montone utility function $u$.
- The dataset $D = (CH(\overline{y^t, z^t}), x^t)_{t \in T}$ is rationalisable by a convex and monotone utility function $u$.

As the sets $CH(\overline{y^t z^t})$ are comprehensive, the revealed preference conditions for answering the condition 2) can be found in the JET paper that you refer to (Revealed preference analysis for convex rationalizations on nonlinear budget sets)

## Solution 2:

The idea is to consider the hyperplane tangent at the indifference curve through $x^t$ as a "linear budget". These linear budgets have to include the set $\overline{y^t, z^t}$. Then making use of these these linear budgets, we can leverage the results of Matzkin & Richter (JET,1991, Testing strictly concave rationality) to obtain a strictly convex rationalisation..

**Afriat style Theorem** Let $D = (\overline{y^t z^t}, x^t)_{t \in T}$ be a dataset. Then the following are equivalent:

$D$ is rationalisable by a strictly monotone utility function with strictly convex indifference curves.

For all $t \in T$, there exists a vector $p^t \in \mathbb{R}^n_{++}$, and a number $u^t \in \mathbb{R}$ such that (i) for all $t, v, \in T$, if $x^v \ne x^t$: $$ u^v - u^t < p^t(x^v - x^t), $$ (ii) for all $t, v \in T$ with $x^v = x^t$, $$ u^t = u^v, $$ and (iii) for all $t \in T$: $$ p^t x^t \ge p^t y^t \text{ and } p^t x^t \ge z^t. $$

For all $t \in T$, there exists a vector $\tilde p^t \in \mathbb{R}^n_{++}$ such that (i) for all $t \in T$: $$ \tilde p^t x^t \ge \tilde p^t y^t \text{ and } \tilde p^t x^t \ge \tilde p^t z^t $$ and (ii): $$ (\tilde p^t, x^t)_{t \in T} \text{ satisfies SARP}. $$

$D$ is rationalisable by a strictly monotone and strictly concave utility function.

($1 \to 3$) The set $\overline{y^t z^t}$ is convex and the upper countour sets $UC(x^t) = \{y \in \mathbb{R}^n: u(y) \ge u(x^t)\}$ are strictly convex. As such, using a suitable supporting hyperplane theorem, we can find vectors $\tilde p^t$ tangent to $UC(x^t)$ at $x_t$ that separates $UC(x^t)$ from $\overline{y^t z^t}$ which means that $\tilde p^t x^t \ge \tilde p^t y^t$ and $\tilde p^t x^t \ge \tilde p^t z^t$. This shows (i). Also from strict monotonicity of $u$, we must have that $\tilde p^t \in \mathbb{R}^n_{++}$. Next, for all $w \ne x^t$ with $\tilde p^t w \le \tilde p^t x^t$, we have $u(x^t) > u(w)$.

Now, define $x^t S x^v$ if $x^t \ne x^v$ and $\tilde p^t x^t \ge \tilde p^t x^v$. Then it follows that $x^t S x^v$ implies $u(x^t) > u(x^v)$, so $S$ has to be acyclic, which shows that $(\tilde p^t, x^t)_{t \in T}$ satisfies SARP.

($2 \iff 3$) Follows from Theorem 2 in Matzkin & Richter (Testing strictly concave rationality, 1991, JET), where we define $p^t = \lambda^t \tilde p^t$.

($3 \to 4$) From Matzkin and Richter it follows that 3 implies the existence of a strictly monotone and strictly concave utility function such that for all $w$ with $\tilde p^t w \le \tilde p^t x^t$, we have $u(x^t) \ge u(w)$. Then if $w \in \overline{y^t z^t}$ we immediately have that $\tilde p^t w \le \tilde p^t x^t$ (as $w$ is a convex combination of $y^t$ and $z^t$) so $u(x^t) \ge u(w)$ as we wanted to show.

($4 \to 1$) Straighforward.

**Remark 1**: The conditions in 2 provide a set of linear inequalities which can be efficiently verified (except maybe for the strict inequality which will require some $\varepsilon$-tweaking).

**Remark 2**: To get some more intuition behind condtion 3-(i), consider the two dimensional case, $y^t, z^t, x^t \in \mathbb{R}^2$. See figure below. there are 3 cases. Here the dashed line gives the slope $\tilde p^t$

- If $y^t > z^t$. Then, due to strict monotonicity, $x^t$ has to be equal to $y^t$. There are (a priori) no restrictions on $\tilde p^t$ (see top left panel of figure). A similar reasoning holds if $z^t > y^t$.
- If $y^t$ and $z^t$ are not ordered and $x^t = y^t$ or $x^t = z^t$. Then the slope $\tilde p^t$ is either flatter, steeper, or equal to the slope from the line $\overline{y^t z^t}$. (panel top right or bottom left).
- If $y^t$ and $z^t$ are not ordered and $x^t$ is between $y^t$ and $z^t$, then $\tilde p^t$ has to equal the slope from the line $\overline{y^t z^t}$. (panel bottom right).