A geometric approach to revealed preference via Hamiltonian cycles

Abstract

It is shown that a fundamental question of revealed preference theory, namely whether the weak axiom of revealed preference (WARP) implies the strong axiom of revealed preference (SARP), can be reduced to a Hamiltonian cycle problem: A set of bundles allows a preference cycle of irreducible length if and only if the convex monotonic hull of these bundles admits a Hamiltonian cycle. This leads to a new proof to show that preference cycles can be of arbitrary length for more than two but not for two commodities. For this, it is shown that a set of bundles satisfying the given condition exists if and only if the dimension of the commodity space is at least three. Preference cycles can be constructed by embedding a cyclic $(L-1)$-polytope into a facet of a convex monotonic hull in $L$-space, because cyclic polytopes always admit Hamiltonian cycles. An immediate corollary is that WARP only implies SARP for two commodities. The proof is intuitively appealing as this gives a geometric interpretation of preference cycles.

Appendix

1.1 Proof of Theorem 1

Proof (Theorem 1)

Without loss of generality, we let $\mathcal{I }= \{1,\ldots ,K\}$ to simplify the exposition of the proof.

(1) $\Rightarrow$ (2) Assume that the conditions in (1) are satisfied. We need to show that

• (i) all $x^i \in T$ are distinct vertices on $\text{ CMH }(T)$, and

• (ii) $x^{i-1}$ is adjacent to $x^{i}$ for all $i \in \mathcal{I }$ and $x^K$ is adjacent to $x^1$.

(i) If $x^i$ is not a distinct vertex, then either

• (i.1) $x^i$ is not distinct (i.e., $x^i = x^j$ for some $i \ne j$), or

• (i.2) $x^i$ is not a vertex because $x^i \in \partial \text{ CMH }(T) \backslash \text{ CH }(T)$ or $x^i \in {{ int}}\text{ CMH }(T)$, or

• (i.3) $x^i$ is not a vertex because $x^i \in \partial \text{ CH }(T) \backslash \text{ CMH }(T)$.

(i.1) Suppose all elements of $T$ are vertices but not all are distinct, i.e. $x^i = x^j$ for some $j \ne i$. Then $\{x^i,x^j\} \subset B^i$, which implies $x^i R^0x^j$, and $\{x^i,x^j\} \subset B^j$, which implies $x^j R^0x^i$. Thus, the set indexed by $\mathcal{I }^{\prime } = \{i,j\}$ has a revealed preference cycle of length $2 < K$, contradicting (1).

(i.2) Suppose $x^i \in T$ is not a vertex because either $x^i \in \partial \text{ CMH }(T) \backslash \text{ CH }(T)$ (a point on the boundary of $\text{ CMH }(T)$ but on the monotonic extension of $\text{ CH }(T)$) or $x^i \in {{ int}}B^i$. In both cases, there exists an $x \in \partial \text{ CH }(T)$ such that $x^i \ge x$ (i.e. $x^i_j \ge x_j$ and $x^i \ne x^j$). Then $x \in {{ int}}B^i$. But by (1), we have $x^{i-1} R^0x^i$ and therefore $x^i \in B^{i-1}$, and therefore $x \in {{ int}}B^{i-1}$.

Suppose first that $x = x^j$ for some $x^j \in T$. Then $x^i R^0x^j$, and because $x^j \in B^{i-1}$, is also true that $x^{i-1} R^0x^j$. Then if $j < i-1$ there exists a set $\mathcal{I }^{\prime } \subset \mathcal{I }$ with $\mathcal{I }^{\prime } = \{j,j+1,j+2,\ldots ,i-2,i-1\}$ with $x^j R^0x^{j+1} \ldots R^0x^{i-1}$, and with $x^{i-1} R^0x^j$, thus the set indexed by $\mathcal{I }^{\prime }$ forms a revealed preference cycle of length $K^{\prime } = K - 1$, contradicting (1). If $j > i$, then the set indexed by $\mathcal{I }^{\prime } = \{i-1,j,j+1,\ldots ,K-1,K,1,\ldots ,j-1,j\}$ forms a revealed preference cycle of length $K^{\prime } = K - 1$, again contradicting (1).

So suppose $x \notin T$. As $x \in \partial \text{ CH }(T)$ it can be expressed as a linear combination of at least two elements of $T \backslash \{x^i\}$ which are on the boundary of $\text{ CH }(T)$. Suppose $x$ is a linear combination of $\{x^j\}_{j \in \mathcal{K }}$ with $\mathcal{K }\subset \mathcal{I }, i \notin \mathcal{K }$, and $x^j \in \partial \text{ CH }(T)$ for all $j \in \mathcal{K }$. Then $\text{ CH }(\{x^j\}_{j \in \mathcal{K }}) \subseteq \partial \text{ CH }(T)$. As $x^j \in B^{i-1}$ implies $x^{i-1} R^0x^j$, in order to have a revealed preference cycle of length $K$ at most one element of $\{x^j\}_{j \in \mathcal{K }}$ can be in $B^{i-1}$. If there is an $x^j \in B^{i-1}, j \in \mathcal{K }$, then $j = i - 1$. As $i \notin \mathcal{K }$, there must be at least one $j \in \mathcal{K }$ with $j \ne i - 1$.

Suppose $i - 1 \notin \mathcal{K }$. Then $x^j \in B^{i+1}$ for all $j \in \mathcal{K }$. As $\partial B^{i-1}$ separates $\mathbb R ^L$ into two half-spaces, all $\{x^j\}_{j \in \mathcal{K }}$ are in the same half-space with respect to $\partial B^{i-1}$. As both $\text{ CH }(\{x^j\}_{j \in K_1})$ and a half-space are convex sets and $x \in B^{i-1}$ and $x^j \notin B^{i-1}, \text{ CH }(\{x^j\}_{j \in \mathcal{K }})$ is contained in the half-space opposite of $x$. But $x \in \text{ CH }(\{x^j\}_{j \in \mathcal{K }})$, a contradiction. Thus for some $j \in \mathcal{K }, x^j \in B^{i-1}$, and therefore $x^{i-1} R^0x^j$ with $j \ne i$. Then there exists a set of indices $\mathcal{I }^{\prime } \subset \mathcal{I }$ with $i \notin \mathcal{I }^{\prime }$ such that the set indexed by $\mathcal{I }^{\prime }$ forms a revealed preference cycle of length $K^{\prime } \le K -1$, contradicting (1).

Suppose $i - 1 \in \mathcal{K }$. As we have assumed that demand is exhaustive, $x^{i-1} \in \partial B^{i-1}$. This time, $x^j \notin B^{i-1}$ for all $j \in \mathcal{K }\backslash \{i-1\}$ for the same reasons as before. Then $\text{ CH }(\{x^j\}_{j \in \mathcal{K }}) \cap B^{i-1} = \{x^{i-1}\}$, but $x \in \text{ CH }(\{x^j\}_{j \in \mathcal{K }})$ and $x \notin B^{i-1}$, a contradiction.

(i.3) Suppose $x^i \in T$ is not a vertex because while it is on the boundary of $\text{ CH }(T)$ and not on the monotonic extension, it is not an extreme point. Then $x^i$ can be expressed as a linear combination of at least two elements of $T \backslash \{x^i\}$ which are on the boundary of $\text{ CH }(T)$. Then we can apply the same arguments as in part (i.2).

This completes the proof that all $x^i \in T$ are distinct vertices on $\text{ CMH }(T)$.

(ii) Suppose the line segment connecting $x^{i-1}$ and $x^{i} \in T$ (the edge joining $\{x^{i-1}\}$ and $\{x^i\}$) is not on the boundary of $\text{ CMH }(T)$, with $x^i \ne x^j$. With $\{x^{i-1},x^i\} \subset B^{i-1}$ and convexity of $B^{i-1}$, every $x = \lambda \,x^{i-1} + (1-\lambda )x^i$ for some $\lambda \in (0,1)$ is in $B^{i-1}$. At least one such $x$ must be in the interior of $\text{ CMH }(T)$ given that the line segment connecting $x^{i-1}$ and $x^i$ is not an edge. Thus, there must be some $\tilde{x} \le x$ on the boundary of $\text{ CMH }(T)$, with $\tilde{x} \in B^{i-1}$. As $\lambda \in (0,1)$ and $x^i \ne x^j, x \ne x^i$ and $x \ne x^j$.

Suppose $\tilde{x} = x^j$ implies $j \notin \{i-1,i\}$. Then $x^j \in B^{i-1}$ and therefore $x^{i-1} R^0x^j$. there exists a set of indices $\mathcal{I }^{\prime } \subset \mathcal{I }$ with $i \notin \mathcal{I }^{\prime }$ such that the set indexed by $\mathcal{I }^{\prime }$ forms a revealed preference cycle of length $K^{\prime } \le K -1$, contradicting (1). Suppose instead $\tilde{x} \notin T$; then $\tilde{x}$ is a linear combination of at least two elements of $T \backslash \{x^i\}$. At least one of those elements must be in $B^{i-1}$, because otherwise $\tilde{x} \le x$ cannot be a linear combination of them. Say, this element is $x^j$; then again, $x^j \in B^{i-1}$ and the same results as before follows.

Suppose the line segment connecting $x^{i-1}$ and $x^{i} \in T$ is on the boundary of $\text{ CMH }(T)$, but intersects with the line segment connecting $x^{j-1}$ and $x^j \in T$ for some $j \ne i$. Then $x = \lambda \,x^{i-1} + (1-\lambda )x^i = \mu \,x^{j-1} + (1-\mu )x^{j}$ for some $\lambda , \mu \in (0,1)$ is the intersection of these two line segments. Then a similar argument as in the preceding paragraph applies (i.e., one of $x^{j-1}$ or $x^j$ must be in $B^{i-1}$, which results in a subset with a revealed preference cycle of length less than $K$).

Thus, the line segments connecting $x^{i-1}$ and $x^i$ and $x^{j-1}$ and $x^j$ do not intersect and are edges of $\text{ CMH }(T)$. Then $x^{i-1}$ and $x^i$ are adjacent, and the sequence $x^1, x^2, \ldots , x^K, x^1$ is a Hamiltonian cycle, which concludes this part of the proof.

(2) $\Rightarrow$ (1) Suppose the conditions in (2) are satisfied. By definition of vertices and Hamiltonian cycles, if $\text{ CMH }(T)$ admits a Hamiltonian cycle involving all elements of $T$, then all elements of $T$ are vertices on $\text{ CMH }(T)$. Furthermore, if the sequence of vertices which constitutes the cycle is $x^1, x^2, \ldots , x^K, x^1$, then by definition $x^{i-1}$ and $x^{i}$ are adjacent for all $i \in \{2,3,\ldots ,K,1\}$. By the supporting hyperplane theorem, there exists a hyperplane $H(q) = \{y \in X: q y = a\}$ such that $x^{i-1} \in H(q)$ and $x^{i} \in H(q)$ for some $q \in \mathbb R ^L$ (where $q$ and $a$ depend on $x^{i-1}$ and $x^i$).

By the conditions (2) the line segment connecting $x^{i-1}$ and $x^i$ is an edge on $\text{ CMH }(T)$; thus, this line segment is a $1$-face on $\text{ CMH }(T)$. Let $F_{i,j}$ be that face. By the definition of faces in Sect. 2, there exists a supporting hyperplane $H^{\prime }$ such that $F_{i,j} = \text{ CMH }(T) \cap H^{\prime }$. Thus, there must exist a $q \in \mathbb R ^L$ such that $\{x^{i-1},x^i\} \subset H(q)$ and $F_{i,j} = H(q) \cap \text{ CMH }(T)$.

Next we show that $a$ can be chosen to be $1$. If $a = 0$, then ${\mathbf{0}}\in H(q)$. But then $x^{i-1} \in H(q)$ and $x^i \in H(q)$ implies that they are on a ray through the origin. Then either $x^{i-1} \le x^i$ or $x^i \le x^{i-1}$, but then either $x^i$ or $x^{i-1}$ is not a vertex of $\text{ CMH }(T)$. Thus $a \ne 0$ and $a = 1$ is possible by normalising $q$.

Next we show that each $q$ can be chosen such that $q > {\mathbf{0}}$. Clearly, if $a = 1$, at least one $q_j, j \in \{1,\ldots ,L\}$ must be strictly positive because all elements of $T$ are in $\mathbb R ^L_+$. Now suppose that for all supporting hyperplanes, $q_k \le 0$ for at least one $k \in \{1,\ldots ,L\}$. Then there is a $y \in \mathbb R _{++}^L$ with $y \ge x^{i-1}$ and $y \in H(q)$. But then $y$ is not a linear combination of $x^{i-1}$ and $x^i$, and therefore $y \notin F_{i,j}$. Then $H(q)$ is not the supporting hyperplane used to define the edge $F_{i,j}$. But there exists a supporting hyperplane $H(q^{\prime })$ for some $q^{\prime }$ with $q^{\prime }_j > 0$ such that $F_{i,j} = H(q^{\prime }) \cap \text{ CMH }(T)$, contradicting the assumption that for all supporting hyperplanes, $q_k \le 0$ for at least one $k \in \{1,\ldots ,L\}$.

Thus if $a = 1$, we must have $q > {\mathbf{0}}$.

Let $q$ be the price vector at which $x^{i-1}$ was chosen, so that $p^i = q, \partial B^{i-1} = \partial B(q) = H(q)$. As $H(q) \cap \text{ CMH }(T)$ is the edge connecting $x^{i-1}$ and $x^i$, we have $x^i \in \partial B^{i-1}$ and $x^j \notin B^{i-1}$ for all $j \notin \{i-1,i\}$. Then $x^{i-1} R^0x^i$ and $[\text{ not }~x^{i-1} R^0x^j]$ for all $j \notin \{i-1,i\}$. Thus we can find budget hyperplanes for each $i \in \mathcal{I }$ such that $\{(x^i,p^i)\}_{i \in \mathcal{I }}$ forms a preference cycle of irreducible length $K$.

1.2 Proof of Proposition 1

Proof (Proposition 1)

Let $T = \{x^i\}_{i \in \mathcal{I }}$, with $\mathcal{I }= \{1,\ldots ,K\}, K > 2$, be a set of bundles, such that each $x^i \in T$ is a distinct vertex on $\text{ CMH }(T)$. Let $a = \arg \max _i \{x^i_1\}_{i \in \mathcal{I }}, b = \arg \max _i \{x^i_2\}_{i \in \mathcal{I }}, y = (x_1^a + 1,x_2^a)$, and $z = (x_1^b,x_2^b+1)$. Any $\mathbf P _2$ with $K$ distinct vertices has at most $K$ edges, and every vertex is incident to at most two edges and is adjacent to at most two vertices (see Brøndsted 1983, Theorems 10.5 and 12.16). In the convex polytope $\text{ CH }(\{x^i\}_{i \in \mathcal{I }} \cup \{y,z\}), x^a$ is adjacent to $y$ and $x^b$ is adjacent to $z$. Thus, on $\text{ CH }(\{x^i\}_{i=1}^K \cup \{y,z\}), x^a$ and $x^b$ can each only be adjacent to one of the elements in $T$. With $\text{ CH }(\{x^i\}_{i \in \mathcal{I }} \cup \{y,z\}) \subset \text{ CMH }(\{x^i\}_{i \in \mathcal{I }})$ it follows that on $\text{ CMH }(\{x^i\}_{i \in \mathcal{I }})$ both $x^a$ and $x^b$ can only be adjacent to one other element of $T$.

We can count the total number of adjacencies: If $x^{i-1}$ and $x^i$ are adjacent, this is one and only one adjacency; thus, if there are $K$ distinct vertices of a $\mathbf P _2$, and each vertex is adjacent to two other vertices, then the total number of adjacencies is $K$. As $x^a$ and $x^b$ are adjacent to only one other vertex, the total number of adjacencies on $\text{ CMH }(\{x^i\}_{i \in \mathcal{I }})$ among the $K$ elements of $T$ can therefore only be at most $(K-1)$, which is one less than is needed for a Hamiltonian cycle. Thus, the conditions of Theorem 1 cannot be satisfied. See also Fig. 2 for an illustration.

1.3 Proof of Proposition 2

Proof (Proposition 2)

It needs to be shown that a set of bundles $T = \{x^i\}_{i \in \mathcal{I }}, T \subset \mathbb R _+^L$ which satisfies the conditions in Theorem 1 always exists. We will do so by embedding a cyclic polytope $\mathbf P _{L-1}$ in $\mathbb R _+^L$. Cyclic polytopes always admit Hamiltonian cycles (1966, Theorem 1.1).

Let $\mathbf A$ be an $L \times (L-1)$ matrix, defined as $\mathbf A _{i,j} = 0$ if $i > j$ and $\mathbf A _{i,j} = 1$ if $i \le j$, that is,

$$\begin{aligned} \mathbf A&= \left( \begin{array}{cccc} 0 &{} 0 &{}\cdots &{} 0 \\ 1 &{} 0 &{} \cdots &{} 0\\ 1 &{} 1 &{} \cdots &{} 0\\ \vdots &{} \vdots &{} \vdots &{} 0\\ 1 &{} 1 &{} \cdots &{} 1 \end{array} \right) \end{aligned}$$

Let $\mathbf a _j$ denote the $j$th row. Let $\tilde{\Delta }^{L-1} = \text{ CH }(\{\mathbf{a }_j\}_{j=1}^{L})$; $\tilde{\Delta }^{L-1}$ is an $(L-1)$-simplex in $\mathbb R ^{L-1}$. For $y \in \tilde{\Delta }^{L-1}$, let $\phi (y) = (1-y_1,y_1-y_2,\ldots ,y_{L-2}-y_{L-1},y_{L-1})$. Then $y = \sum _{j=1}^L \phi (y)_j \mathbf a _j$, that is, every $y \in \tilde{\Delta }^{L-1}$ can be expressed as a linear combination of the row vectors $\mathbf a _j$.

Let $\Delta ^{L-1}$ be the unit $(L-1)$-simplex in $\mathbb R ^{L}$, that is, the convex hull of the unit vectors in $\mathbb R ^L$. Let $\mathbf e _j$ be the $j$th unit vector in $\mathbb R ^L$, that is, $\mathbf e _1 = (1,0,\ldots ), \mathbf e _2 = (0,1,0,\ldots ), \ldots$.

The previously defined $\phi$ is a bijection $\phi : \tilde{\Delta }^{L-1} \rightarrow \Delta ^L$. For all $y \in \tilde{\Delta }^{L-1}$, we have

$$\begin{aligned} \phi (y)&= \phi \left( \sum \limits _{j=1}^L \phi (y)_j \mathbf a _j\right) \\&= \sum \limits _{j=1}^L \phi (y)_j \mathbf e _j. \end{aligned}$$

That is, every element $y \in \tilde{\Delta }^{L-1}$ can be expressed as a linear combination of the $\mathbf a _j$s and corresponds to an element $x \in \Delta ^{L-1}, x = \phi (y)$, which can be expressed as a linear combination of the unit vectors using the same scalars. By construction, for all sets $\{y^1,y^2,\ldots ,y^M\} \subset \tilde{\Delta }^{L-1}$ and for all $\lambda \in [0,1]^{M}$ with $\sum _{j=1}^{M} \lambda _j = 1$, we have

$$\begin{aligned} \phi \left( \sum \limits _{j=1}^{M} \lambda _j y^j \right) = \sum \limits _{j=1}^{M} \lambda _j \phi (y^j), \end{aligned}$$

that is, $\phi$ is an affine map from $\tilde{\Delta }^{L-1}$ to $\Delta ^{L-1}$.

Recall the definition of $f(t)$, the function that parameterises the moment curve $\mathcal M _{L-1}$: $f(t) = (t,t^2,\ldots ,t^{L-1})$, where $1,\ldots ,L-1$ are exponents. Then with $t \in [0,1]$ all $f(t) \in [0,1]^{L-1}$, and all $f(t)$ are contained in $\tilde{\Delta }^{L-1}$ (see Fig. 3a for an illustration). Therefore also the convex hull of all points or of a subset of points from the moment curve are contained in $\tilde{\Delta }^{L-1}$. Consider a set $T^{\prime } = \{f(t_1),\ldots ,f(t_{K})\}$ of $K$ points from $\mathcal M _{L-1}$, such that $\text{ CH }(T^{\prime })$ is a cyclic $\mathbf P _{L-1}$. Let $T = \{\phi (f(t_1)),\ldots ,\phi (f(t_K))\}$; then clearly $\text{ CH }(T)$ is a polytope which is combinatorially equivalent to the cyclic polytope $\text{ CH }(T^{\prime })$, as all adjacencies and the dimension of faces are preserved by the affine map $\phi$: If $F^{\prime } \subset \text{ CH }(T^{\prime })$ is a $k$-face of $\text{ CH }(T^{\prime })$, then $F = \{x \in \mathbb R _+^L: \exists y \in F^{\prime }, x = \phi (y)\}$ is a $k$-face of $\text{ CH }(T)$. See Fig. 3b for an illustration.

Let $H$ be the hyperplane in $\mathbb R ^L$ which contains $\Delta ^{L-1}$ (i.e., the affine hull of $\Delta ^{L-1}$). $\text{ CH }(T)$ is the cyclic polytope in $\mathbb R _+^L$ which we just constructed; we have $\text{ CH }(T) \subset H$. Then $H$ is a supporting hyperplane of $\text{ CMH }(T)$, and $\text{ CH }(T)$ is a facet of $\text{ CMH }(T)$ (i.e. an $(L-1)$-face). By construction, all $x \in T$ are distinct vertices on $\text{ CH }(T)$, thus also on $\text{ CMH }(T)$, and all edges of $\text{ CH }(T)$ are edges on $\text{ CMH }(T)$. Thus, $\text{ CMH }(T)$ allows a Hamiltonian cycle involving all $K$ vertices of $\text{ CH }(T)$ and therefore by Theorem 1 a preference cycle of irreducible length $K$ exists.