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For a problem in revealed preference. Give bundles $x,y\in \mathbb R^n$, must there exist a budget $B\supset\{x,y\}$ and a demand $D(B)\in[x,y]$?

Intuitively, this mean that we have two bundles, and we want to determine a budget set $B$ including both $x$ and $y$, and the demand in $B$ is also in the interval $[x,y]$.

The budget set is determined by a price vector $p\in\mathbb R^n$.

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1 Answer 1

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Let $U$ be a $C^1$, monotone and quasi-concave utility function. Let $\nabla U(z)$ be the gradient of $U$ at the bundle $z$ (If we assume preferences are monotone then $\nabla U(z) \gg 0)$ for all $z \in \mathbb{R}^n$).

There are three cases to consider:

  1. $\nabla U(x) \cdot x \ge \nabla U(x) \cdot y$. In this case, we can take the budget with price $p = \nabla U(x)$ and total income $p \cdot x$. This case corresponds to the setting where $y$ is below the hyperplane of the indifference curve at $x$. In this case $x \in D(B)$ and $y \in B$

  2. $\nabla U(y) \cdot y \ge \nabla U(y) \cdot x$. In this case, we can take the budget with price $p = \nabla U(y)$ and total income $p \cdot y$. This case correponds to the setting where $x$ is below the hyperplane tangent to the indifference curve at $y$. In this case $y \in D(B)$ and $x \in B$.

  3. Assume that case 1 and case 2 do not hold and consider the following problem $$ \alpha^\ast = \max_{\alpha \in [0,1]} U(\alpha x + (1-\alpha) y). $$ The objective function is continuous and the constraint set $[0,1]$ is compact, so an optimal solution exists. Note that the derivative of the objective function is given by: $$ \nabla U(\alpha x + (1-\alpha)y) \cdot (x - y). $$ First, let us exclude $\alpha = 0$ and $\alpha = 1$ as optimal solutions.

    a) if $\alpha = 0$ is an optimal solution then the derivative of the objective function must be negative for $\alpha = 0$ (as it is on the boundary of the feasible set $[0,1]$): $$ \nabla U(y)(x-y) \le 0, $$ But this gives $\nabla U(y)\cdot y > \nabla U(y) \cdot x$, which gives case 2.

    b) if $\alpha = 1$ is an optimal solution, the derivative of the objective function must be positive at $\alpha = 1$: $$ \nabla U(x)(x - y) \ge 0 $$ But this gives $\nabla U(x)\cdot x \ge \nabla U(x) \cdot y$ which gives case 1.

    Conclude that the optimal value of $\alpha$ must be strictly between zero and 1. Let $z = \alpha x + (1-\alpha) y$. As the solution is interior, we have that: $$ \nabla U(z) \cdot (x - y) = 0. $$ In particular $$ \begin{align*} \nabla U(z) \cdot z &= \nabla U(z)\cdot (\alpha x + (1-\alpha) y),\\ &= \alpha \nabla U(z)\cdot(x - y) + \nabla U(z) \cdot y,\\ &= \nabla U(z) \cdot y. \end{align*} $$ and $$ \begin{align*} \nabla U(z) \cdot z &= \nabla U(z)\cdot (\alpha x + (1-\alpha) y),\\ &= (\alpha-1) \nabla U(z)\cdot(x - y) + \nabla U(z) \cdot x,\\ &= \nabla U(z) \cdot x. \end{align*} $$ As such, if we define the budget $B$ with price $p = \nabla U(z)$ and income $p \cdot z$, we have that $x, y \in B$ and that $z \in D(B)$ while $z \in [x,y]$.

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