<p>Let <span class="math-container">$U$</span> be a <span class="math-container">$C^1$</span>, monotone and quasi-concave utility function. Let <span class="math-container">$\nabla U(z)$</span> be the gradient of <span class="math-container">$U$</span> at the bundle <span class="math-container">$z$</span> (If we assume preferences are monotone then <span class="math-container">$\nabla U(z) \gg 0)$</span> for all <span class="math-container">$z \in \mathbb{R}^n$</span>).</p>
<p>There are three cases to consider:</p>
<ol>
<li><p><span class="math-container">$\nabla U(x) \cdot x \ge \nabla U(x) \cdot y$</span>. In this case, we can take the budget with price <span class="math-container">$p = \nabla U(x)$</span> and total income <span class="math-container">$p \cdot x$</span>. This case corresponds to the setting where <span class="math-container">$y$</span> is below the hyperplane of the indifference curve at <span class="math-container">$x$</span>. In this case <span class="math-container">$x \in D(B)$</span> and <span class="math-container">$y \in B$</span></p>
</li>
<li><p><span class="math-container">$\nabla U(y) \cdot y \ge \nabla U(y) \cdot x$</span>. In this case, we can take the budget with price <span class="math-container">$p = \nabla U(y)$</span> and total income <span class="math-container">$p \cdot y$</span>. This case correponds to the setting where <span class="math-container">$x$</span> is below the hyperplane tangent to the indifference curve at <span class="math-container">$y$</span>. In this case <span class="math-container">$y \in D(B)$</span> and <span class="math-container">$x \in B$</span>.</p>
</li>
<li><p>Assume that case 1 and case 2 do not hold and consider the following problem
<span class="math-container">$$
\alpha^\ast = \max_{\alpha \in [0,1]} U(\alpha x + (1-\alpha) y).
$$</span>
The objective function is continuous and the constraint set <span class="math-container">$[0,1]$</span> is compact, so an optimal solution exists.
Note that the derivative of the objective function is given by:
<span class="math-container">$$
\nabla U(\alpha x + (1-\alpha)y) \cdot (x - y).
$$</span>
First, let us exclude <span class="math-container">$\alpha = 0$</span> and <span class="math-container">$\alpha = 1$</span> as optimal solutions.</p>
<p>a) if <span class="math-container">$\alpha = 0$</span> is an optimal solution then the derivative of the objective function must be negative for <span class="math-container">$\alpha = 0$</span> (as it is on the boundary of the feasible set <span class="math-container">$[0,1]$</span>):
<span class="math-container">$$
\nabla U(y)(x-y) \le 0,
$$</span>
But this gives <span class="math-container">$\nabla U(y)\cdot y &gt; \nabla U(y) \cdot x$</span>, which gives case 2.</p>
<p>b) if <span class="math-container">$\alpha = 1$</span> is an optimal solution, the derivative of the objective function must be positive at <span class="math-container">$\alpha = 1$</span>:
<span class="math-container">$$
\nabla U(x)(x - y) \ge 0
$$</span>
But this gives <span class="math-container">$\nabla U(x)\cdot x \ge \nabla U(x) \cdot y$</span> which gives case 1.</p>
<p>Conclude that the optimal value of <span class="math-container">$\alpha$</span> must be strictly between zero and 1. Let <span class="math-container">$z = \alpha x + (1-\alpha) y$</span>. As the solution is interior, we have that:
<span class="math-container">$$
\nabla U(z) \cdot (x - y) = 0.
$$</span>
In particular
<span class="math-container">$$
\begin{align*}
\nabla U(z) \cdot z &amp;= \nabla U(z)\cdot (\alpha x + (1-\alpha) y),\\
&amp;= \alpha \nabla U(z)\cdot(x - y) + \nabla U(z) \cdot y,\\
&amp;= \nabla U(z) \cdot y.
\end{align*}
$$</span>
and
<span class="math-container">$$
\begin{align*}
\nabla U(z) \cdot z &amp;= \nabla U(z)\cdot (\alpha x + (1-\alpha) y),\\
&amp;= (\alpha-1) \nabla U(z)\cdot(x - y) + \nabla U(z) \cdot x,\\
&amp;= \nabla U(z) \cdot x.
\end{align*}
$$</span>
As such, if we define the budget <span class="math-container">$B$</span> with price <span class="math-container">$p = \nabla U(z)$</span> and income <span class="math-container">$p \cdot z$</span>, we have that <span class="math-container">$x, y \in B$</span> and that <span class="math-container">$z \in D(B)$</span> while <span class="math-container">$z \in [x,y]$</span>.</p>
</li>
</ol>

CC BY-SA 4.0
2023-09-28T13:00:47.600Z
2023-09-28T13:00:47.600Z
tdm
<p>For a problem in revealed preference. Give bundles <span class="math-container">$x,y\in \mathbb R^n$</span>,  must there exist a budget <span class="math-container">$B\supset\{x,y\}$</span> and a demand <span class="math-container">$D(B)\in[x,y]$</span>?</p>
<p>Intuitively, this mean that we have two bundles, and we want to determine a budget set <span class="math-container">$B$</span> including both <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, and the demand in <span class="math-container">$B$</span> is also in the interval <span class="math-container">$[x,y]$</span>.</p>
<p>The budget set is determined by a price vector <span class="math-container">$p\in\mathbb R^n$</span>.</p>

<microeconomics><consumer-theory><preferences><demand><theory>
Give bundles $x,y\in \mathbb R^n$, there must exist a budget $B\supset\{x,y\}$ and a demand $D(B)\in[x,y]$?
dodo
# Give bundles $x,y\in \mathbb R^n$, there must exist a budget $B\supset\{x,y\}$ and a demand $D(B)\in[x,y]$?

(Asked by: username_0 on 2023-09-28)

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 $p\in\mathbb R^n$.

**Tags:** microeconomics, consumer-theory, preferences, demand, theory


---

## 1 answers

### Answer by username_1 on 2023-09-28. Score: 1

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 $z \in [x,y]$.


---

