<p>We need to show that:
<span class="math-container">$$
(p - p')\cdot(x(p,1) - x(p', 1)) \le 0.
$$</span>
Note that if <span class="math-container">$x(p,1) = x(p', 1)$</span> then this is obviously satisfied, so assume that <span class="math-container">$x(p,1) \ne x(p', 1)$</span>.</p>
<p>Note that the condition is equivalent to (using <span class="math-container">$1 = p \cdot x(p,1) = p'\cdot x(p',1)$</span>):
<span class="math-container">$$
2 - p' \cdot x(p,1) - p \cdot x(p',1) \le 0. \tag{1}
$$</span></p>
<p>There are 3 cases to consider.</p>
<ol>
<li><p>If <span class="math-container">$1 \le p \cdot x(p',1)$</span> and <span class="math-container">$1 \le p' \cdot x(p,1)$</span> then we can add the two inqualities together to obtain condition <span class="math-container">$(1)$</span>.</p>
</li>
<li><p>If <span class="math-container">$p \cdot x(p,1) = 1 \ge p \cdot x(p', 1)$</span> then let <span class="math-container">$z \ge 1$</span> be such that <span class="math-container">$1 = p \cdot x(p', z)$</span>. (here <span class="math-container">$z = 1/(p \cdot x(p', 1))$</span>. Then WARP requires that <span class="math-container">$p' \cdot x(p', z) = z \le p' \cdot x(p,1)$</span>.
We have the two conditions:
<span class="math-container">$$
1 = p \cdot x(p', z) \text{ and } z \le p' \cdot x(p,1).
$$</span>
Adding both together gives:
<span class="math-container">$$
1 + z - p \cdot x(p', z) - p' \cdot x(p,1) \le 0.
$$</span>
Using <span class="math-container">$x(p', z) = x(p', 1) z$</span> this can be rewritten as:
<span class="math-container">$$
1 + z\underbrace{(1 - p \cdot x(p',1))}_{\ge 0} - p' \cdot x(p,1) \le 0.
$$</span>
As <span class="math-container">$z \ge 1$</span>, we get,
<span class="math-container">$$
2 - p \cdot x(p', 1) - p' \cdot x(p,1) \le 0.
$$</span>
This gives condition <span class="math-container">$(1)$</span></p>
</li>
<li><p>If <span class="math-container">$p' x(p', 1) = 1 \ge p' x(p,1)$</span> we get a similar reasoning as in point <span class="math-container">$2$</span> before (exchanging <span class="math-container">$p'$</span> and <span class="math-container">$p$</span>).</p>
</li>
</ol>

CC BY-SA 4.0
2024-02-22T09:11:02.850Z
2024-02-22T09:11:02.850Z
tdm
<h2>Problem</h2>
<p>I am asked to prove the following result (MWG Exercise 2.F.5):</p>
<blockquote>
<p>The <em>law of demand</em> always holds if the walrasian demand function <span class="math-container">$x(\mathbf{p},w)$</span> satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth.</p>
</blockquote>
<h2>My Attempt So Far</h2>
<p>This is what I have so far:</p>
<blockquote>
<p>Suppose that the Walrasian demand function <span class="math-container">$x(\mathbf{p},w)$</span> satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth <span class="math-container">$w$</span>. We want to prove that, for any price change from <span class="math-container">$\mathbf{p}$</span> to <span class="math-container">$\mathbf{p}'$</span> (without change in wealth), we have
<span class="math-container">\begin{align*}
(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',w) - x(\mathbf{p},w)] \leq 0.\tag1
\end{align*}</span>
By homogeneity of degree 1 in <span class="math-container">$w$</span>, we shall prove
<span class="math-container">\begin{align*}
(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] \leq 0\tag2
\end{align*}</span>
for all <span class="math-container">$\mathbf{p}$</span> and <span class="math-container">$\mathbf{p}'$</span>. We consider the following cases:</p>
<p><em><strong>Case 1:</strong></em> Suppose that <span class="math-container">$x(\mathbf{p},1) = x(\mathbf{p}',1)$</span>. Then the inequality <span class="math-container">$(2)$</span> always holds with <span class="math-container">$(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] = 0$</span>.</p>
<p>From now on, suppose that <span class="math-container">$x(\mathbf{p},1) \neq x(\mathbf{p}',1)$</span>.</p>
<p><em><strong>Case 2:</strong></em> Suppose that <span class="math-container">$\mathbf{p}' \cdot x(\mathbf{p},1) &gt; 1$</span> and <span class="math-container">$\mathbf{p} \cdot x(\mathbf{p}',1) &gt; 1$</span>. Notice that, by Walras' law, <span class="math-container">$1 = \mathbf{p} \cdot x(\mathbf{p},1) = \mathbf{p}' \cdot x(\mathbf{p}',1)$</span>. Then,
<span class="math-container">\begin{align*}
(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] &amp;= \mathbf{p}' \cdot x(\mathbf{p}',1) - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + \mathbf{p} \cdot x(\mathbf{p},1)\\
&amp;= 1 - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + 1\\
&amp;&lt; 0.
\end{align*}</span></p>
</blockquote>
<p>Here is where I got stuck:</p>
<blockquote>
<p><em><strong>Case 3:</strong></em> Suppose that <span class="math-container">$\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$</span> and <span class="math-container">$\mathbf{p} \cdot x(\mathbf{p}',1) &gt; 1$</span>.</p>
<p><em><strong>Case 4:</strong></em> Suppose that <span class="math-container">$\mathbf{p}' \cdot x(\mathbf{p},1) &gt; 1$</span> and <span class="math-container">$\mathbf{p} \cdot x(\mathbf{p}',1) \leq 1$</span>.</p>
</blockquote>
<h2>My Question</h2>
<p>Could someone please help me derive the inequality <span class="math-container">$(2)$</span> for Case 3 and Case 4 above? We could basically just consider Case 3. I want to prove that, if <span class="math-container">$\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$</span> and <span class="math-container">$\mathbf{p} \cdot x(\mathbf{p}',1) &gt; 1$</span>, then
<span class="math-container">\begin{align*}
\mathbf{p}' \cdot x(\mathbf{p},1) + \mathbf{p} \cdot x(\mathbf{p}',1) \geq 2.
\end{align*}</span>
Thanks a lot in advance!</p>

<microeconomics><mathematical-economics><consumer-theory><decision-theory><choice-theory>
Prove: The law of demand holds if WA, Walras' law, homogeneity of degree 0, and homogeneity of degree 1 in wealth hold for Walrasian demand functions
Beerus
# Prove: The law of demand holds if WA, Walras' law, homogeneity of degree 0, and homogeneity of degree 1 in wealth hold for Walrasian demand functions

(Asked by: username_0 on 2024-02-21)

## Problem

I am asked to prove the following result (MWG Exercise 2.F.5):

> The *law of demand* always holds if the walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth.

## My Attempt So Far

This is what I have so far:

> Suppose that the Walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth $w$. We want to prove that, for any price change from $\mathbf{p}$ to $\mathbf{p}'$ (without change in wealth), we have
> \begin{align\*}
> (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',w) - x(\mathbf{p},w)] \leq 0.\tag1
> \end{align\*}
> By homogeneity of degree 1 in $w$, we shall prove
> \begin{align\*}
> (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] \leq 0\tag2
> \end{align\*}
> for all $\mathbf{p}$ and $\mathbf{p}'$. We consider the following cases:
>
> ***Case 1:*** Suppose that $x(\mathbf{p},1) = x(\mathbf{p}',1)$. Then the inequality $(2)$ always holds with $(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] = 0$.
>
> From now on, suppose that $x(\mathbf{p},1) \neq x(\mathbf{p}',1)$.
>
> ***Case 2:*** Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$. Notice that, by Walras' law, $1 = \mathbf{p} \cdot x(\mathbf{p},1) = \mathbf{p}' \cdot x(\mathbf{p}',1)$. Then,
> \begin{align\*}
> (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] &= \mathbf{p}' \cdot x(\mathbf{p}',1) - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + \mathbf{p} \cdot x(\mathbf{p},1)\\
> &= 1 - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + 1\\
> &< 0.
> \end{align\*}

Here is where I got stuck:

> ***Case 3:*** Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$.
>
> ***Case 4:*** Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) \leq 1$.

## My Question

Could someone please help me derive the inequality $(2)$ for Case 3 and Case 4 above? We could basically just consider Case 3. I want to prove that, if $\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$, then
\begin{align\*}
\mathbf{p}' \cdot x(\mathbf{p},1) + \mathbf{p} \cdot x(\mathbf{p}',1) \geq 2.
\end{align\*}
Thanks a lot in advance!

**Tags:** microeconomics, mathematical-economics, consumer-theory, decision-theory, choice-theory

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## 1 answers

### Answer by username_1 on 2024-02-22. Score: 3 | Accepted answer

We need to show that:
$$
(p - p')\cdot(x(p,1) - x(p', 1)) \le 0.
$$
Note that if $x(p,1) = x(p', 1)$ then this is obviously satisfied, so assume that $x(p,1) \ne x(p', 1)$.

Note that the condition is equivalent to (using $1 = p \cdot x(p,1) = p'\cdot x(p',1)$):
$$
2 - p' \cdot x(p,1) - p \cdot x(p',1) \le 0. \tag{1}
$$

There are 3 cases to consider.

1. If $1 \le p \cdot x(p',1)$ and $1 \le p' \cdot x(p,1)$ then we can add the two inqualities together to obtain condition $(1)$.
2. If $p \cdot x(p,1) = 1 \ge p \cdot x(p', 1)$ then let $z \ge 1$ be such that $1 = p \cdot x(p', z)$. (here $z = 1/(p \cdot x(p', 1))$. Then WARP requires that $p' \cdot x(p', z) = z \le p' \cdot x(p,1)$.
   We have the two conditions:
   $$
   1 = p \cdot x(p', z) \text{ and } z \le p' \cdot x(p,1).
   $$
   Adding both together gives:
   $$
   1 + z - p \cdot x(p', z) - p' \cdot x(p,1) \le 0.
   $$
   Using $x(p', z) = x(p', 1) z$ this can be rewritten as:
   $$
   1 + z\underbrace{(1 - p \cdot x(p',1))}\_{\ge 0} - p' \cdot x(p,1) \le 0.
   $$
   As $z \ge 1$, we get,
   $$
   2 - p \cdot x(p', 1) - p' \cdot x(p,1) \le 0.
   $$
   This gives condition $(1)$
3. If $p' x(p', 1) = 1 \ge p' x(p,1)$ we get a similar reasoning as in point $2$ before (exchanging $p'$ and $p$).


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