<p>Let <span class="math-container">$x(p,w)$</span> be the demand at prices <span class="math-container">$p$</span> and income <span class="math-container">$w$</span>.</p>
<p>Let <span class="math-container">$x_0 = x(p_0,w)$</span> and <span class="math-container">$x_1 = x(p_1, w)$</span>. Note that <span class="math-container">$p_0 x_0 = w = p_1 x_1$</span></p>
<p>Assume that WARP is violated, so <span class="math-container">$x_1 \ne x_0$</span>,
<span class="math-container">$$
p_0 x_0 \ge p_0 x_1 \text{ and } p_1 x_0 \le p_1 x_1.
$$</span></p>
<p>Consider any <span class="math-container">$\alpha \in (0,1)$</span>, define <span class="math-container">$p_\alpha = \alpha p_0 + (1-\alpha) p_1$</span> and consider <span class="math-container">$x_\alpha = x(p_\alpha, w)$</span> (which means that <span class="math-container">$p_\alpha x_\alpha = w$</span>).</p>
<p>By convexity of the demand:
<span class="math-container">$$
p_\alpha(\alpha x_0 + (1-\alpha) x_1) &gt; p_\alpha x_\alpha.
$$</span></p>
<ol>
<li><p>if <span class="math-container">$p_\alpha x_0 \le p_\alpha x_1$</span> then we have:
<span class="math-container">$$
p_\alpha x_1 &gt; p_\alpha x_\alpha.
$$</span>
As such,
<span class="math-container">$$
\begin{align*}
w = p_\alpha x_\alpha &amp;&lt; p_\alpha x_1,\\
&amp;= \alpha p_0 x_1 + (1-\alpha) p_1 x_1,\\
&amp;\le \alpha p_0 x_0 + (1-\alpha) w,\\
&amp;= \alpha w + (1-\alpha) w = w,
\end{align*}
$$</span>
a contradiction.</p>
</li>
<li><p>If <span class="math-container">$p_\alpha x_1 \le p_\alpha x_0$</span>, then:
<span class="math-container">$$
p_\alpha x_0 &gt; p_\alpha x_\alpha.
$$</span>
In this case we have the contradiction:
<span class="math-container">$$
\begin{align*}
w = p_\alpha x_\alpha &amp;&lt; p_\alpha x_0,\\
&amp;= \alpha p_0 x_0 + (1-\alpha) p_1 x_0,\\
&amp;\le \alpha w + (1-\alpha) p_1 x_1,\\
&amp;= \alpha w + (1-\alpha) w = w.
\end{align*}
$$</span></p>
</li>
</ol>

CC BY-SA 4.0
2024-02-20T09:06:24.460Z
2024-02-20T09:06:24.460Z
tdm
<p>If you have a Marshallian demand function that is strictly convex, then it satisfies WARP.
How to prove this?</p>

CC BY-SA 4.0
2024-02-19T07:46:21.360Z
You mean strictly convex (Marshallian) demand, or strictly convex preferences?
tdm
CC BY-SA 4.0
2024-02-19T14:57:41.023Z
former case. Strictly convex Marshallian demand.
babededeeptido
CC BY-SA 4.0
2024-02-20T09:28:20.253Z
Convex in which arguments? In prices? In income? In prices and income?
Bertrand
<microeconomics><consumer-theory><choice-theory>
Proof for Marshallian Demand function
babededeeptido
# Proof for Marshallian Demand function

(Asked by: username_0 on 2024-02-19)

If you have a Marshallian demand function that is strictly convex, then it satisfies WARP.
How to prove this?

**Tags:** microeconomics, consumer-theory, choice-theory
> **Comments:**
> **username_1**: You mean strictly convex (Marshallian) demand, or strictly convex preferences?  
> **username_0**: former case. Strictly convex Marshallian demand.  
> **username_2**: Convex in which arguments? In prices? In income? In prices and income?  

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## 1 answers

### Answer by username_1 on 2024-02-20. Score: 2 | Accepted answer

Let $x(p,w)$ be the demand at prices $p$ and income $w$.

Let $x\_0 = x(p\_0,w)$ and $x\_1 = x(p\_1, w)$. Note that $p\_0 x\_0 = w = p\_1 x\_1$

Assume that WARP is violated, so $x\_1 \ne x\_0$,
$$
p\_0 x\_0 \ge p\_0 x\_1 \text{ and } p\_1 x\_0 \le p\_1 x\_1.
$$

Consider any $\alpha \in (0,1)$, define $p\_\alpha = \alpha p\_0 + (1-\alpha) p\_1$ and consider $x\_\alpha = x(p\_\alpha, w)$ (which means that $p\_\alpha x\_\alpha = w$).

By convexity of the demand:
$$
p\_\alpha(\alpha x\_0 + (1-\alpha) x\_1) > p\_\alpha x\_\alpha.
$$

1. if $p\_\alpha x\_0 \le p\_\alpha x\_1$ then we have:
   $$
   p\_\alpha x\_1 > p\_\alpha x\_\alpha.
   $$
   As such,
   $$
   \begin{align\*}
   w = p\_\alpha x\_\alpha &< p\_\alpha x\_1,\\
   &= \alpha p\_0 x\_1 + (1-\alpha) p\_1 x\_1,\\
   &\le \alpha p\_0 x\_0 + (1-\alpha) w,\\
   &= \alpha w + (1-\alpha) w = w,
   \end{align\*}
   $$
   a contradiction.
2. If $p\_\alpha x\_1 \le p\_\alpha x\_0$, then:
   $$
   p\_\alpha x\_0 > p\_\alpha x\_\alpha.
   $$
   In this case we have the contradiction:
   $$
   \begin{align\*}
   w = p\_\alpha x\_\alpha &< p\_\alpha x\_0,\\
   &= \alpha p\_0 x\_0 + (1-\alpha) p\_1 x\_0,\\
   &\le \alpha w + (1-\alpha) p\_1 x\_1,\\
   &= \alpha w + (1-\alpha) w = w.
   \end{align\*}
   $$


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