<p>It is almost true. </p>

<p>There are examples of demand that have a negative definite Slutsky matrix but fails the Weak Axiom.</p>

<p>However, if we ask that
<span class="math-container">$$v \cdot S(p,w) v &lt;0 $$</span>
whenever <span class="math-container">$v \not = \alpha p$</span> for any scalar <span class="math-container">$\alpha$</span> (i.e. <span class="math-container">$S$</span> is negative definite for all vectors except those proportional to price), then the Weak Axiom holds. </p>

CC BY-SA 4.0
2019-11-13T17:33:05.903Z
I see. Is that a stronger condition than negaive semidifinite? 
 Because it could be the case that vS(p,w)v=0 for some v that is not porpotional to p when S(p,w) is negaive semidifinite.
HXW
CC BY-SA 4.0
2019-11-13T18:15:09.400Z
@Huaixin You're correct. If we weaken negative definite to negative semidefinite, again we can obtain an example that fails the Weak Axiom. It does however satisfy a weaker Weak Axiom. This result is by Kihlstrom, Mas-Colell and Sonnenschien (1976), they call it WWA (if you're interested).
Walrasian Auctioneer
CC BY-SA 4.0
2019-11-16T11:14:01.693Z
get it, thanks for the precise answer.
HXW
CC BY-SA 4.0
2019-11-13T17:17:53.107Z
2019-11-13T17:17:53.107Z
Walrasian Auctioneer
<p>We know that if a differentiable Walrasian demand function <span class="math-container">$x(p,w)$</span> satisfies Walras' law (<span class="math-container">$p^Tx=w$</span>), homogeneity of degree zero (<span class="math-container">$x(\alpha p,\alpha w)=x(p,w)$</span>), and the weak axiom of revealed preference, then at any <span class="math-container">$(p,w)$</span>, the Slutsky matrix
<span class="math-container">\begin{equation}
S(p,w)=D_px(p,w)+D_wx(p,w)x(p,w)^T
\end{equation}</span>
is negative semidefinite.</p>

<p><strong>My question</strong> is: if <span class="math-container">$S(p,w)$</span> is negative semidefinite, then what can we say about the demand function <span class="math-container">$x(p,w)$</span>? Can we conclude that <span class="math-container">$x(p,w)$</span> satisfies weak axiom?</p>

<microeconomics><preferences><walrasian><demand><slutsky-equation>
Question about the relationship between Weak Axiom and Slutsky Matrix
HXW
# Question about the relationship between Weak Axiom and Slutsky Matrix

(Asked by: username_0 on 2019-11-12)

We know that if a differentiable Walrasian demand function $x(p,w)$ satisfies Walras' law ($p^Tx=w$), homogeneity of degree zero ($x(\alpha p,\alpha w)=x(p,w)$), and the weak axiom of revealed preference, then at any $(p,w)$, the Slutsky matrix
\begin{equation}
S(p,w)=D\_px(p,w)+D\_wx(p,w)x(p,w)^T
\end{equation}
is negative semidefinite.

**My question** is: if $S(p,w)$ is negative semidefinite, then what can we say about the demand function $x(p,w)$? Can we conclude that $x(p,w)$ satisfies weak axiom?

**Tags:** microeconomics, preferences, walrasian, demand, slutsky-equation

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## 1 answers

### Answer by username_1 on 2019-11-13. Score: 2

It is almost true.

There are examples of demand that have a negative definite Slutsky matrix but fails the Weak Axiom.

However, if we ask that
$$v \cdot S(p,w) v <0 $$
whenever $v \not = \alpha p$ for any scalar $\alpha$ (i.e. $S$ is negative definite for all vectors except those proportional to price), then the Weak Axiom holds.

> **Comments:**
> **username_0**: I see. Is that a stronger condition than negaive semidifinite?   Because it could be the case that vS(p,w)v=0 for some v that is not porpotional to p when S(p,w) is negaive semidifinite.  
> **username_1**: @Huaixin You're correct. If we weaken negative definite to negative semidefinite, again we can obtain an example that fails the Weak Axiom. It does however satisfy a weaker Weak Axiom. This result is by Kihlstrom, Mas-Colell and Sonnenschien (1976), they call it WWA (if you're interested).  
> **username_0**: get it, thanks for the precise answer.  

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