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?


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.

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  • $\begingroup$ 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. $\endgroup$ – Rui Nov 13 '19 at 17:33
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    $\begingroup$ @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). $\endgroup$ – Walrasian Auctioneer Nov 13 '19 at 18:15
  • $\begingroup$ get it, thanks for the precise answer. $\endgroup$ – Rui Nov 16 '19 at 11:14

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