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.

| improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Huaixin Nov 13 '19 at 17:33
  • 1
    $\begingroup$ @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. $\endgroup$ – Huaixin Nov 16 '19 at 11:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.