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Is homogeneity of degree zero necessary in proposition 2.F.1? It seems like the proof does not assume homogeneity of degree zero to establish the proposition. If this is true, it seems that homogeneity is not required to establish that the Slutsky matrix is negative semidefinite (only required assumptions are differentiability and Walras' law?)

Thank you so much for your help!

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  • $\begingroup$ Would be helpful to put here what Prop 2.F.1 is... not everybody has MWG. It'd be great if you could edit the title accordingly to help people searching for it. $\endgroup$ – Art Jan 20 at 1:34
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In between eq. $(2.F.2)$ and $(2.F.3)$ we read

...Walras's law tells us that $w' = p'\cdot x(p',w')$

Assume now that homogeneity of degree zero does not hold. Then we have, $a>0$

$$x(ap,aw) \neq x(p,w)$$

see definition $2.E.1$

Then we can set $p'=ap,w'=aw$ to examine the case $x(p',w') \neq x(p,w)$.

But then Walras' law would imply

$$w' = p'\cdot x(p',w') \implies aw = ap\cdot x(ap,aw) \implies w = p\cdot x(ap,aw)$$

while also, again by Walras' law we have

$$w = p\cdot x(p,w)$$

So under the assumption that Walras' law holds but that we do not have homogeneity of degree zero we end up saying

$$p\cdot x(ap,aw) = p\cdot x(p,w), \,\,\, x(ap,aw) \neq x(p,w)$$

I believe you can contemplate the situation further from here.

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  • $\begingroup$ Just for clarification -- is this supposed to imply that Walras' law implies homogeneity of degree zero? I'm wondering about a situation where p x(ap, aw) = p x(p,w) does not imply that x(ap, aw) = x(p, w)... for instance, we could have p=(1,1,1) , x(ap, aw) = (1,2,1) and x(p,w) = (1,1,2) $\endgroup$ – Michael Kim Jan 18 at 14:40
  • $\begingroup$ Never mind! I see you're referring to the Weak Axiom. Thanks so much for help!!! $\endgroup$ – Michael Kim Jan 18 at 14:58
  • $\begingroup$ @MichaelKim You're welcome. $\endgroup$ – Alecos Papadopoulos Jan 18 at 15:39

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