Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Walras' Law states that summation of pi Ei(p) = 0 for all pi. We define Ei(p) = xi(p) - qi(p) - Ri. What are the next steps that I should take?

share|cite|improve this question

migrated from May 1 '12 at 20:01

This question came from our site for professional and academic economists and analysts.

I don't quite know what your notation is. The proof starts by asserting LNS preferences and claiming walras' law, $\forall p,w$ and , $x \in x(p,w), p\cdot x=w $ The proof is almost always handled by contradiction. You can see most any micro textbook for the full proof. A good start would be to define your assumptions (LNS?) and the various functions you've specified (you'd have to do that for a proper proof anyway.) – Jason B Nov 14 '11 at 4:30
@Jason B - what's LNS? – Beatrice Nov 15 '11 at 17:59
Local non-satiation. It's the claim that, for any point $x$ and any number $\epsilon>0$, there exists a $x'$ in the $\epsilon$-neighbourhood of $x$ such that $x'$ is strictly preferred to $x$. – Zermelo Nov 15 '11 at 18:59
thanks @user68! – Beatrice Nov 15 '11 at 21:56
@Patience the most straightforward proof of Walras' Law requires one to assume LNS preferences and little more (it is implicit in Zermelo's answer). – Jason B Nov 16 '11 at 3:46
up vote 2 down vote accepted

Let $i$ denote an agent; $j$ denote the good.

Walras' law: $p.e(p)=0$ for all $p$.

Start with the budget constraint:

$\sum_{j} p_j.x_{ij}=\sum_{j} p_j.w_{ij}$ where $w_{ij}$ is $i$'s endowment of good $j$, $x_{ij}$ is $i$'s consumption of good $j$.

In other words, $\sum_{j} p_{j}.e_{ij}=0$, where $e_{ij}=x_{ij}-w_{ij}$.

Now just add over all agents $i$. You get $\sum_{j}p_j.e_j=0$, where $e_j=\sum_i e_{ij}$ for each $j$. This is Walras' Law. Note that this applies to ALL $p$ - regardless of whether it's the equilibrium price.

share|cite|improve this answer
Use the dollar sign ($) for Latex. ${p_j * x_{ij}}$ for ${p_j * x_{ij}}$ – BlackJack Nov 15 '11 at 0:33
thanks a lot haha I kept using # instead and got nowhere. – Zermelo Nov 15 '11 at 5:08
by the way, for your specific question (where there is production too) replce $w_j$ for every good $j$ by $w_j+R_j$ – Zermelo Nov 15 '11 at 5:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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