# What's the most straight-forward way to prove Walras's Law?

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?

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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

## 1 Answer

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

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

$\sum_{j} p_j.x_{ij}=\sum_{j} p_j.w_{ij}$ 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.
Use the dollar sign ($) for Latex. `${p_j * x_{ij}}$` for${p_j * x_{ij}}\$ – BlackJack Nov 15 '11 at 0:33
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
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