2
$\begingroup$

I've been reading a book where the author claims that "if there are two goods they are always net substitutes. Along an indifference curve an increase in price of good 2 leads to less of good 2 and more of good 1 being consumed when utility is maximized. When there are more than two goods, the law of diminishing marginal rate of substitution implies that at least one good must be a net substitute for any other good."
The first part can be proved easily using graphs, it's the second that I couldn't prove mathematically.
Source of the quotation has been attached here, Pg-104.

$\endgroup$
2
$\begingroup$

I don't know what is the definition of the "law of DMRS", but I think what the author meant is summarized in these lines of MWG:

Proposition 3.G.2: suppose that $u(\cdot)$ is a continuous utility function representing a locally nonsatiated and strictly convex preference relation $\succsim$ defined on $\mathbb{R}^L_+$. Suppose also that $h(\cdot, u)$ [the hicksian demand function] is continuously differentiable at $(p, u)$, and denote its $L\times L$ derivative matrix [with respect to each $p$] by $D_p h(p, u)$. Then

  1. $D_p h(p, u)$ is a negative semidefinite matrix.

  2. $D_p h(p, u) p = 0$

Well, 1 implies that the elements of the main diagonal of the matrix are all non positive, i.e. $\frac{\partial h_l (p, u)}{\partial p_l} \leq 0$ for each $l$.

We say that two goods are substitutes if $\frac{\partial h_l (p, u)}{\partial p_k} \geq 0$ (a positive variation in the price of $k$ implies an increase in the compensated demand of $l$).

Part 2 of the proposition says that $D_p h(p, u) p = 0$. But as $\frac{\partial h_l (p, u)}{\partial p_l} \leq 0$, for every $l$ we must have a good $m$ such that $\frac{\partial h_l (p, u)}{\partial p_m} \geq 0$, which means exactly that every good has a substitute.

$\endgroup$

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.