I'm puzzled why, in the following optimal-choice problem, good $x$ depends on price $p_y$ but the opposite is not true. Should not the sign of $\frac{\partial x}{\partial p_y}$ be the same of $\frac{\partial y}{\partial p_x}$?

Optimal-choice problem: given prices $p_x$ and $p_y$ and income $m$ find the optimal demand function for $x$ and for $y$: \begin{equation} \max_{x,y} \{x\cdot(y+6)\} \qquad \text{s.t. } p_x x+p_y y \leq m \tag{1} \end{equation} after imposing $MRS_{xy}=p_x/p_y$ and substituting back in the budget constraint I find: \begin{equation} x = \frac{m + 6p_y}{2p_x} \qquad\qquad y = \frac{m}{2p_y}-3 \tag{2} \end{equation} so that $sign\left(\frac{\partial x}{\partial p_y}\right) = sign\left( \frac{6}{2p_x}\right)\neq sign\left(\frac{\partial y}{\partial p_x}\right) = 0$.

What I would like to understand better:

  1. How can this happen? I remember having read somewhere in MWG that the price-derivative matrix should be (even!) symmetric. Can you provide me with some explanation and/or some reference? Alternatively, is there any mistake in the calculations?
  2. Provided the situation described above can happen [i.e. $sign\left(\frac{\partial x}{\partial p_y}\right)\neq sign\left(\frac{\partial y}{\partial p_x}\right)$], I would be very happy if you can come up with some "real world" examples of that.

MWG proposition 3.G.2 is the one you’re thinking of, but that applies to Hicksian demand’s hessian (constant utility). What you solved for is the Walrasian demand, so we want to look at the Slutsky matrix.

$$S_{lk}(p,w) = \partial x_l(p,w) / \partial p_k + \partial x_l(p,w)/ \partial w * x_k(p,w)$$

First term: uncompensated change in demand of good l due to a change in $p_k$ (when multiplied by $\partial p_k$)

Second term: effect of wealth change on demand of good l (when multiplied by $\partial p_k$) [https://scholar.harvard.edu/files/basilico/files/mwg_flashcards.pdf]

If you check the symmetry of this matrix (in our case l and k are x and y), I believe it should work out.


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.