Homothetic preferences and weak separability

With 3 goods (x,y,z), linear Engel curves, where z is separable of x and y, and with all first derivatives are positives and second negatives.

Does the demand of z change with the price of x or y? So, is z independent of $p_{x}$ and $p_y$?

I think so, because I have made some examples where changing prices didn't change the optimal amount of z. However I cannot find a formal proof.

• Maybe I'm wrong again, is it equivalent to ask if homothetic preferences imply independent goods? Then, another question comes up, how can I compute algebraically a demand function without defining the utility function? Once I have the First Order Conditions and the Budget Constrain, I do not know how to proceed in order to get a general demand curve. – Ignasi Mata Pavia Nov 28 '16 at 9:02
• It is not clear to me where the details of your question end and where your own speculations begin. Does the problem state that $MRS_{xy}(x,y,z) = MRS_{xy}(x,y,z')$ for all $z, z'$ or is that something you suspect to be true? – Giskard Nov 28 '16 at 10:13
• Sorry for my misleading question. It is assumed that z is separable from x and y. There are two things, weak separability and homothetic preferences. From such point, I would like to know if demand of z is independent of prices of x and y. – Ignasi Mata Pavia Nov 28 '16 at 10:59
• Please edit your question to show this information. (So other users do not have to read the comments.) – Giskard Nov 28 '16 at 11:05

EDIT: Seems to that after your most recent edit the answer is still no. A relatively simple counterexample is: $$U(x,y,z) = \sqrt{x} + \sqrt{y} + \sqrt{z}.$$
This is clearly seperable and homothetic. But $D_z$ is is not independent of $p_x$ and $p_y$. I think this is clear just by looking at the utility function, as it is not of the Cobb-Douglas type. If you have doubts you can calculate $D_z$ from these equations: \begin{eqnarray*} MRS_{xz}(x,y,z) & = & \sqrt{\frac{z}{x}} = \frac{p_x}{p_z} \\ \\ MRS_{yz}(x,y,z) & = & \sqrt{\frac{z}{y}} = \frac{p_y}{p_z} \\ \\ m & = & p_x \cdot x + p_y \cdot y + p_z \cdot z. \end{eqnarray*} Then \begin{eqnarray*} m & = & p_x \cdot x + p_y \cdot y + p_z \cdot z \\ \\ m & = & p_x \cdot z \cdot \left(\frac{p_z}{p_x}\right)^2 + p_y \cdot z \cdot \left(\frac{p_z}{p_y}\right)^2 + p_z \cdot z \\ \\ \frac{m}{p_z}\frac{1}{\frac{p_z}{p_x} + \frac{p_z}{p_y} + \frac{p_z}{p_z}} & = & z. \end{eqnarray*} As you can see both $p_x$ and $p_y$ appear.