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Prove that if $U(\alpha x)=\alpha U(x)$, then $$\frac{\partial x_i(p,w)}{\partial p_j}=\frac{\partial x_j(p,w)}{\partial p_i}$$ for any $i$ and $j$.

I've proved that $x(p,\alpha w)=\alpha x(p,w)$, but I don't know how to handle partial derivatives.

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  • $\begingroup$ @Giskard, it is a question from my lecturer. Can you provide a counterexample? $\endgroup$
    – Paul R
    Commented Nov 29, 2023 at 7:05
  • $\begingroup$ Oops, I am only now noticing that the equation is about the cross-derivatives (derivative of demand in price of the other good) not 'own-derivatives'. This is actually true. $\endgroup$
    – Giskard
    Commented Nov 29, 2023 at 7:12
  • $\begingroup$ What textbook do you use? You can prove this using the Envelope Theorem or Hotelling's Lemma, but unfortunately neither of these will be accessible at all if you don't know partial derivatives. $\endgroup$
    – Giskard
    Commented Nov 29, 2023 at 7:15
  • $\begingroup$ No actual textbook. Only lectures. Can you provide some proof? $\endgroup$
    – Paul R
    Commented Nov 29, 2023 at 7:26
  • $\begingroup$ Sorry, I don't want to type one in, but you could try looking it up in "Mas-Colell, Whinston Green: Microeconomic Theory*. $\endgroup$
    – Giskard
    Commented Nov 29, 2023 at 7:55

1 Answer 1

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For ease of notation, let's take the setting of two goods $x_1$ and $x_2$. The case is similar when there are more than two goods.

If utility is homogeneous, the demand functions are homogeneous of degree 1 in income, so for all $w$, $x_1(p_1, p_2, w) = x_1(p_1, p_2,1) w$ and $x_2(p_1, p_2, w) = x_2(p_1, p_2, 1)w$. Taking derivatives with respect to $w$ gives: $$ \frac{\partial x_1}{\partial w}(p_1, p_2, w) = x_1(p_1, p_2,1), $$ and $$ \frac{\partial x_2}{\partial w}(p_1, p_2, w) = x_2(p_1, p_2,1). $$

Let $h_1(p_1,p_2, u)$ and $h_2(p_1, p_2,u)$ be the Hicksian demand functions. The Slutsky equation gives: $$ \begin{align*} \frac{\partial h_1}{\partial p_2}(p_1, p_2, u) &= \frac{\partial x_1}{\partial p_2}(p_1, p_2, w) + \frac{\partial x_1}{\partial w}(p_1, p_2, w) x_2(p_1, p_2, w),\\ &=\frac{\partial x_1}{\partial p_2}(p_1, p_2, w) + w\,\, x_1(p_1, p_2,1)x_2(p_1, p_2, 1). \end{align*} $$ Similarly, $$ \begin{align*} \frac{\partial h_2}{\partial p_1}(p_1, p_2, u) &= \frac{\partial x_2}{\partial p_1}(p_1, p_2, w) + \frac{\partial x_2}{\partial w}(p_1, p_2, w) x_1(p_1, p_2, w),\\ &=\frac{\partial x_2}{\partial p_1}(p_1, p_2, w) + w\,\, x_1(p_1, p_2,1)x_2(p_1, p_2, 1). \end{align*} $$ By Symmetry of the Slutsky matrix, we have $\frac{\partial h_1}{\partial p_2} = \frac{\partial h_2}{\partial p_1}$. This gives: $$ \begin{align*} &\frac{\partial x_1}{\partial p_2}(p_1, p_2, w) + w\,\, x_1(p_1, p_2,1)x_2(p_1, p_2, 1) = \frac{\partial x_2}{\partial p_1}(p_1, p_2, w) + w\,\, x_1(p_1, p_2,1)x_2(p_1, p_2, 1),\\ \leftrightarrow &\frac{\partial x_1}{\partial p_2}(p_1, p_2, w) = \frac{\partial x_2}{\partial p_1}(p_1, p_2, w). \end{align*} $$

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