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One of the first assumption is that the demand function is homogeneous of degree zero. The reason and the proof is easy.

It should also be easy why this implies we can normalize the price of one good to 1, but I cannot see exactly what's going on. A counterexample would be of great help i.e. a function with homogeneity different from zero showing that the normalization is not possible/leads to wrong conclusions.

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Demand $x(p, m)$ is the solution to the utility maximization problem:

$\max\limits_{x\in\mathbb{R}^n_+} \ \ u(x) \\ \text{s.t.} \ \ p\cdot x \leq m$

where $p\in\mathbb{R}^n_{++}$ is price vector, and $m$ is the income.

When we multiply both sides of the constraint in problem above by $\lambda > 0$, and look at the revised problem we get

$\max\limits_{x\in\mathbb{R}^n_+} \ \ u(x) \\ \text{s.t.} \ \ \lambda p\cdot x \leq \lambda m$

Since this operation does not affect the constraint, the solution remains unaffected i.e. demand satisfy $x(\lambda p, \lambda m) = x(p, m) $ which shows that demand is homogeneous of degree 0 in $(p, m)$. So, this is always true for demand function. Given that $p_1 > 0$, we can take $\lambda = \frac{1}{p_1}$, and find $x\left(\frac{p}{p_1}, \frac{m}{p_1}\right)$ to get $x(p, m)$.

It is helpful to note that for any function $f(p)$ that is homogeneous of degree $k > 0$, it is the case that $f(\lambda p) = \lambda^k f(p) \neq f(p)$ for $\lambda \neq 1$.

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  • $\begingroup$ I understand all of this but I was more looking for an example where if it's not homogeneous of degree zero then we cannot normalize. I know the general theory, I want to fully understand the concept by looking at example where this fails $\endgroup$
    – Mino
    Mar 8 '17 at 3:13
  • $\begingroup$ Suppose $f(p_1,p_2, m) = \frac{m}{p_1p_2}$. This function is not homogeneous of degree 0. Consider $p_1 > 1$, $f\left(1,\frac{p_2}{p_1}, \frac{m}{p_1}\right) = \frac{m}{p_2} \neq \frac{m}{p_1p_2} = f(p_1,p_2, m)$ $\endgroup$
    – Amit
    Mar 8 '17 at 3:24
  • $\begingroup$ By the way $f$ in not a valid demand because it does not satisfy homogeneity of degree 0 property which every demand function does. $\endgroup$
    – Amit
    Mar 8 '17 at 3:31
  • $\begingroup$ I think you mean $p_1 > 0$ $\endgroup$
    – Mino
    Mar 8 '17 at 3:55

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