Let $u (x) = \prod_{{i\leq n}}$ $(x_i-m_i)^{a_i}$, where $m_i\geq$0 and $a_i\geq$0, Σ$_{{i\leq n}}a_i=1$ show that the expenditure function $e(p,u)$ is linear in $u$.

Based on the definition of the EF, my first guess would be to find the MRS in terms of $x_i$ and $m_i$ to we get $x_i^*$ and $m_i^*$. After we find these two in terms of each other, we substitute it back to the $u$ function to get now $x_i^*$ and $m_i^*$ in terms of $u$ as well. After this we substitute what we found to the budget constraint and prove that the expenditure function is linear in $u$ somehow.

I don't know if I should follow this logic or is there some other way to approach this problem? Any help is appreciated.

  • $\begingroup$ There is no $m_i^*$; $m_i$ is a fixed parameter. $\endgroup$ Nov 29 '20 at 23:25
  • $\begingroup$ It makes sense because I wasn't going anywhere with that :/, but I'm still facing problems deriving it. Do you have any clues as to what may be the first step? The book I have isn't of any help either. Thanks in advance) $\endgroup$ Nov 30 '20 at 11:20
  • $\begingroup$ Having tried it myself, I think it is not true unless we are in the Cobb-Douglas case with $m_i=0$ for all $i$. This is Stone-Geary utility, you can find some background here. $\endgroup$ Nov 30 '20 at 12:43

It is not quite true; the expenditure function is affine in utility, not linear. This is Stone-Geary utility, and I use the result that demand is given by

$$x_i^*=m_i+\frac{a_i}{p_i}\bigg(w-\sum_j m_jp_j\bigg).$$ One can interpret $\sum_j m_jp_j$ as "subsistence expenditure" and $w-\sum_j m_jp_j$ as "non-subsistence expenditure." Plugging this into the utility function, one can show that utility is an increasing linear function of non-subsistence expenditure. The inverse of a linear function is linear again, so we can write non-subsistence expenditure as $l(u)$ with $l$ linear (in the relevant range). This means then that total expenditure is

$$l(p)+\sum_j m_jp_j,$$ which is an affine function as the sum of a linear function and a constant.

  • $\begingroup$ I can see it now. Unfortunately, we never covered this utility function before that's why I got lost. I'll go over it once more to fully grasp it. Thank you for the reference, I really appreciate it. $\endgroup$ Nov 30 '20 at 13:03

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