<p>It is not quite true; the expenditure function is <a href="https://en.wikipedia.org/wiki/Affine_transformation#Over_the_real_numbers" rel="nofollow noreferrer">affine</a> in utility, not linear. This is <a href="https://en.wikipedia.org/wiki/Stone%E2%80%93Geary_utility_function" rel="nofollow noreferrer">Stone-Geary utility</a>, and I use the result that demand is given by</p>
<p><span class="math-container">$$x_i^*=m_i+\frac{a_i}{p_i}\bigg(w-\sum_j m_jp_j\bigg).$$</span>
One can interpret <span class="math-container">$\sum_j m_jp_j$</span> as &quot;subsistence expenditure&quot; and <span class="math-container">$w-\sum_j m_jp_j$</span> as &quot;non-subsistence expenditure.&quot; 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 <span class="math-container">$l(u)$</span> with <span class="math-container">$l$</span> linear (in the relevant range). This means then that total expenditure is</p>
<p><span class="math-container">$$l(p)+\sum_j m_jp_j,$$</span>
which is an affine function as the sum of a  linear function and a constant.</p>

CC BY-SA 4.0
2020-11-30T13:03:32.920Z
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.
Maybeline Lee
CC BY-SA 4.0
2020-11-30T12:58:30.207Z
2020-11-30T12:58:30.207Z
Michael Greinecker
<p>Let <span class="math-container">$u (x) = \prod_{{i\leq n}}$</span> <span class="math-container">$(x_i-m_i)^{a_i}$</span>, where <span class="math-container">$m_i\geq$</span>0 and <span class="math-container">$a_i\geq$</span>0, Σ<span class="math-container">$_{{i\leq n}}a_i=1$</span> show that the expenditure function <span class="math-container">$e(p,u)$</span> is linear in <span class="math-container">$u$</span>.</p>
<p>Based on the definition of the EF, my first guess would be to find the MRS in terms of <span class="math-container">$x_i$</span> and <span class="math-container">$m_i$</span> to we get <span class="math-container">$x_i^*$</span> and <span class="math-container">$m_i^*$</span>. After we find these two in terms of each other, we substitute it back to the <span class="math-container">$u$</span> function to get now <span class="math-container">$x_i^*$</span> and <span class="math-container">$m_i^*$</span> in terms of <span class="math-container">$u$</span> as well. After this we substitute what we found to the budget constraint and prove that the expenditure function is linear in <span class="math-container">$u$</span> somehow.</p>
<p>I don't know if I should follow this logic or is there some other way to approach this problem? Any help is appreciated.</p>

CC BY-SA 4.0
2020-11-30T11:20:05.073Z
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)
Maybeline Lee
CC BY-SA 4.0
2020-11-30T12:43:21.567Z
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](https://samuelwills.files.wordpress.com/2012/11/msced2011_macro-class-3.pdf).
Michael Greinecker
CC BY-SA 4.0
2020-11-29T23:25:29.910Z
There is no $m_i^*$; $m_i$ is a fixed parameter.
Michael Greinecker
<microeconomics><utility>
Expenditure function
Maybeline Lee
# Expenditure function

(Asked by: username_0 on 2020-11-29)

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.

**Tags:** microeconomics, utility

> **Comments:**
> **username_1**: There is no $m_i^*$; $m_i$ is a fixed parameter.  
> **username_0**: 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)  
> **username_1**: 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](https://samuelwills.files.wordpress.com/2012/11/msced2011_macro-class-3.pdf).  

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## 1 answers

### Answer by username_1 on 2020-11-30. Score: 2 | Accepted answer

It is not quite true; the expenditure function is [affine](https://en.wikipedia.org/wiki/Affine_transformation#Over_the_real_numbers) in utility, not linear. This is [Stone-Geary utility](https://en.wikipedia.org/wiki/Stone%E2%80%93Geary_utility_function), 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.

> **Comments:**
> **username_0**: 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.  

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