<p>Yes if you assume that the sub-utility functions are concave. Notice that this is a standard assumption as otherwise, the utility function <span class="math-container">$u = \sum_i f_i$</span> is not guaranteed to be concave (nor quasi-concave).</p>
<p>Let denote by <span class="math-container">$u_i = \dfrac{\partial u}{\partial x_i}$</span> and by <span class="math-container">$u_{i,j} = \dfrac{\partial^2 u}{\partial x_i \partial x_j}$</span>. By additivity <span class="math-container">$u_{i,j} = 0$</span> if <span class="math-container">$i \ne j$</span>.</p>
<p>The first order conditions for the utility maximisation problem give:
<span class="math-container">$$
\begin{align*}
u_i = \lambda p_i, \tag{1}\\
\sum_i p_i x_i = m \tag{2}
\end{align*}
$$</span>
Differentiating both with respect to income <span class="math-container">$m$</span> (and using <span class="math-container">$u_{i,j} = 0$</span>) gives:</p>
<p><span class="math-container">$$
\begin{align*}
&amp;u_{ii} \frac{\partial x_i}{\partial m} = p_i \frac{\partial \lambda}{\partial m} \tag{3}\\
&amp;\sum_i p_i \frac{\partial x_i}{\partial m} = 1 \tag{4}
\end{align*}
$$</span>
Substitute <span class="math-container">$(3$</span>) into <span class="math-container">$(4)$</span>:
<span class="math-container">$$
\begin{align*}
&amp;\sum_i (p_i)^2 \frac{\partial \lambda}{\partial m} \frac{1}{u_{ii}} = 1,\\
\to &amp;\frac{\partial \lambda}{\partial m} = \frac{1}{\sum_i \frac{(p_i)^2}{u_{ii}}} \tag{5}
\end{align*}
$$</span>
Notice that <span class="math-container">$u_{ii} &lt; 0$</span> by concavity of the sub-utility functions. As such, <span class="math-container">$\frac{\partial \lambda}{\partial m} &lt; 0$</span> and also, by <span class="math-container">$(3)$</span>:
<span class="math-container">$$
\frac{\partial x_i}{\partial m} = p_i \frac{1}{u_{ii}} \frac{\partial \lambda}{\partial m} &gt; 0 
$$</span>
as the right hand side is the product of two negative numbers.</p>
<p>This shows that <span class="math-container">$x_i$</span> is a normal good. By symmetry, this holds for all goods.</p>
<p>An alternative quicker way to notice this is to see that</p>
<ol>
<li>by <span class="math-container">$(3)$</span> the sign of <span class="math-container">$\dfrac{\partial x_i}{\partial m}$</span> will be the reverse of the sign of <span class="math-container">$\dfrac{\partial \lambda}{\partial m}$</span>.</li>
<li>This has to be true for all goods, which means that either all goods are normal, or all goods are inferior.</li>
<li>From <span class="math-container">$(4)$</span>, it follows that at least one good should be normal (as otherwise the sum cannot be equal to 1 which is greater than zero).</li>
<li>Conclude that all goods have to be normal.</li>
</ol>

CC BY-SA 4.0
2021-05-15T15:58:15.493Z
You do not need to assume that $u$ is concave: the utility function is ordinal and the demand functions are unchanged after any monotonic transformation of the utility function.
Bertrand
CC BY-SA 4.0
2021-05-15T16:02:59.250Z
@Bertrand I'm not talking about the utility function but the sub-utility functions $f_i$. These better be concave as otherwise $u = \sum_i f_i $ does not even need to be quasi-concave. Notice that although the sum of concave functions is concave, the sum of quasi-concave functions is not necessarily quasi-concave.
tdm
CC BY-SA 4.0
2021-05-15T16:09:59.273Z
I just wanted to mention that assuming concavity is sufficient not necessary: if all $u_{ii}>0$ the demands are still normal.
Bertrand
CC BY-SA 4.0
2021-05-15T16:14:46.787Z
@Bertrand If all sub-utility functions are convex (i.e. $u_{ii} > 0$) then the utility function is (quasi-)convex. In this case, the first order conditions is not going to give you a maximum, but a minimum.
tdm
CC BY-SA 4.0
2021-05-15T16:46:18.243Z
Lovely! My one comment is that you seem to assume an interior solution by assuming (1) holds in equality. This is of course not guaranteed; usually we also assume $x \in \mathbb{R}_+^n$. Any ideas on how to handle the boundary case?
Giskard
CC BY-SA 4.0
2021-05-15T17:09:17.753Z
@Giskard I guess you can argue in the following way. If $x_i = 0$ at the optimum and if income increases, then either $x_i$ goes up (to an interior solution), so it is normal, or it stays at $0$ in which case it is also normal (in the weak sense). As $x_i = 0$, it can never go down if income increases.
tdm
CC BY-SA 4.0
2021-05-15T18:42:42.960Z
I think you could also argue that the right hand side of
$$
\begin{align*}
&\sum_i p_i \frac{\partial x_i}{\partial m} = 1 \tag{4}
\end{align*}
$$
is unchanged if you remove all the indices where $x_i = 0$ and $\frac{\partial x_i}{\partial m} = 0$, and then your interior solution works again.
Giskard
CC BY-SA 4.0
2022-12-02T13:02:03.287Z
Could you please elaborate how to you arrive at the first order conditions for the utility maximisation problem (1):
Barty82
CC BY-SA 4.0
2021-05-15T15:46:45.693Z
2021-05-15T15:46:45.693Z
tdm
<p>Inspired by <a href="https://economics.stackexchange.com/a/43958/1601">this answer</a>.</p>
<hr />
<p>To make it a bit more precise, by normal good I mean demand is (not necessarily strictly) increasing in income, and by additively separable utility function I mean that a monotone transformation exists for which
<span class="math-container">$$
U(x_1,x_2,\dots) = f_1(x_1) + f_2(x_2) + \dots 
$$</span>
(Also assume <span class="math-container">$U$</span> is increasing in all variables.) <br>
Given a linear budget constraint and a utility maximizing consumer, do these goods <span class="math-container">$x_i$</span> exhibit normal behavior for all income levels?</p>

<microeconomics><utility><consumer-theory><giffen-goods>
Are the goods in additively separable utility functions normal goods?
Giskard
# Are the goods in additively separable utility functions normal goods?

(Asked by: username_0 on 2021-05-15)

Inspired by [this answer](https://economics.stackexchange.com/a/43958/1601).

---

To make it a bit more precise, by normal good I mean demand is (not necessarily strictly) increasing in income, and by additively separable utility function I mean that a monotone transformation exists for which
$$
U(x\_1,x\_2,\dots) = f\_1(x\_1) + f\_2(x\_2) + \dots
$$
(Also assume $U$ is increasing in all variables.)   
Given a linear budget constraint and a utility maximizing consumer, do these goods $x\_i$ exhibit normal behavior for all income levels?

**Tags:** microeconomics, utility, consumer-theory, giffen-goods


---

## 1 answers

### Answer by username_1 on 2021-05-15. Score: 5 | Accepted answer

Yes if you assume that the sub-utility functions are concave. Notice that this is a standard assumption as otherwise, the utility function $u = \sum\_i f\_i$ is not guaranteed to be concave (nor quasi-concave).

Let denote by $u\_i = \dfrac{\partial u}{\partial x\_i}$ and by $u\_{i,j} = \dfrac{\partial^2 u}{\partial x\_i \partial x\_j}$. By additivity $u\_{i,j} = 0$ if $i \ne j$.

The first order conditions for the utility maximisation problem give:
$$
\begin{align\*}
u\_i = \lambda p\_i, \tag{1}\\
\sum\_i p\_i x\_i = m \tag{2}
\end{align\*}
$$
Differentiating both with respect to income $m$ (and using $u\_{i,j} = 0$) gives:

$$
\begin{align\*}
&u\_{ii} \frac{\partial x\_i}{\partial m} = p\_i \frac{\partial \lambda}{\partial m} \tag{3}\\
&\sum\_i p\_i \frac{\partial x\_i}{\partial m} = 1 \tag{4}
\end{align\*}
$$
Substitute $(3$) into $(4)$:
$$
\begin{align\*}
&\sum\_i (p\_i)^2 \frac{\partial \lambda}{\partial m} \frac{1}{u\_{ii}} = 1,\\
\to &\frac{\partial \lambda}{\partial m} = \frac{1}{\sum\_i \frac{(p\_i)^2}{u\_{ii}}} \tag{5}
\end{align\*}
$$
Notice that $u\_{ii} < 0$ by concavity of the sub-utility functions. As such, $\frac{\partial \lambda}{\partial m} < 0$ and also, by $(3)$:
$$
\frac{\partial x\_i}{\partial m} = p\_i \frac{1}{u\_{ii}} \frac{\partial \lambda}{\partial m} > 0
$$
as the right hand side is the product of two negative numbers.

This shows that $x\_i$ is a normal good. By symmetry, this holds for all goods.

An alternative quicker way to notice this is to see that

1. by $(3)$ the sign of $\dfrac{\partial x\_i}{\partial m}$ will be the reverse of the sign of $\dfrac{\partial \lambda}{\partial m}$.
2. This has to be true for all goods, which means that either all goods are normal, or all goods are inferior.
3. From $(4)$, it follows that at least one good should be normal (as otherwise the sum cannot be equal to 1 which is greater than zero).
4. Conclude that all goods have to be normal.

> **Comments:**
> **username_2**: You do not need to assume that $u$ is concave: the utility function is ordinal and the demand functions are unchanged after any monotonic transformation of the utility function.  
> **username_1**: @username_2 I'm not talking about the utility function but the sub-utility functions $f_i$. These better be concave as otherwise $u = \sum_i f_i $ does not even need to be quasi-concave. Notice that although the sum of concave functions is concave, the sum of quasi-concave functions is not necessarily quasi-concave.  
> **username_2**: I just wanted to mention that assuming concavity is sufficient not necessary: if all $u_{ii}>0$ the demands are still normal.  
> **username_1**: @username_2 If all sub-utility functions are convex (i.e. $u_{ii} > 0$) then the utility function is (quasi-)convex. In this case, the first order conditions is not going to give you a maximum, but a minimum.  
> **username_0**: Lovely! My one comment is that you seem to assume an interior solution by assuming (1) holds in equality. This is of course not guaranteed; usually we also assume $x \in \mathbb{R}_+^n$. Any ideas on how to handle the boundary case?  
> **username_1**: @username_0 I guess you can argue in the following way. If $x_i = 0$ at the optimum and if income increases, then either $x_i$ goes up (to an interior solution), so it is normal, or it stays at $0$ in which case it is also normal (in the weak sense). As $x_i = 0$, it can never go down if income increases.  
> **username_0**: I think you could also argue that the right hand side of $$ \begin{align*} &\sum_i p_i \frac{\partial x_i}{\partial m} = 1 \tag{4} \end{align*} $$ is unchanged if you remove all the indices where $x_i = 0$ and $\frac{\partial x_i}{\partial m} = 0$, and then your interior solution works again.  
> **username_3**: Could you please elaborate how to you arrive at the first order conditions for the utility maximisation problem (1):  

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