Inspired by this answer.

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?


1 Answer 1


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.
  • $\begingroup$ 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. $\endgroup$
    – Bertrand
    May 15, 2021 at 15:58
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    $\begingroup$ @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. $\endgroup$
    – tdm
    May 15, 2021 at 16:02
  • $\begingroup$ I just wanted to mention that assuming concavity is sufficient not necessary: if all $u_{ii}>0$ the demands are still normal. $\endgroup$
    – Bertrand
    May 15, 2021 at 16:09
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    $\begingroup$ @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. $\endgroup$
    – tdm
    May 15, 2021 at 17:09
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    $\begingroup$ 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. $\endgroup$
    – Giskard
    May 15, 2021 at 18:42

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