# Are the goods in additively separable utility functions normal goods?

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?

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