If preferences are identical and homogeneous, then show that market demand for any good must be independent of the distribution of income.

My workings are as follows: $$q^{d}(p)=\sum_{i=1}^{n}f(p_x,p_{-x},I)$$

$$q^{d}(p)=f(p_{x},p_{-x})\sum_{i=1}^{n}I$$ Here, $x$ is the price of that particular good and $(-x)$ is the price vector of all the rest of the goods except $x$. Market demand is given by the submission of all consumers from $1$ to $n$.
But, I'm still confused how this function is multiplicatively separable in the price vector $P$.Can someone give a proof of it. Thanks!

  • $\begingroup$ That market demand is price-separable seems to follow immediately from its form; that is, $D = f(\mathbf{p}) \sum_{i}M_i.$ See if my answer helps with your understanding of the derivation. $\endgroup$ Oct 12, 2018 at 17:15

1 Answer 1


I take homogeneity of preferences here to mean that preferences are homothetic; that is, preferences can be represented by a utility function that is homogeneous of degree one.

For an arbitrary good $x$ and a vector of prices $\mathbf{p}$, let individual demand be $x_i(\mathbf{p}, M_i)$ for each consumer $i$ with income $M_i$. Thus market demand is $D = \sum_{i} x_i(\mathbf{p},M_i)$. Since preferences are identical,

$$\sum_{i} x_i(\mathbf{p},M_i) = \sum_{i} x(\mathbf{p},M_i).$$

Then, by homogeneity,

$$\sum_{i} x(\mathbf{p},M_i) = \sum_{i} x(\mathbf{p},1) M_i = x(\mathbf{p},1) \sum_{i} M_i = x(\mathbf{p},1)M,$$

where $\sum_i M_i = M$ is aggregate consumer income. Thus $D$ is independent of the distribution of income and is solely a function of aggregate income across consumers, $M$.


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