# Is there a proof for composite commodity theorem?

I have been reading Economics and Consumer Behavior by Angus Deaton and John Muellbauer, specifically reading up on Composite Commodity Theorem, which states:

if prices move in parallel to each other, then the corresponding group of commodities can be treated as a single good. (page 120 in the book)

This is a pretty mind blowing result (well at least for me), however im yet to find an actual proof for the theorem in the literature.

is there a proof for this result?

## 1 Answer

There is this nice expository note:
Carter M., 1995, "An expository note on the composite commodity theorem," Economic Theory, 5, 175-179
and an interesting generalization by
Lewbel, A., 1996, "Aggregation without Separability: A Generalized Composite Commodity Theorem," American Economic Review, 86, 524-543.

EDIT:
One reason why it is difficult to find a proof of the theorem, is that the so called "composite commodity theorem" does not really have the mathematical status of a theorem. It is more a principle directly following from the reparameterization below.
Let $$x\in \mathbb{R}^J_+$$ denotes the disaggregate demand system, which depends on the price vector $$p \in \mathbb{R}^J_+$$, and the budget $$b>0$$. The relative prices are denoted by $$\alpha = p/\textbf{p}$$ where the aggregate price index is $$\textbf{p} \in \mathbb{R_+}$$. If $$\alpha$$ is a vector of constants, we can simplify the demand system such that it depends only upon the aggregate price $$\bf{p}$$: $$x(p,b) = x(\alpha \mathbf{p},b) \equiv \mathbf{x}(\mathbf{p},b).$$ Note that now $$\mathbf{x}:\mathbb{R^2_+} \rightarrow \mathbb{R}^J_+$$. If one is also interested in aggregating the elementary quantities, it is (for instance) possible to define the aggregate composite quantity as: $$\mathbf{X}(\mathbf{p},b) = \alpha^T\mathbf{x}(\mathbf{p},b),$$ where $$\mathbf{X}:\mathbb{R^2_+} \rightarrow \mathbb{R}_+$$. In this case, the elementary and aggregate expenditures are the same, in the sense that: $$\mathbf{p}\mathbf{X}(\mathbf{p},b)=p^Tx(p,b).$$