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Starr’s ’69 paper considered Walrasian equilibria in exchange economies with non-convex preferences i.e., upper contour sets of utility functions are non-convex. Suppose agents and goods with . Starr identified a price vector and a feasible allocation with the property that at most agents did not receiving a utility maximizing bundle at the price vector .

A poetic interlude. Arrow and Hahn’s book has a chapter that describes Starr’s work and closes with a couple of lines of Milton:

A gulf profound as that Serbonian Bog

Betwixt Damiata and Mount Casius old,

Where Armies whole have sunk.

Milton uses the word concave a couple of times in Paradise Lost to refer to the vault of heaven. Indeed the OED lists this as one of the poetic uses of concavity.

Now, back to brass tacks. Suppose is agent ‘s utility function. Replace the upper contour sets associated with for each by its convex hull. Let be the concave utility function associated with the convex hulls. Let be the Walrasian equilibrium prices wrt . Let be the allocation to agent in the associated Walrasian equilibrium.

For each agent let

where is agent ‘s endowment. Denote by the vector of total endowments and let .

Let be the excess demand with respect to and . Notice that is in the convex hull of the Minkowski sum of . By the Shapley-Folkman-Starr lemma we can find for , such that and .

When one recalls, that Walrasian equilibria can also be determined by maximizing a suitable weighted (the Negishi weights) sum of utilities over the set of feasible allocations, Starr’s result can be interpreted as a statement about approximating an optimization problem. I believe this was first articulated by Aubin and Elkeland (see their ’76 paper in Math of OR). As an illustration, consider the following problem :

subject to

Call this problem . Here is an matrix with .

For each let be the smallest concave function such that for all (probably quasi-concave will do). Instead of solving problem , solve problem instead:

subject to

The obvious question to be answered is how good an approximation is the solution to to problem . To answer it, let (where I leave you, the reader, to fill in the blanks about the appropriate domain). Each measures how close is to . Sort the ‘s in decreasing orders. If is an optimal solution to , then following the idea in Starr’s ’69 paper we get:

It states that the Minkowski sum of a large number of sets is approximately convex. The clearest statement as well as the nicest proof I am familiar with is due to J. W. S. Cassels. Cassels is a distinguished number theorist who for many years taught the mathematical economics course in the Tripos. The lecture notes are available in a slender book now published by Cambridge University Press.

This central limit like quality of the lemma is well beyond the capacity of a hewer of wood like myself. I prefer the more prosaic version.

Let be a collection of sets in with . Denote by the Minkowski sum of the collection . Then, every can be expressed as where for all and .

How might this be useful? Let be an 0-1 matrix and with . Consider the problem

Let be a solution to the linear relaxation of this problem. Then, the lemma yields the existence of a 0-1 vector such that and . One can get a bound in terms of Euclidean distance as well.

How does one do this? Denote each column of the matrix by and let . Let . Because and it follows that . Thus, by the Lemma,

where each and . In words, has at most fractional components. Now construct a 0-1 vector from as follows. If , set . If is fractional, round upto 1 with probability and down to zero otherwise. Observe that and the . Hence, there must exist a 0-1 vector with the claimed properties.

The error bound of is to large for many applications. This is a consequence of the generality of the lemma. It makes no use of any structure encoded in the matrix. For example, suppose were an extreme point and a totally unimodular matrix. Then, the number of fractional components of $x^*$ are zero. The rounding methods of Kiralyi, Lau and Singh as well as of Kumar, Marathe, Parthasarthy and Srinivasan exploit the structure of the matrix. In fact both use an idea that one can find in Cassel’s paper. I’ll follow the treatment in Kumar et. al.

As before we start with . For convenience suppose for all . As as has more columns then rows, there must be a non-zero vector in the kernel of , i.e., . Consider and . For and sufficiently small, for all . Increase and until the first time at least one component of and is in . Next select the vector with probability or the vector with probability . Call the vector selected .

Now . Furthermore, has at least one more integer component than . Let . Let be the matrix consisting only of the columns in and consist only of the components of in . Consider the system . As long as has more columns then rows we can repeat the same argument as above. This iterative procedure gives us the same rounding result as the Lemma. However, one can do better, because it may be that even when the number of columns of the matrix is less than the number of rows, the system may be under-determined and therefore the null space is non-empty.

In a sequel, I’ll describe an optimization version of the Lemma that was implicit in Starr’s 1969 Econometrica paper on equilibria in economies with non-convexities.

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