Consider any function of prices and utility,E(p,u), that may or may not be an expendi- ture function. Now suppose thatEsatisfies the expenditure function properties 1 to 7 of
Theorem 1.7, so that it is continuous, strictly increasing, and unbounded above inu, as well as increasing, homogeneous of degree one, concave, and differentiable inp. Thus, E ‘looks like’ an expenditure function. We shall show that E must then be an expendi- ture function. Specifically, we shall show that there must exist a utility function on Rn+ whose expenditure function is preciselyE. Indeed, we shall give an explicit procedure for constructing this utility function.
To see how the construction works, choose(p0,u0)∈Rn++×R+, and evaluate E there to obtain the numberE(p0,u0). Now use this number to construct the (closed) ‘half- space’ in the consumption set,
A(p0,u0)≡ {x∈Rn+|p0ãx≥E(p0,u0)},
illustrated in Fig. 2.1(a). Notice thatA(p0,u0)is a closed convex set containing all points on and above the hyperplane,p0ãx=E(p0,u0). Now choose different pricesp1, keepu0 fixed, and construct the closed convex set,
A(p1,u0)≡ {x∈Rn+|p1ãx≥E(p1,u0)}.
Imagine proceeding like this forallpricesp0and forming the infinite intersection, A(u0)≡
p0
A(p,u0)= {x∈Rn+|pãx≥E(p,u0)for allp0}. (2.1)
The shaded area in Fig. 2.1(b) illustrates the intersection of a finite number of the A(p,u0), and gives some intuition about whatA(u0)will look like. It is easy to imagine that as more and more prices are considered and more sets are added to the intersection, the shaded area will more closely resemble asuperior setfor some quasiconcave real-valued function. One might suspect, therefore, that these sets can be used to construct something
x1 x2
x1 x2
A(p0, u0) 傽A(p, u0)
{x|p0• x E(p0, u0)}
(a) (b)
Figure 2.1. (a) The closed half-spaceA(p0,u0). (b) The intersection of a finite collection of the setsA(p,u0).
very much like a direct utility function representing nice convex, monotonic preferences.
This is indeed the case and is demonstrated by the following theorem.
THEOREM 2.1 Constructing a Utility Function from an Expenditure Function
Let E:Rn++×R+→R+satisfy properties 1 through 7 of an expenditure function given in Theorem 1.7. Let A(u)be as in (2.1). Then the function u: Rn+→R+given by
u(x)≡max{u≥0|x∈A(u)}
is increasing, unbounded above, and quasiconcave.
You might be wondering why we have chosen to defineu(x)the way we have. After all, there are many ways one can employE(p,u) to assign numbers to eachx∈Rn+. To understand why, forget this definition ofu(x)and for the moment suppose that E(p,u) is in fact the expenditure function generated by some utility functionu(x). How might we recoveru(x)from knowledge ofE(p,u)? Note that by the definition of an expendi- ture function,pãx≥E(p,u(x))for all pricesp0, and, typically, there will be equality for some price. Therefore, becauseE is strictly increasing inu,u(x)is the largest value ofusuch thatpãx≥E(p,u)for allp0. That is, u(x)is the largest value ofusuch thatx∈A(u). Consequently, the construction we have given is just right for recovering the utility function that generatedE(p,u)when in factE(p,u)is an expenditure function.
But the preceding considerations give us a strategy for showing that it is: first, show that u(x)defined as in the statement of Theorem 2.1 is a utility function satisfying our axioms.
(This is the content of Theorem 2.1.) Second, show thatEis in fact the expenditure func- tion generated byu(x). (This is the content of Theorem 2.2.) We now give the proof of Theorem 2.1.
Proof:Note that by the definition ofA(u), we may writeu(x)as u(x)=max{u≥0|pãx≥E(p,u) ∀p0}.
The first thing that must be established is thatu(x)is well-defined. That is, it must be shown that the set{u≥0|pãx≥E(p,u) ∀p0}contains a largest element. We shall sketch the argument. First, this set, call itB(x), must be bounded above becauseE(p,u) is unbounded above and increasing inu. Thus,B(x)possesses an upper bound and hence also a least upper bound,u. It must be shown thatˆ uˆ ∈B(x). But this follows becauseB(x) is closed, which we will not show.
Having argued thatu(x)is well-defined, let us consider the claim that it is increasing.
Considerx1≥x2. Then
pãx1≥pãx2 ∀p0, (P.1)
because all components ofx1are at least as large as the corresponding component ofx2. By the definition ofu(x2),
pãx2≥E(p,u(x2)) ∀p0. (P.2) Together, (P.1) and (P.2) imply that
pãx1≥E(p,u(x2)) ∀p0. (P.3) Consequently, u(x2) satisfies the condition: x1∈A(u(x2)). But u(x1) is the largest u satisfyingx1∈A(u). Hence,u(x1)≥u(x2), which shows thatu(x)is increasing.
The unboundedness ofu(ã)onRn+ can be shown by appealing to the increasing, concavity, homogeneity, and differentiability properties ofE(ã)inp, and to the fact that its domain inuis all ofRn+. We shall not give the proof here (although it can be gleaned from the proof of Theorem 2.2 below).
To show that u(ã) is quasiconcave, we must show that for all x1,x2, and convex combinations xt,u(xt)≥min[u(x1),u(x2)]. To see this, suppose that u(x1)= min[u(x1),u(x2)]. Because E is strictly increasing in u, we know that E(p,u(x1))≤ E(p,u(x2))and that therefore
tE(p,u(x1))+(1−t)E(p,u(x2))≥E(p,u(x1)) ∀t∈ [0,1]. (P.4) From the definitions ofu(x1)andu(x2), we know that
pãx1≥E(p,u(x1)) ∀p0, pãx2≥E(p,u(x2)) ∀p0.
Multiplying byt≥0 and(1−t)≥0, respectively, adding, and using (P.4) gives pãxt ≥E(p,u(x1)) ∀p0 and t∈ [0,1].
Consequently, by definition ofu(xt),u(xt)≥u(x1)=min[u(x1),u(x2)]as we sought to show.
Theorem 2.1 tells us we can begin with an expenditure function and use it to construct a direct utility function representing some convex, monotonic preferences. We actually know a bit more about those preferences. If we begin with them and derive the associated expenditure function, we end up with the functionE(ã)we started with!
THEOREM 2.2 The Expenditure Function of Derived Utility,u, isE
Let E(p,u), defined onRn++×Rn+, satisfy properties 1 to 7 of an expenditure function given in Theorem 1.7 and let u(x)be derived from E as in Theorem 2.1. Then for all non-negative
prices and utility,
E(p,u)=min
x pãx s.t. u(x)≥u.
That is, E(p,u)is the expenditure function generated by derived utility u(x).
Proof:Fixp00andu0≥0 and supposex∈Rn+satisfiesu(x)≥u0. Note that because u(ã)is derived fromEas in Theorem 2.1, we must then have
pãx≥E(p,u(x)) ∀p0.
Furthermore, becauseEis increasing in utility andu(x)≥u0, we must have
pãx≥E(p,u0) ∀p0. (P.1)
Consequently, for any given pricesp0, we have established that
E(p0,u0)≤p0ãx ∀x∈Rn+ s.t. u(x)≥u0. (P.2) But (P.2) then implies that
E(p0,u0)≤ min
x∈Rn+p0ãx s.t. u(x)≥u0. (P.3) We would like to show that the first inequality in (P.3) is an equality. To do so, it suffices to find a singlex0∈Rn+such that
p0ãx0≤E(p0,u0) and u(x0)≥u0, (P.4) because this would clearly imply that the minimum on the right-hand side of (P.3) could not be greater thanE(p0,u0).
To establish (P.4), note that by Euler’s theorem (Theorem A2.7), because E is differentiable and homogeneous of degree 1 inp,
E(p,u)= ∂E(p,u)
∂p ãp ∀p0, (P.5)
where we use∂E(p,u)/∂p≡(∂E(p,u)/∂p1, . . . , ∂E(p,u)/∂pn)to denote the vector of price-partial derivatives ofE. Also, becauseE(p,u) is concave inp, Theorem A2.4 implies that for allp0,
E(p,u0)≤E(p0,u0)+∂E(p0,u0)
∂p ã(p−p0). (P.6)
But evaluating (P.5) at(p0,u0)and combining this with (P.6) implies that E(p,u0)≤
∂E(p0,u0)/∂p
ãp ∀p0. (P.7)
Letting x0=∂E(p0,u0)/∂p, note that x0∈Rn+ because E is increasing in p. We may rewrite (P.7) now as
pãx0≥E(p,u0) ∀p0. (P.8)
So, by the definition ofu(ã), we must haveu(x0)≥u0. Furthermore, evaluating (P.5) at (p0,u0)yieldsE(p0,u0)=p0ãx0. Thus, we have established (P.4) for this choice ofx0, and therefore we have shown that
E(p0,u0)= min
x∈Rn+ p0ãx s.t. u(x)≥u0.
Becausep00andu0≥0 were arbitrary, we have shown thatE(p,u)coincides with the expenditure function ofu(x)onRn++×R+.
The last two theorems tell us that any time we can write down a function of prices and utility that satisfies properties 1 to 7 of Theorem 1.7, it will be a legitimate expenditure function for some preferences satisfying many of the usual axioms. We can of course then differentiate this function with respect to product prices to obtain the associated system of Hicksian demands. If the underlying preferences are continuous and strictly increasing, we can invert the function inu, obtain the associated indirect utility function, apply Roy’s identity, and derive the system of Marshallian demands as well. Every time, we are assured that the resulting demand systems possess all properties required by utility maximisation.
For theoretical purposes, therefore, a choice can be made. One can start with a direct utility function and proceed by solving the appropriate optimisation problems to derive the Hicksian and Marshallian demands. Or one can begin with an expenditure function and proceed to obtain consumer demand systems by the generally easier route of inversion and simple differentiation.