Integrability of demand

Given consumption space ${\displaystyle X}$ and a known walrasian demand function ${\displaystyle x:\mathbb {R} _{++}^{L}\times \mathbb {R} _{+}\rightarrow X}$, solving the problem of integrability of demand consists in finding a utility function ${\displaystyle u:X\rightarrow \mathbb {R} }$ such that

${\displaystyle x(p,w)=\operatorname {argmax} \{u(x):p\cdot x\leq w,\;x\in X\}}$

That is, it is essentially "reversing" the consumer's utility maximization problem.

There are essentially two steps in solving the integrability problem for a demand function. First, one recovers an expenditure function ${\displaystyle e(p,u)}$ for the consumer; then with it, one constructs an at-least-as-good set ${\displaystyle V_{u}=\{x\in \mathbb {R} _{+}^{L}:u(x)\geq u\}}$, which is equivalent to finding a utility function ${\displaystyle u(x)}$. If the demand function ${\displaystyle x(p,w)}$ is homogenous of degree zero, satisfies Walras' Law, and has a negative semi-definte substitution matrix ${\displaystyle S(p,w)}$, then it is possible to follow those steps to find a utility function ${\displaystyle u(x)}$ that generates ${\displaystyle x(p,w)}$.[2]

Proof: if the first two conditions (homogeneity of degree zero and Walras' Law) are met, then duality between the expenditure minimization problem and the utility maximization problem tells us that

${\displaystyle x(p,w)=h(p,v(p,w))}$

where ${\displaystyle v(p,w)=u(x(p,w))}$ is the consumers' indirect utility function and ${\displaystyle h(p,u)}$ is the consumers' hicksian demand function. Fix a utility level ${\displaystyle u_{0}=v(p,w)}$ [nb 1]. From Shephard's lemma, and with the identity above we have

${\displaystyle {\frac {\partial e(p)}{\partial p}}=x(p,e(p))}$

(1)

where we omit the fixed utility level ${\displaystyle u_{0}}$ for conciseness. (1) is a system of PDEs in the prices vector ${\displaystyle p}$, and Frobenius' theorem can be used to show that if the matrix

${\displaystyle D_{p}x(p,w)+D_{w}x(p,w)x(p,w)}$

is symmetric, then it has a solution. Notice that the matrix above is simply the substitution matrix ${\displaystyle S(p,w)}$, which we assumed to be symmetric firsthand. So (1) has a solution, and it is (at least theoretically) possible to find an expenditure function ${\displaystyle e(p)}$ such that ${\displaystyle p\cdot x(p,e(p))=e(p)}$.

For the second step, by definition,

${\displaystyle e(p)=e(p,u_{0})=\min\{p\cdot x:x\in V_{u_{0}}\}}$

where ${\displaystyle V_{u_{0}}=\{x\in \mathbb {R} _{+}^{L}:u(x)\geq u_{0}\}}$. By the properties of ${\displaystyle e(p,u)}$, it is not too hard to show [2] that ${\displaystyle V_{u_{0}}=\{x\in \mathbb {R} _{+}^{L}:p\cdot x\geq e(p,u_{0})\}}$. Doing some algebraic manipulation with the inequality ${\displaystyle p\cdot x\geq e(p,u_{0})}$, one can reconstruct ${\displaystyle V_{u_{0}}}$ in its original form with ${\displaystyle u(x)\geq u_{0}}$. If that is done, one has found a utility function ${\displaystyle u:X\rightarrow \mathbb {R} }$ that generates consumer demand ${\displaystyle x(p,w)}$.