Characterizing demand functions with price dependent income

Abstract

We consider the demand function of consumer whose wealth depends on prices. This extends the two traditional cases when the consumer holds a goods bundle, so that his wealth depends linearly on prices, and when his wealth is prescribed, independently of prices. We extend the Slutsky relations to this general case, and we show that they fully characterize the demand functions, as in the traditional cases.

Appendices

Appendix A: Direct utility, indirect utility and smoothness

Lemma 3

If the utility function $U(x)$ satisfies the conditions of Assumption (1) and if $w(p)$ is of class $C^2$ then the map $p\rightarrow x(p)$ and the function $p\rightarrow \lambda (p)$ are of class $C^{2}$.

Proof

We apply the implicit function theorem. The functions $x(p)$ and $\lambda (p)$ are defined implicitly by the $n+1$ conditions

$$\begin{aligned} D_xU(x)-\lambda p&= 0 \\ p^\prime x-w(p)&= 0 \end{aligned}$$

Let $F:\mathbb{R }^n\times \mathbb{R }^n\times \mathbb{R }\rightarrow \mathbb{R } ^{n+1}$ be defined by $F(p,x,\lambda )=\left(F_1(p,x,\lambda ),F_2(p,x,\lambda )\right)$ where $F_1(p,x,\lambda )=D_xU(x)-\lambda p\in \mathbb{R }^n$ and $F_2(p,x,\lambda )=p^\prime x-w(p)\in \mathbb{R }$. To apply the implicit function theorem, it suffices to show that the matrix

$$\begin{aligned} D_{x,\lambda }F=\left(\begin{array}{cc}D_xF_1&D_\lambda F_1 \\ D_x F_2&D_\lambda F_2 \end{array}\right)=\left(\begin{array}{cc} D_{xx}^2U(x)&-p \\ p^\prime&0\end{array}\right) \end{aligned}$$

is nonsingular. Let $\zeta =(\zeta ^n,\zeta ^1)\in \mathbb{R }^n\times \mathbb{R }$. We will show that the linear system $(D_{x,\lambda }F)\zeta =0$ has only the zero solution. This system can be written in an equivalent form as

$$\begin{aligned} D_{xx}^{2}U(x)\zeta ^{n}-p\zeta ^{1}&= 0 \\ p^{\prime }\zeta ^{n}&= 0 \end{aligned}$$

It follows that $\zeta ^{n}\in \{p\}^{\bot }=\{\nabla U\}^{\bot }$. Multiply the first equality by $\zeta ^{n\prime }$ we get

$$\begin{aligned} \zeta ^{n\prime }\left(D_{xx}^{2}U(x)\right)\zeta ^{n}=0 \end{aligned}$$

Since $\zeta ^{n}\in \{\nabla U\}^{\bot }$ and $D_{xx}^{2}U$ is negative definite on this subspace, we conclude that $\zeta ^{n}=0$ and therefore $\zeta =0$ is the only solution. So the matrix $D_{x,\lambda }F$ is nonsingular and we can apply the implicit function theorem which guarantees that $x(p)$ and $\lambda (p)$ are of class $C^{2}$. The proof is complete. $\square$

Lemma 4

Let $V(p)$ be the indirect utility function. Then $V(p)$ has the following properties:

• a: Positively homogenous of degree zero if $w(p)$ is positively homogeneous of degree one.

• b: Quasiconvex if $w(p)$ is convex.

Proof

(a) Suppose $w(p)$ is positively homogenous of degree one. Then, for all $t\ge 0$, changing $p$ to $tp$ does not change the budget set in problem $(\mathcal{P })$, so that $x(tp) =x( p)$ and $V(p) =U\left(x(p) \right)$ is unchanged. To prove (b), we argue as in Varian [13]. Suppose that $V(\hat{p})\le u$ and $V(\bar{p} )\le u$. Let $\tilde{p}=t\hat{p}+(1-t)\bar{p}$. We want to show that $V( \tilde{p})\le \max \{V(\hat{p}),V(\bar{p})\}$. Introduce the following sets

$$\begin{aligned} \hat{S}=\{x\ |\ \hat{p}^{\prime }x\le w(\hat{p})\},\ \ \bar{S}=\{x\ |\ \bar{ p}^{\prime }x\le w(\bar{p})\},\ \ \tilde{S}=\{x\ |\ \tilde{p}^{\prime }x\le w(\tilde{p})\} \end{aligned}$$

We claim that $\tilde{S}\subset \hat{S}\cup \bar{S}$. Indeed, if this is not the case then there exists $x$ such that $\hat{p}^{\prime }x>w(\hat{p})$ and $\bar{p}^{\prime }x>w(\bar{p})$ whereas $\tilde{p}^{\prime }x\le w(\tilde{p} )$. It follows that for any $t\in (0,1)$, $t\hat{p}^{\prime }x>tw(\hat{p})$ and $(1-t)\bar{p}^{\prime }x>(1-t)w(\bar{p})$. Adding up the last two inequalities and using the convexity of $w$, we get

$$\begin{aligned} \tilde{p}^{\prime }x=\left(t\hat{p}^{\prime }+(1-t)\bar{p}^{\prime }\right)x>tw(\hat{p} )+(1-t)w(\bar{p})\ge w\left(t\hat{p}+(1-t)\bar{p}\right)=w(\tilde{p}) \end{aligned}$$

Hence $\tilde{p}^{\prime }x>w(\tilde{p})$ which is a contradiction, so $\tilde{S}\subset \hat{S}\cup \bar{S}$ as announced. This result implies that

$$\begin{aligned} V(\tilde{p})=\max _{x\in \tilde{S}}U(x)\le \max _{x\in \hat{S}\cup \bar{S} }U(x)=\max \{V(\hat{p}),V(\bar{p})\} \end{aligned}$$

Which means that $V(p)$ is quasiconvex. The proof is complete. $\square$

Appendix B: Exterior differential calculus tools

In this section, we give some theorems from exterior differential calculus (see [7], part 2, for a primer, and [5] for a full exposition). The question to be answered is the following. Given a vector field in a space of dimension $n$, when can it be decomposed as a linear combination of $k$ gradients, with $k<n$ ? This question is best rephrased in the language of exterior differential calculus: given a $1$-form

$$\begin{aligned} \omega =\sum _{i=1}^{n}\omega ^{i}dp_{i} \end{aligned}$$

on a space of dimension $n$, when can one find functions $u^{j}( p)$ and $v_{j}(p)$, with $1\le j\le k<n$ such that:

$$\begin{aligned} \omega =\sum _{j=1}^{k}u^{j}dv_{j} \end{aligned}$$

We refer to [7] and the literature therein, notably [1], [2, 8] for the mathematics and the economics of this question. The answer is provided by the following theorems, which go back to Darboux:

Theorem 5

Let $\omega$ be a $1$-form defined in a neighbourhood of $\bar{ p}$ in $R^{n}$. Suppose that, we have

(12)

Then there is a neighbourhood of $\bar{p}$ and smooth functions $v_{1}$ and $( u^{j},v_{j})$, $2\le j\le k$, such that the $dv_{j}$ and the $du^{j}$ do not vanish, and

$$\begin{aligned} \omega =dv_{1}+\sum _{j=2}^{k}u^{j}dv_{j}\ \ \ \ \mathrm{on }\quad \mathcal{V } \end{aligned}$$

(13)

Conversely, if $\omega$ decomposes in the form (13) on $\mathcal{U }$, and the $dv_{j}$ and the $du^{j}$ do not vanish, then it satisfies condition (12) on $\mathcal{U }$.

Proof

The case $k=1$ states that $\omega =dv_{1}$ if and only if $d\omega =0$, which is the well-known Poincaré Lemma. For $k>1$, the converse is easy to prove. Indeed, if $\omega$ decomposes in the form (13), we have:

$$\begin{aligned} d\omega =\sum _{j=2}^{k}du^{j}\wedge dv_{j} \end{aligned}$$

and

$$\begin{aligned} \left( d\omega \right) ^{k-1}=\left( k-1\right) !du^{2}\wedge dv_{2} \wedge \ldots \wedge du^{k}\wedge dv_{k} \end{aligned}$$

Then $(d\omega ) ^{k}=0$ follows immediately (note, however, that $\omega \wedge (d\omega ) ^{k-1}\ne 0$). For the direct part, we refer to [5]. $\square$

We have also the following problem

Theorem 6

Let $\omega$ be a $1$-form defined in a neighbourhood $\mathcal{U }$ of $\bar{ p}$ in $R^{n}$. Suppose that, on $\mathcal{U }$, we have

$$\begin{aligned} \omega \wedge (d\omega )^{k-1}\ne 0,\quad (d\omega )^{k}\ne 0,\quad \omega \wedge (d\omega )^{k}=0 \end{aligned}$$

(14)

Then there is a neighbourhood $\mathcal{V\subset U }$ and smooth functions $(u^{j},v_{j})$, $1\le j\le k$, such that the $dv_{j}$ and the $du^{j}$ do not vanish, and:

$$\begin{aligned} \omega =\sum _{j=1}^{k}u^{j}dv_{j}\ \ \ \ \mathrm{on\ \ }\mathcal V \end{aligned}$$

(15)

Conversely, if $\omega$ decomposes in the form (15) on $\mathcal{U }$, and the $dv_{j}$ and the $du^{j}$ do not vanish, then it satisfies condition (14) on $\mathcal{U }$.

Proof

Again, the converse is easy. If the decomposition (15) holds, then:

$$\begin{aligned} \left( d\omega \right) ^{k}=k!du^{1}\wedge dv_{1}\wedge \ldots \wedge du^{k}\wedge dv_{k} \end{aligned}$$

and $\omega \wedge \left( d\omega \right) ^{k}=0$. For the direct part, we refer to [5] $\square$

Another result, due to Ekeland and Chiappori in the analytic case (the coefficients $\omega ^{i}$ of the $1$-form $\omega$ are supposed to be analytic functions of $p$), and to Ekeland and Nirenberg [10] in the general case, addresses the problem of finding such decompositions when the functions $v_{j}\,\$are required to be convex and the $u^{i}$ positive. We state it as follows:

Theorem 7

Let $\omega$ be a smooth $1$-form in the neighbourhood of some point $\bar{p}$. There exist $2k$ functions $u^{1},\ldots ,u^{k},v_{1},\ldots ,v_{k}$ such that $\omega$ can be decomposed as $\omega =u^{1}dv_{1}+u^{2}dv_{2}+ \cdots +u^{k}dv_{k}$ where the functions $u^{i}$ are positive and the $v_{i}$ are convex if and only if

• (1) $\omega \wedge (d\omega )^{k}=0$.

• (2) There is a $k$-dimensional subspace $E$ of $\mathcal{I }=\{\alpha \ |\ \alpha \wedge \omega \wedge (d\omega )^{k-1}=0\}$ containing $\omega (\bar{p} )$ such that on $E^{\bot }$, the matrix $\omega _{ij}(\bar{p})$ is symmetric and positive definite.

Appendix C: Proofs of main results

Define the differential 1-form $\omega$ as follows:

$$\begin{aligned} \omega =\sum _{i=1}^{n}x^{i}(p)dp_{i} \end{aligned}$$

(16)

If $x(p)$ is a solution of a problem of type $(\mathcal{P })$ then

$$\begin{aligned} \frac{\partial V}{\partial p_i}=\lambda \left(\frac{\partial w}{\partial p_i}-x^i\right) \end{aligned}$$

and the 1-form $\omega$ can be decomposed as

$$\begin{aligned} \omega =\mu dV+dw \end{aligned}$$

(17)

where

$$\begin{aligned} \mu (p)=-\frac{1}{\lambda (p)}<0 \end{aligned}$$

Notice that $\omega$ has a decomposition of the form (13) and that $d\omega =d\mu \wedge dV$, $\omega \wedge d\omega =dw\wedge d\mu \wedge dV\ne 0$ and $d\omega \wedge d\omega =0$. Therefore, the necessary and sufficient condition for this decomposition is fulfilled. However, in our setting the income function $w(p)$ can be found from $w(p)=p^\prime x(p)$ once $x(p)$ is given.

The problem of mathematical integration consists in finding $\mu (p)$ and$\ V(p)$ such that the decomposition (17) holds. The problem of economic integration consists in finding such a decomposition with $\mu (p) <0$ and $V(p)$ quasi-convex (see [7]).

We now proceed to the proof of Theorem (1).

1.1 C.1 Proof of Theorem (1)

Suppose that $\omega =\mu dV+dw$, so that $\omega -dw=\mu dV$. The last condition is equivalent to:

$$\begin{aligned} (\omega -dw)\wedge d\omega =0 \end{aligned}$$

This, in turn, is equivalent to saying that there exists some $1$-form $\beta$ such that

$$\begin{aligned} d\omega =\beta \wedge (\omega -dw) \end{aligned}$$

(18)

We shall determine the 1-form $\beta$ (mod $\omega -dw$) by applying the vector field

$$\begin{aligned} \xi =\sum _{i=1}^{n}p_{i}\frac{\partial }{\partial p_{i}} \end{aligned}$$

(19)

to both sides of Eq. (18). This gives

$$\begin{aligned} d\omega (\xi ,.)=\beta \wedge (\omega -dw)(\xi ,.) \end{aligned}$$

(20)

Expanding the right-hand side we find that

$$\begin{aligned} \beta \wedge (\omega -dw)(\xi ,.)=\ <\beta ,\xi >(\omega -dw)-\beta (w-p^{\prime }Dw) \end{aligned}$$

Equation (20) becomes

$$\begin{aligned} d\omega (\xi ,.)=\ <\beta ,\xi >(\omega -dw)-\beta (w-p^{\prime }Dw) \end{aligned}$$

Solving for $\beta$, we get

$$\begin{aligned} \beta =\frac{1}{w-p^{\prime }Dw} \left(-d\omega (\xi ,.)+\ <\beta ,\xi >(\omega -dw)\right) \end{aligned}$$

Note that the denominator does not vanish, since $w(p)$ has been assumed not to be homogeneous. Plugging this value of $\beta$ into Eq. (18), we obtain

$$\begin{aligned} d\omega =\frac{-1}{w-p^{\prime }Dw}d\omega (\xi ,.)\wedge (\omega -dw) \end{aligned}$$

(21)

However, a direct computation gives:

$$\begin{aligned} d\omega&=\sum _{i<j}\left( \frac{\partial x^{i}}{\partial p_{j}}-\frac{ \partial x^{j}}{\partial p_{i}}\right) dp_{j}\wedge dp_{i} \nonumber \\ d\omega (\xi ,.)&=\sum _{j,k}\frac{\partial x^{j}}{\partial p_{k}} p_{k}dp_{j}-\sum _{j,k}\frac{\partial x^{k}}{\partial p_{j}}p_{k}dp_{j} \end{aligned}$$

(22)

Performing the exterior product in Eq. (21), we get:

$$\begin{aligned} d\omega =\frac{1}{w-p^{\prime }Dw}\sum _{i,j}\sum _{k}\left( \frac{\partial x^{k}}{\partial p_{j}}p_{k}(x^{i}-\frac{\partial w}{\partial p_{i}})-\frac{ \partial x^{j}}{\partial p_{k}}p_{k}(x^{i}-\frac{\partial w}{\partial p_{i}} )\right) dp_{j}\wedge dp_{i} \end{aligned}$$

(23)

Let

$$\begin{aligned} \alpha _{ij}=\sum _{k}\left( \frac{\partial x^{k}}{\partial p_{j}}p_{k}(x^{i}-\frac{\partial w}{\partial p_{i}})-\frac{ \partial x^{j}}{\partial p_{k}}p_{k}(x^{i}-\frac{\partial w}{\partial p_{i}} )\right) \end{aligned}$$

(24)

Then, (23) becomes

$$\begin{aligned} d\omega =\sum _{i,j}\alpha _{ij}dp_j\wedge dp_i \end{aligned}$$

which, in turn, can be written as

$$\begin{aligned} \sum _{i<j}\left( \frac{\partial x^{i}}{\partial p_{j}}-\frac{\partial x^{j}}{ \partial p_{i}}\right) dp_{j}\wedge dp_{i}=\frac{1}{w-p^{\prime }D_{p}w(p)}\sum _{i<j}(\alpha _{ij}-\alpha _{ji})dp_j\wedge dp_i \end{aligned}$$

Using (24), we conclude that the last equality is satisfied if and only if the matrix $s$ defined by

$$\begin{aligned} s_{ij}=\frac{\partial x^{i}}{\partial p_{j}}+\frac{1}{w-p^{\prime }Dw} \sum _{k}\left( \frac{\partial x^{k}}{\partial p_{i}}p_{k}-\frac{\partial x^{i}}{\partial p_{k}}p_{k}\right) (x^{j}-\frac{\partial w}{\partial p_{j}}) \end{aligned}$$

(25)

is symmetric.

We now recall that $w(p) =p^{\prime }x(p)$. Differentiating with respect to $p_{i}$, we get

$$\begin{aligned} \sum _{k}\frac{\partial x^{k}}{\partial p_{i}}p_{k}+x^{i}=\frac{\partial w}{ \partial p_{i}} \end{aligned}$$

(26)

It follows that

$$\begin{aligned} x^{i}-\frac{\partial w}{\partial p_{i}}=-\sum _{k}\frac{\partial x^{k}}{\partial p_{i}}p_{k}=-p^{\prime }D_{p_{i}}x \end{aligned}$$

(27)

Multiplying the last equation by $p_{i}$ and adding up yields

$$\begin{aligned} w-p^{\prime }D_{p}w=-p^{\prime }(D_{p}x)p \end{aligned}$$

(28)

Using (27) and (28), we can write (25) as

$$\begin{aligned} S_{ij}=\frac{\partial x^{i}}{\partial p_{j}}+\frac{1}{p^{\prime }(D_{p}x)p} \sum _{k=1}^{n}\left( \frac{\partial x^{k}}{\partial p_{i}}-\frac{\partial x^{i}}{\partial p_{k}}\right) p_{k}(p^{\prime }D_{p_{j}}x) \end{aligned}$$

and this matrix should be symmetric.

Conversely, let there be a function $x(p)$ such that $S_{ij}$ is symmetric, and set $w(p) =p^{\prime }x(p)$. Differentiating, we get Eqs. (27) and (28) so that $S=S^{\prime }$ is equivalent to $s=s^{\prime }$. But the symmetry condition $s=s^{\prime }$ is equivalent to $(\omega -dw)\wedge d\omega =0$. Therefore, there exist two functions $\mu (p)$ and $V(p)$ such that $\omega -dw=\mu dV$. Finally, after getting the indirect utility function and the Lagrange multiplier we find $U(x)$ using the duality relation

$$\begin{aligned} U(x)=\min _p\{V(p)|p^\prime x\le w\} \end{aligned}$$

(29)

It follows that $D_xU(x)=\lambda p$ and $U(x(p))=V(p)$. This completes the proof. $\square$

1.2 C.2 Proof of Theorem (2)

We apply Theorem (7). Let us define the differential 1-form $\Omega$ by

$$\begin{aligned} \Omega =\omega -dw \end{aligned}$$

Notice that $d\Omega =d\omega$ and $\Omega \wedge d\Omega =0$ since $\Omega =\mu dV$. Using the notation of Theorem (7), we have $k=1$, $\mathcal{I } =\{\alpha |\alpha \wedge \Omega =0\}$ and $E=$ span$\{\Omega \}$. Its clear that $\Omega \in \mathcal{I }$ and that $E^{\bot }=$ span$\{p^{\prime }D_{p}x(p)\}^{\bot }$.

Consider the matrix:

$$\begin{aligned} \Omega _{ij}=\frac{\partial x^{i}}{\partial p_{j}}-\frac{\partial ^{2}w}{\partial p_{j}\partial p_{i}} \end{aligned}$$

The restriction of $(\Omega _{ij})$ to a subspace of codimension one is symmetric and negative definite according to (6) and hypothesis (d). It follows from the Ekeland-Nirenberg Theorem that there exist two functions $\mu (p)$ and $v(p)$ such that $\mu (p)>0$, $v(p)$ is concave and $\Omega =\mu dv$. It follows that $\omega =\mu dv+dw$ which gives

$$\begin{aligned} x^{i}(p)=\mu (p)\frac{\partial v}{\partial p_{i}}+\frac{\partial w}{\partial p_{i}} \end{aligned}$$

Setting $\lambda (p)=\frac{1}{\mu (p)}>0$ and $V(p)=-v(p)$, $V(p)$ is convex since $v(p)$ is concave, we have:

$$\begin{aligned} \frac{\partial V}{\partial p_{i}}=\lambda (p)\left( \frac{\partial w}{\partial p_{i}}-x^{i}(p)\right) \end{aligned}$$

where $V(p)$ is convex and therefore is quasiconvex. The function $V(p)$ is an indirect utility function and $\lambda (p)$ is the Lagrange multiplier. From $V$, we can find the direct utility function $U(x)$ using the duality relation (29). If $V( p)$ is strongly convex then $U(x)$ is quasi-concave (see [9], Proposition 11). The function $U(x)$ defined in this way is a utility function for the individual whose demand function is $x(p)$. This completes the proof. $\square$

1.3 C.3 Proof of Theorem (3)

As above, the necessary and sufficient condition is:

$$\begin{aligned} d\omega =\beta \wedge (\omega -dw) \end{aligned}$$

(30)

Let us apply the vector field $\xi =\sum p_{i}\frac{\partial }{\partial p_{i} }$ to both sides of this equation. Notice first that, since $w(p)$ is homogeneous of degree one and $x(p)$ is homogeneous of degree zero, we have from Euler’s identity

$$\begin{aligned} \sum _{i=1}^{n}\frac{\partial w}{\partial p_{i}}p_{i}=w(p)\quad \text{ and} \quad \sum _{i=1}^{n}\frac{\partial x^{j}}{\partial p_{i}}p_{i}=0,\quad j=1,\ldots ,n \end{aligned}$$

It follows that:

$$\begin{aligned} d\omega (\xi ,.)=<\beta ,\xi >(\omega -dw)-\beta <\omega -dw,\xi > \end{aligned}$$

(31)

But

$$\begin{aligned} <\omega -dw,\xi >\ =\ <\omega ,\xi >-<dw,\xi >\ =w-w=0. \end{aligned}$$

Therefore, equality (31) reduces to $d\omega (\xi ,.)=\ <\beta ,\xi >(\omega -dw)$. Recall that

$$\begin{aligned} d\omega (\xi ,.)=\sum _{i,j}\frac{\partial x^{i}}{\partial p_{j}} p_{j}dp_{i}-\sum _{i,j}\frac{\partial x^{i}}{\partial p_{j}}p_{i}dp_{j} \end{aligned}$$

The first term vanishes because $x(p)$ is homogeneous of degree zero. We end up with

$$\begin{aligned} d\omega (\xi ,.)=-\sum _{i,j}\frac{\partial x^{i}}{\partial p_{j}} p_{i}dp_{j}=<\beta ,\xi >(\omega -dw) \end{aligned}$$

This equation can be written under the form

$$\begin{aligned} d\omega (\xi ,.)=<\beta ,\xi >\sum _{j}\left( x^{j}(p)-\frac{\partial w}{\partial p_{j}}\right) dp_{j} \end{aligned}$$

(32)

Differentiating the budget constraint once, we get $x(p)-D_{p}w(p)=-p^{ \prime }D_{p}x(p)$. Equation (32) can be written as $d\omega (\xi ,.)=\ <\beta ,\xi >d\omega (\xi ,.)$. We conclude that the differential 1-form $\beta$ must satisfy $<\beta ,\xi >\ =1$.

We now go back to (30). Set $\beta =\sum _{j=1}^{n}\beta ^{j}dp_{j}$. Using the fact that $x(p)-D_{p}w(p)=-p^{\prime }D_{p}x(p)$, Eq. (30) can be written as

$$\begin{aligned} \sum _{i<j}\left( \frac{\partial x^{i}}{\partial p_{j}}-\frac{\partial x^{j}}{\partial p_{i}}\right) dp_{j}\wedge dp_{i}=-\sum _{i<j}\left( \beta ^{j}\sum _{k}\frac{\partial x^{k}}{\partial p_{i}}p_{k}-\beta ^{i}\sum _{k} \frac{\partial x^{k}}{\partial p_{j}}p_{k}\right) dp_{j}\wedge dp_{i} \end{aligned}$$

This equality is satisfied if and only if

$$\begin{aligned} \frac{\partial x^{i}}{\partial p_{j}}-\beta ^{i}\sum _{k}\frac{\partial x^{k} }{\partial p_{j}}p_{k}=\frac{\partial x^{j}}{\partial p_{i}}-\beta ^{j}\sum _{k}\frac{\partial x^{k}}{\partial p_{i}}p_{k} \end{aligned}$$

(33)

where $\beta$ is any differential 1-form that satisfies the condition $<\beta ,\xi >\ =1$. The proof is complete. $\square$