Lecture 3. Classical demand theory II.
Equivalence between dual and primal problems
Proposition. Suppose that preferences are continuous and locally non-satiated. For each p ∈ RL + + and each u ≥ u(0),
v(p, e(p, u)) = u.
Proof. We need to show that (a) v(p, e(p, u)) ≥ u and (b) v(p, e(p, u)) ≤ u.
For (a), take
h ∈ h(p, u). By the definition of the dual problem,
u(h) ≥ u and
p⋅h = e(p, u). This implies that
h satisfies the constraints of the primal problem
(p, e(p, u)) (i.e.,
h ∈ B(p, e(p, u)), and thus
v(p, e(p, u)) ≥ u(h) ≥ u.
For (b), take
x ∈ x(p, e(p, u)). Because the assumptions imply that Walras Law holds, we have5
p⋅x = e(p, u).
Suppose that
u(x) > u ≥ u(0) and consider replacing
x by
xε = (1 − ε)x + ε0
for some
ε > 0. Then, for each
ε > 0,
p⋅xε = (1 − ε)p⋅x < e(p, u).
Moreover, by continuity of the utility function, for sufficiently small
ε,
u(xε) > u.
But then,
xε satisfies the constraints of the dual problem and leads to lower expenditure. Contradiction.
Upper contour sets can be deduced from the expenditure function
Assume that preferences are continuous, monotonic and convex. For each
u,
{x:u(x) ≥ u} = {x:p⋅x ≥ e(p, u) for each p ∈ RL + }.
Inclusion “ ⊆ ” follows form the definition of the expenditure function. For the other inclusion, we need to show that for each x = (x1, ..., xL), if u(x) < u, then there exists p ∈ RL + such that p⋅x < e(p, u).
Because the preferences are convex, the set
{x:u(x) ≥ u} is convex. The serparating hyperplane theorem shows that there exists
p ∈ RL, p ≠ 0 (but, possibly,
pl < 0 for some
l) such that
Below, we show that it must be that
pl ≥ 0 for each
l. Given that, notice that the right-hand side of the above inequality is equal to
e(p, u). Thus,
p⋅x < e(p, u).
To finish our proof, we need to show that if (
1↑) holds, then it must be that
pl ≥ 0 for each
l. On the contrary, suppose not and there is
l such that
pl < 0. For each
a > 0 and
ε > 0, define a bundle
xa, ε =
x + a(ε, ..., ε, 1lth position, ε, ..., ε)
=
(x1 + aε, ..., xl − 1 + aε, xl + a, xl + 1 + aε, ..., xL + aε).
Then,
p⋅xa, ε = p⋅x + a(pl + ε ⎲⎳l’ ≠ lpl).
Because
pl < 0, we can find
ε* > 0 small enough so that
pl + ε∑l’ ≠ lpl < 0 and for each
a > 0,
Also, we can find
a* > 0 large enough so that
(Indeed, choose any
y such that
u(y) ≥ u and notice that we can find
a* large enough so that for each
l’,
a*ε* ≥ yl’ − xl’. Then, for each
l’
xa*, ε*l’ ≥ xl’ + a*ε* ≥ xl’ + yl’ − xl’ ≥ yl’,
in which case the monotonicity of preferences implies (
3↑). But then, (
2↑) and (
3↑) contradicts (
1↑).
Generalized Envelope Theorems
-
K ⊆ RL is a compact set,
-
f:RL × [ − 1, 1] → R is continuous in (x, t) and it has partial derivative ft(x, t) that is continuous in (x, t)
Let x*(t) be the set of maximizers of the above problem.
Theorem. (
Generalized Envelope Theorem)
For each t,
limsupt’ → t(V(t’) − V(t))/(t’ − t)
= maxx ∈ x*(t)ft(x, t),
liminft’ → t(V(t’) − V(t))/(t’ − t)
= minx ∈ x*(t)ft(x, t).
In particular, if x*(t) is unique, then V is differentiable at t and (d)/(dt)V(t) = ft(x*(t), t).
Proof. First, we show that
limsupt’ → t(V(t’) − V(t))/(t’ − t) ≥ maxx ∈ x*(t)ft(x, t).
Indeed, because the derivative
ft(x, t) is continuous, there exists
x0 ∈ argmaxx ∈ x*(t)ft(x, t). Moreover, by the definition of the partial derivative,
ft(x0, t) = limt’ → t(f(x0, t’) − f(x0, t))/(t’ − t).
The claim follows from the fact that
V(t’) ≥ f(x0, t’) and V(t) = f(x0, t).
Next, we show that
limsupt’ → t(V(t’) − V(t))/(t’ − t) ≥ maxx ∈ x*(t)ft(x, t).
If not, then, there exists a sequence
tn → t and
xn such that
V(tn) = f(xn, tn)
and
limsupn → ∞(f(xn, tn) − V(t))/(tn − t) > maxx ∈ x*(t)ft(x, t).
By ocmpactness of
K, we can choose a subseqnece
xn → x. By the continuuity of function
f, it must be that
x is an optimal choice for
t, i.e.,
x ∈ x*(t). Because
V(t) ≥ f(xn, t), we have
limsupn → ∞(f(xn, tn) − V(t))/(tn − t) ≤ limsupn → ∞(f(xn, tn) − f(xn, t))/(tn − t) = limsupn → ∞ft(xn, tn’)
for some
tn’ ∈ (t, tn) (the last inequality follows from the Median Value Theorem). The continuity of the partial derivative implies that
limsupn → ∞ft(xn, tn’) = ft(x, t).