Unformatted text preview: Two Properties of Expenditure functions
Proof that e(p, u) is a concave function of p.
Proof: We want to show that for any u and any two price vectors p and p ,
and for any λ between 0 and 1,
λe(p, u) + (1 − λ)e(p , u) ≤ e(λp + (1 − λ)p , u).
Let h = h(p, u) and h = h(p , u), and let hλ = h(λp + (1 − λ)p , u). We note
that u(h) = u(h ) = u(hλ ) since h(p, u) is the cheapest consumption vector
that yields utility u at price vector p. Then e(p, u) = ph(p, u) ≤ phλ (because
u(hλ ) = u) Similarly, e(p , u) = p h(p , u) ≤ p hλ . It follows from these two
inequalities that
λe(p, u) + (1 − λ)e(p , u) ≤ (λp + (1 − λ)p )hλ
= e(λp + (1 − λp , u). Notes on Proving Shepherd’s lemma.
∂e(p, u)
= xi (p, u).
∂pi
Proof: e(p, u) = j pj xh (p, u). Diﬀerentiate this to ﬁnd that
j ∂e(p, u)
= xi (p, u) +
∂pi pj
j ∂xj (p, u)
.
∂pi Note also that u(xh (p, u)) = u for all p. Diﬀerentiate this
uj (xh (p, u))
j ∂xh (p, u)
= 0.
∂pi But uj (xh (p, u)) = λpj . Now ﬁnish proof by substituting uj /λ for pj in the
above equation and noticing that all the complicated stuﬀ disappears. 1 ...
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Full Document
 Fall '09
 Bergstrom
 Derivative, Convex function, Concave function, pj xh, cheapest consumption vector

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