Proving that a Cobb-Douglas function is concave if the sum of

[PDF] multiplicateur de depense publique(definition)

[PDF] revenu d'équilibre en économie fermée

[PDF] élasticité de substitution fonction ces

[PDF] calculer la valeur du revenu national d'équilibre

[PDF] calcul élasticité de substitution

[PDF] exercice corrigé multiplicateur keynésien

[PDF] élasticité de substitution cobb douglas

[PDF] revenu d'équilibre macroéconomique

[PDF] élasticité de substitution microéconomie

[PDF] calcul du revenu de plein emploi

[PDF] revenu d'équilibre définition

[PDF] élasticité de substitution exemple

[PDF] manuel iphone 7

[PDF] indexation et recherche d'image par contenu

[PDF] cours indexation image

Proving that a Cobb-Douglas function is concave

if the sum of exponents is no bigger than 1

Ted Bergstrom, Econ 210A, UCSB

If you tried this problem in your homework, you learned from painful experi- ence that the Hessian conditions for concavity of the Cobb-Douglas function

F(x1;:::xn) =nY

i=1xii fromI propose the following road to a proof.

We rst note the following:

Lemma 1.The function dened by

F(x1;:::xn) =nY

i=1xii is homogeneous of degree

Pni=1i.

You should be able to supply the proof of this lemma. We next note thatFis quasi-concave. To show this, we make use of the fact that any monotone increasing transformation of a concave function is quasi-concave.

Lemma 2.A functionFis quasi-concave ifh(x) =g

(F(x))is a concave function for some strictly increasing functiongfromFor anytsuch that 0< t <1, let =tf(x)tf(x) + (1t)f(x0): Then

1=(1t)f(x0)tf(x) + (1t)f(x0):

Substituting these expressions forand 1into the inequality 1, we have the inequality ftx+ (1t)x0tf(x) + (1t)f(x0) 1 (2)

Sincefis homogeneous of degree 1, it follows

f tx+ (1t)x0tf(x) + (1t)f(x0) =1tf(x) + (1t)f(x0)f(tx+ (1t)x0) and therefore the inequality 2 implies that f(tx+ (1t)x0)tf(x) + (1t)f(x0): which means thatfis a concave function.2 What if 0Sincegis a concave function, it must be that g tf(x) + (1t)f(x0)tg(f(x)) + (1t)gf(x0):(5)

Combining the inequalities 3 and 5, we have

g f(tx+ (1t)x0)tg(f(x)) + (1t)gf(x0)(6) Recalling the denition ofh, we see that the inequality 6 can be written as h tx+ (1t)x0th(x) + (1t)h(x0):(7)

But the inequality 7 is the condition forhto be a concave function.How does Theorem 2 help? Letk=Pi. We can verify that the Cobb-

Douglas functionFmust be homogeneous of degreek. DeneH(x) = F(x)1=k. We note thatHis homogeneous of degree 1 and quasi-concave. ThereforeHis homogeneous of degree 1. From Theorem 1 we know thatH is concave. NowF(x) =g(H(x)) whereg(y) =yk. The second-derivative test shows us thatgis a concave function. So it follows from Theorem 2 thatFis a concave function.

There we are.

This proof was kind of a long road, but I think a very instructive one. Everything that you learned along the way is likely to come in handy some day.

A nal remark

I leave it to you to show that if

P ii>1, thenFis neither concave nor convex. 3quotesdbs_dbs2.pdfusesText_2

[PDF] Proving that a Cobb-Douglas function is concave if the sum of

An Assessment of CES and Cobb-Douglas Production Functions1