Concavity 1

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ES5081 Paul Madden

CONCAVITY & STATIC OPTIMIZATION LECTURE I

Functions A function f : X Y is a rule or mapping that

associates each element of X with one and only one element of Y.

X is the domain and f X = {y Y : y = f x , x X} is the range.

Example: power functions f = R R where f x = x ,

0 . (Notation: R is the set of real numbers, R the non-

negative subset, R the strictly positive subset; R n , R n , R n

are the corresponding sets of n-dimensional vectors.)

See diagram (1.1.1)

All power functions are continuous (the graphs have no “kinks“)

with first derivatives f ' x = x− 1

(= the slope of f x at x),

second derivative f ' ' x = −1 x − 2 (= the slope of f ' x at x).

Some terminology: f is C2 means second derivatives of f exist and

are continuous (power functions are certainly C2 ).

Concavity & Static Optimization1. 1



ES5081 Paul Madden

CONCAVE FUNCTIONS The line joining any 2 points on the

graph lies entirely on or below the graph. Formally, f : X Y is

concave iff:

f x 1− z ≥ f x 1− f z , ∀ x ,z X, ∀ [0,1 ]

If f is C2 then f is concave iff its Hessian matrix H x is negative

semi-definite ∀ x X where

H x =f 11 ' ' x ... f 1n ' ' x

... ... ...f n1 ' ' x ... f nn ' ' x

and f ij ' ' x = ∂2 f ∂ x i∂ x j

at x

n = 1 gives f ' ' x ≤ 0 ∀ x X; n = 2 gives f 11 ' ' x ≤ 0 ,

f 22 ' ' x ≤ 0 and

det H x = f 11 ' ' x ⋅f 22 ' ' x − f 12 ' ' x ⋅f 21 ' ' x ≥ 0 , ∀ x X.

Strict concavity requires that the line joining any 2 points on the

graph lies (ends apart) entirely below the graph;

f x 1− z f x 1− f z , ∀ x , z X, x ≠ z , 0,1 .

The C2 test now requires strict inequalities

(negative definite H x ).




ES5081 Paul Madden

For convex functions the line joining any 2 points on the graph lies

entirely on or above the graph. The C2 test requires a positive

semi-definite H x : for n = 1 , f ' ' x ≥ 0 ∀ x X; for n = 2 ,

f 11 ' ' x ≥ 0 , f 22 ' ' x ≥ 0 and det H x ≥ 0 ∀ x X.

Strict convexity is analogous.

Examples (a) f : R R where f x = x , 0 .

f ' x = x − 1 , f '' x = −1 x − 2 . So f is concave if ≤ 1 ,

strictly concave if 1 , convex if ≥ 1 , strictly convex if 1

(b) f : R 2 R where f x1, x2 = x1 x2 .

f 1 ' = x2 , f 2 ' = x1 : f 11 ' ' = f 22 ' ' = 0 , f 12 ' ' = f 21 ' ' = 1 ⇒ H x = 0 11 0

and det H x = − 1 0 . So f is neither concave nor convex.

UNCONSTRAINED OPTIMIZATION A stationary point of f : X Y

is x X where f i ' x = 0 , i = 1,...,n (all partial derivatives are

zero).




ES5081 Paul Madden

A global maximum (minimum) of f is x X where f x ≥ ≤ f z

∀ z X. An important result is: if f is concave, then x is a global

max of f iff it is a stationary point. [if f is convex, replace max with

min]

Examples (a) f : R R where f x = ln x − x .

f ' x = 1x − 1 = 0 when x = 1 ; f ' ' x = − 1x2 0 ∀ x R ,

So f is concave. So x = 1 is global max of f (& f 1 = 0 )

(b) f : R 2 R where f x1, x2 = x1 x2 − x12 − x2

2

f 1 ' = 1 − 2x 1 , f 2 ' = 1 − 2x 2 ⇒ x = 1

2

, 1

2

is stationary point

f 11 ' ' = f 22 ' ' = − 2 , f 12 ' ' = f 21 ' ' = 0 ⇒ H x = − 2 00 − 2

, n.s.d. so

f is concave and x = 12

, 12

is global max f 12

, 12

= 12

.

CONSTRAINED OPTIMIZATION 1

maxx

f x s. t. g x ≥ 0 (1)

Assume: f, g are concave and ∃ x where g x 0 .




ES5081 Paul Madden

Define the Lagrangian: L x , = f x g x

Then x solves (1) iff ∃ such that;

(A)∂ L∂ xi

= ∂ f ∂ xi

∂ g∂ xi

= 0 ∀ i , (B) ≥ 0 , (C) g x = 0 ,

(D) g x ≥ 0

These conditions are Kuhn-Tucker conditions (& (C) is the

complementary slackness condition)

Note: if f has no stationary points then ≠ 0 from (A), so

(B) – (D) reduce to 0 and g x = 0 .

Examples max x 1 x2 − x12 − x2

2 s. t. 2x 1 2x 2 ≤ 1

2

f is concave (previous exercise), g = 12

− 2x 1 − 2x 2 is concave

(check) and g 0 0 (e.g.).

L = x1 x2 − x12 − x2

2 1

2

− 2x 1 − 2x 2

(A)∂ L∂ x1

= 1 − 2x 1 − 2 = 0 ∂ L∂ x2

= 1 − 2x 2 − 2 = 0 ,

(B) ≥ 0 , (C) = 0 or 2x 1 2x 2 = 12

, (D) 2x 1 2x 2 ≤ 12




ES5081 Paul Madden

Try = 0 : x 1 = x2 = 12

from (A) but 2x 1 2x 2 = 2 12

no

solution. Try 2x 1 2x 2 = 12

: x1 = x2 = 12

1− 2 from (A) so

2 1− 2 = 12

and = 38

0 : x1 = x2 = 18

is solution

therefore.

HOMOGENEOUS FUNCTIONSf : R n R

is homogeneous of

degree r (hod r) iff f tx = t r f x ∀t 0 . A useful result is

Euler's theorem; if f is hod r then ∑i= 1

n

x i⋅ f i ' x = r ⋅ f x .

Examples (a) f x = x ⇒ f tx = tx = t x = t f x and f

is hod

(b) f x1 , x 2 = x11⋅ x2

2⇒ f tx = tx 1

1 tx 22 = t 1 2

⋅ f x and

f is hod 1 2

The Cobb-Douglas functions

f : R n R where f x = ∏i= 1

n

x ii , i 0

n = 1 gives a power function: n = 2 gives example (b) above.




ES5081 Paul Madden

Some properties: (a) f is continuous and C2

(b) f 0 = 0 and f x = 0 when any xi = 0 ; f x 0 if xi 0

(c) f is hod ∑ i

(d) f is strictly concave if ∑ i 1 , concave if ∑ i = 1 , and

neither concave nor convex (if n ≥ 2 ) if ∑ i 1 .

The Cobb-Douglas as a utility function

(u max)max

xu x = x1

1 x22 s. t. p 1 x1 p2 x2 ≤ m

,x1 , x 2 ≥ 0

where m, p 1 , p 2 0 are parameters and ∑ i ≤ 1

All solutions will be “interior“, i.e. xi 0 ∀ i , since any xi = 0

means u x = 0 , and u x 0 can be attained since m 0 .

Hence we ignore non-negativity constraints and the Lagrangian for

the reduced 1 – constraint problem is

L x , = x11 x2

2 m − p1 x1 − p2 x2

Since u i' = i x ii− 1

x j j 0 , u has no stationary points and the K-T

conditions reduce to (requirements satisfied – check)




ES5081 Paul Madden

(I)∂ L∂ x1

= 1 x11− 1

x22 − p 1 = 0 ∂L

∂ x2

= 2 x11 x2

2− 1− p2 = 0

(II) 0 and (III) m = p1 x1 p2 x2

From (I)1

x1

x11 x2

2 = p1 ,2

x2

x11 x2

2 = p2 (⇒ 0 )

Dividing, 1

x1

⋅x2

2

=p1

p2

⇒ p2 x2 = 2

1

p1 x1 ⇒ m = p1 x12

1

p 1 x1

⇒ p1 x1 = 1

1 2

m , p 2 x2 = 2

1 2

m

and the consumer's (Marshallian) demand functions are:

x1 p ,m = 1

1 2

⋅mp1

x2 p,m = 2

1 2

⋅mp2

The value of utility at this solution to (u max) gives the indirect

utility function;

V p , m = u[x1 p ,m , x2 p,m ]= 1

1 2

⋅mp1

12

1 2

⋅mp2

2

Notice that, as prices and income vary, the budget share spent on

good ip i xi

mremains constant = i

1 2. Also the own price




ES5081 Paul Madden

demand elasticity for any good i i =∂ xi

∂ p i

⋅p i

xiis

i = − i

1 2

⋅ mp i

2 ⋅p i

i

1 2

⋅mp i

= − 1

These constant budget shares and unit elastic demands are special

features of Cobb-Douglas consumers.

Consumer theory also studies expenditure minimization:

(E min) minx

∑i= 1

n

p i x i s. t. u x ≥ u , x i ≥ 0 ∀ i where p,

u are parameters. The solutions in x, as functions of p, u , are

Hicksian demand functions, and the resulting expenditure is the

expenditure function.

Note: the problem min f x s. t. g x ≥ 0 has the same x

solutions as max − f x s. t. g x ≥ 0 .




ES5081 Paul Madden

REFERENCES

MADDEN pages 11 – 25, 29 – 30, 34 – 40, 41 – 50, 60 – 74,

75 – 82, 114 – 120

KLEIN pages 29 – 34, 202 – 206, 354 – 357, 235 – 240,

244 – 250

# MAS-COLELL pages 930 – 933, 55 – 56, 63

Note:

# INDICATES A MORE ADVANCED REFERENCE AND MATERIAL


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Concavity 1