UBC Economics 526 October 17, 2012
.1.2.3.4
Section 1
. max f (x) s.t. h(x) = c f : R n R, h : R n R m Draw picture of n = 2 and m = 1 At optimum, constraint tangent to level curve of function Rewrite as ( f )/( x 1 )(x ) ( f )/( x 2 )(x ) =( h)/( x 1)(x ) ( h)/( x 2 )(x ) ( f )/( x 1 )(x ) ( h)/( x 1 )(x ) = ( f )/( x 2)(x ) ( h)/( x 2 )(x ) µ f (x ) µ h (x ) =0 x 1 x 1 (1) f (x ) µ h (x ) =0 x 2 x 2 (2) h(x ) c =0 (3) Lagrangian L(x, µ) f (x) µ(h(x) c).
FOC with equality Theorem Let f : U R and h : U R m be continuously differentiable on U R n. Suppose x interior(u) is a local maximizer of f on U subject to h(x) = c. Also assume that Dh x has rank m. Then there exists µ R m such that (x, µ ) is a critical point of the Lagrangian, i.e. L(x, µ) = f (x) µ T (h(x) c). L (x, µ ) = f µ T h (x ) = 0 x i x i x i L (x, µ ) =h(x ) c = 0 µ j for i = 1,..., n and j = 1,..., m.
max f (x) s.t. g(x) b. x U Binding, g j (x ) = b j are just like equality Df x λ j Dg j,x = 0. Df x is direction f increases, so x + δdf x must violate constraint or x cannot be a maximizer g j (x + δdf T x ) >b j g j (x ) + δdg x Df T x + o(δ2 ) >b j Dg j,x Df T x >0 Combine with first order condition to get λ j > 0 Thus, λ j 0 and λ j = 0 iff g j (x ) < b j (complementary slackness condition)
. FOC with inequality Theorem Let f : U R and g : U R m be continuously differentiable on U R n. Suppose x interior(u) is a local maximizer of f on U subject to g(x) b. Suppose that the first k m, bind g j (x ) = b j for j = 1...k and that the Jacobian for these, has rank k. Then, there exists λ R m such that for we have λ j L(x, λ) = f (x) λ T (g(x) b). L (x, λ ) = f λ T g (x ) = 0 x i x i x i L (x, λ ) =λ j (g(x ) c) = 0 λ j λ j 0 g(x ) b for i = 1,..., n and j = 1,..., m.
Similar result Mixed equality and inequality
Section 2
expansion of f (x) around x. f (x + v) f (x ) =Df x v + v T D 2 f x v + r(v, x ) =v T D 2 f x v + r(v, x ) x + v must satisfy the h(x + v) = h(x ) + Dh x v + r h (v, x ) = c. so Dh x v = 0 x is a local maximizer of f subject to h(x) = c if v T D 2 f x v 0 for all v such that that Dh x v = 0
. condition for constrained maximization Theorem Let f : U R be twice continuously differentiable on U, and h : U R l and g : U R m be continuously differentiable on U R n. Suppose x interior(u) and there exists µ R l and λ R m such that for L(x, λ, µ) = f (x) λ T (g(x) b) µ T (h(x) c). the first order condition is satisfied. Let B be the matrix of the derivatives of the binding evaluated at x. If v T D 2 f x v < 0 for all v 0 such that Bv = 0, then x is a strict local constrained maximizer for f subject to h(x) = c and g(x) b.
Definition Let A be an n by n symmetric matrix and B be m by n, then A is Negative definite on N (B) if x T Ax < 0 for all x N (B) \ {0} Positive definite on N (B) if x T Ax > 0 for all x N (B) \ {0} Indefinite on N (B) if x 1 N (B) \ {0} s.t. x T 1 Ax 1 > 0 and some other x 2 N (B) \ {0} such that x T 2 Ax 2 < 0.
Checking definiteness using determinants Theorem Let A be an n by n symmetric matrix and B be m by n. Then A is negative definite on N (B) iff the last n m leading principal minors of ( 0 ) B B A alternate in sign, and the final (n + m)th principal minor has the same sign as ( 1) n.
. Checking definiteness using eigenvalues Write B = ( B 1 B 2 ) with rankb = rankb1 = m. 0 =Bx = ( ) ( ) x B 1 B 1 2 x 2 x 1 = (B 1 ) 1 B 2 x 2 ( (B1 ) so x N (B) iff x = 1 ) B 2 x 2 for some x 2 R n m x T Ax for x N (B) I n m ( (B1 ) 1 ) T ( ) ( B 2 A1 A 2 (B1 ) 1 ) B 2 x T Ax =x T 2 =x T 2 I n m A T 2 A 3 I n m ) (B 2 T (B1 T ) 1 A 1 (B 1 ) 1 B 2 + B2 T (B1 T ) 1 A 2 + A T 2 (B 1 ) 1 B 2 + A 3 x } {{ } C A negative definite on N (B) iff C is negative definite on R m, i.e. C has negative eigenvalues
Section 3
Theorem ( ) Under the of 2, let x (b, c) denote the solution of the constrained maximization problem, max f (x) s.t. g(x) b x U h(x) = c, and let µ(b, c) and λ(b, c) denote the corresponding Lagrange multipliers. The for each j = 1..m, b j f (x (b, c)) = λ j (b, c) and for each j = 1,..., l, c j f (x (b, c)) = µ j (b, c).
Section 4
Let f : U A R where U R n and A R k. Consider max f (x, α). x U Let x (α) be a local maximizer. Using the chain rule, d dα j f (x (α), α) = n i=1 f x i x i α j + f α j = f α j (x (α), α) where the second line follows from the first order condition.
Let f : U A R and h : U A R l where U R n and A R k. Consider max f (x, α) s.t. h(x, α) = 0. x U Let x (α) be a local maximizer, and let L(x (α), µ (α), α) be the Lagrangian. Using the chain rule, d dα j L(x (α), µ (α), α) = = L α j (x (α), µ (α), α) n i=1 L x i x i α j + l k=1 L µ k µ k α j + L α j