252 Pages·2015·1.87 MB·English

Mathematics for Economists mainly optimization Medvegyev PØter 2013 Medvegyev (CEU) MathematicsforEconomists 2013 1/252 Unconstrained local necessary conditions In the unconstrained case f is de(cid:133)ned on an open set. Use the "variations" g (t) $ f (x0 +th) and chain rule. Theorem If x is a local minimum then 0 1 0 = fx0(x0) = (∂f/∂x1(x0),...,∂f/∂xn(x0)). 2 0 hTfx0x0 (x0)h = ∑∑∂2f/∂xi∂xj (x0)hihj (cid:20) i j Observe that one needs just twice di⁄erentiability at x . 0 Medvegyev (CEU) MathematicsforEconomists 2013 2/252 Unconstrained local su¢ cient conditions Theorem If f is twice di⁄erentiable at x and 0 f (x ) = 0 x0 0 and the second derivative is 1 positive de(cid:133)nite then x0 is a local minimum of f, 2 negative de(cid:133)nite then x0 is a local minimum of f. Observe that one needs just twice di⁄erentiability at x . 0 Medvegyev (CEU) MathematicsforEconomists 2013 3/252 Unconstrained local su¢ cient conditions Lemma If xTHx is a positive de(cid:133)nite quadratic form on Rn then there is an α > 0 such that xTHx α x 2. (cid:21) k k Let α > 0 be the minimum of xTHx over the compact set x = 1. k k Obviously x T x H α > 0 x x (cid:21) k k k k which implies the lemma. Medvegyev (CEU) MathematicsforEconomists 2013 4/252 Unconstrained local su¢ cient conditions As f is twice di⁄erentiable at x 0 1 f (x +h) f (x ) = f (x ),h + hTf (x )h+o h 2 . 0 (cid:0) 0 x0 0 2 x0x0 0 k k (cid:16) (cid:17) (cid:10) (cid:11) Let H = f (x ) and let α > 0 be the constant above. For any ε < α/2 x0x0 0 one has that 2 2 o h ε h k k (cid:20) k k (cid:12) (cid:16) (cid:17)(cid:12) for h small enough. Hence(cid:12)as f (x ) =(cid:12) 0 (cid:12) x0 0 (cid:12) α f (x +h) f (x ) ε h 2 0. 0 0 (cid:0) (cid:21) 2 (cid:0) k k (cid:21) (cid:16) (cid:17) Medvegyev (CEU) MathematicsforEconomists 2013 5/252 Constrained local necessary conditions Theorem Assume that ϕ k = 0,1,2,...,p are di⁄erentiable and assume that ϕ k 0 has a local minimum at x on the set 0 X $ x ϕ (x) = 0,k = 1,...,p . f j k g Then there are multipliers l = (λ ,λ ,...,λ ) Rp+1 0 1 p 2 such that p ∑ λ ϕ (x ) = 0. k k0 0 k=0 If ϕ (x ),k = 1,...,p are linearly independent then λ = 1 is possible. k0 0 0 Medvegyev (CEU) MathematicsforEconomists 2013 6/252 Constrained local necessary conditions Theorem Assume that we are in the regular case and ϕ ,k = 0,1,...,p are twice k di⁄erentiable at x , where x is a local minimum of the constrained 0 0 optimization problem then ϕ (x ),h = 0, k = 1,2,...,p = hTL (x ,λ)h 0. k0 0 ) x00x 0 (cid:21) (cid:10) (cid:11) For local maximums one has ϕ (x ),h = 0, k = 1,2,...,p = hTL (x ,λ)h 0. k0 0 ) x00x 0 (cid:20) (cid:10) (cid:11) Recall that L (x ,λ) is the second derivative of the Lagrange function x00x 0 with respect to x at x . 0 Medvegyev (CEU) MathematicsforEconomists 2013 7/252 Constrained local su¢ cient conditions Theorem Assume L (x ,λ) = 0, that is let x be a stationary point of the x0 0 0 Lagrangian. Assume that we are in the regular case and ϕ ,k = 0,1,...,p are twice di⁄erentiable at x . If k 0 hTL (x ,λ)h > 0, h = 0 x00x 0 6 whenever ϕ (x ),h = 0, k = 1,2,...,p then x is a local minimum of h k0 0 i 0 the constrained optimization problem. If hTL (x ,λ)h < 0, h = 0 x00x 0 6 whenever ϕ (x ),h = 0, k = 1,2,...,p then x is a local maximum of h k0 0 i 0 the constrained optimization problem. Medvegyev (CEU) MathematicsforEconomists 2013 8/252 Constrained local su¢ cient conditions Let F be the vector function of the constraints and let C $ x F (x) = 0 . It is su¢ cient to show that there is a function K f j g such that K has a local minimum on C at x with U and K (x ) = ϕ (x ) 0 0 0 0 and K (x) ϕ (x),x U. In this case if x U C then (cid:20) 0 2 2 \ ϕ (x) K (x) K (x ) = ϕ (x ) 0 (cid:21) (cid:21) 0 0 0 so x is a local minimum of ϕ on C. Let 0 0 K (x) $ ϕ0(x0)(cid:0)hλ,F (x)i(cid:0)γkF (x)k2 where γ is a large enough constant. Obviously if x C then F (x) = 0 2 therefore K (x ) = ϕ (x ) as x C. 0 0 0 0 2 Medvegyev (CEU) MathematicsforEconomists 2013 9/252 Constrained local su¢ cient conditions One must show that ϕ (x) K (x) = ϕ (x ) λ,F (x) γ F (x) 2 0 (cid:21) 0 0 (cid:0)h i(cid:0) k k That is 0 ϕ (x ) (ϕ (x)+ λ,F (x) ) γ F (x) 2 = (cid:21) 0 0 (cid:0) 0 h i (cid:0) k k = ϕ (x ) L(x,λ) γ F (x) 2. 0 0 (cid:0) (cid:0) k k Using the condition on the existence of the second derivative and the stationarity condition L (x ,λ) = 0 and that as x is a feasible solution x0 0 0 L(x ,λ) = ϕ (x ) one must show that 0 0 0 0 (x x )T L (x ,λ)(x x )+o x x 2 0 x00x 0 0 0 (cid:21) (cid:0) (cid:0) (cid:0) k (cid:0) k (cid:0) (cid:16)2 (cid:17) γ F (x )(x x )+o( x x ) . 0 0 0 0 (cid:0) (cid:0) k (cid:0) k (cid:13) (cid:13) (cid:13) (cid:13) Medvegyev (CEU) MathematicsforEconomists 2013 10/252

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