Paul Schrimpf. October 17, UBC Economics 526. Implicit and inverse function theorems. Paul Schrimpf. Inverse functions. Contraction mappings

Transcription

1 and and UBC Economics 526 October 17, 2011

2 and

3 and Section 1

4 and f : R n R m When can we solve f (x) = y for x? Use derivative and what we know about linear equations to get a local answer

5 and If f (a) = b, expand f around a. f (x) = f (a) = Df a (x a) + r 1 (a, x a) = y If r 1 is small, we almost have a system of linear equations Df a x = y f (a) + Df a a Know: Solution exists if rankdf a = rank ( Df a y f (a) + Df a a ) Solution unique and if rankdf a = n

6 and Theorem ( ) Let f : R n R n be continuously differentiable on an open set E. Let a E, f (a) = b, and Df a be invertible. Then 1 there exist open sets U and V such that a U, b V, f is one-to-one on U and f (U) = V, and 2 the of f exists and is continuously differentiable on V with derivative ( Df f 1 (x)) 1.

7 and Proof. Choose λ such that λ Df a 1 = 1/2. Since Df a is continuous, there is an open neighborhood U of a such that Df x Df a < λ for all x U. For any y R n, consider ϕ y (x) = x + Dfa 1 (y f (x)). Note that Dϕ y x =I Dfa 1 Df x =Df 1 a (Df a Df x ) Df 1 a Then, by the mean value theorem for x 1, x 2 U, λ = 1 2 (1) ϕ y (x 1 ) ϕ y (x 2 ) = Dϕ ȳ x (x 1 x 2 ) 1 2 x 1 x 2.. ϕ y is a contraction, so it has a unique fixed point. When ϕ y (x) = x, it must be that y = f (x). Thus for each y f (U), there is at most one x such that f (x) = y. That is, f is one-to-one on U. This proves the first part of the theorem and that f 1 exists.

8 and Section 2

9 and Definition Let f : R n R n. f is a contraction mapping on U R n if for all x, y U, f (x) f (y) c x y for some 0 c < 1. If f is a contraction mapping, then an x such that f (x) = x is called a fixed point of the contraction mapping.

10 and Lemma Let f : R n R n be a contraction mapping on U R n. If x 1 = f (x 1 ) and x 2 = f (x 2 ) for some x 1, x 2 U, then x 1 = x 2. Proof. Since f is a contraction mapping, f (x 1 ) f (x 2 ) c x 1 x 2. f (x i ) = x i, so x 1 x 2 c x 1 x 2. Since 0 c < 1, the previous inequality can only be true if x 1 x 2 = 0. Thus, x 1 = x 2.

11 and Lemma Let f : R n R n be a contraction mapping on U R n, and suppose that f (U) U. Then f has a unique fixed point. Proof. Pick x 0 U. As in the discussion before the lemma, construct the sequence defined by x n = f (x n 1 ). Each x n U because x n = f (x n 1 ) f (U) and f (U) U by assumption. Since f is a contraction on U, x n+1 x n c n x 1 x 0, so lim n x n+1 x n = 0, and {x n } is a Cauchy sequence. Let x = lim n x n. Then x f (x) x x n + f (x) f (x n ) x x n + c x x n x n x, so for any ɛ > 0 N, such that if n N, then x x n < ɛ 1+c. Then, x f (x) < ɛ

12 and Section 3

13 and Cannot always write conditions of a model as f (x) = y Often only f (x, y) = c. Using same sort of idea, can get x as a of y.

14 and f : R n+m R n, x R n, y R m Have a model that requires f (x, y) = c Know that f (x 0, y 0 ) = c Expand f around x 0 and y 0 f (x, y) =f (x 0, y 0 ) + D x f (x0,y 0 )(x x 0 ) + D y f (x0,y }{{} 0 )(y y 0 ) + r }{{} n n n m If r small enough, f (x 0, y 0 ) + D x f (x0,y 0 )(x x 0 ) + D y f (x0,y 0 )(y y 0 ) c a system of linear equations D x f (x0,y 0 )(x x 0 ) ( c f (x 0,

15 and Theorem ( ) Let f : R n+m R n be continuously differentiable on some open set E and suppose f (x 0, y 0 ) = c for some (x 0, y 0 ) E, where x 0 R n and y 0 R m. If D x f (x0,y 0 ) is invertible, then there exists open sets U R n and W R n k with (x 0, y 0 ) U and y 0 W such that 1 For each y W there is a unique x such that (x, y) U and f (x, y) = c. 2 Define this x as g(y). Then g is continuously differentiable on W, g(y 0 ) = x 0, f (g(y), y) = c for all y W, and Dg y0 = ( D x f (x0,y 0 )) 1 Dy f (x0,y 0 )

16 and Section 4

17 and Application: Roy s identity V (m, p) an indirect utility V (m, p) = max U(c) s.t. pc m. (2) c expenditure, E(u, p) E(u, p) = min c pc s.t. U(c) u. (3) Observe that V (E(u, p), p) = u (if U continuous and p 0)

18 and Differentiate Application: Roy s identity V E (E(u, p), p) (u, p) + V (E(u, p), p) =0 m p i p i Shephard s lemma is E p i (u, p) = V m (E(u, p), p V p i (E(u, p), p c i (u, p) = E p i (u, p), Roy s identity is V ci m (m, p) (m, p) = V p i (m, p).

19 and Finite horizon macro model. Production y t = A t kt α Budget c t + k t+1 = (1 δ)k t + A t kt α. Social planner s problem max {c t,k t} T t=0 T t=0 β t c1 γ t 1 γ s.t. c t + k t+1 = (1 δ)k t + A t k α t.

20 and Lagrangian max {c t,k t,λ t} T t=0 T t=0 First order conditions β t c1 γ t 1 γ + λ t(c t + k t+1 (1 δ)k t A t kt α ) [c t ] : β t ct γ =λ t ( [k t ] : λ t 1 =λ t (1 δ) + At αkt α 1 ) [λ t ] : c t + k t+1 =(1 δ)k t + A t kt α

21 and Suppose A t changes unexpectedly at time T 1 Want to find the change in c T 1, c T, and k T Relevant first order conditions 0 =F (c T, c T 1, k T, A T, A T 1, c T 2, k T 1 ) c T 1 + k T (1 δ)k T 1 A T 1 k α T 1 = c T (1 δ)k T A T kt α c γ T 1 c γ T β ( (1 δ) + A T αk α 1 ) T

22 and The implicit theorem says that c T 1 A T 1 c T A T 1 k T A T 1 = = F 1 c T 1 F 1 c T F 1 k T F 2 c T 1 F 2 c T F 2 k T F 3 c T 1 F 3 c T F 3 k T 1 F 1 A T 1 F 2 A T 1 F 3 A T β ( (1 δ) + A T αk α 1 ) γc γ 1 T 1 γc γ 1 T T

23 and Gaussian elimination: (1 δ) γc γ 1 T 1 γc γ 1 T β ( (1 δ) + A T αk α 1 ) T c γ T βa T α( (1 δ) A T 0 γc γ 1 T β ( (1 δ) + A T αk α 1 ) T c γ T βa T α(α 1)k k α T (1 δ) A T αk α 1 T E γc γ 1 t 1 k α T 1 where ( E = γc γ 1 T β ( (1 δ) + A T αk α 1 T c γ T βa T α(α 1)k α 2 T + γc γ 1 T 1. ) ) ( (1 δ) + AT αk α 1 ) T

24 and Assume α 1, so that E > 0. Then, c T A T 1 = c T 1 A T 1 =k α T 1 k T = γc γ 1 T 1 A T 1 E kα T 1 > 0 k T ( (1 δ) + AT α T k α 1 ) A T, T 1 k T A T 1 = kα T 1 E γc γ 1 T 1 kα T 1 ( E kt α 1 = γc γ 1 T 0 c T 1 A T 1 < k α T 1 β ( (1 δ) + A T αk α 1 ) ) ( T (1 δ) + A E