The Implicit Function Theorem (IFT): key points

Transcription

1 The Implicit Function Theorem (IFT): key points 1 The solution to any economic model can be characterized as the level set corresponding to zero of some function 1 Model: S = S(p;t), D = D(p), S = D; p =price; t =tax; 2 Level Set: LS(p;t) = S(p;t) D(p) = 0. 2 When you do comparative statics analysis of a problem, you are studying the slope of the level set that characterizes the problem. 1 Intuitive comp. stat. question: change t, what happens to p? 2 Intuitive idea: t; how much p is needed to keep on LS( ; )? 3 Math answer is the slope of LS 3 The implicit function theorem tells you 1 when this slope is well defined 2 if it is well-defined, what are the derivatives of the implicit function 4 It s an extremely powerful tool 1 explicit function p(t) could be nasty; no closed form E.g., : LS(p;t) = tp 15 + t 13 + p 95 p = 0; what s p(t)? 2 don t need to know p(t) in order to know dp(t) 3 can compute dp(t) dt from partials of LS( ; ). dt.

2 Level Sets locally well-defined functions diff able functions x g(α) α 0 Case 1: x is a well-defined, differentiable function of α α x g(α) (ᾱ, x) α 0 Case 2: x is a well-defined, but not differentiable function of α α x g(α) (ᾱ, x) α 0 α

3 Implicit function theorem (single variable version) I Theorem: Given f : R 2 R 1, f C 1 and (ᾱ, x) R 2 f (ᾱ, x), if 0, x nbds U α of ᾱ, U x of x & a unique g : U α U x, g C 1 s.t. α U α, f(α,g(α)) = f(ᾱ, x) i.e., (α,g(α)) is on the level set of f through (ᾱ, x) g (α) = f(α,g(α)) / f(α,g(α)) α x Trivial Proof of the second line: f(α,g(α)) α f(α,x) := f(α,g(α)) = 0 + f(α,g(α)) dg(α) x dα dg(α) dα = 0 = f(α,g(α)) α f(α,g(α)) x

4 Implicit function theorem (single variable version) II Example: 1 f(p;t) = tp 15 + t 13 + p 95 p 2 f t = p t 12 ; f = p 15tp p 94 1/2 1 p 3 if f dp 0, = ( p t 12)/( ) 15tp p 94 1/2 1 p dt p Note that the following is not true: if f(ᾱ, x) 0, nbd U α of ᾱ, & a x unique g : U α R, g C 1 s.t. α U α, f(α,g(α)) = f(ᾱ, x). Also true (if you are a mathematician ( but not an economist): )/ (p 1 if f 0, dt = 15tp p 94 1/ t dp p + 13t 12)

5 Implicit function theorem (single variable version) III Example: (slightly less dramatic) 1 LS(x;α) = x 2 + α LS α = 2α; 3 if LS x LS x = 2x dx 0, dα = dls dα / dls dx = α/x. Suppose ᾱ = 0.9, x = Use IFT to estimate effect of a 100% increase in α to 1.8 dx dα = 0.9/0.44 2; x(1.8) ( 2 0.9) 1.36 i.e., way off, true answer: no real x solves LS hazardous to base policy decisions on comp stat analysis

6 Implicit function theorem (single variable version) IV x α

7 Implicit function theorem (intermediate version) Theorem: Given f : R n+1 R 1, f C 1 and (ᾱα, x) R n R 1 f (ᾱ, x), if 0, x j, nbds U j of ᾱ j, U x of x & a unique g : U j U x, g C 1 s.t. α j U j, f(α,g(α)) = f(ᾱα, x) i.e., g puts us on the level set of f containing (ᾱα, x) dg(α) = f(α,g(α)) / f(α,g(α)) or in vector form dα j α j x g(α) = α f(α,g(α))/f x (α,g(α)). This is a straightforward extension of the first theorem. Example: 1 Model: S = S(p;t), D = D(p;y), S = D; t =tax, y =income; 2 Level Set: LS(p;t,y) = S(p;t) D(p;y) = 0. 3 Can use IFT to compute p = (p t,p y ) 4 Once you have gradient, can get any directional derivative what part of the theorem guarantees this? 5 Use IFT to approx impact of any combination of param changes

8 Equilibrium price as a function of tax and income S(p, t ) P S(p, t) p(t, y ) p(t, y) D(p, y ) D(p, y) Q

9 The zero level set of S(p,t) D(p,y) Zero Level set of Supply demand model equilibrium price income tax

10 Implicit function theorem (multivariate version) Theorem: Given f : R n+m R m, f C 1 & (ᾱα, x) R n R m, if det(jf x (ᾱα, x)) 0, U α of ᾱα, U x of x &!g : U α U x, g C 1 s.t. α U α, f(α,g(α)) = f(ᾱα, x) i.e., g puts us on level set of f containing (ᾱα, x) g 1 (α) g α 1 1 (α) f 1 (α,g(α)) α n α 1 α n g 2 (α) g α 1 2 (α) f 2 (α,g(α)) α n = Jf x (α,g(α)) 1 α 1 α n g m (α) f α 1 m (α,g(α)) α 1 g m (α) α n f 1 (α,g(α)) f 2 (α,g(α)) f m (α,g(α)) α n where Jf x (α,g(α)) is the Jacobian of f treating α as parameters

11 Inverse function theorem Theorem: (Inverse function theorem): Given η : R m R m, η C 1 and x R n, if η( x) = ᾱα and det(jη( x)) 0, then U x of x and U α of ᾱα s.t. η is a one-to-one and onto map from U x to U α. Moreover, there exists a C 1 inverse function η 1 : U α U x, defined by, for η(x) U α, η 1 (η(x)) = x, with Jacobian defined by Jη 1 (η(x)) = (Jη(x)) 1.

12 What some other people do Many economists don t apply IFT directly; they totally differentiate The IFT gives a formula for the Jacobian of the implicit funct g Total differentiators end up with an expression for differential of g Relationship between Jacobian and differential is as it always is Recall: for every n n matrix, unique L : R n R n. dx = Jg(α; x)dα Summary: IFT route ends up at: Jg(α;x) Total differentiation route ends up at: dx = Jg(α; x)dα

13 Total Differentiation Example: Perloff I max L,K plα K β wl rk (1) Requirement: stay on the zero level set of f, where f(w,r;l,k ) = [ ] [ π L pl π K = K β ] w pl α K β 1 r Totally differentiate: 0 = πl L 0 = πk L dl + πl K dl + πk K = 0 (2) πl dk + dw (3) w dk + πk r dr (4) Move exog variables to right hand side; put in matrix form: [ ] π L π [dl ] [ ] L π L [dw ] [ ][ ] 0 L K = w 1 0 dw = π K π K dk π 0 K dr 0 1 dr L K r (5)

14 Total Differentiation Example: Perloff II Invert [ ] dl = dk [ π L L π K L π L K π K K ] 1 [ π L General form: dg 1 dg 2. = Jf x(α,g(α)) 1 dg m 0 w π 0 K r f 1 (α,g(α)) ] [dw ] dr α 1 f 2 (α,g(α)) α 1. f m (α,g(α)) α 1 f 1 (α,g(α)) α n f 2 (α,g(α)) α n f m (α,g(α)) α n (6) dα 1 dα 2. (7) dα n or, noting that the two matrices on the r.h.s. of the last equality are just Jg(α), we can rewrite the above more economically as (8) dg = Jg(α)dα IFT goes directly from step (2) to (7); skips steps in between. To actually solve system, typically use Cramer s rule

15 Cramer s Rule: technique Cramer s rule is a method to solve Ax = b. Could invert A and computer x = A 1 b. But inversion involves effort Cramer s rule involves much less. Let A i denote the matrix A with the i th column replaced by b. Then for i = 1,...n, x i = det(a i ) / det(a). much easier to compute a few determinants than an inverse. how on earth can this work????

16 Some intuition for Cramer s rule Recall For i = 1,...n, x i = det(a i ) / det(a). column interpretation of x when Ax = b. x i s are the weights on columns of A so that b can be written written as linear combination of these columns volume interpretation of det(a) det(a) is the volume of parallelepiped defined by columns of A. In the figure below, A = [v 1,v 2 ], and b is close to v 1, so x 1 close to 1, x 2 close to 0 now A 1 = [b,v 2 ], A 2 = [v 1,b], what can you say about A 1 vs A? what can you say ratio of their determinants? what can you say about A 2 vs A? Also have explain why the signs work out properly

17 Cramer s Rule: example a 1 a 1 b a 2 a 2 The columns of A and vector b det(a) = area parallelogram defined by cols of A b a 1 b a 2

18 Multivariate IFT: graphical representation dx 2 dx 1 Level set of f 1 corresponding to 0 α The two tangent planes combined ᾱ x2 x1 Level set of f 2 corresponding to 0 ᾱ x2 x1 The tangent plane to the 0-level set of f 1 The two tangent planes: enlarged dα dα dx 2 dx 1 The tangent plane to the 0-level set of f 2 dα f : R 1+2 R 2, i.e., two endog variables x, one exog α 1 x is in the horizontal plane; α on vertical plane 2 (ᾱ, x) LS 1 LS 2 is a soln, i.e., belongs to both zero level sets 3 as α changes, x changes to keep vector in LS 1 LS 2 4 (α, x) slides along the groove in the bottom right panel 1