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 (LS) corresponding to zero of some function 1 Model: S = S(p;t), D = D(p), S = D; p =price; t =tax; 2 f(p;t) = S(p;t) D(p) =. Level Set (LS): {(p,t) : f(p;t) = }. 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., : f(p;t) = tp 15 + t 13 + p 95 p = ; 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 f( ; ). () October 13, / 34 dt.

2 Level Sets locally well-defined diff able functions back to slide 1 x g(α) α Case 1: x is a well-defined, differentiable function of α α x g(α) (ᾱ, x) α Case 2: x is a well-defined, but not differentiable function of α α x g(α) (ᾱ, x) α α () October 13, / 34

3 Implicit function theorem (single variable version) Theorem: Given f : R 2 R 1, f C 1 and (ᾱ, x) R 2 f (ᾱ, x), if, 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) and g (α) = f(α,g(α)) / f(α,g(α)) α x Trivial Proof of the second line: back to slide 8 back to slide 11 f(α,x) := f(α,g(α)) = d(f,g(α)) dα dg(α) dα = f(α,g(α)) α = f (α,g(α)) α f (α,g(α)) x + f(α,g(α)) dg(α) x dα = Note that the following is not true: if f(ᾱ, x), nbd U α of ᾱ, & and a unique x function g : U α R, g C 1 s.t. α U α, f(α,g(α)) = f(ᾱ, x). go to slide 4 () October 13, / 34

4 U α and U x back to slide 3 x g(α) U x (ᾱ, x) U α α α Note that There are two C 1 functions g 1 and g 2 mapping U α to R s.t. for i = 1,2, for α U α f(α,g i (α)) = f(ᾱ, x). but only one of these maps U α into some nbd of x. () October 13, / 34

5 Why the implicit in implicit function theorem Example: Illustrates why it s called the implicit function theorem closed-form explicit function relating α and x doesn t exist t is time, p is an equilibrium price that depends on t; assume p > f(p;t) = tp 15 + t 13 + p 95 p ; no closed form expression for p(t) no way to isolate p on the left-hand-side, to obtain p(t), but who cares? all qualitative properties of p( ) (e.g., derivative signs) are inherited from f once we have p ( ), we can compute all higher order derivatives and approximate p( ) arbitrarily closely with a high order Taylor epxansion f t = p t 12 ; f = p 15tp p 94 1/2 1 p if f dp, = ( p t 12)/( 15tp p 94 1/2 1 p dt p ) Also true (if you are a mathematician ( but not an economist): )/ (p if f, dt = 15tp p 94 1/ t dp p + 13t 12) only makes sense in a world where time depends on prices. endogenous vs exogenous distinction is economic not mathematical () October 13, / 34

6 IFT guaranteed to work only for dx sufficiently small Example: illustrating that U α needs to be chosen carefully: can t be too big 1 f(x;α) = x 2 + α 2 1 = ; LS = {(x;α) : f(x;α) = }. 2 f(x;α) α = 2α; 3 if f(x;α) x f(x;α) x = 2x, then g (α) = dx dα = df(x;α) dα / df(x;α) dx = α/x. Suppose ᾱ =.9, x = go to slide 7 Use IFT to estimate effect of a 1% increase in α to 1.8 g (α) = dx dα =.9/.44 2; First order Taylor: g(α + dα) g(α) + g (α)dα g(1.8).44 + ( 2.9) 1.36 Clearly, no x exists such that (x;α +.9) LS conclude: hazardous to base policy decisions on comp stat analysis real world dα s rarely small enough for the theorem to be applicable. () October 13, / 34

7 U α can t be too large back to slide x α What s the max diameter of U α s.t. theorem properties hold (recall ᾱ =.9)? exactly what property would fail if U α were larger? () October 13, / 34

8 Implicit function theorem (intermediate version) Theorem: Given f : R n+1 R 1, f C 1 and (ᾱα, x) R n R 1 f (ᾱα, x), if, x nbds U α of ᾱα & U x of x, & a unique 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 the level set of f containing (ᾱα, x) and dg(α) = f(α,g(α)) / f(α,g(α)), j = 1, n or in vector form dα j α j x g(α) = α f(α,g(α))/f x (α,g(α)). go to slide 3 This is a straightforward extension of the previous theorem. Example: 1 Model: S = S(p;t), D = D(p;y), S = D; t =tax, y =income; f(p;t,y) = S(p;t) D(p;y), The function that s implicitly defined is p(t, y) If f p, can use IFT to compute p = (p t,p y ) From p = (p t,p y ), can compute directional derivatives and the differential: dp = p t ( t,ȳ)dt + py ( t,ȳ)dy ( = dt f( t,ȳ, p) t / f( t,ȳ, p) + dy f( t,ȳ, p) p y / f( t,ȳ, p) p () October 13, / 34 )

9 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 () October 13, / 34

10 Zero level set of S(p,t) D(p,y) (abstractly, the graph of g(t,y)) go to slide 2 Zero Level set of Supply demand model equilibrium price income tax () October 13, / 34

11 Implicit function theorem (multivariate version) back to slide 15 Theorem: Given f : R n+m R m, f C 1 & (ᾱα, x) R n R m, if det(jf x (ᾱα, x)), 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) and g 1 (α) g α 1 1 (α) f 1 (α,g(α)) f α n α 1 1 (α,g(α)) α n g 2 (α) g α 1 2 (α) f 2 (α,g(α)) f α n..... = Jf.. x (α,g(α)) 1 α 1 2 (α,g(α)) α n g m (α) f α 1 m (α,g(α)) α 1 g m (α) α n f m (α,g(α)) α n where Jf x (α,g(α)) is Jacobian of f treating α as parameters go to slide 3 () October 13, / 34

12 Multivariate IFT: 1x2 case Theorem: Given f : R 1+2 R 2, f C 1 & (ᾱ, x) R 1 R 2, if det(jf x (ᾱ, x)), U α of ᾱ, U x of x &!g : U α U x, g C 1 s.t. α U α, f(α,g(α)) = f(ᾱ, x) i.e., g puts us on level set of f containing (ᾱ, x) ] and [ g 1 (α) α g 2 (α) α = [ f 1 (α,g(α)) x 1 f 2 (α,g(α)) x 1 ] f 1 1 (α,g(α)) x 2 f 2 (α,g(α)) x 2 } {{ } Jf x (α,g(α)) 1 [ f 1 (α,g(α)) α f 2 (α,g(α)) α ] When dα is added to α for i = 1,2, there is a continuum of dx s s.t. f i (ᾱ + dα, x + dx) = f i (ᾱ, x) but provided that Jf x (α,g(α)) is nonsingular a unique dx s.t. (ᾱ+dα, x+dx) belongs to both level sets () October 13, / 34

13 Multivariate IFT (1x2 case): graphical representation Level set of f 1 corresponding to Level set of f 2 corresponding to ᾱ ᾱ x2 x1 x2 x1 The tangent plane to the -level set of f 1 The tangent plane to the -level set of f 2 dα dα dx 2 dx 1 dx 2 dx 1 The two tangent planes combined The two tangent planes: enlarged dα 1 f : R 1+2 R 2, i.e., there are two endogenous variables x, and one exogenous 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 (tangent plane to LS) 1 (tangent plane to LS) 2 4 (α,x) slides along the groove in the bottom right panel, which is the intersection of the two tangent planes () October 13, / 34

14 , Multivariate IFT (1x2 case): when determinant condition fails Level set of f Level set of f x One (of many) reasons why Jf x (α,g(α)) could be singular; tangent planes parallel. When dα is added to α for i = 1,2, there is a continuum of dx s s.t. f i (ᾱ + dα, x + dx) f i (ᾱ, x) since tangent planes are identicial any shift dx such that f 1 (ᾱ + dα, x + dx) f 1 (ᾱ, x) necessarily satisfies f 2 (ᾱ + dα, x + dx) f 2 (ᾱ, x) conclude: a unique function g with properties specified in the theorem x () October 13, / 34

15 Example: apply IFT to first order conditions Problem max L,K π(l,k ) = plα K β wl rk Soln: ( L, K ) In this case, the level set LS we stay on is the FOC, where [ ] [ πl pl FOC(w,r;L,K ) = = α 1 K β ] w pl α K β 1 r π K = Match concepts: general expression for IFT vs this example go to slide 11 f is FOC i.e., π = (π L ( L, K ),πk ( L, K )) x is ( L, K ). α is (w,r). Jf x (ᾱα, x) is Jacobian of π( L, K ) w.r.t. (L,K ), or Hπ, the Hessian of π Jf α (ᾱα, x) is Jacobian of π( L, K ) w.r.t. (w,r). ] [ L(w,r) g(α) is. K (w,r) () October 13, / 34

16 Example: apply IFT to first order conditions (cont) Plug in all the pieces... Jf x (ᾱα, x) is Jacobian of π( L, K ) w.r.t. ( L, K ), Jf x (ᾱα, x) = Hπ( L, K ) = L 2 L K [ 2 π( L, K ) 2 π( L, K ) 2 π( L, K ) L K Jf α (ᾱα, x) is Jacobian of π( L, K ) w.r.t. (w,r). g(α) is Jf α (ᾱα, x) = Jπ w,r = ] [ L(w,r) K (w,r) [ 2 π( L, K ) L w 2 π( L, K ) K w Jg(α) = Jf x (ᾱα, x) 1 Jf α (ᾱα, x) ] [ [ L(w,r) 2 π( L, K ) J = K (w,r) 2 π( L, K ) 2 π( L, K ) K 2 2 π( L, K ) L r 2 π( L, K ) K r, which is implicitly but not explicitly defined. = 2 π( L, K ) L 2 L K 2 π( L, K ) L K K [ 2 2 π( L, K ) 2 π( L, K ) L 2 L K 2 π( L, K ) 2 π( L, K ) L K K 2 ] ] ] 1 [ 2 π( L, K ) L w 2 π( L, K ) K w ] 1 [ π( L, K ) L r 2 π( L, K ) K r () October 13, / 34 ] ]

17 The Cobb Douglas case, in detail max π(l,k ) = pf(l,k ) wl rk, f(l,k ) = Lα K β,α + β 1 L,K [ ] f(l,k ) πl pα L w FOC(w,r;L,K ) = = = π K pβ f(l,k ) K r α(α 1) αβ L 2 LK Hπ( L, K ) = pf(l,k ) det(hπ x) = αβ LK β(β 1) K 2 pf(l,k )αβ (1 α β) (LK ) 2 Note that we can apply the IFT iff α + β < 1; (what happens if α + β = 1?) [ ] 1 J α( xπ) = 1 Apply IFT using Cramer s Rule L w L r = det 1 = det α(α 1) L 2 αβ LK β(β 1) K 2 αβ LK / / 1 det(hπ x) = / β(β 1) K 2 det(hπ x) < det(hπ x) = αβ / det(hπ x) < LK In general x i p i = π jj / det(hπx) < x i p j = π ij / det(hπx); sign is negative of sign of cross-partial () October 13, / 34

18 What some other people do Many economists don t apply IFT directly; two differences they totally differentiate each line of f separately they end up at a slightly different place from where the IFT ends up The IFT gives a formula for the Jacobian of the implicit function 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 () October 13, / 34

19 IFT via total differentiation max L,K = plα K β wl rk Requirement: satisfy FOC (stay on zero level set of f ) where [ ] [ πl pl f(w,r;l,k ) = = α 1 K β w pl α K β 1 r π K Totally differentiate w.r.t. exogenous variables: = π L L dl + π L K dk + π L w dw = π K L dl + π K K dk + π K r dr Move exog variables to right hand side; put in matrix form: [ ] πl π [dl L ] [ ] πl [dw ] [ L K π K π K = w 1 dk π K = dr 1 L K r ] = ][ ] dw dr Invert [ ] [ πl dl = L dk π K L ] 1 π L [ ][ ] K 1 dw π K 1 dr K () October 13, / 34

20 Apply IFT to an equilibrium system Question: (from ARE prelim, 28): A monopolist sells in two countries: 1 and 2. It produces a good at a constant marginal cost of c and cannot produce more than a total of Q units. Country 1 imposes a per unit tax of τ units on the good sold in that country. Assume that the constraint binds. 1 For a given value of τ, show how the equilibrium in the two countries are determined. 2 Show how these equilibrium prices change as τ increases. () October 13, / 34

21 Answer to ARE prelim, 28 question: KKT method Answer: Assume that inverse demand curves are concave in price. The monopolist s optimization problem is max (p 1 τ)d 1 (p 1 ) + p 2 D 2 (p 2 ) c(d 1 (p 1 ) + D 2 (p 2 )) p 1,p 2 s.t. D 1 (p 1 ) + D 2 (p 2 ) Q Since the constraint is assumed to be binding, the KKT conditions are specifically f(p 1,p 2 ;τ) λ g(p 1,p 2 )) = D 1 (p 1 ) + D 2 (p 2 ) = Q D 1 (p 1 ) + (p 1 τ c λ)d 1(p 1 ) = D 2 (p 2 ) + (p 2 c λ)d 2(p 2 ) = (1) D 1 (p 1 ) + D 2 (p 2 ) = Q Call (1) the zero level of function KKT (p 1,p 2,λ;τ); apply IFT to KKT () October 13, / 34

22 Answer to ARE prelim, 28 question: Lagrangian method Answer: Assume that inverse demand curves are concave in price. The monopolist s optimization problem is The Lagrangian is max (p 1 τ)d 1 (p 1 ) + p 2 D 2 (p 2 ) c(d 1 (p 1 ) + D 2 (p 2 )) p 1,p 2 s.t. D 1 (p 1 ) + D 2 (p 2 ) Q L(p 1,p 2,λ;τ) = (p 1 τ)d 1 (p 1 ) + p 2 D 2 (p 2 ) c(d 1 (p 1 ) + D 2 (p 2 )) + λ(q D 1 (p 1 ) D 2 (p 2 )) Assuming capacity constraint binds, the first order conditions are L p1 = D 1 (p 1 ) + (p 1 τ c λ)d 1(p 1 ) = L p2 = D 2 (p 2 ) + (p 2 c λ)d 2(p 2 ) = L λ = Q D 1 (p 1 ) D 2 (p 2 ) = Solution is a triple (p 1,p 2,λ ) which solves this system, given τ. () October 13, / 34

23 The monopolist s optimization problem, more The Lagrangian again: L p1 = D 1 (p 1 ) + (p 1 τ c λ)d 1(p 1 ) = = D 1 (p 1 ) + (p 1 τ)d 1(p 1 ) }{{} Marginal revenue good 1 L p2 = D 2 (p 2 ) + (p 2 c λ)d 2(p 2 ) = = D 2 (p 2 ) + p 2 D 2(p 2 ) }{{} Marginal revenue good 2 L λ = Q D 1 (p 1 ) D 2 (p 2 ) = (c + λ)d 1(p 1 ) }{{} Marginal cost (incl. shadow cost) (c + λ)d 2(p 2 ) }{{} Marginal cost (incl. shadow cost) Get to exactly the same place as the KKT, but with a lot more ink. = = () October 13, / 34

24 The monopolist s optimization problem, graphical; τ > τ Marginal revenue MR 2( ) c + λ (τ) MR 1(, τ) MR 1(, τ ) D 2(p 2) Q c + λ (τ ) c D 1(p 1) As tax increases Marginal revenue curve for good 1 shifts down Output shifts from good 1 to good 2 Price of good 2 has to fall: to sell more product price of good 1 has to rise: to sell less product Shadow price falls () October 13, / 34

25 Apply IFT to the FOC of the Lagrangian L p1 = D 1 (p 1 ) + (p 1 τ c λ)d 1(p 1 ) = L p2 = D 2 (p 2 ) + (p 2 c λ)d 2(p 2 ) = (2) L λ = Q D 1 (p 1 ) D 2 (p 2 ) = The Hessian of the Lagrangian w.r.t. endog vars is: HL p,λ = HL τ = 2D 1 (p 1 ) + (p 1 τ c λ )D 1 (p 1 ) D 1 (p 1 ) 2D 2 (p 2 ) + (p 2 c λ )D 2 (p 2 ) D 2 (p 2 ) D 1 (p 1 ). D 1 (p 1 ) D 2 (p 2 ) Hence from the implicit function theorem, we have dp 1 dτ dp 2 = HL 1 D 1 (p 1 ) (3) dτ (p,λ) dλ dτ Alternatively, you (my colleagues) could totally (2) and end up with dp 1 = First row of r.h.s. of (3) dτ, etc () October 13, / 34

26 Apply IFT to the zero level set of the function KKT The Jacobian of KKT w.r.t. endog vars is: JKKT p,λ = JKKT τ = D 1 (p 1 ) + (p 1 τ c λ)d 1(p 1 ) = D 2 (p 2 ) + (p 2 c λ)d 2(p 2 ) = Q D 1 (p 1 ) D 2 (p 2 ) = 2D 1 (p 1 ) + (p 1 τ c λ )D 1 (p 1 ) D 1 (p 1 ) 2D 2 (p 2 ) + (p 2 c λ )D 2 (p 2 ) D 2 (p 2 ) D 1 (p 1 ). Hence from the implicit function theorem, we have dp 1 dτ dp 2 dτ dλ dτ D 1 (p 1 ) D 2 (p 2 ) = JKKT 1 (p,λ) D 1 (p 1 ) () October 13, / 34

27 It s straightforward to check that the determinant of HL (p,λ) is [ D { 2 (p 2 )2 2D 1 (p 1 ) + (p 1 τ c λ )D 1 (p 1 )} +D 1 (p 1 )2 { 2D 2 (p 2 ) + (p 2 c λ )D 2 (p 2 )}] > Applying Cramer s rule, we have dp D 1 dτ = 1 (p 1 ) D 1 (p 1 ) / det 2D 2 (p 2 ) + (p 2 c λ )D 2 (p 2 ) D 2 (p 2 ) D 2 (p 2 ) det(hl (p,λ) ) = D 1 (p 1 )D 2 (p 2 )2/ det(hl (p,λ) ) > dp 2 dτ = det 2D 1 (p 1 ) + (p 1 c τ λ )D 1 (p 1 ) D 1 (p 1 ) D 1 (p 1 ) / D 2 (p 1 ) D 1 (p 2 ) det(hl (p,λ) ) = D 2 (p 1 )D 1 (p 2 )2/ det(hl (p,λ) ) < 2D dλ dτ = 1 (p 1 ) + (p 1 τ c λ )D 1 (p 1 ) D 1 (p 1 ) / det 2D 2 (p 2 ) + (p 2 c λ )D 2 (p 2 ) D 1 (p 1 ) D 2 (p 2 ) det(hl (p,λ) ) = { 2D 2 (p 2 ) + (p 2 c λ )D 2 (p 2 )} D 1 (p 2 )2/ det(hl (p,λ) ) < () October 13, / 34

28 Interpreting the solution to a Linear Equation System Let x solve Ax = b, where A is n n, x,b R n. 1 Column interpretation: b is a x-weighted linear combination of the columns of A. i.e., b = n i=1 A(:,i)x i, where A(:,i) denotes i th column of A if columns of A form a basis set for R n, any b R n can be written as a linear combination of columns of A. x is the unique vector of combination weights if columns of A are a linear dependent set x s.t. b = Ax only if b is in subspace spanned by A. if a solution exists (unlikely), there s necessarily a continuum of them 2 Row interpretation: x belongs to the intersection of the level sets defined by the rows of A and b. i.e., for i = 1,...n, x {y : A(i,:)ẏ = b i }, A(i,:) denotes i th row of A each level set is a line or hyperplane if rows of A form a basis set for R n, there will be a unique intersection if rows of A are a linear dependent set solution exists iff hyperplanes coincide (unlikely); if so, there s necessarily a continuum of solutions () October 13, / 34

29 Row intepretation of solution to two equations () October 13, / 34

30 Row intepretation of solution to three equations () October 13, / 34

31 Cramer s Rule: technique Cramer s rule is a method to solve Ax = b, A is n n, det(a). Brute force: invert A and compute x = A 1 b. But if n > 2, 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. but how on earth can this work???? () October 13, / 34

32 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 now A 1 = [b,v 2 ], and A 2 = [v 1,b], what can you say about A 1 vs A? A 1 and A are almost the same matrix what can you say about A 2 columns of A 2 are close to linear dependent? what can you say ratio of their determinants? Also need to explain why the signs work out properly (but won t) () October 13, / 34

33 Cramer s Rule: example v 1 v 1 b v 2 v 2 The columns of A and vector b Parallelogram/Determinant of A b v 1 b v 2 Parallelogram/Determinant of A 1 Parallelogram/Determinant of A 2 () October 13, / 34

34 Inverse function theorem Special case of the implicit function theorem General Impl FT: find g(α) s.t. f(α,g(α)) = f(ᾱα, x) = Define f(ᾱα, x) = ᾱα η(x). InvFT: find g(α) s.t. f(α,g(α)) = α g(α) = f(ᾱα, x) = ᾱα η( x) = In this case, we have special notation for g(α): η 1 (α) 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)), 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. () October 13, / 34