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Econ 509, Introduction to Mathematical Economics I Professor Ariell Reshef University of Virginia Lecture notes based on Chiang and Wainwright, Fundamental Methods of Mathematical Economics. 1 Mathematical economics Why describe the world with mathematical models, rather than use verbal theory and logic? After all, this was the state of economics until not too long ago (say, 1950s). 1. Math is a concise, parsimonious language, so we can describe a lot using fewer words. 2. Math contains many tools and theorems that help making general statements. 3. Math forces us to explicitly state all assumptions, and help preventing us from failing to acknowl- edge implicit assumptions. 4. Multi dimensionality is easily described. Math has become a common language for most economists. It facilitates communication between econo- mists. Warning: despite its usefulness, if math is the only language for economists, then we are restricting not only communication among us, but more importantly we are restricting our understanding of the world. Mathematical models make strong assumptions and use theorems to deliver insightful conclusions. But, remember the A-A’C-C’Theorem: Let C be the set of conclusions that follow from the set of assumptions A. Let A’be a small perturbation of A. There exists such A’ that delivers a set of conclusions C’ that is disjoint from C. Thus, the insightfullness of C depends critically on the plausibility of A. The plausibility of A depends on empirical validity, which needs to be established, usually using econo- metrics. On the other hand, sometimes theory informs us on how to look at existing data, how to collect new data, and which tools to use in its analysis. Thus, there is a constant discourse between theory and empirics. Neither can be without the other (see the inductivism v deductivism debate). Theory is an abstraction of the world. You focus on the most important relationships that you consider important a priori to understanding some phenomenon. This may yield an economic model. 2 Economic models Some useful notation: 8 for all, 9 exists, 9! exists and is unique. If we cross any of these, or pre…x by : or , then it means "not". 1 2.1 Ingredients of mathematical models 1. Equations: De…nitions : =R C : Y =C +I +G+E M : Kt+1 = (1 ) Kt + It Behavioral/Optimization : qd = p : MC = MR : MC = P Equilibrium : q d = q s 2. Parameters: e.g. , , from above. 3. Variables: exogenous, endogenous. Parameters and functional forms govern the equations, which determine the relationships between variable. Thus, any complete mathematical model can be written as F ( ; Y; X) = 0 ; where F is a set of functions (say, demand and supply), is a set of parameters (say, elasticities), Y are endogenous variables (price and quantity) and X are exogenous, predetermined variables (income, weather). Some models will not have explicit X variables. 2.2 From chapter 3: equilibrium analysis s One general de…nition of a model’ equilibrium is "a constellation of selected, interrelated variables so adjusted to one another that no inherent tendency to change prevails in the model which they constitute". Selected: there may be other variables. This implies a choice of what is endogenous and what is exogenous, but also the overall set of variables that are explicitly considered in the model. Changing the set of variables that is discussed, and the partition to exogenous and endogenous will likely change the equilibrium. Interrelated: all variables must be simultaneously in a state of rest, i.e. constant. And the value of each variable must be consistent with the value of all other variables. Inherent: this means that only the relationships within the model are setting the equilibrium. It implies that the exogenous variables and parameters are all …xed. 2 Since all variables are at rest, an equilibrium is often called a static. Comparing equilibria is called therefore comparative statics. An equilibrium can be de…ned as Y that solves F ( ; Y; X) = 0 ; for given and X. This is one example for the usefulness of mathematics for economists: see how much is described by so little notation. We are interested in …nding an equilibrium for F ( ; Y; X) = 0. Sometimes, there will be no solution. Sometimes it will be unique and sometimes there will be multiple equilibria. Each of these situations is interesting in some context. In most cases, especially when policy is involved, we want a model to have a unique equilibrium, because it implies a function from ( ; X) to Y (the implicit function theorem). But this does not necessarily mean that reality follows a unique equilibrium; that is only a feature of a model. Warning: models with unique equilibrium are useful for many theoretical purposes, but it takes a leap of faith to go from model to reality –as if the unique equilibrium pertains to reality. Students should familiarize themselves with the rest of chapter 3 on their own. 2.3 Numbers Natural, N: 0; 1; 2::: or sometimes 1; 2; 3; ::: Integers, Z: ::: 2; 1; 0; 1; 2; ::: Rational, Q: n=d where both n and d are integers and d is not zero. n is the numerator and d is the denominator. p Irrational numbers: cannot be written as rational numbers, e.g., , e, 2. Real, R: rational and irrational. The real line: ( 1; 1). This is a special set, because it is dense, in the sense that there are just as many real numbers between 0 and 1 (or any other real numbers) as on the entire real line. Complex: an extension of the real numbers, where there is an additional dimension in which we add p to the real numbers imaginary numbers: x + iy, where i = 1. 2.4 Sets We already described some sets above (N, Q, R, Z). A set S contains elements e: S = fe1 ; e2 ; e3 ; e4 g ; where ei may be numbers or objects (say: car, bus, bike, etc.). We can think of sets in terms of the number of elements that they contain: 3 Finite: S = fe1 ; e2 ; e3 ; e4 g. Countable: there is a mapping between the set and N. Trivially, a …nite set is countable. In…nite and countable: Q. Despite containing in…nitely many elements, they are countable. Uncountable: R, [0; 1]. Membership and relationships between sets: e 2 S means that the element e is a member of set S. Subset: S1 S2 : 8e 2 S1 ; e 2 S2 . Sometimes denoted as S1 S2 . Sometimes a strict subset is = de…ned as 8e 2 S1 ; e 2 S2 and 9e 2 S2 ; e 2 S1 . Equal: S1 = S2 : 8e 2 S1 ; e 2 S2 and 8e 2 S2 ; e 2 S1 . The null set, ?, is a subset of any set, including itself, because it does not contain any element that is not in any subset (it is empty). Cardinality: there are 2n subsets of any set of magnitude n = jSj. Disjoint sets: S1 and S2 are disjoint if they do not share common elements, i.e. if there does not exist an element e such that 8e 2 S1 and e 2 S2 . Operations on sets: Union (or): A [ B = feje 2 A or e 2 Bg. Intersection (and): A \ B = feje 2 A and e 2 Bg. Complement: de…ne as the universe set. Then A or Ac = feje 2 = and e 2 Ag. Minus: for B = A, AnB = feje 2 A and e 2 Bg. E.g., A = nA. Rules: Commutative: A[B = B[A A\B = B\A Association: (A [ B) [ C = A [ (B [ C) (A \ B) \ C = A \ (B \ C) 4 Distributive: A [ (B \ C) = (A [ B) \ (A [ C) A \ (B [ C) = (A \ B) [ (A \ C) Do Venn diagrams. 2.5 Relations and functions Ordered pairs: whereas fx; yg = fy; xg because they are sets, but not ordered, (x; y) 6= (y; x) unless x = y (think of the two dimensional plane R2 ). Let X and Y be two sets. Then Cartesian product of X and Y is a set that is given by X Y = f(x; y) jx 2 X; y 2 Y g : For example, the n-dimensional Euclidian space is Rn = R R ::: R = f(x1 ; x2 ; :::xn ) jxi 2 Rg : Cartesian products are relations between sets: 8x 2 X; 9y 2 Y such that (x; y) 2 X Y ; so that the set Y is related to the set X. Any subset of a Cartesian product also has this trait. Note that each x 2 X may have more than one y 2 Y that is related to it. If 8x 2 X; 9!y 2 Y such that (x; y) 2 S X Y ; then y is a function of x. We write this in shorthand notation as y = f (x) or f :X!Y : The second term also is called mapping, or transformation. Note that although for y to be a function of x we must have 8x 2 X; 9!y 2 Y , it is not necessarily true that 8y 2 Y; 9!x 2 X. In fact, there need not exist any such x at all. For example, y = a + x2 , a > 0. In y = f (x), y is the value or dependent variable; x is the argument or independent variable. The set of all permissible values of x is called domain. For y = f (x), y is the image of x. The set of all possible images is called the range. 5 2.6 Functional forms Students should familiarize themselves with polynomials, exponents, logarithms, "rectangular hyperbolic" functions (unit elasticity). 2.7 Functions of more than one variable z = f (x; y) means that 8 (x; y) 2 domain X Y; 9!z 2 Z such that (x; y; z) 2 S X Y Z : This is a function from a plane in R2 to R or a subset of it. y = f (x1 ; x2 ; :::xn ) is a function from the Rn hyperplane or hypersurface to R or a subset of it. 3 Equilibrium analysis Students cover independently. Conceptual points reported above in 2.2. 4 Matrix algebra 4.1 De…nitions Matrix: 2 3 a11 a12 ::: a1n 6 a21 a22 a2n 7 6 7 Am n =6 . . 7 = [aij ] i = 1; 2; :::m; j = 1; 2; :::n : 4 . . . . 5 am1 am2 ::: amn Notation: usually matrices are denoted in upper case. m and n are called the dimensions. Vector: 2 3 x1 6 x2 7 6 7 xm 1 =6 . 7 : 4 . . 5 xm Notation: usually lowercase. Sometimes called a column vector. A row vector is x0 = x1 x2 xm : 4.2 Matrix operations Equality: A = B i¤ aij = bij 8ij. Clearly, the dimensions of A and B must be equal. Addition/subtraction: A B = C i¤ aij bij = cij 8ij. Scalar multiplication: B = cA i¤ bij = c aij 8ij. 6 Matrix multiplication: Let Am n and Bk l be matrices. – if n = k then the product Am n Bn l exists and is equal to a matrix Cm l of dimensions m l. – if m = l then the product Bk m Am n exists and is equal to a matrix Ck n of dimensions k n. – If product exists, then 2 3 ! 2 3 6 7 b11 b12 ::: b1l 6 a11 a12 ::: a1n 7 6 b21 7 6 76 b22 b2l 7 Am n Bn l = 6 a21 a22 a2n 7 6# . . 7 6 . . 74 . . . . 5 4 . . . . 5 bn1 bn2 ::: bnl am1 am2 ::: amn " n # X = cij = aik bkj i = 1; 2; :::m; j = 1; 2; :::l : k=1 Transpose: Let Am n = [aij ]. Then A0 n m = [aji ]. Also denoted AT . Properties: 0 – (A0 ) = A 0 – (A + B) = A0 + B 0 0 – (AB) = B 0 A0 Operation rules – Commutative addition: A + B = B + A. – Distributive addition: (A + B) + C = A + (B + C). – NON commutative multiplication: AB 6= BA. – Distributive multiplication: (AB) C = A (BC). – Associative: premultiplying A (B + C) = AB +AC and postmultiplying (A + B) C = AC +BC. Identity matrix: 2 3 1 0 ::: 0 6 0 1 0 7 6 7 In = 6 . .. . 7 : 4 . . . 5 . . 0 0 ::: 1 AI = IA = A (of course, dimensions must conform). Zero matrix: all elements are zero. 0 + A = A, 0A = A0 = 0 (of course, dimensions must conform). Idempotent matrix: AA = A. Ak = A ; k = 1; 2; ::. 7 Example: the linear regression model is yn 1 = Xn k k 1 + "n 1 and the estimated model by OLS 1 1 y = Xb+e, where b = (X 0 X) X 0 y. Therefore we have y = Xb = X (X 0 X) X 0 y and e = y y = y Xb = b b h i 1 1 1 y X (X 0 X) X 0 y = I X (X 0 X) X 0 y. We can de…ne the projection matrix as P = X (X 0 X) X 0 and the residual generating matrix as R = [I P ]. Both P and R are idempotent. What does it mean that P is idempotent? And R? Singular matrices: even if AB = 0 does NOT imply that A = 0 or B = 0. E.g., 2 4 2 4 A= ; B= : 1 2 1 2 Likewise, CD = CE does NOT imply D = E. E.g., 2 3 1 1 2 1 C= ; D= ; E= : 6 9 1 2 3 2 This is because A, B and C are singular: there is one (or more) row or column that is a linear combination of the other rows or columns, respectively. 4.3 Vector products Scalar multiplication: Let xm 1 be a vector. Then the scalar product cx is 2 3 cx1 6 cx2 7 6 7 cxm 1 = 6 . 7 : 4 . 5 . cxm Inner product: Let xm 1 and ym 1 be vectors. Then the inner product is a scalar m X x0 y = xi yi : i=1 This is useful for computing correlations. Outer product: Let xm 1 and yn 1 be vectors. Then the outer product is a matrix 2 3 x1 y1 x1 y2 : : : x1 yn 6 x2 y1 x2 y2 x2 yn 7 6 7 xy 0 = 6 . . 7 : 4 . . . . 5 xm y1 xm y2 ::: xm yn m n This is useful for computing the variance/covariance matrix. Geometric interpretations: do in 2 dimensions. All extends to n dimensions. – Scalar multiplication. – Vector addition. – Vector subtraction. – Inner product and orthogonality (xy = 0 means x?y). 8 4.4 Linear dependence De…nition 1: a set of k vectors x1 ; x2 ; :::xk are linearly independent i¤ neither one can be expressed as a linear combination of all or some of the others. Otherwise, they are linearly dependent. De…nition 2: a set of k vectors x1 ; x2 ; :::xk are linearly independent i¤ :9 a set of scalars c1 ; c2 ; :::ck such Pk that ci 6= 0 8i and i=1 ci xi = 0. Otherwise, they are linearly dependent. Consider R2 . All vectors that are multiples are linearly dependent. If two vectors cannot be expressed as multiples then they are linearly independent. If two vectors are linearly independent, then any third vector can be expressed as a linear combination of the two. It follows that any set of k > 2 vectors in R2 must be linearly independent. 4.5 Vector spaces and metric spaces The complete set of vectors of n dimensions is a space, or vector space. If all elements of these vectors are real numbers (2 R), then this space is Rn . Any set of n linearly independent vectors is a base for Rn . A base spans the space to which it pertains. This means that any vector in Rn can be expressed as a linear combination of of the base (it is spanned by the base). Bases are not unique. Bases are minimal: they contain the smallest number of vectors that span the space. Example: unit vectors. Consider the vector space R3 . Then 2 3 2 3 2 3 1 0 0 e1 = 4 0 5 ; e2 = 4 1 5 ; e3 = 4 0 5 0 0 1 is a base. Distance metric: Let x; y 2 S, some set. De…ne the distance between x and y by a function d: d = d (x; y), which has the following properties: – d (x; y) = 0 , x = y. – d (x; y) 0. d (x; y) > 0 , x 6= y. – d (x; y) d (x; z) + d (z; y) 8x; y; z (triangle inequality). 9 A metric space is given by a vector space + distance metric. The Euclidean space is given by Rn + the following distance function v u n q uX d (x; y) = t 2 0 (xi yi ) = (x y) (x y) : i=1 But you can imagine other metrics that give rise to other di¤erent metric space. 4.6 Inverse matrix De…nition: if for some square (n n) matrix A there exists a matrix B such that AB = In , then B is the 1 1 inverse of A, and is denoted A , i.e. AA = I. Properties: 1 Not all square matrices have an inverse. If A does not exist, then A is singular. Otherwise, A is nonsingular. 1 A is the inverse of A and vice versa. The inverse is square. 1 1 The inverse, if it exists, is unique. Proof: suppose not, i.e. AB = I and B 6= A . Then A AB = 1 1 A I, IB = B = A , a contradiction Operation rules: 1 1 – A =A 1 1 1 1 1 1 – (AB) =B A . Proof: Let (AB) = C. Then (AB) (AB) = I = C (AB) = CAB ) CABB = 1 1 1 1 1 CA = IB =B ) CAA =C=B A 1 1 0 1 1 0 – (A0 ) = A . Proof: Let (A0 ) = B. Then (A0 ) A0 = I = BA0 ) (BA0 ) = AB 0 = I 0 = 1 1 0 I ) A AB 0 = A 1 I ) B0 = A 1 ) B= A Conditions for nonsingularity: Necessary condition: matrix is square. Given square matrix, a su¢ cient condition is that the rows or columns are linearly independent. It does not matter whether we use the row or column croterion because matrix is square. 1 A is square + linear independence , A is nonsingular , 9A | {z } necessary and su¢ cient conditions How do we …nd the inverse matrix? Soon...Why do we care? See next section. 10 4.7 Solving systems of linear equations We seek a solution x to the system Ax = c 1 An n xn 1 = cn 1 ) x = cA ; where A is a nonsingular matrix and c is a vector. Each row of A gives coe¢ cients to the elements of x: n X row 1 : a1i xi = c1 i=1 Xn row 2 : a2i xi = c2 i=1 Many linear models can be solved this way. We will learn clever ways to compute the solution to this system. We care about singularity of A because (given c) it tells us something about the solution x. 4.8 Markov chains We introduce this through an example. Let x denote a vector of employment and unemployment rates: x0 = e u , where e + u = 1. De…ne the matrix P as a transition matrix that gives the conditional probabilities for transition from the state today to a state next period, pee peu P = ; pue puu where Pij = Pr (state j tomorrowjstate i today). Clearly, pee + peu = 1 and pue + puu = 1. Now add a time dimension to x: x0 = t et ut . We ask: what is the employment and unemployment rates going to be in t + 1 given xt ? Answer: pee peu x0 = x0 P = t+1 t et ut = et pee + ut pue et peu + ut puu : pue puu What will they be in t + 2? Answer: x0 = x0 P 2 . More generally, x0 0 +k = x0 0 P k . t+2 t t t A transition matrix, sometimes called stochastic matrix, is de…ned as a square matrix whose elements are non negative and all rows sum to 1. This gives you conditional transition probabilities starting from each state, where each row is a starting state and each column is the state in the next period. Steady state: a situation in which the distribution over the states is not changing over time. How do we …nd such a state, if it exists? Method 1: Start with some initial condition x0 and iterate forward x0 = x0 P k , taking k ! 1. k 0 Method 2: de…ne x as the steady state value. Solve x0 = x0 P . I.e., P 0 x = x. 11 5 Matrix algebra continued and linear models 5.1 Rank De…nition: The number of linearly independent rows (or, equivalently, columns) of a matrix A is the rank of A: r = rank (A). If Am n then rank (A) min fm; ng. If a square matrix An n has rank n, then we say that A is full rank. Multiplying a matrix A by a another matrix B that is full rank does not change the rank of A. If rank (A) = rA and rank (B) = rB , then rank (AB) = min frA ; rB g. Finding the rank: the echelon matrix method. First de…ne elementary operations: 1. Multiply a row by a non zero scalar: c Ri , c 6= 0. 2. Adding c times of one row to another: Ri + cRj . 3. Interchanging rows: Ri $ Rj . All these operations alter the matrix, but do not change its rank (in fact, they can all be expressed by multiplying matrices, which are all full rank). De…ne: echelon matrix. 1. Zero rows appear at the bottom. 2. For non zero rows, the …rst element on the left is 1. 3. The …rst element of each row on the left (which is 1) appears to the left of the row directly below it. The number of non zero rows in the echelon matrix is the rank. We use the elementary operations in order to change the subject matrix into an echelon matrix, which has as many zeros as possible. A good way to start the process is to concentrate zeros at the bottom. Example: 2 3 2 3 2 1 3 0 11 4 4 1 0 1 0 1 4 A=4 2 6 2 5 R1 $ R 3 : 4 2 6 2 5 R1 : 4 2 6 2 5 4 4 1 0 0 11 4 0 11 4 2 1 3 2 1 3 2 1 3 1 0 1 4 0 1 0 4 2 4 R2 2R1 : 4 0 512 2 5 R3 + 2R2 : 4 0 5 1 2 2 5 R2 : 4 0 1 4=11 5 11 0 11 4 0 0 0 0 0 0 There is a row of zeros: rank (A) = 2. So A is singular. 12 5.2 Determinants and nonsingularity Denote the determinant of a square matrix as jAn n j. This is not absolute value. If the determinant is zero then the matrix is singular. 1. jA1 1j = a11 . 2. jA2 2j = a11 a22 a12 a21 . 3. Determinants for higher order matrices. Let Ak k be a square matrix. The i-j minor jMij j is the determinant of the matrix given by erasing row i and column j from A. Example: 2 3 a b c e f A=4 d e f 5 ; jM11 j = : h i g h i The Laplace expansion of row i gives the determinant of A: k X k X i+j jAk kj = aij ( 1) jMij j = aij Cij ; j=1 j=1 i+j where Cij = ( 1) jMij j is called the cofactor. Example: expansino by row 1 a b c d e f = aC11 + bC12 + cC13 g h i = a jM11 j b jM12 j + c jM13 j e f d f d e = a b +c h i g i g h = a (ei f h) b (di f g) + c (dh eg) : In doing this, it is useful to choose the expansion with the row that has the most zeros. Properties of determinants 1. jA0 j = jAj ips 2. Interchanging rows or columns ‡ the sign of the determinant. 3. Multiplying a row or column by a scalar c multiplies the determinant by c. 4. Ri + cRj does not change the determinant. 5. If a row or a column are multiples of another row or column, respectively, then the determinant is zero: linear dependence. 6. Changing the minors in the Laplace expansion by alien minors will give zero, i.e. k X i+j aij ( 1) jMnj j = 0 ; i 6= n : j=1 13 6 Determinants and singularity: jAj = 0 , A is nonsingular , columns and rows are linearly independent 1 , 9A 1 , for Ax = c ; 9!x = A c , the column (or row) vectors of A span the vector space : 5.3 Finding the inverse matrix Let A be a nonsingular matrix, 2 3 a11 a12 ::: a1n 6 a21 a22 a2n 7 6 7 An n =6 . . 7 : 4 . . . . 5 an1 an2 ::: ann The cofactor matrix of A is CA : 2 3 C11 C12 ::: C1n 6 C21 C22 C2n 7 6 7 CA = 6 . . 7 ; 4 . . . . 5 Cn1 Cn2 ::: Cnn i+j 0 where Cij = ( 1) jMij j. The adjoint matrix of A is adjA = CA : 2 3 C11 C21 : : : Cn1 6 C12 C22 Cn2 7 0 6 7 adjA = CA = 6 . . 7 : 4 . . . . 5 C1n C2n ::: Cnn 0 Consider ACA : 2 Pn Pn Pn 3 Pn j=1 a1j C1j j=1 a1j C2j Pn ::: j=1 a1j Cnj Pn 6 j=1 a2j C1j j=1 a2j C2j j=1 a2j Cnj 7 0 6 7 ACA = 6 . . 7 4 . . 5 Pn . Pn Pn . j=1 anj C1j j=1 anj C2j ::: j=1 anj Cnj 2 Pn 3 j=1 a1j C1j Pn 0 ::: 0 6 0 j=1 a2j C2j 0 7 6 7 = 6 . . 7 4 . . . 5 Pn . 0 0 ::: j=1 anj Cnj 2 3 jAj 0 : : : 0 6 0 jAj 0 7 6 7 = 6 . . 7 = jAj I ; 4 .. . . 5 0 0 : : : jAj 14 where the o¤ diagonal elements are zero due to alien cofactors. It follows that 0 ACA = jAj I 0 1 ACA = I jAj 1 0 1 adjA A = CA = : jAj jAj Example: 1 2 4 3 0 4 2 1 2 1 A= ; CA = ; CA = ; jAj = 2; A = 3 1 : 3 4 2 1 3 1 2 2 5.4 s Cramer’ rule For the system Ax = c and nonsingular A, we have 1 adjA x=A c= c: jAj Denote by Aj the matrix A with column j replaced by c. Then it turns out that jAj j xj = : jAj 5.5 Homogenous equations: Ax = 0 Let the system of equations be homogenous: Ax = 0. If A is nonsingular, then only x = 0 is a solution. If A is singular, then there are in…nite solutions, including x = 0. 5.6 Summary of linear equations: Ax = c For nonsingular A: 1. c 6= 0 ) 9!x 6= 0 2. c = 0 ) 9!x = 0 For singular A: 1. c 6= 0 ) 9x, in…nite solutions 6= 0. If there is inconsistency –linear dependency in A, the elements of c do not follow the same linear combination –there is not solution. 2. c = 0 ) 9x, in…nite solutions, including 0. One can think of the system Ax = c as defrining a relation between c and x. If A is nonsingular, then there is a function (mapping/transformation) between c and x. In fact, when A is nonsingular, this transformation is invertible. 15 5.7 Leontief input/output model We are interested in computing the level of output that is required from each industry in an economy that is required to satisfy …nal demand. This is not a trivial question, because output of some industries are inputs for other industries, while also being consumed in …nal demand. These relationships constitute input/output linkages. Assume 1. Each industry produces one homogenous good. 2. inputs are used in …xed proportions. 3. Constant returns to scale. This gives rise to the Leontief (…xed proportions) production function. The second assumption can be relaxed, depending on the interpretation of the mode.. If you only want to use the framework for accounting purposes, then this is not critical. De…ne aio as the unit requirement of inputs from industry i used in the production of output o. I.e., in order to produce on unit of output o you need aio units of i. For n industries An n = [aio ] is a technology matrix. Each column tells you how much of each input is required to produce one unit of output o. If some industry i does not require its own output for production, then aii . If all industries were used as inputs as well as output, then there would be no primary inputs, i.e. labor, entrepreneurial talent, land, natural resources. To accommodate primary inputs, we add an open sector. If the aio are denominated in monetary values, i.e., in order to product $1 of output o you need $aio Pn Pn of input i, then we must have i=1 aio 1. And if there is an open sector, then we must have i=1 aio < 1. This simply means that the cost of producing $1 is less than $1. By CRS and competitive economy, we have the zero pro…t condition, which means that all revenue is paid out to inputs. So primary inputs receive Pn (1 i=1 aio ) from each industry o. Equilibrium implies supply = demand = demand for intermediate inputs + …nal demand . For some output o we have n X xo = aio xi + do i=1 = a1o x1 + a2o x2 + ::: + ano xn + do : | {z } |{z} interm ediate dem and …nal This implies a1o x1 a2o x2 + ::: (1 aoo ) xo ao+1;o xo+1 ::: ano xn = do : 16 In matrix notation 2 32 3 2 3 (1 a11 ) a12 a13 a1n x1 d1 6 a21 (1 a22 ) a23 a2n 76 x2 7 6 d2 7 6 76 7 6 7 6 a31 a32 (1 a33 ) a3n 76 x3 7 6 d3 7 6 76 7=6 7 : 6 . . . . . . .. . . 76 . . 7 6 . . 7 4 . . . . . 54 . 5 4 . 5 an1 an2 an3 (1 ann ) xn dn Or (I A) x = d : (I A) is the Leontief matrix. This implies that you need more x than just …nal demand because some x is used as intermediate inputs ("I A < I"). 1 x = (I A) d: You need nonsingular (I A). But even then the solution to x might not be positive. We need to …nd conditions for this. 5.7.1 Existence of non negative solution Consider 2 3 a b c A=4 d e f 5 : g h i De…ne Principal minors: the minors that arise from deleting the i-th row and i-th column. E.g. e f a c a b jM11 j = ; jM22 j = ; jM33 j = : h i g i d e k-th order principal minor: is a principal minor that arises from a matrix of dimensions k k. If the dimensions of the original matrix are n n, then a k-th order principal minor is obtained after deleting the same n k rows and columns. E.g., the 1-st order principal minors of A are jaj ; jej ; jij : The 2-nd order principal minors are jM11 j, jM22 j and jM33 j given above. Leading principal minors: these are the 1-st, 2-nd, 3-rd (etc.) order principal minors, where we keep the upper most left corner of the original matrix in each one. E.g. a b c a b jM1 j = jaj ; jM2 j = ; jM3 j = d e f : d e g h i 17 Simon-Hawkins Condition (Theorem): consider the system of equations Bx = d. If (1) all non diagonal elements of Bn n are non positive, i.e. bij 0; 8i 6= j; (2) all elements of dn 1 are non negative, i.e. di 0; 8i; Then 9x 0 such that Bx = d i¤ (3) all leading principal minors are strictly positive, i.e. jMi j > 0; 8i. In our case, B = I A, the Leontief matrix. Economic meaning of SHC. To illustrate, use a 2 2 example: 1 a11 a12 I A= : a21 1 a22 From (3) we have jM1 j = j1 a11 j = 1 a11 > 0, i.e. a11 < 1. This means that less than the total output of x1 is used to produce x1 , i.e. viability. Next, we have jM2 j = jI Aj = (1 a11 ) (1 a22 ) a12 a21 = 1 a11 a22 + a11 a22 a12 a21 > 0 It follows that (1 a11 ) a22 + a11 + a12 a21 < 1 | {z } >0 a11 + a12 a21 < 1 |{z} | {z } direct use indirect use This means that the total amount of x1 demanded (for production of x1 and for production of x2 ) is less than the amount produced (=1), i.e. the resource constraint is kept. 5.7.2 Closed model version The closed model version treats the primary sector as any industry. Suppose that there is only one primary input: labor. Then one interpretation is that the value of consumption of each good is in …xed proportions (these preferences can be represented by a Cobb-Douglas utility function). In this model …nal demand, as de…ned above, must equal zero. Since income accrues to primary inputs (think of labor) and this income is captured in x, then it follows that the d vector must be equal to zero. We know that …nal demand equals income. If …nal demand was positive, then we would have to have an open sector to pay for that demand (from its income). I.e. we have a homogenous system: (I A) x = 0 2 32 3 2 3 (1 a00 ) a01 a02 x0 0 4 a10 (1 a11 ) a12 5 4 x1 5 = 4 0 5 ; a20 a21 (1 a22 ) x2 0 where 0 denotes the primary sector (there could be more than one). Each row in the original A matrix must sum to 1, i.e. a0o + a2o + ::: + ano = 1; 8o, because all of the input is exhausted in production. But then each column in I A can be expressed as minus the sum of all 18 other columns. It follows that I A is singular, and therefore x is not unique! It follows that you can scale up or down the economy with no e¤ect. In fact, this is a general property of CRS economies with no outside sector or endowment. One way to pin down the economy is to set some xi to some level, as an endowment. 6 Derivatives and limits Teaching assistant covers. 7 Di¤erentiation and use in comparative statics 7.1 Di¤erentiation rules dy 1. If y = f (x) = c, a constant, then dx =0 d n 2. dx ax = anxn 1 d 1 3. dx ln x = x d 4. dx [f (x) g (x)] = f 0 (x) g 0 (x) d 5. [f (x) g (x)] = f 0 (x) g (x) + f (x) g 0 (x) dx h i 0 0 0 0 6. dx f (x) = f (x)g(x)+f (x)g (x) = f (x) f (x) g (x) d g(x) [g(x)]2 g(x) g(x) g(x) d df dg 7. dx f [g (x)] = dg dx (chain rule) 1 8. Inverse functions. Let y = f (x) be strictly monotone. Then an inverse function, x = f (y), exists and dx df 1 (y) 1 1 = = = ; dy dy dy=dx df (x) =dx 1 where x and y map one into the other, i.e. y = f (x) and x = f (y). Strictly monotone means that x1 > x2 ) f (x1 ) > f (x2 ) (strictly increasing) or f (x1 ) < f (x2 ) 1 (strictly decreasing). It implies that there is an inverse function x = f (y) because 8y 2Range 9!x 2Domain (recall: 8x 2Domain 9!y 2Range de…nes f (x)). 7.2 Partial derivatives Let y = f (x1 ; x2 ; :::xn ). De…ne the partial derivative of f with respect to xi : @y f (xi + xi ; x i ) f (xi ; x i ) = lim : @xi xi !0 xi Operationally, you derive @y=@xi just as you would derive dy=dxi , while treating all other x i as constants. Example. Consider the following production function 1=' y = z [ k ' + (1 ) l' ] ; ' 1: 19 De…ne the elasticity of substitution as the percent change in relative factor intensity (k=l) in response to a 1 percent change in the relative factor returns (r=w). What is the elasticity of substitution? If factors are paid their marginal product, then 1 1 1 yk = z [ ]' ' k' 1 =r ' 1 1 1 yl = z [ ]' ' (1 ) l' 1 =w: ' Thus ' 1 r k = w 1 l and then 1 1 k 1 ' r 1 ' = : l 1 w 1 1 The elasticity of substitution is 1 '. It is constant, = 1 '. This production function exhibits constant elasticity of substitution, denoted a CES production function. 7.3 Gradients y = f (x1 ; x2 ; :::xn ) The gradient is de…ned as rf = (f1 ; f2 ; :::fn ) ; where @f fi = : @xi We can use this in …rst order approximations: f jx0 = rf (x0 ) x 3 02 2 31 x1 x01 B6 . 7 6 . 7C f (x) f (x0 ) (f1 ; f2 ; :::fn )jx0 @4 . 5 . 4 . 5A : . xn x0n Application to open input/output model: (I A) x = d 1 x = (I A) d=Vd 2 3 2 32 3 x1 v11 v1n d1 6 . 7 6 . .. . 76 . 7 : 4 . 5 = 4 . . . . . 54 . . 5 . xn vn1 vnn dn Think of x as a function of d: rx1 = v11 v12 v1n @xi vij = : @dj 20 7.4 Jacobian and functional dependence Let there be two functions y1 = f (x1 ; x2 ) y2 = g (x1 ; x2 ) The Jacobian determinant is y1 @ @y1 @y1 @y y2 @x1 @x2 jJj = = = @y2 @y2 : @x0 @ (x1 ; x2 ) @x1 @x2 Theorem (functional dependence): jJj = 0 8x i¤ the functions are dependent. Example: y1 = x1 x2 and y2 = ln x1 + ln x2 . x2 x1 jJj = 1 1 =0: x1 x2 Example: y1 = x1 + 2x2 and y2 = ln x1 + 2x2 . 2 2 1 4x2 jJj = 1 4x2 =0: x1 +2x2 2 x1 +2x2 2 Another example: x = V d, 2 3 2 32 3 2 P 3 x1 v11 v13 d1 v1i di 6 . 7 6 . .. . 76 . 7=6 . 7 4 . 5=4 . . . . . 54 . . 5 4 . . 5 : P . xn vn1 vnn dn vni di So jJj = jV j. It follows that linear dependence is equivalent to functional dependence for a system of linear equations. If jV j = 0 then there are 1 solutions for x and the relationship between d and x cannot be inverted. 8 Total di¤erential, total derivative and the implicit function the- orem 8.1 Total derivative Often we are interested in the total rate of change in some variable in response to a change in some other variable or some parameter. If there are indirect e¤ects, as well as direct ones, you want to take this into account. Sometimes the indirect e¤ects are due to general equilibrium constraints and can be very important. Example: consider the utility function u (x; y) and the budget constraint px x + py y = I. Then the total e¤ect of a small change in x on utility is du @u @u dy = + : dx @x @y dx 21 More generally: F (x1 ; :::xn ) X @F n dF dxj = ; dxi j=1 @xj dxi where we know that dxi =dxi = 1. Example: z = f (x; y; u; v), where x = x (u; v) and y = y (u; v). dz @f dx @x dv @f dy @y dv @f @f dv = + + + + + : du @x du @v du @y du @v du @u @v du dz If we want to impose that v is constant, then this is denoted as du v and then all terms that involve dv=du are zero: dz @f dx @f dy @f = + + : du v @x du @y du @u 8.2 Total di¤erential Now we are interested in the change (not rate of...) in some variable or function if all its arguments change a bit, perturbed. For example, if the saving function for the economy is S = S (y; r), then @S @S dS = dy + dr : @y @r More generally, y = F (x1 ; :::xn ) n X @F dy = dxj : j=1 @xj One can view the total di¤erential as a linearization of the function around a speci…c point. The same rules that apply to derivatives apply to di¤erentials; just simply add dx after each partial derivative: 1. dc = 0 for constant c. @(cun ) 2. d (cun ) = cnun 1 du = @u du. @(u v) @(u v) 3. d (u v) = du dv = @u du + @v dv: @(u v w) @(u v w) @(u v w) d (u v w) = du dv dw = @u du + @v dv + @w dw: @(uv) @(uv) 4. d (uv) = vdu + udv = @u du + @v dv: @(uvw) @(uvw) @(uvw) d (uvw) = vwdu + uwdv + uvdw = @u du + @v dv + @w dw: vdu udv @(u=v) @(u=v) 5. d (u=v) = v2 = @u du + @v dv Example: suppose that you want to know how much utility, u (x; y), changes if x and y are perturbed. Then @u @u du = dx + dy : @x @y 22 Now, if you imposed that utility is not changing, i.e. you are interested in an isoquant, then this implies that du = 0 and then @u @u du = dx + dy = 0 @x @y and hence dy @u=@x = : dx @u=@y This should not be understood as a derivative, but rather as a ratio of perturbations. We will see soon conditions under which this is actually a derivative. Log linearization. Suppose that you want to log-linearize z = f (x; y) around some point, say (x ; y ; z ). This means …nding the percent change in z in response to a percent change in x and y. We have @z @z dz = dx + dy : @x @y Divide through by z to get dz x @z dx y @z dy = + z z @x x z @y y x @z y @z b z = b x+ b y; z @x z @y where dz b z= d ln z z is approximately the percent change. Another example: Y = C +I +G dY = dC + dI + dG dY C dC I dI G dG = + + Y Y C Y I Y G b Cb I b Gb Y = C+ I+ G: Y Y Y 8.3 The implicit function theorem This is a useful tool to study the behavior of an equilibrium in response to a change in an exogenous variable. Consider F (x; y) = 0 : We are interested in characterizing the implicit function between x and y, if it exists. We already saw one implicit function when we computed the utility isoquant. In that case, we had u (x; y) = u 23 for some constant level of u. This can be rewritten in the form above as u (x; y) u=0: From this we derived a dy=dx slope. But this can be more general and constitute a function. Another example: what is the slope of a tangent line at any point on a circle? x2 + y 2 = r2 x2 + y 2 r2 = 0 F (x; y) = 0 Taking the total di¤erential Fx dx + Fy dy = 2xdx + 2ydy = 0 dy x = ; y 6= 0 : dx y p p For example, the slope at r= 2; r= 2 is 1. The implicit function theorem: Let the function F (x; y) 2 C 1 on some open set and F (x; y) = 0. Then there exists a (implicit) function y = f (x) 2 C 1 that satis…es F (x; f (x)) = 0, such that dy Fx = dx Fy on this open set. More generally, if F (y; x1 ; x2 ; :::xn ) 2 C 1 on some open set and F (y; x1 ; x2 ; :::xn ) = 0, then there exists a (implicit) function y = f (x1 ; x2 ; :::xn ) 2 C 1 that satis…es F (x; f (x)) = 0, such that n X dy = fi dxi : i=1 @y If we allow only one speci…c xi to be perturbed, then fi = @xi = Fxi =Fy . From F (y; x1 ; x2 ; :::xn ) = 0 and y = f (x1 ; x2 ; :::xn ) we have @F @F @F dy + dx1 + ::: + dxn = 0 @y @x1 @xn dy = f1 dx1 + ::: + fn dxn so that @F (f1 dx1 + ::: + fn dxn ) + Fx1 dx1 + ::: + Fxn dxn = (Fx1 + Fy f1 ) dx1 + ::: + (Fxn + Fy fn ) dxn = 0 : @y Since we only allow xi to be perturbed, dx i = 0 then (Fxi + Fy fi ) = 0 and so fi = Fxi =Fy . 24 8.3.1 A more general version of the implicit function theorem Let F (x; y) = 0 be a set of n functions where xm 1 (exogenous) and yn 1 (endogenous). If 1. F 2 C 1 and @F 2. jJj = @y 0 6= 0 at some point (x0 ; y0 ) (no functional dependence), then 9y = f (x), a set of n functions in a neighborhood of (x0 ; y0 ) such that f 2 C 1 and F (x; f (x)) = 0 in that neighborhood of (x0 ; y0 ). We further develop this. From F (x; y) = 0 we have @F @F @F @F dyn 1 + dxm 1 =0 ) dy = dx : @y 0 n n @x0 n m @y 0 @x0 From y = f (x) we have @y dyn 1 = dxm 1 @x0 n m Combining we get @F @y @F dxm 1 = dxm 1 : @y 0 n n @x0 n m @x0 n m Now suppose that only x1 is perturbed, so that dx0 = dx1 0 0 . Then we get only the …rst column in the set of equations above: @F 1 @y1 @F 1 @y2 @F 1 @yn @F 1 + + ::: + dx1 = dx1 @y1 @x1 @y2 @x1 @yn @x1 @x1 . . . @F n @y1 @F n @y2 @F n @yn @F n + + ::: + dx1 = dx1 @y1 @x1 @y2 @x1 @yn @x1 @x1 By eliminating the dx1 terms we get @F 1 @y1 @F 1 @y2 @F 1 @yn @F 1 + + ::: + = @y1 @x1 @y2 @x1 @yn @x1 @x1 . . . @F n @y1 @F n @y2 @F n @yn @F n + + ::: + = @y1 @x1 @y2 @x1 @yn @x1 @x1 and thus @F @y @F = : @y 0 n n @x1 n 1 @x1 n 1 h i h i @F @F @y Since we required jJj = @y 0 6= 0 it follows that the @y 0 matrix is nonsingular, and thus 9! @x1 , n n n 1 s a solution to the system. This can be obtained by Cramer’ rule: @yj jJj j = ; @x1 jJj 25 h i @F where jJj j is obtained by replacing the j-th column in jJj j by @x1 . Why is this useful? We are often interested in how a model behaves around some point, usually an equilibrium or perhaps a steady state. But models are typically nonlinear and the behavior is hard to characterize without implicit functions. 2 @F 1 @F 1 @F 1 32 3 2 32 3 @F 1 @F 1 @F 1 @y1 @y2 @yn dy1 @x1 @x2 @xm dx1 6 @F 2 @F 2 @F 2 76 7 6 @F 2 @F 2 @F 2 76 7 6 76 dy2 7 6 76 dx2 7 6 @y1 @y2 @yn 76 . 7+6 @x1 @x2 @xm 76 . 7=0 6 . . . . .. . . 74 . 5 6 . . . . . . 74 . 5 4 . . . . 5 . 4 . . . 5 . n n @F n @F n @F n dyn @F @F @F n dxm @y1 @y2 @yn @x1 @x2 @xm 2 @F 1 @F 1 @F 1 32 3 2 32 3 @F 1 @F 1 @F 1 @y1 @y2 @yn dy1 @x1 @x2 @xm dx1 6 @F 2 @F 2 @F 2 76 7 6 @F 2 @F 2 @F 2 76 7 6 76 dy2 7 6 7 6 dx2 7 6 @y1 @y2 @yn 76 . 7= 6 @x1 @x2 @xm 76 . 7 6 . . . . .. . . 74 . 5 6 . . . . . . 74 . 5 4 . . . . 5 . 4 . . . 5 . n n @F n @F n @F n dyn @F @F @F n dxm @y1 @y2 @yn @x1 @x2 @xm 2 3 2 @y 1 @y 1 @y 1 32 3 dy1 @x1 @x2 @xm dx1 6 dy2 7 6 @y 2 @y 2 @y 2 76 dx2 7 6 7 6 76 7 7=6 76 @x1 @x2 @xm 6 . . 7=0 4 . . 5 6 4 . . . . . . . . . 74 5 . . 5 dyn @y n @y n @y n dxm @x1 @x2 @xm 2 @F 1 @F 1 @F 1 32 32 3 2 32 3 @y1 @y1 @y1 @F 1 @F 1 @F 1 @y1 @y2 @yn @x1 @x2 @xm dx1 @x1 @x2 @xm dx1 6 @F 2 @F 2 @F 2 76 @y2 @y2 @y2 76 7 6 @F 2 @F 2 @F 2 76 7 6 76 @x1 @x2 @xm 76 dx2 7 6 76 dx2 7 6 @y1 @y2 @yn 76 . . . 76 . 7= 6 @x1 @x2 @xm 76 . 7 6 . . . . .. . . 76 . . . 74 . 5 6 . . . . . . 74 . 5 4 . . . . 54 . . . 5 . 4 . . . 5 . @F n @F n @F n @yn @yn @yn dxm @F n @F n @F n dxm @y1 @y2 @yn @x1 @x2 @xm @x1 @x2 @xm 8.3.2 Example: demand-supply system + demand : q d = d( p; y) + supply : q s = s( p) equilibrium : q d = q s : Let d; s 2 C 1 . By eliminating q we get + + s( p) d( p; y) = 0 ; which is an implicit function F (p; y) = 0 ; where p is endogenous and y is exogenous. We are interested in how the endogenous price responds to income. By the implicit function theorem 9p = p (y) such that dp Fy dy dy = = = >0 dy Fp sp dp sp dp 26 because dp < 0. An increase in income unambiguously increases the price. To …nd how quantity changes we apply the total derivative approach to the demand function: dq @d dp @d = + dy @p dy @y | {z } |{z} "substitution" e¤ect<0 incom e e¤ect>0 so the sign here is ambiguous. But we can show that it is positive by using the supply side: dq @s dp = >0: dy @p dy Draw demand-supply system. This example is simple, but the technique is very powerful, especially in nonlinear general equilibrium models. 8.3.3 Example: demand-supply system –continued Now consider the system by writing it as a system of two implicit functions: F (p; q; y) = 0 F 1 (p; q; y) = d (p; y) q=0 F 2 (p; q; y) = s (p) q=0: Apply the general theorem. Check for functional dependence in the endogenous variables: @F dp 1 jJj = = = dp + sp > 0 : @ (p; q) sp 1 So there is no functional dependence. Thus 9p = p (y) and 9q = q (y). We now wish to compute the derivatives with respect to the exogenous argument y. Since dF = 0 we have @F 1 @F 1 @F 1 dp + dq + dy = 0 @p @q @y @F 2 @F 2 @F 2 dp + dq + dy = 0 @p @q @y Thus " # " # @F 1 @F 1 @F 1 @p @q dp @y dy @F 2 @F 2 = @F 2 dq @p @q @y dy Use the following @p dp = dy @y @q dq = dy @y 27 to get " #" # " # @F 1 @F 1 @p @F 1 @p @q @y dy @y dy @F 2 @F 2 @q = @F 2 @p @q @y dy @y dy " #" # " # @F 1 @F 1 @p @F 1 @p @q @y @y @F 2 @F 2 @q = @F 2 @p @q @y @y Using the expressions for F 1 and F 2 we get " #" # @d @p @d @p 1 @y @y @s @q = : @p 1 @y 0 @p @q We seek a solution for @y and @y . s This is a system of equations, which we solve using Cramer’ rule: @d @y 1 @d @p jJ1 j 0 1 @y = = = >0 @y jJj jJj jJj and @d @d @p @y @s @d @s @q jJ2 j @p 0 @y @p = = = >0: @y jJj jJj jJj 8.3.4 Example: demand-supply system –continued again Now we use the total derivative approach. We have s (p) d (p; y) = 0 : Take the total derivative with respect to y: @s dp @d dp @d =0 @p dy @p dy @y Thus dp @s @d @d = dy @p @p @y and so @d dp @y = @s @d >0: dy @p @p We could even do the total di¤erential approach... 9 Optimization with one variable and Taylor expansion A function may have many local minimum and maximum. A function may have only one global minimum and maximum, if it exists. 28 9.1 Local maximum, minimum First order necessary conditions (FONC): Let f 2 C 1 on some open convex set (will be de…ned 0 properly later) around x0 . If f (x0 ) = 0, then x0 is a critical point, i.e. it could be either a maximum or minimum. 1. x0 is a local maximum if f 0 (x0 ) changes from positive to negative as x increases around x0 . 2. x0 is a local minimum if f 0 (x0 ) changes from negative to positive as x increases around x0 . 3. Otherwise, x0 is an in‡ection point (not max nor min). Second order su¢ cient conditions (SOC): Let f 2 C 2 on some open convex set around x0 . If 0 f (x0 ) = 0 (FONC satis…ed) then: 1. x0 is a local maximum if f 00 (x0 ) < 0 around x0 . 2. x0 is a local minimum if f 00 (x0 ) > 0 around x0 . 3. Otherwise (f 00 (x0 ) = 0) we cannot be sure. Extrema at the boundaries: if the domain of f (x) is bounded, then the boundaries may be extrema without satisfying any of the conditions above. Draw graphs for all cases. Example: y = x3 12x2 + 36x + 8 FONC: f 0 (x) = 3x2 24x + 36 = 0 x2 8x + 12 = 0 x2 2x 6x + 12 = 0 x (x 2) 6 (x 2) = 0 (x 6) (x 2) = 0 x1 = 6; x2 = 2 are critical points and both satisfy the FONC. f 00 (x) = 6x 24 f 00 (2) = 12 ) maximum f 00 (6) = +12 ) minimum 29 9.2 The N -th derivative test If f 0 (x0 ) = 0 and the …rst non zero derivative at x0 is of order n, f (n) (x0 ) 6= 0, then 1. If n is even and f (n) (x0 ) < 0 then x0 is a local maximum. 2. If n is even and f (n) (x0 ) > 0 then x0 is a local minimum. 3. Otherwise x0 is an in‡ection point. Example: 4 f (x) = (7 x) : 3 f 0 (x) = 4 (7 x) ; so x = 7 is a critical point (satis…es the FONC). 2 f 00 (x) = 12 (7 x) ; f 00 (7) = 0 f 000 (x) = 24 (7 x) ; f 000 (7) = 0 f 0000 (x) = 24 > 0 ; so x = 7 is a minimum: f (4) is the …rst non zero derivative. 4 is even. f (4) > 0. Understanding the N -th derivative test is based on the Maclaurin expansion and the Taylor ex- pansion. 9.3 Maclaurin expansion Terms of art: Expansion: express a function as a polynomial. Around x0 : in a small neighborhood of x0 . Consider the following polynomial f (x) = a0 + a1 x + a2 x2 + a3 x3 + ::: + an xn f (1) (x) = a1 + 2a2 x + 3a3 x2 + ::: + nan xn 1 f (2) (x) = 2a2 + 2 3a3 x + ::: + (n 1) nan xn 2 . . . f (n) (x) = 1 2 ::: (n 1) nan : 30 Evaluate at x = 0: f (0) = a0 f (1) (0) = a1 f (2) (0) = 2a2 . . . f (n) (0) = 1 2 ::: (n 1) nan = n!an : Maclaurin expansion around 0: f (0) f (1) (0) f (2) (0) 2 f (3) (0) f (n) (0) n f (x)jx=0 = + x+ x + x + ::: x : 0! 1! 2! 3! n! 9.4 Taylor expansion Example: quadratic equation. f (x) = a0 + a1 x + a2 x2 : De…ne x = x0 + , where we …x x0 as an anchor and allow to vary. This is essentially relocating the origin to (x0 ; f (x0 )). 2 g( ) a0 + a1 (x0 + ) + a2 (x0 + ) = f (x) : Note that g ( ) = f (x) : Taking derivatives g0 ( ) = a1 + 2a2 (x0 + ) = a1 + 2a2 x0 + 2a2 g 00 ( ) = 2a2 : s Use Maclaurin’ expansion for g ( ) around = 0: g (0) g (1) (0) g (2) (0) 2 g ( )j =0 = + + : 0! 1! 2! Using = x x0 and the fact that x = x0 when = 0, we get a Maclaurin expansion for f (x) around x = x0 : f (x0 ) f (1) (x0 ) f (2) (x0 ) 2 f (x)jx=x0 = + (x x0 ) + (x x0 ) : 0! 1! 2! More generally, we have the Taylor expansion for an arbitrary C n function: f (x0 ) f (1) (x0 ) f (2) (x0 ) 2 f (n) (x0 ) n f (x)jx=x0 = + (x x0 ) + (x x0 ) + ::: + (x x0 ) + Rn 0! 1! 2! n! = Pn + Rn ; where Rn is a remainder. Theorem: 31 As we choose higher n, then Rn will be smaller and in the limit vanish. As x is farther away from x0 Rn may grow. The Lagrange form of Rn : for some point p 2 [x0 ; x] (if x > x0 ) or p 2 [x; x0 ] (if x < x0 ) we have 1 n+1 Rn = f (n+1) (p) (x x0 ) : (n + 1)! Example: for n = 0 we have f (x0 ) f (x)jx=x0 = + Rn = f (x0 ) + Rn = f (x0 ) + f 0 (p) (x x0 ) : 0! Rearranging this we get f (x) f (x0 ) = f 0 (p) (x x0 ) for some point p 2 [x0 ; x] (if x > x0 ) or p 2 [x; x0 ] (if x < x0 ). This is the Mean Value Theorem: 9.5 Taylor expansion and the N -th derivative test De…ne: x0 is a maximum (minimum) of f (x) if the change in the function, f f (x) f (x0 ), is negative (positive) in a neighborhood of x0 , both on the right and on the left of x0 . The Taylor expansion helps determining this. f (2) (x0 ) 2 f (n) (x0 ) n 1 n+1 f = f (1) (x0 ) (x x0 ) + (x x0 ) + ::: + (x x0 ) + f (n+1) (p) (x x0 ) : 2 n! (n + 1)! | {z } rem ainder 0 1. Consider the case that f (x0 ) 6= 0, i.e. the …rst non zero derivative at x0 is of order 1. Choose n = 0, so that the remainder will be of the same order of the …rst non zero derivative and evaluate f = f 0 (p) (x x0 ) : Using the fact that p is very close to x0 , so close that f 0 (p) 6= 0, we have that f changes signs around x0 . 32 2. Consider the case of f 0 (x0 ) = 0 and f 00 (x0 ) 6= 0. Choose n = 1, so that the remainder will be of the same order of the …rst non zero derivative (2) and evaluate f 00 (p) 2 1 00 2 f = f 0 (x0 ) (x x0 ) + (x x0 ) = f (p) (x x0 ) : 2 2 2 Since (x x0 ) > 0 always and f 00 (p) 6= 0 we get f is either positive (minimum) or negative (maximum) around x0 . 3. Consider the case of f 0 (x0 ) = 0, f 00 (x0 ) = 0 and f 000 (x0 ) 6= 0. Choose n = 2, so that the remainder will be of the same order of the …rst non zero derivative (3) and evaluate f 00 (p) 2 f 000 (p) 3 1 000 3 f = f 0 (x0 ) (x x0 ) + (x x0 ) + (x x0 ) = f (p) (x x0 ) : 2 6 6 3 Since (x x0 ) changes signs around x0 and f 000 (p) 6= 0 we get f is changing signs and therefore not an extremum. 4. In the general case f 0 (x0 ) = 0, f 00 (x0 ) = 0, ... f (n 1) (x0 ) = 0 and f (n) (x0 ) 6= 0. Choose n 1, so that the remainder will be of the same order of the …rst non zero derivative (n) and evaluate f (2) (x0 ) 2 f (n 1) (x0 ) n 1 1 (n) n f = f (1) (x0 ) (x x0 ) + (x x0 ) + ::: + (x x0 ) + f (p) (x x0 ) 2 (n 1)! n! 1 (n) n = f (p) (x x0 ) : n! In all cases f (n) (p) 6= 0. n If n is odd, then (x x0 ) changes signs around x0 and f changes signs and therefore not an extremum. n If n is even, then (x x0 ) > 0 always and f is either positive (minimum) or negative (maximum). Warning: in all the above we need f 2 C n at x0 . For example, 1 2 e 2x x 6= 0 f (x) = 0 x=0 is not C 1 at 0. 10 Exponents and logs These are used a lot in economics due to their useful properties, some of which have economic interpretations, in particular in dynamic problems that involve time. 10.1 Exponent function y = f (t) = bt ; b > 1 : (the case of 0 < b < 1 can be dealt with similarly.) 33 f (t) 2 C 1 . f (t) > 0 8t 2 R. f 0 (t) > 0, f 00 (t) > 0, therefore strictly increasing and so 9t = f 1 (y) = logb y, where y 2 R++ . Any y > 0 can be expressed as an exponent of many bases. Make sure you know how to convert bases: loga y logb y = : loga b 10.2 The constant e y = Aert captures growth rates. d t e = et dt d Aert = rAert : dt It turns out that m 1 1=n lim 1+ = lim (1 + n) = e = 2:71828::: m!1 m n!0 Think of 1=m = n as time. Use a Taylor expansion of ex around zero: 1 x 0 1 x 00 2 1 x 000 3 ex = e0 + (e ) x=0 (x 0) + (e ) x=0 (x 0) + (e ) x=0 (x 0) + ::: 1! 2! 3! 1 1 = 1 + x + x2 + x3 + ::: 2! 3! Evaluate this at x = 1: 1 1 e1 = e = 1 + 1 + + + ::: = 2:71828::: 2! 3! 10.3 Examples 10.3.1 Interest compounding Suppose that you are o¤ered an interest rate r on your savings after a year. Then the return after one year is 1 + r. If you invested A, then at the end of the year you have A (1 + r) : r r Now suppose that an interest m is o¤ered for 1=m of a year. In that case you get to compound m m times throughout the year. In that case an investment of A will be worth at the end of the year r m r m=r r A 1+ =A 1+ : m m 34 r Now suppose that you get a instant rate of interest m where m ! 1 (n ! 0), compounded m ! 1 times throughout the year. In that case an investment of A will be worth at the end of the year r m r m=r r r m=r r r 1=u lim A 1 + = lim A 1+ =A lim 1+ =A lim (1 + u) = Aer : m!1 m m!1 m m!1 m u=r=m!0 Thus, r is the instantaneous rate of return. Suppose that we are interested in an arbitrary period of time, t, where, say t = 1 is a year (but this is arbitrary). Then the same kind of math will lead us to …nd the value of an investment A after t time to be r m r m=r r A 1+ =A 1+ : m m if m is …nite, and Aert if we get continuous compounding (m ! 1). 10.3.2 Growth rates The interest rate example tells you how much the investment is worth when it growth at a constant, instan- taneous rate: dV =dt rAert growth rate = = = r per instant (dt): V Aert Any discrete growth rate can be described by a continuous growth rate: t A (1 + i) = Aert ; where (1 + i) = er : 10.3.3 Discounting The value today of X t periods in the future is X NPV = t ; (1 + i) t where 1= (1 + i) is the discount factor. This can also be represented by continuous discounting X rt NPV = t = Xe ; (1 + i) t t rt where the same discount factor is 1= (1 + i) = (1 + i) =e . 35 10.4 Logarithms Log is the inverse function of the exponent. y = bt , t = logb y : b > 1, t 2 R, y 2 R + +. This is very useful, e.g. for regressions analysis. E.g., 16 = 24 , 4 = log2 16 : Also, y = blogb y : Convention: loge x = ln x : Rules: ln (uv) = ln u + ln v ln (u=v) = ln u ln v ln aub = ln a + b ln u loga x logb x = loga b , where a; b; x > 0 ln e 1 – Corollary: logb e = ln b = ln b Some useful properties of logs: 1. Log di¤erences approximate growth rates: X2 X2 X2 X1 ln X2 ln X1 = ln = ln 1+1 = ln 1 + = ln (1 + x) ; X1 X1 X1 where x is the growth rate of X. Take a …rst order Taylor approximation of ln (1 + x) around ln (1): 0 ln (1 + x) ln (1) + (ln (1)) (1 + x 1) = x : So we have ln X2 ln X1 x: 2. Logs "bend down": their image relative to the argument below the 45 degree line. Exponents do the opposite. 3. The derivative of log is always positive, but ever diminishing. 36 4. Nevertheless, lim logb x = 1. Also, lim logb x = 1. Therefore the range is R. x!1 x!0 5. Suppose that y = Aert . Then ln y = ln A + rt. Therefore ln y ln A t= : r This answers the question: how long will it take to grow from A to y, if growth is at an instantaneous rate of r. 6. Converting y = Abct into y = Aert : bc = er , therefore c ln b = r, therefore y = Aert = y = Ae(c ln b)t . 10.5 Derivatives of exponents and logs d 1 ln t = dt t d d ln t 1 logb t = = dt dt ln b t ln b d t e = et dt Let y = et , so that t = ln y: d t d 1 1 e = y= = = y = et : dt dt dt=dy 1=y By chain rule: d u du e = eu dt dt d du=dt ln u = dt u Higher derivatives: dn t t ne = e (dt) d 1 d2 1 d3 2 ln t = ; ln t = ; 3 ln t = t3 ::: dt t (dt)2 t 2 (dt) d t d2 t 2 d3 t 3 b = bt ln b ; t 2 b = b (ln b) ; t 3 b = b (ln b) ::: dt (dt) (dt) 10.6 Application: optimal timing The value of k bottles of wine is given by p t V (t) = ke : rt Discounting: D (t) = e . The present value of V (t) today is p p rt t t rt P V = D (t) V (t) = e ke = ke : 37 p t rt p Choosing t to maximize P V = ke is equivalent to choosing t to maximize ln P V = ln k + t rt. FONC: 0:5 0:5t r = 0 0:5 0:5t = r Marginal bene…t to wait one more instant = marginal cost of waiting one more instant. t = 1=4t2 . SOC: 1:5 0:25t <0 so t is a maximum. 10.7 Growth rates again Denote d _ x=x: dt So the growth rate at some point in time is dx=dt _ x = : x x So in the case x = Aert , we have _ V =r : V And since x (0) = Aer0 = A, we can write without loss of generality x (t) = x0 ert . Growth rates of combinations: 1. For y (t) = u (t) v (t) we have _ y _ u v _ = + y u v gy = gu + gv Proof: ln y (t)ln u (t) + ln v (t) = d d d ln y (t) = ln u (t) + ln v (t) dt dt dt 1 dy 1 du 1 dv = + y (t) dt u (t) dt v (t) dt 2. For y (t) = u (t) =v (t) we have _ y u_ _ v = y u v gy = gu gv Proof: similar to above. 3. For y (t) = u (t) v (t) we have u u gy = gu gv u v u v 38 10.8 Elasticities An elasticity of y with respect to x is de…ned as dy=y dy x y;x = = : dx=x dx y Since @ ln x dx d ln x = dx = @x x we get d ln y y;x = : d ln x 11 Optimization with more than one choice variable 11.1 The di¤erential version of optimization with one variable This helps developing concepts for what follows. Let z = f (x) 2 C 1 , x 2 R. Then dz = f 0 (x) dx : FONC: an extremum may occur when f 0 (x) = 0, or equivalently, when dz = 0, since dx 6= 0. SOC: d2 z = d [dz] = d [f 0 (x) dx] = f 00 (x) dx2 : A maximum occurs when f 00 (x) < 0 or equivalently when d2 z < 0. A minimum occurs when f 00 (x) > 0 or equivalently when d2 z > 0. 11.2 Extrema of a function of two variables Let z = f (x; y) 2 C 1 , x; y 2 R. Then dz = fx dx + fy dy : FONC: dz = 0 for arbitrary values of dx and dy, not both equal to zero. A necessary condition that gives this is fx = 0 and fy = 0 : As before, this is not a su¢ cient condition for an extremum, not only because of in‡ection point, but also due to saddle points. Note: in matrix notation @f dx dx dz = = rf dx = fx fy = fx dx + fy dy : @ (x; y) dy dy If x 2 Rn then 2 3 dx1 n @f 6 . 7 X dz = dx = rf dx = f1 fn 4 . 5= . fi dxi : @x0 i=1 dxn 39 De…ne @2f fxx = @x2 @2f fyy = @y 2 @2f fxy = @x@y @2f fyx = @y@x Young’ Theorem: If both fxy and fyx are continuous, then fxy = fyx . s Now we apply this d2 z = d [dz] = d [fx dx + fy dy] = fxx dx2 + 2fxy dxdy + fyy dy 2 : In matrix notation fxx fxy dx d2 z = dx dy : fxy fyy dy And more generally, if x 2 Rn then @2f d2 z = dx0 dx : @x@x0 | {z } Hessian SONC (second order necessary conditions): for arbitrary values of dx and dy d2 z 0 gives a maximum. d2 z 0 gives a minimum. SOSC (second order su¢ cient conditions): for arbitrary values of dx and dy d2 z < 0 gives a maximum. In the two variable case 2 d2 z < 0 i¤ fxx < 0, fyy < 0 and fxx fyy > fxy : d2 z > 0 gives a minimum. In the two variable case 2 d2 z > 0 i¤ fxx > 0, fyy > 0 and fxx fyy > fxy : Comments: SONC is necessary but not su¢ cient, while SOSC are not necessary. 2 If fxx fyy = fxy a point can be an extremum nonetheless. 2 If fxx fyy < fxy then this is a saddle point. 2 2 If fxy > 0, then fxx fyy > fxy > 0 implies sign(fxx ) =sign(fyy ). 40 11.3 Quadratic form and sign de…niteness This is a tool to help analyze SOCs. Relabel terms for convenience: z = f (x1 ; x2 ) d2 z = q ; dx1 = d1 ; dx2 = d2 f11 = a ; f22 = b ; f12 = h Then d2 z = f11 dx2 + 2f12 dx1 dx2 + f22 dx2 1 2 q = ad2 + 2hd1 d2 + bd2 1 2 a h d1 = d2 d1 : h b d2 This is the quadratic form. Note: d1 and d2 are variables, not constants, as in the FONC. We require the SOCs to hold 8d1 ; d2 , and in particular 8d1 ; d2 6= 0. Denote the Hessian by @2f H= @x@x0 The quadratic form is q = d0 Hd De…ne 8 9 8 9 > > positive de…nite > > > >0 > > > < = < = positive semide…nite 0 q is if q is invariably ; > negative semide…nite > > > > > 0 > > : ; : ; negative de…nite <0 regardless of values of d. Otherwise, q is inde…nite. Consider the determinant of H, jHj, which we call here the discriminant of H: positive de…nite jaj > 0 q is i¤ and jHj > 0 : negative de…nite jaj < 0 jaj is (the determinant of) the …rst ordered minor of H. In the simple two variable case, jHj is (the determinant of) the second ordered minor of H. In that case jHj = ab h2 : If jHj > 0, then a and b must have the same sign, since ab > h2 > 0. 41 11.4 Quadratic form for n variables and sign de…niteness n XXn q = d0 Hd = hij di dj : i=1 j=1 q is positive de…nite i¤ all (determinants of) the principal minors are positive h11 h12 jH1 j = jh11 j > 0; jH2 j = > 0; ::: jHn j = jHj > 0 : h21 h22 q is negative de…nite i¤ (determinants of) the odd principal minors are negative and the even ones are positive: jH1 j < 0; jH2 j > 0; jH3 j < 0; ::: 11.5 Characteristic roots test for sign de…niteness Consider some n n matrix Hn n. We look for a characteristic root r (scalar) and characteristic vector xn 1 (n 1) such that Hx = rx : Developing this expression: Hx = rIx ) (H rI) x = 0 : De…ne (H rI) as the characteristic matrix: 2 3 h11 r h12 h1n 6 h21 h22 r h2n 7 6 7 (H rI) = 6 . . .. . 7 4 . . . . . . . 5 hn1 hn2 ::: hnn r If (H rI) has a non trivial solution (x 6= 0), then (H rI) must be singular, so that jH rIj = 0. This is an equation that we can solve for r. The equation jH rIj = 0 is the characteristic equation, and is an n degree polynomial in r, with n non trivial solutions (some of the solutions can be equal). Some properties: If H is symmetric, then we will have r 2 R. This is useful, because most applications in economics will deal with symmetric matrices, like Hessians and variance-covariance matrices. For each characteristic root that solves jH rIj = 0 there are many characteristic vectors x such that 0 Hx = rx. Therefore we normalize: x x = 1. Denote the normalized characteristic vectors as v. Denote the characteristic vectors (eigenvector) of the characteristic root (eigenvalue) as vi and ri . 0 0 The set of eigenvectors is orthonormal, i.e. orthogonal and normalized: vi vj = 0 8i 6= j and vi vi = 1. 42 11.5.1 Application to quadratic form Let V = (v1 ; v2 ; :::vn ) be the set of eigenvectors of the matrix H. De…ne the vector y that solves d = V y. We use this in the quadratic form q = d0 Hd = y 0 V 0 HV y = y 0 Ry ; where V 0 HV = R. It turns out that 2 3 r1 0 0 6 0 r2 0 7 6 7 R=6 . .. 7 4 . . . 5 0 0 rn Here is why: 2 0 3 2 0 0 0 3 v1 r1 v1 v1 r1 v1 v2 r1 v1 vn 6 v20 7 6 0 r2 v2 v1 0 r2 v2 v2 0 r2 v2 vn 7 6 7 6 7 V 0 HV = V 0 Hv1 Hv2 Hvn =6 . 7 r1 v1 r2 v2 rn vn =6 . . .. . 7=R; 4 . . 5 4 . . . . . . . 5 0 0 0 0 vn rn vn v1 rn vn v2 rn vn vn 0 0 where the last equality follows from vi vj = 0 8i 6= j and vi vi = 1. It follows that sign(q) depends only on Pn the characteristic roots: q = y 0 Ry i=1 ri yi . 2 11.5.2 Characteristic roots test for sign de…niteness 8 9 8 9 > > positive de…nite > > > >0 > > > < = < = positive semide…nite 0 q is i¤ all ri ; > negative semide…nite > > > > > 0 > > : ; : ; negative de…nite <0 regardless of values of d. Otherwise, q is inde…nite. When n is large, …nding the roots can be hard, because it involves …nding the roots of a polynomial of degree n. But the computer can do it for us. 11.6 Global extrema, convexity and concavity We seek conditions for a global maximum or minimum. If a function has a "hill shape" over its entire domain, then we do not need to worry about boundary conditions and the local extremum will be a global extremum. Although the global maximum can be found at the boundary of the domain, it will not be detected by the FONC. Strictly concave: the global maximum is unique. at Concave, but not strictly: this allows for ‡ regions, so the global maximum may not be unique. 43 Let z = f (x) 2 C 2 , x 2 Rn . 8 9 8 9 > > positive de…nite > > > strictly convex > > > < = < = 2 positive semide…nite convex If d z is 8x in the domain, then f is ; > negative semide…nite > > > > > concave > > : ; : ; negative de…nite strictly concave When an objective function is general, then we must assume convexity or concavity. If a speci…c functional form is used, we can check whether it is convex or concave. 11.6.1 Convexity and concavity de…ned De…nition 1: A function f is concave i¤ 8x; y 2graph of f the line between x and y lies on or below the graph. If 8x 6= y the line lies strictly below the graph, then f is strictly concave. For convexity replace "below" with "above". De…nition 2: A function f is concave i¤ 8x; y 2domain of f , which is assumed to be a convex set, and 8 2 (0; 1) we have f (x) + (1 ) f (y) f [ x + (1 ) y] : For strict concavity replace " " with "<" and add 8x 6= y. For convexity replace " " with " " and "<" with ">". Draw …gures. The term x + (1 ) y, 2 (0; 1) is called a convex combination. Properties: 1. If f is linear, then f is both concave and convex, but not strictly. 2. If f is (strictly) concave, then f is (strictly) convex. Proof: f is concave. Therefore 8x; y 2domain of f and 8 2 (0; 1) we have f (x) + (1 ) f (y) f [ x + (1 ) y] = ( 1) [ f (x)] + (1 ) [ f (y)] f [ x + (1 ) y] 3. If f and g are concave functions, then f + g is also concave. If one of the functions is strictly concave, then f + g is strictly concave. 44 Proof: f and g are concave, therefore f (x) + (1 ) f (y) f [ x + (1 ) y] g (x) + (1 ) g (y) g [ x + (1 ) y] [f (x) + g (x)] + (1 ) [f (y) + g (y)] f [ x + (1 ) y] + g [ x + (1 ) y] [(f + g) (x)] + (1 ) [(f + g) (y)] (f + g) [ x + (1 ) y] The proof for strict concavity is identical. 11.6.2 Example Is z = x2 + y 2 concave or convex? Consider …rst the LHS of the de…nition: (i) : f (x1 ; y1 ) + (1 ) f (x2 ; y2 ) = 2 x2 + y1 + (1 1 ) x2 + y2 2 2 : Now consider the RHS of the de…nition: 2 2 (ii) : f [ x1 + (1 ) x2 ; y1 + (1 ) y2 ] = [ x1 + (1 ) x2 ] + [ y1 + (1 ) y2 ] 2 2 = x2 + y1 + (1 1 2 ) 2 x2 + y2 + 2 (1 2 ) (x1 x2 + y1 y2 ) : Now subtract (i) (ii): h i 2 2 (1 2 ) x2 + y1 + x2 + y2 1 2 2 2 (1 ) (x1 x2 + y1 y2 ) = (1 ) (x1 x2 ) + (y1 y2 ) 0: So this is a convex function. Moreover, it is strictly convex, since 8x1 6= x2 and 8y1 6= y2 we have (i) (ii)> 0. Clearly, x2 + y 2 is strictly concave. 11.6.3 Example 2 Is f (x; y) = (x + y) concave or convex? Use the same procedure from above. 2 2 (i) : f (x1 ; y1 ) + (1 ) f (x2 ; y2 ) = (x1 + y1 ) + (1 ) (x2 + y2 ) : Now consider 2 (ii) : f [ x1 + (1 ) x2 ; y1 + (1 ) y2 ] = [ x1 + (1 ) x2 + y1 + (1 ) y2 ] 2 = [ (x1 + y1 ) + (1 ) (x2 + y2 )] 2 2 2 2 = (x1 + y1 ) + 2 (1 ) (x1 + y1 ) (x2 + y2 ) + (1 ) (x2 + y2 ) : Now subtract (i) (ii): h i 2 2 (1 ) (x1 + y1 ) + (x2 + y2 ) 2 (1 ) (x1 + y1 ) (x2 + y2 ) 2 = (1 ) [(x1 + y1 ) (x2 + y2 )] 0: So convex but not strictly. Why not strict? Because when x + y = 0, i.e. when y = x, we get f (x; y) = 0. The shape of this function is a hammock, with the bottom at y = x. 45 11.7 Di¤erentiable functions, convexity and concavity Let f (x) 2 C 1 and x 2 R. Then f is concave if 8x1 ; x2 2domain of f f x1 f x2 f 0 x1 ; x1 x2 i.e. the slope is smaller than the derivative at x1 . Think of x1 as the point of reference and x2 as a target point. For convex replace " " with " ". Another way to write this is f x1 f x2 + f 0 x1 x1 x2 : If x 2 Rn . Then f is concave if 8x1 ; x2 2domain of f f x1 f x2 + rf x1 x1 x2 : Another way to write this is f rf x1 x: Let z = f (x) 2 C 2 and x 2 Rn . Then f is concave i¤ 8x 2domain of f we have d2 z is negative semide…nite. If d2 z is negative de…nite, then f is strictly concave (not only if). Replace "negative" with "positive" for convexity. 11.8 Convex sets in Rn This is related, but distinct from convex and concave functions. De…ne: convex set in Rn . Let the set S Rn . If 8x; y 2 S 8 and 2 [0; 1] we have x + (1 )y 2 S then S is a convex set. (this de…nition actually holds in other spaces too.) Essentially, a set is convex if it has no "holes" (no doughnuts) and the boundary is not "dented" (no bananas). 46 11.8.1 Relation to convex functions 1 The concavity condition 8x; y 2domain of f and 8 2 (0; 1) we have f (x) + (1 ) f (y) f [ x + (1 ) y] assumes that the domain is convex: 8x; y 2domain of f and 8 2 (0; 1) x + (1 ) y 2 domain of f ; because f [ x + (1 ) y] must be de…ned. 11.8.2 Relation to convex functions 2 If f is a convex function, then the set S = fx : f (x) kg ; k 2 R is a convex set (but not only if, i.e. this is a necessary condition, not su¢ cient). If f is a concave function, then the set S = fx : f (x) kg ; k 2 R is a convex set (but not only if, i.e. this is a necessary condition, not su¢ cient). 47 This is why there is an intimate relationship between convex preferences and concave utility functions. 11.9 Example: input decisions of a …rm =R C = pq wl rk : Let p, w, r be given, i.e. the …rm is a price taker in a competitive economy. To simplify, let output, q, be the numeraire, so that p = 1 and everything is then denominated in units of output: =q wl rk : Production function with decreasing returns to scale: q=k l ; < 1=2 Choose fk; lg to maximize . FONC: @ 1 = k l r=0 @k @ 1 = k l w=0: @k SOC: @2 ( 1) k 2 l 2 k l 1 1 jHj = = 2 : k k 1l 1 ( 1) k l 2 @ @ k l l 2 2 jH1 j = ( 1) k l < 0 8k; l > 0. jH2 j = jHj = (1 2 ) k 2( 1) 2( l 1) > 0 8k; l > 0. Therefore is a concave function and the extremum will be a maximum. From the FONC: 1 q k l = =r k 1 q k l = =w l 48 so that rk = wl = q. Thus q k = r q l = : w Using this in the production function: q q 2 1 2 1 1 2 q=k l = = q2 = 1 2 ; r w rw rw so that 1 1 1 1 2 1 1 2 k = 1 2 r w 1 1 1 1 2 1 1 2 l = 1 2 : r w 12 Optimization under equality constraints 12.1 Example: the consumer problem Objective : Choose fx; yg to maximize u (x; y) Constraint(s) : s.t. (x; y) 2 B = f(x; y) : x; y 0; xpx + ypy 0g (draw the budget set, B). Under some conditions, which we will explore soon, we will get the result that the consumer chooses a point on the budget line, s.t. xpx + ypy = 0, and that x; y 0 is trivially satis…ed (nonsatiation and strict convexity of u). So we state a simpler problem: Objective : Choose fx; yg to maximize u (x; y) Constraint(s) : s.t. xpx + ypy = 0 : The optimum will be denoted (x ; y ). The value of the problem is thus u (x ; y ). Constraints can only hurt the unconstrained value (although they may not). This will happen when the unconstrained optimum point is not in the constraint set. E.g., Choose fx; yg to maximize x x2 + y y2 has a maximum at (x ; y ) = (1=2; 1=2), but this point is not on the line x+y = 2, so applying this constraint will move us away from the unconstrained optimum and hurt the objective. 12.2 Lagrange method: one constraint, two variables Let f; g 2 C 1 . Suppose that (x ; y ) is the solution to Choose fx; yg to maximize z = f (x; y) , s.t. g (x; y) = c 49 and that (x ; y ) is not a critical point of g (x; y) = c, i.e. both gx 6= 0 and gy 6= 0 at (x ; y ). Then there exists a number such that (x ; y ; ) is a critical point of L = f (x; y) + [c g (x; y)] ; i.e. @L = c g (x; y) = 0 @ @L = fx gx = 0 @x @L = fy gy = 0 : @y From this it follows that at (x ; y ; ) g (x ; y ) = c = fx =gx = fy =gy : The last equations make it clear why we must check the constraint quali…cations, that both gx 6= 0 and gy 6= 0 at (x ; y ), i.e. check that (x ; y ) is not a critical point of g (x; y) = c. For linear constraints this will be automatically satis…ed. Always write + [c g (x; y)]. Recall that for unconstrained maximum, we must have dz = fx dx + fy dy = 0 ; and thus dy fx = dx fy But even in the constrained problem this still holds –as we will see below –except that now dx and dy are not arbitrary: they must satisfy the constraint, i.e. gx dx + gy dy = 0 : Thus dy gx = : dx gy From both of these we obtain gx fx = ; gy fy i.e. the objective and the constraint are tangent. This follows from fy fx = = : gy gx 50 A graphic interpretation. Think of the gradient as a vector that points in a particular direction. This direction is where to move in order to increase the function the most, and is perpendicular to the isoquant of the function. Notice that we have rf (x ; y ) = rg (x ; y ) (fx ; fy ) = (gx ; gy ) : This means that the constraint and the isoquant of the objective at the optimal value are parallel. They may point in the same direction if > 0 or in opposite directions if < 0. 12.3 is the shadow cost of the constraint tells you how much f would increase if we relax the constraint by one unit, i.e. increase or decrease c by 1 (for equality constraints, it will be either-or). For example, if the objective is utility and the constraint is your budget in money, $, then is in terms of utils/$. It tells you how many more utils you would get if you had one more dollar. Write the system of equations that de…ne the optimum as identities F 1 ( ; x; y) = c g (x; y) = 0 F 2 ( ; x; y) = fx gx = 0 F 2 ( ; x; y) = fy gy = 0 : 6 This is a system of functions of the form F ( ; x; y; c) = 0. If all these functions are continuous and jJj = 0 at ( ; x ; y ), where 0 gx gy @F jJj = = gx fxx gxx fxy gxy ; @ ( x y) gy fxy gxy fyy gyy then by the implicit function theorem we have at = (c), x = x (c) and y = y (c) with derivatives de…ned that can be found as we did above. The point is that such functions exist and that they are di¤erentiable. It follows that there is a sense in which d =dc is meaningful. 51 Now consider the value of the Lagrangian L = L( ; x ; y ) = f (x ; y ) + [c g (x ; y )] ; where we remember that (x ; y ; ) is a critical point. Take the derivative w.r.t. c: dL dx dy d dx dy = fx + fy + [c g (x ; y )] + 1 gx gy dc dc dc dc dc dc dx dy d = [fx gx ] + [fy gy ] + [c g (x ; y )] + dc dc dc = : Therefore dL @L = = : dc @c This is a manifestation of the envelope theorem (see below). But we also know that at the optimum we have c g (x ; y ) = 0 : So at the optimum we have L (x ; y ; ) = f (x ; y ) ; and therefore dL df = = : dc dc 12.4 The envelope theorem Let x be a critical point of f (x; ). Then df (x ; ) @f (x ; ) = : d @ Proof: since at x we have fx (x ; ) = 0, we have df (x ; ) @f (x ; ) dx @f (x ; ) @f (x ; ) = + = d @x d @ @ Drawing of an "envelope" of functions and optima for f (x ; 1 ), f (x ; 2 ), ... 12.5 Lagrange method: one constraint, many variables Let f (x) ; g (x) 2 C 1 and x 2 Rn . Suppose that x is the solution to Choose x to maximize f (x) , s.t. g (x) = c : and that x is not a critical point of g (x) = c. Then there exists a number such that (x ; ) is a critical point of L = f (x) + [c g (x)] ; 52 i.e. @L = c g (x; y) = 0 @ @L = fi gi = 0 ; i = 1; 2; :::n : @xi The constraint quali…cation is similar to above: rg = (g1 (x ) ; g2 (x ) ; :::gn (x )) 6= 0 : 12.6 Lagrange method: many constraints, many variables Let f (x) ; g j (x) 2 C 1 j = 1; 2; :::m, and x 2 Rn . Suppose that x is the solution to Choose x to maximize f (x) , s.t. g 1 (x) = c1 ; g 2 (x) = c2 ; :::g m (x) = cm : and that x satis…es the constraint quali…cations. Then there exists m numbers 1; 2 ; ::: m such that (x ; ) is a critical point of m X L = f (x) + j cj g j (x) ; j=1 i.e. @L = cj g j (x) = 0 ; j = 1; 2; :::m @ j @L = fi gi = 0 ; i = 1; 2; :::n : @xi The constraint quali…cation now requires that @g rank =m; @x0 m n which is as large as it can possibly be. This means that we must have m n, because otherwise the maximal rank would be n < m. This constraint quali…cation, as all the others, means that there exists an m dimensional tangent hyperplane (a Rn m vector space). Loosely speaking, it ensures that we can construct tangencies freely enough. 12.7 Constraint quali…cations in action This example shows that when the constraint quali…cation is not met, the Lagrange method does not work. Choose fx; yg to maximize x, s.t. x3 + y 2 = 0 : The constraint set is given by y2 = x3 ) y = x3=2 for x 0; i.e. n o C = (x; y) : x 0; y = x3=2 ; y = x3=2 53 Notice that (0; 0) is the maximum point. Let us check the constraint quali…cation: rg = 3x2 2y rg (0; 0) = (0; 0) : This violates the constraint quali…cations, since (0; 0) is a critical point of g (x; y). Now check the Lagrangian L = x+ x3 y2 L = x3 y2 = 0 Lx = 1 3x2 = 0 ) = 1=3x2 Ly = 2y = 0 ) either = 0 or y = 0 : Suppose x = 0. Then = 1 –not admissible. Suppose x 6= 0. Then > 0 and thus y = 0. But then from the constraint set x = 0 –a contradiction. 12.8 Constraint quali…cations and Fritz John Theorem Let f (x) ; g (x) 2 C 1 , x 2 Rn . Suppose that x is the solution to Choose x to maximize f (x) , s.t. g (x) = c Then there exists two numbers 0 and 1 such that ( 1; x ) is a critical point of L= 0f (x; y) + 1 [c g (x; y)] ; i.e. @L = c g (x; y) = 0 @ @L = 0 fi 1 gi = 0 ; i = 1; 2; :::n @xi 54 and 0 2 f0; 1g f 0; 1g 6 = (0; 0) : This generalizes to multi constraint problems. 12.9 Second order conditions We want to know whether d2 z is negative or positive de…nite on the constraint set. Using the Lagrange method we …nd a critical point (x ; ) of the problem L = f (x) + [c g (x)] : But this is not a maximum of the L problem. In fact, (x ; ) is a saddle point: perturbations of x around x will hurt the objective, while perturbations of around will increase the objective. If (x ; ) is a critical point of the L problem, then: holding constant, x maximizes the problem; and holding x constant, maximizes the problem. This makes sense: lowering the shadow cost of the constraint as much as possible at the point that maximizes the value. This complicates characterizing the second order conditions, to distinguish maxima from minima. We want to know whether d2 z is negative or positive de…nite on the constraint set. Consider the two variables case dz = fx dx + fy dy From g (x; y) = c we have gx gx dx + gy dy = 0 ) dy = dx ; gy i.e. dy is not arbitrary. We can treat dy as a function of x and y when we di¤erentiate dz the second time: @ (dz) @ (dz) d2 z = d (dz) = dx + dy @x @y @ @ [fx dx + fy dy] dx + [fx dx + fy dy] dy @x @y @ (dy) @ (dy) = fxx dx + fyx dy + fy dx + fxy dx + fyy dy + fy dy @x @y = fxx dx2 + 2fxy dxdy + fyy dy 2 + fy d2 y ; where we use @ (dy) @ (dy) d2 y = d (dy) = dx + dy : @x @y This is not a quadratic form, but we use g (x; y) = c again to transform it into one, by eliminating d2 y. Di¤erentiate dg = gx dx + gy dy = 0 ; 55 using dy as a function of x and y again: @ (dg) @ (dg) d2 g = d (dg) = dx + dy @x @y @ @ = [gx dx + gy dy] dx + [gx dx + gy dy] dy @x @y @ (dy) @ (dy) = gxx dx + gyx dy + gy dx + gxy dx + gyy dy + gy dy @x @y = gxx dx2 + 2gxy dxdy + gyy dy 2 + gy d2 y = 0 Thus 1 d2 y = gxx dx2 + 2gxy dxdy + gyy dy 2 : gy Use this in the expression for d2 z to get gxx gxy gyy d2 z = fxx fy dx2 + 2 fxy fy dxdy + fyy fy dy 2 : gy gy gy From the FONCs we have fy = gy and by di¤erentiating the FONCs we get Lxx = fxx gxx Lyy = fyy gyy Lxy = fxy gxy : We use all this to get d2 z = Lxx dx2 + 2Lxy dxdy + Lyy dy 2 : This is a quadratic form, but not a standard one, because, dx and dy are not arbitrary. As before, we want to know the sign of d2 z, but unlike the unconstrained case, dx and dy must satisfy dg = gx dx + gy dy = 0. Thus, we have second order necessary conditions (SONC): If d2 z is negative semide…tine s.t. dg = 0, then maximum. If d2 z is positive semide…tine s.t. dg = 0, then minimum. The second order su¢ cient conditions are (SOSC): If d2 z is negative de…nite s.t. dg = 0, then maximum. If d2 z is positive de…nite s.t. dg = 0, then minimum. These are less stringent conditions relative to unconstrained optimization, where we required conditions on d2 z for all values of dx and dy. Here we consider only a subset of those values, so the requirement is less stringent, although slightly harder to characterize. 56 12.10 Bordered Hessian and constrained optimization Using the notations we used before for a Hessian, a h H= h b we can write d2 z = Lxx dx2 + 2Lxy dxdy + Lyy dy 2 as d2 z = adx2 + 2hdxdy + bdy 2 : We also rewrite gx dx + gy dy = 0 as dx + dy = 0 : The second order conditions involve the sign of d2 z = adx2 + 2hdxdy + bdy 2 s.t. 0 = dx + dy : Eliminate dy using dy = dx to get 2 dx2 d2 z = a 2h +b 2 2 : The sign of d2 z depends on the square brackets. For a maximum we need it to be negative. It turns out that 0 2 2 a 2h +b = a h H : h b Notice that H contains the Hessian, and is bordered by the gradient of the constraint. Thus, the term "bordered Hessian". The n-dimensional case with one constraint Let f (x) ; g (x) 2 C 2 , x 2 Rn . Suppose that x is a critical point of the Lagrangian problem. Let @2L Hn n = @x@x0 be the Hessian of L evaluated at x . Let rg be the set of linear constraints on dn 1 (= dxn 1 ), evaluated at x : rgd = 0 : 57 We want to know what is the sign of d2 z = q = d0 Hd such that rgd = 0 : The sign de…niteness of the quadratic form q depends on the following bordered Hessian 0 rg1 n H (n+1) (n+1) = 0 : rgn 1 Hn n Recall that sign de…niteness of a matrix depends on the signs of the determinants of the leading principal minors. Therefore positive de…nite (min) H 3 ; H 4 ; ::: H n < 0 d2 z is s.t. dg = 0 i¤ ; negative de…nite (max) H 3 > 0; H 4 < 0; H 5 > 0; ::: Note that in the text they start from H 2 , which they de…ne as the third leading principal minor and is an abuse of notation. We have one consistent way to de…ne leading principal minors of a matrix and we should stick to that. The general case Let f (x) ; g j (x) 2 C 2 j = 1; 2; :::m, and x 2 Rn . Suppose that x is a critical point of the Lagrangian problem. Let @2L Hn n = @x@x0 be the Hessian of L evaluated at x . Let @g Am n = @x0 be the set of linear constraints on dn 1 (= dxn 1 ), evaluated at x : Ad = 0 : We want to know what is the sign of d2 z = q = d0 Hd such that Ad = 0 : The sign de…niteness of the quadratic form q depends on the bordered Hessian 0m m Am n H (m+n) (m+n) = : A0 n m Hn n The sign de…niteness of H depends on the signs of the determinants of the leading principal minors. 58 For a maximum (d2 z negative de…nite) we require that H 2m ; H 2m+1 ::: H m+n alternate signs, m where sign H 2m = ( 1) (Dixit). An alternative formulation for a maximum (d2 z negative de…nite) requires that H n+m ; H n+m 1 ::: n alternate signs, where sign H n+m = ( 1) (Simon and Blume). Anyway, the formulation in the text is wrong. For a minimum...? We know that searching for a minimum of f is like searching for a maximum of f . So one could set up the problem that way and just treat it like a maximization problem. 12.11 Quasiconcavity and quasiconvexity This is a less restrictive condition on the objective function. De…nition: a function f is quasiconcave i¤ 8x1 ; x2 2domain of f , which is assumed to be a convex set, and 8 2 (0; 1) we have f x2 f x1 ) f x1 + (1 ) x2 f x1 or, more simply put f x1 + (1 ) x2 min f x2 ; f x1 : In words: the image of the convex combination is larger than the lower of the two images. De…nition: a function f is quasiconvex i¤ 8x1 ; x2 2domain of f , which is assumed to be a convex set, and 8 2 (0; 1) we have f x2 f x1 ) f x1 + (1 ) x2 f x2 or, more simply put f x1 + (1 ) x2 max f x2 ; f x1 : In words: the image of the convex combination is smaller than the higher of the two images. For strict quasiconcavity and quasiconvexity replace the second inequality with a strict inequality, but at not the …rst. Strict quasiconcavity or convexity rule out ‡ segments. = 2 f0; 1g. 59 at Due to the ‡ segment, the function on the left is not strictly quasiconcave. Note that neither of these functions is convex nor concave. Thus, this is a weaker restriction. The following function, while not convex nor concave, is both quasiconcave and quasiconvex. Compare de…nition of quasiconcavity to concavity and then compare graphically. Properties: 1. If f is linear, then it is both quasiconcave and quasiconvex. 2. If f is (strictly) quasiconcave, then f is (strictly) quasiconvex. 3. If f is concave (convex), then it is quasiconcave (quasiconvex) –but not only if. Note that unlike concave functions, the sum of two quasiconcave functions is NOT quasiconcave. 60 Alternative de…nitions: Let x 2 Rn . f is quasiconcave i¤ 8k 2 R the set S = fx : f (x) kg ; k 2 R is a convex set (for concavity it is only "if", not "only if"). f is quasiconvex i¤ 8k 2 R the set S = fx : f (x) kg ; k 2 R is a convex set (for convexity it is only "if", not "only if"). These may be easier to verify. Recal that for concavity and convexity the conditions above were necessary, but not su¢ cient. Consider a continuously di¤erentiable function f (x) 2 C 1 and x 2 Rn . Then f is quasiconcave i¤ 8x1 ; x2 2domain of f , which is assumed to be a convex set, we have f x2 f x1 ) rf x1 x2 x1 0: In words: the function does not change the sign of the slope (more than once). quasiconvex i¤ 8x1 ; x2 2domain of f , which is assumed to be a convex set, we have f x2 f x1 ) rf x2 x2 x1 0: In words: the function does not change the sign of the slope (more than once). at For strictness, change the last inequality to a strict one, which rules out ‡ regions. Consider a twice continuously di¤erentiable function f (x) 2 C 2 and x 2 Rn . As usual, the Hessian is denoted H and the gradient as rf . De…ne 01 1 rf1 n B= 0 : rfn 1 Hn n (n+1) (n+1) Conditions for quasiconcavity and quasiconvexity in the positive orthant, x 2 Rn involve the leading + principal minors of B. Necessary condition: f is quasiconcave on Rn if (but not only if) 8x 2 R, the leading principal + minors of B follow this pattern jB2 j 0; jB3 j 0; jB4 j 0; ::: Su¢ cient condition: f is quasiconcave on Rn only if 8x 2 R, the leading principal minors of B follow + this pattern jB2 j < 0; jB3 j > 0; jB4 j < 0; ::: Finally, there are also explicitly quasiconcave functions. 61 De…nition: a function f is explicitly quasiconcave if 8x1 ; x2 2domain of f , which is assumed to be a convex set, and 8 2 (0; 1) we have f x2 > f x1 ) f x1 + (1 ) x2 > f x1 : at This rules out ‡ segments, except at the top of the hill. Ranking of quasi concavity, from strongest to weakest: 1. strictly quasiconcave f x2 f x1 ) f x1 + (1 ) x2 > f x1 : 2. explicitly quasiconcave f x2 > f x1 ) f x1 + (1 ) x2 > f x1 : 3. quasiconcave f x2 f x1 ) f x1 + (1 ) x2 f x1 : 12.12 Why is quasi-concavity important? Global maximum Quasi-concavity is important because it allows arbitrary cardinality in the utility function, while maintaining ordinality. Concavity imposes decreasing marginal utility, which is not necessary for characterization of convex preferences and convex indi¤erence sets. Only when dealing with risk do we need to impose concavity. We do not need concavity for global extrema. Suppose that x is the solution to Choose x to maximize f (x) , s.t. g (x) = c : If 1. the set fx : g (x) = cg is a convex set; and 2. f is explicitly quasiconcave, then f (x ) is a global (constrained) maximum. If in addition f is strictly quasiconcave, then this global maximum is unique. 12.13 Application: cost minimization We like apples (a) and bananas (b), but want to reduce the cost of any (a; b) bundle for a given level of + + utility (U ( a; b)) (or fruit salad, if we want to use a production metaphor). Choose fa; bg to minimize cost C = apa + bpb , s.t. U (a; b) = u 62 Set up the appropriate Lagrangian L = apa + bpb + [u U (a; b)] : Here is in units of $/util: it tells you how much an additional util will cost. FONC: @L = u U (a; b) = 0 @ @L = pa Ua = 0 ) pa =Ua = >0 @a @L = pb Ub = 0 ) pb =Ub = >0: @b Thus Ua pa M RS = = Ub pb So we have tangency. Let the value of the problem be C = a pa + b pb : Take a total di¤erential at the optimum to get db pa dC = pa da + pb db = 0 ) = <0: da pb We could also obtain this result from the implicit function theorem, since C (a; b) ; U (a; b) 2 C 1 and jJj = 0. 6 Yet another way to get this is to see that since U (a; b) = u, a constant, db Ua dU (a; b) = Ua da + Ub db = 0 ) = <0: da Ub At this stage all we know is that the isoquant for utility slopes downward, and that it is tangent to the isocost line at the optimum. SOC: recall 2 3 0 Ua Ub H = 4 Ua Uaa Uab 5 : Ub Uab Ubb We need positive de…niteness of dC for a minimum, so that we need H 2 < 0 and H 3 = H < 0. 0 Ua 2 H2 = = Ua < 0 ; Ua Uaa so this is good. But H3 = 0 Ua [Ua ( Ubb ) ( Uab ) Ub ] + Ub [Ua ( Uab ) ( Uaa ) Ub ] 2 2 = Ua Ubb Ua Uab Ub Ub Ua Uab + Ub Uaa 2 2 = Ua Ubb 2Ua Uab Ub + Ub Uaa 63 Without further conditions on U , we do not know whether the expression in the parentheses is negative or not ( > 0). The curvature of the utility isoquant is given by d db d2 b d Ua d Ua (a; b) = = = = da da da2 da Ub da Ub (a; b) db db Uaa + Uab da Ub Ua Ubb da + Uab = 2 Ub h i h i Ua Ua Uaa + Uab Ub Ub Ua Ubb Ub + Uab = 2 Ub 2 Uaa Ub Ua Uab + Ua Ubb =Ub Ua Uab = 2 Ub 2 2 Uaa Ub 2Ua Uab Ub + Ua Ubb = 3 Ub 1 2 2 = 3 Uaa Ub 2Ua Uab Ub + Ua Ubb : Ub d2 b This involves the same expression in the parentheses. If the indi¤erence curve is convex, then da2 0 and thus the expression in the parentheses must be negative. This coincides with the positive semi-de…niteness of dC . Thus, convex isoquants and existence of a global minimum in this case come together. This would d2 b ensure a global minimum, although not a unique one. If da2 > 0, then the isoquant is strictly convex and the global minimum is unique, as dC is positive de…nite. If U is strictly quasiconcave, then indeed the isoquant is strictly convex and the global minimum is unique. 12.14 Further topics Expansion paths. Homotheticity. Elasticity of substitution. Constant elasticity of substitution and relation to Cobb-Douglas. 13 Optimization with inequality constraints 13.1 One inequality constraint Let f (x) ; g (x) 2 C 1 , x 2 Rn . The problem is Choose x to maximize f (x) , s.t. g (x) c: 64 Write the constraint in a "standard way" g (x) c 0: Suppose that x is the solution to Choose x to maximize f (x) , s.t. g (x) c 0 and that x is not a critical point of g (x) = c, if this constraint binds. Write down the Lagrangian function L = f (x) + [c g (x)] : Then there exists a number such that @L (1) : = fi gi = 0 ; i = 1; 2; :::n @xi (2) : [c g (x; y)] = 0 (3) : 0 (4) : g (x) c: The standard way: write g (x) c ip 0 and then ‡ it in the Lagrangian function [c g (x)]. Conditions 2 and 3 are called complementary slackness conditions. If the constraint is not binding, then changing c a bit will not a¤ect the value of the problem; in that case = 0. The constraint quali…cations are that if the constraint binds, i.e. g (x ) = c, then rg (x ) 6= 0. Conditions 1-4 in the text are written di¤erently, although they are an equivalent representation: @L (i) : = fi gi = 0 ; i = 1; 2; :::n @xi @L (ii) : = [c g (x; y)] 0 @ (iii) : 0 (iv) : [c g (x; y)] = 0 : Notice that from (ii) we get g (x) c. If g (x) < c, then L > 0. 13.2 One inequality constraint and one non-negativity constraint There is really nothing special about this problem, but it is worthwhile setting it up, for practice. Let f (x) ; g (x) 2 C 1 , x 2 Rn . The problem is Choose x to maximize f (x) , s.t. g (x) c and x 0: 65 I rewrite this as Choose x to maximize f (x) , s.t. g (x) c 0 and x 0: Suppose that x is the solution to this problem and that x is not a critical point of the constraint set (to be de…ned below). Write down the Lagrangian function L = f (x) + [c g (x)] + ' [x] : Then there exists two numbers and ' such that @L (1) : = fi gi + ' = 0 ; i = 1; 2; :::n @xi (2) : [c g (x; y)] = 0 (3) : 0 (4) : g (x) c (5) : ' [x] = 0 (6) : ' 0 (7) : x 0 ,x 0: The constraint quali…cation is that is not a critical point of the constraints that bind. If only g (x) = c binds, then we require rg (x ) 6= 0. See the general case below. The text gives again a di¤erent – and I argue less intuitive – formulation. The Lagrangian is set up without explicitly mentioning the non-negativity constraints Z = f (x) + ' [c g (x)] : In the text the FONCs are written as @Z (i) : = fi 'gi 0 @xi (ii) : xi 0 @Z (iii) : xi = 0 ; i = 1; 2; :::n @xi @Z (iv) : = [c g (x)] 0 @' (v) : ' 0 @Z (vi) : ' =0: @' The unequal treatment of di¤erent constraints is confusing. My method treats all constraints consistently. A non-negativity constraint is just like any other. 66 13.3 The general case Let f (x) ; g j (x) 2 C 1 , x 2 Rn , j = 1; 2; :::m. The problem is Choose x to maximize f (x) , s.t. g j (x) cj ; j = 1; 2; :::m : Write the the problem in the standard way Choose x to maximize f (x) , s.t. g j (x) cj 0 ; j = 1; 2; :::m : Write down the Lagrangian function m X L = f (x) + j cj g j (x) : j=1 Suppose that x is the solution to the problem above and that x does not violate the constraint quali…cations (see below). Then there exists m numbers j, j = 1; 2; :::m such that m X @L j (1) : = fi j gi (x) = 0 ; i = 1; 2; :::n @xi j=1 (2) : j cj g j (x) = 0 (3) : j 0 (4) : g j (x) cj ; j = 1; 2; :::m : The constraint quali…cations are as follows. Consider all the binding constraints. Count them by jb = 1; 2; :::mb . Then we must have that the rank of 2 @g1 (x ) 3 @x0 6 2 @g (x ) 7 @g B (x ) 6 7 0 =6 6 @x0 . 7 7 @x 4 . . 5 @g mb (x ) @x0 mb n is mb , as large as possible. 13.4 Minimization It is worthwhile to consider minimization separately, although minimization of f is just like maximization of f . We compare to maximization. Let f (x) ; g (x) 2 C 1 , x 2 Rn . The problem is Choose x to maximize f (x) , s.t. g (x) c: Rewrite as Choose x to maximize f (x) , s.t. g (x) c 0 67 Write down the Lagrangian function L = f (x) + [c g (x)] : FONCs @L (1) : = fi gi = 0 ; i = 1; 2; :::n @xi (2) : [c g (x; y)] = 0 (3) : 0 (4) : g (x) c: Compare this to Choose x to minimize h (x) , s.t. g (x) c: Rewrite as Choose x to minimize h (x) , s.t. g (x) c 0 Write down the Lagrangian function L = h (x) + [c g (x)] : FONCs @L (1) : = hi gi = 0 ; i = 1; 2; :::n @xi (2) : [c g (x; y)] = 0 (3) : 0 (4) : g (x) c: Everything is the same. Just pay attention to the inequality setup correctly. This will be equivalent to Compare this to Choose x to maximize h (x) , s.t. g (x) c: Rewrite as Choose x to maximize h (x) , s.t. c g (x) 0: and set up the proper Lagrangian function for maximization L= h (x) + [g (x) c] : This will give the same FONCs as above. 68 13.5 Example Choose fx; yg to maximize min fax; byg , s.t. xpx + ypy I ; where a; b; px ; py > 0. Convert this to the following problem Choose fx; yg to maximize ax, s.t. ax by; xpx + ypy I 0: This is equivalent, because given a level of y, we will never choose ax > by, nor can the objective exceed by by construction. Choose fx; yg to maximize ax, s.t. ax by 0; xpx + ypy I 0: Set up the Lagrangian L = ax + [I xpx ypy ] + ' [by ax] : FONC: Lx = a px a' = 0 Ly = py + b' = 0 [I xpx ypy ] = 0 0 xpx + ypy I ' [by ax] = 0 ' 0 ax by : The solution process is a trial and error process. The best way is to start by checking which constraints do not bind. 1. Suppose ' = 0. Then py = 0 ) =0 ) a a' = 0 ) ' = 1 > 0 –a contradiction. Therefore ' > 0 must hold. Then ax = by ) y = ax=b. 2. Suppose = 0. Then b' = 0 ) ' = 0 – a contradiction (even without ' > 0, we would reach another contradiction: a = 0). Therefore > 0. Then xpx + ypy = I ) xpx + axpy =b = I ) x (px + apy =b) = I ) x = I= (px + apy =b), y = aI= (bpx + apy ). Solving for the multipliers (which is an integral part of the solution) involves solving px + a' = a py b' = 0: 69 This can be written in matrix notation px a a = : py b ' 0 The solution requires nonsingular matrix: px a = bpx apy < 0 : py b s Solving by Cramer’ Rule: a a 0 b ab = = >0 bpx apy bpx + apy px a py 0 apy ' = = >0: bpx apy bpx + apy Finally, we check the constraint quali…cations. Since both constraints bind ( > 0, ' > 0), we must have a rank of two for the matrix xpx + ypy @ ax by px py = : @ [x y] a b In this case we can verify that the rank is two by the determinant, since this is a square 2 2 matrix: px py = bpx apy < 0 : a b 13.6 Example Choose fx; yg to maximize x2 + x + 4y 2 , s.t. 2x + 2y 1 ; x; y 0 Rewrite as Choose fx; yg to maximize x2 + x + 4y 2 , s.t. 2x + 2y 1 0; x 0; y 0 Consider the Jacobian of the constraints 2 3 2x + 2y @4 x 5 2 3 y 2 2 =4 1 0 5 : @ [x y] 0 1 This has rank 2 8 (x; y) 2 R2 , so the constraint quali…cations are never violated. The constraint set is a triangle, all the constraints are linear, the the constraint quali…cation will not fail. Set up the Lagrangian function L = x2 + x + 4y 2 + [1 2x 2y] + ' [x] + [y] 70 FONCs Lx = 2x + 1 2 +'=0 Ly = 8y 2 + =0 [1 2x 2y] = 0 0 2x + 2y 1 'x = 0 ' 0 x 0 y=0 0 y 0 1. From Lx = 0 we have 2x + 1 + ' = 2 > 0 with strict inequality, because x 0 and ' 0. Thus > 0 and the constraint 2x + 2y = 1 binds, so that y = 1=2 x or x = 1=2 y: 2. Suppose ' > 0. Then x = 0 ) y = 1=2 ) =0 ) = 2 ) ' = 3. A candidate solution is (x ; y ) = (0; 1=2). 3. Suppose ' = 0. Then 2x + 1 = 2 : From Ly = 0 we have 8y + =2 : Combining the two we get 2x + 1 = 8y + 2 (1=2 y) + 1 = 8y + 2 2y = 8y + 10y + = 2: The last result tells us that we cannot have both = 0 and y = 0, because we would get 0 = 2 – a contradiction (also because then we would get = 0 from Ly = 0). So either = 0 or y = 0 but not both. (a) Suppose y = 0. Then x = 1=2 ) =1 ) = 2. A candidate solution is (x ; y ) = (1=2; 0). (b) Suppose y > 0. Then = 0 ) y = 0:2 ) x = 0:3 ) = 0:8. A candidate solution is (x ; y ) = (0:3; 0:2). Eventually, we need to evaluate the objective function with each candidate to see which is the global maximizer. 71 13.7 The Kuhn-Tucker su¢ ciency theorem Let f (x) ; g j (x) 2 C 1 , j = 1; 2; :::m. The problem is Choose x 2 Rn to maximize f (x) , s.t. x 0 and g j (x) cj ; j = 1; 2; :::m : Theorem: if 1. f is concave on Rn , 2. g j are convex on Rn , 3. x satis…es the FONCs, then x is a global maximum –not necessarily unique. We know: convex g j (x) cj gives a convex set. One can show that the intersection of convex sets is also a convex set, so that the constraint set is also convex. SO the theorem actually says that trying to maximize a concave function on a convex set give a global maximum, if it exists. It could exist on the border or not –the FONCs will detect it. But these are strong conditions on our objective and constraint functions. The next theorem relaxes things quite a bit. 13.8 The Arrow-Enthoven su¢ ciency theorem Let f (x) ; g j (x) 2 C 1 , j = 1; 2; :::m. The problem is Choose x 2 Rn to maximize f (x) , s.t. x 0 and g j (x) cj ; j = 1; 2; :::m : Theorem: if 1. f is quasi concave on Rn , + 2. g j are quasi convex on Rn , + 3. x satis…es the FONCs of the Kuhn-Tucker problem, 4. Any one of the following conditions on f holds: (a) 9i such that fi (x ) < 0. (b) 9i such that fi (x ) > 0 and xi > 0 (i.e. it does not violate the constraints). 72 (c) rf (x ) 6= 0 and f 2 C 2 around x . (d) f (x) is concave. then x is a global maximum. Arrow-Enthoven constraint quali…cation test for a maximization problem If 1. g j (x) 2 C 1 are quasi convex, 2. 9x0 2 Rn such that all constraints are slack, + 3. Any one of the following holds: (a) g j (x) are convex. (b) @g (x) =@x0 6= 0 8x in the constraint set. then the constraint quali…cation is not violated. 13.9 Envelope theorem for constrained optimization Recall the envelope theorem for unconstrained optimization: if x is a critical point of f (x; ). Then df (x ; ) @f (x ; ) = : d @ @f (x ; ) This was due to @x = 0. Now we face a more complicated problem: Choose x 2 Rn to maximize f (x; ) , s.t. g (x; ) = c : For a problem with inequality constraints we simply use only those constraints that bind. We will consider small perturbations of , so small that they will not a¤ect which constraint binds. Set up the Lagrangian function L = f (x; ) + [c g (x; )] : FONCs L = c g (x; ) = 0 Lxi = fi (x; ) gi (x; ) = 0 ; i = 1; 2; :::n We apply the implicit function theorem to this set of equations to get 9x = x ( ) and = ( ) for which there well de…ned derivatives around ( ; x ). We know that at the optimum we have that the value of the problem is the value of the Lagrangian function f (x ; ) = L = f (x ; ) + [c g (x ; )] = f (x ( ) ; ) + ( ) [c g (x ( ) ; )] : 73 De…ne the value of the problem as v ( ) = f (x ; ) = f (x ( ) ; ) : Take the derivative with respect to to get dv ( ) dL dx d dx = = fx +f + [c g (x ( ) ; )] gx +g d d d d d dx d = [fx gx ] + [c g (x ( ) ; )] + f g d d = f g : Of course, we could have just applied this directly using the envelope theorem: dv ( ) dL @L = = =f g : d d @ 13.10 Duality We will demonstrate the duality of utility maximization and cost minimization. But the principles here are more general than consumer theory. The primal problem is Choose x 2 Rn to maximize U (x) , s.t. p0 x = I : (this should be stated with but we focus on preferences with nonsatiation and strict convexity so the solution lies on the budget line and x > 0 is satis…ed). The Lagrangian function is L = u (x) + [I p0 x] FONCs: Lxi = ui pi = 0 ) = ui =pi ; i = 1; :::n L = [I p0 x] = 0 We apply the implicit function theorem to this set of equations to get Marshallian demand xm i = xm (p; I) i m m i = i (p; I) for which there well de…ned derivatives around ( ; x ). We de…ne the indirect utility function v (p; I) = u [xm (p; I)] : The dual problem is Choose x 2 Rn to minimize px s.t. u (x) = u ; 74 where u is a level of promised utility. The Lagrangian function is Z = p0 x + ' [u u (x)] : FONCs: Zxi = pi 'ui = 0 ) ' = pi =ui ; i = 1; :::n Z' = u (x) = 0 We apply the implicit function theorem to this set of equations to get Hicksian demand xh i = xh (p; u) i 'h i = 'h (p; u) i for which there well de…ned derivatives around (' ; x ). We de…ne the expenditure function e (p; u) = p0 xh (p; u) : Duality: all FONCs imply the same thing: ui pi = ; uj pj Thus, at the optimum xm (p; I) i = xh (p; u) i e (p; u) = I v (p; I) = u: Moreover, 1 '= and this makes sense given the interpretation of ' and . Duality relies on unique global extrema. We need to have all the preconditions for that. Make drawing. 13.10.1 s Roy’ identity v (p; I) = u (xm ) + (I p0 xm ) : Taking the derivative with respect to a price, 2 3 n X n X @xm @v @xm j @ j = uj + (I p0 xm ) 4 pj + xm 5 i @pi j=1 @pi @pi j=1 @pi n X @xm j @ = (uj pj ) + (I p0 xm ) xm i j=1 @pi @pi = xm : i 75 An increase in pi will lower demand by xm , which decreases the value of the problem, as if by the decreasing i income by xm times the utils/$ per dollar of pseudo lost income. In other words, income is now worth xm i i less, and this taxes the objective by xm . Taking the derivative with respect to income, i 2 3 n X @xm n X @xm @v j @ j 5 = uj + (I p0 xm ) + 41 pj @I j=1 @I @I j=1 @I n X @xm j @ = (uj pj ) + (I p0 xm ) + j=1 @I @I = : An increase in income will increase our utility by , which is the standard result. In fact, we could get these results applying the envelpe theorem directly: @v = xm i @pi @v = : @I Roy’ identity is thus s @v=@pi = xm : i @v=@I Why is this interesting? Because this is the amount of income needed to compensate consumers for (that will leave them indi¤erent to) an increase in the price of some good xi . To see this, …rst consider v (p; I) = u ; where u is a level of promised utility (as in the dual problem). By the implicit function theorem 9I = I (pi ) in a neighbourhood of xm , which has a well de…ned derivative dI=dp. This function is de…ned at the optimal bundle xm . Now consider the total di¤erential of v, evaluated at the optimal bundle xm : vpi dpi + vI dI = 0 : This di¤erential does not involve other partial derivatives because it is evaluated at the the optimal bundle xm (i.e. the envelope theorem once again). And we set this di¤erential to zero, because we are considering keeping the consumer exactly indi¤erent, i.e. her promised utility and optimal bundle remain unchanged. Then we have dI vpi @v=@pi = = = xm : i dpi vI @v=@I This result tells you that if you are to keep the consumer indi¤erent to a small change in the price of good i, i.e. not changing the optimally chosen bundle, then you must compensate the consumer by xm units of i s income. We will see this again in the dual problem, using Sheppard’ lemma, where keeping utility …xed 76 @e is explicit. We will see that @pi = xh = xm is exactly the change in expenditure that results from keeping i i utility …xed, while increasing the price of good i. To see this graphically, consider a level curve of utility. The slope of the curve at (p; I) (more generally, the gradient) is xm . 13.10.2 s Sheppard’ lemma e (p; u) = p0 xh + ' u u xh : Taking the derivative with respect to a price, @e n X @xp j @' n X @xh j = xh + i pj + u u xh ' uj @pi j=1 @pi @pi j=1 @pi n X @xh j @' = (pj 'uj ) + u u xh + xh i j=1 @pi @pi = xh : i An increase in pi will increases cost by xh while keeping utility …xed at u (remember that this is a minimiza- i tion problem so increasing the value of the problem is "bad"). Note that this is exactly the result of Roy’s Identity. Taking the derivative with respect to promised utility, 2 3 n X n X @e @xh j @' @xh j 5 = pj + u u xh + ' 41 uj @u j=1 @u @u j=1 @u n X @xh j @' = (pj 'uj ) + u u xh +' j=1 @u @u = ': An increase in utility income will expenditures by ', which is the standard result. 77 In fact, we could get these results applying the envelpe theorem directly: @e = xh i @pi @e = ': @u 14 Integration 14.1 Preliminaries Consider a general function x = x(t) and its derivative with respect to time dx _ x: dt _ This is how much x changes during a very short period dt. Suppose that you know x at any point in time. We can write down how much x changed from some initial point in time, say t = 0, until period t as follows: Z t _ xdt : 0 This is the sum of all changes in x from period 0 to t. The term of art is integration, i.e. we are integrating all the increments. But you cannot say what x (t) is, unless you have the value of x at the initial point. This is the same as saying that you know what the growth rate of GDP is, but you do not know the level. But given x0 = x (0) we can tell what x (t) is: Z t x (t) = x0 + _ xdt : 0 E.g. x = t2 _ Z t Z t 1 xdt = _ u2 du = t3 + c : 0 0 3 The constant c is arbitrary and captures the fact that we do not know the level. Suppose that the instant growth rate of y is a constant r, i.e. _ y =r : y This can be written as _ y ry = 0 ; which is an ordinary di¤erential equation. We know that y = ert gives y=y = r. But so does y = kert . _ So once again, without having additional information about the value of y at some initial point, we cannot say what y (t) is. 78 14.2 Inde…nite integrals Denote dF (x) f (x) = : dx Therefore, dF (x) = f (x) dx : Summing over all small increments we get Z Z dF (x) = f (x) dx = F (x) + c ; where the constant of integration, c, denotes that the integral is correct up to an indeterminate constant. This is so because knowing the sum of increments does not tell you the level. Another way to see this is d d F (x) = [F (x) + c] : dx dx Integration is the opposite operation of di¤erentiation. Instead of looking at small perturbations, or incre- ments, we look for the sum of all increments. Commonly used integrals R xn+1 1. xn dx = n+1 +c R R 2. f 0 (x) ef (x) dx = ef (x) + c ; ex dx = ex + c R f 0 (x) R 1 R dx 3. f (x) dx = ln [f (x)] + c ; x dx = x = ln x + c Operation rules R R R 1. Sum: [f (x) + g (x)] dx = f (x) dx + g (x) dx R R 2. Scalar multiplication: k f (x) dx = kf (x) dx 3. Substitution/change of variables: Let u = u (x). Then Z Z Z du f (u) dx = f (u) u0 dx = f (u) du = F (u) + c : dx E.g. Z Z Z Z 2 3 3 1 4 2x x + 1 dx = 2 x + x dx = 2 x dx + 2 xdx = x + x2 + c 2 Alternatively, de…ne u = x2 + 1, hence u0 = 2x, and so Z Z Z 2 du 1 2x x + 1 dx = udx = udu = u2 + c0 dx 2 1 2 2 1 4 = x + 1 + c0 = x + 2x2 + 1 + c0 2 2 1 4 1 1 = x + x2 + + c0 = x4 + x2 + c : 2 2 2 79 4. Integration by parts: Since d (uv) = udv + vdu we have Z Z Z d (uv) = uv = udv + vdu : Thus the integration by part formula is Z Z udv = uv vdu : To reduce confusion denote V = V (x) ; v (x) = dV (x) =dx U = U (x) ; u (x) = dU (x) =dx Then we write the formula as Z Z U (x) dV (x) = U (x) V (x) V (x) dU (x) Z Z U (x) v (x) dx = U (x) V (x) V (x) u (x) dx : 'x E.g., let f (x) = 'e . Then Z Z 'x 'x 'x x'e dx = xe e dx In the notation above, we have Z Z 'x 'x 'x |{z} 'e{z }dx = |{z} | e } x | x {z 1 |{z} | e {z } dx U v U V u V 14.3 De…nite integrals The area under the f curve, i.e. between the f curve and the horizontal axis, from a to b is Z b f (x) dx = F (b) F (a) : a This is also called the fundamental theorem of calculus. The Riemann Integral: create n rectangles that lie under the curve, that take the minimum of the heights: ri , i = 1; 2:::n. Then create n rectangles with height the maximum of the heights: Ri , i = 1; 2:::n. As the number of these rectangles increases, the sums of the rectangles may converge. If they do, then we say that f is Reimann-integrable. I.e. if n X n X lim ri = lim Ri n!1 n!1 i=1 i=1 80 then Z b f (x) dx : a is well de…ned. Properties: Rb Ra 1. Minus/switching the integration limits: a f (x) dx = b f (x) dx = F (b) F (a) = [F (a) F (b)] Ra 2. Zero: a f (x) dx = F (a) F (a) = 0 3. Partition: for all a < b < c Z c Z b Z c f (x) dx = f (x) dx + f (x) dx : a a b Rb Rb 4. Scalar multiplication: a kf (x) dx = k a f (x) dx ; 8k 2 R Rb Rb Rb 5. Sum: a [f (x) + g (x)] dx = a f (x) dx + a g (x) dx Rb Rb Rb 6. By parts: a U vdx = U V jb a a uV dx = U (b) V (b) U (a) V (b) a uV dx Suppose that we wish to integrate a function from some initial point x0 until some inde…nite point x. Then Z x f (t) dt = F (x) F (x0 ) : x0 and so Z x F (x) = F (x0 ) + f (t) dx : x0 s Leibnitz’ Rule Let f 2 C 1 (i.e. F 2 C 2 ). Then b( Z ) b( Z ) @ @b ( ) @a ( ) @ f (x; ) dx = f (b ( ) ; ) f (a ( ) ; ) + f (x; ) dx : @ @ @ @ a( ) a( ) 81 Proof: let f (x; ) = dF (x; ) =dx. Then b( Z ) @ @ b( ) f (x; ) dx = [F (x; )ja( ) @ @ a( ) @ = [F (b ( ) ; ) F (a ( ) ; )] @ @b ( ) @a ( ) = Fx (b ( ) ; ) + F (b ( ) ; ) Fx (a ( ); ) F (a ( ) ; ) @ @ @b ( ) @a ( ) = f (b ( ) ; ) f (a ( ) ; ) + [F (b ( ) ; ) F (a ( ) ; )] @ @ b( Z ) @b ( ) @a ( ) d = f (b ( ) ; ) f (a ( ) ; ) + F (x; ) dx @ @ dx a( ) b( Z ) @b ( ) @a ( ) @ = f (b ( ) ; ) f (a ( ) ; ) + f (x; ) dx : @ @ @ a( ) s The last line follows by Young’ Theorem. Clearly, if the integration limits do not depend on , then Zb Zb @ @ f (x; ) dx = f (x; ) dx ; @ @ a a and if f does not depend on , then b( Z ) @ @b ( ) @a ( ) f (x) dx = f (b ( )) f (a ( )) : @ @ @ a( ) 14.4 Improper integrals 14.4.1 In…nite integration limits Z 1 Z b f (x) dx = lim f (x) dx = F (1) F (a) : a b!1 a 'x 'x E.g., X exp('): F (x) = 1 e , f (x) = 'e for x 0. Z 1 Z b 'x 'x 'b '0 'e dx = lim 'e dx = lim e +e =1: 0 b!1 0 b!1 Also Z 1 Z 1 Z 1 'x 'x 1 'x E (x) = xf (x) x = x'e dx = xe 0 e dx 0 0 0 1 '1 '0 1 'x 1 '1 1 '0 = " 1e " + 0e + e =0 e + e ' 0 ' ' 1 = : ' 82 14.4.2 In…nite integrand E.g., sometimes the integral is divergent, even though the integration limits are …nite: Z 1 Z b 1 1 1 dx = lim dx = [ln (x)j1 = ln (1) ln (1) = 1 1=1 1 x b!1 1 x Z 1 Z 1 1 1 1 dx = lim dx = [ln (x)j0 = ln (1) ln (0) = 0 + 1 = 1 : 0 x b!0 b x Suppose that for some p 2 (a; b) lim f (x) = 1 : x!p Then the integral from a to b is convergent i¤ the partitions are also convergent: Z b Z p Z b f (x) dx = f (x) dx + f (x) dx : a a p E.g. 1 lim =1: x!0 x3 Therefore, the integral Z 1 Z 0 Z 1 0 1 1 1 1 1 1 dx = dx + dx = + 1 x3 1 x3 0 x3 2x2 1 2x2 0 does not exist, because neither integral converges. 14.5 Example: investment and capital formation In discrete time we have the capital accumulation equation Kt+1 = (1 ) Kt + It ; where It is gross investment at time t. Rewrite as Kt+1 Kt = It Kt : g We want to rewrite this in continuous time. In this context, investment, It , is instantaneous and capital depreciates at an instantaneous rate of . Consider a period of length . The accumulation equation is Kt+ Kt = It Kt : Divide by to get Kt+ Kt = It Kt : Now take ! 0 to get _ Kt = It Kt ; 83 where it is understood that It is instantaneous investment at time t, and Kt is the amount of capital available at that time. Kt is the amount of capital that vanishes due to depreciation. Write _ n Kt = It ; n n where It is net investment. Given a functional form for It we can tell how much capital is around at time t, given an initial amount at time 0, K0 . n Let It = ta . then Z t Z t Z t t _ n ua+1 ta+1 Kt K0 = Kdt = Iu du = ua du = = : 0 0 0 a+1 0 a+1 14.6 s Domar’ growth model Domar was interested in answering: what must investment be in order to satisfy the equilibrium condition at all times. Structure of the model: _ _ 1. Fixed saving rate: It = sYt , s 2 (0; 1). Therefore I = sY . And so _ 1_ Y = I ; s i.e. there is a multiplier e¤ect of investment on output. 2. Potential output is given by a CRS production function t = Kt ; therefore _ _ = K= I : 3. Long run equilibrium is given when potential output is equal to actual output =Y ; therefore _ _ =Y : We have three equations: _ 1_ (i) output demand : Y = I s (ii) potential output : _ = I (iii) equilibrium : _ _ =Y : 84 Use (iii) in (ii) to get _ I=Y and then use (i) to get 1_ I= I ; s which gives I_ = s: I Now integrate in order to …nd the level of investment at any given time: Z _ Z I dt = sdt I ln I = st + c It = e( s)t+c = e( e = I0 e( s)t c s)t : The larger is productivity, , and the higher the saving rate, s, the more investment is required. This is the amount of investment needed for output to keep output in check with potential output. Now suppose that output is not equal to its potential, i.e. 6= Y . This could happen if the investment is not growing at the correct rate of s. Suppose that investment is growing at rate a, i.e. It = I0 eat : De…ne the utilization rate Yt u = lim : t!1 t Compute what the capital stock is at any moment: Z t Z t Z t _ 1 K t K0 = Kd + I d = I0 ea d = I0 eat 0 0 0 a (the constant of integration is absorbed in K0 .) Now compute 1 Yt s It 1 It 1 I0 eat a I0 eat a u = lim = lim = lim = lim 1 at = lim at + aK = : t!1 t t!1 Kt s t!1 Kt s t!1 a I0 e + K0 s t!1 I0 e 0 s The last equality follows from using L’Hopital’ rule. If a > s then u > 1 there is a shortage of capacity, s excess demand. If a > s then u < 1 there is a excess of capacity, excess supply. Thus, in order to keep output demand equal to output potential we must have a = s and thus u = 1. In fact, this holds at any point in time: _ d I = I0 eat = aI0 eat : dt Therefore _ 1 _ a at Y = I = I0 e s s _ = I = I0 eat : 85 So _ Y a I0 eat a = s at = =u: _ I0 e s _ If the utilization rate is too high u > 1, then demand growth outstrips supply, Y > _ . If the utilization rate _ is too low u < 1, then demand growth lags behind supply, Y < _ . Thus, the razor edge condition: only a = s keeps us at a sustainable equilibrium path: If u > 1, i.e. a > s , there is excess demand, investment is too high. Entrepreneurs will try to invest even more to increase supply, but this implies an even larger gap between the two. If u < 1, i.e. a < s , there is excess supply, investment is too low. Entrepreneurs will try to cut investment to lower demand, but this implies an even larger gap between the two. This model is clearly unrealistic and highly stylized. 15 First order di¤erential equations _ We deal with equations that involve y. The general form is _ y + u (t) y (t) = w (t) The goal is to characterize y (t) in terms of u (t) and w (t). dy d2 y First order means dt , not dt2 . _ No products: y y is not permitted. In principle, we can have dn y=dtn , where n is the order of the di¤erential equation. In the next chapter we will deal with up to d2 y=dt2 . 15.1 Constant coe¢ cients 15.1.1 Homogenous case _ y + ay = 0 This gives rise to _ y = a y which has solution at y (t) = y0 e : We need an additional condition to pin down y0 . 86 15.1.2 Non homogenous case _ y + ay = b ; where b 6= 0. The solution method involves splitting the solution into two: y (t) = yc (t) + yp (t) ; where yp (t) is a particular solution and yc (t) is a complementary function. yc (t) solves the homogenous equation _ y + ay = 0 ; so that at yc (t) = Ae : _ yp (t) solves the original equation for a stationary solution, i.e. y = 0, which implies that y is constant and thus y = b=a, where a 6= 0. The solution is thus at b y = yc + yp = Ae + : a Given an initial condition y (0) = y0 , we have a0 b b b y0 = Ae + =A+ ) A = y0 : a a a The general solution is b at b at b at y (t) = y0 e + = y0 e + 1 e : a a a One way to think of the solution is a linear combination of two points: the initial condition y0 and the at particular, stationary solution b=a. (If a > 0, then for t 0 we have 0 e 1, which yields a convex combination). Verify this solution: 2 3 b 6 b b b7 at 6 at 7 y _ = a y0 e = a 6 y0 e + 7= ay + b a 4 a a a5 | {z } y _ ) y + ay = b : Yet a di¤erent way to look at the solution is b at b y (t) = y0 e + a a at b = ke + ; a 87 for some arbitrary point k. In this case at _ y= ake ; and we have at at b _ y + ay = ake + a ke + =b: a When a = 0, we get _ y=b so y = y0 + bt : This follows directly from Z Z _ y = b y = bt + c ; _ where c = y0 . We can also solve this using the same technique as above. yc solves y = 0, so that this is a constant yc = A. yp should solve 0 = b, but this does not work unless b = 0. So try a di¤erent _ particular solution, yp = kt, which requires k = b, because then yp = k = b. So the general solution is y = yc + yp = A + bt : Together with a value for y0 , we get A = y0 . E.g. _ y + 2y = 6 : _ yc solves y + 2y = 0, so 2t yc = Ae : _ yp solves 2y = 6 (y = 0), so yp = 3 : Thus 2t y = yc + yp = Ae +3 : 20 Together with y0 = 10 we get 10 = Ae + 3, so that A = 7. This completes the solution: 2t y = 7e +3 : Verifying this solution: 2t _ y= 14e and 2t 2t _ y + 2y = 14e + 2 7e +3 =6 : 88 15.2 Variable coe¢ cients The general form is _ y + u (t) y (t) = w (t) : 15.2.1 Homogenous case w (t) = 0: _ y _ y + u (t) y (t) = 0 ) = u (t) : y Integrate both sides to get Z Z _ y dt = u (t) dt y Z ln y + c = u (t) dt R R c u(t)dt u(t)dt y = e = Ae ; c where A = e . Thus, the general solution is R u(t)dt y = Ae : Together with a value for y0 and a functional form for u (t) we can solve explicitly. E.g. y + 3t2 y _ = 0 y + 3t2 y _ = 0: Thus _ y = 3t2 y Z Z _ y dt = 3t2 dt y Z ln y + c = 3t2 dt R c 3t2 dt t3 y = e = Ae : 15.2.2 Non homogenous case w (t) 6= 0: _ y + u (t) y (t) = w (t) : The solution is Z R R u(t)dt u(t)dt y=e A+ w (t) e dt : 89 Obtaining this solution requires some elaborate footwork, which we will do. But …rst, see that this works: e.g., y + t2 y = t2 ) u (t) = t2 ; w (t) = t2 : _ Z Z 1 u (t) dt = t2 dt = t3 3 Z R Z 3 3 u(t)dt w (t) e dt = t2 et =3 dt = et =3 ; since Z f 0 (y) ef (y) dy = ef (y) : Thus h i t3 =3 3 t3 =3 y=e A + et =3 = Ae +1 : Verifying this solution: t3 =3 y= _ t2 Ae so h i t3 =3 t3 =3 y + u (t) y (t) _ = t2 Ae + t2 Ae +1 t3 =3 t3 =3 = t2 Ae + t2 Ae + t2 = t2 = w (t) : 15.3 Solving exact di¤erential equations Suppose that the primitive di¤erential equation can be written as F (y; t) = c so that dF = Fy dy + Ft dt = 0 : We use the latter total di¤erential to obtain F (y; t), from which we obtain y (t). We set F (y; t) = c to get initial conditions. De…nition: the di¤erential equation M dy + N dt = 0 s is an exact di¤erential equation i¤ 9F (y; t) such that M = Fy and N = Ft . By Young’ Theorem we have @M @2F @N = = : @t @t@y @y The latter is what we will be checking in practice. 90 E.g., let F (y; t) = y 2 t = c. Then dF = Fy dy + Ft dt = 2ytdy + y 2 dt = 0 : Set M = 2yt ; N = y 2 : Check: @2F @M = = 2y @t@y @t @2F @N = = 2y @t@y @y So this is an exact di¤erential equation. Before solving, one must always check that the equation is indeed exact. Step 1: Since dF = Fy dy + Ft dt we can integrate both sides. But instead, we write Z F (y; t) = Fy dy + ' (t) Z = M dy + ' (t) ; where ' (t) is a residual function. Step 2: Take the derivative N = Ft from step 1 to identify ' (t). Step 3: Solve for y (t), taking into account F (y; t) = c. Example: 2yt dy + y 2 dt = 0 : |{z} |{z} M N Step 1: Z Z F (y; t) = M dy + ' (t) = 2ytdy + ' (t) = y 2 t + ' (t) : Step 2: @F (y; t) @ 2 = y t + ' (t) = y 2 + '0 (t) : @t @t Since N = y 2 we must have '0 (t) = 0, i.e. ' (t) is a constant function, ' (t) = k, for some k. Thus F (y; t) = y 2 t + k = c ; 91 so we can ignore the constant k and write F (y; t) = y 2 t = c : Step 3: We can now solve for y (t): 1=2 y (t) = (ct) : Example: (t + 2y) dy + y + 3t2 dt = 0 : So that M = (t + 2y) N = y + 3t2 : Check that this equation is exact: @M @N =1= ; @t @y so this is indeed an exact di¤erential equation. Step 1: Z Z F (y; t) = M dy + ' (t) = (t + 2y) dy + ' (t) = ty + y 2 + ' (t) : Step 2: @F (y; t) @ = ty + y 2 + ' (t) = y + '0 (t) = N = y + 3t2 ; @t @t so that '0 (t) = 3t2 and Z Z ' (t) = '0 (t) dt = 3t2 dt = t3 : Thus F (y; t) = ty + y 2 + ' (t) = ty + y 2 + t3 : Step 3: we cannot solve this analytically for y (t), but using the implicit function theorem, we can characterize it. Example: Let T F (t) be the time until some event occurs, T 0. De…ne the hazard rate as f (t) h (t) = ; 1 F (t) which is the "probability" that the event occurs at time t, given that it has not occurred by time t. We can write 0 R (t) h (t) = ; R (t) 92 where R (t) = 1 F (t). We know how to solve such di¤erential equations: 0 R (t) + h (t) R (t) = 0 : Rt h(s)ds R (t) = Ae : Since R (0) = 1 (the probability that the event occurs at all), then we have A = 1: Rt h(s)ds R (t) = e : It follows that Rt Z t Rt Rt 0 h(s)ds @ h(s)ds h(s)ds f (t) = R (t) = e h (s) ds = e [ h (t)] = h (t) e : @t Suppose that the hazard rate is constant: h (t) = : In that case Rt ds t f (t) = e = e ; which is the p.d.f. of the exponential distribution. Now suppose that the hazard rate is not constant, but 1 h (t) = t : In that case Rt 1 1 s ds 1 t f (t) = t e = t e ; which is the p.d.f. of the Weibull distribution. This is useful if you want to model an increasing hazard ( > 1) or decreasing hazard ( < 1). When = 1 or we get the exponential distribution. 15.4 Integrating factor and the general solution Sometimes we can turn a non exact di¤erential equation into an exact one. For example, 2tdy + ydt = 0 is not exact: M = 2t N = y and Mt = 2 6= Ny = 1 : But if we multiply the equation by y, we get an exact equation: 2ytdy + y 2 dt = 0 ; which we saw above is exact. 93 15.4.1 Integrating factor We have the general formulation _ y + uy = w ; where all variables are functions of t and we wish to solve for y (t). Write the equation above as dy + uy = w dt dy + uydt = wdt dy + (uy w) dt = 0: The integrating factor is Rt u(s)ds e : If we multiply the equation by this factor we always get an exact equation: Rt Rt u(s)ds u(s)ds e dy + e (uy w) dt = 0 : To verify this, write Rt u(s)ds M = e Rt u(s)ds N = e (uy w) and @M @ R t u(s)ds Rt = e = e u(s)ds u (t) @t @t @N @ R t u(s)ds Rt = e (uy w) = e u(s)ds u (t) : @y @y So @M=@t = @N=@y. This form can be recovered from the method of undetermined coe¢ cients. We seek some A such that A |{z}dy + A (uy w)dt = 0 | {z } M N and @M @A _ = =A @t @t @N @ = [A (uy w)] = Ay @y @y are equal. This means _ A = Au _ A=A = u Rt u(s)ds A = e : 94 15.4.2 The general solution We have some equation that is written as _ y + uy = w : Rewrite as dy + (uy w) dt = 0 : Multiply by the integrating factor to get an exact equation Rt Rt u(s)ds u(s)ds e {z }dy + e | (uy w)dt = 0 : | {z } M N Step 1: Z Z Rt Rt u(s)ds u(s)ds F (y; t) = M dy + ' (t) = e dy + ' (t) = ye + ' (t) : Step 2: @F @ h R t u(s)ds i Rt = ye + ' (t) = ye u(s)ds u (t) + '0 (t) = N : @t @t Using N from above we get Rt Rt u(s)ds N = ye u (t) + '0 (t) = e u(s)ds (uy w) ; so that Rt '0 (t) = e u(s)ds w and so Z Rt u(s)ds ' (t) = e wdt : Now we can write Z Rt Rt u(s)ds u(s)ds F (y; t) = ye e wdt = c Step 3, solve for y: Z Rt Rt u(s)ds u(s)ds y=e c+ e wdt : 15.5 First order nonlinear di¤erential equations of the 1st degree In general, _ y = h (y; t) will yield an equation like this f (y; t) dy + g (y; t) dt = 0 : In principle, y and t can appear in any degree. First order means y, not y (n) . _ n _ _ First degree means y, not (y) . 95 15.5.1 Exact di¤erential equations See above. 15.5.2 Separable variables f (y) dy + g (t) dt = 0 : Then just integrate Z Z f (y) dy = g (t) dt and solve for y (t). Example: 3y 2 dy tdt = 0 Z Z 2 3y dy = tdt 1 2 y3 = t +c 2 1=3 1 2 y (t) = t +c : 2 Example: 2tdy ydt = 0 dy dt = y 2t Z Z dy dt = y 2t 1 ln y = ln t + c 2 1 1=2 y = e 2 ln t+c = eln t ec = ec t1=2 : 15.5.3 Reducible equations Suppose that _ y = h (y; t) can be written as y + Ry = T y m ; _ where R = R (t) T = T (t) 96 are functions only of t and m 6= 0; 1 : This is a Bernoulli equation, which can be reduced to a linear equation and solved as such. Here’s how: y + Ry _ = T ym 1 y + Ry 1 _ m = T ym Use a change of variables z = y1 m so that m _ z = (1 m) y _ y _ y _ z = : ym 1 m Plug this in the equation to get _ z + Rz = T 1 m 2 3 dz + 4(1 m) Rz (1 m) T 5 dt = 0 | {z } | {z } w w dz + [uz + w] = 0: This is something we know how to solve: Rt Z Rt u(s)ds u(s)ds z (t) = e A+ e wdt : from which we get the original 1 y (t) = z (t) 1 m : Example: y + ty = 3ty 2 _ In this case R = t T = 3t m = 2: Divide by y 2 and rearrange to get 2 1 y _ y + ty 3t = 0 : 97 Change variables 1 z = y 2 _ z = y _ y so that we get _ z + tz 3t = 0 dz + ( tz + 3t) dt = 0: so that we set u = t w = 3t : Using the formula we get Rt Z R t u(s)ds z (t) = e A + e u(s)ds wdt Rt Z Rt = e sds A 3 e sds tdt Z 2 2 = et =2 A 3 e t =2 tdt 2 h 2 i = et =2 A + 3e t =2 2 = Aet =2 +3 : So that 1 2 1 y (t) = = Aet =2 + 3 : z Example: y + y=t = y 3 : _ In this case R = 1=t T = 1 m = 3: Divide by y 3 and rearrange to get 3 1 2 y _ y+t y 1=0: 98 Change variables 2 z = y 3 _ z = 2y _ y so that we get _ z z + 1 = 0 2 t z _ z+ 2 +2 = 0 t z dz + 2 + 2 dt = 0 : t so that we set u = 2=t w = 2: Using the formula we get Rt Z Rt u(s)ds u(s)ds z (t) = e A+ e wdt Rt Z Rt 1 1 = e2 t ds A 2 e 2 t ds dt Z = e2 ln t A 2 e 2 ln t dt = t2 A 2t 2 = At2 2: So that 1 2 y (t) = = At2 2 : z2 15.6 The qualitative graphic approach Given _ y = f (y) _ we can plot y as a function of y. This is called a phase diagram. This is an autonomous di¤erential equation, since t does not appear explicitly as an argument. We have three cases: _ 1. y > 0 : y is growing, so we shift to the right. _ 2. y < 0 : y is decreasing, so we shift to the left. _ 3. y = 0 : y is stationary, an equilibrium. 99 _ System A is dynamically stable: the y curve is downward sloping; any movement away from the stationary point y will bring us back there. _ System B is dynamically unstable: the y curve is upward sloping; any movement away from the stationary point y take farther away. For example, consider _ y + ay = b with solution b at b y (t) = y0 e + a a at at b = e y0 + 1 e : a This is a linear combination between the initial point y0 and b=a. at System A happens when a > 0: lim e ! 0, so that lim y (t) ! b=a = y . t!1 t!1 at System B happens when a < 0: lim e ! 1, so that lim y (t) ! 1. t!1 t!1 15.7 The Solow growth model (no long run growth version) 1. CRS production function Y = F (K; L) y = f (k) where y = Y =L, k = K=L. Given FK > 0 and FKK < 0 we have f 0 > 0 and f 00 < 0. 100 _ 2. Constant saving rate: I = sY , so that K = sY K. _ 3. Constant labor force growth: L=L = n. K_ = sF (K; L) K = sLf (k) K _ K = sf (k) k: L Since d K _ KL _ KL _ K _ KL _ K _ k= = = = kn ; dt L L2 L LL L we get _ k = sf (k) (n + ) k : This is an autonomous di¤erential equation in k. Since f 0 > 0 and f 00 < 0 we know that 9k such that sf (k) < (n + ) k. And given the Inada conditions _ _ –f 0 (0) = 1 and f (0) = 0 –9k such that sf (k) > (n + ) k. Therefore, k > 0 for low levels of k; and k > 0 _ for high levels of k. Given the continuity of f we know that 9k such that k = 0, i.e. the system is stable. 101 16 Higher order di¤erential equations We will discuss here only second order, since it is very rare to …nd higher order di¤erential equations in economics. The methods introduced here can be extended to higher order di¤erential equations. 16.1 Second order, constant coe¢ cients y 00 + a1 y 0 + a2 y = b ; where y = y (t) y0 = dy=dt y 00 = d2 y=dt2 ; 102 and a, b, and c are constants. The solution will take the form y = yp + yc ; where the particular solution, yp , characterizes a stable point and the complementary function, yc , charac- terizes dynamics/transitions. The particular solution. We start with the simplest solution possible; if this fails, we move up in the degree of complexity. If a2 6= 0, then yp = b=a2 is solution, which implies a stable point. b If a2 = 0 and a1 6= 0, then yp = a1 t : b If a2 = 0 and a1 = 0, then yp = 2 t2 : In the latter solutions, the "stable point" is moving. Recall that this solution is too restrictive, because it constrains the coe¢ cients in the di¤erential equation. The complementary function solves the homogenous equation y 00 + a1 y 0 + a2 y = 0 : We "guess" y = Aert which implies y0 = rAert y 00 = r2 Aert and thus y 00 + a1 y 0 + a2 y = A r2 + a1 r + a2 ert = 0 : Unless A = 0, we must have r2 + a1 r + a2 = 0 : The roots are p a1 a2 1 4a2 r1;2 = : 2 For each root ri there is a potentially di¤erent Ai coe¢ cient. So there are two possible solutions, y1 = A1 er1 t y2 = A2 er2 t : 103 But we cannot just chose one solution, because this will restrict the coe¢ cients in the original di¤erential equation. Thus, we have yc = A1 er1 t + A2 er2 t : Given two conditions on y – i.e. two values of either one of y, y 0 or y 00 at some point in time – we can pin down A1 and A2 . There are three options for the composition of the roots: Two distinct real roots: r1 ; r2 2 R and r1 6= r2 . This will give us values for A1 and A2 , given two conditions on y. yc = A1 er1 t + A2 er2 t : Repeated real root: r1 = r2 2 R, r = a1 =2. It might seem that we can just add up the solution as before, but this will restrict the coe¢ cients in the original di¤erential equation. This is so because in yc = (A1 + A2 ) er2 t we cannot separately identify A1 from A2 . We guess again: y1 = A1 ert y2 = A2 t ert : This turns out to work, because both solve the homogenous equation. You can check this. Thus for repeated real root the complementary function is yc = A1 er1 t + A2 ter2 t : p Complex roots, r1;2 = r bi, i = 1, a2 < 4a2 . This gives rise to oscillating dynamics 1 yc = ert [A1 cos (bt) + A2 sin (bt)] We do not discuss in detail here. Stability: does yc ! 0? r1 ; r2 2 R: need both r1 ; r2 < 0. r1 = r2 = r 2 R: need r < 0. Complex roots: need r < 0. 104 16.2 Di¤erential equations with moving constant y 00 + a1 y 0 + a2 y = b (t) ; where a1 and a2 are constants. We require that b (t) takes a form that combines a …nite number of "elementary functions", e.g. ktn , ekt , etc. We …nd yc in the same way as above, because we consider the homogenous equation where b (t) = 0. We …nd yp by using some educated guess and verify our guess by using the method of undetermined coe¢ cients. There is no general solution procedure for any type of b (t). Example: polynomial b (t): y 00 + 5y 0 + 3y = 6t2 t 1: Guess: y p = ' 2 t2 + ' 1 t + ' 0 : This implies 0 yp = 2'2 t + '1 00 yp = 2'2 : Plug this into yp to get y 00 + 5y 0 + 3y = 2'2 + 5 (2'2 t + '1 ) + 3 '2 t2 + '1 t + '0 = 3'2 t2 + (10'2 + 3'1 ) t + (2'2 + 5'1 + 3'0 ) : we need to solve 3'2 = 6 10'2 + 3'1 = 1 2'2 + 5'1 + 3'0 = 1: This gives '2 = 2, '1 = 7, '0 = 10. Thus yp = 2t2 7t + 10 : But this may not always work. For instance, if y 00 + a1 y 0 + a2 y = t 1 : 1 Then no guess of the type yp = 't or yp = ' ln t will work. Example: missing y (t) and polynomial b (t) y 00 + 5y 0 = 6t2 t 1: 105 The former type of guess, y p = ' 2 t2 + ' 1 t + ' 0 ; will not work, because '2 will never show up in the equation, so cannot be recovered. Instead, try y p = t ' 2 t2 + ' 1 t + ' 0 : If this fails, try y p = t2 '2 t2 + '1 t + ' 0 ; and so on. Example: exponential b (t) y 00 + a1 y 0 + a2 y = Bert : Guess: yp = Atert with the same r and look for solutions for A. The guess yp = Aert will not work. E.g. y 00 + 3y 0 4y = 2e 4t : Guess: 4t yp = Ate 0 4t 4t 4t yp = Ae + 4Ate = Ae (1 4t) 00 4t 4t 4t yp = 4Ae (1 4t) + 4Ae = Ae ( 8 + 16t) : Plug in the guess y 00 + 3y 0 4y = Ae 4t ( 8 + 16t) + 3Ae 4t (1 4t) + 4Ate 4t 4t = Ae ( 8 + 16t + 3 12t 4t) 4t = 5Ae We need to solve 4t 4t 5Ae = 2e so A = 0:4 and 4t yp = 0:4te : 106 17 First order di¤erence equations yt+1 + ayt = c : As with di¤erential equations, we wish to trace out a path for some variable y over time, i.e. we seek y (t). But now time is discrete, which gives rise to some peculiarities. De…ne yt yt+1 yt ; (not the standard notation) which is like yt yt+ yt = ; t where = 1. 17.1 Backward iteration 1. yt+1 + yt = c. y1 = y0 + c y2 = y1 + c = y0 + c + c = y0 + 2c y3 = y2 + c = y0 + 2c + c = y0 + 3c . . . yt = y0 + ct : 2. ayt+1 byt = 0, a 6= 0. Then yt+1 = kyt , where k = b=a. y1 = ky0 y2 = ky1 = k 2 y0 . . . yt = k t y0 : 17.2 General solution yt+1 + ayt = c ; where a 6= 0. The solution method involves splitting the solution into two: y (t) = yc (t) + yp (t) ; where yp (t) is a particular solution and yc (t) is a complementary function. 107 yc (t) solves the homogenous equation yt+1 + ayt = 0 : Guess yt = Abt so that yt+1 + ayt = 0 implies Abt+1 + aAbt = 0 b+a = 0 b = a: t yc (t) = A ( a) ; where a 6= 0. a 6= 1. yp (t) solves the original equation for a stationary solution, yt = k, a constant. This implies k + ak = c c k = : 1+a So that c yp = ; a 6= 1: 1+a a= 1. Guess yp (t) = kt. This implies k (t + 1) kt = c k = c: So that yp = ct ; a = 1: The general solution is t c A ( a) + 1+a if a 6= 1 yt = yc (t) + yp (t) = : A + ct if a= 1 Given an initial condition y (0) = y0 , then for a 6= 1 c c y0 = A + ) A = y0 : 1+a 1+a 108 for a = 1 y0 = A : The general solution is ( h i c t c y0 1+a ( a) + 1+a if a 6= 1 yt = : y0 + ct if a= 1 For a 6= 1 we have h i c t t yt = y0 ( a) + 1 ( a) ; 1+a c which is a linear combination of the initial point and the stationary point 1+a . And if a 2 ( 1; 1), then this process is stable. Otherwise it is not. For a = 1 and c 6= 1 the process is never stable. Example: yt+1 5yt = 1 : First, notice that a 6= 1 and a 6= 0. yc solves yt+1 5yt = 0 : Let yc (t) = Abt , so that Abt+1 5Abt = 0 Abt (b 5) = 0 b = 5; so that yc (t) = A5t : yp = k solves k 5k = 1 k = 1=4 ; so that yp = 1=4. yt = yc (t) + yp (t) = A5t 1=4 : Given y0 = 7=4 we have A = 2, which completes the solution. 109 17.3 Dynamic stability Given c c y t = y0 bt + ; 1+a 1+a the dynamics are governed by b (= a). 1. b < 0 will give rise to oscillating dynamics. c 1 < b < 0: oscillations diminish over time. In the limit we converge on the stationary point 1+a . b= 1: constant oscillations. b< 1: growing oscillations over time. The process is divergent. 2. b = 0 and b = 1: no oscillations, but this is degenerate. b = 0 means a = 0, so yt = c. b = 1 means a = 1, so yt = y0 + ct. c 3. 0 < b < 1 gives convergence to the stationary point 1+a . 4. b > 1 gives divergent dynamics. Only jbj < 1 gives convergent dynamics. 17.4 Application: cobweb model This is an early model of agriculture markets. Farmers determined supply last year based on the prevailing price at that time. Consumers determine demand based on current prices. Thus, three equations complete the description of this model s supply : qt+1 = s (pt ) = + pt d demand : qt+1 = d (pt+1 ) = pt+1 s d equilibrium : qt+1 = qt+1 ; where ; ; ; > 0. Imposing equilibrium: + pt = pt+1 + pt+1 + pt = : | {z } | {z } a c The solution to this di¤erence equation is t + + pt = p0 + : + + 110 The process is convergent (stable) i¤ j j < j j. Since both are positive, we need < . Interpretation: what are and ? These are the slopes of the demand and supply curves, respectively. If follows that if the slope of the supply curve is lower than that of the demand curve, then the process if convergent. I.e., as long as the farmers do not "overreact" to current prices next year, the market will converge on a happy stable equilibrium price and quantity. Conversely, as long as consumers are not "insensitive" to prices, then... 17.5 Nonlinear di¤erence equations We will use only a qualitative/graphic approach and restrict to autonomous equations, in which t is not explicit. Let yt+1 = ' (yt ) : Draw a phase diagram with yt+1 on the vertical axis and yt on the horizontal axis and the 45 degree ray starting from the origin. For simplicity, y > 0. A stationary point satis…es y = ' (y). But sometimes the stationary point is not stable. If j'0 (y)j < 1 at the stationary point, then the process is stable. More generally, as long as j'0 (yt )j < 1 the process is stable, i.e. it will converge to some stationary point. When j'0 (yt )j 1 the process will diverge. 111 Example: Galor and Zeira (1993), REStud. 112 18 Phase diagrams with two variables (19.5) We now analyze a system of two autonomous di¤erential equations: _ x = F (x; y) _ y = G (x; y) : _ _ First we …nd the x = 0 and y = 0 loci by setting F (x; y) = 0 G (x; y) = 0: Apply the implicit function theorem separately to the above, which gives rise to two (separate) functions: _ x = 0 : y = fx=0 (x) _ _ y = 0 : y = gy=0 (x) ; _ 113 where Fx f0 = Fy Gx g0 = : Gy Now suppose that we have enough information about F and G to characterize f and g. And suppose that f and g intersect, which is the interesting case. This gives rise to a stationary point, in which both x and y are constant: fx=0 (x ) = gy=0 (x ) _ _ ) y : There are two interesting cases, although you can characterize the other ones, once you do this. Case 1: dynamic stability. Fx < 0; Fy > 0 Gx > 0; Gy < 0 : Both f and g are upward sloping. First, notice that _ _ in all points above fx=0 x < 0 and in all points below fx=0 x > 0. _ _ _ _ in all points above gy=0 y < 0 and in all points below gy=0 y > 0. _ _ Given an intersection, this gives rise to four regions in the (x; y) space: _ _ 1. Below fx=0 and above gy=0 : x < 0 and y < 0. _ _ _ _ 2. Above fx=0 and above gy=0 : x > 0 and y < 0. _ _ _ _ 3. Above fx=0 and below gy=0 : x > 0 and y > 0. _ _ _ _ 4. Below fx=0 and below gy=0 : x < 0 and y > 0. _ _ This gives rise to a stable system. From any point in the (x; y) space we converge to (x ; y ). 114 _ _ Given the values that x and y take (given the direction in which the arrows point in the …gure), we can draw trajectories. In this case, all trajectories will eventually arrive at the stationary point at the intersection _ _ of x = 0 and y = 0. _ Notice that at the point in which we cross the x = 0 the trajectory is vertical. Similarly, at the point _ in which we cross the y = 0 the trajectory is horizontal. Case 2: saddle point. Fx > 0; Fy < 0 Gx < 0; Gy > 0 : Both f and g are still upward sloping, but now the pattern is di¤erent, because gy=0 crosses fx=0 at a steeper _ _ slope. Notice that _ _ in all points above fx=0 x < 0 and in all points below fx=0 x > 0. _ _ _ _ in all points above gy=0 y > 0 and in all points below gy=0 y < 0. _ _ Given an intersection, this gives rise to four regions in the (x; y) space: _ _ 1. Below fx=0 and above gy=0 : x > 0 and y > 0. _ _ _ _ 2. Above fx=0 and above gy=0 : x < 0 and y > 0. _ _ _ _ 3. Above fx=0 and below gy=0 : x < 0 and y < 0. _ _ 115 _ _ 4. Below fx=0 and below gy=0 : x > 0 and y < 0. _ _ This gives rise to an unstable system. However, there is a stationary point at the intersection, (x ; y ). But in order to converge to (x ; y ) there are only two trajectories that bring us there, one from the region above fx=0 and below gy=0 , the other from the region below fx=0 and above gy=0 . These trajectories are _ _ _ _ called stable branches. If we are not on those trajectories, then we are on unstable branches. Note that being in either region does not ensure that we are on a stable branch, as the …gure illustrates. 19 Optimal control Like in static optimization problems, we want to maximize (or minimize) an objective function. The di¤erence is that the objective is the sum of a path of values at any instant in time; therefore, we must choose an entire path as a maximizer.1 The problem is generally stated as follows: Z T Choose u (t) to maximize F (y; u; t) dt 0 s.t. Law of motion _ : y = g (y; u; t) Initial condition : y (0) = y0 rT transversality condition : y (T ) e 0: 1 Thetheory behind this relies on "calculus of variations", which was …rst developed to compute trajectories of missiles (to the moon and elsewhere) in the U.S.S.R. 116 where r is some average discount rate that is relevant to the problem. To this we need to sometimes add Terminal condition : y (T ) = yT Constraints on control : u (t) 2 U The function y (t) is called the state variable. The function u (t) is called the control variable. It is useful ow to think of of the state as a stock (like capital) and the control as a ‡ (like investment or consumption). Usually we will have F; g 2 C 1 , but in principle we could do without di¤erentiability with respect to u. I.e., we only need that the functions F and g are continuously di¤erentiable with respect to y and t. The transversality condition immediately implies that y (T ) 0, but also something more. It tells you rT that if y (T ) > 0, then its value at the end of the problem, y (T ) e must be zero. This will become clearer below, when we discuss the Lagrangian approach. If there is no law of motion for y, then we can solve the problem separately at any instant as a static problem. The value would just be the sum of those static values. There is no uncertainty here. To deal with uncertainty, wait for your next course in math. To ease notation we will omit time subscripts when there is no confusion. Example: the saving/investment problem for individuals. 1. Output: Y = F (K; L). 2. Investment/consumption: I = Y C = F (K; L) C. _ 3. Capital accumulation: K = I K. We want to maximize the present value of instantaneous utility from now (at t = 0) till we die (at some distant time T ). The problem is stated as Z T t Choose C (t) to maximize e U [C (t)] dt 0 s.t. _ K = I K K (0) = K0 K (T ) = KT : 117 19.1 s Pontryagin’ maximum principle and the Hamiltonian function De…ne the Hamiltonian function: H (y; u; t; ) = F (y; u; t) + g (y; u; t) : The function (t) is called the co-state function and also has a law of motion. Finding is part of the solution. The FONCs of this problem ensure that we maximize H at every point in time, and as a whole. If u is a maximizing plan then : H (y; u ; t; ) H (y; u; t; ) 8u 2 U (i) @H or : = 0 if F; g 2 C 1 @u @H State equation (ii) : _ _ = y ) y = g (y; u; t) @ @H Costate equation (iii) : = _ ) _ + Fy + gy = 0 @y Transversality condition (iv) : (T ) = 0 : Conditions (ii)+(iii) are a system of …rst order di¤erential equations that can be solved explicitly if we have functional forms and two conditions: y (0) = y0 and (T ) = 0. Note that has the same interpretation as the Lagrange multiplier: it is the shadow cost of the constraint is at any instant. We adopt the convention that y (0) = y0 is always given. There are a few way to introduce terminal conditions, which gives the following taxonomy 1. When T is …xed, (a) (T ) = 0, y (T ) free. (b) y (T ) = yT , (T ) free. (c) y (T ) ymin (or y (T ) ymax ), (T ) free. Add the following complementary slackness conditions: y (T ) ymin (T ) 0 (T ) (y (T ) ymin ) = 0 2. T is free and y (T ) = yT . Add H (T ) = 0. 3. T Tmax (or T Tmin ) and y (T ) = yT . Add the following complementary slackness conditions: H (T ) 0 T Tmax H (T ) (T Tmax ) = 0 118 19.2 The Lagrangian approach The problem is Z T Choose u (t) to maximize F (y; u; t) dt 0 s.t. _ y = g (y; u; t) rT y (T ) e 0 y (0) = y0 : _ _ You can think of y = g (y; u; t) as an inequality y g (y; u; t). We can write this up as a Lagrangian. For this we need Lagrange multipliers for the law of motion constraint at every point in time, as well as an additional multiplier for the transversality condition: Z T Z T rT L = F (y; u; t) dt + (t) [g (y; u; t) _ y] dt + y (T ) e 0 0 Z T Z T rT = [F (y; u; t) + (t) g (y; u; t)] dt _ (t) y (t) dt + y (T ) e : 0 0 Using integration by parts we have Z Z ydt = _ y+ _ ydt so that Z T Z T T _ (t) y (t) dt + y (T ) e rT L = [F (y; u; t) + (t) g (y; u; t)] dt [ (t) y (t)j0 + 0 0 Z T Z T [F (y; u; t) + (t) g (y; u; t)] dt (T ) y (T ) + (0) y (0) + _ (t) y (t) dt + y (T ) e rT : 0 0 Before writing down the FONCs for the Lagrangian, recall that H (y; u; t; ) = F (y; u; t) + (t) g (y; u; t) : The FONCs for the Lagrangian are (i) : Lu = Fu + gu = 0 (ii) : L = g _ y = 0 (in the original Lagrangian) (iii) : Ly = Fy + gy + _ = 0 : These imply (are consistent with) (i) : Hu = Fu + gu = 0 _ (ii) : H = g = y (iii) : Hy = Fy + gy = _ : 119 The requirement that y (0) = y0 can also be captured in the usual way, as well as y (T ) = yT , if it is required. The transversality condition is captured by the complementary slackness conditions rT y (T ) e 0 0 rT y (T ) e = 0: rT We see here that if y (T ) e > 0, then its value, , must be zero. 19.3 Autonomous problems In these problems t is not an explicit argument. Z T Choose u to maximize _ F (y; u) dt s.t. y = g (y; u) 0 plus boundary conditions. The Hamiltonian is thus H (y; u; ) = F (y; u) + g (y; u) : These problems are easier to solve and are amenable to analysis by phase diagrams. Example: the cake eating problem Objective: You want to eat your cake in an optimal way, maximizing your satisfaction from eating it, starting now (t = 0) and …nishing before bedtime, at T . The cake starts at size S0 . _ When you eat cake, the size diminishes by the amount that you ate: S = C. You like cake, but less so when you eat more: U 0 (C) > 0, U 00 (C) < 0. The problem is Z T Choose C to maximize U (C) dt s.t. 0 _ S = C S (0) = S0 S (T ) 0: This is an autonomous problem. The Hamiltonian is H (C; S; ) = U (C) + [ C] : 120 FONCs: @H (i) : = U 0 (C) =0 @C @H _ (ii) : = C=S @ @H (iii) : =0= _ @S (iv) : S (T ) 0; (T ) 0; S (T ) (T ) = 0 : From (iii) it follows that is constant. From (i) we have U 0 (C) = , and since is constant, C is constant too. Then given a constant C we get from (ii) that S=A Ct : And given S (0) = S0 we have S = S0 Ct : But we still do not know what is C, except that it is constant. So we solve for the complementary slackness conditions, i.e., will we leave leftovers? Suppose > 0. Then S (T ) = 0. Therefore 0 = S0 CT ; which gives S0 C= : T Suppose = 0. Then it is possible to have S (T ) > 0. But then we get U 0 = 0 –a contradiction. The solution is thus C (t) = S0 =T S (t) = S0 (S0 =T ) t 1 (t) = (U 0 ) (S0 =T ) : at If we allowed a ‡ part in the utility function after some satiation point, then we could have a solution with leftovers S (T ) > 0. In that case we would have more than one optimal path: all would be global at because with one ‡ part U is still quasi concave. 19.3.1 Anecdote: the value of the Hamiltonian is constant in autonomous problems We demonstrate that on the optimal path the value of the Hamiltonian function is constant. H (y; u; t; ) = F (y; u; t) + g (y; u; t) : The derivative with respect to time is 121 dH = H u u + Hy y + H _ + Ht : _ _ dt The FONCs were Hu = 0 Hy = _ H _ = y: Plugging this into dH=dt gives dH @H = : dt @t This is in fact a consequence of the envelope theorem, although not in a straightforward way. If time is not @H explicit in the problem, then @t = 0, which implies the statement above. 19.4 Current value Hamiltonian Many problems in economics involve discounting (as we saw above), so the problem is not autonomous. However, usually the only place that time is explicit is in the discount factor, Z T Z T F (y; u; t) dt = e rt G (y; u) dt : 0 0 You can try to solve those problems "as-is", but an easier way (especially if the costate is of no particular interest) is to use the current value Hamiltonian: e H = ert H = G (y; u) + 'g (y; u) ; where ' = ert : A maximizing plan u satis…es the following FONCs: e (i) e : H (y; u ; ') H (y; u; ') 8u 2 U @He or : e = 0 if H; g 2 C 1 @u @He State equation (ii) : _ _ = y ) y = g (y; u) @' @He Costate equation (iii) : _ _ = ' + r' ) ' r' + Fy + gy = 0 @y e Transversality condition (iv) : ' (T ) = 0 or H (T ) = 0 or other. Example: the cake eating problem with discounting 122 We now need to choose a functional form for the instantaneous utility function. The problem is Z T rt Choose C to maximize e ln (C) dt s.t. 0 _ S = C S (0) = S0 S (T ) 0: We write the present value Hamiltonian e H = ln C + ' [ C] FONCs: e @H 1 = '=0 @C C e @H = _ C=S @' e @H = 0= _ ' + r' @S S (T ) 0; ' (T ) 0; S (T ) ' (T ) = 0 : From (iii) we have _ ' =r ; ' hence ' = Bert ; for some B. From (i) we have 1 1 rt C= = e : ' B From (ii) we have _ S = C Z t Z t _ Sdt = Cdt 0 0 Z t S (t) = A+ Cdt ; 0 which, together with S0 implies Z t S (t) = S0 Cdt ; 0 123 1 rt which makes sense. Now, using C = B e we get Z t 1 rz S (t) = S0 B e dz 0 t 1 1 rz = S0 B e r 0 1 1 rt 1 r0 = S0 B e + e r r 1 rt = S0 1 e rB Suppose ' (T ) = 0. Then B = 0 and C (T ) = 1 –not possible. So ' (T ) > 0, which implies S (T ) = 0. Therefore 1 rT 0 = S0 1 e rB r 1 e rT B = S0 Therefore S0 rt C= rT ] e ; r [1 e which is decreasing, and rT r 1 e '= ert ; S0 which is increasing. And …nally rt 1 e S (t) = S0 1 rT : 1 e This completes the characterization of the problem. 19.5 In…nite time horizon s When the problem’ horizon is in…nite, i.e. never ends, we need to modify the transversality condition. These are lim (T ) y (T ) = 0 T !1 for the present value Hamiltonian, and rT lim ' (T ) e k (T ) = 0 T !1 for the current value Hamiltonian. 19.6 The neoclassical growth model 1. Preferences: u (C), u0 > 0, u00 < 0. Inada conditions: u (0) = 0, u0 (0) = 1, u0 (C) ! 0 as C ! 1. 124 2. Aggregate production function: Y = F (K; L), CRS, Fi > 0, Fii < 0. Given this we can write the per-worker version y = f (k), where f 0 > 0, f 00 < 0 and y = Y =L, k = K=L. Inada conditions: f (0) = 0, f 0 (0) = 1, f 0 (k) ! 0 as k ! 1. _ 3. Capital accumulation: K = I K=Y C K. As we saw in the Solow model, we can write this _ in per worker terms k = f (k) c (n + ) k, where n is the constant growth rate of labor force. 4. There cannot be negative consumption, but also, once output is converted into capital, we cannot eat it. This can be summarized in 0 C F (K; L). This is an example of a restriction on the control variable. s 5. A social planner chooses a consumption plan to maximize everyone’ welfare, in equal weights. The objective function is Z Z 1 1 nt t rt V = L0 e e U (c) dt = e U (c) dt ; 0 0 where we normalize L0 = 1 and we set r = n > 0, which ensures integrability. Notice that everyone gets the average level of consumption c = C=L. The problem is Choose c to maximize V s.t. _ k = f (k) c (n + ) k 0 c f (k) k (0) = k0 Write down the current value Hamiltonian H = u (c) + ' [f (k) c (n + ) k] : FONCs: Hc = u0 (c) '=0 H' = [f (k) c _ (n + ) k] = k Hk = ' [f 0 (k) (n + )] = r' _ ' rT lim ' (T ) e k (T ) = 0 T !1 Ignore for now 0 c f (k). The transversality condition here is a su¢ cient condition for a maximum, although in general this speci…c condition is not necessary. If this was a present value Hamiltonian the same transversality condition would be limT !1 (T ) k (T ) = 0, which just means that the value of an additional unit of capital in the limit is zero. 125 From Hc we have u0 (c) = '. From Hk we have _ ' = [f 0 (k) (n + + r)] : ' We want to characterize the solution qualitatively using a phase diagram. To do this, we need two equations: one for the state, k, and one for the control, c. Notice that ' = u00 (c) c ; _ _ so u00 (c) c _ = [f 0 (k) (n + + r)] : u0 (c) Rearrange to get c _ u0 (c) 0 = [f (k) (n + + r)] : c cu00 (c) Notice that cu00 (c) u0 (c) is the coe¢ cient of relative risk aversion. Let c1 u (c) = : 1 This is a class of constant relative relative risk aversion, or CRRA, utility functions, with coe¢ cient of RRA = . Eventually, our two equations are _ k = f (k) c (n + ) k _ c 1 0 = [f (k) (n + + r)] : c From this we derive _ k = 0 : c = f (k) (n + ) k c = _ 0 : f 0 (k) = n + + r : _ The c = 0 locus is a vertical line in the (k; c) space. Given the Inada conditions and diminishing returns to _ capital, we have that the k = 0 locus is hump shaped. Since r > 0, the peak of the hump is to the right of _ the vertical c = 0 locus. _ The phase diagram features a saddle point, with two stable branches. If k is to the right of the c = 0 _ _ _ = 0 locus we have locus, then c < 0 and vice versa for k to the left of the c = 0 locus. For c above the k _ _ k < 0 and vice versa for c below the k = 0 locus. See textbook for …gure. De…ne the stationary point as (k ; c ). Suppose that we start with k0 < k . Then the optimal path for consumption must be on the stable branch, i.e. c0 is on the stable branch, and will eventually go to c . 126 The reason is that any other choice is not optimal. Higher consumption will eventually lead to depletion of the capital stock, which eventually leads to no output and therefore no consumption (U.S.A.). Too little consumption will lead …rst to an increase in the capital stock and an increase in output, but eventually this is not sustainable as the plan requires more and more consumption forgone to keep up with e¤ective depreciation (n + ) and eventually leads to zero consumption as well (U.S.S.R.). One can do more than just analyze the phase diagram. First, given functional forms we can compute the exact paths for all dynamic variables. Second, we could linearize (a …rst order Taylor expansion) the system of di¤erential equations around the saddle point to compute dynamics around that point. 127