ECON0702: Mathematical Methods in Economics

Transcription

1 ECON0702: Mathematical Methods in Economics Yulei Luo SEF of HKU January 12, 2009 Luo, Y. (SEF of HKU) MME January 12, / 35

2 Course Outline Economics: The study of the choices people (consumers, rm managers, and governments) make to attain their goals, given their scarce resources. Economic model: Simpli ed version of reality used to analyze real-world economic situations. This course will mainly focus on how to use mathematical methods to solve economic models. After solving the models, you may use economic data to test the hypothesis derived from models. If you are interested in testing economic models, you may take Econ0701: Econometrics. Luo, Y. (SEF of HKU) MME January 12, / 35

3 Course Outline Static Analysis Main objective: characterize partial and general equilibrium: demand=supply and solve for equilibrium prices and quantities. Mathematical tools: matrix algebra Economic applications: market equilibrium models and national-income models Comparative-Static Analysis Main objective: compare di erent equilibrium states that are associated with di erent sets of values of parameters or exogenous variables (i.e., they are given and not determined by the model structure). Mathematical tools: calculus (di erentiation, partial di erentiation, etc.) Economic applications: market equilibrium models and national-income models Luo, Y. (SEF of HKU) MME January 12, / 35

4 (Continued.) Static Optimization Analysis Main objective: characterize the optimizing behavior of consumers and rms behind equilibrium at one period (static, atemporally). Mathematical tools: unconstrained and constrained static optimization Economic applications: utility maximization, pro t maximization, and cost minimization. Dynamic Analysis Main objective: characterize the evolution of economic variables over time (dynamic, intertemporally). Mathematical tools: rst-order and higher-order di erence (discrete-time) or di erential (continuous-time) equations. Economic applications: economic growth models, cobweb model, etc. Dynamic Optimization Main objective: characterize the optimal behavior of agents over time. Mathematical tools: optimal control theory (the Lagrange multiplier method) or dynamic programming (The Bellman equation method) Economic applications: optimal economic growth models, life-cycle optimal consumption-saving models, optimal investment models. Luo, Y. (SEF of HKU) MME January 12, / 35

5 The meaning of equilibrium In essence, an equilibrium for a speci c model is a situation that is characterized by a lack of tendency to change. It is for this reason that the analysis of equilibrium is referred to as statics. We will discuss two examples of equilibrium. One is the equilibrium attained by a market under given demand and supply conditions. The other is the equilibrium of national income under given conditions of total consumption and investment patterns. Luo, Y. (SEF of HKU) MME January 12, / 35

6 Partial equilibrium model (linear) In a static equilibrium model, the standard problem is to nd the set of values of the endogenous variables (i.e., they are not given and are determined by the model structure) which will satisfy the equilibrium conditions of the model. Partial equilibrium market model: a model of price determination in a single market. Three variables: Q d = the quantity demanded of the commodity Q s = the quantity supplied of the commodity P = the price of commodity. And the equilibrium condition is The model setup is then Q d = Q s. (1) Q d = Q s (2) Q d = a bp (3) Q s = c + dp (4) where a, b, c, and d are all positive. Luo, Y. (SEF of HKU) MME January 12, / 35

7 One way to nd the equilibrium is by successive elimination of variables and equations through substitution. From Q d = Q s, we have Since b + d 6= 0, the equilibrium price is then a bp = c + dp. (5) P = a + c b + d, (6) and the equilibrium quantity can be obtained by substituting P into either Q d or Q s : Q = ad bc b + d, (7) where we assume that ad meaningful. bc > 0 to make the model economically Luo, Y. (SEF of HKU) MME January 12, / 35

8 Partial equilibrium model (Nonlinear) The partial equilibrium market model can be nonlinear. The model setup is then Q d = Q s (8) Q d = 4 P 2 (9) Q s = 1 + 4P (10) As before, the three-equation system can be reduced to a single equation by substitution: P 2 + 4P 5 = 0, (11) which can be solved by applying the quadratic formula: P = 1 or 5. (12) Note that only the rst root is economically meaningful. Luo, Y. (SEF of HKU) MME January 12, / 35

9 General market equilibrium In the above, we only consider a single market. In the real world, there would normally exist many goods. Thus, a more realistic model for the demand and supply functions of a commodity should take into account the e ects not only of the price of the commodity itself but also of the prices of other commodities. As a result, the price and quantity variables of multiple commodities must enter endogenously into the model. Consequently, the equilibrium condition of an n commodity market model will involve n equations, one for each commodity, in the form where i = 1,, n, E i = Q i,d Q i,s = 0, (13) Q i,d = Q i,d (P 1,, P n ) and Q i,s = Q i,s (P 1,, P n ) (14) are the demand and supply functions of commodity i. Thus solving n equation system for P : if a solution does exist. E i (P 1,, P n ) = 0 (15) Luo, Y. (SEF of HKU) MME January 12, / 35

10 Two commodity market model The model setup is as follows Q 1,d = Q 1,s (16) Q 1,d = a 0 + a 1 P 1 + a 2 P 2 (17) Q 1,s = b 0 + b 1 P 1 + b 2 P 2 (18) Q 2,d = Q 2,s (19) Q 2,d = α 0 + α 1 P 1 + α 2 P 2 (20) Q 2,s = β 0 + β 1 P 1 + β 2 P 2, (21) which implies that (a 0 b 0 ) + (a 1 b 1 ) P 1 + (a 2 b 2 ) P 2 = 0 (22) (α 0 β 0 ) + (α 1 β 1 ) P 1 + (α 2 β 2 ) P 2 = 0 (23) Luo, Y. (SEF of HKU) MME January 12, / 35

11 (Continued.) Denote that c i = a i b i ; d i = α i β i, (24) where i = 0, 1, 2. The above linear equations can be written as c 1 P 1 + c 2 P 2 = c 0 (25) d 1 P 1 + d 2 P 2 = d 0, (26) which can be solved by further elimination of variables: P 1 = c 2d 0 c 0 d 2 c 1 d 2 c 2 d 1 ; P 2 = c 0d 1 c 1 d 0 c 1 d 2 c 2 d 1. (27) Note that for the n commodity linear market model, we have n linear equations and n unknowns. However, an equal number of equations and unknowns doesn t necessarily guarantee the existence of a unique solution. Only when the number of unknowns equals to the number of functional independent equations, the solution exists and is unique. We thus need systematic methods to test the existence of a unique solution. Luo, Y. (SEF of HKU) MME January 12, / 35

12 Equilibrium National-Income Model Consider a simple Keynesian national-income model, Y = C + I 0 + G 0 (Equilibrium condition) (28) C = a + by (The consumption function), (29) where Y and C stand for the endogenous variables national income and consumption expenditure, respectively, and I 0 and G 0 are the exogenously determined investment and government spending. Following the same procedure, the equilibrium income and consumption can solved as follows Y = a + I 0 + G 0 1 b (30) C = a + b (I 0 + G 0 ). (31) 1 b Luo, Y. (SEF of HKU) MME January 12, / 35

13 Why Matrix Algebra? As more and more commodities are incorporated into the model, it becomes more and more di cult to solve the model by substitution. A method suitable for handling a large system of simultaneous equations is matrix algebra. Matrix algebra provides a compact way of writing an equation system; it leads to a way to test the existence of a solution by evaluation of a determinant a concept closely related to that of a matrix; it also gives a method to nd that solution if it exists. Luo, Y. (SEF of HKU) MME January 12, / 35

14 Quick Review of Matrix Algebra Matrix: a matrix is simply a rectangular array of numbers. So any table of data is a matrix. 1 The size of a matrix is indicated by the number of its rows and the number of its columns. A matrix A with k rows and n columns is called a k n matrix. The number in row i and column j is called the (i, j) entry and is written as a ij. 2 Two matrices are equal if they both have the same size and if the corresponding entries in the two matrices are equal. 3 If a matrix contains only one column (row), it is called a column (row) vector. a11 a Example: 12 a1 is a 2 2 matrix. is a 2 1 matrix or a a 21 a 22 a 2 column vector. Luo, Y. (SEF of HKU) MME January 12, / 35

15 (Continued.) Addition: One can add two matrices with the samesize. E.g., a11 a a a + = 12. a 21 a a 12 a Subtraction: subtract matrices of the same size simply by subtracting their corresponding entries. E.g., a11 a a11 1 a = 12. a 21 a a 12 a 22 1 Scalar multiplication: matrices can be multiplied by ordinary numbers a11 a (scalar). E.g., α 12 αa11 αa = 12. a 12 a 22 αa 12 αa 22 Luo, Y. (SEF of HKU) MME January 12, / 35

16 (Continued.) Matrix multiplication: Not all pairs of matrices can be multiplied together, and the order in which matrices are multiplied can matter. We can de ne the matrix product AB if and only if Number of columns of A = Number of rows of B. (32) For the matrix product to exist, A must be k m and B must be m n. To obtain the (i, j) entry of AB, multiply the ith row of A and the jth column of B. If A is k m and B is m n, then the product AB is k n. That is, the product matrix AB inherits the number of its rows from A and the number of its columns from B. E.g., a a1 a 1 2 = a1 2 + a2 2. (33) a11 a 12 a 12 a 22 a 2 α1 α 2 = α1 a 11 + α 2 a 12 α 1 a 12 + α 2 a 22 (34) Luo, Y. (SEF of HKU) MME January 12, / 35

17 (Continued.) Linear dependence of vectors. A set of vectors v 1,, v n is said to be linearly dependent if and only if any one of them can be expressed as a linear combination of the remaining vectors; otherwise, they are linearly independent. For any vectors v 1,, v n to be linear independent, for any set of scalars k i, n i =1 k i v i = 0 if and only if k i = 0 for all i. Example: the three vectors v 1 = 2 7 1, v 2 = 8 4, v 3 = 5 are linearly dependent since v 3 is a linear combination of v 1 and v 2 : (35) 3v 1 2v 2 = v 3 (36) Luo, Y. (SEF of HKU) MME January 12, / 35

18 (Continued.) Laws of matrix algebra (think of matrices as generalized numbers) Associative laws: (A + B) + C = A + (B + C ) (37) (AB)C = A(BC ) (38) Communication law for addition A + B = B + A (39) Distributive laws A(B + C ) = AB + AC (40) Note that Communication law for multiplication does NOT hold: AB 6= BA in general. (41) Luo, Y. (SEF of HKU) MME January 12, / 35

19 (Continued.) Transpose: the transpose of a k m matrix A is the m k matrix obtained by interchanging the rows and columns of A. It is also written as A T. E.g., T a1 a1 a 2 = (42) a 2 Square matrix: a matrix with the same numbers of rows and columns. E.g., a11 a 12. a 21 a 22 Determinant of a square matrix: For 1 1 matrix a, det(a) = a.for 2 2 a11 a matrix A = 12, a 21 a 22 det(a) = jaj = a 11 a 22 a 12 a 21 (43) which is just the product of two diagonal entries minus the product of two o -diagonal entries. Luo, Y. (SEF of HKU) MME January 12, / 35

20 (Continued.) Inverse: Let A be a square matrix. The matrix B is an inverse for A if 1 0 AB = BA = I where I = is called identity matrix (a square matrix 0 1 with ones in its principal diagonal and zeros everywhere else).if the matrix B exists, we say that A is invertible and denote B = A 1. Squareness is a necessary but not su cient condition for the existence of an inverse. If a square matrix A has an inverse, A is said to be nonsingular. If A has no inverse, it is said to be a singular matrix. Luo, Y. (SEF of HKU) MME January 12, / 35

21 Some Properties of The Transpose and the Inverse The operation of matrix transpose and matrix inverse satisfy (AB) 0 = B 0 A 0 (44) (A 0 ) 0 = A (45) A 0 A = 0 () A = A 0 = 0 (46) (A + B) 0 = A 0 + B 0 (47) (A 1 ) 0 = (A 0 ) 1 (48) (A 1 ) 1 = A (49) (AB) 1 = B 1 A 1 (50) provided that the objects above are well de ned. Luo, Y. (SEF of HKU) MME January 12, / 35

22 (Continued.) System of equations in matrix form: Consider the following two-equation (two unknown variables) system a 11 x 1 + a 12 x 2 = b 1 (51) a 21 x 1 + a 22 x 2 = b 2 (52) a11 a Let A = 12 x1 be the coe cient matrix of the system, x =, a 21 a 22 x 2 b1 and b =. The above equations can be written in a compact form: b 2 a11 a 12 x1 b1 = or Ax = b. (53) a 21 a 22 x 2 b 2 The solution to the above equations can be written as if A 1 exists. x =A 1 b, (54) Luo, Y. (SEF of HKU) MME January 12, / 35

23 Test for the Existence of the Inverse and Find the Inverse (Continued.) The necessary and su cient conditions for nonsingularity are that the matrix satis es the squareness and linear independence conditions (that is, its rows or equivalently, its columns are linearly independent.). An n n nonsingular matrix A has n linearly independent rows (or columns); consequently, it must have rank n. Consequently, an n n matrix having rank n must be nonsingular. The concept of determinant can be used to not only determine whether a square matrix is nonsingular, but also calculate the inverse of the matrix. Luo, Y. (SEF of HKU) MME January 12, / 35

24 (Continued.) The value of a determinant of order n can be found by the Laplace expansion of any row or any column as follows jaj = n a ij ( 1) i +j jm ij j, (55) j=1 where jm ij j is the minor of the element a ij and can be obtained by deleting the ith row and jth column of the determinant jaj and jc ij j = ( 1) i +j jm ij j is called cofactor. Example: for a 3 3 matrix A, its determinant has the value: a 11 a 12 a 13 jaj = a 21 a 22 a 23 a 31 a 32 a 33 = a 11 a 22 a 23 a 32 a 33 (56) a a 21 a a 31 a 33 + a 13 a 21 a 22 a 31 a 32. (57) Luo, Y. (SEF of HKU) MME January 12, / 35

25 Determinantal Criterion for Non-singularity Given a linear equation system Ax = b, where A is a square coe cient matrix, we have jaj 6= 0 () A is row or column independent () A is nonsingular () A 1 exists () A unique solution x =A 1 b exists. Luo, Y. (SEF of HKU) MME January 12, / 35

26 Finding the Inverse Matrix Each element a ij of n n matrix A has a cofactor jc ij j. A cofactor matrix C = jc ij j is also n n. The transpose C 0 is referred to as the adjoint of A and denoted by adj A. The procedure to calculate A 1 : 1 Find jaj 2 Find the cofactors of all elements of A and form C = jc ij j 3 Form C 0 = adj A 4 Determine A 1 = adj A jaj (58) Luo, Y. (SEF of HKU) MME January 12, / 35

27 Cramer s Rule Another way to calculate the inverse of a matrix A nn is to use the Cramer rule. Let B j be the n n matrix formed by taking A and substituting its j th column, a j, with the constants vector b. For example, for j = 2, B 2 = a 1 b a 3 a n (59) 2 3 a 11 b 1 a 13 a 1n = 6 a 21 b 2 a 23 a 2n (60) a n1 b n a n3 a nn Let jaj 6= 0 and jb j j be the determinants of A and B j, respectively. See any linear algebra textbook or pages in CW for proof. Cramer rule then says that the j th element of the solution x =A 1 b is given by x j = jb j j 8j = 1,, n. (61) jaj Luo, Y. (SEF of HKU) MME January 12, / 35

28 Reconsider the Two-commodity Model which can be rewritten as c1 c 2 P1 d 1 d 2 {z P 2 } {z } A x which implies that P1 P 2 = = c 1 P 1 + c 2 P 2 = c 0 (62) d 1 P 1 + d 2 P 2 = d 0, (63) = c 0, (64) d 0 {z } b 1 c1 c 2 c 0 = 1 d2 c 2 c 0 d 1 d 2 d 0 jaj d 1 c 1 d 0 " # c2 d 0 c 0 d 2 c 1 d 2 c 0 d 1 c 2 d 1 c 1 d 0. (65) c 1 d 2 c 2 d 1 Luo, Y. (SEF of HKU) MME January 12, / 35

29 Reconsider Equilibrium National-Income Model The model Y = C + I 0 + G 0 (Equilibrium condition) (66) C = a + by (The consumption function), (67) can be rewritten in a compact matrix form 1 1 Y I0 + G = 0, (68) b 1 C a {z } {z } {z } A x b which implies that Y C = 1 1 b 1 1 b 1 I0 + G 0 a Applying Cramer rule gives: Y = I0 + G 0 1 / 1 1 a 1 b 1 " = a+i 0 +G 0 1 b a+b(i 0 +G 0 ) 1 b #. (69) = a + I 0 + G 0. (70) 1 b Luo, Y. (SEF of HKU) MME January 12, / 35

30 IS-LM Model: Closed Economy Consider another linear model of macroeconomy: the IS-LM model made up of two sectors: the real goods sector and the monetary sector. The goods market involves the following equations: Y = C + I + G (71) C = a + b(1 t)y (72) I = d ei (73) G = G 0, (74) where Y, C, I, and i (the interest rate) are endogenous variables, G 0 is the exogenous variable, and a, b, d, e, and t are structural parameters. In the newly introduced money market, we have M d = M s (Equilibrium condition in the money market) (75) M d = ky li (Aggregate money demand) (76) M s = M 0, where M 0 is the exogenous money supply and k and l are parameters. Note that these three equations can be reduced to (77) M 0 = ky li (78) Luo, Y. (SEF of HKU) MME January 12, / 35

31 (Continued.) Together, the two sectors give the following system of equations which can be written in a compact matrix form: Y 6 b(1 t) C e 54 I 5 k 0 {z 0 l i } {z } A x Y C I = G 0 (79) b(1 t)y C = a (80) I + ei = d (81) ky li = M 0, (82) = G 0 a d M {z } b Luo, Y. (SEF of HKU) MME January 12, / 35

32 (Continued.) Applying Cramer rule gives: 2 G Y = 6 a d 0 1 e M l / b(1 t) e k 0 0 l = l (a + d + G 0) + em 0 ek + l [1 b(1 t)]. (83) Similarly, we may derive the expressions for other endogenous variables (C, I, i) in terms of both exogenous variables and structural parameters. It is clear that Cramer s rule is elegant, but if you have to invert a numerical matrix (e.g., a = 1, t = 0.1, b = 0.4, etc.), you d better to use Matlab, Gauss, Mathematica, etc. Luo, Y. (SEF of HKU) MME January 12, / 35

33 Note on Homogeneous-equation System The equation system Ax = b considered before can have any constants in the vector b. If b = 0, that is, every elements in the b vector is zero, the equation system becomes Ax = 0, (84) where 0 is a zero vector. This special case is referred to as a homogeneous-equation system. homogeneous here means that if all the variables x 1,, x n are multiplied by the same number, the equation system will remain valid. This is possible only if b = 0. Note that if the coe cient matrix A is nonsingular (that is, it is invertible), a homogeneous-equation system can yield only a trivial solution, that is, x = 0. (check it yourself) Luo, Y. (SEF of HKU) MME January 12, / 35

34 (Continued.) The only way to get a nontrivial solution from the above HE system is to have jaj = 0, that is, to have a singular coe cient matrix A. (85) No unique nontrivial solution exists for this HE system. For example, consider a 2 2 case: a 11 x 1 + a 12 x 2 = 0 (86) a 21 x 1 + a 22 x 2 = 0, a11 a assume that A = 12 is singular, so that jaj = 0. This implies that a 21 a 22 the row vector a 11 a 12 is a multiple of the row vector a21 a 22 ; consequently, one of the two equations is redundant. By deleting the second equation from the system, we end up with one equation in two variables and the solution is x 1 = ( a 12 /a 11 ) x 2, (87) which is well-de ned if a 11 6= 0 and represents an in nite number of solutions. Luo, Y. (SEF of HKU) MME January 12, / 35

35 Solution Outcomes for a Linear-equation System Ax = b Vector b b 6= 0 jaj 6=0 There exists a unique, nontrivial solution x 6= 0 jaj = 0 In nite numbers of solutions if equations are dependent No solution if equations are inconsistent b = 0 jaj 6=0 There exists a unique, trivial solution x = 0 jaj = 0 In nite numbers of solutions if equations are dependent Not possible if equations are inconsistent Luo, Y. (SEF of HKU) MME January 12, / 35