SS 2018 02.06.2018 1289BMMM01 Basismodul Mathematics/Methods Block 1: Mathematics for Economists Prüfer: Prof. Dr. Rainer Dyckerhoff Bitte füllen Sie die nachfolgende Zeile aus! Matrikelnummer (student ID number) Initialen (initials) Umfang des Aufgabenhefts: Bearbeitungsdauer: 12 Seiten 90 Minuten You can find an english translation of the following text on page 12. In dieser Klausur sind maximal 90 Punkte zu erreichen. Sie ist in der Regel bestanden, wenn mindestens 45 Punkte erreicht wurden. Bitte tragen Sie Ihre Matrikelnummer sowie Ihre Initialen in die oben dafür vorgesehenen Felder ein. Die Klausur besteht aus fünf Aufgaben. Alle fünf Aufgaben sind zu bearbeiten. Die Lösungen sind auf den jeweiligen Aufgabenblättern vollständig anzugeben. Außer den Endresultaten müssen auch Zwischenschritte angegeben werden. Sollte der Platz nicht reichen, setzen Sie die Bearbeitung auf einem Zusatzblatt fort. Kennzeichnen Sie das Zusatzblatt mit Matrikelnummer und Initialen und weisen Sie an der entsprechenden Stelle im Aufgabenheft auf das Zusatzblatt hin. Es darf ein eigenhändig handgeschriebenes DIN-A4-Blatt als Formelsammlung verwendet werden. Dabei ist nur das bereitgestellte Formular (Matrikelnummer und Initialen eintragen!) zulässig. Die Rückseite darf auch beschrieben sein. Die Formelsammlung muss mit der Klausur abgegeben werden. Abweichungen von diesen Vorgaben werden als Täuschungsversuch gewertet. Den auf den Klausurplätzen ausgelegten Mantelbogen können Sie als Konzeptpapier benutzen. Eintragungen darauf werden bei der Bewertung nicht berücksichtigt. Dieses Aufgabenheft einschließlich Ihrer Formelsammlung ist nach Bearbeitung der Klausuraufgaben in den auf den Klausurplätzen ausgelegten Mantelbogen einzulegen. Aufgabe Punkte 1 2 3 4 5 Gesamt c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 1
Problem 1 (8+5+5=18) Consider the optimization problem min/max x 2 + 2y 2 subject to x 2 y = 1. a) By using the Lagrange method, find all stationary points. b) For (at least) one of the stationary points found in part a) holds x > 0. Using the secondorder conditions, check whether there is a local maximum, a local minimum or no extremum at that point. c) For the stationary point considered in part b), calculate the approximate change of the optimum value if the constraint is changed to 1.2x 2 y = 1. Hint: If all of your stationary points have x 0, then you can choose any of your stationary points in parts b) and c). Solution: a) The Lagrangian is given by L(x, y, λ) = x 2 + 2y 2 + λ(1 x 2 + y). Thus, the first-order necessary conditions are given by: L x = 2x 2xλ = 0 2x(1 λ) = 0 L y = 4y + λ = 0 λ = 4y L λ = 1 x 2 + y = 0 From the first equation follows x = 0 or λ = 1. Case 1: If x = 0, then it follows from the constraint that y = 1 which implies λ = 4. Case 2: If λ = 1, then it follows from the second equation that y = 1/4. From the constraint then follows x 2 = 1 + y = 1 1 4 = 3 3 x = ± 4 2. All in all, we have three stationary points: (x, y) = (0, 1) with multiplier λ = 4 and (x, y) = (± 3/2, 1/4) with multiplier λ = 1. b) The modified Hessian is given by 0 2x 1 H L (λ, x, y) = 2x 2 2λ 0. 1 0 4 There is one stationary point with x > 0, namely (x, y) = ( 3/2, 1/4) with multiplier λ = 1. Inserting this point in the modified Hessian gives 3 ) 0 H L (1, 2, 1 3 1 = 3 0 0. 4 1 0 4 Here, n = 2 (two variables) and m = 1 (one constraint). Since n m = 2 1 = 1, there is only one minor to check, namely 3. Using Sarrus rule we get 3 = 4 3 ( 3) = 12 > 0. Since 3 > 0, there is a local minimum at ( 3/2, 1/4). c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 2
c) We consider the constraint in the form 1 αx 2 +y = 0. We have to compute the approximate change of the optimum value when α is changed from α = 1 to α = 1.2, i.e, by α = 0.2. The Lagrangian is given by Its derivative w.r.t. α is given by L(x, y, λ, α) = x 2 + 2y 2 + λ(1 αx 2 + y). L α (x, y, λ, α) = λx2. Inserting the stationary point considered in part b) and α = 1 gives L( 3 ) α 2, 1 4, 1, 1 = 1 ( 3 Therefore, the approximate change of the optimum value is 2 ) 2 = 3 4. L α α = 3 4 1 5 = 3 20 = 0.15. Hence, the optimum value decreases by about 0.15. c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 3
Problem 2 (5+4+7+2=18) Consider the inhomogeneous second-order linear difference equation y t + 6y t 1 + 25y t 2 = 8t + 2. a) Find the general solution to the associated homogeneous difference equation. For the representation of the solution, use only real basis functions. b) Find the solution to the homogeneous difference equation for the initial conditions y 0 = 3 and y 1 = 5. c) Find a particular solution to the inhomogeneous difference equation. d) Is the given inhomogeneous difference equation stable? Is the associated homogeneous difference equation stable? Justify your answers. Solution: a) The characteristic polynomial is given by z 2 + 6z + 25 = 0. Using the p-q-formula it follows that z 1,2 = 3 ± 9 25 = 3 ± 16 = 3 ± 4i. Computing the polar form of z 1 = 3 + 4i gives and r = z = ϕ = arg(z) = arccos Thus, the general solution is given by ( 3) 2 + 4 2 = 25 = 5 ( ) 3 = arccos ( 0.6) = 2.2143. 5 y t = γ 1 5 t cos(2.2143 t) + γ 2 5 t sin(2.2143 t), where γ 1, γ 2 R. b) From the initial condition y 0 = 3 follows y 0 = γ 1 5 0 cos(0) + γ 2 5 0 sin(0) = γ 1! = 3. From this and the initial condition y 1 = 5 we get and thus y 1 = 3 5 1 cos(2.2143) + γ 2 5 1 sin(2.2143)! = 5 γ 2 = 5 15 cos(2.2143) 5 sin(2.2143) = 7 2 = 3.5. Hence, the unique solution to the given initial conditions is y t = 3 5 t cos(2.2143 t) + 3.5 5 t sin(2.2143 t). c) Since the forcing function is b t = 8t + 2, we use as a trial solution a general polynomial of degree one, i.e., y t = a + bt. Inserting this in the difference equation gives Multiplying out gives (a + bt) + 6 ( a + b(t 1) ) + 25 ( a + b(t 2) ) = 8t + 2. a + bt + 6a + 6bt 6b + 25a + 25bt 50b = 8t + 2. c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 4
Collecting like powers of t gives Equating coefficients gives (32a 56b) + 32bt = 8t + 2. 32b = 8 and 32a 56b = 2. From the first equation follows b = 1/4. Inserting this in the second equation gives a = 2 + 56b 32 = 16 32 = 1 2. Therefore, a particular solution is given by y t = 0.5 + 0.25t. d) The homogeneous equation is not stable, since at least one zero (in fact both) of the characteristic polynomial has absolute value z 1,2 = 5 1. The inhomogeneous equation is not stable, since the inhomogeneous equation is stable if and only if the homogeneous equation is stable. c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 5
Problem 3 (2+4+3+2+7=18) I. For x > 0 consider the first-order linear differential equation y + 4 y x = x2 where y(1) = 1 2 a) Find the general solution to the associated homogeneous differential equation. b) Find a particular solution to the inhomogeneous differential equation. c) Find the unique solution to the inhomogeneous differential equation for the given initial condition. II. Consider the inhomogeneous system of first-order LDEs, y t = Ay t 1 + b t, where [ ] [ ] 3.4 6 3.3 A = and b 1.8 3.2 t = 2 t. 0.3 Solution: Hint: The eigenvalues of A are λ 1 = 0.2 and λ 2 = 0.4 with corresponding eigenvectors u 1 = [ 5, 3] T and u 2 = [2, 1] T respectively. a) Find the general solution to the homogeneous system. b) Find a particular solution to the inhomogeneous system. I. a) With the notation of the lecture we have a(x) = 4/x. Therefore, an antiderivative is given by A(x) = 4 ln(x). Hence, the general solution is given by y(x) = γe A(x) = γe 4 ln(x) = γx 4 = γ x 4, γ R. b) A particular solution can be found by variation of constants. With b(x) = x 2 we get γ(x) = b(x)e A(x) dx = x 2 x 4 dx = A particular solution is therefore given by c) The general solution is given by y p (x) = γ(x)e A(x) = x7 7 1 x 4 = x3 7. y(x) = x3 7 + γ x 4. x 6 dx = x7 7. From the initial condition we therefore get the following equation y(1) = 1 7 + γ! = 1 2 γ = 1 2 1 7 = 5 14. Thus, the unique solution is y(x) = x3 7 + 5 14x 4. II. a) The general solution to the homogeneous system is given by [ ] [ ] 5 y t = γ 1 0.2 t 2 + γ 3 2 ( 0.4) t, γ 1 1, γ 2 R. c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 6
b) Since the forcing function has the form b t = u2 t, we use y t = a2 t as a trial solution. Inserting this in the system gives Dividing by 2 t 1 yields a2 t = Aa2 t 1 + u2 t. 2a = Aa + 2u (A 2I)a = 2u. Solving the latter system with the Gauß-Jordan algorithm gives [ 5.4 6 6.6 1.8 1.2 0.6 ] [ ] [ ] 1.8 1.2 0.6 1.8 1.2 0.6 0 2.4 4.8 0 1 2 [ ] [ ] 1.8 0 1.8 1 0 1. 0 1 2 0 1 2 Therefore a = [ 1, 2] T and a particular solution of the system is given by y t = [ ] 1 2 t. 2 c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 7
Problem 4 (5+3+2+5+3=18) I. Consider the matrix 4 6 3 A = 0 2 0. 6 6 5 a) Find all eigenvalues of A and indicate the multiplicity of each eigenvalue. b) For one of the eigenvalues, compute its eigenspace: If there is a multiple eigenvalue, compute the eigenspace for this eigenvalue. If all the eigenvalues are simple eigenvalues, you can choose which eigenspace you compute. c) Is A diagonalizable? Give a short justification for your answer. d) By using the Cayley-Hamilton theorem, express the matrix A 5 as a linear combination of I, A and A 2. II. Let A be a real and symmetric n n matrix. What is known about the eigenvalues and eigenvectors of A? In particular, comment on Solution: the number of real eigenvalues, the dimensions of the associated eigenspaces, and the relation between eigenvectors that belong to different eigenvalues. I. a) The characteristic polynomial is the determinant of A λi. Using Sarrus rule we get 4 λ 6 3 det(a λi) = det 0 2 λ 0 6 6 5 λ = ( 4 λ)(2 λ)(5 λ) + 18(2 λ) = (2 λ)[( 4 λ)(5 λ) + 18] = (2 λ)[λ 2 λ 2]. Thus, λ 1 = 2. The zeros of the quadratic polynomial in brackets are given by λ 2,3 = 1 2 ± 1 4 + 2 = 1 2 ± 9 4 = 1 2 ± 3 2. Therefore, λ 2 = 2 and λ 3 = 1. Thus, there is a double eigenvalue λ 1,2 = 2 and a simple eigenvalue λ 3 = 1. b) We have to find the eigenspace associated with the eigenvalue λ 1,2 = 2. To find this eigenspace, we have to solve the homogeneous linear equation (A 2 I)x = 0. Applying the Gauß-Jordan algorithm gives 6 6 3 0 0 0 6 6 3 1 1 1 2 0 0 0 0 0 0 [ 1 1 1 2 Hence, the eigenspace is given by 1 1 V (2) = c 2 1 1 1 1 + c 2 0 : c 1, c 2 R 0 1 = c 1 1 + c 2 0 : c 1, c 2 R 0 2. c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 8 ]
c) From b) we know that the geometric multiplicity of λ 1,2 is equal to 2. The eigenvalue λ 3 = 1 has an algebraic multiplicity of 1 which implies that the geometric multiplicity is also 1. Therefore, the sum of the geometric multiplicities is 2+1 = 3 and thus equal to the dimension of the matrix A. Therefore, A is diagonalizable. d) The linear combination can be found by dividing λ 5 by the characteristic polynomial of A. Because of part a) the characteristic polynomial is given by Polynomial long division gives p A (λ) = (2 λ)(λ 2 λ 2) = λ 3 + 3λ 2 4. ( ) ( ) λ 5 : λ 3 + 3λ 2 4 = λ 2 3λ 9 + 23λ2 12λ 36 λ 3 + 3λ 2 4 λ 5 + 3λ 4 4λ 2 3λ 4 4λ 2 3λ 4 + 9λ 3 12λ 9λ 3 4λ 2 12λ 9λ 3 + 27λ 2 36 Therefore, A 5 = 23A 2 12A 36I. 23λ 2 12λ 36 II. For a symmetric n n matrix, the following holds: A symmetric matrix has exactly n real eigenvalues, where, as usual, eigenvalues have to be counted with their multiplicities. For each eigenvalue the geometric multiplicity of an eigenvalue is equal to its algebraic multiplicity. Eigenvectors belonging to different eigenvalues are orthogonal. c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 9
Problem 5 (4+2+3+1+2+6=18) Please enter your answers in the designated boxes. Only the answers in the boxes will be evaluated. An argumentation or intermediate steps are not necessary in this problem. a) Compute (for arbitrary t) the Casorati determinant of the following pairs of functions and simplify as much as possible. In the left box, enter the corresponding matrix, in the right box enter the value of the Casorati determinant. The Casorati determinant of the functions 2 t and 3 t is C(t) = det 2t 3 t 2 t+1 3 t+1 = 6 t The Casorati determinant of the functions t 2 and t 3 is C(t) = det t 2 t 3 (t + 1) 2 (t + 1) 3 = t 2 (t + 1) 2 b) Find the rank of the following matrices. 2 3 1 5 7 3 rk 3 1 2 = 3 rk 3 4 2 = 2 1 2 3 2 0 4 c) Find all solutions of the equation z 3 = i in complex numbers. Give the solutions in the form a + bi. z i = i, z 2 = 3 2 1 2 i, z 3 = 3 2 1 2 i d) Find the inverse of the following matrix 0 0.5 0 0 0 0 0 0.5 A = 0.5 0 0 0. A 1 = 0 0 0.5 0 0 0 2 0 2 0 0 0 0 0 0 2 0 2 0 0 c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 10
e) Consider the following final tableau of the Gauß-Jordan algorithm for solving a linear equation Ax = b. State the general solution in basis representation. x 4 x 2 x 3 x 1 1 0 5 3 9 0 1 2 1 4 L = 0 4 0 9 + λ 1 0 2 1 5 + λ 2 1 1 0 3 : λ 1, λ 2 R f) Consider the problem of minimizing the objective f(x, y) = 6(x 2)(y 1) subject to the constraints x 1 and y 2. Then, the Lagrangian is given by The KKT conditions are: L(x, y, µ 1, µ 2 ) = 6(x 2)(y 1) + µ 1 (1 x) + µ 2 (y 2). Stationarity: L x = 6(y 1) µ 1 = 0 L y = 6(x 2) + µ 2 = 0 Primal feasibility: x 1, y 2 Dual feasibility: µ 1, µ 2 0 Complementary slackness: µ 1 (1 x) = 0, µ 2 (y 2) = 0 Find all points (x, y) with corresponding multipliers µ 1, µ 2, at which the KKT conditions for a minimum are satisfied. Hint: There is at least one such point. There are two points at which the KKT conditions are satisfied: (x, y) = (2, 1) with multipliers (µ 1, µ 2 ) = (0, 0) and (x, y) = (1, 2) with multipliers (µ 1, µ 2 ) = (6, 6). c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 11
English translation of the text on page 1. You can reach at most 90 points in this exam. Normally, the exam is passed if at least 45 points are reached. Please enter your student ID number as well as your initials in the designated fields on page 1. The exam consists of five problems. All five problems have to be solved correctly to get the maximum number of points. For each problem you have to give the complete solution in this booklet at the designated places. Besides the final results you also have to write down the intermediate steps of your calculations. If there is not enough room, you may continue on a separate sheet that is marked with your student ID number and initials. In that case, give a hint to that additional sheet in the booklet. You may use a handwritten formulary. You may use only a single printout (in DIN A4) of the PDF file provided in the Ilias course (enter your student ID number and initials). The formulary has to be written by yourself. You may write on both sides of the printout. You have to hand-in the formulary together with this booklet. Any deviation from these regulations will be considered as an attempt at deception. The cover sheet that you will find on your table can be used as scratch paper. All entries on this cover sheet will not be assessed. After completing the exam, you have to insert this booklet as well as your formulary in the cover sheet. c 2018, Institut für Ökonometrie und Statistik, Universität zu Köln S. 12