EC400B Math for Micro: Lecture 1Francesco Nava September 2014 Nava (LSE) EC400B – Lecture 1 September 2014 1 / 27 Course Outline Lecture 1: Tools for optimization (Quadratic forms). Lecture 2: Tools for optimization (Taylor’s expansion) and Unconstrained optimization. Lecture 3: Concavity, convexity, and quasi-concavity. Lecture 4: Constrained Optimization I: Equality Constraints, Lagrange Theorem. Lecture 5: Constrained Optimization II: Inequality Constraints, Kuhn-Tucker Theorem. Lecture 6: Constrained Optimization III: The Maximum Value Function, Envelope Theorem, Implicit Function Theorem. Nava (LSE) EC400B – Lecture 1 September 2014 2 / 27 Course Outline Lecture 1: Tools for optimization (Quadratic forms). Lecture 2: Tools for optimization (Taylor’s expansion) and Unconstrained optimization. Lecture 3: Concavity, convexity, and quasi-concavity. Lecture 4: Constrained Optimization I: Equality Constraints, Lagrange Theorem. Lecture 5: Constrained Optimization II: Inequality Constraints, Kuhn-Tucker Theorem. Lecture 6: Constrained Optimization III: The Maximum Value Function, Envelope Theorem, Implicit Function Theorem. Nava (LSE) EC400B – Lecture 1 September 2014 2 / 27 Course Outline Lecture 1: Tools for optimization (Quadratic forms). Lecture 2: Tools for optimization (Taylor’s expansion) and Unconstrained optimization. Lecture 3: Concavity, convexity, and quasi-concavity. Lecture 4: Constrained Optimization I: Equality Constraints, Lagrange Theorem. Lecture 5: Constrained Optimization II: Inequality Constraints, Kuhn-Tucker Theorem. Lecture 6: Constrained Optimization III: The Maximum Value Function, Envelope Theorem, Implicit Function Theorem. Nava (LSE) EC400B – Lecture 1 September 2014 2 / 27 Nava (LSE) EC400B – Lecture 1 September 2014 2 / 27 .Course Outline Lecture 1: Tools for optimization (Quadratic forms). Lagrange Theorem. Lecture 4: Constrained Optimization I: Equality Constraints. convexity. Implicit Function Theorem. and quasi-concavity. Kuhn-Tucker Theorem. Lecture 6: Constrained Optimization III: The Maximum Value Function. Lecture 5: Constrained Optimization II: Inequality Constraints. Lecture 3: Concavity. Envelope Theorem. Lecture 2: Tools for optimization (Taylor’s expansion) and Unconstrained optimization. Lecture 5: Constrained Optimization II: Inequality Constraints. Implicit Function Theorem. Lecture 6: Constrained Optimization III: The Maximum Value Function. Nava (LSE) EC400B – Lecture 1 September 2014 2 / 27 . Lecture 3: Concavity. Lecture 4: Constrained Optimization I: Equality Constraints. Envelope Theorem.Course Outline Lecture 1: Tools for optimization (Quadratic forms). convexity. Kuhn-Tucker Theorem. Lecture 2: Tools for optimization (Taylor’s expansion) and Unconstrained optimization. and quasi-concavity. Lagrange Theorem. Lecture 3: Concavity. Lecture 2: Tools for optimization (Taylor’s expansion) and Unconstrained optimization. Implicit Function Theorem. Lecture 5: Constrained Optimization II: Inequality Constraints. and quasi-concavity. Kuhn-Tucker Theorem. Lagrange Theorem. Envelope Theorem. convexity.Course Outline Lecture 1: Tools for optimization (Quadratic forms). Lecture 6: Constrained Optimization III: The Maximum Value Function. Nava (LSE) EC400B – Lecture 1 September 2014 2 / 27 . Lecture 4: Constrained Optimization I: Equality Constraints. ac.m. Course Material: available on Moodle Nava (LSE) EC400B – Lecture 1 September 2014 3 / 27 . Fri 12 September — 10:00-11:20 a.m.nava@econ. Wed 17 September — 10:00-11:20 a.ac.Administration Contact Info: LIF 3.m. Thu 18 September — 10:00-11:20 a.uk Office Hours: Thu 11 September — 10:00-11:20 a. f.20.m.m.m.lse. Tue 16 September — 10:00-11:20 a. Extension 6353. Wed 24 September — 11:00-12:20 a.uk). Mon 15 September — 10:00-11:20 [email protected]. or by appointment (e-mail f. Carl P. Simon and Lawrence E. Nava (LSE) EC400B – Lecture 1 September 2014 4 / 27 . Blume Mathematics for Economists. Morton I. Chiang Fundamental Methods of Mathematical Economics. Alpha C. Peter Hammond. Kamien and Nancy L. Atle Seierstad and Arne Strom Further Mathematics for Economic Analysis.Suggested Textbooks Knut Sydsaeter. Schwartz Dynamic Optimization: The Calculus of Variations and Optimal Control in Economics and Management. Akira Takayama Mathematical Economics. such as risk minimization problems in finance. where riskiness is measured by the quadratic variance of the returns from investments. Nava (LSE) EC400B – Lecture 1 September 2014 5 / 27 . (ii) Conditions for optimization techniques are stated in terms of quadratic forms.What is a quadratic form? Quadratic forms are useful because: (i) The simplest functions after linear ones. (iii) Economic optimization problems have a quadratic objective function. the simplest functions with a unique global extremum are the quadratic forms: y = x2 and y = −x 2 The level curve of a general quadratic form in R2 is a11 x12 + a12 x1 x2 + a22 x22 = b and can take the form of an ellipse. Nava (LSE) EC400B – Lecture 1 September 2014 6 / 27 . a point. or possibly.What is a Quadratic Form? Among the functions of one variable. the empty set. a pair of lines. a hyperbola. the empty set.What is a Quadratic Form? Among the functions of one variable. a pair of lines. Nava (LSE) EC400B – Lecture 1 September 2014 6 / 27 . a hyperbola. a point. or possibly. the simplest functions with a unique global extremum are the quadratic forms: y = x2 and y = −x 2 The level curve of a general quadratic form in R2 is a11 x12 + a12 x1 x2 + a22 x22 = b and can take the form of an ellipse. ..Definition of Quadratic Form Definition A quadratic form on Rn is a real valued function X Q(x1 .. xn ) = aij xi xj i≤j The general quadratic form of a11 x12 + a12 x1 x2 + a22 x22 can be written (non uniquely) as: x1 x2 Nava (LSE) a11 a12 0 a22 EC400B – Lecture 1 x1 x2 September 2014 7 / 27 . . x2 . x2 .. .. xn ) = aij xi xj i≤j The general quadratic form of a11 x12 + a12 x1 x2 + a22 x22 can be written (non uniquely) as: x1 x2 Nava (LSE) a11 a12 0 a22 EC400B – Lecture 1 x1 x2 September 2014 7 / 27 ..Definition of Quadratic Form Definition A quadratic form on Rn is a real valued function X Q(x1 . . ..Definition of Quadratic Form Each quadratic form can be represented as Q(x) = xT A x where A is a unique symmetric matrix: a11 a12 /2 .. . a1n /2 a2n /2 .. .. then the real valued function Q(x) = xT A x. is a quadratic form.. a1n /2 a12 /2 a22 ... ann Conversely. . . a2n /2 . Nava (LSE) EC400B – Lecture 1 September 2014 8 / 27 . .. if A is a symmetric matrix... a min. or neither. ax 2 is non negative and equals 0 only when x = 0. A Simple Example: Consider the quadratic form: y = ax 2 If a > 0. the quadratic form is negative definite and 0 is a global max. If so. If a < 0.Definiteness of Quadratic Forms A quadratic form always takes the value 0 when x = 0. We want to know whether x = 0 is a max. Nava (LSE) EC400B – Lecture 1 September 2014 9 / 27 . the quadratic form is positive definite and 0 is a global min. Nava (LSE) EC400B – Lecture 1 September 2014 10 / 27 .Examples in Two Dimensions In two dimensions. for instance: a positive definite quadratic form is: x12 + x22 a negative definite quadratic form is: −x12 − x22 an indefinite quadratic form is: x12 − x22 since it can take both positive and negative values. it is called negative semidefinite.Semidefiniteness of Quadratic Forms There are two intermediate cases: 1 if a quadratic form is always non negative. but also equals zero for some x 6= 0. 2 if a quadratic form is never positive. but can be zero at points other than the origin. it is called positive semidefinite. An example of a positive semidefinite quadratic form is: (x1 + x2 )2 which can be 0 for points such that x1 = −x2 . Nava (LSE) EC400B – Lecture 1 September 2014 11 / 27 . Then A is: positive definite if xT A x > 0 for all x ∈ Rn \{0} positive semidefinite if xT A x ≥ 0 for all x ∈ Rn \{0} negative definite if xT A x < 0 for all x ∈ Rn \{0} negative semidefinite if xT A x ≤ 0 for all x ∈ Rn \{0} indefinite if xT A x > 0 and yT A y < 0 for some x. such that: Q(x) = xT A x Definition Let A be an (n × n) symmetric matrix.Definiteness and Semidefiniteness of Symmetric Matrixes We apply the same terminology to the symmetric matrix A. y ∈ Rn \{0} Nava (LSE) EC400B – Lecture 1 September 2014 12 / 27 . Application (More on This Later) A function f (x) of one variable is concave on some interval if its second derivative f 00 (x) ≤ 0 on that interval. Nava (LSE) EC400B – Lecture 1 September 2014 13 / 27 . The generalization of this result to higher dimensions states that a function is concave on some region if its matrix of second derivatives (Hessian matrix) is negative semidefinite for all x in the region. The determinant of a (2 × 2) matrix: A= a11 a12 a21 a22 is computed as follows: det(A) = |A| = a11 a22 − a12 a21 Nava (LSE) EC400B – Lecture 1 September 2014 14 / 27 .The Determinant Definition The determinant of a matrix is a unique scalar associated with the matrix. The Determinant The determinant of a (3 × 3) matrix: a11 a12 a13 A = a21 a22 a23 a31 a32 a33 is computed as follows: . . a a |A| = a11 . . 22 23 a32 a33 Nava (LSE) . . . . . − a12 . a21 a23 . . a31 a33 EC400B – Lecture 1 . . . . . + a13 . a21 a22 . . a31 a32 . . . . September 2014 15 / 27 . is called a k th order principal submatrix of A. i1 . Nava (LSE) EC400B – Lecture 1 September 2014 16 / 27 .. Definition The determinant of a (k × k) principal submatrix is called a k th order principal minor of A. say columns i1 ...Principal Submatrices and Principal Minors Definition Let A be a (n × n) matrix... . i2 . i2 . in−k . The (k × k) submatrix of A formed by deleting n − k columns.. in−k and the same n − k rows from A. . Example Principal Minors Consider a general (3 × 3) matrix A: There is one third order principal minor: |A|. There are three second order principal minors: . . . . . a11 a12 . . a11 a13 . . . .. . . a21 a22 . . a31 a33 . . . . a22 a23 . . a32 a33 . . . . There are three first order principal minors: a11 . a22 and a33 . Nava (LSE) EC400B – Lecture 1 September 2014 17 / 27 . Definition The determinant of the k th order leading principal submatrix is called the k th order leading principal minor of A denoted |Ak |.Leading Submatrices and Leading Minors Definition Let A be a (n × n) matrix. The k th order principal submatrix of A obtained by deleting the last n − k rows and columns from A is called the k th order leading principal submatrix of A denoted Ak . Nava (LSE) EC400B – Lecture 1 September 2014 18 / 27 . |A3 | > 0 . A is negative definite if and only if all its n leading principal minors alternate in sign as follows: |A1 | < 0. If so: A is positive definite if and only if all its n leading principal minors are strictly positive: |A1 | > 0. Nava (LSE) EC400B – Lecture 1 September 2014 19 / 27 . |A2 | > 0. the k th order leading principal minor having the same sign of (−1)k . |A3 | < 0 .Testing Definiteness Let A be an (n × n) symmetric matrix. . |A2 | > 0. . . . . . A is negative definite if and only if all its n leading principal minors alternate in sign as follows: |A1 | < 0. . If so: A is positive definite if and only if all its n leading principal minors are strictly positive: |A1 | > 0. |A3 | > 0 . the k th order leading principal minor having the same sign of (−1)k . |A2 | > 0. |A2 | > 0. . |A3 | < 0 . Nava (LSE) EC400B – Lecture 1 September 2014 19 / 27 .Testing Definiteness Let A be an (n × n) symmetric matrix. |A3 | ≥ 0 . |A3 | ≤ 0 .Testing Definiteness A is positive semidefinite if and only if every principal minor of A is non-negative: |A1 | ≥ 0. . |A2 | ≥ 0. A is negative semidefinite if and only if every principal minor of odd order is non-positive and every principal minor of even order is non-negative: |A1 | ≤ 0. . . Nava (LSE) EC400B – Lecture 1 September 2014 20 / 27 . |A2 | ≥ 0. . Nava (LSE) EC400B – Lecture 1 September 2014 20 / 27 . |A3 | ≤ 0 .Testing Definiteness A is positive semidefinite if and only if every principal minor of A is non-negative: |A1 | ≥ 0. A is negative semidefinite if and only if every principal minor of odd order is non-positive and every principal minor of even order is non-negative: |A1 | ≤ 0. . . |A2 | ≥ 0. . |A3 | ≥ 0 . . |A2 | ≥ 0. Special Case: Diagonal Matrixes Consider the following (3 × 3) diagonal a1 0 A = 0 a2 0 0 matrix A: 0 0 a3 A also corresponds to the simplest quadratic forms: a1 x12 + a2 x22 + a3 x32 . Nava (LSE) EC400B – Lecture 1 September 2014 21 / 27 . negative semidefinite if and only if all the ai s are non-positive. positive semidefinite if and only if all the ai s are non-negative. negative definite if and only if all the ai s are negative. indefinite if there are two ai s of opposite sign. Nava (LSE) EC400B – Lecture 1 September 2014 22 / 27 .Special Case: Diagonal Matrixes Such quadratics forms are: positive definite if and only if all the ai s are positive. Nava (LSE) EC400B – Lecture 1 September 2014 22 / 27 .Special Case: Diagonal Matrixes Such quadratics forms are: positive definite if and only if all the ai s are positive. indefinite if there are two ai s of opposite sign. negative semidefinite if and only if all the ai s are non-positive. positive semidefinite if and only if all the ai s are non-negative. negative definite if and only if all the ai s are negative. indefinite if there are two ai s of opposite sign. Nava (LSE) EC400B – Lecture 1 September 2014 22 / 27 . negative semidefinite if and only if all the ai s are non-positive. positive semidefinite if and only if all the ai s are non-negative.Special Case: Diagonal Matrixes Such quadratics forms are: positive definite if and only if all the ai s are positive. negative definite if and only if all the ai s are negative. Special Case: (2 × 2) Matrixes Consider the (2 × 2) symmetric matrix: a b A= b c The associated quadratic form is: Q(x1 . x2 ) = xT A x = x1 x2 a b b c x1 x2 = ax12 + 2bx1 x2 + cx22 Nava (LSE) EC400B – Lecture 1 September 2014 23 / 27 . x2 ) = xT A x = x1 x2 a b b c x1 x2 = ax12 + 2bx1 x2 + cx22 Nava (LSE) EC400B – Lecture 1 September 2014 23 / 27 .Special Case: (2 × 2) Matrixes Consider the (2 × 2) symmetric matrix: a b A= b c The associated quadratic form is: Q(x1 . If a 6= 0.Special Case: (2 × 2) Matrixes If a = 0. 0) = 0. add and subtract b 2 x22 /a to get: b2 b2 Q(x1 . x2 ) = ax12 + 2bx1 x2 + cx22 + x22 − x22 a a 2 2 b 2bx1 x2 b 2 + 2 x2 − x22 + cx22 = a x12 + a a a 2 ac − b 2 b = a x1 + x2 + x22 a a Nava (LSE) EC400B – Lecture 1 September 2014 24 / 27 . then Q cannot be definite since Q(1. then Q cannot be definite since Q(1. add and subtract b 2 x22 /a to get: b2 b2 Q(x1 . 0) = 0.Special Case: (2 × 2) Matrixes If a = 0. x2 ) = ax12 + 2bx1 x2 + cx22 + x22 − x22 a a 2 2 b 2bx1 x2 b 2 + 2 x2 − x22 + cx22 = a x12 + a a a 2 ac − b 2 b = a x1 + x2 + x22 a a Nava (LSE) EC400B – Lecture 1 September 2014 24 / 27 . If a 6= 0. Special Case: (2 × 2) Matrixes If coefficients a and (ac − b 2 )/a are positive. Q cannot be negative. x2 ) is positive definite since: . Thus Q(x1 . It will equal 0 only when x1 + b x2 /a = 0 and x2 = 0 (ie x1 = x2 = 0). . . a b . . . >0 |a| > 0 and det(A) = . b c . in order for Q to be positive definite. Conversely. we need both: . . . a b . . . >0 |a| > 0 and det(A) = . b c . Nava (LSE) EC400B – Lecture 1 September 2014 25 / 27 . Q cannot be negative. x2 ) is positive definite since: .Special Case: (2 × 2) Matrixes If coefficients a and (ac − b 2 )/a are positive. It will equal 0 only when x1 + b x2 /a = 0 and x2 = 0 (ie x1 = x2 = 0). Thus Q(x1 . . . a b . . . >0 |a| > 0 and det(A) = . b c . in order for Q to be positive definite. Conversely. we need both: . . . a b . . . >0 |a| > 0 and det(A) = . b c . Nava (LSE) EC400B – Lecture 1 September 2014 25 / 27 . x2 ) is positive definite since: .Special Case: (2 × 2) Matrixes If coefficients a and (ac − b 2 )/a are positive. Q cannot be negative. It will equal 0 only when x1 + b x2 /a = 0 and x2 = 0 (ie x1 = x2 = 0). Thus Q(x1 . . . a b . . . >0 |a| > 0 and det(A) = . b c . we need both: . Conversely. in order for Q to be positive definite. . . a b . . . >0 |a| > 0 and det(A) = . b c . Nava (LSE) EC400B – Lecture 1 September 2014 25 / 27 . when the leading principal minors alternative in sign. If ac − b 2 < 0. x2 ) will be indefinite. which occurs if and only if: a < 0 and ac − b 2 > 0 That is. Q is negative definite if and only if both coefficients are negative. Nava (LSE) EC400B – Lecture 1 September 2014 26 / 27 .Special Case: (2 × 2) Matrixes Similarly. the two coefficients will have opposite signs and Q(x1 . Nava (LSE) EC400B – Lecture 1 September 2014 26 / 27 . when the leading principal minors alternative in sign. If ac − b 2 < 0. which occurs if and only if: a < 0 and ac − b 2 > 0 That is.Special Case: (2 × 2) Matrixes Similarly. Q is negative definite if and only if both coefficients are negative. x2 ) will be indefinite. the two coefficients will have opposite signs and Q(x1 . Examples of (2 × 2) Matrices Consider a symmetric matrix: A= 2 3 3 7 Since |A1 | = 2 and |A2 | = 5. B is indefinite. Nava (LSE) EC400B – Lecture 1 September 2014 27 / 27 . Consider a symmetric matrix: B= 2 4 4 7 Since |B1 | = 2 and |B2 | = −2. A is positive definite. Nava (LSE) EC400B – Lecture 1 September 2014 27 / 27 . B is indefinite. Consider a symmetric matrix: B= 2 4 4 7 Since |B1 | = 2 and |B2 | = −2. A is positive definite.Examples of (2 × 2) Matrices Consider a symmetric matrix: A= 2 3 3 7 Since |A1 | = 2 and |A2 | = 5.