ECN115 Mathematical Methods in Economics and Finance Lecture 1 (Queen Mary University of London) 2024-2025 PDF

ECN115 Mathematical Methods in Economics and Finance Evgenii Safonov 2024-2025. LECTURE 1 1 About the Module Module Information Module Code: ECN115 Module Title: Mathematical Methods in Economics and Finance Lecture: Wednesd...

ECN115 Mathematical Methods in Economics and Finance Evgenii Safonov 2024-2025. LECTURE 1 1 About the Module Module Information Module Code: ECN115 Module Title: Mathematical Methods in Economics and Finance Lecture: Wednesday 9:00-11:00 in GCG10 Peston LT. Thursday 12:00-14:00 in GCG10 Peston LT. Lecturer: Evgenii Safonov email: [email protected] office: GC332 office hours: TBA Classes: See MyTimetable on QMplus 2 Teaching Assistants Mahmoud Shahin [email protected] Zhenghao Liu [email protected] Tallon Trent Howie [email protected] 3 Module Content The module refreshes the existing mathematical knowledge and introduces mathematical tools used in the undergraduate economics and finance classes Themes: ◁ Introduction: sets, numbers, Mathematical Induction, Binomial formula ◁ Elements of one-variable mathematical analysis and calculus ◁ Constrained optimisation (one variable) ◁ Introduction to multivariate analysis and optimisation ◁ (If time permits) other topics including linear algebra 4 Assessment Three types: ◁ Online quizzes (15%). You should work alone ◁ Graded problem sets (15%). You can discuss problem sets and work in groups, but each student should submit an individual solution ◁ Final exam (70%). Individual work, centrally administrated Timing: 14 October. Quiz I (7.5%). One hour (you choose) from 00:00 to 23:59 3 November. Problem Set I (7.5%). Available 1 week in advance or earlier 2 December. Quiz II (7.5%). One hour (you choose) from 00:00 to 23:59 10 December. Problem Set II (7.5%). Available 1 week in advance or earlier In January. Final exam (70%). 2 hours, only paper and pen 5 Important component: weekly homeworks (NOT graded) ◁ It is almost impossible to learn mathematics without practice ◁ Each week except when graded problem sets are due, you are expected to try solving a homework ◁ A homework is a mix of somewhat basic and more advanced problems. If you cannot solve some problem, it is completely OK ◁ The most challenging problems are analysed during classes ◁ Finally, the solutions are posted Lecture −→ Homework −→ Class −→ Solutions online 6 Diagnostic Quiz ◁ Friday 27 September ◁ Does not count towards the grade ◁ Objective: access the background level of mathematics ◁ Format: 20 questions, mostly multiple choice, some ask to calculate a number; 80 minutes for the whole test 7 Last years materials ◁ You can find exam topics and exams themselves (2 per year) for the last 2 years in QM+: ECN115 —> Key Module Information —> Syllabus —> Previous years’ exam materials ◁ Note that this year, topics might be slightly different! ◁ I’ll post the solutions as well closer to your exams 8 Introduction Outline ◁ Elementary logic ◁ Sets ◁ Number systems and properties of real numbers ◁ Basic inequalities ◁ Summation operator ◁ Next week’s material: Mathematical Induction 9 Elementary logic Vocabulary ◁ A proposition is a statement that may be either true or false. ◁ A proof is a demonstration that one proposition follows from one or more other propositions. ◁ An axiom is a proposition that is accepted without proof. ◁ A theorem is a proposition that has been proved on the basis of axioms and previously established theorems. Logical implication and equivalence. Given propositions P and Q, we write P =⇒ Q for the proposition that P implies Q (i.e., “if P, then Q”). When P =⇒ Q and Q =⇒ P, we write P ⇐⇒ Q and say that P and Q are equivalent. Negation. Proposition “not P” is true when proposition P is false, and “not P” is false when P is true. 10 Elementary logic—illustration ◁ Proposition A: Charlie likes all fruits ◁ Proposition B: Charlie likes apples ◁ We have A =⇒ B. Proof: since apples are fruits, and Charlie likes all fruits, therefore, Charlie likes apples ◁ It is not true that B =⇒ A. To show this, we can construct a counterexample. Suppose Charlie likes apples (proposition B is true), but he does not like blueberries. If A is true, then Charlie likes all fruits and hence, likes blueberries as well, in contradiction. Therefore, in this example, proposition A is not true, and hence, B does not imply A ◁ Since B does not imply A, propositions A and B are not equivalent—which is intuitively obvious ◁ One thing to note: all our arguments rely on the fact that there are fruits other than apples (i.e. blueberries). If the only fruit in the world would be apple, propositions A and B would be equivalent! 11 Outline ◁ Elementary logic ◁ Sets ◁ Number systems and properties of real numbers ◁ Basic inequalities ◁ Summation operator ◁ Next week’s material: Mathematical Induction 12 Sets ◁ A set is a collection (what is a collection?) of objects. If the set A consists of the objects x, y , z,..., then we write A = {x, y , z,...} ◁ If object x is in set A, then we write x ∈ A and call x an element of A. If not, then we write x ∈ /A ◁ The two sets A and B are equal if every element of A is also an element of B, and every element of B is an element of A. We denote it by A = B ◁ A set that contains no objects is called an empty set. We denote it by the symbol ∅; sometimes, people denote it by {} Remark: not everything can be described as a set. For example, there is no “set of all sets” and some similar entities otherwise we can come to a contradiction (see Russell’s Paradox). 13 Sets—relations and operations ◁ If each element of A is also an element of B, then we write A ⊆ B (in some textbooks, A ⊂ B) and call A a subset of B, and B a superset of A ✓ Note that A = B is equivalent to (A ⊆ B and B ⊆ A) ◁ The collection of objects in both set A and set B is written A ∩ B and called the intersection of the two sets. ✓ Note that B ∩ A = A ∩ B ◁ The collection of objects in either set A, or set B, or in both sets A and B is written A ∪ B and called the union of the two sets. ✓ Note that B ∪ A = A ∪ B 14 Set relations and operations—examples Emma likes E = {oranges, strawberries} Bob likes B = {oranges, apples, bananas} Hannah likes H = {bananas, blueberries, apples} Sam likes S = {bananas, oranges, apples, blueberries} Then, for example, B ∩H = {apples, bananas} B ∪H = {oranges, apples, bananas, blueberries} = S S ∩E = {oranges} = B ∩ E E ∩H = ∅ Also, B⊆S H⊆S Note also that any set is a subset of itself. For instance, E ⊆ E , B ⊆ B,... 15 Set construction by conditioning Definition. Suppose that a set A might contain some elements that have property P. Then we can define the set {x ∈ A | x has property P} to be the subset of A that consists of all elements of the set A that have property P Example ◁ Let A be the set of students in QMUL, and P be the property of being born after 2003 year. Then the set {x ∈ A | x was born after 2003} is the set of students in QMUL who were born after 2003 year 16 Set substraction Definition. Given two sets A and B, the set A∖B = {x ∈ A | x ̸∈ B} is called the difference between sets A and B; we also say that A∖B is the result of substraction of set B from set A Thus, A∖B is the subset of elements of A that are NOT elements of B Examples ◁ {apples, oranges, blueberries}∖{bananas, oranges} = {apples, blueberries} ◁ {2, 4, 6, 8, 10, 12, 14,...}∖{3, 6, 9, 12, 15,...} = {2, 4, 8, 10, 14,...} ◁ A ⊆ B implies A∖B = ∅ 17 Outline ◁ Elementary logic ◁ Sets ◁ Number systems and properties of real numbers ◁ Basic inequalities ◁ Summation operator ◁ Next week’s material: Mathematical Induction 18 Number systems Definitions ◁ The natural numbers are the set N = {1, 2, 3,...} ✓ Archimedean property: for any n ∈ N, we have n + 1 ∈ N ◁ The integers are the setZ = {... , −3, −2, −1, 0, 1, 2, 3,...} ◁ The rational numbers are the set Q consisting of all expressions p/q with p ∈ Z and q ∈ N ◁ The real numbers are (informally) the set R of points on the number line Observations ◁ We have N Z Q R ⊆ ⊆ ⊆. Equality (=) and inequality (≥, ≤, >, b, or a = b, or a < b. ✓ a < b is the same as b > a ✓ We denote by a ≥ b if a > b or a = b. Similarly, a ≤ b if a < b or a = b O2. Transitivity: if a > b and b > c, then a > c. Addition operation: for numbers a, b, assigns a unique number a + b A1. Commutativity: a + b = b + a. A2. Associativity: a + (b + c) = (a + b) + c. A3. Properties of the number zero: a + 0 = a. A4. Existence of the opposite number: a + (−a) = 0. A5. Additivity of the order: a > b =⇒ a + c > b + c. 20 Properties of real numbers Multiplication operation: for numbers a, b, assigns a unique number a · b M1. Commutativity: a · b = b · a. M2. Associativity: a · (b · c) = (a · b) · c. M3. Properties of the number unit: 1 ̸= 0, a · 1 = a. M4. Existence of the inverse of a non-zero number: a · (1/a) = 1 for a ̸= 0. M5. Distributive law: a · (b + c) = a · b + a · c. M6. Multiplicative properties of the order relation: ✓ if a > b and c > 0, then a · c > b · c ✓ if a > b and c < 0, then a · c < b · c 21 Least Upper Bound property of real numbers (extra material) ◁ Note that the rational numbers Q also satisfy all the above properties ◁ The real numbers satisfy one more property that rational numbers do not (here comes the extra material): Least Upper Bound property: if a non-empty subset of real numbers A ⊆ R R has an upper bound—that is, if there is y ∈ such that for all x ∈ A, R x ≤ y —then A has a least upper bound—that is, there is z ∈ such that for all x ∈ A, x ≤ z, and if y < z, there is x ∈ A such that x > y. 22 Intervals of real numbers ◁ Closed interval: [a, b] = {x ∈ R | a ≤ x ≤ b} ◁ Open interval: (a, b) = {x ∈ R | a < x < b} ◁ Half-open intervals: (a, b] = {x ∈ R | a < x ≤ b} [a, b) = {x ∈ R | a ≤ x < b} ◁ Infinite and half-infinite intervals: (a, ∞) = R | a < x} {x ∈ (−∞, b) = {x ∈ R | x < b} (−∞, ∞) = R Similarly, [a, ∞) = {x ∈ R | a ≤ x} and (−∞, b] = {x ∈ R | x ≤ b} 23 Outline ◁ Elementary logic ◁ Sets ◁ Number systems and properties of real numbers ◁ Basic inequalities ◁ Summation operator ◁ Next week’s material: Mathematical Induction 24 Basic inequalities Note: below, we discuss the ideas behind solving basic inequalities and how these ideas rely on the properties of the real numbers. When solving problems, it suffices just to do correct calculations. Example 1. Let us find all x that satisfy inequality 2x + 1 > 0 ◁ Using properties A5, and A4, add number −1 to both sides of the inequality, and use A3 to find 0 + (−1) = −1 2x + 1 > 0 =⇒ 2x + 1 + (−1) > 0 + (−1) =⇒ 2x > −1 ◁ The reverse implication works as well; to see this, we can add number 1 to both sides of the inequality 2x > −1: 2x > −1 =⇒ 2x + (1) > −1 + (1) =⇒ 2x + 1 > 0 ◁ Therefore, the two inequalities are equivalent: 2x + 1 > 0 ⇐⇒ 2x > −1 25 Basic inequalities ◁ By property M4, there is an inverse 1/2 of the non-zero number 2; moreover, one can see that 1/2 > 0 (can you show it using properties of the real numbers?). Using property M6, multiply both sides of the previous inequality by 1/2: 2x > −1 =⇒ (1/2)·2x > (1/2)·(−1) =⇒ x > −1/2 ◁ Similarly, we can find that the reverse implication holds: x > −1/2 =⇒ 2·x > 2·(−1/2) =⇒ 2x > −1 ◁ Therefore, all inequalities are equivalent: 2x + 1 > 0 ⇐⇒ 2x > −1 ⇐⇒ x > −1/2 and we can express the solution of the inequality 2x + 1 > 0 as x ∈ (−1/2, ∞) 26 Basic inequalities Example 2. Let us find all x that satisfy inequality (2x + 1) · (x − 1) > 0 ◁ Note that neither (2x + 1) nor (x − 1) can be equal to zero, otherwise, (2x + 1) · (x − 1) = 0. Hence, it suffices to consider 4 cases: if 2x +1>0 and x − 1 > 0, then (2x + 1) · (x − 1) > 0 if 2x +1>0 and x − 1 < 0, then (2x + 1) · (x − 1) < 0 if 2x +1 0, then (2x + 1) · (x − 1) < 0 if 2x +1 0 ◁ Therefore, (2x+1)·(x−1) > 0 ⇐⇒ [2x+1 > 0 and x−1 > 0] or [2x+1 < 0 and x−1 < 0] ◁ Since 2x + 1 > 0 is equivalent to x > −1/2, x − 1 > 0 is equivalent to x > 1, and x > 1 =⇒ x > −1/2, then [2x + 1 > 0 and x − 1 > 0] ⇐⇒ x > 1 ◁ Similarly, [2x + 1 < 0 and x − 1 < 0] ⇐⇒ x < −1/2 ◁ Thus, the solution is x ∈ (−∞, −1/2) ∪ (1, ∞) 27 Quadratic inequalities Example 3. Let us find all x that satisfy inequality 3x 2 − 5x + 2 > 0 ◁ Idea: represent 3x 2 − 5x + 2 as a product of two linear terms (if possible), i.e. 3x 2 − 5x + 2 = e · (x + f ) · (x + g ) where e, f , g are some real numbers ◁ We might expand e · (x + f ) · (x + g ) = ex 2 + e(f + g )x + fg from which e = 3. However, we only know that f + g = −5/3 and fg = 2/3, but not the exact values of f and g ◁ Observation: substitute x = −f , then e · (x + f ) · (x + g ) = 0. Thus, x = −f is a solution (root) of the equation 3x 2 − 5x + 2 = 0. And similarly, x = −g is a root of the same equation ◁ Solving 3x 2 − 5x + 2 = √ 0, we get the discriminant D = (−5)2 − 4 · 3 · 2 = 1, −(−5) ± 1 hence x1,2 = are the roots, that is, x1 = 2/3, x2 = 1 6 2 ◁ Thus, 3x − 5x + 2 = 3 · (x − 2/3) · (x − 1) ◁ Solving 3 · (x − 2/3) · (x − 1) > 0 as in Example 2, we find that inequality 3x 2 − 5x + 2 > 0 is equivalent to x ∈ (−∞, 2/3) ∪ (1, ∞) 28 Outline ◁ Elementary logic ◁ Sets ◁ Number systems and properties of real numbers ◁ Basic inequalities ◁ Summation operator ◁ Next week’s material: Mathematical Induction 29 Summation operator ◁ We can consider sums of many numbers, such as 3+5+2+9+5 ◁ When there are several numbers ak , ak+1 ,..., an , indexed by some integer j that takes consecutive values k, k + 1,... up to n, we use a special short notation for the sum of those numbers, the summation operator: n ∑︁ aj = ak + ak+1 +... + an−1 + an j=k We can read this notation as follows: as index j changes from k to n, we sum up all corresponding numbers aj. n ∑︁ ∑︀n ◁ Sometimes the notation is j=k instead of. It means the same sum j=k 30 Summation operator: examples For example, if aj = j, then 7 ∑︁ 7 ∑︁ aj = j = 3 + 4 + 5 + 6 + 7 = 25 j=3 j=3 3 ∑︁ 3 ∑︁ aj = j = −2 − 1 + 0 + 1 + 2 + 3 = 3 j=−2 j=−2 If bj = j 2 , then 7 ∑︁ 7 ∑︁ bj = j 2 = 32 + 42 + 52 + 62 + 72 = 135 j=3 j=3 3 ∑︁ 3 ∑︁ bj = j 2 = (−2)2 + (−1)2 + 02 + 12 + 22 + 32 = 19 j=−2 j=−2 31 Summation operator: properties n ∑︁ When n = k, it follows that consists of one term with index j = n = k: j=k k ∑︁ aj = ak j=k By analogy, when n = k − 1, we define the value of the summation operator as a sum of zero terms, which we assume to be zero: k−1 ∑︁ aj = 0 j=k Note that if aj = a for all j, then it is easy to calculate a sum of ak : n ∑︁ n ∑︁ aj = a = a +... + a = (n − k + 1) · a ⏟ ⏞ j=k j=k n−k+1 terms 32 Summation operator: properties The properties of the real numbers imply the following properties of the summation operator: n ∑︁ n ∑︁ n ∑︁ (P1) (aj + bj ) = aj + bj j=k j=k j=k n ∑︁ n ∑︁ (P2) c · aj = c · aj j=k j=k And the following property comes from its definition: n ∑︁ m ∑︁ m ∑︁ (P3) aj + aj = aj for k ≤ n ≤ m j=k j=n+1 j=k 33 Next week’s material Outline ◁ Elementary logic ◁ Sets ◁ Number systems and properties of real numbers ◁ Basic inequalities ◁ Summation operator ◁ Next week’s material: Mathematical Induction 34 Mathematical Induction ◁ A proof technique applied when there are several similar propositions P1 , P2 ,...,Pn ,... that depend on the parameter n ∈N 35 Mathematical Induction ◁ A proof technique applied when there are several similar propositions P1 , P2 ,...,Pn ,... that depend on the parameter n ∈ N ◁ The idea is: instead of proving each of the propositions P1 , P2 , P3 to prove only two propositions: ✓ The induction base is the first proposition P1 ✓ The induction step is the proposition that for any index n, Pn =⇒ Pn+1 35 Mathematical Induction ◁ A proof technique applied when there are several similar propositions P1 , P2 ,...,Pn ,... that depend on the parameter n ∈ N ◁ The idea is: instead of proving each of the propositions P1 , P2 , P3 to prove only two propositions: ✓ The induction base is the first proposition P1 ✓ The induction step is the proposition that for any index n, Pn =⇒ Pn+1 ◁ When both these the induction step and the induction base are proven, all propositions P1 ,...,Pn ,... are proven P1 , P2 , P3 , P4 , P5 ,... 35 Mathematical Induction ◁ A proof technique applied when there are several similar propositions P1 , P2 ,...,Pn ,... that depend on the parameter n ∈ N ◁ The idea is: instead of proving each of the propositions P1 , P2 , P3 to prove only two propositions: ✓ The induction base is the first proposition P1 ✓ The induction step is the proposition that for any index n, Pn =⇒ Pn+1 ◁ When both these the induction step and the induction base are proven, all propositions P1 ,...,Pn ,... are proven P1 ; P2 , P3 , P4 , P5 ,... 35 Mathematical Induction ◁ A proof technique applied when there are several similar propositions P1 , P2 ,...,Pn ,... that depend on the parameter n ∈ N ◁ The idea is: instead of proving each of the propositions P1 , P2 , P3 to prove only two propositions: ✓ The induction base is the first proposition P1 ✓ The induction step is the proposition that for any index n, Pn =⇒ Pn+1 ◁ When both these the induction step and the induction base are proven, all propositions P1 ,...,Pn ,... are proven P1 =⇒ P2 ; P3 , P4 , P5 ,... 35 Mathematical Induction ◁ A proof technique applied when there are several similar propositions P1 , P2 ,...,Pn ,... that depend on the parameter n ∈ N ◁ The idea is: instead of proving each of the propositions P1 , P2 , P3 to prove only two propositions: ✓ The induction base is the first proposition P1 ✓ The induction step is the proposition that for any index n, Pn =⇒ Pn+1 ◁ When both these the induction step and the induction base are proven, all propositions P1 ,...,Pn ,... are proven P1 =⇒ P2 =⇒ P3 ; P4 , P5 ,... 35 Mathematical Induction ◁ A proof technique applied when there are several similar propositions P1 , P2 ,...,Pn ,... that depend on the parameter n ∈ N ◁ The idea is: instead of proving each of the propositions P1 , P2 , P3 to prove only two propositions: ✓ The induction base is the first proposition P1 ✓ The induction step is the proposition that for any index n, Pn =⇒ Pn+1 ◁ When both these the induction step and the induction base are proven, all propositions P1 ,...,Pn ,... are proven P1 =⇒ P2 =⇒ P3 =⇒ P4 ; P5 ,... 35 Mathematical Induction ◁ A proof technique applied when there are several similar propositions P1 , P2 ,...,Pn ,... that depend on the parameter n ∈ N ◁ The idea is: instead of proving each of the propositions P1 , P2 , P3 to prove only two propositions: ✓ The induction base is the first proposition P1 ✓ The induction step is the proposition that for any index n, Pn =⇒ Pn+1 ◁ When both these the induction step and the induction base are proven, all propositions P1 , P2 , P3 ,... are proven P1 =⇒ P2 =⇒ P3 =⇒ P4 =⇒ P5... 35 Mathematical Induction ◁ A proof technique applied when there are several similar propositions P1 , P2 ,...,Pn ,... that depend on the parameter n ∈ N ◁ The idea is: instead of proving each of the propositions P1 , P2 , P3 to prove only two propositions: ✓ The induction base is the first proposition P1 ✓ The induction step is the proposition that for any index n, Pn =⇒ Pn+1 ◁ When both these the induction step and the induction base are proven, all propositions P1 , P2 , P3 ,... are proven P1 =⇒ P2 =⇒ P3 =⇒ P4 =⇒ P5... ◁ When proving the induction step Pn =⇒ Pn+1 , the statement that proposition Pn is true is called the induction hypothesis. 35 Example: sum of first n natural numbers ◁ For example, let the propositions Pn for n = 1, 2, 3,... be n ∑︁ n(n + 1) j = j=1 2 Put it differently, the proposition Pn states that the sum of all natural n(n + 1) numbers from 1 to n is equal to 2 ◁ The induction base is the first proposition, P1 ; thus, n = 1. In our example, it is 1 ∑︁ 1 · (1 + 1) j = j=1 2 Since the left hand side is 1 and the right hand side is 1, then P1 is true 36 Example: sum of first n natural numbers The induction step is the proposition that for any natural n, Pn =⇒ Pn+1 In our example, the proposition Pn+1 is n+1 ∑︁ (n + 1)((n + 1) + 1) j = j=1 2 To prove the induction step in our example: n+1 n ∑︁ ∑︁ n(n + 1) j = j + (n + 1) = + (n + 1) = j=1 j=1 2 (︁ n )︁ (n + 2)(n + 1) (n + 1)((n + 1) + 1) = + 1 (n + 1) = = 2 2 2 which we wanted to prove. Note that we used proposition Pn (the induction hypothesis) in the proof. 37 Alternative way to prove the induction step in our example Equivalently, the induction step is the proposition that for any natural n ≥ 2, Pn−1 =⇒ Pn. In this formulation, the induction hypothesis is the proposition Pn−1 which is n−1 ∑︁ (n − 1)((n − 1) + 1) (n − 1)n j = = j=1 2 2 To prove the induction step in our example: n n−1 ∑︁ ∑︁ (n − 1)n j = j +n = +n = j=1 j=1 2 (︂ )︂ n−1 (n + 1)n = +1 n = 2 2 hence Pn holds, and since this argument works for all n ≥ 2, Pn−1 =⇒ Pn , proving the induction step. 38 Alternative formulation of the principle ◁ Want to prove propositions P1 , P2 ,...,Pn ,... ✓ The induction base is the first proposition P1 ✓ The induction step is the proposition that for any index n, {P1 ,..., Pn } =⇒ Pn+1 ✓ Remark: we’ve formulated the induction step originally as Pn =⇒ Pn+1 for any n. The two formulations are, in fact, equivalent ◁ When proving the induction step Pn =⇒ Pn+1 , the statement that propositions P1 ,..., Pn are true is called the induction hypothesis. 39