Gianni Bosi and Magalì E. Zuanon. Basic Optimization and Financial Mathematics

Transcription

1 Gianni Bosi and Magalì E. Zuanon Basic Optimization and Financial Mathematics

2

3 i Preface It isn t that they can t see the solution. They can t see the problem. G.K. CHESTERTON The Scandal of Father Brown The point of a pin" This booklet contains the notes for a short course in Optimization and Financial Mathematics. It is far from being exhaustive but nevertheless it is entirely self-contained, following a classical course of calculus. The main goal of these notes is to provide students with the essential information about these topics in a primarily theoretical framework. The first part (Chapters 1, 2 and 3) illustrates the basic tools of optimization in several variables, passing through essential linear algebra and quadratic forms. The second part (Chapters 4, 5, 6 and 7) contains Theory of Interest as a mathematical problem topic, which is rather unlike what is done in typical finance courses. In this part we consider the deterministic approach only. In the last Chapter we introduce some arguments of mathematics of life insurance, the understanding of the basic principles of this part being a solid foundation for further study of the theory in a more general setting. The authors apologize in advance for any typos and mistakes. They are grateful to the readers who will be so kind to point out them. Trieste, March 21st, 2012 Gianni BOSI Dipartimento di Scienze Economiche, Aziendali, Matematiche e Statistiche, Università di Trieste, Piazzale Europa 1, Trieste, Italy. [email protected] Magalì E. ZUANON Dipartimento di Metodi Quantitativi, Università degli Studi di Brescia, Contrada Santa Chiara 50, Brescia, Italy. [email protected]

4 Contents pag. 1 Some concepts of matrix algebra Basic definitions Determinants The algebraic structure of R n The topological structure of R n Real functions of n real variables General concepts Quadratic forms Differentiability and optimization Convex and concave functions Constrained optimization with equality constraints Exercises The classical financial regimes General concepts Regime of the linear interest Regime of the commercial discount Regime of the compound interest or exponential regime Exercises The regime of the compound interest The final value under the regime of the compound interest Equivalence of interest rates Force of Interest: Continuous compounding Axiomatization of the regime of the compound interest Exercises Annuities and perpetuities Annual payments Non-annual payments Amortization and capital construction by constant instalments 65

5 2 Contents 5.4 Exercises Amortization of a debt General concepts Amortization by constant annual principal shares Amortization by constant instalments Negotiation of a debt: Makeham formula Exercises Actuarial mathematics Probability and Lifetimes Expected Present Values of Insurance Contracts Exercises Solutions to selected exercises 87 Bibliography 97

6 CHAPTER 1 Some concepts of matrix algebra 1.1 Basic definitions In this section we first present the basic concepts concerning matrices. Then we introduce the operations of sum of two matrices, product of a matrix and a real number and finally product of two matrices. Definition (Real Matrix). An m n real matrix (m and n are positive integers) is a collection a a 1n a a 2n A = [a ij ] 1 i m,1 j n =..... a m1... a mn of real numbers a ij (1 i m, j j n) double ordered by row and by column. So, the matrix A = [a ij ] 1 i m,1 j n has m rows and n columns and a ij is precisely the element that belongs both to the i-th row and to the j-th column. In the sequel a real matrix will be simply referred to as a matrix for the sake of simplicity. Example Consider the following matrix ( ) A = Then A is a 2 4 matrix. For example, we have that a 13 = 3 and a 22 = 5.

7 4 Chapter 1. Some concepts of matrix algebra We now present the basic definitions of a square matrix and an identity matrix. Definition (Square Matrix). An n n matrix A = [a ij ] 1 i n,j j n is said to be a square matrix of order n (in this case there are n rows and n columns). In the particular case when n = 1 (there is only one column), we have to do with a column vector a of R m, that is a = a 1 a 2.. a m. Definition (Identity matrix). The identity matrix of order n is the square matrix I n of order n that is defined as follows: I n = so that the elements on the main diagonal of A (i.e. the elements a ii with 1 i n) are all equal to 1 while all the other elements are equal to 0. So, for example, we have I 2 = ( ), I 3 = ,, I 4 = It is possible to define the operations of sum of two matrices and product of a matrix and a real number. Definition (Sum of two matrices). Let two m n matrices A = [a ij ] 1 i m,1 j n and B = [b ij ] 1 i m,1 j n be given. Then the sum C = A+B of the matrices A and B is the m n matrix C = [c ij ] 1 i m,1 j n whose generic element c ij is the sum of the corresponding elements a ij and b ij of A and B, respectively (that is, c ij = a ij + b ij with 1 i m, 1 j n)..

8 1.1. Basic definitions 5 Example Consider the following 2 4 matrices ( ) ( A =, B = ). We have that ( ) A + B = = ( Definition (Product of a matrix and a real number). Let λ be a real number and consider an m n matrix A = [a ij ] 1 i m,1 j n. Then the product λa of λ and A is the m n matrix λa = [λa ij ] 1 i m,1 j n. The following example concerns a linear combination of two matrices. Both the sum and the product by a real number are involved. ). Example Consider the following 2 4 matrices ( ) ( A =, B = ). We want to determine the 2 4 matrix defined as 3A 2B. We have that ( ) ( ) A 2B = 3 2 = ( ) ( ) = + = ( ) = Definition (Product of two matrices). Let A = [a ij ] 1 i m,1 j n be an m n matrix and let B = [b ij ] 1 i n,1 j r be an n r matrix (so A has n columns and B has n rows). Then we define the product row by column of A and B as the m r matrix AB = C = [c ij ] 1 i m,1 j r with m rows and r columns whose generic element c ij is obtained by multiplying the i-th row a i of the matrix A and the j-th column b j of the matrix B, that is n c ij = a ih b hj (i = 1,..., m, j = 1,..., r). h=1 Remark The product of two matrices A and B is not defined whenever the number of columns of A is different from the number of rows of B. We present an example concerning the product of two matrices.

9 6 Chapter 1. Some concepts of matrix algebra Example Consider the following matrices ( ) ( A =, B = ). Notice that BA is not defined since B has 3 columns and A has 2 rows. We have that ( ) ( ) AB = = ( ) ( 2) 1 ( 4) = = ( 1) ( 1) ( 2) 2 ( 4) + ( 1) 6 ( ) = Remark It should be noted that for every square matrix A of order n we have that AI n = I n A = A. Remark The product of two matrices is not commutative. Even if for two matrices A and B both the products AB and BA are defined, it is not true in general that AB = BA. In order to illustrate this fact, consider the following two square matrices of order 2: A = ( ), B = ( ). We have that AB = ( ), BA = ( ). 1.2 Determinants We present the definition and the basic properties of the determinants. We consider determinants as tools in order to establish sufficient conditions for the existence of points of (relative) maximum and minimum. This will be done in the following chapter. In practice, we shall limit ourselves to the consideration of determinants of square matrices of order 2 or 3. Nevertheless, the definition concerns the most general case of a square matrix of arbitrary dimension.

10 1.2. Determinants 7 Definition (Determinant). Let A be a square matrix of order n, A = [a ij ] 1 i n,j j n. Then the determinant of A is the real number det(a) = a a 1n a a 2n a n a nn = ( 1) d(π) a 1π(1) a nπ(n), π where the previous summation is over all permutations (bijective mappings) π : {1,..., n} {1,..., n} and d(π) is the degree of the permutation π (the number of exchanges needed to give the fundamental permutation (1, 2,..., n) starting from the permutation (π(1),..., π(n))). In the sequel we shall consider in particular determinants of square matrices of order 2 and of order 3. Therefore, we need a fast procedure in order to calculate the determinants in these cases. Remark (Determinant of a 2 2 matrix). Given a square matrix of order 2 ( ) a11 a A = 12, a 21 a 22 we have that det(a) = a 11 a 22 a 12 a 21 since the permutation {2, 1} requires one exchange. Remark (Determinant of a 3 3 matrix). Given a square matrix of order 3 we have that A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33, det(a) = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 a 13 a 22 a 31 a 11 a 23 a 32 a 12 a 21 a 33 since the permutations {1, 2, 3}, {2, 3, 1} and {3, 1, 2} require an even number of exchanges while the permutations {3, 2, 1}, {1, 3, 2} and {2, 1, 3} require an odd number of exchanges. There is a fast rule which allows us to immediately arrive at the determinant of a 3 3 matrix. It is called the Sarrus rule".

11 8 Chapter 1. Some concepts of matrix algebra Remark (Sarrus rule). Given a square matrix of order 3 A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 write the matrix A and the first two columns of A subsequently on the right side, that is a 11 a 12 a 13 a 11 a 12 a 21 a 22 a 23 a 21 a 22 a 31 a 32 a 33 a 31 a 32. Then consider the sum of the products of the elements of the 3 diagonals starting above to the left and ending below to the right (i.e., a 11 a 22 a 33, a 12 a 23 a 31, a 13 a 21 a 32 ) and subtract the sum of the products of the elements of the 3 diagonals starting below to the right and ending above to the left (i.e., a 13 a 22 a 31, a 11 a 23 a 32, a 12 a 21 a 33 ). The resulting number is precisely the determinant of A (i.e., det(a) = a 11 a 22 a 33 +a 12 a 23 a 31 +a 13 a 21 a 32 a 13 a 22 a 31 a 11 a 23 a 32 a 12 a 21 a 33 ). Example (Determinant of some 3 3 matrix). If we consider the square matrix of order 3 A = then we have that det(a) = ( 2) ( 2) + ( 4) 1 3 ( 2) 0 ( 4) 3 ( 2) = = 8. Remark (Determinant of the product λa). It is easily seen that the following property of the determinants holds true :,, (i) for every real number λ and for every square matrix A of order n we have that det (λa) = λ n deta. Definition (Transpose matrix). Let A = [a ij ] 1 i m,1 j n be an m n matrix. Then the transpose of A is the n m matrix A = [a ij ] 1 i n,1 j m whose generic element a ij is defined as follows a ij = a ji 1 i n, 1 j m. It should be noted that the the transpose A of any matrix A is obtained by writing the rows of A as the columns of A or equivalently by writing the columns of A as the rows of A.

12 1.3. The algebraic structure of R n 9 Example Consider the 2 4 matrix ( ) A = Then its transpose A is defined as follows: A = Remark (Determinant of the transpose of a matrix). From the definition of the determinant it is immediate to see that for every square matrix A the determinant of A is equal to the determinant of A (i.e., det(a) = det(a )). Definition (Symmetric matrix). If A = [a ij ] 1 i n,1 j n is a square matrix of order n and we have that A = A, then A is said to be a symmetric matrix. Example Consider the following matrices of order 3: A = 1 0 4, B = We have that A is symmetric while B is not symmetric. 1.3 The algebraic structure of R n In this section we present the basic properties of the space R n with respect to the operations of sum of two elements and product of an element and a real number. As usual, we shall denote by R the set of all real numbers and by R n (n N + ) the n-dimensional Euclidean space consisting of all n-tuples x = (x 1, x 2,..., x n ) of real numbers. We consider on R n the following operations: sum of two elements of R n (+ : R n R n R n ) (x 1, x 2,..., x n ) + (y 1, y 2,..., y n ) = (x 1 + y 1, x 2 + y 2,..., x n + y n );

13 10 Chapter 1. Some concepts of matrix algebra product of a real number and an element of R n ( : R R n R n ) α (x 1, x 2,..., x n ) = (αx 1, αx 2,..., αx n ) The algebraic structure (R n, +, ) (consisting of the underlying space R n together with the operations + and ) is said to be a real vector space. Therefore, we can refer to x R n as a vector of R n. We present the properties according to which (R n, +, ) is a real vector space. Just notice that the definition of (real) vector space is much more general, but goes beyond our purposes. Definition (Vector space). We refer to (R n, +, ) as a real vector space since the following properties are verified: (1) (x + y) + z = x + (y + z) for every x, y, z R n ; (2) There exists a (unique) element 0 = (0,..., 0) such that x+0 = 0+x = x for every x R n ; (3) For every x R n there exists a so called opposite element x = ( x 1,..., x n ) such that x + ( x) = 0; (4) x + y = y + x for every x, y R n ; (5) α(βx) = (αβ)x for every α, β R, for every x R n ; (6) 1x = x for every x R n ; (7) α(x + y) = αx + αy for every α R, for every x, y R n ; (8) (α + β)x = αx + βx for every α, β R, for every x R n. Remark The algebraic structure (R n, +) satisfies the following properties listed above: (1) associative property, (2) existence of a neutral element, (3) existence of an opposite element of any element of R n, (4) commutative property. (R n, +) is said to be a commutative (abelian) group.

14 1.4. The topological structure of R n 11 Definition (Euclidean distance). Given two vectors x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) of R n, the Euclidean distance between x and y is the real number defined as follows: d(x, y) = n (x i y i ) 2. i=1 Remark (Properties of the Euclidean distance). The Euclidean distance satisfies the following properties: (0) d(x, y) 0 for every x, y R n ; (1) d(x, y) = 0 if and only if x = y for every x, y R n ; (2) d(x, y) = d(y, x) for every x, y R n (symmetric property); (3) d(x, y) + d(y, z) d(x, z) for every x, y, z R n (triangle inequality). The pair (R n, d) is referred to as a metric space (in the sense that there is a metric (i.e., a distance) d : R n R n R that associates to every pair of vectors of R n a real number in a such a way that the previous properties are verified). Also in this case, the concept of a metric space is much more general, as well as the concept of norm presented in the following definition. Definition (Euclidean norm). The Euclidean norm of any vector x R n is defined to be the real number x = d(x, 0) = n x 2 i. i=1 Remark (Properties of the norm). The (Euclidean) norm satisfies the following properties: (0) x 0 for every x R n ; (1) x = 0 if and only if x = 0 for every x R n ; (2) αx = α x for every α R, for every x R n ; (3) x + y x + y for every x, y R n.

15 12 Chapter 1. Some concepts of matrix algebra 1.4 The topological structure of R n We now present the basic definition of an open ball centered at a point of point of R n. The basic topological concepts will follow. Definition (Open ball centered at a point). Let x 0 be a vector of R n and let r be a positive real number. We denote by B r (x 0 ) the open ball with center at x 0 and radius r, that is defined to be the set of all vectors (points) of R n whose distance from x 0 is smaller than r, that is to say B r (x 0 ) = {x R n : d(x, x 0 ) < r} = x Rn : n (x i x 0 i )2 < r. i= Figure 1.1 Open ball centered at (0, 0) with radius 1.

16 1.4. The topological structure of R n 13 Definition (Accumulation point). A point x 0 R n is a said to be an accumulation point of a set A R n if every open ball with center at x 0 contains infinitely many points of A. Definition (Interior point of a set). A point x 0 R n is said to be an interior point of a set A R n if there exists an open ball centered at x 0 that is contained in A (that is, r R, r > 0 such that B r (x 0 ) A). Definition (Open sets and closed sets). Let A be a subset of R n. Then A is said to be (1) open if every point of A is an interior point of A (that is, for every x 0 A there exists r R, r > 0 such that B r (x 0 ) A); (2) closed if the complement of A, denoted by CA = R n \ A is open (that is, for every x 0 CA there exists r R, r > 0 such that B r (x 0 ) A = ). Example It is easy to check (but here we state without proof) that any open ball B r (x 0 ) is an open set. Further, every closed ball is a closed set. C r (x 0 ) = {x R n : d(x, x 0 ) r} = x Rn : n (x i x 0 i )2 r Remark (properties of the open sets). If we denote by τ the family consisting of all the open subsets of R n, then we have that τ satisfies the following properties: (i) X τ, τ; i=1 (ii) the union of every family of open sets is an open set; (iii) the intersection of any finite family of open sets is an open set. We say that τ is a topology on R n. Definition (Convergent sequence). We say that a sequence {x h } h N + of points of R n converges to a point x 0 R n (that is, x h x 0 ) if for every real number ǫ > 0 there exists a natural number h N + such that d(x h, x 0 ) < ǫ for every h > h (i.e., lim h + d(xh, x 0 ) = 0).

17 14 Chapter 1. Some concepts of matrix algebra Example (Example of a convergent sequence). Consider the sequence of points of R 2 defined as follows: {( 1 h, h )} h N +. We claim that ( 1 h, 1+ 1 h ) (0, 1). Indeed, consider that d(( 1 h, 1+ 1 h ), (0, 1)) = 1 h h = 2 2 h. For any fixed positive real number ǫ we have that 2 2 h < ǫ 2 as soon as h > such that h > 2 ǫ 2 ǫ.. So we only have to consider any natural number h N+ Definition (Bounded set). A subset A of R n is said to be bounded if it is contained in an open ball B r (0) centered at 0 (i.e., if there exists r R, r > 0 such that A B r (0)). Definition (compact set). A subset A of R n is said to be compact if it is closed and bounded.

18 CHAPTER 2 Real functions of n real variables 2.1 General concepts We shall denote by f : (R n )D R a real function of n real variables that is defined on the domain D. So f is a law (o prescription) that associates to every point (vector) x D a well determined real number f(x), that is called the value of f at the point x. From now on (unless a different specification) a vector x R n is thought of as a column vector x 1 x 2 x =... x n Example (Linear functions). A function f : R n R is said to be a linear function if f satisfies the following two conditions; (i) f(x + y) = f(x) + f(y) for all x, y R n ; (ii) f(αx) = αf(x) for all x R n and for all α R. It can be shown that a function f is linear if and only if there exists a (column) vector a R n such that for every x R n n f(x) = a x = (a 1,..., a n ) x = a i x i. i=1

19 16 Chapter 2. Real functions of n real variables Definition (Graph). If f : (R n )D R is a real function of n real variables then the graph of f is the subset of R n+1 that is defined as follows: G f = {(x 1, x 2,..., x n, f(x 1, x 2,..., x n )) : x = (x 1, x 2,..., x n ) D}. It should be noted that in the particular case when n = 2, the graph of a function of two real variables appears as a surface in the three-dimensional space Figure 2.1 Graph of f(x, y) = x 2 + y 2. Definition (Level curve). If f : (R n )D R is a real function of n real variables and c is any real number, then we refer to the level curve (corresponding to c) as the subset D f c of D defined as follows: D f c = {x D : f(x) = c}. Example (f(x, y) = x 2 +y 2 ). Consider the function (positive definite quadratic form) f : R 2 R, f(x, y) = x 2 +y 2. Then for every positive real number c the c-level curve D f c = {x D : x2 + y 2 = c} is a circumference centered at (0, 0) with radius c. We now present, for the sake of completeness, the basic definitions of continuity of a function at a point and finite limit at a point.

20 2.2. Quadratic forms Figure 2.2 Level curves of f(x, y) = x 2 + y 2. Definition Continuity at a point] Let f : (R n )D R be a real function of n real variables and consider a point x 0 D. Then f is said to be continuous at x 0 if f verifies the following condition: for every real number ǫ > 0 there exists a real number δ > 0 such that f(x) f(x 0 ) < ǫ for every x D such that d(x, x 0 ) < δ. Definition (Continuity on a domain). A function f : (R n )D R is said to be continuous (on D) if f is continuous at every point x 0 D. Definition (Finite limit at a point). Let f : (R n )D R be a real function of n real variables and let x 0 be an accumulation point of its domain D. We say that f has finite limit l ( R) as x approaches x 0 ( lim = l) if x x0f(x) for every real number ǫ > 0 there exists a real number δ > 0 such that f(x) l) < ǫ for every x D with 0 < d(x, x 0 ) < δ.

21 18 Chapter 2. Real functions of n real variables Figure 2.3 Level curves of f(x, y) = x 2 + y. 2.2 Quadratic forms In this section we introduce the quadratic forms and study their representations and sign by using the considerations in the previous chapters. Definition (Definition of a quadratic form). A quadratic form q on R n is a function q : R n R of the form n j q(x 1,..., x x ) = a ij x i x j. j=1 i=1

22 2.2. Quadratic forms 19 Definition (Definite and semidefinite quadratic forms). A quadratic form q on R n is said to be (i) positive definite (negative definite) if q(x) > 0 (q(x) < 0) for every x 0 (x R n ); (ii) positive semidefinite (negative semidefinite) if q(x) 0 (q(x) 0) for every x R n ; (iii) indefinite of sign if q(x) > 0 for some x R n and also q(x) < 0 for some x R n. Remark (Associated symmetric matrices). A quadratic form q(x 1,..., x x ) = n j=1 j i=1 a ijx i x j can be represented by means of a symmetric matrix A (of order n) in such a way that q(x) = x Ax for every x R n. To this aim it suffices to consider the following symmetric matrix: A = [a ij ] 1 i n,j j n = 1 a 11 2 a a 1 12 a a 1n 2 a 2n 1 2 a 1n a nn We say that A is the symmetric matrix associated to the quadratic form q. Example (Quadratic form on R 2 ). Consider the quadratic form q on R 2 defined as q(x, y) = x 2 2xy + y 2. If we define x = ( x y then we have that q(x) = x Ax. ), A = ( Example (Quadratic form on R 3 ). Consider the quadratic form on R 3 defined as q(x, y, z) = x 2 + 2y 2 + 3z 2 + 4xy 6xz + 8yz. If we define x = x y z then we have that q(x) = x Ax., A = ) We now introduce the different kinds of definition of a symmetric matrix by using the corresponding notions of definitions of a quadratic form.,.

23 20 Chapter 2. Real functions of n real variables Definition (Positive definite symmetric matrix). The symmetric matrix A of order n is said to be (i) positive definite (negative definite) if the quadratic form q(x) = x Ax is positive definite (negative definite); (ii) positive semidefinite (negative semidefinite) if the quadratic form q(x) = x Ax is positive semidefinite (negative semidefinite);; (iii) indefinite of sign if the quadratic form q(x) = x Ax, q : R n R is indefinite of sign. Example (Matrices definite of sign). (i) The symmetric matrix A = ( ) is indefinite of sign since q(x 1, x 2 ) = x Ax = x 2 1 x2 2 (q(1, 0) = 1 > 0) or negative q(0, 1) = 1 < 0). may be positive (ii) The symmetric matrix A = ( ) is positive definite since q(x 1, x 2 ) = x Ax = x x2 2 all x 0. is greater than zero for (iii) The symmetric matrix A = ( ) is positive semidefinite since q(x 1, x 2 ) = x Ax = x x 1x 2 + x 2 2 nonnegative but it it is equal to zero whenever x 1 = x 2. is always The following concept of principal minor is very useful in order too analyze the definition of a quadratic form.

24 2.2. Quadratic forms 21 Definition (Principal minors). Let A = [a ij ] 1 i n,1 j n be a square matrix of order n. Define, for every 1 h n, A h = [a ij ] 1 i h,1 j h = a a 1h a a 2n a h a hh. The principal minor of order h is defined to be the determinant of A h. We do not prove that following theorem, which contains a characterization of the positive/definition of a symmetric matrix, and therefore of a quadratic form, since we can always refer to the associated symmetric matrix. Theorem (Conditions for positive definitions). Let A be a symmetric matrix of order n. Then we have that: (i) A is positive definite if and only if all the n principal minors are positive; (ii) A is negative definite if and only if all its n principal minors alternate in sign with, in particular, det(a 1 ) < 0, det(a 2 ) > 0, det(a 3 ) < 0,..., and so on; (iii) if at least one principal minor of A is different from 0 but its sign doesn t agree with any of the previous cases, then A is indefinite of sign; (iv) A is positive semidefinite if and only if all its n principal minors are nonnegative and at least one of them is equal to zero; (v) A negative semidefinite if and only if all its n principal minors of odd order are smaller or equal to zero, all its principal minors of even order are greater or equal to zero and at least one of them is equal to zero.

25 22 Chapter 2. Real functions of n real variables Example (Positive definite matrices). (i) The symmetric matrix A = ( is positive definite since det(a 1 ) = 2 > 0 and det(a) = 2 1 = 1 > 0. ) (ii) The symmetric matrix A = is indefinite of sign since det(a 1 ) = 1 > 0 e det(a) = 7 < 0. (iii) The symmetric matrix A = is negative semidefinite since det(a 1 ) = 1 < 0, det(a 2 ) = 0 e det(a) = Differentiability and optimization We introduce the basic concepts related to the differential calculus in n variables. Definition (Partial derivative). Let f : (R n )D R be a real function of n real variables and let x 0 D be an accumulation point of D. If the limit f (x 0 f(x 0 1 ) = lim,..., x0 i 1, x0 i + h, x0 i+1,..., x0 n ) f(x0 ) x i h 0 h exists and it is finite then such a limit is said to be the partial derivative of f with respect to x i at the point x 0.

26 2.3. Differentiability and optimization 23 Definition (Gradient vector). Let f : (R n )D R be a real function of n real variables and let x 0 D be an accumulation point of D. If the partial derivatives f x i (x 0 ) exist for all i {1,..., n}, then we define the gradient vector D 1 f(x 0 ) of f at the point x 0 as follows: D 1 f(x 0 ) = f x 1 (x 0 )... f x n (x 0 ). Theorem (Derivative of compound functions). Let z = f(x 1,..., x n ) be a real function of n real variables such that the partial derivatives f x i (x) exist for every x belonging to the domain of f and for every i {1,..., n}. Consider m real functions of n real variables x i = g i (t 1,..., t m ) (i = 1,..., n) such that the partial derivatives g i t j (t) exist for every t belonging to the common domain of the functions g i. Then, for every j {1,..., m}, we have that z = z x 1 + z x z x n. t j x 1 t j x 2 t j x n t j Example (Derivative of a compound function). Consider the function z = f(x, y) = e x 2y, x = t t 2, y = 2t 2. From Theorem 2.3.3, we have that z = z x + z y = 2t 1 e x 2y = 2t 1 e t2 1 3t 2, dt 1 x t 1 y t 1 z = z dt 2 x x + z t 2 y y t 2 = e x 2y 4e x 2y = 3e t2 1 3t 2. It is clear that, equivalently, we could have considered directly the partial derivatives with respect to t 1 and t 2 of the function f(t 1, t 2 ) = e t2 1 3t 2. Definition (Continuous differentiability). A function f : (R n )D R is said to be continuously differentiable or C 1 on D if the partial derivative f x i (x) exists and it is continuous at every point x D for all i {1,..., n}.

27 24 Chapter 2. Real functions of n real variables Definition (Partial derivatives of higher order). If a function f : (R n )D R is C 1 on D and the partial derivative with respect to x j (j {1,..., n}) of f x i (x) exists for every x D, then we shall write f x j ( ) f (x) = 2 f (x). x i x j x i Further, if 2 f x j x i (x) is continuous at every point x D and for every i, j {1,..., n}, then we shall say that f is C 2 or twice continuously differentiable on D. Example Consider the real function of two real variables defined as Then we have that f x (x, y) = log(x y2 ) + f(x, y) = xlog(x y 2 ). x x y 2, f y (x, y) = 2xy x y 2, 2 f 1 (x, y) = x2 x y 2 y 2 (x y 2 ) 2, 2 f x + y2 (x, y) = 2x y2 (x y 2 ) 2, 2 f x y (x, y) = 2y 3 (x y 2 ) 2 = 2 f (x, y). y x Definition (Hessian matrix). Let f : (R n )D R be a real function of n real variables. If f is C 2 then at every point x 0 D we can define the hessian matrix of f D 2 f(x 0 ) = 2 f x f (x 0 ) 2 f 2 f x n x 1 (x 0 ) x 2 x 1 (x 0 )... 2 f (x 0 ) x x 1 x 2 (x 0 ) f x 1 x n 1 (x 0 ) f x 1 x n (x 0 ).... Hence we have that D 2 f(x 0 ) = [a ij ] 1 i n,1 j n with a ij = 2 f x j x i (x 0 ) (i, j {1,..., n}). 2 f x 2 n (x 0 ).

28 2.3. Differentiability and optimization 25 Remark (symmetric property of the hessian matrix). If f : (R n )D R is C 2 then, according to Schwartz Theorem, we have that for every x D and for every pair (i, j) of indexes the following property is verified: 2 f x j x i (x) = 2 f x i x j (x). Therefore the hessian matrix D 2 f(x) is symmetric for every x D. Definition (points of minimum and maximum). Let f : (R n )D R be a real function of n real variables. Then a point x 0 D is said to be (i) a point of maximum (minimum) if f(x) f(x 0 ) (f(x) f(x 0 )) for every x D; (ii) a point of strict maximum (minimum) if f(x) < f(x 0 ) (f(x) > f(x 0 )) for every x D, x x 0 ; (iii) a point of relative maximum (minimum) if there exists an open ball B r (x 0 ) such that f(x) f(x 0 ) (f(x) f(x 0 )) for every x B r (x 0 ) D; (iv) a point of strict relative maximum (minimum) if there exists an open ball B r (x 0 ) such that f(x) < f(x 0 ) (f(x) > f(x 0 )) for every x B r (x 0 ) D, x x 0. The famous Weierstrass theorem guarantees the existence of a maximum and a minimum of a continuous function on a compact set. We present the statement without the proof. Theorem (Weierstrass Theorem). Let f : (R n )D R be a real function of n real variables. If f is continuous and D is compact, then there exists a point of minimum and a point of maximum for f.. Theorem (Necessary condition for a relative minimum or maximum). If f : (R n )D R is C 1 and x 0 D is a point of relative minimum or maximum that is in addition an interior point of D then it must be f x i (x 0 ) = 0 for i = 1,..., n, or equivalently D 1 f(x 0 ) = 0.

29 26 Chapter 2. Real functions of n real variables Definition (Stationary point). Let f : (R n )D R be a real function of n real variables. We say that x 0 D is a stationary point of f if or equivalently D 1 f(x 0 ) = 0. f x i (x 0 ) = 0 for i = 1,..., n, In order to present sufficient conditions for the existence of relative maxima or minima, we need the Taylor s formula of order 2. Theorem (Taylor s formula of order 2). Let f : (R n )D R be a real function of n real variables which is defined on an open set D. If f is C 2 on D and if x 0 D then there exists a real function of n real variables R 2 (h; x 0 ) such that for every point x 0 + h D such that the segment with extremes x 0 and x 0 + h is contained in D it holds that f(x 0 + h) = f(x 0 ) + [D 1 f(x 0 )] h h D 2 f(x 0 )h + R 2 (h; x 0 ), where the function R 2 (h; x 0 ) is such that R 2 (h; x 0 ) lim h 0 h 2 = 0. We are now ready to prove the following theorem. Theorem (Sufficient condition for relative maximum or minimum). Let f : (R n )D R be a real function of n real variables that is defined on an open set D. If f is C 2 and x 0 D is a stationary point for f then x 0 is (i) a point of strict relative maximum if the hessian matrix D 2 f(x 0 ) is negative definite; (ii) a point of strict relative minimum if the hessian matrix D 2 f(x 0 ) is positive definite. If the hessian matrix D 2 f(x 0 ) is indefinite of sign, then x 0 is neither a point of minimum nor a point of maximum for f. Proof. We shall only prove that statement (ii) holds. The proof of statement (i) is perfectly analogous. Let x 0 D be a stationary point for a real-valued function f which is C 2 on an open set D R n. Assume that the hessian matrix D 2 f(x 0 ) is positive definite. From the Taylor s formula (see Theorem ), since x 0 D is a stationary point of f (hence, D 1 (x 0 ) = 0) we have

30 2.3. Differentiability and optimization 27 that, for every h R n such that the segment with extremes x 0 and x 0 + h is contained in D, f(x 0 + h) = f(x 0 ) h D 2 f(x 0 )h + R 2 (h; x 0 ), and, dividing by h 2, h f(x 0 + h) f(x 0 ) h 2 = 1 2 h D2 f(x 0 h ) h + R 2(h; x 0 ) h 2. The quadratic form q(h) = h D 2 f(x 0 )h is continuous on R n and therefore, from the Weierstrass theorem , attains a minimum value a on the unit circle {v R n : v = 1}, which is a closed and bounded set. On the other hand, since the hessian matrix D 2 f(x 0 ) (and therefore the quadratic form q(h) = h D 2 f(x 0 )h) is positive definite, such a minimum value a is greater than 0. Hence, we have that h 0 < a 2 < 1 2 h D2 f(x 0 h ) h. R 2 (h; x 0 ) Since lim h 0 h 2 = 0, there exists a real number r > 0 such that, if 0 < h < r, then a 4 < R 2(h; x 0 ) h 2 < a 4. Therefore, as soon as 0 < h < r, we have that h f(x 0 + h) f(x 0 ) h 2 = 1 2 h D2 f(x 0 h ) h + R 2(h; x 0 ) h 2 > a 2 a 4 > 0, which implies that x 0 is a point of strict relative minimum for f. Definition (Saddle point). Let f : (R n )D R be a real function of n real variables. Then a point x 0 D is said to be a saddle point of f provided that x 0 is a stationary point and the hessian matrix D 2 f(x 0 ) is indefinite of sign. In this case every open ball centered at x 0 contains points x D such that f(x) < f(x 0 ) and points x D such that f(x) > f(x 0 ). In the following remark we consider the sufficient conditions for the existence of relative maxima or minima in the case of two variables.

31 28 Chapter 2. Real functions of n real variables Figure 2.4 Graph of f(x, y) = x 2 y 2, saddle point x 0 = (0, 0). Remark (Sufficient conditions in the case of two variables). Let f be a real function of two real variables and assume that its domain D is open. From the previous theorem, if f is C 2 and x 0 D is a stationary point for f then x 0 is (i) a point of strict relative maximum if 2 f x 2 (x 0 ) < 0, 2 f x 2 (x 0 ) 2 f y 2 (x 0 ) (ii) a point of strict relative minimum if 2 f x 2 (x 0 ) > 0, 2 f x 2 (x 0 ) 2 f y 2 (x 0 ) [ 2 f x y (x0 )] 2 > 0; [ 2 f x y (x0 )] 2 > 0; (iii) a saddle point for f if 2 f x 2 (x 0 ) 2 f y 2 (x 0 ) [ 2 f x y (x0 )] 2 < 0. Nothing can be said in general if det(d 2 f(x 0 )) = 0. Example (Extremal points in case of 3 variables). Consider the real function of 3 real variables defined as We have that f x (x, y, z) = 4x(x2 +y 2 ) y, f(x, y, z) = (x 2 + y 2 ) 2 + z 2 xy. f y (x, y, z) = 4y(x2 +y 2 ) x, f (x, y, z) = 2z. z

32 2.3. Differentiability and optimization 29 In order to determine the stationary points of f, consider the system 4x(x 2 + y 2 ) y = 0 4y(x 2 + y 2 ) x = 0. z = 0 The previous system is equivalent to the following one: (4x 2 + 4y 2 + 1)(x y) = 0 4y(x 2 + y 2 ) x = 0, z = 0 that in turn is equivalent to the following pair of systems: 4x 2 + 4y = 0 4y(x 2 + y 2 ) x = 0 z = 0, x y = 0 4y(x 2 + y 2 ) x = 0 z = 0. The first one has no solutions in in R 3, while the second one, that can be written as follows: x(8x 2 1) = 0 y = x, z = 0 has solutions (0, 0, 0), ( 2 of the second order are 4, 2 2 f x 2 (x, y, z) = 4(3x2 + y 2 ), 2 f x y (x, y, z) = 2 f (x, y, z) = 8xy 1, y x We have that 2 f x (x, y, z) 2 (x, y, z) Since 2 f y x 4, 0) and ( 2 4, 2 4, 0). The partial derivatives 2 f y 2 (x, y, z) = 4(x2 + 3y 2 ), 2 f y z (x, y, z) = 2 f (x, y, z) = 0. z y 2 f x y (x, y, z) 2 f y 2 (x, y, z) D 2 (f(x, y, z)) = 2 f (x, y, z) = 2, z2 2 f x z (x, y, z) = 2 f (x, y, z) = 0, z x = 16(3x2 + y 2 )(x 2 + 3y 2 ) (8xy 1) 2. 4(3x 2 + y 2 ) 8xy 1 0 8xy 1 4(x 2 + 3y 2 ) we have that det(d 2 f(x, y, z) = 2[16(3x 2 + y 2 )(x 2 + 3y 2 ) (8xy 1) 2 ]). Hence, 2 f x 2 (0, 0, 0) = 0 = det(a 1 ), det(a 2 ) = 1, det(d 2 f(0, 0, 0)) = 2.,

33 30 Chapter 2. Real functions of n real variables The point (0, 0, 0) is a saddle point (therefore it is not an extremal point). If we consider the points ( 2 4, 2 4, 0) and ( 2 4, 2 4, 0), we obtain that D 2 (f(x, y, z)) is positive definite at those points and therefore these are points of relative minimum. 2.4 Convex and concave functions WE present some arguments concerning convexity. Definition (Convex set). A subset D of R n is said to be convex if for all x, y D and for every real number t [0, 1] we have that tx + (1 t)y D. If t [0, 1], then tx + (1 t)y D is said to be a convex linear combination of the points x and y. In this case, tx + (1 t)y is a point of the segment whose extremes are x and y. Definition (Convex and concave functions). A function f : (R n )D R defined on a convex set D R n is said to be (i) convex (concave) if f(tx + (1 t)y) tf(x) + (1 t)f(y) (respectively, f(tx + (1 t)y) tf(x) + (1 t)f(y)) for all x, y D and for every real number 0 < t < 1; (ii) strictly convex (strictly concave) if f(tx + (1 t)y) < tf(x)+(1 t)f(y) (respectively, f(tx + (1 t)y) > tf(x) + (1 t)f(y)) for all x, y D and for every real number 0 < t < 1. The convexity/concavity of a quadratic form is simple to be analyzed by means of the following proposition, which is not proven here. Proposition (Convex quadratic forms). A quadratic form q on R n is convex (concave) if and only if q is positive semidefinite (negative semidefinite). Further, q is strictly convex (concave) if q is positive definite (negative definite).

34 2.4. Convex and concave functions Figure 2.5 The function f(x, y) = sin(x + y) is neither convex nor concave. Example (Strictly convex quadratic form). Consider the following quadratic form on R 2 : q(x, y) = x 2 4xy + 8y 2. Then we have that q(x, y) = x Ax with x = (x, y) and A = ( Since D 2 f(x) A is positive definite, we have that q is strictly convex. We present the proof of the following nice result. Theorem (LocalRelative minimum and convexity). Let f : (R n )D R be a convex (concave) function. Then every point x 0 D that is a point of relative minimum (maximum) is also a point of minimum (maximum) for f on D. Proof. Let f be a convex function on a set D R n and let x 0 be a point of relative minimum for f. Then, from the definition of a point of relative minimum, there exists an open ball B r (x 0 ) such that f(x) f(x 0 ) for every x B r (x 0 ) D. Let x D be a generic point of the domain of f. Consider the convex linear combination z t = tx 0 + (1 t)x (0 < t < 1). Since f is ).

35 32 Chapter 2. Real functions of n real variables convex, we have that f(z t ) = f(tx 0 +(1 t)x) tf(x 0 )+(1 t)f(x) for every real number t such that 0 < t < 1. On the other hand, if t sufficiently near 1 then z t belongs to B r (x 0 ) and therefore f(x 0 ) f(z t ) tf(x 0 )+(1 t)f(x) implies that f(x 0 ) tf(x 0 )+(1 t)f(x). This last inequality can be written as (1 t)f(x 0 ) (1 t)f(x), which is equivalent to f(x 0 ) f(x). Hence, since such a condition is verified for every x D, we have that x 0 is a point of minimum for f. Theorem (Condition for convexity). Let f : (R n )D R be C 2 on an open and convex set D. Then f is convex (concave) if and only if the hessian matrix D 2 f(x) is positive (negative) semidefinite for all x D. If the hessian matrix D 2 f(x) is positive (negative) definite for all x D, then f is strictly convex (concave). Example (Convex function). Consider the function f(x, y) = e x y on R 2. We have that ( ) D 2 e x y e f(x) = x y. e x y e x y Since D 2 f(x) is positive semidefinite, we have that f is convex. Theorem (Convexity and extremal points). Let f : (R n )D R be C 1 on an open and convex set D. If f is convex (concave) and x 0 is a stationary point for f, then x 0 is a point of relative minimum (maximum). If in addition f is strictly convex (concave), then the stationary point x 0 is the unique point of strict minimum (maximum). 2.5 Constrained optimization with equality constraints We now present the some essential material concerning the constrained optimization with equality constraints. Consider m + 1 real functions f, g 1,..., g m : (R n )D R, with m < n. In the [constrained optimization problem with equality constraints we look for the points of minimum and maximum of the function f subject to the constraints g i (x 1,..., x n ) = 0 with i = 1,..., m. Since it is clear that min f(x) = max ( f(x)), we can limit ourselves to the consideration of the following problem:

36 2.5. Constrained optimization with equality constraints 33 max f(x 1,..., x n ) sub g 1 (x 1,..., x n ) = 0. g m (x 1,..., x n ) = 0 (2.5.1) Since we want to reduce ourselves to the consideration of a problem of unconstrained optimazation, let us introduce the Lagrangean function (or simply the Lagrangean) L : D R m R, L(x 1,..., x n, λ 1,..., λ m ) = m f(x 1,..., x n ) λ i g i (x 1,..., x n ). i=1 where the m real variables λ 1,..., λ m are said to be the Lagrange multipliers. Let us now define the set m D 0 = {(x 1,..., x n ) R n : g i (x 1,..., x m ) = 0}, i=1 and let us assume that the functions f, g 1,..., g m are C 1 on D. If a point (x 0, λ 0 ) D 0 R m is a stationary point of the Lagrangean L, then x 0 is a candidate to be a point of maximum for f constrained to g i (x 1,..., x n ) = 0 (i = 1,..., m). A possible justification of such an assertion can be found in the following Theorem, which immediately goes after the definition of a saddle point of the Lagrangean. Let us first notice that, with the following definitions: λ = λ 1. λ m, g(x) = g 1 (x). g m (x) the pervious problem of constrained maximum with equality constraints can be formulated as follows: max f(x), sub g(x) = λ. (2.5.2) Definition (Saddle point of the Lagrangean). If we consider the constrained maximization problem with equality constraints 2.5.2, a point (x 0, λ 0 ) D R m is said to be a saddle point of the Lagrangean L(x, λ) = f(x) λ g(x) if L(x 0, λ 0 ) = max x min L(x, λ). λ

37 34 Chapter 2. Real functions of n real variables Theorem (saddle points and constrained maxima). Consider the constrained maximization problem with equality constraints If (x 0, λ 0 ) is a saddle point of the Lagrangean L(x, λ), then x 0 is a solution of the problem Proof. If (x 0, λ 0 ) is a saddle point of the Lagrangean L(x, λ), then for every λ R m and for every vector x D we have that L(x, λ 0 ) L(x 0, λ 0 ) L(x 0, λ). The last inequality is equivalent to the requirement according to which, for every λ R m, [λ λ 0 ] g(x 0 ) 0, that in turn implies that g(x 0 ) = 0 (that is, x 0 must satisfy the equality constraints). Then the first inequality becomes f(x) λ 0 g(x 0 ) f(x 0 ), which is equivalent to f(x) f(x 0 ) since x satisfies the equality constraints. Therefore x 0 is actually a point of (absolute) maximum as soon we require that the constraints g(x) = 0 are satisfied. Example (extremal points with equality constraints). Consider the following real function of two real variables: f(x, y) = x 2 + y2 2. We want to determine the maximum and the minimum of f conditional to the equality constraints x 2 + y 2 6y = 0. The Lagrangean is L(x, y, λ) = x 2 + y2 2 λ(x2 + y 2 6y). The system of the partial derivatives with respect to the variables x, y e λ equal to zero is x 2 + y 2 6y = 0 2x 2λx = 0 x(1 λ) = 0. y 2λy + 6λ = 0 From the second equation we get x = 0 or else λ = 1. If λ = 1, then we have that x = 0 and y = 6. If λ 1, then we must have x = 0 and y = 0 or y = 6. If y = 0, then we have that λ = 0 and therefore the solutions are x = 0, y = 0 e λ = 0. If y = 6, then it must be λ = 1 and we arrive at a contradiction. Finally, since f(0, 6) = 18 and f(0, 0) = 0, we have that the points (0, 6) e (0, 0) are the points of maximum and respectively minimum given the contraints.

38 2.6. Exercises Exercises Quadratic forms Study the definition of the following quadratic forms on R n : 1. q(x, y, z) = x 2 + y 2 + z 2 2xy + 2yz; 2. q(x, y, z) = x 2 + 2y 2 + z 2 + 2xy 2xz; 3. q(x, y, z) = x 2 + y 2 + z 2 2xz; 4. q(x, y, z) = x 2 + 4y 2 + 2z 2 4xy; 5. q(x, y, z) = x 2 + y 2 + z 2 4xy 4xz; 6. q(x, y, z) = x 2 + 4y 2 + 2z 2 4xy: 7. q(x, y, z) = x 2 + 3y 2 z 2 2xy + 2yz; 8. q(x, y, z) = x 2 + 5y 2 + z 2 2xy 4yz; 9. q(x, y, z) = x 2 + 2y 2 + z 2 2xy + 2xz; 10. q(x, y, z) = x 2 + y 2 + 2z 2 xy xz. Answer the following multiple choice questions concerning a quadratic form q(x) on R n with an associated symmetric matrix A: 11. (a) q(x) is positive definite if there exists a positive minor of A; (b) q(x) is negative definite if det(a 1 ) < 0 and det(a 2 ) < 0; (c) indefinite of sign if det(a 1 ) < 0 and det(a 2 ) < 0; (d) none of the previous assertions is true. 12. (a) q(x) is positive semidefinite if there exists a nonnegative minor of A; (b) q(x) is negative semidefinite if there exists a non positive minor of A; (c) q(x) is indefinite of sign if det(a 1 ) 0 and det(a 2 ) < 0; (d) none of the previous assertions is true.

39 36 Chapter 2. Real functions of n real variables 13. (a) q(x) is negative semidefinite if all the principal minors of A are nonnegative; (b) q(x) is positive definite if det(a 2 ) < 0; (c) indefinite of sign if det(a 1 ) < 0; (d) none of the previous assertions is true. 14. (a) q(x) is positive definite if all the principal minors of A are nonnegative; (b) q(x) is negative definite if all the principal minors of A are nonnegative; (c) indefinite of sign if det(a 2 ) < 0; (d) none of the previous assertions is true. 15. (a) q(x) is positive definite if all the principal minors of A are nonnegative; (b) q(x) is indefinite of sign if det(a 2 ) < 0; (c) q(x) is negative definite if det(a 1 ) < 0; (d) none of the previous assertions is true. 16. (a) q(x) is positive definite if all the principal minors of A are nonnegative; (b) q(x) is negative definite if det(a 1 ) < 0; (c) indefinite of sign if det(a 2 ) < 0; (d) none of the previous assertions is true. Maxima and minima Determine the nature of the stationary points of the following functions: 1. f(x, y) = x 2 x y + 2xy; 2. f(x, y) = x(x y 1); 3. f(x, y) = x 2 + y 2 (x + 1); 4. f(x, y) = x(x + y + 1); 5. f(x, y) = x 2 2xy + y 3 ; 6. f(x, y) = x 2 y(1 2x);

40 2.6. Exercises f(x, y) = x 3 xy + x + y; 8. f(x, y) = 1 x + 1 y + xy. Answer the following multiple choice questions concerning a C 2 real function f : (R n )D R of n real variables and a point x 0 D: 9. (a) a point of strict relative maximum if the gradient at x 0 is D 1 f(x 0 ) = 0; (b) a saddle point of f if the hessian matrix D 2 f(x 0 ) is indefinite of sign; (c) a point of strict relative minimum if the gradient at x 0 is D 1 f(x 0 ) = 0 and the hessian matrix D 2 f(x 0 ) is negative definite; (d) none of the previous assertions is true. 10. (a) if x 0 is an interior point of strict relative maximum then the gradient at x 0 is D 1 f(x 0 ) = 0; (b) (c) (d) if the hessian matrix D 2 f(x 0 ) is positive definite then x 0 is a point of strict relative maximum; if the hessian matrix D 2 f(x 0 ) is positive definite then x 0 is a point of strict relative minimum; none of the previous assertions is true. 11. (a) a point of strict relative minimum if the gradient at x 0 is D 1 f(x 0 ) = 0 and the hessian matrix D 2 f(x 0 ) is positive definite; (b) a saddle point of f if the hessian matrix D 2 f(x 0 ) is indefinite of sign; (c) a point of strict relative maximum if the gradient at x 0 is D 1 f(x 0 ) = 0 and the hessian matrix D 2 f(x 0 ) is positive definite; (d) none of the previous assertions is true. 12. (a) a point of strict relative minimum if the gradient at x 0 is D 1 f(x 0 ) = 0 and the hessian matrix D 2 f(x 0 ) is negative definite; (b) a saddle point of f if the gradient at x 0 is D 1 f(x 0 ) = 0; (c) (d) a saddle point of f if the hessian matrix D 2 f(x 0 ) is negative definite; none of the previous assertions is true. 13. (a) a point of strict relative minimum if the gradient at x 0 is D 1 f(x 0 ) = 0;

41 38 Chapter 2. Real functions of n real variables (b) a saddle point of f if the hessian matrix D 2 f(x 0 ) is positive definite; (c) a point of strict relative maximum if the gradient at x 0 is D 1 f(x 0 ) = 0 and the hessian matrix D 2 f(x 0 ) is positive definite; (d) none of the previous assertions is true.

42 CHAPTER 3 The classical financial regimes 3.1 General concepts We consider in this chapter the basic notions concerning the transfer of money between two different agents at different times. Then we present the classical financial regimes of the linear interest, the commercial discount and the compound interest. Assume that a time unit is fixed. We shall assume that time is measured in years unless a different specification. We start by presenting the basic concepts of present value, final value, interest and discount in the deterministic case and in the classical framework. It is known that the use of the money is compensated by means of the payment of an interest. So the interest is the cost of using money. To analyse financial transactions, a clear understanding of the concept of interest is required. In the most common context, interest is an amount charged to the borrower for the use of the lender s money over a period of time. Looking at this from the lender s perspective, the money the lender is investing is changing value with time due to the interest being added. For this reason, interest is sometimes referred to as the time value of money. We assume that 0 is the initial date at which a certain amount C is invested. So C is the money invested in financial transactions. It is sometimes referred to as the principal. Then the amount C has grown to a future date t > 0 is referred to as the final value F. Then F is the sum of the initial value C and the interest I corresponding to C, that is F(t) = C + I(t). (3.1.1)

43 40 Chapter 3. The classical financial regimes Therefore, C can be referred to as the present value of F. Interest expressed as a percent of the principal C, will be referred to as an interest rate. Finally, we refer to D(t) = F(t) C (3.1.2) as the discount over the amount F(t) at time 0 since it is precisely the difference between the amount F(t) (due at a future date t > 0) and the present value of F(t) calculated at 0. Please notice that this difference is considered at 0, because if such a difference is considered at time t then clearly we have that F(t) C = I(t) from the basic formula (3.1.1). The different ways of calculating I give rise to the classical financial regimes. Example Let F(t) denote the final value value of an investment at time t years. (a) (b) Write an expression giving the amount of interest earned from time t to time t + s in terms of F only. Use (a) to find the annual interest rate, i.e., the interest rate from time t years to time t + 1 years. Remark Interest rates are most often computed on an annual basis, but they can be determined for non-annual time periods as well. For example, a bank offers you for your deposits an annual interest rate of 10% compounded semi-annually. Imagine a fund growing at interest. It would be very convenient to have a function representing the accumulated value of an invested principal at any time. Unless stated otherwise, we will assume that the change in the fund is due to interest only. If t is the length of time, measured in years, for which the principal has been invested, then the amount of money at that time will be denoted by F(t). We now present the classical financial regimes. 3.2 Regime of the linear interest Consider an investment ofe1such that the interest earned in each year is constant and equals to i. Then, at the end of the first year, the accumulated value (final value) is F(1) = 1 + i, at the end of the second year it is F(2) = 1 + 2i and at the end of the n th year it is F(n) = 1 + in, n 0.

44 3.2. Regime of the linear interest 41 Thus, we consider a linear accumulation function, F(n) = 1 + in, n 0. The accruing of interest according to this function is called simple interest. Note that the effective rate of interest, i = F(1) 1 is also called the simple annual interest rate. In the financial regime of the linear interest (or simple interest) the interest is proportional both to the initial capital C that is invested at time 0 and to the length t of the investment period, that is I(t) = Cit, (3.2.1) where i is the factor of proportionality that is defined to be as the annual interest rate. Indeed, if in the previous formula we put C = 1 and t = 1, we get precisely I(1), that is therefore the interest yielded when an amount of e 1 is invested for one year. We can have, for example i = 2%, or i = 4% and so on. The final value and the present value in this regime are determined by means of the following laws of capitalization and respectively actualization: F(t) = C(1 + it), (3.2.2) F(0) = C = F(t) 1 + it. (3.2.3) If we put C = 1 and t = 1 in formula (3.2.2), then we get F(1) = 1+i. This is the final value ofe1invested for one year. We define u = 1 + i (3.2.4) and we refer to u as the capitalization factor. We further define the actualization factor (1) v as follows: v = 1 u = i. (3.2.5) Hence, v is the present value at 0 ofe1due at t = 1. Since necessarily we have that i > 0, it must be u > 1 and 0 < v < 1. Remark Even though the rate of simple interest is constant over each period of time, i, the effective rate of interest per period (i n ) is not constant, it is decreasing from each period to the next. 2. Because of this fact, simple interest is less favorable to the investor as the number of periods increases. For later use we define the annual discount rate d as follows: (1) This is said to be fattore di attualizzazione in the Italian language.

45 42 Chapter 3. The classical financial regimes f(t) 1+i f(t) = 1+it 1 1 t Figure 3.1 Graph of the final value in the regime of the simple interest. d = 1 v. (3.2.6) Then d is precisely the discount overe1invested for one year (see definition (3.1.2)). Please notice that d = 1 v = i = i = iv, (3.2.7) 1 + i so that d is the present value at 0 of i due at time 1. We have that d < i. Finally, notice that we have introduced the (pure) numbers i, v, u and d. We have that the assignment of one of these numbers automatically determines all the other numbers. 3.3 Regime of the commercial discount According the factor of proportionality d, in this regime the discount D on a final amount F(t) (see definition (3.1.2)) is thought of as proportional to the length t of the investment period and to F(t), that is D(t) = F(t)dt. (3.3.1) Notice that the factor of proportionality must be equal to the annual discount rate d since if we put F = 1 and t = 1 we get precisely the discount on an amount ofe1invested for one year. In this regime we have that the laws of actualization and respectively capitalization are the following ones: C = F(t) D(t) = F(t)(1 dt); (3.3.2) F(t) = C 1 dt. (3.3.3) Clearly we must have 1 dt > 0, so that, when d is assigned, the previous laws are valid only whenever the time t is such that 0 t < 1 d.

46 3.4. Regime of the compound interest or exponential regime 43 f(t) f(t) = 1 1 dt 1+i 1 1 1/d t Figure 3.2 Graph of the final value in the regime of the commercial discount. 3.4 Regime of the compound interest or exponential regime Simple interest has the property that the interest earned is not invested to earn additional interest. In contrast, compound interest has the property that the interest earned at the end of one year is automatically invested in the next year to earn additional interest. Let us start with an investment ofe1and with compound interest rate i per year. At the end of the first year, the accumulated value is 1 + i. At the end of the second year, the accumulated value is (1 + i) + i(1 + i) = (1 + i) 2, so that, after t years, F(t) = (1 + i) t for integral t 0. Interest accruing according to this function is called compound interest and we call i the rate of compound interest. This is the only regime that is fully justified from a mathematical viewpoint. In general, if i is the annual interest rate, then the future value F(t) at t of an initial amount C due at 0 is determined as follows: F(t) = C(1 + i) t = Cu t. (3.4.1) Therefore, the expression (3.4.1) is the capitalization law in the regime of the compound interest. The present value (at 0) of an amount F(t) due at t is the following: C = F(t) (1 + i) t = F(t)vt. (3.4.2) Hence, the expression (3.4.2) is the actualization law in the regime of the compound interest.

47 44 Chapter 3. The classical financial regimes The interest I(t) (on an initial amount C invested for t years) is therefore determined by means of If we define I(t) = F(t) C = C((1 + i) t 1) = C(u t 1). (3.4.3) δ = log(1 + i), (3.4.4) then we have that the previous laws of capitalization and actualization (3.4.1) and (3.4.2) can be written as follows: F(t) = Ce δt, C = F(t)e δt. We say that δ is the force of interest. Observe that (2) δ i i2 2. f(t) f(t) = (1+i) t 1+i 1 1 t Figure 3.3 Graph of the final value in the regime of the compound interest. The comparison among the final values in the different regimes is illustrated in the following proposition. Proposition Let 0 < i < 1. We have (a) 1 1 dt < (1 + i)t < 1 + it for 0 < t < 1; (b) 1 1 dt = (1 + i)t = 1 + it for t = 0 or t = 1; (c) 1 1 dt > (1 + i)t > 1 + it for t > 1; (2) This is a consequence of the Taylor s formula relative to the function f(x) = log(1+x) with initial point x 0 = 0, according to which we have that log(1 + x) = x x2 + 2 o(x2 ) o(x 2 ) with lim = 0. x 0 x 2

48 3.5. Exercises 45 f(t) f(t) = 1 1 dt f(t) = (1+i) t 1+i f(t) = 1+it 1 1 1/d t Figure 3.4 Comparison of the graphs of the final value corresponding to the initial amount C = 1 in the different regimes. Remark It was already observed that i, d, u and v are pure numbers". Hence we have that the regime of the compound interest (or exponential regime) is the only regime that is correct from a dimensional viewpoint. Indeed, in this case F(t) = C(1 + i) t = Ce δt has the dimension of an amount of money, since δ has the dimension of the reciprocal of time (that is, number / time ). On the other hand, the other regimes are not correct from a dimensional point of view. For example, in the regime of the linear interest, F(t) = C(1 + it) has not the dimension of an amount of money, since 1 + it is not a pure number (it has the dimension of time). The same happens in the case of the regime of the commercial discount. Example It is known thate600 invested for 2 years will earne264 in interest. Find the accumulated value ofe2000 invested at the same rate of compound interest for 3 years. 2. At an annual compound interest rate of 5%, how long will it take you to triple your money? 3.5 Exercises Solve the following exercises: 1. Determine in the regime of the linear interest the present value of e 600 due after four years and seven months when the annual interest rate is i = 0.06;

49 46 Chapter 3. The classical financial regimes 2. Determine in the regime of the linear interest the present value of e 1000 due after four years and eight months when the annual discount rate is d = 0.07; 3. Determine in the regime of the commercial discount the present value ofe800 due after three years and three months when the annual interest rate is i = Answer to the following multiple choice questions: 4. Consider the regime of the linear interest with annual interest rate i. Then the interest generated by the investment of a capital C from 0 to a time t is (a) (b) I(t) = Ci; I(t) = Cit; (c) I(t) = C[it 1]; (d) none of the previous assertions is true. 5. Consider the regime of the compound interest and let δ be the force of interest. Then (a) e δ = 1 i; (b) log(1 + δ) = i; (c) e δ = 1 + i; (d) none of the previous assertions is true. 6. Consider the regime of the compound interest corresponding to a force of interest δ (annual interest rate i). Then the future value F of an initial amount C after t years is (a) F = Cte δt ; (b) F = C(1 + i) δt ; (c) F = Ce tlog(1+i) ; (d) none of the previous assertions is true. 7. Let i be the annual interest rate and d be the annual discount rate. Then (a) d = (b) 1+i ; d = i(1 + i);

50 3.5. Exercises 47 (c) d = 1 1 (d) 1+i ; none of the previous assertions is true. 8. Denote by F l (t) and F e (t) the future value at a time t > 0 of a capital C due at time 0 in the regime of the linear interest and respectively in the exponential regime. Then (a) F l (t) > F e (t) if t > 1; (b) F l (t) > F e (t) if t < 1; (c) F l (t) < F e (t) if t < 1; (d) none of the previous assertions is true. 9. Let i be the annual interest rate and δ be the force of interest. Then (a) e δ 1 = i; (b) e δ + i = 1; (c) e δ i = 1; (d) none of the previous assertions is true. 10. Consider the regime of the compound interest corresponding to a force of interest δ (annual interest rate i). Then the present value an amount F due after t years is (a) C = Fte δt ; (b) C = F(1 + i) δt ; (c) C = Fe δt ; (d) none of the previous assertions is true. 11. Consider the regime of the compound interest and let δ be the force of interest and i the annual interest rate. Then (a) e δ = i; (b) e δ 1 = i; (c) (d) e δ + 1 = log(1 + i); none of the previous assertions is true.

51

52 CHAPTER 4 The regime of the compound interest 4.1 The final value under the regime of the compound interest Assume that the initial amount that is due at t = 0 is equal toe 1. Our aim is to recover the capitalization law (3.4.1) by imposing acceptable conditions on the the function F(t) that represents the final value of 1 at time t 0. From the definition of F(t), we have that F(0) = 1, F(1) = 1 + i. If we consider the change of F between t and t + t, that is F(t) = F(t + t) F(t), we have that F(t) can be thought of as the interest on the initial amount F(t) when this amount is invested on t until the date t + t. Assume that F(t) = δf(t) t + o( t), (4.1.1) where o( t) is any function of t that is an infinitesimal of order greater than 1 as t 0, that is to say o( t) lim = 0. t 0 t This means that the interest on the amount F(t) is proportional to F(t) itself and to the length t of the investment period according to a proportionality factor δ that is constant with respect to t. Dividing both members in (4.1.1) by t we get F(t) t = δf(t) + o( t). t

53 50 Chapter 4. The regime of the compound interest If we pass to the limit as t 0 we arrive at the following differential equation (1) : F (t) = δf(t), that is equivalent to d logf(t) = δ. dt It is immediate to establish that the (general) solution is determined as follows: logf(t) = dt + k F(t) = e dt+k = e δt e k, where k is any real number. Using the initial condition F(0) = 1, we have that F(0) = 1 = e k, that obviously implies k = 0. Hence, it must be F(t) = e δt. In order to check that δ is actually the force of interest, just observe that, since F(1) = 1 + i, we have that F(1) = 1 + i = e δ, or equivalently δ = log(1 + i). In a more general setting, the force of interest may depend on time. Remark A more realistic hypothesis would require that the force of interest depends on time, hence it is actually a function δ(t). In this case, condition (4.1.1) becomes that leads to the differential equation F(t) = δ(t)f(t) t + o( t), (4.1.2) F (t) = δ(t)f(t), whose solution, considering the initial condition F(0) = 0, is the following: In this case the condition F(1) = 1 implies that 1 0 t F(t) = e δ(u)du 0. (4.1.3) δ(u)du = log(1 + i). 4.2 Equivalence of interest rates The annual interest rate i was defined to be the interest overe1that is invested for precisely one year. On the other hand, sometimes it is useful to introduce interest rates relative to a fraction of the year. In this case, for the (1) A differential equation is an equation involving a function F and its derivatives. The function F itself appears as the unknown.

54 4.3. Force of Interest: Continuous compounding 51 sake of clarity, we refer to the effective interest rate relative to such a fraction. When we speak of effective rate of interest we mean interest is paid once per measurement period at the end of the period. So, for example, we can refer to i [2] as a the the semiannual (compound) interest rate, or the interest rate relative to a period of six months), i [3] as the interest rate relative to a period of four months (one third of the year), i [4] as the interest rate relative to a period of three months (one fourth of the year), and so on. Hence, we have that i = i [1], and in general if the year is divided into k periods (where k is a positive integer), we denote by i = i [k] the corresponding (compound) interest rate. Since the regime of the compound interest is taken into consideration, for a fixed positive integer k, the following identity must be true: 1 + i = (1 + i [k] ) k. (4.2.1) Indeed, the final value ofe1invested for one year must be equal to the final value of e 1 invested (say) k times subsequently" using the interest rate i [k]. Hence, the conversion formulas that for a fixed positive integer k allows us to recover i from i [k] or the converse are the following: i [k] = (1 + i) 1 k 1, (4.2.2) i = (1 + i [k] ) k 1. (4.2.3) A nominal rate of interest j [k] payable k times per period, where k is a positive integer, represents k times the effective rate of compound interest used for each of the k th of a period. In this case, it is clear that j[k] k is the effective rate of interest for each k th period. We say that j [k] is the interest rate convertible k times per year. Example Find the final value ofe 3000 to be paid at the end of 8 years with a rate of compound interest of 5% per year; convertible quarterly; convertible monthly. 2. Given the annual interest rate of 12%, compounded monthly, find the equivalent effective annual interest rate.

55 52 Chapter 4. The regime of the compound interest 4.3 Force of Interest: Continuous compounding Effective rates of interest and discount measure interest over one full measurement period (typically, the year), while nominal rates of interest and discount measure interest over k ths of a period. Now we want to measure interest at any particular moment of time. This measure is called force of interest (see Definition (3.4.4)). We consider the case of a nominal compound interest j [k] converted k times a period. We can think of the force of interest, denoted by δ = log(1 + i), where i is the effective annual interest rate, as the limit of j [k] as the number k of times we credit the compounded interest goes to infinity. Indeed we have that (2) lim k + j[k] = lim k + ki[k] = Hence, = lim δ e δ k 1 = δ, k + δ k lim k ((1 + i) 1 k 1) = k + e δ k 1 lim k + 1 = k δ = lim k j[k]. (4.3.1) Intuitively, δ represents a nominal interest rate which is converted continuously, a notion of more theoretical importance. However, δ can be used in practice as an approximation to interest converted very frequently, such as daily. Example Given the nominal interest rate of 12%, compounded monthly, find the equivalent force of interest δ. 2. Using a constant force of interest of 4.2%, calculate the present value of a payment of to be made in 8 years time. Now, using the so called accumulation function of compound interest f(t) = (1 + i) t (4.3.2) representing the final value F(t) in the regime of the compound interest corresponding to the initial capital F(0) = C = 1, we notice that δ = log(1 + i) = f (t) f(t). (2) e t 1 We use the following well known relevant limit: lim = 1. t 0 t

56 4.3. Force of Interest: Continuous compounding 53 The definition of force of interest in terms of a compound interest accumulation function can be extended to any accumulation function. Note that, in general, the force of interest can be a function of t (see Remark (4.1.1)). However, in the case of compound interest δ(t) is constant. Also, we notice that, since F(t) = F(0)f(t), we can write δ(t) = F (t) F(t), and therefore we have that t F(t) = F(0)e δ(u)du 0. Example You are given that F(t) = at 2 + bt + c for 0 t 2, and that F(0) = 100, F(1) = 110 and F(2) = 136. Determine the force of interest at time t = 1/2. 2. Find δ(t) in the case of simple interest, that is, when f(t) = 1 + it. 3. A deposit ofe10 is invested at time 2 years. Using a force of interest of δ(t) = t, find the final value of this payment at the end of 5 years. 4. Show that if δ(t) = δ for all t then f(t) = (1 + i) t for some i. Example Find the accumulated value ofee200 invested for 5 years if the force of interest is δ(t) = 1 8+t. Find the accumulation value ofe500 invested for 9 years if the rate of interest is 5% for the first 3 years, 5.5% for the second 3 years and 6.25% for the third 3 years. Find the effective rate over the first four-year period from now if the effective rate of interest is 5% for the first three years from now, 4.5% for the next three years, and 4% for the last three years. What about the six-year period?

57 54 Chapter 4. The regime of the compound interest Example In return for a payment ofe1,200 at the end of 10 years, a lender agrees to paye200 immediately,e400 at the end of 6 years, and a final amount at the end of 15 years. Find the amount of the final payment at the end of 15 years if the nominal rate of interest is 9% converted semiannually. Investor A deposits e 1,000 into an account paying 4% compounded quarterly. At the end of three years, he deposits an additionale1,000. Investor B deposits X into an account with force of interest δ(t) = 1 6+t. After 6 years, investors A and B have the same amount of money. Find X. Example A single payment ofe3, will pay off a debt whose original repayment plan wase1,000 due on January 1 of each of the next 4 years, beginning January 1, If the effective annual rate is 8%, on what date must the payment ofe3, must paid? 4.4 Axiomatization of the regime of the compound interest In this section we want to present a formal derivation of the final value under the regime of the compound interest by using an axiomatic approach. We start from a general value function representing the equivalence of sums of money exigible at different epochs. If s and t are any two epochs such that 0 s t, we shall denote by F(s, C; t) the final value at t of the initial amount C > 0 due at s. Clearly, we have to require that F(s, C; s) = C for every amount C and every epoch s 0. We have that F(s, C; t) is expressed as the sum of C and the interest I(s, C; t) (due at t) generated by the investment of C from s until t, that is F(s, C; t) = C + I(s, C; t). (4.4.1) The following assumptions can be considered concerning the function F(s, C; t) of the three variables s, C and t defined in (4.4.1): homogeneity with respect to the amount C: for every positive real number λ, for every amount C > 0, and for every choice of the epochs s and t we have that F(s, λc; t) = λf(s, C; t);

58 4.4. Axiomatization of the regime of the compound interest 55 uniformity with respect to time: for every fixed chosen amount C, the function F(s, C; t) depends on the epochs s and t (for every s, t such that 0 s t) only through the difference t s (the length of the investment interval [s, t]), and it does not depend on the particular choice of s and t; time divisibility (3) : for every choice of the amount C and of three epochs s, s 1 and t con 0 s < s 1 < t, we have that F(s 1, F(s, C; s 1 ); t) = F(s, C; t); strict monotonicity: F(s, C; t) < F(s, C; t 1 ) for every amount C and for every choice of three epochs s, t and t 1 with 0 s t < t 1 ; normalization condition: F(s, C; s) = C for every amount C and every epoch s 0. The following theorem fully justifies the adoption of the exponential regime. Theorem The following conditions are equivalent: (i) The final value F(s, C; t) admits finite partial derivatives with respect to the variables s and t and satisfies the properties of homogeneity with respect to the amount C, uniformity with respect to time, time divisibility, strict monotonicity and the normalization condition; (ii) There exists a positive constant δ such that F(s, C; t) = Ce δ(t s) for every nonnegative amount C and for all epochs s and t with 0 s t. Proof. Let us first prove the sufficiency part (i) (ii). Assume that the final value F(s, C; t) is continuous and satisfies the properties of homogeneity with respect to the amount C, uniformity with respect to time, time divisibility, strict monotonicity and the normalization condition. Since F(s, C; t) is homogeneous with respect to the amount C, we can limit ourselves to the consideration of the function u(s, t) = F(s, 1; t) of the two temporal variables s and t (0 s t). Indeed, we have that F(s, C; t) = Cu(s, t) for all 0 s t. From the property of uniformity with respect to time, actually we can reduce ourselves to consider u(s, t) as a function of only one variable τ = t s (the length of the investment period. Therefore, we set u(s, t) = f(t s) for some function f of only one nonnegative (temporal) variable. Hence, it remains to show that f(τ) = e δτ (3) This property is referred to as scindibilità in the italian language.

59 56 Chapter 4. The regime of the compound interest for some positive constant δ. Since the original function F(s, C; t) admits finite partial derivatives with respect to the variables s and t, we have that f is differentiable. From both time divisibility and homogeneity with respect to the amount, we have that the function f satisfies the following property for every τ, h > 0: f(τ + h) = f(τ)f(h). Therefore, if we consider the natural logarithm of both members, then have that log f(τ + h) = log f(τ) + log f(h). If we further consider the derivatives with respect to τ and h, then we arrive at log f(τ + h) = τ τ log f(τ) = f (τ) f(τ), log f(τ + h) = h h log f(h) = f (h) f(h). Hence, the ratio f (τ) f(τ) is constant and in particular the axiom of strict monotonicity implies the existence of some positive real number δ such that f (τ) f(τ) = δ. We solve this elementary differential equation by following a perfectly analogous procedure to that presented in Section 5.1 and from the general solution we arrive at the particular solution f(τ) = ke δτ, f(τ) = e δτ, which is obtained by considering he normalization condition f(0) = 1. Finally, we go back toi the definition of the function f in order to recover F(s, C; t) = Ce δ(t s) for every nonnegative amount C and for all epochs s and t with 0 s t. This consideration finishes the first part of the proof. We now prove the necessity part (ii) (i). Assume that there exists a positive constant δ such that F(s, C; t) = Ce δ(t s) for every nonnegative amount C and for all epochs s and t with 0 s t. It is clear that the conditions of homogeneity with respect to time and the

60 4.5. Exercises 57 normalization condition are verified. Further, F(s, C; t) depends on the epochs s and t only through the difference t s and therefore the property of uniformity with respect to time is also verified. In order to show that the property of time divisibility also holds, consider three epochs s, s 1 and t such that 0 s < s 1 < t. Then we have that F(s 1, F(s, C; s 1 ); t) = Ce δ(s 1 s) e δ(t s 1) = Ce δ(t s) = F(s, C; t). In order to show that F(s, C; t) is strictly monotone, consider that if 0 s t < t 1 then F(s, C; t 1 ) = e δ(t 1 s) = e δ(t 1 t) e δ(t s) = e δ(t 1 t) F(s, C; t) > F(s, C; t). Remark It should be noted, that all the financial regimes (linear interest, commercial discount and compound interest) satisfy the axioms of homogeneity with respect to the amount of money, uniformity with respect to time, strict monotonicity and the normalization condition. On the other hand, the regime of the compound interest (exponential regime) is the only regime which satisfy the condition of time divisibility. 4.5 Exercises Solve the following exercises: 1. Determine (in the exponential regime) the four-months interest rate i [3] corresponding to the force of interest δ = 0.03; 2. Determine (in the exponential regime) the force of interest δ corresponding to the three-months interest rate i [4] = 0.01; 3. Determine (in the exponential regime) the two-months interest rate i [6] corresponding to the force of interest δ = 0.07; 4. Determine (in the exponential regime) the force of interest δ corresponding to the two-months interest rate i [6] = 0.01; 5. Determine (in the exponential regime) the force of interest δ corresponding to the six-months interest rate i [2] = 0.02; 6. Determine (in the exponential regime) the six-months interest rate i [2] corresponding to the force of interest δ = 0.05;

61 58 Chapter 4. The regime of the compound interest 7. Determine (in the exponential regime) the force of interest δ corresponding to the four-months interest rate i [3] = 0.01; Answer to the following multiple choice questions: 8. Consider the regime of the compound interest corresponding to a force of interest δ (semiannual interest rate i [2] ). Then the future value F of an initial amount C after t years is (a) F = C(1 + i [2] ) 2δt ; (b) F = C(1 + i [2] ) 2t ; (c) F = C(1 + i [2] ) t ; (d) none of the previous assertions is true. 9. Consider the regime of the compound interest and let δ be the force of interest and i the annual interest rate. Then (a) e δ = i; (b) e δ 1 = i; (c) (d) e δ + 1 = log(1 + i); none of the previous assertions is true. 10. Consider the regime of the compound interest and let i be the annual interest rate and i [3] the equivalent four-months interest rate. Then (a) (1 + i) 3 = 1 + i [3] ; (b) i [3] = (1 + i) 1 3 1; (c) i = (1 + i [3] ) 3 ; (d) none of the previous assertions is true. 11. Consider the regime of the compound interest and let δ be the force of interest and i [3] the equivalent four-months interest rate. Then (a) e δ = (1 + i) 1 3 1; (b) e δ 1 = (1 + i) 1 3 1; (c) e δ + 1 = (1 + i) 1 3 1; (d) none of the previous assertions is true. 12. Consider the following assertions.

62 4.5. Exercises 59 (a) (b) (c) (d) The regime of the commercial discount satisfies time divisibility; The regime of the linear interest does satisfy homogeneity with respect to the amount; The regime of the compound interest does not satisfy time divisibility; none of the previous assertions is true. 13. Consider the following assertions. (a) (b) (c) (d) The regime of the commercial discount satisfies time divisibility; The regime of the linear interest satisfies time divisibility; The regime of the commercial discount satisfies homogeneity with respect to the amount; none of the previous assertions is true. 14. Consider the following assertions. (a) (b) (c) (d) The regime of the commercial discount does not satisfy strict monotonicity; The regime of the compound interest does not satisfy time divisibility; The regime of the linear interest satisfies time divisibility; none of the previous assertions is true. 15. Consider the following assertions. (a) (b) (c) (d) The regime of the linear interest satisfies time divisibility; The regime of the compound interest does not satisfy strict monotonicity; The regime of the commercial discount satisfies time divisibility; none of the previous assertions is true.

63

64 CHAPTER 5 Annuities and perpetuities 5.1 Annual payments From now on, we shall only consider the exponential regime corresponding to some choice of the annual interest rate i. Generally speaking, an annuity is refereed to be as any terminating stream of fixed payments (or incomes) over a specified period of time. A perpetuity is a stream of cash payments that continues forever. Any payment is an instalment that is due at a specified maturity. In particular, we shall refer to the case when the period between two consecutive maturities is always the same, and the instalments are all equal each other. We refer to a unitary annuity if all the instalments are equal toe1. We say that an annuity is ordinary if the instalments are due at the end of each period. We refer to an annuity-due as an annuity such that the instalments are due at the beginning of each period. Consider the case of an ordinary unitary annuity with n instalments due at the end of each year (so, the first instalment is due at time t = 1 and the last instalment at time t = n). We shall denote by a n i the present value (at 0) and by s n i the final value at t = n. Further, we denote by a i the present value (at 0) of an ordinary perpetuity payinge1at each date t = 1, 2,..., n, n 1 n t Figure 5.1 Unitary ordinary annuity with n instalments.

65 62 Chapter 5. Annuities and perpetuities The present values of such an annuity is determined by using the following formula: 1 (1 + i) n a n i = = 1 vn. (5.1.1) i i Indeed, we have that a n i = v + v v n 1 + v n = v(1 + v + v v n 1 ) = v 1 v (1 v + v v2 + v 2 v v n 1 + v n 1 v n ) = v iv (1 vn ) = 1 vn. i There is a much more elegant way of recovering formula (5.1.1) avoiding any algebraic computations. Indeed, assume that the amortization of e 1 borrowed at t = 0 is performed by paying the annual interest over e 1 (that is, the annual interest rate i) at the end of each year for n years. At the (final) maturity n we also give backe1(the amount borrowed at t = 0). If we consider this financial operation at t = 0 we have that it must be (1) 1 = ia n i + v n, so that, solving for a n i the previous equation, we get precisely a n i = 1 vn i. From (5.1.1) we immediately recover the present value of the ordinary perpetuity: a i = lim a n i = n + 1 v n lim n + i = 1 i. (5.1.2) The final value of an ordinary unitary annuity with n instalments due at the end of each year is therefore the following: s n i = u n a n i = un 1. (5.1.3) i Consider now the case of a unitary annuity-due with n instalments due at the beginning of each year. So, the first instalment is due at t = 0 and the last instalment at t = n n 1 n t Figure 5.2 Unitary annuity-due with n instalments. (1) Just consider that, as it is apparent, ia n i is the present value at t = 0 of an ordinary annuity paying i at the end of each year for n years, while v n is the present value at t = 0 ofe1due at t = n.

66 5.2. Non-annual payments 63 The present value and the final value will be denoted by ä n i and s n i, respectively. Further, we denote by ä i the present value (at 0) of a perpetuity-due payinge1at each date t = 0, 1, 2,..., n,... We first consider the present value of such an annuity. From the definition, we have that ä n i = 1 + v v n 1 = ua n i = u 1 vn i Therefore, we get the following formula: = 1 vn. d ä n i = 1 + v v n 1 = 1 vn. (5.1.4) d From (5.1.4) we immediately recover the present value of the perpetuity-due: ä i = lim ä 1 v n n i = lim n + n + d The final value of an annuity-due is determined as follows: = 1 d. (5.1.5) s n i = u n ä n i = un 1. (5.1.6) d We can consider also the deferred cases, that happen when the first instalment is due after m years. The deferred period of m years is denoted by the subscript m to the left, so that the symbols m/ a n i, m/ ä n i, m/ a i and m/ ä i stand, in the presence of a deferred period of m years, for the present value of an ordinary annuity, an annuity-due, an ordinary perpetuity and a perpetuity-due, respectively. We have that m/a n i = v m a n i, m/ä n i = v m ä n i, m/a i = vm i, m/ä i = vm d. Clearly, the future value is unchanged in the presence of a deferred period, so that m/ s n i = s n i and m/ s n i = s n i (the meaning of the symbols is obvious). 5.2 Non-annual payments Consider now the more general case of an ordinary annuity that consists of constant instalments due at each epoch m + h k with 0 m n 1 and 1 h k, where n (the fixed number of years) and k (the fixed number of fractions of the year) are positive integers. So the first payment is due at 1 k

67 64 Chapter 5. Annuities and perpetuities and the period between two consecutive maturities is also equal to 1 k. We have to do with nk instalments and we assume that all the instalments are equal to 1 k by analogy with the unitary case (indeed, we paye1per year if we do not consider any financial argument). 0 1 k 1 k 1 k 2 k 1 k 1 1 k 1+ 1 k 1 k n 1 k 1 k n t Figure 5.3 Fractioned ordinary annuity with nk instalments all equal to 1 k. The present value of such an annuity is the following (consider the analogy with the expression (5.1.1)): a (k) n i = 1 k 1 (1 + i [k] ) nk i [k] = 1 k 1 v n i [k] = i 1 v n ki [k] i = i ki [k] a n i. Indeed from the definition of the equivalent interest rate i [k] we have that (1 + i [k] ) k = v (see (4.2.1)). Hence we have shown that a (k) n i = i ki [k] a n i. (5.2.1) The limit of the previous expression as k + (denoted by a ( ) n i ) is the present value of a continuous stream of payments lasting n years and paying e 1 per year. If we consider that, from the relation (4.3.1), we arrive at the following expression: a ( ) n i = lim k + j[k] = lim k + ki[k] = δ, lim k + a(k) n i = lim k + i ki [k] a n i = i δ a n i. (5.2.2) Therefore, the present value a ( ) i of a continuous ordinary perpetuity" payinge1per-year is a ( ) i = lim n a( ) n i = lim i n δ a n i = 1 δ. (5.2.3) Remark Generally speaking, if the period between two consecutive maturities is 1 k and there are exactly m instalments (equal each other) starting from 1 k, then in order to determine the present value we use a formula that is perfectly analogous to expression (5.1.1) and that is obtained by considering the number m and the equivalent interest rate i [k]. If the instalments are all equal to R, then the present value of such an ordinary annuity is P = R 1 (1 + i[k] ) m i [k].

68 5.3. Amortization and capital construction by constant instalments Amortization and capital construction by constant instalments If C is any fixed amount of money that is due at t = 0, n is a fixed integer and i is the annual interest rate, then the equation C = Ra n i (5.3.1) defines the constant instalment R of the ordinary annuity with n annual (constant) payments at the end of each year that provides the amortization of the debt C. Please notice that the ordinary annuity consisting of n instalments due at 1, 2,..., n and all equal to α n i = 1 a n i is equivalent to the amount 1 at t = 0. Therefore, α n i can be refereed to as the instalment providing the amortization of the unitary debt. On the other hand, if C is any fixed amount of money that is due at the maturity n (with n a fixed integer) then the equation C = Rs n i defines the constant instalment R of the ordinary annuity with n annual (constant) payments at the end of each year that provides the construction of the capital C. Please notice that ordinary annuity consisting of n instalments due at 1, 2,..., n and all equal to σ n i = 1 s n i is equivalent to the amount 1 at t = n. Therefore, σ n i can be refereed to as the instalment providing the construction of the unitary capital. Remark It should be noted that the following identity holds for every positive integer n and for every annual interest rate i: α n i = i + σ n i. (5.3.2) We only have to divide by a n i both members of the basic identity 1 = ia n i + v n.

69 66 Chapter 5. Annuities and perpetuities 5.4 Exercises Solve the following exercises: 1. Determine the present value of an ordinary annuity consisting of 7 constant instalments all equal to e 400 and payable every six months in case that the first instalment is due after two years and the annual interest rate in the regime of the compound interest is i = 5%; 2. Determine the present value of an ordinary annuity consisting of 7 constant instalments all equal to e 200 and payable every two months in case that the first instalment is due after two months and the annual interest rate in the regime of the compound interest is i = 3%; 3. Determine the present value of an ordinary annuity consisting of 10 instalments (the first is due after one year) due every year in case that the first 6 instalments are equal toe400 and the other 4 instalments are equal toe200. The annual interest rate in the regime of the compound interest is i = 6%; 4. Determine the present value of an ordinary annuity consisting of 10 instalments (the first is due after one year) due every year in case that the first 4 instalments are equal toe200 and the other 6 instalments are equal toe300. The annual interest rate in the regime of the compound interest is i = 5%; 5. Determine the present value of an ordinary annuity consisting of 8 constant instalments all equal to e 300 and payable every three months in case that the first instalment is due after three months and the annual interest rate in the regime of the compound interest is i = 4%; 6. Determine the present value of an ordinary annuity consisting of 10 instalments (the first is due after one year) due every year in case that the first 4 instalments are equal toe200 and the other 6 instalments are equal toe300. The annual interest rate in the regime of the compound interest is i = 5%; 7. Determine the present value of an ordinary annuity consisting of 8 constant instalments all equal toe400 and payable every four months in case that the first instalment is due after three years and the annual interest rate in the regime of the compound interest is i = 4%; 8. Determine the present value of an ordinary annuity consisting of 12 constant instalments all equal to e 200 and payable every six months

70 5.4. Exercises 67 in case that the first instalment is due after one year and one half the annual interest rate in the regime of the compound interest is i = 4%; 9. Determine the present value of an ordinary annuity consisting of 10 instalments (the first is due after one year) due every six months in case that the first 5 instalments are equal toe200 and the other 5 instalments are equal to 100 euros. The annual interest rate in the regime of the compound interest is i = 6%;

71

72 CHAPTER 6 Amortization of a debt 6.1 General concepts For the sake of simplicity, we are only concerned with the amortization of an initial debt C considered at t = 0 by means of n instalments R 1, R 2,..., R n, where R h is payable at the epoch h (h = 1,..., n). The generic instalment R h is expressed as the sum of the principal share C h, which is paid as a mere repayment of the capital debt C, and the interest share I h, which is paid as the interest on the residual debt Therefore, we have that Please notice that D h = C h+1 + C h C n = n k=h+1 C k. (6.1.1) R h = C h + I h h = 1,..., n. (6.1.2) D 0 = C 1 + C C n = n C h = C. (6.1.3) We agree on the fact that every quantity we are going to consider at some not necessarily integer epoch t (0 t n) is computed after the payment (if any) of the instalment due at t. Clearly, we consider the compound interest regime corresponding to some fixed annual interest rate i. Therefore, i is used in order to compute the interest. Further, we consider an evaluation annual interest rate i 1 (that is in general different from i). The residual value of the debt (or the value of the debt for the sake of brevity) at a certain time t (with 0 t n) is therefore the present value at t (computed by using the evaluation interest rate i 1 ) of the future" instalments. h=1

73 70 Chapter 6. Amortization of a debt Since there is no payment at t = 0, the value V (0) of the debt at 0 is interpreted as the issue price of the debt. The price of the debt is said to be at par if C = V (0), under par if V (0) < C and over par if V (0) > C. For every fixed time t we denote by s = s(t) the maximum positive integer that is smaller or equal to t. Then the value of the debt at t, denoted by V (t), has the following expression: V (t) = R s+1 (1 + i 1 ) (s+1 t) + R s+2 (1 + i 1 ) (s+2 t) R n (1 + i 1 ) (n t) = n s h=1 R s+h (1 + i 1 ) (s+h t). (6.1.4) If we denote by U(t) and by P(t) the present value of the future" interest shares and respectively the present value of the future" principal shares, it must be V (t) = U(t) + P(t), U(t) = I s+1 (1 + i 1 ) (s+1 t) + I s+2 (1 + i 1 ) (s+2 t) n s + I n (1 + i 1 ) (n t) = I s+h (1 + i 1 ) (s+h t), (6.1.5) h=1 P(t) = C s+1 (1 + i 1 ) (s+1 t) + C s+2 (1 + i 1 ) (s+2 t) C n (1 + i 1 ) (n t) = n s h=1 C s+h (1 + i 1 ) (s+h t). (6.1.6) 6.2 Amortization by constant annual principal shares In this case, for any initial debt C at t = 0 we define C h = C n h = 1,..., n. Therefore, the residual debt is defined as D h = n k=h+1 C n = n h C h = 1,..., n, n and therefore, since the the interest is paid at the end of each year, we have that I h = id h 1 = ic n h + 1 h = 1,..., n. n

74 6.3. Amortization by constant instalments Amortization by constant instalments Given the initial debt C at t = 0, the constant instalment R due at t = 1,..., n is determined, by using formula (5.3.1), as follows: R = C a n i. (6.3.1) Observe that the principal shares increase in a geometric progression with common ratio 1 + i. Indeed, we have that and therefore R = C n (1 + i) = C n 1 (1 + i) 2 =... = C 1 (1 + i) n, C h = R(1 + i) (n h+1) = Cα n i (1 + i) (n h+1). Therefore, the interest share I h is defined as follows: I h = R C h = Cα n i [1 (1 + i) (n h+1) ]. 6.4 Negotiation of a debt: Makeham formula Consider the amortization of a debt in the case when the interest is paid at the end of each year (for n years). In this case, the present value U h at a generic integer epoch h ( {1,...n}) of the future interest shares can be expressed as a function of the present value P h of the future principal shares and of the residual debt D h. Indeed, from the definition (1) of U h, we have that U h = I h+1 (1 + i 1 ) 1 + I h+2 (1 + i 1 ) I n (1 + i 1 ) (n h) = = i(c h C n )(1 + i 1 ) 1 + i(c h C n )(1 + i 1 ) ] + ic n (1 + i 1 ) (n h) = i [C h+1 a 1 i1 + C h+2 a 2 i C n a n h i1 = i[c h+1 1 (1 + i 1 ) 1 i 1 1 (1 + i 1 ) (n h) + C n ] = i i 1 i 1 = i i 1 [D h P h ]. 1 (1 + i 1 ) 2 + C h i 1 [ n h k=1 n h C h+k + k=1 C h+k (1 + i 1 ) k ] (1) Please notice that for the sake of convenience, since h is a generic integer epoch, we write U h and P h instead of U(h) and respectively P (h).

75 72 Chapter 6. Amortization of a debt If we sum P h to both members of the previous identity, we therefore obtain the following expression of the value of the debt at h: V h = P h + i i 1 [D h P h ]. It is easy to show that this last formula is equivalent to the following one, which is referred to as the Makeham formula: In particular, the issue price is V h = D h i 1 i i 1 [D h P h ]. (6.4.1) V 0 = C i 1 i i 1 [C P 0 ]. Therefore, since it is clear that C > P 0, we have that if i 1 > i then V 0 < C, i 1 = i then V 0 = C, and if i 1 < i then V 0 > C. 6.5 Exercises Solve the following exercises: 1. The amortization of a debt ofe5000 is made by means of the payment of 10 constant instalments (the first is due after one year) payed each year. Determine the value of the debt just after four years and two months in case that the annual interest rate is i = 5% and the evaluation rate is i 1 = 6%; 2. The amortization of a debt of e 1000 is made by means of the payment of 10 instalments payed at the end of each year (the first is paid just after one year). Assume that the principal shares are constant and the annual interest rate is 8% in the exponential regime. Determine the second interest share I 2 ; 3. The amortization of a debt ofe4000 is made by means of the payment of 10 constant instalments (the first is due after one year) payed each year. Determine the value of the debt just after four years and four months in case that the annual interest rate is i = 7% and the evaluation rate is i 1 = 8%; 4. The amortization of a debt ofe10000 is made by means of the payment of 15 constant instalments payed each year (the first is due just after two years). Determine an evaluation of the loan just after four years and six months in case that the annual interest rate is i = 4% and the evaluation rate is i 1 = 5%.

76 6.5. Exercises The amortization of an initial debt amounting toe10000 is made by means of the payment of 40 constant instalments to be paid every 3 months (the first is due just after 3 months). Determine the constant instalment if the 3 months interest rate i [4] is equal to 2%; 6. The amortization of a debt ofe5000 is made by means of the payment of 10 constant instalments (the first is due after one year) payed each year. Determine the value of the debt just after four years and four months in case that the annual interest rate is i = 5% and the evaluation rate is i 1 = 6%; 7. The amortization of a debt of e 5000 is made by means of the payment of 8 constant instalments (the first is due after one year) payed each year. Determine an evaluation of the loan just after two years and two months in case that the annual interest rate is i = 8% and the evaluation rate is i 1 = 9%; 8. The amortization of a debt of e 4000 is made by means of the payment of 9 constant instalments (the first is due after one year and four months) payed each year. Determine the value of the debt just after four years and four months in case that the annual interest rate is i = 6% and the evaluation rate is i 1 = 7%.

77

78 CHAPTER 7 Actuarial mathematics 7.1 Probability and Lifetimes In this Chapter we are concerned with the part of Actuarial Mathematics currently called Life Insurance Mathematics. While Actuarial Mathematics concerns, generally speaking, the Theory of Insurance, which can be viewed as the theory of contingent payments, in Life Insurance Mathematics the insured capital depends on the life or death of one person (the insured), while the benefits (if any) are paid to the policy holder (which is possibly different from the insured). Therefore, the central difficulty of issuing life insurance is that of determining the length of the future life of the insured. In this context, the answer to the following central question: What is the value today of a random sum of money (benefit) which will be paid at a random time in the future? is provided (in absence of surcharges) by the so called Net Single Premium which is reasonable to view as the average present value of the actual payments. After presenting the very basic notation describing the probabilities of life of one individual, we introduce the main types of life insurance contracts together with the corresponding premia. The material is far from being exhaustive, but nevertheless we are convinced that it may furnish a first and useful information about this topic. In the cohort life-table model, imagine a number l 0 of individuals born simultaneously and followed until death, resulting data d x, l x for each age x = 0, 1, 2, where l x is the number of lives aged x (i.e. alive at birthday x); d x = l x l x+1 is the number dying between x and x + 1.

79 76 Chapter 7. Actuarial mathematics Now, allowing the age-variable x to take all real values, not just whole numbers, treat S(x) = l x l 0 as a piecewise continuously differentiable non-increasing function called survivor or survival function. Then, for all positive reals x and t, S(x) S(x + t) is the fraction of the initial cohort which fails between times x and x + t, and S(x) S(x + t) S(x) = l x l x+t l x denotes the fraction of those alive at exact age x who fail before x + t. It should be noted that, according to the previous description, the lifetimes covered by the life table are understood to be governed by an identical "mechanism" of failure, and that any probability question about single lifetime is really a question concerning the fraction of those lives about which the question is asked whose lifetimes will satisfy the started property. Example What is the probability of life aged 29 dies between exact ages 35 and 41 or between 52 and 60? Note that a life aged 29 is one of the cohort surviving to the 29th birthday. The life-table data, and the mechanism by which members of the population die, are summarized first through the survivor function S(x) which at integer values of x agrees with the ratios lx l 0. Note that S(x) has values between 0 and 1, and can be interpreted as the probability for a single individual to survive at least x times units. Since fewer people are alive at larger ages, S(x) is a decreasing function of x, and in applications S(x) should be piecewise continuously differentiable. In addition, by definition, S(0) = 1. Another way of summarizing the probabilities of survival given by this function is to define the density function as the (absolute) rate of decrease of the function S, that is to say f(x) = ds(x) dx = S (x). (7.1.1) Then, denoting by P the underlying probability, we have that, for any ages a < b, P(life aged 0 dies between ages a and b) = l a l b l 0 = S(a) S(b) = b a ( S (x))dx = b a f(x)dx. The terminal age ω of a life table is an integer value large enough that S(ω) is negligibly small but no value S(t) for t < ω is zero. For practical purposes,

80 7.1. Probability and Lifetimes 77 no individual lives to the ω birthday. While ω is finite in real life-tables and in some analytical survival models, most theoretical forms for S(x) have no finite age ω at which S(ω) = 0, and in those forms, ω = by convention. An equivalent representation is S(x) = x f(t)dt (7.1.2) where f(t) S (t) is called the failure density. If T denotes the random variable which is the age at death for a newly born individual governed by the same causes of failure as the life-table cohort, then P(T x) = S(x) and P(x T x + ǫ) x+ǫ lim = lim f(u)du = f(x) (7.1.3) ǫ 0 + ǫ ǫ 0 + x as long as the failure density is a continuous function. Two further useful actuarial notations, often used to specify the theoretical lifetime distribution are: t p x = P(T > x + t T x) = S(x + t)/s(x); /t q x = 1 t p x = P(T x + t T x) = [S(x) S(x + t)]/s(x). The quantity /t q x is referred to as the age-specific death rate for periods of length t. In the most usual case where t = 1 and x is an integer age, let us define Therefore, q x = /1 q x, p x = 1 p x. q x = 1 p x = P(T x + 1 T x) = [S(x) S(x + 1)]/S(x). The rate q x would be estimated from the cohort life table as the ratio d x /l x of those who die between ages x and x+1 as a fraction of those who reached age x. We now present the answer to the first question posed in Example We have that P(35 < T 41 T 29) = 6 p 29 /6 q 35 = l 29+6 l 29 l35 l 35+6 l 35 = l 35 l 41 l 29. The way in which the quantity q x varies with x is one of the most important topics of study in actuarial science.

81 78 Chapter 7. Actuarial mathematics Definition The limiting death-rate /ǫ q x /ǫ per unit time as ǫ ց 0 is called by actuaries the force of mortality µ(x). The same function is also called failure intensity, failure rate, or hazard intensity. The reasoning above shows that, for small ǫ, /ǫq x ǫ f(x) S(x) and therefore µ(x) = d dx log(s(x)), Then we have that x+t x µ(y)dy = ln S(x) log S(x + t). tp x = x S(x) = e S(x + t) S(x) 0 µ(y)dy, x+t = e x µ(y)dy. The most popular examples of failure distribution are: 1. the uniform failure distribution: S(x) = (ω x)/ω for 0 x ω and then µ(x) = (ω x) 1 ; 2. the exponential failure distribution: S(x) = e µx for x 0, then µ(x) = µ; 3. the Gompertz-Makeham failure distribution, which is characterized by µ(x) directly defined as A + Bc x. Example Suppose that the force of mortality µ(y) is specified for exact ages y ranging from 5 to 55 as µ(y) = 10 4 ( y ). Then find analytical expressions for the survival probabilities S(y) assuming S(5) = The following remark, containing expressions of the probabilities of death and survival, is a consequence of standard properties of any probability P. Remark We have that n p x = p x p x+1 p x+2 p x+n 1 ; /n q x = q x + p x q x p x q x n 1 p x q x+n 1 ; m/n q x = m p x /n q x+m with m/n q x = P(m < T m + n T x).

82 7.2. Expected Present Values of Insurance Contracts Expected Present Values of Insurance Contracts We are now ready to draw together the expectations of random variables defined as functions of a life-table waiting time T until death and discounting of future payment based on interest rate assumptions. The net single premium or net single risk premium of the contract is the single cash payment made by the insured at the beginning of the insurance period which would exactly compensate for the average of the future payments which the insurer will have to make. There are three types of contracts to consider: endowment, life annuities and insurance. More complicated kinds of contracts can be obtained by combining these in various ways. Let x denote the initial age of the holder of the insurance, life annuity or endowment and assume for convenience that the contract is initiated on the holder s birthday. Fix a non random effective interest rate i, and retain the notation v = (1 + i) 1 that we have already adopted. Definition An endowment contract is a contract which pays a contractual face amount at the end of n policy years if the policy holder initially aged x survives to age x + n. x P x+n C Ð Ú ¼ Figure 7.1 Benefits in an endowment contract The random variable X, present value of the benefits in an endowment contract, is described in Table 7.1. Table 7.1 X: Present value of the benefits in an endowment contract present value Cv n probability np x 0 /nq x The endowment is the simplest contract, since neither the amount nor the time of payment is uncertain. The pure endowment contract commits the insurer to pay an amount of C = F(0) at time n if T x + n.

83 80 Chapter 7. Actuarial mathematics The net present value for a pure endowment contract with face amount F(0) = C is P = E( X) = Cv n np x. (7.2.1) The net present value for a pure endowment contract with face amount F(0) = 1 is defined to be ne x = v n np x, (7.2.2) where n E x is called actuarial present value. It is easy to see that Consider that ne x < v n. If we define then the expression (7.2.3) becomes ne x = v n np x = v n l x+n l x = v n l x+n l x v x v x = vx+n l x+n v x l x (7.2.3) v x l x = D x, (7.2.4) ne x = D x+n D x. (7.2.5) The Commutation Functions are a computational device to ensure that net single premiums from the same life table and figured at the same interest rate can all be obtained from a single table lookup. The first commutation function is the function D x defined above (see equation 7.2.4). Definition A life annuity contract is an agreement to pay a scheduled payment to the policy holder at the end of each year while the annuitant is alive. The life annuity contract requires the insurer to pay an amount F(k) at each epoch k T x until the insured is alive.

84 7.2. Expected Present Values of Insurance Contracts 81 In the case of ordinary life annuities-immediate, which have payments commencing at time 1 and continuing until death, the expected-present value can be calculated as the composition of more endowment contracts. The situation of an aged x insured in case that the instalment to be paid is constant, F(k) = R, is represented in Figure 7.2. x P x+1 R Ð Ú ¼ ω 1 R Ð Ú ¼ Figure 7.2 Benefits in a life annuity contract If the particular case when each payment is unitary (R = F(k) = 1), the discrete random variable X, present value of the benefits of the insurance company, is represented in Table 7.2. Table 7.2 X: Present value of the benefits present value probability 0 /1q x v = a 1 i v + v 2 = a 2 i v + v 2 + v 3 = a 3 i v + v 2 + v v ω x 1 = a ω x 1 i 1/1 q x 2/1 q x 3/1 q x (ω x 1)/1 q x The net premium P payed by a x-aged insured is determined as follows: P = E( X) = v 1/1 q x + (v + v 2 ) 2/1 q x (v + v v ω x 1 ) (ω x 1)/1 q x = v l x+1 l x+2 l x + (v + v 2 ) l x+2 l x+3 l x (v + v v ω x 1 ) l ω 1 l ω l x = v l x+1 + v 2 l x v ω x 1 l ω 1 l x l x l x = a x. (7.2.6) Multiplying and dividing each term by v x and using the commutation function D x, we arrive at a x = D x+1 + D x D ω 1 D x D x D x

85 82 Chapter 7. Actuarial mathematics If we define the second commutation function we obtain the expression = 1 E x + 2 E x ω x 1 E x. (7.2.7) a x = N x = ω 1 t=x D t, (7.2.8) ω 1 t=x+1 D t D x = N x+1 D x. (7.2.9) Let us now consider other cases life annuities. In the life annuity-due contract, the first instalment is due at time k = 0. Therefore the situation, in the general case of constant instalments all equal to R, is represented in Figure 7.3. x P R x+1 R Ð Ú ¼ ω 1 R Ð Ú ¼ Figure 7.3 Benefits in a life annuity-due contract The net premium of a unitary life annuity-due is therefore ä x = 1 + a x = D x + N x+1 D x = N x D x. (7.2.10) Another situation is represented by the life annuity m-year deferred contact (see Figure 7.4), when the first payment is due precisely after m + 1 years if the insured is alive at age x + m + 1. Therefore, the net premium in the case R 1 is: or equivalently m/a x = D x+m D x m/a x = m E x a x+m, (7.2.11) N x+m+1 D x+m = N x+m+1 D x. (7.2.12) x P x+m x+m+1 R Ð Ú ω 1 R Ð Ú ¼ ¼ Figure 7.4 Benefits in a life annuity m-year deferred contract In a temporary life annuity of n years contract the insurer pays an amount F(k) at each epoch 1 k n until the insured is alive (see Figure 7.5). Hence, the net premium when F(k) R 1 is

86 7.2. Expected Present Values of Insurance Contracts 83 /na x = a x n/ a x = N x+1 N x+n+1 D x. (7.2.13) x P x+1 R Ð Ú x+n R Ð Ú ¼ ¼ Figure 7.5 Benefits in a temporary life annuity of n years Definition An insurance contract is an agreement to pay a face amount if the insured, a life aged x, dies at any time during a specified period, the term of the policy, with payment to be made at the end of the year within the which the death occurs. The first form of the insurance contract is to pay a face amount F(0) = C at x + h + 1 if the death occurs between x + h and x + h + 1. x P x+h x+h+1 C ØÛ Ò x+h x+h+1 ¼ ÓØ ÖÛ Figure 7.6 Benefits in an elementary insurance contract Please notice that there is no benefit if the insured dies before age x + h or survives at age x + h + 1. The random variable, X, present value of the benefits, in case that the face amount C is equal to 1, is represented in Table 7.3. Table 7.3 X: Present value of the benefits present value v h+1 probability hp x q x+h 0 1 ( h p x q x+h ) The expected present value is therefore E( X) = v h+1 hp x q x+h = v h+1 l x+h l x = v h+1 d x+h l x = h/1 A x d x+h l x+h

87 84 Chapter 7. Actuarial mathematics If we define the first commutation function in the death case we have C x = v x+1 d x, (7.2.14) h/1a x = vx+h+1 d x+h v x l x = C x+h D x. (7.2.15) The most general form of an insurance contract is to pay a face amount F(k) at the end of each year if the death occurs between x and x + ω. In this case, the net premium is A x = /1 A x + 1/1 A x + + ω x 1/1 A x = C x + C x C ω 1. D x D x D x If we define M x = ω 1 t=x the last commutation function, we have C t, (7.2.16) A x = M x D x. (7.2.17) The following other forms of insurance contracts can be considered: the m-year deferred insurance contract with a net premium equal to: the n-year temporary contract: m/a x = m E x A x+m = D x+m D x M x+m D x+m = M x+m D x ; (7.2.18) /na x = A x n/ A x = M x M x+n D x. (7.2.19)

88 7.3. Exercises 85 Instead of paying a single premium, the policy holder may decide to pay a net periodical premium P, which will be determined, as usual, on the basis of the equivalence principle, according to which the premium should be set in such a way that the actuarial present value of the benefits paid is equal to the actuarial present value of the premiums received. Therefore, if the policy holder pays for h years a constant net premium P starting from now, based on the specific contract that is considered, it must be in any case E( X) = P /h ä x, where E( X) is the average present value of the benefits. Finally, the expense loaded premium P (T ) includes the provisions for expenses and therefore is determined as follows: P (T ) = P + c, where P is the net premium and c the charge. 7.3 Exercises 1. Use the life table to calculate the net single premium for a life annuity with 7,500 payment per year if the individual is 51 aged. 2. Find the probability for a 35 aged individual to be alive between 55 and 70 if µ(x) = The Mathematical Association of America offers the following alternative to members aged 60. You can pay the annual dues and subscription rate of 90, or you can become a life member for a single fee of 675. Life members are entitled to all benefits of ordinary members, including subscriptions. Should one become a life member? 4. A 37-aged individual contract a policy to receive 60,000 at 70 if alive and 40,000 to the son otherwise. Find the net single premium if i = A 30-aged individual contracts to receive 10,000 at 40 if alive. He pays 2 net premium, the first one, amount P, today and the second one, amount 2P, next year if he is alive. Use the equivalence principle to find P if i = 10% and l x = 80, 000 1, 000(x 30). 6. A 37-aged individual pays today 10,750 of gross premium to receive F at 70 if he is alive. Find F if i = 0.05 and taxes are 3.5% of gross premium.

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90 CHAPTER 8 Solutions to selected exercises Chapter 2 Quadratic forms 1. q(x, y, z) = x 2 + y 2 + z 2 2xy + 2yz is indefinite of sign. A = , det(a 1 ) = 1, det(a 2 ) = 0, det(a) = q(x, y, z) = x 2 + 2y 2 + z 2 + 2xy 2xz is indefinite of sign. A = , det(a 1 ) = 1, det(a 2 ) = 1, det(a) = q(x, y, z) = x 2 + y 2 + z 2 2xz is positive semidefinite. A = , det(a 1 ) = 1, det(a 2 ) = 1, det(a) = 0.

91 88 Chapter 8. Solutions to selected exercises 4. q(x, y, z) = x 2 + 4y 2 + 2z 2 4xy is positive semidefinite. A = , det(a 1 ) = 1, det(a 2 ) = 0, det(a) = q(x, y, z) = x 2 + y 2 + z 2 4xy 4xz is indefinite of sign. A = , det(a 1 ) = 1, det(a 2 ) = 3, det(a) = q(x, y, z) = x 2 + 4y 2 + 2z 2 4xy is positive semidefinite. A = , det(a 1 ) = 1, det(a 2 ) = 0, det(a) = q(x, y, z) = x 2 + 3y 2 z 2 2xy + 2yz is indefinite of sign. A = , det(a 1 ) = 1, det(a 2 ) = 2, det(a) = q(x, y, z) = x 2 + 5y 2 + z 2 2xy 4yz is positive semidefinite. A = , det(a 1 ) = 1, det(a 2 ) = 4, det(a) = q(x, y, z) = x 2 + 2y 2 + z 2 2xy + 2xz is indefinite of sign. A = , det(a 1 ) = 1, det(a 2 ) = 1, det(a) = 1.

92 q(x, y, z) = x 2 + y 2 + 2z 2 xy xz is positive definite. A = , det(a 1 ) = 1, det(a 2 ) = 3 4, det(a) = (c). 12. (c). 13. (d). 14. (c). 15. (b). 16. (c). Chapter 2 Maxima and minima 1. f(x, y) = x 2 x y + 2xy. D 1 f(x, y) = ( 2x 1 + 2y 1 + 2x ), D 2 f(x, y) = ( ). Stationary point is ( 1 2, 0) (saddle point). 2. f(x, y) = x(x y 1). ( D 1 2x y 1 f(x, y) = x ), D 2 f(x, y) = ( ). Stationary point is (0, 1) (saddle point).

93 90 Chapter 8. Solutions to selected exercises Figure 8.1 Graph of f(x, y) = x 2 x y + 2xy Figure 8.2 Graph of f(x, y) = x(x y 1). 3. f(x, y) = x 2 + y 2 (x + 1). ( D 1 2x + y 2 f(x, y) = 2y(x + 1) ), D 2 f(x, y) = ( 2 2y 2y 2(x + 1) ). Stationary points are (0, 0) (relative minimum), ( 1, 2) (saddle point), ( 1, 2) (saddle point). 4. f(x, y) = x(x + y + 1). D 1 f(x, y) = ( 2x 2y 2x + 3y 2 ) (, D f(x, y) = 2 6y ). Stationary point is ( 1, 2) (saddle point).

94 Figure 8.3 Graph of f(x, y) = x 2 + y 2 (x + 1) Figure 8.4 Graph of f(x, y) = x(x + y + 1). 5. f(x, y) = x 2 2xy + y 3. ( ) ( D 1 2x 2y f(x, y) = 2x + 3y 2, D f(x, y) = 2 6y ). Stationary points are (0, 0) (saddle point), ( 2 3, 2 3 ) (relative minimum). 6. f(x, y) = x 2 y(1 2x). ( ) D 1 2x + 2y f(x, y) =, D 2 f(x, y) = 1 + 2x ( ). Stationary point is ( 1 2, 1 2 ) (saddle point). 7. f(x, y) = x 3 xy + x + y. ( D 1 3x f(x, y) = 2 y + 1 x + 1 ), D 2 f(x, y) = ( 6x ).

95 92 Chapter 8. Solutions to selected exercises Figure 8.5 Graph of f(x, y) = x 2 2xy + y Figure 8.6 Graph of f(x, y) = x 2 y(1 2x).

96 Figure 8.7 Graph of f(x, y) = x 3 xy + x + y Figure 8.8 Graph of f(x, y) = 1 x + 1 y + xy. Stationary point is (1, 4) (saddle point). 8. f(x, y) = 1 x + 1 y + xy. ( ) D y f(x, y) = x 2 1, D + x 2 f(x, y) = y 2 ( 2 x y 3 ). Stationary point is (1, 1) (relative minimum). 9. (d). 10. (a). 11. (a). 12. (d). 13. (d).

97 94 Chapter 8. Solutions to selected exercises Chapter 3 1. Present value = ( ) = i = 1 1 d 1 = , Present value = ( ) = d = i 1+i = , Present value = 800( ( )) = (b). 5. (c). 6. (c). 7. (c). 8. (b). 9. (a). 10. (c). 11. (b). Chapter 4 1. i [3] = (1 + i) = e = δ = log(1 + i [4] ) 4 = i [6] = (1 + i) = e δ 6 1 = δ = log(1 + i) = log(1 + i [6] ) 6 ) = δ = log(1 + i) = log(1 + i [2] ) 2 ) = i [2] = (1 + i) = e δ 2 1 = δ = log(1 + i) = log(1 + i [3] ) 3 ) = (b). 9. (b).

98 (b). 11. (d). 12. (b). 13. (c). 14. (d). 15. (d). Chapter 5 1. i [2] = (1 + i) = , Present value = 1 (1+i[2] ) 7 i [2] 400 (1 + i) 3 2 = i [6] = (1 + i) = , Present value = 1 (1+i[6] ) 7 i [6] 200 = Present value = 400 a (1.06) 6 a = Present value = 200 a (1.05) 4 a = Chapter 6 1. R = 5000 a = , V ( ) = Ra = I 2 = ic n 1 n = R = 4000 a = , V ( ) = Ra =

99

100 Bibliography [1] Ken Binmore and Joan Davies, Calcolo differenziale di più variabili, Casa Editrice Ambrosiana, [2] Luciano Daboni and Claudio de Ferra, Elementi di Matematica finanziaria, Edizioni Lint, Trieste, [3] Lorenzo Peccati, Sandro Salsa and Annamaria Squellati, Mathematics for Economics and Business, Egea, [4] Eric V. Slud, Actuarial Mathematics and Life-Table Statistics, Mathematics Department, University of Maryland, College Park, USA, 2006.