Valuing Contingent Payments - Lecture Slides | MATH 3630, Study notes of Mathematics

Download Valuing Contingent Payments - Lecture Slides | MATH 3630 and more Study notes Mathematics in PDF only on Docsity! Valuing Contingent Payments Lecture: Weeks 4-6 Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 1 / 33 An introduction An introduction Central theme: to quantify the value today of a (random) sum of money to be paid at a random time in the future. main application is in life insurance contracts, but could be applied in other contexts. Generally computed in two steps: 1 take the present value (PV) random variable, bT vT ; and 2 calculate the expected value E(bT vT ) for the average value - this value is referred to as the Actuarial Present Value (APV). In general, we want to understand the entire distribution of the PV random variable bT vT : it could be highly skewed, in which case, there is danger to use expectation. other ways of summarizing the distribution such as variances and percentiles/quantiles would be useful. Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 2 / 33 The present value random variable The present value random variable Denote by Z, the present value random variable. This gives the value, at policy issue, of the benefit payment. In the case where the benefit is payable at the moment of death, Z clearly depends on the time-until-death T . It is Z = bT vT where: bT is called the benefit payment function vT is the discount function Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 5 / 33 Fixed term life insurance Fixed term life insurance An n-year term life insurance provides payment if the insured dies within n years from issue. For a unit of benefit payment we have bT = { 1, T ≤ n 0, T > n and vT = vT . The present value random is therefore Z = { vT , T ≤ n 0, T > n and E(Z) is called the actuarial present value (APV) of the insurance. Actuarial notation: Ā 1x: n = E(Z) = ∫ n 0 v t pt xµx+tdt. Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 6 / 33 Fixed term life insurance Rule of moments Rule of moments The j-th moment of the distribution of Z can be expressed as: E ( Zj ) = ∫ n 0 vtj pt xµx+tdt = ∫ n 0 e−(jδ)t pt xµx+tdt. This is actually equal to the APV but evaluated at the force of interest jδ. In general, we have the following rule of moment: E ( Zj ) @ δt = E (Z) @ jδt . For example, the variance can be expressed as Var (Z) = Ā2 1x: n − ( Ā 1x: n )2 . Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 7 / 33 Pure endowment insurance Pure endowment insurance For an n-year pure endowment insurance, a benefit is payable at the end of n years if the insured survives at least n years from issue. Here, we have bT = { 0, T ≤ n 1, T > n and vT = vn so that the PV r.v. is Z = { 0, T ≤ n vn, T > n . APV for pure endowment: A 1x: n = En x = v n pn x. Variance (using rule of moments): Var (Z) = v2n pn x · qn x = A2 1x: n − ( A 1x: n )2 . Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 10 / 33 Endowment insurance Endowment insurance For an n-year endowment insurance, a benefit is payable if death is within n years or if the insured survives at least n years from issue, whichever occurs first. Here, we have bT = 1 and vT = { vT , T ≤ n vn, T > n so that the PV r.v. is Z = { vT , T ≤ n vn, T > n . APV endowment: Āx: n = Ā 1 x: n + A 1 x: n . Variance (using rule of moments): Var (Z) = Ā2 x: n − ( Āx: n )2 . Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 11 / 33 Endowment insurance Variance of endowment insurance Variance of endowment insurance Let Z1, Z2, and Z3 be the PV r.v.’s for the term, pure endowment and the endowment insurances: Z1 = { vT , T ≤ n 0, T > n , Z2 = { 0, T ≤ n vn, T > n , and Z3 = { vT , T ≤ n vn, T > n . Clearly, Z3 = Z1 + Z2 and the expectation is the sum of the two expectations (as in previous slides). For the variance, use: Var (Z3) = Var (Z1) + Var (Z2) + 2Cov (Z1, Z2) .1 (In class lecture), we show: Cov (Z1, Z2) = −Ā 1x: n ·A 1x: n 1correction made 10 Oct 2009 Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 12 / 33 De Moivre’s law De Moivre’s law Find expressions for the APV for the same types of insurances in the case where you have: De Moivre’s law. Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 15 / 33 Varying benefit insurance Varying benefit insurance Type bT Z A.P.V. Incr. WL bT + 1c bT + 1c vT ( IĀ ) x WL incr. m-thly bTm + 1c /m vT bTm + 1c /m ( I(m)Ā ) x Cont. incr. WL T TvT ( ĪĀ ) x Decr. n-year term { n− bT c , T ≤ n 0, T > n { (n− bT c) vT , T ≤ n 0, T > n ( DĀ ) 1 x: n These items will be discussed in class. Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 16 / 33 Insurances payable at EOY of death Insurances payable at EOY of death For insurances payable at the end of the year (EOY) of death, the PV r.v. Z clearly depends on the curtate future lifetime Kx. It is Z = bK+1vK+1. To illustrate, consider an n-year term insurance which pays benefit at the end of year of death: bK+1 = { 1, K = 0, 1, ..., n− 1 0, otherwise , vK+1 = vK+1, and therefore Z = { vK+1, K = 0, 1, ..., n− 1 0, otherwise . Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 17 / 33 (Discrete) whole life insurance Example wk4.2 Example wk4.2 For a whole life insurance of 1 on (41) with death benefit payable at the end of the year of death, let Z be the present value random variable for this insurance. You are given: i = 0.05; p40 = 0.9972; A41 −A40 = 0.00822; and A2 41 − A2 40 = 0.00433. Calculate Var(Z). Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 20 / 33 (Discrete) endowment life insurance (Discrete) endowment life insurance The APV of a (discrete) endowment life insurance is the sum of the APV of a (discrete) term and a pure endowment: Ax: n = A 1 x: n + A 1 x: n The policy pays a death benefit of $1 at the end of the year of death, if death is prior to the end of n years, and a benefit of $1 if the insured survives at least n years. In effect, we have bK+1 = 1 and vK+1 = { vK+1, K ≤ n vn, K > n so that the PV r.v. is Z = { vK+1, K ≤ n vn, K > n . One can also apply the rule of moments to evaluate the corresponding variance. Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 21 / 33 Recursive relationships Recursive relationships For most situations in (discrete) insurances, we have recursion formula of the form u (x) = vqx + vpxu (x + 1) The following will be derived/discussed in class: whole life insurance: Ax = vqx + vpxAx+1 term insurance: A 1x: n = vqx + vpxA 1 x + 1: n−1 endowment insurance: Ax: n = vqx + vpxA x+1: n−1 Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 22 / 33 Annual cost of insurance - continued - continued The following formulas immediately follow from the previous development: Ax+1 −Ax = iAx − qx (1−Ax+1) Ax = ∑∞ y=x v y+1−xqy (1−Ay+1) Interpretation - to be discussed in class. Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 25 / 33 Other forms of insurance Other forms of insurance Deferred insurances Varying benefit insurances Very similar to the continuous cases You are expected to read and understand these other forms of insurances. It is also useful to understand the various (possible) recursion relations resulting from these various forms. Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 26 / 33 Insurances payable m-thly Insurances payable m-thly Consider the case where we have just one-year term and the benefit is payable at the end of the m-th of the year of death. We thus have A (m) 1 x: 1 = m−1∑ r=0 v(r+1)/m · pr/m x · q1/m x+r/m. We can show that under the UDD assumption, this leads us to: A (m) 1 x: 1 = i i(m) A 1 x: 1 . In general, we can generalize this to: A (m) 1 x: n = i i(m) A 1x: n . Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 27 / 33 Relationships - in general Relationships - in general Consider the case where bT can be written as a function of only K, i.e. bT = b∗K+1. We then write PV r.v. as Z = bT vT = b∗K+1vT = b ∗ K+1v K+1 (1 + i)1−S . Under UDD assumption, K and S are independent, and that 1− S is uniform on (0, 1). So that: E (Z) = E ( b∗K+1v K+1 ) E [ (1 + i)1−S ] = i δ E ( b∗K+1v K+1 ) Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 30 / 33 Differential equations for continuous insurances Differential equations for continuous insurances Recursive-type expressions can be established for insurances payable at the moment of death. For example, in the case of whole life insurance, one can show that d dx Āx = δĀx − µx ( 1− Āx ) . proof to be done in class Continuous analogue of the discrete recursion formula. Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 31 / 33 Example wk4.3 Example wk4.3 Each of 100 independent lives purchases a single premium 5-year deferred whole life insurance of 10 payable at the moment of death. Using a Normal approximation, calculate F such that the probability the insurer has sufficient funds to pay all claims is 0.95. You are given: µ = 0.004; δ = 0.006; F is the aggregate amount the insurer receives from the 100 lives; and the 95th percentile of the standard Normal distribution is 1.645. Lecture: Weeks 4-6 (Math 3630) Valuing Contingent Payments Fall 2009 - Valdez 32 / 33