Stat 475 Life Contingencies I Chapter 2: Survival models
The future lifetime random variable Notation We are interested in analyzing and describing the future lifetime of an individual. We use (x) to denote a life age x for x 0. Then the random variable describing the future lifetime, or time until death, for (x) is denoted by T x for T x 0. We also have notation for the density, distribution and survival functions of T x : Density: f x (t) = d dt F x(t) Distribution: F x (t) = Pr[T x t] Survival: S x (t) = Pr[T x > t] = 1 F x (t) 2
The future lifetime random variable Relationships We now consider the set of random variables {T x } x 0 and the relationships between them: Important Assumption Pr[T x t] = Pr[T 0 x + t T 0 > x] Important Consequence S 0 (x + t) = S 0 (x) S x (t) This idea can be generalized: Important General Relationship S x (t + u) = S x (t) S x+t (u) 3
Survival function conditions and assumptions We will require any survival function corresponding to a future lifetime random variable to satisfy the following conditions: Necessary and Sufficient Conditions 1 S x (0) = 1 2 lim t S x (t) = 0 3 S x (t) must be a non-increasing function of t In addition, we will make the following assumptions for the survival functions used in this course: Assumptions 1 S x (t) is differentiable for all t > 0. 2 lim t t S x (t) = 0 3 lim t t 2 S x (t) = 0 4
Future lifetime random variable example Assume that the survival function for a newborn is given by S 0 (t) = 8(t + 2) 3 1 Verify that this is a valid survival function. 2 Find the density function associated with the future lifetime random variable T 0, that is, find f 0 (t). 3 Find the probability that a newborn dies between the ages of 1 and 2. 4 Find the survival function corresponding to the random variable T 10. 5
Force of Mortality We will define the force of mortality as µ x = lim dx 0 + Pr [T 0 x + dx T 0 > x] dx Then for small values of dx, we can use the approximation µ x dx Pr [T 0 x + dx T 0 > x] We can relate the force of mortality to the survival function: µ x = d dx S 0(x) S 0 (x) and { t } S x (t) = exp µ x+s ds 0 6
Some commonly used forms for the force of mortality It is sometimes convenient to describe the future lifetime random variable by explicitly modeling the force of mortality. Some of the parametric forms that have been used are: Gompertz: µ x = Bc x, 0 < B < 1, c > 0 Makeham: µ x = A + Bc x, 0 < B < 1, c > 0 de Moivre 1 : µ x = 1 ω x, 0 x < ω Exponential 1 : µ x = k, 0 x If there is a limiting age, it is typically denoted by ω. 1 When these models are used to model human mortality, it is primarily for mathematical convenience rather than realism. 7
Force of mortality example Again, assume that the survival function for a newborn is given by S 0 (x) = 8(x + 2) 3 1 Find the force of mortality, µ x. 2 Consider the shape of µ x as a function of x. Does it seem to be a realistic survival model for describing human lifetimes? What would you expect the force of mortality corresponding to humans to look like? 8
Actuarial Notation Thus far we have described the properties of the future lifetime random variable using statistical notation. Actuaries have also developed their own (different) notation to describe many of the same concepts: tp x = Pr [T x > t] = S x (t) tq x = Pr [T x t] = F x (t) u t q x = Pr [u < T x u + t] = F x (u + t) F x (u) In any of the above expressions, the subscript t may be omitted when its value is 1. The density of the random variable T x can be written as f X (t) = t p x µ x+t. 9
Some Basic Relationships We can express some relationships in this actuarial notation: tp x = 1 t q x u t q x = u p x u+t p x u t q x = u p x t q x+u t+up x = t p x u p x+t { t } tp x = exp µ x+s ds 0 tq x = t 0 sp x µ x+s ds 10
Mean of the Future Lifetime Random Variable One quantity that is often of interest is the mean of the future lifetime random variable, E [T x ], which is called the complete expectation of life. E [T x ] = e x = = = 0 0 0 t f x (t) dt t t p x µ x+t dt tp x dt We can also find the term expectation of life. E[min(T x, n)] = e x:n = = n 0 n 0 t t p x µ x+t dt + tp x dt n n t p x µ x+t dt 11
Variance of the Future Lifetime Random Variable In order to calculate the variance of T x, we need its second moment: E [ Tx 2 ] = t 2 f x (t) dt = = 2 0 0 0 t 2 tp x µ x+t dt t t p x dt Then we can calculate the variance in the usual way: V [T x ] = E [ Tx 2 ] (E [Tx ]) 2 = E [ Tx 2 ] ( ex ) 2 12
Summary of notation for the future lifetime RV We can summarize the statistical and actuarial notation for the properties of T x covered thus far: Concept Statistical Notation Actuarial Notation Expected Value E [T x ] e x Distribution Function F x (t) tq x Survival Function S x (t) tp x Force of Mortality λ(x) or λ x µ x Density f x (t) tp x µ x+t Deferred Mortality Prob. Pr [u < T x u + t] u t q x 13
Curtate Future Lifetime Random Variable We are often interested in the integral number of years lived in the future by an individual. This discrete random variable is called the curtate future lifetime and is denoted by K x. We can consider the pmf (probability mass function) of K x : Pr [K x = k] = Pr [k T x < k + 1] = k q x = k p x q x+k We also have the relation K x = T x. 14
Moments of the Curtate Future Lifetime Random Variable We can find the mean (called the curtate expectation of life and denoted by e x ) and variance of K x : E [K x ] = e x = kp x k=1 E [ Kx 2 ] = 2 k k p x e x k=1 V [K x ] = E [ Kx 2 ] (ex ) 2 Using the trapezoidal rule, we can also find an approximate relation between the means of T x and K x : e x e x + 1 2 15
Exercise 2.2 S 0 (x) = 18000 110x x 2 18000 (a) What is the implied limiting age? [90] (b) Is it a valid survival function? (c) Calculate 20 p 0. [0.8556] [ ] (d) S 20 (t) 15400 150t t 2 15400 (e) 10 10 q 20 [0.1169] (f) µ 50 [0.021] 16
Exercise 2.1/2.6 F 0 (t) = { 1 ( 1 t 120) 1/6 for 0 t 120 1 for t > 120 (a) 30 p 0 (b) 20 q 30 (c) 25 p 40 (d) Derive µ x (e) Calculate e x and Var[T x ] for (30) (f) Calculate e x and Var[T x ] for (80) [0.9532] [0.041] [0.9395] [ ] 1 720 6x [ 77.143, 21.396 2 ] [ 34.286, 9.509 2 ] 17
Exercise 2.5 F 0 (t) = 1 e λt (a) S 0 (t) e λt (b) µ x λ 1 (c) e x e λ 1 (d) Is this reasonable for human mortality? 18
Exercise 2.6 p x = 0.99 p x+1 = 0.985 3p x+1 = 0.95 q x+3 = 0.02 (a) p x+3 0.98 (b) 2 p x 0.97515 (c) 2 p x+1 0.96939 (d) 3 p x 0.95969 (e) 1 2 q x 0.03031 19
SOA #32 1 for 0 t < 1 S 0 (t) = 1 et 100 for 1 t < 4.5 0 for 4.5 t Calculate µ 4. [1.20] 20
SOA #200 The graph of a piecewise linear survival function, S 0 (t), consists of 3 line segments with endpoints (0, 1), (25, 0.50), (75, 0.40), (100, 0). Calculate 20 55 q 15 55q 35. [0.6856] 21