T 0 = time (age) to failure random variable for a new entity, where the space of T 0 is {t | 0 < t <  } and  =  is possible F 0 (t) = Pr(T 0  t) is the CDF (cumulative distribution function) for T 0 Note that F 0 (t) = 1  S 0 (t) and S 0 (t) = 1  F 0 (t). Section 5.1 S 0 (t) = Pr(T 0 > t) is the SDF (survival distribution function) for T 0 f 0 (t) = F 0 (t) =  S 0 (t) is the PDF (probability density function) for T 0 d — dt d — dt

1. Chapter 5 Class Exercises Let  and a be positive constants. Suppose T 0 is a time (age) to failure random variable with CDF (cumulative distribution function) (a)Find S 0 (t), the SDF (survival distribution function) for T 0. 1 – e – t /  F 0 (t) =———— for 0 < t < a 1 – e – a /  e – t /  – e – a /  S 0 (t) =—————— for 0 < t < a 1 – e – a /  (b)Find f 0 (t), the PDF (probability density function) for T 0. e – t /  f 0 (t) =—————— for 0 < t < a  (1 – e – a /  )

1. Chapter 5 Class Exercises Let  and a be positive constants. Suppose T 0 is a time (age) to failure random variable with CDF (cumulative distribution function) (c)Let b and c be positive constants less than a. Find Pr[b < T 0 < c]. 1 – e – t /  F 0 (t) =———— for 0 < t < a 1 – e – a /  e – b /  – e – c /  —————— 1 – e – a /  (d)Find the mode of the distribution for T 0. a

T 0 = time (age) to failure random variable for a new entity, where the space of T 0 is {t | 0 < t <  } and  =  is possible F 0 (t) = Pr(T 0  t) is the CDF (cumulative distribution function) for T 0 Note that F 0 (t) = 1  S 0 (t) and S 0 (t) = 1  F 0 (t). Section 5.1 S 0 (t) = Pr(T 0 > t) is the SDF (survival distribution function) for T 0 0 (t) =  t = is the HRF (hazard rate function) for T 0 f 0 (t) —— S 0 (t) also called force of mortality for T 0 f 0 (t) = F 0 (t) =  S 0 (t) is the PDF (probability density function) for T 0 d — dt d — dt

 0 (t) = 0 (y) dy is the CHF (cumulative hazard function) for T 0 t0 t0 Note that  0 (t) = t0 t0 S0(y)S0(y) d — dy S0(y)S0(y)  ln[S 0 (y)] =  ln[S 0 (t)], 0 t which implies that S 0 (t) = dy = e =  0(t) 0(t) exp  0 (y) dy. t0 t0

1. Chapter 5 Class Exercises Let  and a be positive constants. Suppose T 0 is a time (age) to failure random variable with CDF (cumulative distribution function) (e) Find 0 (t), the HRF (hazard rate function) for T 0. 1 – e – t /  F 0 (t) =———— for 0 < t < a 1 – e – a /  e – t /  ———————  (e – t /  – e – a /  ) (f) Find  0 (t), the CHF (cumulative hazard function) for T 0. ln(1 – e – a /  )  ln(e – t /  – e – a /  )

2. Chapter 5 Class Exercises Let 0 (t) =  t = abt b  1 for 0  t, where a and b are constants. (a) Suppose the values of a and b are such that 0 (t) =  t is the HRF for a survival distribution; find the survival function S 0 (t) and the possible values for a and b. S 0 (t) = exp(  at b ) a > 0 and b > 0 (b)Find f 0 (t), the PDF (probability density function) for T 0. f 0 (t) = abt b  1 exp(  at b )

2. Chapter 5 Class Exercises Let 0 (t) =  t = abt b  1 for 0  t, where a and b are constants. (c)Find the mode of the distribution for T 0. The mode of the distribution is 0 when b ≤ 1 and is [(b  1)/(ab)] 1/b when b > 1. (d)Suppose b = 1. Name the type of distribution that the lifetime random variable T 0 has, and state what the mean of T 0 is. T 0 has an exponential distribution with mean 1/a.

Chapter 5 Class Exercises 3. Let 0 (t) =  t = for 0  t  1/a where a is a constant. (a) Suppose the value of a is such that 0 (t) =  t is the HRF for a survival distribution; find the survival function S 0 (t) and the possible values for a. S 0 (t) = 1  ata > 0 (b)Find f 0 (t), the PDF (probability density function) for T 0. f 0 (t) = a a ——— 1 – at

Chapter 5 Class Exercises 3. Let 0 (t) =  t = for 0  t  1/a where a is a constant. a ——— 1 – at (c) Find the mode of the distribution for T 0. There is no mode for a uniform distribution. (d) Name the type of distribution that the lifetime random variable T 0 has, and state what the mean of T 0 is. T 0 has a uniform(0, 1/a) distribution; the mean of T 0 is 1/(2a).

4. Chapter 5 Class Exercises Let 0 (t) =  t = e t for 0  t. Decide whether or not 0 (t) =  t could be the HRF for a survival distribution. If not, demonstrate why; if yes, find the survival function S 0 (t), and find the PDF. 0 (t) =  t can be an HRF, since S 0 (t) = exp(1  e t ), which would be the corresponding survival function, satisfies the three required properties: S 0 (0) = 1,as t   S(t)  0,S(t) is a decreasing function The PDF is f 0 (t) = e t exp(1  e t ).

5. Chapter 5 Class Exercises Let 0 (t) =  t = e  t for 0  t. Decide whether or not 0 (t) =  t could be the HRF for a survival distribution. If not, demonstrate why; if yes, find the survival function S 0 (t), and find the PDF. 0 (t) =  t can not be an HRF, since S 0 (t) = exp(e  t  1) would be the corresponding survival function, but as t   S(t)  e  1  Let 0 (t) =  t =be the HRF for a lifetime random (a)Find the survival function S 0 (t) for T 0. a for 0  t < c b for c  t variable T 0 where a, b, and c are positive constants. S 0 (t) = exp(  at)for 0  t < c exp(  c(a  b)  bt)for c  t

Chapter 5 Class Exercises 6. Let 0 (t) =  t =be the HRF for a lifetime random (b)Suppose a = 1/25, b = 1/50, and c = 40; find each of the probabilities listed. a for 0  t < c b for c  t variable T 0 where a, b, and c are positive constants. S 0 (t) = exp(  t/25)for 0  t < 40 exp(  4/5  t/50)for 40  t Pr[T 0 < 30] Pr[T 0 < 40] 1  exp(  1.2) 1  exp(  1.6)

Chapter 5 Class Exercises 6. Let 0 (t) =  t =be the HRF for a lifetime random (b)Suppose a = 1/25, b = 1/50, and c = 40; find each of the probabilities listed. a for 0  t < c b for c  t variable T 0 where a, b, and c are positive constants. S 0 (t) = exp(  t/25)for 0  t < 40 exp(  4/5  t/50)for 40  t Pr[40 < T 0 < 50] Pr[30 < T 0 < 50] exp(  1.6)  exp(  1.8) exp(  1.2)  exp(  1.8)

Chapter 5 Class Exercises 7.Decide which of the functions listed cannot be a survival function, and explain why not in each case. (a) cos(t) for 0  t  3  /2 The function is negative for some t and increasing for some t. (b) cos(t) for 0  t   /2 The function could be a survival function. (c) sin(t) for 0  t   /2 The function is increasing; also, the value is 0  1 when t = 0, and is 1  0 when t =  /2. (d) for 0  t  3  The function is increasing for some t. 1 + cos(t) ———— 2

8. Chapter 5 Class Exercises Let a, b, n, and  be constants, with n > 0 and  > 0. Suppose T 0 is a time (age) to failure random variable with SDF (survival distribution function) S 0 (t) = at n + b for 0  t  . (a)Find the value of b. (b) Find a in terms of  and n. b = 1 a =  1/  n (c)Find f 0 (t), the PDF (probability density function) for T 0 in terms of  and n. f 0 (t) = nt n  1 /  n

8. Chapter 5 Class Exercises Let a, b, n, and  be constants, with n > 0 and  > 0. Suppose T 0 is a time (age) to failure random variable with SDF (survival distribution function) S 0 (t) = at n + b for 0  t  . (d) Find E(T 0 ) in terms of  and n. (e)Suppose n = 1. Name the type of distribution that the lifetime random variable T 0 has. E(T 0 ) = n  /(n + 1) uniform(0,  )

9. Chapter 5 Class Exercises Let a, b, n, and  be constants, with n > 0 and  > 0. Suppose T 0 is a time (age) to failure random variable with SDF (survival distribution function) S 0 (t) = (at + b) n for 0  t  . (a)Find the value of b. (b) Find a in terms of  and n. b = 1 a =  1/  (c)Find f 0 (t), the PDF (probability density function) for T 0 in terms of  and n. f 0 (t) = (n/  )(1  t/  ) n  1

9. Chapter 5 Class Exercises Let a, b, n, and  be constants, with n > 0 and  > 0. Suppose T 0 is a time (age) to failure random variable with SDF (survival distribution function) S 0 (t) = at n + b for 0  t  . (d) Find E(T 0 ) in terms of  and n. (e)Suppose n = 1. Name the type of distribution that the lifetime random variable T 0 has. E(T 0 ) =  /(n + 1) uniform(0,  )

Section 5.2 For a time-to-failure random variable with a uniform distribution, the HRF (hazard rate function) is the SDF (survival distribution function) is and the PDF (probability density function) is The mean and variance for this distribution are S 0 (t) =for 0  t    – t ——  f 0 (t) =for 0  t   1 —  respectively and  — 2  2 — 12 0 (t) =  t = for 0  t   1 ——  – t

Section 5.2 For a time-to-failure random variable with an exponential distribution, the HRF (hazard rate function) is the SDF (survival distribution function) is and the PDF (probability density function) is The mean and variance for this distribution are S 0 (t) = e  t for 0  t f 0 (t) = e  t for 0  t respectively and 1 — 1 — 2 0 (t) =  t = for 0  twhere > 0

Section 5.2 For a time-to-failure random variable with a Gompertz distribution, the HRF (hazard rate function) is the SDF (survival distribution function) is and the PDF (probability density function) is The mean and variance for this distribution are S 0 (t) = exp for 0  t not able to be written in a convenient mathematical form 0 (t) =  t = Bc t for 0  twhere B > 0 and c > 1 B —— (1  c t ) ln c Note that if we were to let c = 1, then this would be the HRF for distribution an exponential

Section 5.2 For a time-to-failure random variable with a Makeham distribution, the HRF (hazard rate function) is the SDF (survival distribution function) is and the PDF (probability density function) is The mean and variance for this distribution are S 0 (t) = exp for 0  t not able to be written in a convenient mathematical form 0 (t) =  t = A + Bc t for 0  twhere B > 0, c > 1, and A >  B B —— (1  c t )  At ln c Note that if we were to let A = 0, then this would be the HRF fordistribution a Gompertz

T x = time to failure random variable for an entity known to exist at age x, where the space of T x is {t | 0 < t <   x} and  =   x =  is possible Note that F x (t) = 1  S x (t) and S x (t) = 1  F x (t) Section 5.3 Observe that T x = T 0  x and that Pr(T x > t) = Pr(T 0 > x + t | T 0 > x) t p x = Pr(T x > t) = S x (t) is the SDF for T x t q x = 1  t p x = Pr(T x  t) = F x (t) is the CDF for T x t p x = Pr(T 0 > x + t | T 0 > x) = Pr(T 0 > x + t  T 0 > x) ————————— = Pr(T 0 > x) Pr(T 0 > x + t) —————— = Pr(T 0 > x) S 0 (x + t) ———— = S 0 (x) 1  F 0 (x + t) ————— 1  F 0 (x) Note that

t q x = Pr(T 0  x + t | T 0 > x) = Pr(T 0  x + t  T 0 > x) ————————— = Pr(T 0 > x) Pr(x < T 0  x + t) ——————— = Pr(T 0 > x) S 0 (x)  S 0 (x + t) ——————— S 0 (x) F 0 (x + t)  F 0 (x) ——————— = 1  F 0 (x) Note that 0 (t) =  t = is the HRF (hazard rate function) for T 0 f 0 (t) —— S 0 (t) also called force of mortality for T 0 f x (t) = F x (t) =  S x (t) is the PDF (probability density function) for T x d — dt d — dt Note that f x (t) = d — dt t q x = d — dt F 0 (x + t)  F 0 (x) ——————— = 1  F 0 (x) F 0 (x + t)  F 0 (x) ——————— = ????????1  F 0 (x)

a –– 40 | 0.03 = 1 – v n —— = 0.03 With v = 1 / 1.03, R = 5000 ——— = $ (Note: On the TI-83 calculator, the | 2nd | | FINANCE | keys should be used in place of the | APPS | key and Finance option.) (c)Calculate the value of R on the TI-84 calculator by doing the following: We must have 5000 = R a –– 40 | 0.03 R = (a)Let R be the amount withdrawn (i.e., the payments) at each half- year, and write a formula for R using complete actuarial notation. (b)Calculate the value of R using the appropriate formula. a –– 40 | ———

Press the | APPS | key, select the Finance option, and select the TVM_Solver option. Enter the following values for the variables displayed: N = 40 I% = 3 PV = –5000 PMT = 0 FV = 0 P/Y = 1 C/Y = 1 Select the END option for PMT, press the | APPS | key, and select the Finance option. Select the tvm_Pmt option, and after pressing the | ENTER | key, the desired result should be displayed. $

(d)Calculate the value of R in Excel by entering the following formula in any cell: $ =5000/PV(0.03,40,-1,0,0) This is the balance remaining (generally 0) A 0 implies payments at the end of each period. A 1 implies payments at the beginning of each period.

3. Chapter 3 Class Exercises Find the total amount of interest that would be paid on a $3000 loan over a 6-year period with an effective rate of interest of 7.5% per annum, under each of the following repayment plans: (a)The entire loan plus accumulated interest is paid in one lump sum at the end of 6 years. 3000(1.075) 6 =$ Total Interest Paid =$ (b)Interest is paid each year as accrued, and the principal is repaid at the end of 6 years. Each year, the interest on the loan is3000(0.075) =$225 Total Interest Paid =$1350 (c)The loan is repaid with level payments at the end of each year over the 6-year period = R a –– 6 | R =$ Total Interest Paid =6(639.13) – 3000 =$ Let R be the level payments.

a – n| s – n| 1 – v n = ——  i (1 + i) n – 1 = ————  i a – n| 1 = i + v n The right hand side can be interpreted as the sum of the “present value of the interest payments” and the “present value of 1 (the original investment)” 1 = (1 + i) n  i The right hand side can be interpreted as the “accumulated value of 1 (the original investment)” minus the “accumulated value of the interest payments” s – n| Observe that s – n| a – n| =(1 + i) n Also, 1 —— + i = s – n| i ———— + i = (1 + i) n – 1 i(1 + i) n ———— = (1 + i) n – 1 i —— = 1 – v n 1 —— a – n| This identity will be important in a future chapter.

An annuity under which payments of 1 are made at the beginning of each period for n periods is called an annuity-due. 012…n – 1n Payments Periods 1111 The present value of the annuity at time 0 is denoted, where the interest rate i is generally included only if not clear from the context. The accumulated value of the annuity at time n is denoted, where the interest rate i is generally included only if not clear from the context. = 1 + v + v 2 + … + v n–1 = 1 – v n —— = 1 – v 1 – v n —— d = (1 + i) n + (1 + i) n–1 + … + (1 + i) = (1 + i) n – 1 (1 + i)———— = (1 + i) – 1 (1 + i) n – 1 ———— d.. a – n|i.. s – n|i.. a – n|.. s – n|

Observe that = (1 + i) n Also, 1 —— + d = 1 ——.. a – n|.. s – n| In addition, observe that a – n| =(1 + i).. a – n| =(1 + i).. s – n| s – n| a ––– n–1| =1 +.. a – n| =– 1.. s – n| s ––– n+1| These last four formulas demonstrate that annuity-immediate and annuity-due are really just the same thing at two different points in time, as is illustrated graphically in Figure 3.3 of the textbook... s – n|.. a – n| d ———— + d = (1 + i) n – 1 d(1 + i) n ———— = (1 + i) n – 1 d —— = 1 – v n

4. Chapter 3 Class Exercises An investor wishes to accumulate $3000 at the end of 15 years in a fund which earns 8% effective. To accomplish this, the investor plans to make deposits at the end of each year, with the final payment to be made one year prior to the end of the investment period. How large should each deposit be? (a)Let R be the payments each year, and write a formula for R using complete actuarial notation. We must have (b)Calculate the value of R using the appropriate formula = R.. s –– 14 | 0.08 R = 3000 ———— =.. s –– 14 | —————— s –– 15 | 0.08 – 1

(a)Let R be the payments each year, and write a formula for R using complete actuarial notation. We must have (b)Calculate the value of R using the appropriate formula = R.. s –– 14 | 0.08 R = 3000 ———— =.. s –– 14 | —————— s –– 15 | 0.08 – 1 ( ) 14 – 1 —————— = 0.08/1.08 Using either.. s –– = 14 | or s –– = 15 | 0.08 ( ) 14 – 1 —————— = 0.08 we have R = $ (Note: On the TI-83 calculator, the | 2nd | | FINANCE | keys should be used in place of the | APPS | key and Finance option.) (c)Calculate the value of R on the TI-84 calculator by doing the following:

Press the | APPS | key, select the Finance option, and select the TVM_Solver option. Enter the following values for the variables displayed: N = 14 I% = 8 PV = 0 PMT = 0 FV = –3000 P/Y = 1 C/Y = 1 Select the BEGIN option for PMT, press the | APPS | key, and select the Finance option. Select the tvm_Pmt option, and after pressing the | ENTER | key, the desired result should be displayed. $

(d)Calculate the value of R in Excel by entering the following formula in any cell: $ =3000/FV(0.08,14,-1,0,1) This is the balance remaining (generally 0) A 0 implies payments at the end of each period. A 1 implies payments at the beginning of each period.