Transcript lycofs01.lycoming.edu

Section 5.1
T0 = time (age) to failure random variable for a new entity,
where the space of T0 is {t | 0 < t < } and  =  is possible
F0(t) = Pr(T0  t) is the CDF (cumulative distribution function) for T0
S0(t) = Pr(T0 > t) is the SDF (survival distribution function) for T0
Note that F0(t) = 1  S0(t) and S0(t) = 1  F0(t).
d
d
f0(t) = — F0(t) =  — S0(t) is the PDF (probability density function) for
dt
dt
T
0
Chapter 5 Class Exercises
1. Let  and a be positive constants. Suppose T0 is a time (age) to failure
random variable with CDF (cumulative distribution function)
1 – e –t/
F0(t) = ————
for 0 < t < a
–
a
/

1–e
(a) Find S0(t), the SDF (survival distribution function) for T0 .
e –t/ – e –a/
S0(t) = ——————
for 0 < t < a
–
a
/

1–e
(b) Find f0(t), the PDF (probability density function) for T0 .
e –t/
f0(t) = ——————
for 0 < t < a
–
a
/

(1 – e
)
Chapter 5 Class Exercises
1. Let  and a be positive constants. Suppose T0 is a time (age) to failure
random variable with CDF (cumulative distribution function)
1 – e –t/
F0(t) = ————
for 0 < t < a
–
a
/

1–e
(c) Let b and c be positive constants less than a. Find Pr[b < T0 < c].
e –b/ – e –c/
——————
1 – e –a/
(d) Find the mode of the distribution for T0 .
a
Section 5.1
T0 = time (age) to failure random variable for a new entity,
where the space of T0 is {t | 0 < t < } and  =  is possible
F0(t) = Pr(T0  t) is the CDF (cumulative distribution function) for T0
S0(t) = Pr(T0 > t) is the SDF (survival distribution function) for T0
Note that F0(t) = 1  S0(t) and S0(t) = 1  F0(t).
d
d
f0(t) = — F0(t) =  — S0(t) is the PDF (probability density function) for
dt
dt
T
0
f0(t)
0(t) = t = —— is the HRF (hazard rate function) for T0
S0(t)
also called force of mortality for T0
t
0(t) =
  (y) dy is the CHF (cumulative hazard function) for T
0
0
0
t
Note that 0(t) =

0
t
d
— S0(y)
dy
dy =  ln[S0(y)] =  ln[S0(t)] ,
S0(y)
0
t

which implies that S0(t) = e  0(t) = exp  0(y) dy
0
.
Chapter 5 Class Exercises
1. Let  and a be positive constants. Suppose T0 is a time (age) to failure
random variable with CDF (cumulative distribution function)
1 – e –t/
F0(t) = ————
for 0 < t < a
–
a
/

1–e
(e) Find 0(t), the HRF (hazard rate function) for T0 .
e –t/
———————
(e – t /  – e – a / )
(f) Find 0(t), the CHF (cumulative hazard function) for T0 .
ln(1 – e – a / )  ln(e – t /  – e – a / )
Chapter 5 Class Exercises
2. Let 0(t) = t = abtb  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 S0(t) and the
possible values for a and b.
S0(t) = exp(atb)
a > 0 and b > 0
(b) Find f0(t), the PDF (probability density function) for T0 .
f0(t) = abtb  1 exp( atb)
Chapter 5 Class Exercises
2. Let 0(t) = t = abtb  1 for 0  t, where a and b are constants.
(c) Find the mode of the distribution for T0 .
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 T0 has, and state what the mean of T0 is.
T0 has an exponential distribution with mean 1/a.
Chapter 5 Class Exercises
a
3. Let 0(t) = t = ——— for 0  t  1/a where a is a constant.
1 – at
(a) Suppose the value of a is such that 0(t) = t is the HRF for a
survival distribution; find the survival function S0(t) and the
possible values for a.
S0(t) = 1  at
a>0
(b) Find f0(t), the PDF (probability density function) for T0 .
f0(t) = a
Chapter 5 Class Exercises
a
3. Let 0(t) = t = ——— for 0  t  1/a where a is a constant.
1 – at
(c) Name the type of distribution that the lifetime random variable T0
has, and state what the mean of T0 is.
T0 has a uniform(0, 1/a) distribution; the mean of T0 is 1/(2a).
(d) Find the mode of the distribution for T0 .
There is no mode for a uniform distribution.
Chapter 5 Class Exercises
4. Let 0(t) = t = et 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 S0(t), and find the PDF.
0(t) = t can be an HRF, since S0(t) = exp(1  e t ), which would be
the corresponding survival function, satisfies the three required
properties:
S0(0) = 1,
as t   S(t)  0,
The PDF is f0(t) = e t exp(1  e t ).
S(t) is a decreasing function
Chapter 5 Class Exercises
5. 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 S0(t), and find the PDF.
0(t) = t can not be an HRF, since S0(t) = exp(et  1) would be the
corresponding survival function, but as t   S(t)  e1  0.
6. Let 0(t) = t =
a
b
for 0  t < c
for c  t
be the HRF for a lifetime random
variable T0 where a, b, and c are positive constants.
(a) Find the survival function S0(t) for T0 .
S0(t) =
exp( at)
exp( c(a  b)  bt)
for 0  t < c
for c  t
Chapter 5 Class Exercises
6. Let 0(t) = t =
a
b
for 0  t < c
for c  t
be the HRF for a lifetime random
variable T0 where a, b, and c are positive constants.
(b) Suppose a = 1/25, b = 1/50, and c = 40; find each of the
probabilities listed.
S0(t) =
exp( t/25)
exp( 4/5  t/50)
Pr[T0 < 30]
1  exp( 1.2)
Pr[T0 < 40]
1  exp( 1.6)
for 0  t < 40
for 40  t
Chapter 5 Class Exercises
6. Let 0(t) = t =
a
b
for 0  t < c
for c  t
be the HRF for a lifetime random
variable T0 where a, b, and c are positive constants.
(b) Suppose a = 1/25, b = 1/50, and c = 40; find each of the
probabilities listed.
S0(t) =
exp( t/25)
exp( 4/5  t/50)
Pr[40 < T0 < 50]
exp( 1.6)  exp( 1.8)
Pr[30 < T0 < 50]
exp( 1.2)  exp( 1.8)
for 0  t < 40
for 40  t
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.
1 + cos(t)
(d) ————
for 0  t  3
2
The function is increasing for some t.
Chapter 5 Class Exercises
8. Let a, b, n, and  be constants, with n > 0 and  > 0. Suppose T0 is a
time (age) to failure random variable with SDF (survival distribution
function) S0(t) = atn + b for 0  t   .
(a) Find the value of b.
b=1
(b) Find a in terms of  and n.
a = 1/n
(c) Find f0(t), the PDF (probability density function) for T0 in terms of
 and n.
f0(t) = ntn  1/n
Chapter 5 Class Exercises
8. Let a, b, n, and  be constants, with n > 0 and  > 0. Suppose T0 is a
time (age) to failure random variable with SDF (survival distribution
function) S0(t) = atn + b for 0  t   .
(d) Find E(T0) in terms of  and n.
E(T0) = n/(n + 1)
(e) Suppose n = 1. Name the type of distribution that the lifetime
random variable T0 has.
uniform(0, )
Chapter 5 Class Exercises
9. Let a, b, n, and  be constants, with n > 0 and  > 0. Suppose T0 is a
time (age) to failure random variable with SDF (survival distribution
function) S0(t) = (at + b)n for 0  t   .
(a) Find the value of b.
b=1
(b) Find a in terms of  and n.
a = 1/
(c) Find f0(t), the PDF (probability density function) for T0 in terms of
 and n.
f0(t) = (n/)(1  t/)n  1
Chapter 5 Class Exercises
9. Let a, b, n, and  be constants, with n > 0 and  > 0. Suppose T0 is a
time (age) to failure random variable with SDF (survival distribution
function) S0(t) = atn + b for 0  t   .
(d) Find E(T0) in terms of  and n.
E(T0) = /(n + 1)
(e) Suppose n = 1. Name the type of distribution that the lifetime
random variable T0 has.
uniform(0, )
Section 5.2
For a time-to-failure random variable with a uniform distribution,
the HRF (hazard rate function) is
1
0(t) = t = —— for 0  t  
–t
the SDF (survival distribution function) is
–t
S0(t) = —— for 0  t  

and the PDF (probability density function) is
1
f0(t) = —
for 0  t  

The mean and variance for this distribution are
2

respectively — and —
12
2
Section 5.2
For a time-to-failure random variable with an exponential distribution,
the HRF (hazard rate function) is
0(t) = t =  for 0  t
where  > 0
the SDF (survival distribution function) is
S0(t) = et for 0  t
and the PDF (probability density function) is
f0(t) = et for 0  t
The mean and variance for this distribution are
1
1
respectively — and —2


Section 5.2
For a time-to-failure random variable with a Gompertz distribution,
the HRF (hazard rate function) is
0(t) = t = Bct for 0  t
where B > 0 and c > 1
Note that if we were to let c = 1, then this would
be the HRF for an exponential distribution
the SDF (survival distribution function) is
B
S0(t) = exp —— (1  ct) for 0  t
ln c
and the PDF (probability density function) is
not able to be written in a convenient mathematical form
The mean and variance for this distribution are
not able to be written in a convenient mathematical form
Section 5.2
For a time-to-failure random variable with a Makeham distribution,
the HRF (hazard rate function) is
0(t) = t = A + Bct for 0  t where B > 0, c > 1, and A > B
Note that if we were to let A = 0, then this would
be the HRF for a Gompertz distribution
the SDF (survival distribution function) is
B
S0(t) = exp —— (1  ct)  At for 0  t
ln c
and the PDF (probability density function) is
not able to be written in a convenient mathematical form
The mean and variance for this distribution are
not able to be written in a convenient mathematical form
Section 5.3
Tx = time to failure random variable for an entity known to exist at age x,
where the space of Tx is {t | 0 < t <   x} and  =   x =  is possible
Observe that Tx = T0  x and that Pr(Tx > t) = Pr(T0 > x + t | T0 > x)
t px
= Pr(Tx > t) = Sx(t) is the SDF for Tx
t qx
= 1  t px = Pr(Tx  t) = Fx(t) is the CDF for Tx
Note that Fx(t) = 1  Sx(t) and Sx(t) = 1  Fx(t)
Pr(T0 > x + t  T0 > x)
Note that t px = Pr(T0 > x + t | T0 > x) = ————————— =
Pr(T0 > x)
1  F0(x + t)
Pr(T0 > x + t)
S0(x + t)
—————— = ———— = —————
1  F0(x)
Pr(T0 > x)
S0(x)
Pr(T0  x + t  T0 > x)
Note that t qx = Pr(T0  x + t | T0 > x) = ————————— =
Pr(T0 > x)
F0(x + t)  F0(x)
S0(x)  S0(x + t)
Pr(x < T0  x + t)
——————— = ——————— = ———————
1  F0(x)
S0(x)
Pr(T0 > x)
d
d
fx(t) = — Fx(t) =  — Sx(t) is the PDF (probability density function) for
dt
dt
T
x
d F0(x + t)  F0(x)
F0(x + t)  F0(x)
d
Note that fx(t) = — t qx = — ——————— = ——————— =
dt
1  F0(x)
????????1  F0(x)
dt
f0(t)
0(t) = t = —— is the HRF (hazard rate function) for T0
S0(t)
also called force of mortality for T0
(a) Let R be the amount withdrawn (i.e., the payments) at each halfyear, and write a formula for R using complete actuarial notation.
We must have 5000 = R a ––
40 | 0.03
5000
R = ———
a ––
40 | 0.03
(b) Calculate the value of R using the appropriate formula.
With v = 1 / 1.03,
a ––
40 | 0.03
1 – vn
= —— = 23.1148
0.03
5000
R = ——— = $ 216.31
23.1148
(c) Calculate the value of R on the TI-84 calculator by doing the
following:
(Note: On the TI-83 calculator, the | 2nd | | FINANCE | keys
should be used in place of the | APPS | key and Finance option.)
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.
$ 216.31
(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)
$ 216.31
A 0 implies payments at the end of
each period. A 1 implies payments
at the beginning of each period.
Chapter 3 Class Exercises
3. 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 = $4629.90
Total Interest Paid = $1629.90
(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 is 3000(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.
Let R be the level payments.
3000 = R a ––
R = $639.13
6 | 0.075
Total Interest Paid = 6(639.13) – 3000 = $834.78
1 – vn
a –n| = ——
i
 1 = i a –n| + vn
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 + i)n – 1
s n|
– = ————  1 = (1 + i)n  is –
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”
– = a –n| (1 + i)n
Observe that s n|
1
i
i(1 + i)n
i
1
+ i = ————
+ i = ————
= ——n = ——
Also, ——
n
n
s n|
–
a –n|
(1 + i) – 1
(1 + i) – 1
1–v
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.
Payments
1
1
1
Periods
0
1
2 … n–1
1
n
The present value of ..the annuity
The accumulated value of the
..
a
–
at time 0 is denoted n|i , where
annuity at time n is denoted s –n|i ,
the interest rate i is generally
where the interest rate i is
included only if not clear from
generally included only if not
clear from the context.
..the context. 2
..
n–1
a–
– = (1 + i)n + (1 + i)n–1 + … + (1 + i) =
n| = 1 + v + v + … + v = s n|
1 – vn
1 – vn
(1 + i)n – 1
(1 + i)n – 1
—— = ——
(1 + i) ———— = ————
1–v
d
(1 + i) – 1
d
.. = ..
–
a –n| (1 + i)n
Observe that s n|
1
d
d(1 + i)n
d
1
+ d = ————
= ——n = ——
.. + d = ————
..
Also, ——
n
n
(1 + i) – 1
(1 + i) – 1
1–v
s n|
–
a –n|
..
In addition, observe that a –n| = a –n| (1 + i)
..
– (1 + i)
s n|
– = s n|
..
a –n| = 1 + a –––
n–1|
..
––– – 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.
Chapter 3 Class Exercises
4. 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 3000 = R ..s ––
14 | 0.08
3000
3000
= ——————
R = ————
..
s ––
s ––
–1
14 | 0.08
15 | 0.08
(b) Calculate the value of R using the appropriate formula.
(a) Let R be the payments each year, and write a formula for R using
complete actuarial notation.
We must have 3000 = R ..s ––
14 | 0.08
3000
3000
= ——————
R = ————
..
s ––
s ––
–1
14 | 0.08
15 | 0.08
(b) Calculate the value of R using the appropriate formula.
(1 + 0.08)14 – 1
..
Using either s ––
= —————— = 26.1521
14 | 0.08
0.08/1.08
or s ––
15 | 0.08
(1 + 0.08)14 – 1
= —————— = 27.1521 we have R = $ 114.71
0.08
(c) Calculate the value of R on the TI-84 calculator by doing the
following:
(Note: On the TI-83 calculator, the | 2nd | | FINANCE | keys
should be used in place of the | APPS | key and Finance option.)
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.
$ 114.71
(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)
$ 114.71
A 0 implies payments at the end of
each period. A 1 implies payments
at the beginning of each period.