Actuarial Science and Risk Management
Lifetime Distributions
Solutions
1. Calculate the Kaplan Meier estimate of the hazard function for the
lifetime of an ipod touch given the following data based on a study of
50 ipods over a period of 365 days.
1
Step 1 here is to re express the information given in terms of a table
setting out the number dj and timing tj of the ipod breakdown’s.
We will need to also note the number exposed to risk at each time
nj .The table of data calculating the distribution steps is given by:
Step 2 is then to write out the Kaplan Meier estimate of the hazard
function as:


0
0 < t ≤ 40





0.0612 41 < t ≤ 106





0.1049 107 < t ≤ 109
F (t) = 0.1267 110 < t ≤ 209



0.1491 210 < t ≤ 299





0.1734 300 < t ≤ 333



0.1985 t > 333
2. Calculate the Nelson Aalen estimate of the integrated hazard function
using the same data as above,
2
3
The Nelson Aalen estimate is given by:
Step 2 is then to write out the Nelson Aalen estimate of the integrated
hazard function as:


0
0 < t ≤ 40





0.0594 41 < t ≤ 106




0.1021 107 < t ≤ 109

F (t) = 0.1238 110 < t ≤ 209



0.1460 210 < t ≤ 299





0.1700 300 < t ≤ 333



0.1948 t > 333
3. Calculate using Greenwoods formula the 95% confidence intervals for
the NA and the KM estimates of F(300) the probability of the ipod
being broken by day 300.
4
Greenwoods formula for the Nelson Aalen estimate is given by the
final column in the table below:
The variance at time 300 is therefore given by
(1 − 0.1700)2 ∗ 0.004218 = 0.002906
A 95% confidence interval is therefore given by:
³
´ ³
´
√
√
0.1700 − 1.96 0.002906, 0.1700 + 1.96 0.002906 = 0.0643, 0.2756
Greenwoods formula for the Kaplan Meier estimate is given by the
final column in the table below:
The variance at time 300 is therefore given by
(1 − 0.1734)2 5∗ 0.00459 = 0.003136
A 95% confidence interval is therefore given by:
³
´ ³
´
√
√
0.1734−1.96 0.003136, 0.1734+1.96 0.003136 = 0.06366, 0.2832
4. You are assessing the quality of a new kitchen implement over a 30 day
period. You observe 25 appliances and are given the following data
relating to the timing of breakdowns. Those times marked with an
asterisk are censored items.
14, 13, 15∗, 7∗, 23, 18∗, 5, 18, 6∗, 22, 19∗, 15, 4, 11, 14∗, 18, 10, 8∗, 17, 15∗, 22, 25, 26∗, 28.
Write out the Kaplan Meier estimate for the hazard function.
The Kaplan Meier estimate is given by:
.
6
Expressing this as a distribution function we have:


0
0<t≤3





0.0400 4 < t ≤ 5





0.0800 5 < t ≤ 10





0.1260 10 < t ≤ 11




0.1720 11 < t ≤ 13





0.2180 13 < t ≤ 14



0.2640 14 < t ≤ 15
F (t) =

0.3131 15 < t ≤ 17





0.3703 17 < t ≤ 18





0.4848 18 < t ≤ 22




0.6136 22 < t ≤ 23





0.6909 23 < t ≤ 25





0.7682 25 < t ≤ 28



0.8841 28 < t ≤ 30
5. Using the data above calculate the Nelson Aalen estimate of the hazard
function.
7
The Nelson Aalen estimate is given by:
.
Expressing this as a distribution function we have:


0
0<t≤3





0.0392 4 < t ≤ 5





0.0784 5 < t ≤ 10




0.1234 10 < t ≤ 11




0.1683 11 < t ≤ 13





0.2133 13 < t ≤ 14



0.2582 14 < t ≤ 15
F (t) =

0.3060 15 < t ≤ 17





0.3615 17 < t ≤ 18




0.4677 18 < t ≤ 22




0.5854 22 < t ≤ 23





0.6606 23 < t ≤ 25





0.7357 25 < t ≤ 28



0.8397 28 < t ≤ 30
8
6. Calculate using Greenwoods formula the 95% confidence intervals for
the NA and the KM estimates of F(25) the probability of the appliance
being broken by day 25.
9
The Greenwood estimate for the Kaplan Meier estimates is given by:
The variance at t=25 is therefore given by
(1 − 0.7682)2 ∗ 0.2235 = 0.0120
The 95% confidence interval is then given by:
(0.7682 − 1.96 ∗ 0.01200.5 , 0.7682 + 1.96 ∗ 0.01200.5 ) = (0.5533, 0.9830)
The Greenwood estimate for the Nelson Aalen estimates is given by:
The variance at t=25 is therefore given by
(1 − 0.7357)2 ∗ 0.1407 = 0.0098
The 95% confidence interval is then given by:
10
(0.7357 − 1.96 ∗ 0.00980.5 , 0.7357 + 1.96 ∗ 0.00980.5 ) = (0.5413, 0.9300)
7. A life insurance company has carried out a mortality investigation. It
followed a sample of independent policyholders aged between 50 and 55
years. Policyholders were followed from their 50th birthday until they
died, withdrew from the investigation while still alive, or reached their
55th birthday (whichever occurred first).An extract from the data for
12 policyholders is shown below. Use this data to calculate the Kaplan
Meier estimate of the survival function.
11
The Kaplan Meier estimate is given by:
The Kaplan Meier estimate, written as a distribution function where t
represents the number of months after age 50 is therefore:


0
0<t≤8





0.0833 9 < t ≤ 18
F (t) = 0.2677 18 < t ≤ 51



0.4133 51 < t ≤ 54



0.5600 54 < t
12