MATH3085 Survival Models SEMESTER 1 EXAMINATION 2018/19
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SEMESTER 1 EXAMINATION 2018/19
MATH3085 Survival Models
1. [17 marks] A biologist investigating survival times of a particular species of fly larva proposes to model lifetimes as observations of a random variable T, and proposes
ST (t) =
for some parameter β > 0, as the survival function for T.
(a) [3 marks] Giving your reasons, describe why this function is valid as a survival function.
(b) [4 marks] Find the density and hazard functions for T.
A sample of nsuch larvae are observed over a period of 168 hours (one week),
during which time m of the larvae were observed to die, at timest1 , t2 , . . . , tm ,
measured in hours. The remaining n - m larvae were still alive at 168 hours when observation ended.
(c) [6 marks] Show that the maximum likelihood estimator β(ˆ) of the parameter β is
obtained as the solution to the equation
= 2 +
[You are not required to solve this equation].
(d) [4 marks] Find an expression for the standard error of β(ˆ) in terms of
t1 , t2 , . . . , tm , n, m, and β(ˆ). [Hint: use the observed information rather than the
Fisher information.]
2. [16 marks] A study of a new lubricant was undertaken to assess whether it increased the length of time between engine failures. Four different engines, two of type A and two of type B were studied for 12 months. A Cox proportional hazards model is
proposed for the data, where the times-to-failure are modelled as observations of independent random variables Ti , i = 1, . . . , 4, with hazard functions
hTi (t) = exp(β1xi + β2 zi ) h0 (t)
where xi = 1 for the new lubricant and 0 for the existing lubricant, and zi = 1 for engine type A and 0 for engine type B. For each engine, i = 1, . . . , 4, the time under
observation ti is recorded together with an indicator variable di which is given the
valuedi = 1 if failure is observed at ti and di = 0 if observation ended for any other reason.
Observation (i) 1 2 3 4
Recorded time (ti) 12 1 4 3
Lubricant (xi) 1 0 1 0
Engine type (zi) 1 1 0 0
Indicator variable (di) 0 1 1 1
(a) [6 marks] Derive an expression for the partial log-likelihood `(β1 , β2 ) for these data.
The maximum (partial) likelihood estimate of β1 is β(ˆ)1 = -0.941 and its standard error is s.e.(β(ˆ)1 ) = 1.443.
(b) [4 marks] Calculate a 95% confidence interval for β1. What conclusions do you draw about the effect of changing the lubricant?
(c) [4 marks] An alternative, accelerated failure time model, based on a Weibull baseline, is proposed to model the data. Write down this model carefully, and explain how it differs from the Cox proportional hazards model above.
(d) [2 marks] Giving your reasoning, explain whether the censoring in this example is informative or non-informative
3. [18 marks] An insurer underwriting the warranty of a particular product uses a three state model with states:
1: Product is operational
2: Product has suffered mechanical failure
3: Product has suffered electrical failure
States 2 and 3 are assumed to be absorbing.
(a) [2 marks] Illustrate the state space, and possible transitions for this process, by a diagram.
The insurer proposes to model transitions between the states using a
time-homogeneous Markov process, with transition intensities μij , i, j = 1, 2, 3.
(b) [6 marks] Using the Kolmogorov forward equation, show that
p11 (x, t) = -(μ12 + μ13 )p11 (x, t).
and hence that p11 (x, t) = exp[-(μ12 + μ13 )t].
(c) [6 marks] Using the Kolmogorov forward equation, show that
p12 (x, t) = μ12p11 (x, t)
and hence use the result obtained in (b) to find an expression for p12 (x, t).
The insurer has collected data on 15 product histories in order to estimate the transition intensities. The data are summarised in the table below.
Total time observed in state (months) |
1 220 |
2 180 |
3 300 |
|
1 ! 2 |
1 ! |
3 |
Observed number of transitions |
5 |
8 |
|
(d) [4 marks] Calculate maximum likelihood estimates and standard errors for the transition intensities μ12 and μ13 .
4. [12 marks] In the product failure example of question 3, a more detailed investigation of the distribution of time to electrical failure is proposed. The 8 observed electrical
failures occurred at times 2,5,9,9,15,15,15 and 20 months. The other seven products did not have observed electrical failure and so are censored observations, at times 1,5,8,8,17,43 and 48 months.
(a) [6 marks] Calculate the Kaplan-Meier estimate for the survival function of the
random variable representing time to electrical failure. Sketch the estimate on a suitable set of axes (a very accurate sketch is not required, but you should label axes and show any points of discontinuity clearly).
(b) [4 marks] Write down the estimate of the probability of survival for 1 year, and
use Greenwood’s formula to calculate its standard error. Hence calculate a 95% confidence interval for this probability.
(c) [2 marks] How might you transform the estimate in (a) so that it can be used to assess the suitability of the time-homogeneous model (constant hazard) used in Question 3.
5. [19 marks] Data on eight lives, aged between 90 and 91 at some point during the calendar year 2017, are available as follows:
Life (i) |
Age on 1.1.2017 |
Date of death (if during 2017) |
Life (i) |
Age on 1.1.2017 |
Date of death (if during 2017) |
1 |
89 years 2 months |
– |
5 |
90 years 2 months |
1.4.17 |
2 |
89 years 7 months |
1.11.17 |
6 |
90 years 6 months |
– |
3 |
89 years 10 months |
– |
7 |
90 years 9 months |
– |
4 |
89 years 10 months |
1.4.17 |
8 |
90 years 10 months |
1.6.17 |
(a) [2 marks] In the standard notation, what probabilities are denoted by qx and b-a qx+a for 0 a < b 1?
(b) [4 marks] For each life i in the table above, calculate ai ≥ 0, the time (after age 90) at which observation in the interval [90, 91) begins, bi 1, the time (after age 90) at which observation may be censored, and di, the indicator of whether death was observed (di = 1) in the interval [90 + ai , 90 + bi] or if the
observation was censored at bi (di = 0) .
(c) [6 marks] Let Di be the random variable of which observed value is di. Write
down an expression for E ( Σi(8)=1 Di) in terms of the bi-aiq90+ai for these data.
Hence, using the approximation
b-a qx+a = (b - a)qx for 0 a < b 1,
show that a moment-based estimator for q90 is given by
˜(q)90 = Σi(8)=1 di
Σi(8)=1 (bi - ai )
and calculate this estimate for the data in the table above.
A larger data set, of census records from the same population, shows 2600 deaths aged 90 (last birthday) in 2017. Census data on exposure are available as mid-year (July 1) population counts, which record 8300 individuals aged 90 (last birthday) on July 1, 2016, 8750 on July 1, 2017 and 8600 on July 1, 2018.
(d) [4 marks] Use the census approximation to obtain the central exposed to risk, aged 90, in 2017.
(e) [3 marks] Hence, calculate the actuarial estimate of q90, in 2017.
6. [18 marks]
The table below present the data from a mortality investigation in a small European
country, together with a set of corresponding standard mortality rates for a neighbouring country m9(S)0 , . . . , m9(S)4 .
Age (x) |
Observed deaths |
EC |
Central mortality rate |
lx |
|
Observed |
Standard |
||||
90 |
16 |
398.9 |
0.0401 |
0.0247 |
10000 |
91 |
19 |
406.9 |
0.0467 |
0.0394 |
9756 |
92 |
27 |
321.0 |
0.0841 |
0.0644 |
9379 |
93 |
38 |
294.9 |
0.1286 |
0.0998 |
8794 |
94 |
44 |
272.6 |
0.1614 |
0.1156 |
7958 |
95 |
|
|
|
|
7088 |
(a) [6 marks] Use a chi-squared test to assess whether the observed data are consistent with the standard mortality rates.
(b) [5 marks] The life table column lx has been computed from the standard mortality rates, assuming a uniform distribution of deaths over each year of age. Calculate the alternative lx values (for x = 91, . . . , 95) under the assumption of constant force of mortality over the age ranges [x, x + 1) for x 2 f90, . . . , 94g.
(c) [4 marks] Given that the whole life expectancy at age 95, e(。)95, is 2.615 years, use the life table column in the table above to calculatee(。)90 , the whole life expectancy
at age 90.
(d) [3 marks] Describe briefly a suitable process by which the observed rates can be used to obtain a set of graduated mortality rates in this example.
2024-01-11