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LANCASTER UNIVERSITY
2008 EXAMINATIONS
PART II (Third year)
MATHEMATICS & STATISTICS 1 12 hours
Math 353: Survival and Reliability
In this module you should answer All Section A questions and 1 Section B question. In Section A there are questions worth a total of 50 marks, but the maximum mark that you can gain there is capped at 40.
SECTION A
A1. Let T be a positive continuous lifetime random variable with probability density function f (t). (a) Give a real-life example of a positive continuous lifetime random variable T. [3] (b) Define the survivor function S(t) of T in terms of f (t). [2] (c) Define the hazard function h(t) of T in terms of f (t). [2] (d) State briefly how h(t) may be interpreted in terms of a conditional probability. [3] (e) Show that S(t) can be expressed as
S(t) = exp
∫ (^) t 0
h(u) du
[3]
(f) Define the cumulative hazard function H(t) of T in terms of f (t). [3] (g) Using the expression in (f), show that H(T ) has an exponential distribution. [4] (h) Show that h(t) is constant, λ, if only if T is an Exponential(λ) random variable. [6]
Question A1 continued over the page
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SECTION A continued Economists are interested in business mortality, for which data on the length of life for business enterprises are collected over years. In the economic theory, it is considered that the early years of the business are the most difficult with most types of business. The longer a business survives, generally, other things being equal, the smaller becomes the probability of failure. It is noted that the phenomenon is best described using the notion of hazard function and the following form of hazard function is proposed.
h(t) = (^) t +b a
Sketch the hazard function and determine appropriate range for a and b. Explain why this hazard function would be appropriate to model business failures. [3]
A2. The data below are the failure times of bulb in two conditions A and B. Censored observations are denoted by *. (a) Calculate and draw the Kaplan–Meier estimate of Sˆ(t) for each of the two groups. Group A :5, 7 , 8 ∗, 9 , 11 , 14 Group B :9, 9 ∗, 14 , 17 , 19 ∗, 27 [6] (b) Carry out a log-rank test for any differences between the treatments. [6] (c) State your conclusion by combining results from the above analyses. [3]
A3.
(a) Failure times Ti follow a Weibull distribution with survivor function exp{−(λit)α}, where λi are rate parameters dependent on covariates xi associated with Ti. Explain why this class of lifetime distributions may be described as belonging to a proportional hazard family. [4] (b) What would you take as the baseline hazard function for this family? [2]
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SECTION B continued
Model Covariates Max. log-lik β 0 β 1 β 2 β 3 1 1 -144 4. 2 1+type -143.9 4.631 0. 3 1+start -144 4.810 -0. 4 1+group -140.9 2.508 1. 5 1+type+start -143.9 4.560 0.086 0. 6 1+type+group -140.8 2.504 -0.036 1. 7 1+group+start -140.9 2.463 1.239 0. 8 1+type+group+start -140.8 2.512 -0.037 -0.0002 1.
Table 2. Summary of Exponential survival models for call connection times.
(ii) Interpret the fitted parameters of your chosen model. [2] (iii) How would you diagnose the appropriateness of your chosen parametric model? Suggest a diagnostic method and explain how you would apply your diagnostics. [3]
(c) For a more flexible model, an Extreme Value distribution is proposed to describe this data set. This distribution has survivor function
S(t) = exp
− exp
( (^) t − μ σ
, t ∈ (−∞, ∞),
with parameters μ ∈ (−∞, ∞) and σ ∈ (0, ∞). (i) Derive and sketch the hazard function and hence find the density of this distribution. Comment on the roles of μ and σ in this model. [4] (ii) Suggest a diagnostic which could be used to assess the appropriateness of this model. [3] (iii) Let V = exp(T /σ). Find the distribution of V , expressing the parameters of this distribution in terms of μ and σ. Suggest a diagnostic plot based on the distribution of V that could be used to assess the appropriateness of an Extreme Value model. [6]
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SECTION B
B2. (a) A dataset from a brain cancer study includes survival times of 58 patients that were treated in two different groups. The data contains for each patient the additional fol- lowing variables. Variable Description group 0=group with no prior radiation, 1=group with prior radiation sex gender (male=0, female=1) age age in years kps.pre Karnofsky performance score before the treatment, numerical value 0-
Table 3. Description of explanatory variables for brain cancer study. (i) Figure 1 shows Kaplan-Meier estimates for the survivor functions for two treatment groups. At the distinct failure times ti, di subjects fail, out of ri which were at risk just prior to this time. Explain how the Kaplan-Meier estimates Sˆ(t) uses this information at the failure times ti, given its value just prior to this time. Briefly justify your argument in terms of conditional probability. [3]
0 2 4 6 8 10 12
0.^ 0.^ 0.^ 0.^ 0.^
t
S(t)
group=0group=
Figure 1: Kaplan-Meier Estimates stratified by treatment group.
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0.02 0.05 0.20 0.50 2.00 5.
−^
−^
−^
−^
0
1
t
log(−log(S(t)))
group=0group=
0.02 0.05 0.20 0.50 2.00 5.
−^
−^
−^
−^
0
1
t
log(−log(S(t)))
malefemale
Figure 2: Diagnostic plots for brain cancer study.
(c) Consider an accelerated life model where the survivor function S(t; x) = P (T > t|x) to be of the form S(t; xi) = S 0
t/ exp(γ′xi)
where S 0 (t) is some baseline hazard survivor function with hazard function h 0 (t). Write Ti for the survival time of the i-th subjet with covariate xi. (i) Show that the accelerated failure model can be described as
log Ti = γ′xi + log Wi
where the Wi has survivor function S 0. What is an interpretation of this model? [5] (ii) If S 0 (t) corresponds to the Weibull(λ, α) distribution, show that the accelerated life model corresponds to Cox’s proportional hazards model with
γ = − β α ,
where β are coefficients from Cox’s proportional hazards model. [5]
end of exam
The χ^2 distribution
Values of x for which P (X^2 > x) = p where X^2 has χ^2 distribution with r degrees of freedom.
- 0.20 0.10 0.05 0.01 0. p - 1 1.642 2.706 3.841 6.635 10. - 2 3.219 4.605 5.991 9.210 13. - 3 4.642 6.251 7.815 11.345 16. - 4 5.989 7.779 9.488 13.277 18. - 5 7.289 9.236 11.070 15.086 20. - 6 8.558 10.645 12.592 16.812 22. - 7 9.803 12.017 14.067 18.475 24. - 8 11.030 13.362 15.507 20.090 26. - 9 12.242 14.684 16.919 21.666 27. - 10 13.442 15.987 18.307 23.209 29. - 11 14.631 17.275 19.675 24.725 31. - 12 15.812 18.549 21.026 26.217 32. - 13 16.985 19.812 22.362 27.688 34. - 14 18.151 21.064 23.685 29.141 36. - 15 19.311 22.307 24.996 30.578 37. - 16 20.465 23.542 26.296 32.000 39.
- r 17 21.615 24.769 27.587 33.409 40. - 18 22.760 25.989 28.869 34.805 42. - 19 23.900 27.204 30.144 36.191 43. - 20 25.037 28.412 31.410 37.566 45. - 21 26.171 29.615 32.671 38.932 46. - 22 27.301 30.813 33.924 40.289 48. - 23 28.429 32.007 35.172 41.638 49. - 24 29.553 33.196 36.415 42.980 51. - 25 30.675 34.382 37.652 44.314 52. - 26 31.795 35.563 38.885 45.642 54. - 27 32.912 36.741 40.113 46.963 55. - 28 34.027 37.916 41.337 48.278 56. - 29 35.139 39.087 42.557 49.588 58. - 30 36.250 40.256 43.773 50.892 59. - 40 47.269 51.805 55.758 63.691 73. - 50 58.164 63.167 67.505 76.154 86. - 60 68.972 74.397 79.082 88.379 99. - 70 79.715 85.527 90.531 100.425 112. - 80 90.405 96.578 101.879 112.329 124. - 90 101.054 107.565 113.145 124.116 137.
- 100 111.667 118.498 124.342 135.807 149.
