MATH5916: Survival Analysis Term 1 2022 Partial Solutions
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MATH5916:
Survival Analysis
Term 1 2022
Partial Solutions
Week 1
1. (a) ❼ Time origin: time of heart attack ❼ Time scale: calendar time
❼ Endpoint: death
(b) ❼ Left truncation: patients who die before reaching the heart attack ward for
treatment will be excluded from the study.
❼ Right censoring: for patients who survive beyond the end of follow-up (which
might, for example, be when they are discharged from hospital), we only know their survival time exceeds the period from heart attack to end of follow-up. Also if the study is only interested in death related to the heart attack, then deaths from another cause would generate right-censored observations.
2. Design 1: simple type I right censoring at 20 years
Design 2: marriages not ending in divorce are right truncated
3. (a) S(t) = e −λt (b) h(t) = λ
(c) H(t) = λt
4. (a) E(T0j) = and E(T1j) =
(b) h0 (t) = λ and h1 (t) = λβ , β is the hazard ratio
(c)
n0 n1
L(λ,β) = λe−λt0j λβe−λβt1j
j=1 j=1
(d)
βˆ(λˆ) = n(n)t111(0)j(j)
I(λ,β) = n01 1β12
Asymptotically normal with mean (λ,β)′ and variance matrix I−1(λ,β)
Week 2
1. (a) Left truncation implies individuals survive up to a certain point after their time origin, say, Ti > vi . Therefore, the contributions to the likelihood for uncensored and right censored individuals is
L = f(ti |Ti > vi ) S(ti |Ti > vi )
U C
= U C
(b) Assuming left truncation such that Ti > vi is given, the MLE for λ is
= d
i (ti − vi )
2. Basic ideas given in lecture, use delta method for variance in part (c)
3. (a) interval censoring for women who developed breast cancer, right censoring for those that did not
(b)
4 8
L = (S(tℓi − S(tui )) S(ti )
i=1 i=5
= (e−λ55 − e−λ56)(e−λ58 − e−λ59)(e−λ52 − e−λ53)(e−λ59 − e−λ60)(e−λ60)4
4. (a) 0 = 0.01123596,se(0 ) = 0.002808989, and 1 = 0.004564315,se(1 ) = 0.001376193
(b) For treatment group 0, the CIs are
Method lower CL upper CL
Exact |
0.006422319 |
0.017373749 |
Asymptotic |
0.005730337 |
0.016741573 |
Likelihood ratio |
0.006577653 |
0.01766719 |
(c) Using the exact interval above for 0 , the median survival is 61.6901 (95% CI: 39.89623, 107.92786)
(d) For treatment group 0, median survival is 61.6901 (95% CI: 37.79298, 100.6977)
5. (a) = 0.2078368,se() = 0.02399893 (b) same as part (a), = 0.2078368
(c) with no left truncation, = 0.1721407
Week 3
1. The Weibull has S(t) = exp(−λtγ )
(a) Letting 1 − S(tα ) = 1 − exp(−λ(tα )γ ) and solving for tα will give the result (b) Letting tα = λ−1/γ and solving for α gives 1 − 1/e
(c) θˆ = 33.766 and
var γˆ(θˆ) =
2. Starting from the hint for using h(t|V) = ρ0 + ρ2 V , the individual survival function is S(t|V) = e −(ρ0 +ρ2V)t
which has population survival function
S(t) = e−ρ0 tE(e−ρ2Vt) = e−ρ0 t MV (−ρ2t)
for MV the moment generating function of V . The population hazard is then
d
dt
= ρ0 − (MV (−ρ2t))
and equating this to the Gompertz-Makeham hazard gives
MV (t) = exp (et − 1)
which is the moment generating function for a Poisson with parameter ρ 1 /ρ2
3. If done correctly, your plot should look similar to
|
Week 4
1. (a) Input data and use survfit function, or compute manually (b) From fit model = 2.736, so (t) = exp(−exp(−2.736)t);
(c) KM: se( (10)) = 0.144; Exp: se( (10)) = 0.12812 (hint: derive delta method variance for transformation S(10) = e−10λ)
(d) –
(tℓ ) = j:tℓ = × · · ·2. (a) Write out
and then simplify
(b) Write out relationship:
vaˆr( (tℓ )) vaˆr( (tℓ−1)) 1 − (tℓ ) 1 − (tℓ−1)
(tℓ )2 − (tℓ−1)2 = n (tℓ ) − n (tℓ−1)
vaˆr( (tℓ−1)) vaˆr( (tℓ−2)) 1 − (tℓ−1) 1 − (tℓ−2)
(tℓ−1)2 − (tℓ−2)2 = n (tℓ−1) − n (tℓ−2)
finishing at t0 . Then add up terms and simplify.
(c) P(T < 12) = 1 − (11) = 1 − = and
var(1 − (11)) = var( (11)) = = = 0.009718173
(c) X2 = 2.17,p = 0.14, extra Weibull parameter does not improve fit significantly
(d) –
Strata
|
(f) Similar to above, just specify dist = ‘lognormal’ and dist = ‘extreme’ (g) Fit the usual way, for example, the log-normal would be:
> hep .lnorm <- survreg(Surv(smonths, event) ~ treat,
data = hepdat, dist = "lognormal")
> predict(hep .lnorm, data .frame(treat = c(0, 1)),
type = "quantile", p = 0 .5)
1 2
47.09753 164.31348
(h) X2 = 4.7,p = 0.03
4. (a) –
Strata
|
(b) X2 = 0.1,p = 0.7
(c) will need to make stage a factor variable first
Strata
0 1000 2000 3000 4000 5000 Time |
(d) X2 = 70.1,df = 3,p < 0.0001
Week 5
1. Since the KM estimate is a step function and t = 26 occurs between event times, i.e., within [19, 30), the estimates are
(26) = (1 − 1/18)(1 − 1/15) = 0.881
var( (26)) = (0.881)2 · = (0.0790)2
2. (a) For ψi = exp(xi(T)β)
ψ 1 ψ3 ψ4 ψ6
3. (a)
exp(9β) + exp(10β) + exp(5β) + exp(6β) + exp(3β) + exp(4β)
exp(5β) + exp(6β) + exp(3β) + exp(4β)
exp(6β)
×
exp(5β) + exp(6β) + exp(3β) + exp(4β)
(b) Similar to Breslow but “discount“ the denominator for ties
(c) Since there are 3 ties, there are 3! = 6 unique orderings and so there are 6 terms in the partial likelihood
2022-04-13