Module 1 · Homework 1

Worked answers with reasoning. References point to the lecture notes PDF and the notes pages.

Q1 — Relationship between $\lambda(t)$ and $S(t)$

Answer: Knowing $\lambda(t)$ over time uniquely determines $S(t)$, as $S(t) = \exp\left(-\int_0^t \lambda(u)\,du\right)$.

This is the key identity $S(t) = e^{-\Lambda(t)}$ with $\Lambda(t) = \int_0^t \lambda(u)\,du$ — any one of the five functions ($f$, $F$, $S$, $\lambda$, $\Lambda$) determines the rest.

Why the others are wrong:

”$\lambda(t) + S(t) = N(t)$” — dimensional nonsense: $\lambda(t)$ is a rate (not a count), $S(t)$ is a probability in $[0,1]$, $N(t)$ is a count of subjects.
”$\lambda(t)$ is the pdf” — no: $\lambda(t) = f(t)/S(t)$, the rate among those still at risk, not among everyone. And “probability of surviving at most to time $t$” describes $F(t)$, not $S(t) = P(T > t)$.
“$S(t)$ conditional on survival to $t$; $\lambda(t)$ the derivative of $S$” — backwards on both counts: $S(t)$ is unconditional, and $\frac{d}{dt}S(t) = -f(t)$, not $\lambda(t)$. The correct derivative relation is $\lambda(t) = -\frac{d}{dt}\log S(t)$.

Q2 — Informative censoring

Answer: A patient is lost to follow-up because they are moved into hospice care at a nursing home due to rapidly declining health.

Censoring is informative when the censoring time carries information about the event time ($U_i$ and $T_i$ dependent). Hospice transfer due to rapidly declining health censors exactly the patients closest to death — a patient censored this way is not representative of the others who survived to that time, and survival would be overestimated.

The other three are noninformative:

5-year follow-up limit — administrative censoring, set by study design, independent of any patient’s health.
Arrested for tax fraud — dropout reason unrelated to cancer prognosis.
Study runs out of funding — study-level mechanism; who gets cut off does not depend on how sick they are.

Q3 — Exponential: $P(T > t+s \mid T > t)$

Answer: $\exp{-\lambda s}$.

Since ${T > t+s} \subset {T > t}$, the intersection is just ${T > t+s}$:

\[P(T > t+s \mid T > t) = \frac{P(T > t+s,\; T > t)}{P(T > t)} = \frac{P(T > t+s)}{P(T > t)} = \frac{S(t+s)}{S(t)}\]

With the exponential $S(t) = e^{-\lambda t}$:

\[\frac{e^{-\lambda(t+s)}}{e^{-\lambda t}} = e^{-\lambda s}\]

This is the memoryless property: having already survived $t$ changes nothing — the constant hazard means the clock effectively restarts.

Distractors: $e^{-\lambda t}$ is the unconditional $S(t)$; $\frac{\lambda^s}{s!}e^{-\lambda}$ is the Poisson pmf; the normal-density expression is irrelevant; and the information given is sufficient.

Q4 — Exponential discharge times (memoryless in action)

Answer: 0.60

Setup: $T \sim \text{Exponential}(\lambda)$, with $P(T \le 4) = 0.6$. Asked: $P(T < 10 \mid T > 6)$.

No need to even solve for $\lambda$: the conditioning window is $10 - 6 = 4$ days — the same width as the given probability’s window. By the memoryless property (Q3),

\[P(T < 10 \mid T > 6) = 1 - P(T > 10 \mid T > 6) = 1 - P(T > 4) = P(T \le 4) = 0.6\]

Having already stayed 6 days changes nothing; the clock restarts, and “discharged within the next 4 days” has the same probability as it did on day zero.

Verification the long way (q4.R): solve $1 - e^{-4\lambda} = 0.6 \Rightarrow \lambda = -\log(0.4)/4 \approx 0.2291$, then

\[P(T < 10 \mid T > 6) = \frac{S(6) - S(10)}{S(6)} = 0.6 \quad\checkmark\]

Rounded to 2 decimals: 0.60.

Q5 — Weibull hazard shapes

Answer: The hazard function is increasing for $k > 1$, decreasing for $k < 1$, and constant for $k = 1$.

The Weibull hazard is

\[\lambda(t) = k\lambda(\lambda t)^{k-1}\]

and its time-dependence sits entirely in the factor $(\lambda t)^{k-1}$:

Shape	Exponent $k-1$	Hazard	Interpretation
$k > 1$	positive	increasing	wear-out / ageing
$k = 1$	zero	constant $= \lambda$	reduces to the exponential
$0 < k < 1$	negative	decreasing	early failures

The other options claim a single behavior “for all $k$” — impossible: the shape parameter’s entire purpose is to let the hazard move either way (that flexibility over the exponential’s constant hazard is why the Weibull is used at all).

Q6 — Advantage of Weibull over exponential

Answer: The Weibull distribution can model both increasing and decreasing hazard rates, whereas the exponential can only model a constant hazard.

Follows directly from Q5: the shape parameter $k$ frees the hazard to rise ($k > 1$) or fall ($k < 1$), while the exponential is locked to $\lambda(t) = \lambda$ — and the Weibull keeps simple closed forms for $S$, $\lambda$, $\Lambda$ while doing it.

Why the others are wrong:

Censoring claim — censoring handling lives in the likelihood ($f^{\delta}S^{1-\delta}$), not in the distribution family; both handle right or left censoring identically.
“Weibull is nonparametric” — both are fully parametric (exponential: $\lambda$; Weibull: $\lambda, k$). Nonparametric is Kaplan–Meier territory.
“Special case, so no advantage” — true premise, inverted conclusion: being a strict generalization is exactly the advantage — the Weibull contains the exponential plus everything it cannot do.

Q7 — Censored likelihood

Answer:

\[L(\theta) = \prod_i f(t_i \mid \theta)^{\delta_i}\, S(t_i \mid \theta)^{1-\delta_i}\]

Observed event ($\delta_i = 1$): the event happened at $t_i$ → density $f(t_i \mid \theta)$.
Censored ($\delta_i = 0$): all we know is $T_i > t_i$ → survival $S(t_i \mid \theta)$.

The exponents are a switch ($f^1 S^0 = f$ vs. $f^0 S^1 = S$), and independent subjects multiply.

Why the others are wrong:

Product-plus-product — likelihoods of independent observations multiply, never add; a sum of two products is not a joint probability of the data.
“Censored data provide no information” — false: $T_i > t_i$ is real evidence about $\theta$. Dropping censored subjects discards exactly the long survivors and biases the estimate (the “delete censored rows” mistake, in likelihood form).
Sum of densities and survivals — sums again, gives every subject both contributions regardless of $\delta_i$, and is not a probability at all.

Q8 — Exponential MLE after a censoring correction

Answer: $-0.05$

The censored exponential MLE is

\[\hat\lambda = \frac{\sum_i \delta_i}{\sum_i x_i} = \frac{\text{\# events}}{\text{total observed time}}\]

Correcting patient 5 from event to censored leaves the denominator unchanged — $\sum x_i = 20.9$ either way, since censored subjects still contribute their observation time — and drops the event count from 3 to 2:

\[\hat\lambda_{new} - \hat\lambda_{old} = \frac{2}{20.9} - \frac{3}{20.9} = -\frac{1}{20.9} = -0.0478 \approx \mathbf{-0.05}\]

Sign check: an event reclassified as censored means less observed failure over the same exposure → lower estimated hazard.

Script: q8.R — includes a survreg cross-check ($\lambda = e^{-\text{Intercept}}$), matching to all decimals.

Q9 — Median survival time

Answer: the time $k$ at which $S(k) = 0.5$.

Median survival $\tau$ is defined by $S(\tau) = 0.5$: half the population still event-free.

Why the others are wrong:

$\lambda(k) = 0.5$ — the hazard is a rate, not a probability; no median interpretation.
$\int_0^k x f(x)dx = 0.5$ — that’s a partial mean (to $\infty$ it gives mean survival $\mu$), not probability mass. The probability version $\int_0^k f(x)dx = 0.5$ would be equivalent to $S(k)=0.5$ — but that is not what is written.
$\Lambda(k) = 0.5$ — at the median, $\Lambda(\tau) = -\log(0.5) = \log 2 \approx 0.693$, not $0.5$.

Q10 — Nonparametric estimation of survival

Answer: Kaplan–Meier estimator.

The most commonly used nonparametric estimate of $S(t)$ — no distributional assumption, built from risk sets via the product-limit construction.

Pearson correlation — linear association between two variables; unrelated.
Weibull survival model — survival, yes, but parametric (scale + shape).
HBKR estimator — nonparametric but for empirical dependence/copulas, not survival. (The course’s other nonparametric survival estimator would be Nelson–Aalen / Fleming–Harrington, not listed.)

Q11 — Interval censoring

Answer: the exact moment the event occurred is unknown; it is only known that it occurred within a certain interval.

The lecture example: diabetes trials with fasting glucose measured every 6 months — a diagnosis at test time $R_i$ places the true onset within $(L_i, R_i)$, between the previous and current tests. The event is “silent”; only the bracketing interval is observed.

“Events later than we can observe (loss to follow-up, drop-out, study termination)” — that is right-censoring, the near-verbatim distractor.
Seasonal measurement windows / vacation gaps — data-collection gaps, not a censoring type.
Intentionally ignoring slow events — not censoring at all; that is selection bias by discarding data.

Q12–Q14 — KM, Nelson–Aalen, Fleming–Harrington

Data: $x = (190, 47, 5, 84, 89, 103, 311)$, $\delta = (1,1,0,1,0,1,0)$. Events at 47, 84, 103, 190; censored at 5, 89, 311. Risk-set table:

$t_{(k)}$	$n_k$	$d_k$	$1 - d_k/n_k$	$\hat{S}_{KM}$	$d_k/n_k$	$\hat\Lambda_{NA}$
47	6	1	5/6	0.8333	1/6	0.1667
84	5	1	4/5	0.6667	1/5	0.3667
103	3	1	2/3	0.4444	1/3	0.7000
190	2	1	1/2	0.2222	1/2	1.2000

Q12 — Kaplan–Meier $\hat{S}(100)$: 0.67

Events at or before 100: 47 and 84.

\[\hat{S}_{KM}(100) = \tfrac{5}{6}\cdot\tfrac{4}{5} = \tfrac{2}{3} = 0.6667 \to \mathbf{0.67}\]

Q13 — Nelson–Aalen $\hat\Lambda(85)$: 0.37

\[\hat\Lambda_{NA}(85) = \sum_{k:\,t_{(k)}\le 85} \frac{d_k}{n_k} = \tfrac{1}{6} + \tfrac{1}{5} = 0.3667 \to \mathbf{0.37}\]

Q14 — Fleming–Harrington $\hat{S}(190)$: 0.30

\[\hat\Lambda_{NA}(190) = \tfrac{1}{6} + \tfrac{1}{5} + \tfrac{1}{3} + \tfrac{1}{2} = 1.2 \qquad \hat{S}_{FH}(190) = e^{-1.2} = 0.3012 \to \mathbf{0.30}\]

Close to but not equal to $\hat{S}_{KM}(190) = 0.222$ — as expected of the two estimators.

Script: q12_14.R — hand computations plus survfit cross-checks (KM; ctype=1 cumhaz for Nelson–Aalen; type="fh" for Fleming–Harrington), all matching.

Q15 — Exponential with mean 5 years: $P(T \le 1)$

Answer: 0.18

The parametrization trap: for the exponential, mean survival is $\mu = 1/\lambda$ — so mean 5 years gives $\lambda = 1/5 = 0.2$, not $5$.

\[P(T \le 1) = F(1) = 1 - e^{-0.2 \cdot 1} = 0.1813 \to \mathbf{0.18}\]

Script: q15.R (with pexp check).