Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 207 (Statistics in Medicine)

The permitted transitions in the continuous-time multi-state model areState 5 is absorbing. The absence of arrows from dementia states 2 and 4 back to nondementia states expresses irreversibility; stroke likewise remains in the history after entry to states 3 or 4.

Write $P_{rs}(t;q)$ for the transition probabilities generated by the constant intensity matrix. The panel-observed multi-state likelihood contributions are:

The homogeneous Markov assumption can fail because mortality and disease incidence depend on age, calendar time, and duration since stroke or dementia; duration dependence would call for a Semi-Markov multi-state model. The common-intensity assumption can fail through patient heterogeneity in sex, genetics, vascular risk, treatment, and frailty. In addition, clinic attendance may be informative: patients with worsening cognition may attend sooner or may be too ill to attend, violating an observation process independent of the latent disease process.

After stroke, absence of a dementia effect on the death intensity is the restrictionFit the unrestricted model and the nested model subject to this equality, then use a likelihood-ratio test. Under regularity conditions, twice the log-likelihood difference is asymptotically chi-squared with one degree of freedom.

Because the measurement can cross the threshold in both directions while the model declares dementia irreversible, one needs the diagnostic rule's sensitivityand specificityThese quantify false negative and false positive state classifications and permit a hidden-state or misclassification model instead of treating every threshold crossing as the true irreversible onset.

If $T$ is future time to dementia from age 50, its survival function isThe tail-sum formula for expectation gives the expected future time alive without dementia:

Education is a causal risk factor if an intervention that changes educational exposure changes the distribution of the corresponding potential outcome for myopia. An observed statistical association need not have that interpretation. Shared causes such as socioeconomic circumstances, parental behavior, genetic traits, or preference for indoor near work can confound education and myopia. Reverse causality is also possible: children with impaired distance vision may select more reading and indoor study, which can increase educational attainment without education causing the impairment.

The decisive experiment would be a sufficiently large Randomized controlled trial assigning children to meaningfully different durations or intensities of education, maintaining the assigned contrast, measuring refractive error before treatment and repeatedly afterward, and using blinded outcome assessment with complete follow-up. Randomization makes baseline causes of myopia exchangeable between arms in expectation, while assignment precedes the outcome; under adherence, no interference, and consistent measurement, the difference in outcome distributions estimates the causal effect of the assigned education policy. Such a trial would be ethically unacceptable because it deliberately withholds or imposes education, which explains the appeal of quasi-experimental and genetic approaches.

The core instrumental variable assumptions are:

For a local average effect one additionally uses instrumental-variable monotonicity. Because the valid instrument supplies variation in education independent of the confounders and has no direct route to the outcome, the instrument-outcome association can be attributed to the education changed by the instrument without measuring every confounder.

Among the education-associated variants in the left panel, larger genetic effects on time in education are associated with more negative refractive error, hence greater myopia. Under the stated assumptions, the fitted negative slope supports a causal effect of more education on increased myopia. In the reverse direction, the myopia-associated variants in the right panel show a much weaker slope whose confidence band includes zero, so the figure gives little evidence that genetically increased myopia causally increases time in education. The asymmetry supports education-to-myopia causation over reverse causation, subject to the Mendelian-randomization assumptions.

Three useful extensions are:

Since $n_0=q{n}_k$ and the $k$ experimental arms each contain $n_k$ patients,ConsequentlyDifferentiation gives $V^{\prime }(q)=(1-k/q^2)/n_{\mathrm{tot}}$, so the stationary point is $q=\sqrt{k}$. Sincethis is the unique minimum. Thus a shared standard arm should be $\sqrt{k}$ times the size of each new-treatment arm.

For $k=4$, the optimum is $q=2$. The supplied two-sided sample size formula givesso round upward to $12$ patients in each new-treatment arm and take $n_0=24$. The analyzed total is thereforeAllowing for ten percent attrition requirespatients to be recruited.

A group sequential design can stop early for efficacy or futility, reducing expected sample size, cost, and time and limiting exposure to an inferior treatment. Its disadvantages include greater design and operational complexity, a potentially larger maximum sample size, and the need for adjusted stopping boundaries and inference to preserve the type I error after repeated looks.

The canonical joint distribution for group sequential test statistics is

Let ${\phi }_2(z;\mu ,\Sigma )$ be the bivariate normal density with the mean and covariance from part ii. Continuing and then not rejecting means $f_1<Z_1<e_1$ and $Z_2<e_2$, so its probability is

The trial uses $N_1$ observations per arm if it stops at the first analysis and $N_2$ if it continues. Since $Z_1\sim N(\delta \sqrt{I_1},1)$,Therefore the expected per-arm sample size is

$N(t)$ is the event counting process, while $Y(t)$ is the at-risk process. Their key relationship is that, conditionally on the observed past, the expected event increment is $Y(t)dH(t)$. It separates exposure to risk from event occurrence and naturally incorporates right censoring.

Before the observed endpoint, $Y+N=1$. If censoring occurs before the event, both processes are zero after the censoring time, so their sum is zero. With the inclusive endpoint conventions in the question, at an observed event time $t=T\le C$ both are one, so the sum is two at that single instant.

The cumulative hazard function is defined through the conditional mean incrementEquivalently, $dH(t)=\mathbb{E}\{dN(t)\mid Y(t)=1\}$ and $H(t)={\int }_0^{t}dH(u)$.

Aggregate the individual increments. At an event time $x_i$, their conditional expectation is the numberat risk times $dH(x_i)$. Replacing expectation by the observed event increment gives $d\hat{H}(x_i)=v_i/r_i$. Thus the estimator is the Nelson–Aalen estimator

The quantities $v_i-\tilde{H}(x_i)$ act as event-count residuals. Requiring their sum to vanish calibrates the fitted total number of events to the observed total, just as an intercept score equation calibrates fitted means. This is the aggregate zero-residual property of a martingale residual.

Put $D_j=\tilde{H}(x_j)-\tilde{H}(x_{j-1})$, with $\tilde{H}(x_0)=0$. ThenThe natural local calibration is thereforeTaking $\tilde{H}$ to be the right-continuous step function with these increments givesBecause $n-j+1$ is exactly the risk-set size $r_j$, this is the estimator from part c and automatically satisfies ${\sum }_i\{v_i-\tilde{H}(x_i)\}=0$.

A proportional hazards model has $h(t\mid g)=h_0(t)r(g)$, with a time-independent hazard ratio. In the Cox proportional-hazards model, $r(g)=e^{\beta g}$ and the baseline hazard $h_0$ is unspecified. At each untied event time, conditioning on the identity of the person who fails given the current risk set produces the factorMultiplying these factors gives the Cox partial likelihood. It is called partial because it uses the conditional event ordering information that identifies $\beta$ while discarding the part of the full likelihood involving the unspecified baseline hazard.

Write $\theta =e^{\beta }$ for the group-1 to group-0 hazard ratio. At $t_1$, all four subjects are at risk, so subject 1 contributes $1/(2+2\theta )$. At $t_2$, subjects 2, 3, and 4 are at risk, so subject 2 contributes $\theta /(1+2\theta )$.

If $t_2<c<t_4$, subject 3 leaves before $t_4$ and subject 4, if it fails, is alone in its risk set. Henceindependently of $k$. Its log derivative vanishes when $1-2{\theta }^2=0$, giving

If $c=t_4$, subject 3 remains in the risk set at a failure of subject 4. ThereforeFor $k=0$, $\hat{\theta }=1/\sqrt{2}$. For $k=1$, the score equation is $1-\theta -4{\theta }^2=0$, soThus $\hat{\theta }$ is constant at $1/\sqrt{2}$ for $t_2\le c<t_4$. At $c=t_4$ it remains there when $k=0$ and drops to approximately $0.39$ when $k=1$, because only then does the censoring time change an event's risk set.

For independent exponential survival times with total observed person-time at risk $T_{\bullet }$ and $d$ events, the maximum-likelihood rate is $\hat{h}=d/T_{\bullet }$. A two-group exponential model is a proportional-hazards model because both hazards are constant, so their ratio is constant.

HereThus the group-1 to group-0 hazard-ratio estimate isIt decreases continuously as $c$ increases. The parametric exponential likelihood uses exact exposure times through each arm's person-time, whereas the Cox partial likelihood uses only which subjects belong to each event's risk set. For $t_2<c<t_4$, changing $t_4$ alters group 0 person-time and hence $\hat{\lambda }$, but subject 4 is alone if it fails at $t_4$, so that event contributes one to the partial likelihood and $\hat{\theta }$ is unchanged.

At each distinct event time $t_j$, let $d_j$ events occur among $r_j$ people in the risk set. The Kaplan–Meier estimator iscensorings remove people from later risk sets but create no factor.

Left truncation, or delayed entry, means that an individual is observed only after surviving to an entry time. A registry assembled from patients alive when a clinic opens is a practical example. Adapt Kaplan–Meier by admitting each person to the risk set only at their entry time.

Period survival analysis is useful when recent prognosis is desired but complete long-term follow-up of a recent diagnosis cohort is unavailable. For a chosen calendar year, intersect every patient's observed follow-up with that year. Express the surviving pieces on the time-since-diagnosis scale, treat the beginning of the calendar window as delayed entry and its end as right censoring, and apply Kaplan–Meier with those entry and exit times.

Take the 2017 calendar window to be $[1,13]$ on the supplied month scale. Intersecting each follow-up interval with this window and translating to time since diagnosis gives:

Patient 2 died before the period and patient 10 entered after it, so neither contributes. At duration 6, patients 4, 5, 6, 7, and 8 are at risk, giving factor $1-1/5=4/5$. At duration 15, patients 1 and 3 are at risk, giving factor $1-1/2=1/2$. Therefore the period Kaplan–Meier estimator isover the range supported by the period data.