Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 207 (Biostatistics)

Write $y_{WP}$ and $y_{AP}$ for the estimated log odds ratios relative to the common control. Indirect treatment comparison uses the consistency relation ${\delta }_{WA}={\delta }_{WP}-{\delta }_{AP}$. For independent trials, subtraction adds their variances, givingThe substantive assumption is transitivity in network meta-analysis: the study populations, outcome definitions and follow-up are sufficiently comparable, particularly in their distributions of effect modifiers, that the two control comparisons estimate effects applicable to the same target population. The common control's name alone does not justify that assumption. This also assumes comparable marginal odds ratios; these need not equal covariate-adjusted odds ratios even without confounding.

Using the supplied rounded variances and the given standard normal quantile, the approximate confidence interval isOn the odds ratio scale this is $e^{-1.26}\simeq 0.284$, with approximate limits $(0.104,0.771)$. The estimated odds of recurrent stroke under treatment $W$ are about 28% of those under $A$, a reduction of about 72% in odds. The interval excludes equal odds, so this comparison provides evidence of lower odds under $W$ if the indirect treatment comparison assumptions hold. These are odds, not a 72% reduction in probability; a confidence interval describes the long-run coverage of the procedure, not a posterior probability for this particular interval.

Let ${\hat{\delta }}_D$ and ${\hat{\delta }}_I$ be independent, approximately unbiased estimates of one consistent treatment effect, with variances $v_D,v_I>0$. The inverse-variance weighted mean givesIndeed, weights $w,1-w$ give variance $w^2{v}_D+(1-w{)}^2{v}_I$; differentiating gives $w=v_I/(v_D+v_I)$, the weight on the direct estimate. Here the three displayed trials give ${\hat{\delta }}_D=-1.33$, $v_D=0.04$, and hence $\hat{\delta }\simeq -1.3203$ with variance $0.03448$. This numerical pooling uses the displayed evidence, not the unavailable data from all twelve trials. A shared study or shared control would introduce covariance, requiring the corresponding correlated-estimator formula; inconsistency would invalidate the common-effect interpretation.

Under the null hypothesis of consistency, ${\hat{\delta }}_D-{\hat{\delta }}_I$ has approximately zero expectation and variance $v_D+v_I$, since the displayed trials are independent. Thus a two-sided Wald test usesThe observed difference is $-0.07$, and $Z=-0.07/\sqrt{0.29}\simeq -0.130$. Its two-sided p-value is approximately $0.897$, so there is no detectable discrepancy here. Failure to reject is not proof of consistency: this test has limited statistical power, and transitivity in network meta-analysis also needs substantive assessment. If evidence sources overlap, replace $v_D+v_I$ by $v_D+v_I-2\mathrm{Cov}({\hat{\delta }}_D,{\hat{\delta }}_I)$.

A network meta-analysis can offer three distinct benefits, assuming a connected treatment network and credible transitivity in network meta-analysis.

These are potential benefits, not guarantees: heterogeneous effect modifiers, inconsistent evidence, disconnected networks or imprecise rankings can defeat them.

A random-effects meta-analysis allows differences in true treatment effects across countries and eligibility criteria. For trial $i=1,\dots ,6$, let $y_i$ be its estimated log odds ratio and $v_i$ its estimated within-trial variance. A usual approximate statistical model isindependently across trials, with $b_i$ independent of sampling errors. Here $\mu$ is the mean true log odds ratio in the population of comparable trials, and ${\tau }^2\ge 0$ is between-trial heterogeneity, not additional sampling error. Therefore $y_i\sim N(\mu ,v_i+{\tau }^2)$, and for a fitted heterogeneity value,One can estimate ${\tau }^2$ by restricted maximum likelihood or another justified method; $({\sum }_i{w}_i{)}^{-1}$ is a plug-in conditional variance and does not fully account for estimating heterogeneity. With only six studies that uncertainty matters. Different eligibility rules motivate this random effect but do not by themselves establish exchangeability or remove bias; known effect modifiers may warrant stratification or regression.

Use leave-one-study-out influence analysis: fit all six trials, then omit Cohen 1989 and refit both $\mu$ and ${\tau }^2$. Report $\hat{\mu }-{\hat{\mu }}_{(-C)}$, the corresponding change in the odds ratio, and changes in the confidence interval and heterogeneity. As a diagnostic with heterogeneity held fixed, writing $W={\sum }_i{w}_i$ givesThe fitted Cohen weight would be $(0.15+{\hat{\tau }}^2{)}^{-1}$ and $y_C=-1.95$. A precise but discordant trial can have large influence; refitting assesses additional influence through heterogeneity. The other five trials' data are absent, so a numerical six-trial influence assessment is not identifiable from the displayed table.

A Type I error is rejecting a true null hypothesis. Here it means declaring a statistical association between the biomarker and bowel cancer when, in the population and under the association model being tested, there is no statistical association. It is a false positive association finding, not merely a large estimation error. A test with level $0.05$ limits this probability to $0.05$ under the null hypothesis, subject to valid calibration; the p-value is not the probability that the null hypothesis is true.

Under the intended additional assumption of mutually independent, continuously calibrated null p-values, each test fails to reject with probability $0.95$. The familywise error rate is thereforeThere is a qualification to the printed use of “not correlated”: zero pairwise correlation coefficients do not generally imply mutual independence. A multivariate normal distribution with zero off-diagonal covariances does give independence, but an arbitrary biomarker distribution need not. The formula above requires independent rejection events, not just uncorrelated measurements.

The usual intuition for positively dependent tests is that they reject together, reducing the effective number of opportunities for a false positive. A rigorous sufficient assumption is positive association of random variables for the rejection indicators $I_1,\dots ,I_{10}$. Products of their non-rejection indicators are decreasing functions; association and induction giveThus under this stronger positive-dependence assumption the probability is no larger, and perfect dependence gives $0.05$ instead of $0.4013$. Strict inequality is not forced by every form of positive dependence.

Pairwise positive correlation coefficients alone, as printed, is insufficient. For a counterexample choose a vector of ten exchangeable random variables that are rejection indicators as follows: with probabilities $0.545,0.450,0.005$, respectively, reject none, reject one uniformly selected test, or reject all ten. Thenbut the familywise error rate is $0.455>1-0.95^{10}$. This can be realized with valid null p-values: conditional on the indicators, independently draw $p_i$ uniformly on $(0,0.05)$ if $I_i=1$ and on $(0.05,1)$ otherwise. Every $p_i$ is uniform on $(0,1)$, and $\mathrm{Cov}(p_i,p_j)=0.25\mathrm{Cov}(I_i,I_j)>0$. Taking biomarker statistics $X_i=1-p_i$ gives pairwise positively correlated null statistics with exactly these one-sided tests. Therefore the unconditional larger-or-smaller claim needs a specified dependence model.

The Bonferroni correction compares every p-value with $0.05/100000=5\times 10^{-7}$. Only the first two displayed values lie below that threshold; all undisplayed values are at least the tenth displayed value. HenceFor the Benjamini-Hochberg procedure at false discovery rate level $q=0.1$, the rank-$i$ threshold is $qi/100000=i\times 10^{-6}$. The first nine displayed ranks meet their thresholds, while rank ten does not: $1.1\times 10^{-5}>10^{-5}$. Consequently nine is the intended count if every later rank also fails its threshold. The rule is a step-up rule, however:It does not stop at the first failed comparison. The full rejection count is therefore not determined by ten order statistics. For example, completing the list with $99990$ values equal to $1$ gives $k=9$, whereas completing it with $99990$ values equal to $0.02$ gives $k=100000$, since $p_{(100000)}=0.02\le 0.1$. Both completions respect the printed ten smallest p-values. Thus at least nine are rejected; exactly nine requires an extra assumption about the undisplayed ranks. In the second completion even the tenth marker is rejected, despite failing its own rank threshold. The false discovery rate guarantee also requires the usual validity and independence or appropriate positive-dependence assumptions, unlike the arbitrary-dependence Bonferroni correction.

Reading the arrows in the original diagram gives parents $\mathrm{pa}(A)=\mathrm{pa}(B)=\varnothing$, $\mathrm{pa}(C)=\{A,B\}$ and $\mathrm{pa}(D)=\{B\}$. The Bayesian network factorization isFor binary variables, each conditional probability table has one free probability per parent configuration, since its two entries sum to one. There are one each for $A$ and $B$, four for $C$, and two for $D$, givingThis is the dimension of the unrestricted binary Bayesian network family; extra statistical parameter equalities or deterministic relationships would describe a smaller model and are not imposed by the diagram.

The original PDF shows the chain $A\to C\to B$, not a collider. Its Bayesian network factorization is $p(a,b,c)=p(a)p(c\mid a)p(b\mid c)$. For any $c$ with positive marginal probability,Integrating or summing over $a$ also gives the same conditional marginal $p(b\mid c)$. ThusThis is conditional independence; the graph does not generally imply marginal independence, since summing $p(a)p(c\mid a)p(b\mid c)$ over $c$ can transmit dependence from $A$ to $B$. Special statistical parameter choices can make marginal independence hold as well, so the claim is that only the conditional statement is guaranteed by the graph. Conditional distributions on null values of $C$ are immaterial to this assertion.

Let $\mathcal{D}=(x^{(1)},\dots ,x^{(n)})$ contain independent complete observations of the $p$-dimensional vector, and let $G$ be a Directed acyclic graph. A Gaussian Bayesian network uses, for node $j$,Here ${\alpha }_j$ is an intercept, ${\gamma }_j$ the vector of parent coefficients, and ${\sigma }_j^2>0$ the conditional variance; the vector ${\theta }_G$ collects these statistical parameters. Its likelihood function factors asChoose a proper statistical parameter prior distribution $\pi ({\theta }_G\mid G)$ and a graph prior distribution $\pi (G)$. Bayes' theorem gives the Bayesian network structure score, up to the normalization common to graphs,The integral is Bayesian model evidence: it averages over nuisance parameters rather than substituting their best-fitting values. A graph prior can favor sparse graphs, while the evidence balances fit against the amount of prior statistical parameter space that predicts the observations well. Compare these scores over admissible Directed acyclic graphs, using enumeration when feasible or a search procedure otherwise; search need not find a global maximum, and observational data need not identify a unique causal orientation.

A conjugate prior makes the integral analytic. For example, for ${\eta }_j=({\alpha }_j,{\gamma }_j)$ take ${\eta }_j\mid {\sigma }_j^2\sim N(m_j,{\sigma }_j^2{V}_j)$, ${\sigma }_j^2\sim \mathrm{IG}(a_j,b_j)$, where $V_j$ is positive-definite matrix and $a_j,b_j>0$; these are independent local normal-inverse-gamma priors. Under prior independence between nodes, the evidence is a product of local regression evidences. Each posterior has the same family, so the local integral is the ratio of prior and posterior normalization constants, with the likelihood's constants included. This avoids costly numerical integration and makes local graph updates inexpensive. Hyperparameters must be specified coherently if score equivalence between observationally equivalent graphs is desired; arbitrary local priors do not automatically have that property.

The first departure from the no-symptoms state has an exponential distribution with rate $\lambda$. Spending the entire interval there means no departure at all, soThis is the holding time survival probability. The transition probability $p_{11}(t)$ only says that the patient is symptom-free at the endpoint, allowing an onset and recovery in between, and is therefore not the answer. The formula also covers $\lambda =0$, when state 1 is absorbing.

Let $A_2(t)={\int }_0^{t}1\{X(s)=2\}ds$ be the occupation time of a continuous-time Markov chain in the symptoms state. Tonelli theorem permits exchanging the nonnegative integral and expectation. Starting from state 1, put $q=\lambda +\mu >0$. ThensoFor small $t$ the answer is $\lambda t^2/2+O(t^3)$, as expected when onset must first occur. It is between zero and $t$. If $\lambda =\mu =0$, no transition can occur and the answer is zero; the displayed quotient is interpreted separately in that degenerate case.

Dividing the expected occupation time of a continuous-time Markov chain by $t$ and letting $t\to \infty$ givesFor $\lambda ,\mu >0$ this is also the almost-sure long-run time fraction, not just the limit of an expectation. The two-state continuous-time Markov chain is irreducible and positive recurrent. Alternatively, a regeneration cycle consists of a mean $1/\lambda$ symptom-free holding time followed by a mean $1/\mu$ symptomatic holding time; the renewal-reward theorem gives $(1/\mu )/(1/\lambda +1/\mu )=\lambda /(\lambda +\mu )$. This agrees with the stationary distribution solving ${\pi }_1\lambda ={\pi }_2\mu$ and ${\pi }_1+{\pi }_2=1$.

Starting in state 1, if $\lambda =0$ the fraction is zero; if $\lambda >0$ and $\mu =0$ the patient eventually enters state 2 forever, and the fraction is one. If both rates vanish, the fraction remains zero from this initial state and the quotient $0/0$ is undefined.

Condition on each patient's initial observed state and assume independent patient trajectories and observation times that do not carry additional information about the unobserved path. The Markov property factors each trajectory's likelihood function into its successive one-year transition probabilities. ThusMore explicitly, with $q=\lambda +\mu >0$ and $E=e^{-q}$,Successive transitions within one patient need not be independent unconditionally: this is a product of conditional factors justified by the Markov property. If initial-state probabilities are modeled rather than conditioned on, their likelihood contributions must also be included. Zero-count terms contribute zero; a positive count attached to a zero model probability makes the log-likelihood equal to $-\infty$.

Set $S=n_{11}+n_{22}$, $R=n_{12}+n_{21}$ and $N=S+R$. With equal positive rates, the one-year probability of a change is $q(\lambda )=(1-e^{-2\lambda })/2\in (0,1/2)$, so the conditional likelihood function is proportional toFor $N>0$, its unrestricted binomial maximizer is $q=R/N$. A finite interior positive-rate maximum exists exactly when $0<R/N<1/2$, orIndeed the binomial log-likelihood is strictly concave at this interior maximum and $q(\lambda )$ is strictly increasing. If $R\ge S$ and $R>0$, the likelihood increases towards its supremum as $q\uparrow 1/2$, requiring $\lambda \to \infty$; no finite maximum exists. Equality $R=S>0$ also has only this limiting maximum.

If $R=0<S$, allowing the boundary $\lambda =0$ gives a finite maximum there; insisting on $\lambda >0$ gives only a supremum as $\lambda \downarrow 0$. Thus if “valid” includes zero rates, the condition is $R<S$ with $N>0$; if rates must be strictly positive, it is $0<R<S$. If $N=0$, every rate has the same empty-data likelihood and the statistical parameter is not identifiable. These boundary qualifications prevent a negative or infinite plug-in rate from being called a valid interior estimate.

Visits can be missed or rescheduled because of illness, clinic availability or patient choice; follow-up may end through dropout or right censoring. The correct panel-observed multi-state likelihood uses each observed interval length ${\Delta }_{ik}$:Thus the four aggregated one-year counts are generally no longer sufficient; records must retain actual durations. Numerical maximum likelihood estimation uses the corresponding transition probabilities, or a matrix exponential of the generator when a larger model is fitted. One must not treat a two-year interval as a single one-year transition or as two known transitions through an unobserved intermediate state.

This conditional calculation is valid when visit scheduling and loss to follow-up are ignorable given the modeled information. If symptomatic patients are systematically examined earlier or leave the study differently, observation itself may be informative; the visit or dropout process may require a joint statistical model. Simply replacing one year by the actual interval does not correct informative observation.

For a treated patient the recovery transition intensity is ${\mu }_T=0.25\times 2=0.5$ per year. A symptoms spell is a holding time with an exponential distribution of rate ${\mu }_T$, soThe treatment multiplier for onset affects how often symptoms begin, not the duration of a spell once the patient is in state 2, under this homogeneous continuous-time Markov chain.

The untreated rates are ${\lambda }_U=0.2$ and ${\mu }_U=0.25$; treatment changes them to ${\lambda }_T=0.2\times 0.8=0.16$ and ${\mu }_T=0.25\times 2=0.5$. Their stationary distributions give symptoms fractionsConsequentlyThus the estimated long-run symptoms fraction under treatment is about 54.5% of the untreated fraction. This combines less frequent onset with faster recovery; it is not either individual hazard ratio alone.

Use one joint hypothesis test for $H_0:{\beta }_{\lambda }={\beta }_{\mu }=0$. Fit the unrestricted model with statistical parameters $({\lambda }_0,{\mu }_0,{\beta }_{\lambda },{\beta }_{\mu })$, where ${\lambda }_i={\lambda }_0{e}^{{\beta }_{\lambda }{z}_i}$ and ${\mu }_i={\mu }_0{e}^{{\beta }_{\mu }{z}_i}$ for treatment indicator $z_i$. Refit under $H_0$, estimating the two common baseline intensities; do not fix them at their unrestricted estimates. The likelihood-ratio test statistic isUnder regular identifiable positive-rate models, sufficient independent patient trajectories, and the null hypothesis, Wilks theorem gives two degrees of freedom because two coefficients are constrained. Reject if $D$ exceeds the chosen upper chi-squared distribution quantile. Alternatively a two-dimensional Wald test uses ${\hat{\beta }}^T\widehat{\mathrm{Cov}}(\hat{\beta }{)}^{-1}\hat{\beta }$, including the off-diagonal covariance. Two separate tests do not provide this single calibrated assessment. The point estimates alone cannot produce a numerical p-value without fitted likelihoods or a coefficient covariance matrix.

For a nonnegative continuous event time, the survivor function is $F(t)=\mathbb{P}(T>t)$; here the printed $F$ denotes survival, not the usual cumulative distribution function. A lower median isFor a continuous strictly decreasing survivor function this is the unique solution of $F(m)=1/2$. If the curve has a flat stretch at $1/2$, the distribution has a nonunique median and the infimum specifies the lower endpoint.

The tail integral formula for moments follows from $T={\int }_0^{\infty }1\{T>t\}dt$ and Tonelli theorem. Therefore, when survival vanishes beyond $t^{\ast }$,If $F(t)>0$ at every finite time, the mean is finite precisely when ${\int }_0^{\infty }F(t)dt<\infty$. For a concrete sufficient condition, require $F(t)\le C{t}^{-1-\epsilon }$ for all $t\ge M>0$, with $C<\infty$ and $\epsilon >0$. The integral over $[0,M]$ is at most $M$, and the tail integral is at most $C{M}^{-\epsilon }/\epsilon$. An exponential tail bound also suffices. Merely having $F(t)\to 0$, or even $tF(t)\to 0$, does not suffice: a properly completed decreasing tail proportional to $1/(t\log t)$ for $t\ge e$ has infinite mean.

Let $a_1<a_2<\cdots$ be the distinct observed event times, $r_j$ the risk set size immediately before $a_j$, and $d_j$ the events there. Under independent censoring, the Kaplan–Meier estimator isIt starts at one, jumps down at events, and is unchanged by a pure censoring time; right censoring removes people from subsequent risk sets. If events and censorings coincide, the convention here processes events before censoring, so those censored at the recorded time are included just before the event. The estimated median is the first time the curve reaches or falls below $1/2$. If it never does during observation, the median is not estimable from the observed curve without tail assumptions.

With one remaining individual censored at $t_1$, no event factor is introduced. HenceThis is the terminal observed step value. With an empty risk set after $t_1$, continued horizontal plotting is a convention and supplies no information about the true later survivor function. The tail rests on one person and is imprecise; this is terminal censoring and survival-mean identifiability. If ${\varphi }_0>0$, the unrestricted mean cannot be obtained nonparametrically from this censored tail: extending the last positive step to infinity would give an infinite area, which does not establish an infinite population mean. A restricted mean survival time up to a suitably supported finite cutoff remains estimable.

If the final individual experiences an event at $t_1$, the Kaplan–Meier estimator acquires the factor $1-1/1=0$. ThusThe conventional fitted curve is zero thereafter. Its area gives the finite plug-in estimateThe area is a sum of rectangles: if $a_0=0<a_1<\cdots <a_m=t_1$ are the event times, it equals ${\sum }_{j=0}^{m-1}\hat{F}(a_j)(a_{j+1}-a_j)$. Censoring times within a constant segment do not alter its area.

Both terminal estimates have poor reliability because the last risk set has size one: one observed outcome determines whether the entire remaining fitted tail stays positive or collapses to zero. The zero tail is an empirical endpoint convention, not evidence that every future member of the population must fail by $t_1$. With earlier right censoring, the displayed finite area can be especially sensitive to this last event and does not remove uncertainty about the population tail. In the final-censoring case the full mean lacks a determined tail area; in the final-event case the conventional full fitted mean is finite and calculable, but should be reported with this limitation. A restricted mean survival time at a prespecified cutoff within reliable follow-up often avoids the unstable terminal extrapolation.

A proportional hazards model assigns individual $i$ the hazard function $h_i(t)=h_0(t){\varphi }_i$, where $h_0$ is the baseline hazard for multiplier one and ${\varphi }_i>0$ is a time-constant hazard multiplier. In a Cox proportional-hazards model, ${\varphi }_i=e^{x_i^T\beta }$ for a covariate vector $x_i$ and coefficient vector $\beta$; ratios $h_i(t)/h_k(t)={\varphi }_i/{\varphi }_k$ are constant whenever the baseline is nonzero.

Assume independent individuals, independent censoring conditional on the covariates, and no tied events. Immediately before an event at $t$, condition on its observed risk set $R(t)$ and one event in a small time interval. The individual event probability is $h_0(t){\varphi }_{i}dt+o(dt)$, while the total is $h_0(t){\sum }_{k\in R(t)}{\varphi }_{k}dt+o(dt)$. Their limiting ratio is ${\varphi }_i/{\sum }_{k\in R(t)}{\varphi }_k$. Multiplication over observed events gives the Cox partial likelihoodThe unspecified baseline hazard cancels. This is the usual successive conditional event contribution defining the partial likelihood, not the full event-time likelihood conditioned simultaneously on all event times.

For the four individuals the successive event risk sets are $\{a,b,c,d\}$, $\{c,d\}$ and $\{d\}$. Individual $b$ leaves at its right censoring time and contributes no event numerator. ConsequentlyThe last event contributes one and carries no further relative-risk information.

Because $t^{\ast }>t_b>t_a$, the three possible complete event orders are $t_a<t^{\ast }<t_c<t_d$, $t_a<t_c<t^{\ast }<t_d$, and $t_a<t_c<t_d<t^{\ast }$. Write $K={\varphi }_a/({\varphi }_a+{\varphi }_b+{\varphi }_c+{\varphi }_d)$ and $V={\varphi }_b+{\varphi }_c+{\varphi }_d$. Their partial likelihoods, with the final singleton factor omitted, are respectivelyAdding first $L_2$ and $L_3$ gives $K{\varphi }_c/V$, and thereforeThusThis is Cox rank-likelihood deletion consistency: summing out the unobserved position of $b$ leaves the relative order information in the observed events. It is an algebraic marginalization of complete-order probabilities under time-constant hazard multipliers. It does not include the probability density function of the actual censoring time, establish the distribution of $t^{\ast }$ given all observed times, or justify informative censoring. Ignoring the censoring mechanism in the survival analysis still requires independent censoring given the modeled covariates.

The ordinary Nelson–Aalen estimator adds $d_j/r_j$ at event time $t_j$. Under a fitted proportional hazards model, replace the number at risk by the sum of fitted hazard multipliers:This is the Breslow estimator of the integrated baseline hazard, using fitted regression coefficients obtained by maximizing the Cox partial likelihood. The denominator arises because the total event intensity is $h_0(t){\sum }_{i\in R(t)}{\varphi }_i$; with every multiplier equal to one it reduces to the Nelson–Aalen estimator.

The estimator is right-continuous, so ${\hat{H}}_0(t_a)$ already includes the event at $t_a$. Subtracting it leaves only the events at $t_c$ and $t_d$, whose risk sets are $\{c,d\}$ and $\{d\}$. HenceThere is no increment at the censoring time $t_b$. If the lower endpoint had instead been $t_a-$, an additional $1/({\hat{\varphi }}_a+{\hat{\varphi }}_b+{\hat{\varphi }}_c+{\hat{\varphi }}_d)$ would be present. The final singleton event can increase the baseline hazard estimate even though its partial likelihood contribution is one.

In a competing risks model, the first event has time $T$ and type $J$; observing one type prevents another type from being the first event for that individual. Its cause-specific hazard isFor absolutely continuous event times the total hazard function is ${\sum }_j{h}_j(t)$, giving the event-free survivor function $S(t)=\exp [-{\int }_0^t{\sum }_j{h}_j(u)du]$. Since an event of type $j$ near $t$ has unconditional density $S(t)h_j(t)$, its cumulative incidence function satisfiesThis cause-specific hazard to cumulative incidence formula does not require independent latent times for the competing causes. In general $F_j(t)$ is not $1-e^{-{\int }_0^t{h}_j}$, since that expression ignores removal by other causes.

Put $a={\theta }_A\ge 0$, $b={\theta }_B\ge 0$ and $q=a+b$, with $\tau \ge 0$. HereIntegrating $aS(u)$ separately before and after $\tau$ gives, for $q>0$,After $\tau$, only type $A$ can occur, so the second term is the fraction event-free at $\tau$ multiplied by their subsequent event probability. The expressions agree at $\tau$. If $a=b=0$, both incidence functions are identically zero; if $a=0<b$, this formula gives $F_A=0$.

Only times up to $\tau$ contribute to the cumulative incidence function for type $B$, because its cause-specific hazard vanishes afterwards. With $q={\theta }_A+{\theta }_B>0$,soThis increases before $\tau$ and then remains constant; it includes the protection from type $B$ events produced by earlier type $A$ events. If both intensities vanish, $F_B=0$; if ${\theta }_B=0$, it is also identically zero.

The two event types are disjoint and exhaustive among first events, so their cumulative incidence functions sum to the probability of any event by $t$. Equivalently their derivatives add to $S(t)(h_A(t)+h_B(t))=-S^{\prime }(t)$. ThereforeFor the usual positive disease-death intensity ${\theta }_A>0$, the limit is one: eventually every individual has an event, even though treatment-side-effect events cease after $\tau$. If the permitted boundary ${\theta }_A=0$ is included, however,Then a fraction $e^{-{\theta }_B\tau }$ remains event-free forever, because all hazards vanish after $\tau$. This includes the no-event case ${\theta }_A={\theta }_B=0$. Thus an assertion that the limit always equals one requires ${\theta }_A>0$.

By conditional probability and continuity of event times,When $\tau >0$ and $q={\theta }_A+{\theta }_B>0$, the common factor $1-e^{-q\tau }$ cancels, givingBefore $\tau$, the constant cause-specific hazards divide the total event intensity in these proportions, regardless of how long the interval is. If $\tau =0$ or both intensities are zero, the conditioning event has probability zero and this conditional probability is undefined, not a ratio assigned by convention.

A frailty model represents unobserved heterogeneity using a nonnegative frailty random variable $U$ that multiplies an individual's baseline hazard. With $H_0(t)={\int }_0^t{h}_0(s)ds$, the individual and population survivor functions areThus the population survivor function is the Laplace transform of the frailty distribution at $H_0(t)$; at points where differentiation is justified, $\overline{h}(t)=h_0(t)\mathbb{E}[U\mid T>t]$. The last identity follows by differentiating the integral and dividing $-{\overline{S}}^{\prime }$ by $\overline{S}$.