Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 33 / 5

For a nonnegative random variable, $T={\int }_0^{\infty }1\{T>t\}dt$. Tonelli theorem permits interchanging expectation and this nonnegative integral, giving the tail integral formula for moments:The equality also holds when the mean is infinite. Thus mean completion time is the total area under the survival function, while a restricted mean survival time integrates only to a specified follow-up horizon.

For a continuous positive survival time, $S^{\prime }(t)=-f(t)$ and the hazard function is $h(t)=f(t)/S(t)$ wherever $S(t)>0$. ThereforeSince $S(0)=1$, integrating gives the cumulative hazard function identityAt an endpoint where survival reaches zero, the corresponding cumulative hazard is interpreted as $+\infty$ and the exponential as zero.

Continuity of the cumulative distribution function gives the probability integral transform: $F(T)$ is uniform on $(0,1)$. Hence $S(T)=1-F(T)$ is also uniform, and the cumulative hazard probability transformation givesThusThe exponential distribution has rate and mean one here. This proof does not require a strictly increasing cumulative hazard: intervals with zero density carry no probability mass. It assumes a proper continuous finite survival time, without an atom at infinity.

The Kaplan–Meier estimator is $\hat{S}(t)={\prod }_{t_j\le t}(1-d_j/r_j)$, with event count $d_j$ and risk set size $r_j$. Its median is the first event time with estimated survival at most $1/2$. Since $\hat{S}(2)=0.5749$ and $\hat{S}(3)=0.4286$, the estimated median completion time is $3$ minutes.

The original last event exhausts the final risk set, so the fitted survival function drops to zero at $18$ and its conventional full area is finite. Integrate the right-continuous step function: its height is one before time one, and after each event the height is the newly computed survival value. There are three-minute gaps from $12$ to $15$ and from $15$ to $18$. ThusUsing the exact risk-set products rather than rounded printed survival values gives $4.187856$. This is not the sample mean of the observed event-or-censoring times; right censoring changes the estimated survival weights.

If the final observation at $18$ is instead censored, every earlier Kaplan–Meier estimator value is unchanged and no event factor is inserted at $18$. The median remains $3$ minutes, but the curve stays at $\hat{S}(15)\simeq 0.0107$ at the end of follow-up. By terminal censoring and survival-mean identifiability, a full unrestricted mean is no longer determined nonparametrically by these observations. Extending the final positive step forever would give an infinite integral, but that plotting convention does not establish an infinite population mean. Extra assumptions about the unobserved tail would be needed for a finite full-mean estimate.

The restricted mean survival time through $18$ is still $4.187856$ minutes in either version: the two curves differ only at or after the endpoint, which does not affect the area up to it. All these interpretations require independent censoring. Giving up may be related to how long a person would have taken to finish, so that assumption needs scientific justification in this study.

Let $a_i$ be age in years, $g_i$ the male indicator and $m_i$ the indicator for the Messiah group. The fitted Cox proportional-hazards model iswith an unspecified common baseline hazard. Its coefficients were estimated by Cox partial likelihood, using the Breslow approximation for tied event times. There are $100$ participants, $96$ recorded completions and four censored observations.

A higher hazard function means a greater instantaneous chance of completion among those not yet completing, so it describes faster completion rather than greater mortality. Holding gender and group fixed, an extra year of age multiplies the completion hazard by $1.0066$, about a $0.66\%$ increase. The 95% confidence interval is $(0.9865,1.0270)$ and the Wald test has $p=0.5241$, so there is little evidence of an age association.

Holding age and group fixed, male participants have a hazard ratio $0.5469$ relative to female participants, about a $45.3\%$ lower completion hazard. Its 95% confidence interval is $(0.3560,0.8401)$ and $p=0.00586$. Under the fitted model this corresponds to slower completion for males. It is not a ratio of mean or median completion times.

Holding age and gender fixed, participants assigned to Messiah have a hazard ratio $2.1283$ relative to The Kingdom, with 95% confidence interval $(1.3653,3.3176)$ and $p=0.000854$. This is strong evidence of a group difference, with faster completion in the Messiah group under the fitted model. Random assignment supports interpreting a group contrast as an effect of the assigned condition, subject to the assumptions about follow-up and censoring; age and gender contrasts remain observational associations.

Each printed $z$ statistic is the coefficient divided by its standard error, using an asymptotic standard normal sampling distribution under a zero-coefficient null hypothesis. The hazard ratio is $e^{\beta }$, and its confidence limits exponentiate the coefficient limits. Proportional hazards assumes these covariate-specific hazard multipliers stay constant over follow-up, together with the specified linear age effect and independent censoring conditional on covariates.

The command cox.zph checks the proportional hazards assumption test through time dependence of scaled Schoenfeld residuals. Under constant coefficients, their expected trend against transformed event time should be flat. A significant covariate-specific test would suggest that its coefficient varies with time; the global test checks the coefficients jointly.

The p-values are $0.461$ for age, $0.538$ for gender and $0.882$ for group, with global $p=0.846$. Thus these tests find no evidence against proportional hazards. They do not prove the assumption, establish the correct age functional form, or test the censoring mechanism.

The separate diagnostic plot targets the broader fitted survival distribution. The code evaluates a fitted cumulative hazard function at each observed time and multiplies by the subject's fitted hazard multiplier, forming Cox–Snell residuals:The curve returned without newdata by survfit is a reference-profile curve, not necessarily the all-zero-covariate baseline. The multiplier must use the same centering as that reference curve; the fitted linear predictors use the model's reference convention. With this consistency, $-\log {\hat{S}}_{\mathrm{ref}}$ times the multiplier is the fitted subject-specific cumulative hazard.

Part (a) shows why complete transformed event times should be approximately unit exponential if the Cox proportional-hazards model is correct. The Q–Q plot compares ordered fitted residuals with an independent random exponential sample. Most points are near the diagonal, with some upper-tail departures, so it gives no obvious indication of a gross distributional failure. A random reference sample adds simulation noise, and estimated parameters and tied recorded times prevent this from being an exact distribution-free test.

Importantly, four observed times are censored. Their transformed times retain the censoring indicators and are not complete exponential observations. Simply plotting all $100$ residuals as uncensored therefore gives only a rough visual check. A more appropriate Cox–Snell residual survival diagnostic fits a Kaplan–Meier estimator or Nelson–Aalen estimator to $(r_i,D_i)$, retains the censoring flags, and checks whether the estimated cumulative hazard is close to $r$, equivalently whether estimated survival is close to $e^{-r}$. Taken together, the supplied checks are broadly compatible with the fitted model, while none resolves potentially informative censoring from giving up.