Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 29 (Biostatistics)

The area for the pedigrees on page 7 is blank in the official Part III PDF and PostScript edition, and in the corresponding MPhil PDF and MPhil PostScript edition. All four copies have been checked visually. The family-specific calculations in Question 6 remain dependent on those unavailable drawings.

Write $S(t)=\mathbb{P}(T>t)$ for the survival function; this is the function denoted $F_T$ in the survival notation. For a proper continuous event-time distribution with hazard function $h$, its cumulative hazard function is $H(t)={\int }_0^{t}h(s)ds=-\log S(t)$. The probability integral transform makes $S(T)$ uniform on $(0,1)$, including when $S$ has flat intervals: those intervals have zero event probability. ConsequentlyThus the cumulative hazard probability transformation gives $\boxed{H(T)\sim \mathrm{Exponential}(1)}$. No strict monotonicity of $H$ is needed on intervals that carry no event mass. The usual proper continuous survival law is essential; an atom of subjects who never fail would instead require separate treatment of the mass at infinity.

Under independent censoring, order the distinct observed failure times as $t_1<\cdots <t_m$. Let $r_j=\#\{i:x_i\ge t_j\}$ be the size of the risk set just before $t_j$, and let $d_j=\#\{i:x_i=t_j,v_i=1\}$ be the number of failures there. The estimated conditional probability of surviving that event time is $1-d_j/r_j$. Multiplying these conditional survival probabilities gives the Kaplan–Meier estimatorA censored observation leaves the risk set after its censoring time but does not create a downward survival jump. With tied failure and censoring times, the displayed risk-set convention includes those censored at the time while processing failures, then removes them.

Since $H=-\log S$, the Kaplan–Meier estimator of the cumulative hazard function isThis is not exactly the Nelson–Aalen estimator ${\sum }_{t_j\le t}{d}_j/r_j$, although the two are close when the individual fractions are small. If a jump exhausts the risk set, $\hat{S}$ becomes zero and ${\hat{H}}_{\mathrm{KM}}$ becomes infinite; transformed residuals beyond that point require a finite-tail modelling convention or restriction of follow-up.

The transformed values are Cox–Snell residuals. Under a correct common survival function and a good fitted cumulative hazard function, the complete transformed failure times should have unit exponential distribution. Transform the observed times and retain their original censoring indicators, then estimate residual survival with a Kaplan–Meier estimator. Its target isso a plot of its logarithm against $u$ should lie approximately on a line of slope $-1$. Censored residuals must not be treated as fully observed failure times.

There is a useful caution for this particular common nonparametric fit. The shared-fit Cox–Snell residuals reproduce the fitted survival curve property means that the pooled residual plot is largely built into the transformation. With event ordering and compatible tie processing preserved, at a transformed failure time $u_j=\hat{H}(t_j)$ its residual Kaplan–Meier estimator equals $\hat{S}(t_j)=e^{-u_j}$. Finite steps, coalesced censoring times and estimation uncertainty still matter, and an infinite final residual arises if the fitted survivor becomes zero. Separate group plots are more informative about a common-distribution assumption than this pooled fit alone.

If both groups share the same survival function and the common fit is adequate, their complete Cox–Snell residuals have the same unit exponential distribution. Under independent censoring in each group, both groupwise Kaplan–Meier estimators should therefore be close to $e^{-u}$ and to each other, subject to ordinary sampling variation and the number of subjects still at risk. Different censoring patterns can change precision without changing the residual survival target.

Let $S_A,S_B$ be the distinct group survival functions, and let $H_{\ast }$ denote the common pooled cumulative hazard function used for the residuals. At a point where its inverse is defined, residual survival in group $g$ isThus the group residual curves generally differ. On intervals where $S_A(t)>S_B(t)$, the same common increasing transformation gives $S_{u,A}(u)>S_{u,B}(u)$: the better-surviving group has more large residual failure times. If the original survival curves cross, the residual curves can also cross, so no fixed ordering or proportional hazards assumption is implied.

Distinct group distributions generally produce distinct residual survival curves, even when the pooled residual curve looks exponential. Separate correctly fitted group hazards would restore the unit-exponential target within each group, but that is a different transformation from the common fit under discussion.

For the population identity, write $X=\min (T,C)$, $V=1_{\{T\le C\}}$, and $G(t)=\mathbb{P}(C\ge t)$. Assume independent censoring, with $f_T(t)=h(t)S(t)$. Thenwhile the Tonelli theorem and the cumulative hazard function giveThis is mean accumulated hazard under independent censoring, proving $\mathbb{E}[V-H(X)]=0$. For the fitted martingale residual $Y=V-\hat{H}(X)$,Hence its expectation is approximately zero when the fitted hazard error at the observed time is small in mean. This spells out the required sense of a good estimate; pointwise consistency alone does not control a divergent tail. In particular, a Kaplan–Meier estimator with zero final survival gives infinite transformed times there and cannot supply finite residuals without a tail restriction. Neither exact finite-sample zero mean nor validity under informative censoring is being assumed.

To check an omitted explanatory variable $Z$, plot the martingale residuals against $Z$ or compare groupwise mean residuals, using a smooth trend or appropriate uncertainty intervals. Under a correct conditional survival model and conditionally independent censoring, $\mathbb{E}[Y\mid Z]$ should be approximately zero. A systematic positive trend indicates more observed failures than the fitted accumulated hazard predicts; a negative trend indicates fewer. Such patterns suggest adding $Z$, a nonlinear term or an interaction and reassessing the fit. The bounded-above, often long negative tail of martingale residuals means that the diagnostic is a mean-pattern check, not a requirement for symmetric Gaussian residuals.

For a continuous nonnegative event time with hazard function $h$, the cumulative hazard function isIts derivative is the instantaneous hazard wherever the derivative exists.

Writing $F$ for the survival function here, the relation $F^{\prime }(t)=-h(t)F(t)$ with $F(0)=1$ givesThis notation uses $F$ for survival rather than for the cumulative distribution function.

The continuous failure-time probability density function is the negative derivative of the survival function:The minus sign is essential because survivor probability decreases with time.

Assume independent individuals and independent censoring conditional on their explanatory variables, with the censoring mechanism carrying no parameter $\theta$. A failure contributes its failure-time probability density function $f_i(x_i;\theta )$, while a censored observation contributes the survival function $F_i(x_i;\theta )$ because its failure time exceeds $x_i$. Terms from the censoring distribution then factor out of the survival-parameter likelihood. Thus the survival likelihood and its logarithm areThe individual subscripts allow the same parameter vector to act through different covariate values.

Insert $f_i=h_i{F}_i$ and $F_i=e^{-H_i}$ into the survival likelihood. Each individual's log contribution simplifies toThereforeFailures contribute a log hazard function term as well as exposure, whereas both failures and censored subjects contribute the negative cumulative hazard function over their observed follow-up.

For individual $i$, let ${\pi }_i={\pi }_i(\psi )$, $q_i(t)$ be the known baseline hazard function, and $r_i(t;\varphi )$ the extra hazard. PutThe class-specific survival functions are $e^{-Q_i(t)}$ and $e^{-Q_i(t)-R_i(t)}$. Consequently the unconditional survivor and failure density areMixture probabilities must be summed before taking a logarithm; a weighted average of class log-likelihoods would be a different, complete-data calculation. Substitution into the right-censored survival likelihood givesEquivalently, this additive excess-hazard mixture likelihood has contributionHere $q_i$ and $q_i+r_i$ must be nonnegative hazards. The factor $-Q_i(x_i)$ can be dropped when maximizing over $\varphi ,\psi$ because $q_i$ is known. Factoring out censoring presumes a common parameter-free censoring mechanism independent of failure and latent class, conditional on the observed covariates; unmodelled class-specific censoring would change these mixture contributions.

Treat each release as an independent Bernoulli observation with a two-week event probability, giving two independent binomial distributions. Under the specified effect, the probabilities are $p_0=0.0016$ and $p_1=0.0008$ with $n=15000$ in each period. The expected event counts are therefore 24 and 12. For a conventional two-sided 5% comparison of proportions, let $D={\hat{p}}_0-{\hat{p}}_1$, $\delta =p_0-p_1=0.0008$ and $\bar{p}=(p_0+p_1)/2=0.0012$. UseThe approximate null rejection region is $|D|>1.96s_0$, and under the alternative $D$ is approximately $N(\delta ,s_1^2)$. Thus the power of a two-sample rare-event comparison isHere $\Phi$ is the standard normal distribution function. Hence the normal approximation gives $\boxed{\text{two-sided power approximately }52\%}$: detecting the proposed halving is far from assured.

For the small counts, a discrete calculation is also appropriate. Approximate the two counts by independent Poisson distributions with means 24 and 12. Conditional on their total $K$, the pre-intervention count is $\mathrm{Binomial}(K,1/2)$ under equal rates and $\mathrm{Binomial}(K,2/3)$ under the halving alternative; $K\sim \mathrm{Poisson}(36)$ under that alternative. If ${\mathcal{R}}_k$ is the rejection set of the symmetric exact two-sided 5% binomial test, its power isThis conservative discrete test has about 45% power, lower than the normal approximation because its achieved level can be below 5%. If a decrease-only one-sided 5% test had been specified in advance, replace 1.96 by 1.645; the normal-approximation power is about 64%. The tail convention and the chosen test must accompany any quoted statistical power.

Use a concurrent randomized comparison of the new leaflet with the existing release information. Random allocation makes the groups comparable in expectation and avoids attributing a secular change in drug supply, release populations or post-release services to the leaflet. Both arms should use the same outcome ascertainment and two-week follow-up. This improves the causal interpretation over a historical before-and-after comparison; adequate sample size remains necessary given the low statistical power calculated above. If contamination prevents individual allocation, a carefully designed cluster randomized controlled trial is an alternative, with its sample-size calculation allowing for within-cluster dependence.

The one-day slaughter-inspection yield isor 500 detections per million inspected adult cattle. The routine field detections correspond, using a stable herd size of five million, toper adult-cattle day, or 8 detections per million adult-cattle days. Therefore the numerical one-day comparison isThe annual field rate is $14600/(5\times 10^6)=0.00292$ per cattle-year; comparing the inspection probability directly with this annual proportion would mix observation periods. Applying the random inspection-sample proportion to all 150000 slaughtered adults would predict about 75 detected cases in that slaughter cohort. These arithmetic comparisons describe detection yields, not a 62.5-fold biological disease-risk difference: a cross-sectional inspection estimates a prevalence yield, whereas daily field reporting is an ascertainment rate over animal-time.

Two reasons are selection bias and unequal ascertainment.

A beta distribution is a convenient prior for a probability $p\in (0,1)$ and is conjugate to the binomial distribution of the death count. Match its mean $m=0.05$ and variance $v=0.02^2=0.0004$. For $p\sim \mathrm{Beta}(\alpha ,\beta )$,The moment matching for a beta prior formula givesConditional on $p$, use $D\sim \mathrm{Binomial}(90,p)$ and the observed count $D=9$. Multiplying its likelihood $p^9(1-p{)}^{81}$ by the prior density gives the Bayesian posterior through Beta-binomial conjugacy:Under squared-error loss, take its posterior mean as the point estimate. A 95% equal-tail credible interval is given by its 0.025 and 0.975 quantiles. Numerically,Matching two moments does not uniquely determine a prior distribution; the beta family is a justified convenient choice, not a conclusion forced by those moments. Its transfer to this hospital presumes that the historical between-hospital distribution is relevant to the present risk and case mix.

The posterior mean is a weighted average of the sample proportion and prior mean:Thus it is pulled from 10% toward the prior's 5%, giving about 7.17%. This is an affine shrinkage estimator for a binomial proportion. The prior carries substantial concentration relative to the 90 observations, so the posterior is also more precise under this model: its standard deviation is about $0.01785$, compared with the binomial plug-in standard error $\sqrt{0.1(0.9)/90}=0.03162$. The posterior 95% credible interval $[0.04076,0.11035]$ is narrower and shifted down compared with the simple Wald interval $[0.03802,0.16198]$ based on $9/90$ alone.

The point estimate shrinks toward the historical mean, and the interval is narrower under the informative prior. A credible interval and a frequentist confidence interval have different probability interpretations; the comparison does not make the prior's relevance automatic.

For any prior density $\pi (p)$, the Bayes estimate under squared-error loss is the posterior mean:For the matched beta distribution, the integrals are beta integrals and their ratio isThe posterior mean minimizes posterior expected squared error because $\mathbb{E}[(p-a{)}^2\mid D]=\mathrm{Var}(p\mid D)+(a-\mathbb{E}[p\mid D]{)}^2$. The phrase Bayes point estimate depends on the loss: under absolute-error loss it would be a posterior median, while the posterior mode is a different summary and here equals $(14.8875-1)/(207.75-2)\approx 0.06750$.

Exchangeability means that the joint prior law of the hospital risks is unchanged by relabelling hospitals; it permits different realized $p_j$. We should not assume unconditional exchangeability when known hospital characteristics predict systematic risk differences. Examples include severity and age of patients, elective versus emergency operations, procedure mix, specialist referral status, or a relationship between surgical volume and risk. Mixing such groups in one unconditional prior confounds these differences with unexplained hospital variation.

Use covariates or clinically comparable strata first; only remaining hospital effects might reasonably be exchangeable random variables. This is conditional exchangeability of hospital risks. Different sample sizes alone do not refute exchangeability of the underlying risks, although a known association between size and risk would matter. Nor does variation in the observed death proportions refute exchangeability: sampling variability and genuine random heterogeneity are both allowed.

Use a hierarchical Bayesian model with binomial observation errors and a common distribution for the underlying risks:Give $(\alpha ,\beta )$ a shared hyperprior, or parameterize them by mean $m=\alpha /(\alpha +\beta )$ and concentration $\kappa =\alpha +\beta$ and put priors on $m\in (0,1)$ and $\kappa >0$. Conditional on these hyperparameters, the risks are independent and identically distributed; after integrating them out they are exchangeable random variables with shared uncertainty. The data estimate the overall risk level and between-hospital variation. Conditional Beta-binomial conjugacy givesand full Bayes point estimates under squared-error loss areAn Empirical Bayes method instead estimates the hyperparameters from the marginal beta-binomial distribution likelihoodand inserts their fitted values in the conditional posterior means. Both approaches provide partial pooling, stronger for smaller hospitals. Full Bayes also propagates hyperparameter uncertainty into the credible intervals. If systematic covariates are needed, one can place exchangeable residual effects in a logistic regression model rather than assuming a common unconditional risk distribution.

Let $I,J$ denote the two ordered alleles at the genetic locus. Under Hardy-Weinberg equilibrium they are independent with probabilities ${\pi }_i,{\pi }_j$. Write the multiplicative penetrance as $\mathbb{P}(D\mid I=i,J=j)=k{\psi }_i{\psi }_j$, with parameters chosen so these are valid probabilities, and set $Z={\sum }_u{\pi }_u{\psi }_u$. ThenThe Bayes' theorem givesThus multiplicative penetrance preserves Hardy-Weinberg equilibrium: the affected subjects still have two independent allele draws, with tilted allele frequencies. The displayed product uses ordered allele slots. For the usual unordered genotypes, the corresponding probabilities areThe factor of two is required for heterozygotes and must not be dropped when translating the ordered notation into genotype counts.

For genetic association analysis, the case genotype probabilities factor into allele frequencies, so under this model an allelic comparison with representative population controls is appropriate; separate dominance departures are not required by the penetrance model. The ratio of case allele frequency to population allele frequency is proportional to ${\psi }_i$, and relative risk multipliers satisfy ${\psi }_i/{\psi }_j=({\pi }_i^{\ast }/{\pi }_i)/({\pi }_j^{\ast }/{\pi }_j)$. Population controls are important to this exact statement. Among specifically unaffected controls,which generally does not factor. For a rare disease it is close to the population genotype law, but no rare-disease assumption was needed for the case factorization itself. Population stratification and sampling dependence must also be addressed before using an ordinary allelic chi-squared test.

Label each parent's two allele copies by its realized transmission: let $(A,B)$ be the transmitted and untransmitted alleles of the first parent and $(C,D^{\prime })$ those of the second. Under random mating and Hardy-Weinberg equilibrium, the four parental allele draws are independent with population frequencies $\pi$. Random Mendelian segregation merely swaps the two independent copies within each parent, so the transmitted/untransmitted labelled pair still has probability ${\pi }_a{\pi }_b$; the second pair has probability ${\pi }_c{\pi }_d$.

The child receives $A,C$. If $\mathcal{D}$ is its disease event, its penetrance is $k{\psi }_A{\psi }_C$ and its overall disease probability is $k{Z}^2$. ThereforeThis proves the transmitted and untransmitted alleles under multiplicative penetrance factorization. Under these assumptions, the transmitted pair $a/c$ has the affected-case distribution, while the complementary pair $b/d$ has the population distribution and is independent of the transmitted pair. It is the family pseudo-control genotype.

As in part (a), the formula records labelled transmissions. When observed genotypes are unordered, sum over every compatible parental-origin transmission. This includes multiplicities for overlapping parental alleles or two heterozygous parents producing a heterozygous child; it avoids interpreting one product as the total probability of all observationally identical configurations. The homogeneous random-mating model is also essential to the unconditional independence: mixing ancestry strata can correlate the case and pseudo-control through their shared family background.

The usual transmission disequilibrium test conditions on the observed parental genotypes, rather than modelling their population probabilities. For a heterozygous parent carrying $i,j$, the two transmission outcomes are equally likely under no association. With multiplicative penetrance, ascertainment through an affected child weights them by ${\psi }_i,{\psi }_j$, soThe other parent's penetrance factor cancels. This is a special case of disease-ascertained transmission probability. Under the null ${\psi }_i={\psi }_j$, the conditional probability is $1/2$. Count the discordant transmitted/untransmitted pairs from heterozygous parents and test their balance, using McNemar's test or its exact conditional binomial form. Homozygous parental transmissions do not distinguish the alternatives.

This conditional analysis removes the nuisance population allele frequencies and retains the family matching. It is therefore robust to population stratification, which can otherwise make allele frequencies differ between cases and controls without a within-family transmission effect. It does not require the unconditional Hardy-Weinberg equilibrium and random-mating assumptions used for the population pseudo-control factorization. Independence of suitably sampled families, valid genotypes and the Mendelian null transmission model still matter.

A case chromosome is transmitted and a pseudo-control chromosome is untransmitted. Summing the paired table by columns for cases and by rows for controls givesFor the ordinary unpaired chi-squared test of independence, the expected cells are 30,30,70,70. Its uncorrected Pearson statistic isFor the paired McNemar's test, only the discordant transmissions matter, with counts 24 and 18. ThusBoth uncorrected statistics happen to coincide and have the same approximate one-degree-of-freedom chi-squared distribution reference, giving $p\approx 0.355$. The data do not show persuasive transmission imbalance at the conventional 5% level.

The equality is a numerical coincidence, not a justification for ignoring pairing. To see the equality criterion for paired and unpaired allele tests, write the original matched cells as $a,b,c,d$ and $n=a+b+c+d$. The unpaired marginal-table statistic iswhereas McNemar's is $(b-c{)}^2/(b+c)$. Here $n=100$, $b+c=42$, and $(2a+b+c)(2d+b+c)=60\times 140=2n(b+c)$, giving equality. Changing concordant counts can change the unpaired statistic without changing McNemar's. The transmission disequilibrium test should retain its conditional paired interpretation.

If a continuity correction is used, state it explicitly: McNemar's corrected statistic is $(|24-18|-1{)}^2/42=25/42\approx 0.5952$; the corresponding Yates-corrected marginal statistic also coincides here. An exact conditional test uses $\mathrm{Binomial}(42,1/2)$ and gives the two-sided probability $2\mathbb{P}(K\le 18)\approx 0.4408$.

The recombination fraction is the probability that a transmitted gamete is recombinant between the two loci in the ordinary two-locus genetic linkage model. Its conventional range isZero corresponds to complete linkage and $1/2$ to independent assortment. Multiple physical crossovers can restore parental phase, so the recombination fraction is not the expected number of crossovers or an unrestricted crossover probability.

The individuals' haplotypes are not identifiable from the supplied source: the pedigree drawings are missing. The original PDF has a blank figure area, and the accompanying TeX contains no family genotypes. Even the two-locus genotypes of individuals 3 and 4 are not stated elsewhere, so numerical allele pairs cannot be supplied honestly.

The method, once the drawings are available, is to use parental or grandparental genotypes to assign the allele inherited at each locus on the same chromosome. A double heterozygote with alleles $A_1,A_2$ and $B_1,B_2$ has the two candidate haplotype pairs $A_1{B}_1/A_2{B}_2$ and $A_1{B}_2/A_2{B}_1$; ancestor and offspring transmissions determine or restrict its phase. These symbols describe the general phasing problem, not invented genotypes for the missing individuals. If pedigree transmissions leave more than one phase possible, retain all compatible phases rather than selecting one without evidence.

For known parental haplotypes, count $r_1$ recombinant transmissions among $m_1$ informative meioses. The conditional pedigree likelihood has the formwhere $C_1$ contains parameter-independent gamete and genotype factors. Differentiating its logarithm gives $r_1/\theta -(m_1-r_1)/(1-\theta )$, so the constrained maximum-likelihood estimate isWhen $r_1=0$ the maximum is at zero; if no meiosis is informative, the likelihood is constant and the parameter is not estimable from that family. This formula presumes the family phase has been resolved; otherwise use phase averaging in a linkage likelihood first.

The family-specific value cannot be calculated because its pedigree, phases and transmission counts are absent from the supplied PDF. The missing data cannot be reconstructed from the requested estimate alone.

Let $s$ index compatible parental phases in family 2, with fixed prior conditional weights ${\pi }_s$. For a given phase, let $r_s$ be the recombinant count and $m_s$ the informative-transmission count, absorbing any fixed genotype factors into $C_s$. The appropriate pedigree likelihood isThis is phase averaging in a linkage likelihood; maximizing over a phase after observing the offspring would not be the same likelihood. In the common special case of two equally likely phases that exchange recombinant and nonrecombinant labels across $m_2$ transmissions, it reduces, up to a constant, toThe constrained maximum-likelihood estimate compares all stationary points and endpoints on $[0,1/2]$.

The actual phase set, weights, counts and numerical maximum for family 2 cannot be determined without its missing pedigree. The special two-phase expression is a conditional example, not an assertion about this unidentified family.

For independent families, multiply their pedigree likelihoods. If family 1 has the known-phase counts above, the combined log-likelihood derivative at an interior point isThus the score equation for a phase-averaged linkage likelihood isFor the general phase sum, let $w_s(\theta )={\pi }_s{C}_s{\theta }^{r_s}(1-\theta {)}^{m_s-r_s}/L_2(\theta )$. Since the weights sum to one, the equivalent interior equation isThe finite likelihood sum is continuous on the compact parameter interval $[0,1/2]$, so a constrained maximum exists. This does not by itself prove an interior score root: complete nonrecombinant data, for example, maximize at zero. To show the requested root lies in the appropriate range for the two particular families, one must insert their transmission counts and evaluate the score or the resulting polynomial at the relevant endpoints.

The source lacks those pedigree counts, so the data-specific equation and its asserted interior-root verification remain undetermined. No universal interior-root claim is substituted for the missing calculation.

The LOD score is the base-ten log likelihood ratio against independent assortment. For the two families,For a known-phase first family this becomeswith endpoint terms interpreted by limits, such as $0\log 0=0$ when the corresponding count is zero. Positive values favour genetic linkage, and the likelihood ratio is $10^{Z_{\max }}$. A conventional large positive threshold such as 3 represents a likelihood ratio of 1000, not automatically a posterior probability or a universal 5% test.

For a test calibrated to this study, use the distribution of the maximized statistic under $\theta =1/2$, for example by simulating transmissions conditional on the same parental genotype and ascertainment scheme. Reject for sufficiently large values with the chosen significance threshold; phase uncertainty and the boundary null prevent assuming an unqualified ordinary interior Wilks reference. The missing pedigrees prevent the numerical value or their explicit likelihood polynomial from being supplied, but the likelihood-ratio definition and testing procedure apply once those data are recovered.