Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 40 (Biostatistics)

For $t\ge 0$ and $\theta >0$, the exponential distribution has probability density function $f(t)=\theta e^{-\theta t}$. Integrating the probability density function gives the survivor function $S(t)=P(T>t)=e^{-\theta t}$. The hazard function divides instantaneous failure density by the probability of having survived, and the integrated hazard accumulates this rate. ThusThe probability density function is zero for $t<0$, and the survivor function is one there.

Write $X=\min (T,c)$ and $\Delta =1_{\{T\le c\}}$ for the right-censored survival time and observed-failure indicator. Since the exponential distribution is continuous, including or excluding equality at $c$ makes no difference. Using the tail-integral expression for an expectation,Consequently the expected integrated hazard at the observed endpoint satisfiesAdministrative censoring therefore stops both the accumulated exposure and the opportunity to record a failure.

For individual $i$, let $X_i=\min (T_i,c_i)$ and ${\Delta }_i=1_{\{T_i\le c_i\}}$. The preceding expectation calculation applies separately at each fixed censoring time:Summing and using linearity of expectation, with $D={\sum }_i{\Delta }_i$ the failure count, givesThis identity needs the stated marginal exponential distributions, but not independence between individuals; independence will be needed for the product likelihood.

Replace the expected total person-time and failure count in the preceding identity by their observed values, $E={\sum }_i{x}_i$ and $d$. The resulting estimating equation is $\theta E=d$, soThis events divided by exposure estimator is failures per unit observed person-time. We assume positive total exposure; if $d=0$, it gives the boundary estimate zero.

For independent individuals under fixed administrative censoring, an observed failure contributes the probability density function $\theta e^{-\theta x_i}$, while a censored observation contributes the survivor function $e^{-\theta x_i}$. Hence the survival likelihood and log-likelihood areFor $d>0$, the score function ${\ell }^{\prime }(\theta )=d/\theta -E$ vanishes at $d/E$, and ${\ell }^{\prime \prime }(\theta )=-d/{\theta }^2<0$. This is the unique maximum over $\theta >0$:Thus maximum likelihood estimation agrees with the observed-exposure estimating equation. If there are no failures, the likelihood decreases with $\theta$: its supremum is approached as $\theta \downarrow 0$, or attained at zero if that boundary is admitted.

Differentiate the log-likelihood again. For $d>0$, the positive interior maximum-likelihood estimate givesThe observed information is $-{\ell }^{\prime \prime }(\hat{\theta })=d/{\hat{\theta }}^2$, giving the usual local standard error approximation $\hat{\theta }/\sqrt{d}$. For $d=0$ there is no positive interior maximum, so the displayed substitution and this standard error approximation do not apply; at every positive $\theta$ the second derivative is zero.

Let $N(t)$ count observed failures and let $Y(t)$ be the at-risk process, counting subjects under observation and still alive immediately before $t$. Under independent censoring, a short interval of length $dt$ has conditional expected failure count $Y(t)h(t)dt=Y(t)dH(t)$. The estimating equation obtained by replacing the count expectation by its observation is $d\hat{H}(t)=dN(t)/Y(t)$. Each distinct failure gives $\Delta N(a_m)=1$ and $Y(a_m)=r_m$, so the Nelson–Aalen estimator isThe same increment maximizes the local working likelihood for a Poisson distribution with mean $r_m\Delta H_m$: differentiating $\Delta N_m\log (\Delta H_m)-r_m\Delta H_m$ gives $\Delta H_m=\Delta N_m/r_m$. The counting-process derivation explains why the risk set, rather than the initial sample size, supplies the denominator.

Collect distinct observed failure times as $a_m$, with $d_m$ failures and $r_m$ individuals in the risk set just before $a_m$. The grouped Nelson–Aalen estimator isAll failures recorded at one time use the same pre-event risk set. When failures and censoring share a recorded time, the usual convention includes those censored at that time in the risk set and processes failures before removing censored individuals. A known different ordering should instead be honored. If ties arise solely through rounding of a continuous-time process, an alternative is to resolve the unobserved ordering: $d_m$ successive failures without intervening censoring would contribute ${\sum }_{s=0}^{d_m-1}(r_m-s{)}^{-1}$. That is not the grouped jump $d_m/r_m$; the difference represents a choice about what the recorded ties mean.

At a distinct event time $a_m$, estimate the conditional probability of surviving that time by $1-d_m/r_m$. Multiplication of these conditional survival probabilities gives the Kaplan–Meier estimator, and taking minus its logarithm gives an integrated hazard estimate:The conditional failure count has a binomial likelihood whose maximizing event fraction is $d_m/r_m$. Tied failures therefore enter directly as one count. Even ordering $d_m$ failures arbitrarily, with no intervening censoring, produces the same survivor function factor becauseThe failure-versus-censoring ordering convention still matters when both have the same recorded time. If $d_m=r_m$, the survivor function becomes zero and the logarithmic integrated hazard is infinite.

Put $u_m=d_m/r_m$. The Taylor series of the logarithm gives the small-jump comparison of cumulative hazard estimators:Consequently, if ${\max }_m{u}_m\le \epsilon <1$, the two integrated hazard estimates obeyThey are close when every event fraction is small. In particular, with distinct single failures and large risk sets, $u_m=1/r_m$ is small. With tied data, large risk sets alone are insufficient if a substantial fraction of the risk set fails together.

At the last observed failure time $a_{\ast }$, any individual still in the risk set must either fail there or be censored there or later. If there are no such censored observations, every remaining individual fails, so $d_{\ast }=r_{\ast }$. Thus the final Kaplan–Meier estimator factor is zero, while the final Nelson–Aalen estimator jump is one:There is no small-jump approximation at this endpoint. Without ties the final risk set has size one, so the same discrepancy follows from its last single failure.

Use the binomial distribution for independent outcomes in each nonoverlapping batch, and let the CUSUM accumulate the log-likelihood ratio favoring a specified change. The two numbered calculations below treat the batch score and the initial-response design separately.

Let $p_1$ be the alternative death probability. The odds ratio specifiesAssume $0<p_0<1$ and $k>0$. For observed batch count $y_t$, the common combinatorial factor in the two binomial likelihoods cancels, giving the CUSUM incrementThereforeFor $k>1$ large death counts add positive evidence of deterioration; for $k<1$ unusually small death counts add evidence for improvement. The score has nonpositive expectation under the null and nonnegative expectation under the specified alternative, as follows from the nonnegativity of Kullback-Leibler divergence. Independent, nonoverlapping batches avoid counting the same patient's outcome repeatedly.

A classical sequential probability ratio test addresses a fixed-start choice between two simple statistical hypotheses. It sums log-likelihood ratio increments until an upper boundary accepts the alternative or a lower boundary accepts the null; it then terminates that test. A CUSUM instead monitors indefinitely for a change at an unknown time. Its reset at zero discards sustained evidence from a previously satisfactory period, so a late change need not first undo all earlier negative scores. An upper boundary $h$ signals when the accumulated change evidence is sufficiently large.

A fast initial response CUSUM takes $0<X_0<h$, often $X_0=h/2$, in place of a zero start. Until its first reset or signal, it acts like a fixed-start sequential probability ratio test with a lower boundary at $-X_0$ and an upper boundary at $h-X_0$ for the subsequent cumulative increments. Upon reaching the lower boundary it resets and continues as an ordinary CUSUM. The head start gives faster detection if a change is already present when monitoring begins or restarts. It is useful after an intervention, a shutdown or a previous alarm when immediate renewed surveillance is desired. Its initial false-alarm behavior differs from the zero-start scheme, so the head start and signalling boundary must be calibrated together; the preceding numerical choice is a design convention, not a universal optimum.

Four distinct objections, each with a corresponding improvement, are as follows.

A confidence interval containing the national rate does not demonstrate equivalence, and a raw ranking does not establish a causal performance difference. The improved presentation should convey precision and uncertainty as well as the estimated outcome.

Use individual stratified randomization within each prison, with concealed centrally generated permuted-block randomization and varying block sizes. Prison can affect both baseline risk and intervention delivery, so balancing assignments within prison protects against imbalance in that important covariate. Varying block sizes and allocation concealment prevent staff from predicting the next allocation and selecting participants accordingly. Prespecify how incomplete final blocks are treated. To enforce the overall 4000:4000 allocation exactly, coordinate the stratum totals centrally, allocating odd final places through a concealed random rule; exact half allocation within every prison is impossible if some prison totals are odd.

Individual randomization stratified by prison is preferable to randomizing only twenty whole prisons for this design. A cluster-randomized trial would have few independent units and could lose substantial precision through intraclass correlation. Sharing or selling the intervention can cause treatment contamination, so record it and retain an intention-to-treat analysis; substantial spillover might justify a separately powered cluster design rather than being ignored.

Use a conventional two-sided test at significance level $0.05$. Let $n=4000$ per arm, $p_C=0.005$, $p_T=0.0025$ and $\Delta =p_C-p_T=0.0025$. Under the specified alternative, the normal approximation to the difference of independent binomial proportions has standard deviationFor planning the pooled null standard error, use $\bar{p}=(p_C+p_T)/2=0.00375$, giving ${\sigma }_0=\sqrt{2\bar{p}(1-\bar{p})/n}=0.00136674$. The rejection threshold is approximately $1.96{\sigma }_0=0.00267880$, exceeding the mean alternative difference. With $\Phi$ the standard normal distribution function, the approximate statistical power isNo: the usual two-sided 5% calculation gives about 45% power, below 50%. The expected death counts are only 20 and 10, so an exact discrete design calculation could refine this approximation.

The question does not specify the tail convention. If a beneficial direction is prespecified and a one-sided test at 5% is justified, replace $1.96$ by approximately $1.645$; the resulting statistical power is about $0.573$, and the answer is then yes. This change of testing convention must be declared.

Assume individuals who immediately sell the intervention retain the untreated risk $0.005$, while those who keep it have the posited halved risk $0.0025$. The law of total expectation gives the expected count per thousand assigned to treatment:The assignment risk is therefore $0.003$. This is an intention-to-treat mixture under the stated adherence-specific risks, not an assertion that receipt or sale is randomized. It also assumes that selling does not itself change the seller's baseline risk and that effects on recipients outside this group do not alter this calculation.

For equal allocation, let $n$ be the required sample size per arm, $p_C=0.0045$, $p_T=0.003$ and $\Delta =0.0015$. In the standard sample size for comparing two proportions calculation, a two-sided test at 5% with 80% statistical power requiresUsing $z_{0.975}=1.95996$ and $z_{0.8}=0.84162$ gives $n\simeq 26063.64$, which is rounded upward for actual recruitment. ThusSince the risks are small, dropping the $(1-p)$ corrections and using $1.96+0.84$ gives $N_{\mathrm{total}}\simeq 2(2.8{)}^2(0.0075)/(0.0015{)}^2=52266.7$, conventionally about 52.3 thousand. These are slightly different approximations to the same planning requirement. Add an allowance if outcome ascertainment is incomplete or the allocation is clustered. A prespecified one-sided test would need a different, smaller sample size; the calculation here uses the conventional two-sided interpretation consistently with part (b).

Two useful questions separate intervention retention from actual opportunities to use it.

Responses can explain implementation, but cannot replace the randomized intention-to-treat analysis. Reincarcerated survivors are a selected subgroup, and recall errors and missing follow-up may affect their reports.

Genetic linkage analysis asks whether a genetic marker and a disease genetic locus cosegregate within families more often than would be expected under independent assortment. It uses pedigrees and Mendelian segregation to estimate the recombination fraction $\theta$, with evidence for linkage arising when the data favor $\theta <1/2$. Even when founders are in linkage equilibrium, linked loci can show familial cosegregation.

Genetic association analysis instead asks whether observed allele or genotype frequencies differ with disease status in a sampled population, often comparing cases with controls. It can detect a causal variant or a correlated genetic marker through linkage disequilibrium; population structure and selection can also create association. Linkage is about familial transmission; association is about population dependence. Neither an association alone nor a low family recombination estimate establishes that the marker itself causes disease.

Distinguish the disease alleles $D$ (normal) and $d$ (recessive disease) from the marker alleles $A,a$. The affected mother and two affected children are $dd$. The unaffected father must be $Dd$, since he has affected children; the unaffected child is $Dd$, since the mother always transmits $d$. The mother is $AA$ at the marker, so the marker genotypes identify the father's transmissions to the three children as $a,A,A$. Their disease transmissions from the father are $d,d,D$.

The father's unphased double heterozygosity has two possible pairs of haplotypes:Under $s_1$, the three paternal transmissions $(ad,Ad,AD)$ are respectively nonrecombinant, recombinant and nonrecombinant. Under $s_2$, they are recombinant, nonrecombinant and recombinant. Each particular gamete has probability $(1-\theta )/2$ or $\theta /2$, giving the conditional pedigree likelihoodsAssuming linkage equilibrium among founder haplotypes, the two phases have equal prior weights conditional on the observed paternal genotype. Phase averaging in a linkage likelihood therefore givesAny factors for the parental genotypes omitted by this conditioning are constant in $\theta$. The derivative is $(1-2\theta )/16$, so the maximum likelihood estimate is $\hat{\theta }=1/2$ and there is no evidence for linkage: the maximum LOD score is zero. Choosing the best phase after observing the offspring would give a different and inappropriate calculation; the unknown phase must be averaged. The linkage-equilibrium assumption is modified explicitly in part (v).

Arrange rows by marker genotype, with cases and controls as the two columns. Both column totals are 100, and the three row totals are 100, 60 and 40. Under the null of no genetic association, estimate each common row probability from its pooled proportion. The expected count is row total times column total divided by the grand total. Thus the expected contingency table isThe Pearson chi-squared test of independence usesUnder independent sampled individuals and the no-association null, its reference chi-squared distribution has $(3-1)(2-1)=2$ degrees of freedom. All expected cells are well above five. The null is overwhelmingly rejected: the approximate p-value is $e^{-93.44/2}\simeq 5.13\times 10^{-21}$. This establishes association in the sampling model, not by itself a causal role for the marker.

Each diploid individual contributes two marker alleles. Count $A$ copies as twice the $AA$ count plus the $Aa$ count, and count $a$ copies analogously. The observed chromosome-level contingency table, followed by its null expected table, isThere are 200 chromosomes in each group; pooled totals are 140 $A$ and 260 $a$. The usual allelic Pearson chi-squared test of independence givesWith the ordinary independent-chromosome null model, use one degree of freedom; its p-value is about $1.04\times 10^{-25}$, providing very strong evidence of genetic association.

The two alleles from the same person are not automatically two independent observations. The usual one-degree-of-freedom calibration is justified, for example, for unrelated individuals with Hardy-Weinberg equilibrium under the null. If that assumption is not suitable, retain individuals as the sampling units and test their allele dosages using an appropriate variance or permutation of case-control labels, or use the genotype-level test from part (iii). No Hardy-Weinberg equilibrium assumption was needed for that genotype-level test.

Using cases as disease haplotypes and controls as control haplotypes in the model requested here, the chromosome counts estimateThese are conditional marker allele frequencies, not the reverse conditional probabilities $P(d\mid A)$. Under independent founder chromosomes, conditional on the father's $Dd,Aa$ genotype, the phase $AD/ad$ has weight proportional to $0.60\cdot 0.90=0.54$, and $Ad/aD$ has weight proportional to $0.10\cdot 0.40=0.04$. Their normalized weights are $27/29$ and $2/29$. This conditions on the model's observed parental genotypes; the omitted disease-allele frequencies cancel in the phase-weight ratio.

Using the two phase-specific pedigree likelihoods already derived,The normalizing factor is independent of $\theta$, so maximize $g(\theta )=0.54\theta -1.04{\theta }^2+0.50{\theta }^3$. Its derivative gives exactlyOnly the smaller root lies in $[0,1/2]$:On this interval $g^{\prime \prime }(\theta )=-2.08+3\theta <0$, so this stationary point is the unique maximum. The maximum LOD score, retaining the same linkage disequilibrium frequencies in numerator and denominator, isThis is only a likelihood ratio of about $1.145$ against $\theta =1/2$, hence very weak evidence. The population frequencies are treated as fixed plug-in estimates, as requested; uncertainty in those estimates and possible disease-model differences between controls and normal-allele chromosomes would matter in a fuller analysis.

The kinship coefficient ${\varphi }_{XY}$ is the probability that one allele copy sampled uniformly from each of individuals $X,Y$, independently conditional on their copies, is identical by descent. The inbreeding coefficient $F_X$ is the probability that the two homologous allele copies within $X$ are identical by descent. For distinct offspring and older individual, Mendelian segregation givesFor the last identity, two independent selections choose the same copy with probability $1/2$, and choose distinct copies with probability $1/2$, in which case identity has probability $F_X$. An unrelated, noninbred founder has $F=0$ and self-kinship $1/2$, while distinct such founders have zero mutual kinship coefficient. The source's statement of zero founder kinship must be read as zero kinship between distinct founders; zero self-kinship would contradict the definition.

To identify the unlabelled individuals in the original pedigree, call the three children of $A,B$ from left to right $U,V,W$. Then $G$ has parents $C,U$, $D$ has parents $V,W$, and $E,F$ have parents $G,D$. The three full siblings $U,V,W$ are noninbred because $A,B$ are unrelated noninbred founders in a pedigree. Their pairwise kinship coefficients areSince $D$ is the child of the sibling pair $V,W$, its inbreeding coefficient is

The founder $C$ is unrelated to the descendants of $A,B$, while the kinship coefficient between $D$ and its parents' sibling $U$ isAs $G$ has parents $C,U$, the kinship coefficient between $D,G$ isThe parents of $E$ are $D,G$, soThe same calculation gives $F_F=1/8$.

The self-kinship coefficient of the inbred individual $D$ is ${\varphi }_{DD}=(1+F_D)/2=5/8$. Since $E$ receives a uniformly chosen parental allele from each of $D,G$,The ${\varphi }_{DD}$ term is essential: self-kinship coefficient is not zero, even for a noninbred individual.

The full siblings $E,F$ have parents $G,D$. Since $G$ is noninbred, ${\varphi }_{GG}=1/2$, whereas ${\varphi }_{DD}=5/8$ and ${\varphi }_{GD}=1/8$. Applying the parental kinship coefficient recursion to both siblings givesThe familiar outbred-sibling value $1/4$ does not apply to this inbred pedigree.

For the usual no-linkage reference model with unrelated, noninbred parents, define $J$ as the number of homologous allele copies shared identical by descent by two full siblings at the marker. The two siblings receive the same paternal copy with probability $1/2$, and independently receive the same maternal copy with probability $1/2$. Therefore Mendelian IBD sharing of full siblings givesHenceFor affected siblings, absence of genetic linkage makes the marker transmissions independent of selection by disease, so these are the relevant null moments. This is a fresh outbred reference calculation, not an application of these sharing probabilities to the inbred siblings $E,F$ in part (b). Without the outbred-parent assumption the three ordinary sharing probabilities require modification.

Use the prespecified excess-sharing alternative: affected full siblings linked to a disease genetic locus are expected to share more marker copies identical by descent. The total sharing count is $0(42)+1(98)+2(60)=218$, so $\bar{J}=218/200=1.09$. Under the no-linkage null, independent sibling pairs have $E[J]=1$ and $\mathrm{Var}(J)=1/2$, giving the mean allele-sharing test for affected siblingsThe upper 5% standard normal critical value is approximately $1.64$, so there is evidence of excess sharing, and hence linkage, at the one-sided 5% level. The normal approximation gives upper-tail p-value $1-\Phi (1.8)\simeq 0.0359$.

The conclusion depends on the intended directional test. A two-sided test of the sharing mean uses $1.96$, so it would not reject at 5%. An omnibus Pearson chi-squared goodness-of-fit test of the three counts against $(50,100,50)$ gives $64/50+4/100+100/50=3.32$ with two degrees of freedom and also would not reject. The directional mean-sharing test targets the scientifically specified excess-sharing alternative; these different tests should not be selected after seeing which rejects. Independence of the 200 pairs is assumed: if pairs overlap within families, the standard error needs adjustment.