Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 207 / 3

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.