Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 37 / 5 / ii

The permitted transitions form the following diagram. Each arrow is labelled by its fitted transition intensity in units of inverse years; the absence of an arrow is a structural restriction, not an estimated small effect.

State 1 can move directly to 2, 3 or 5; state 2 can move to 4 or 5; state 3 can move to 4 or 5; and state 4 can move only to 5. There are no recovery transitions, and state 5 is an absorbing state. The rates of leaving the four living states are the sums of their outgoing arrows:The diagonal generator entries are their negatives, so every row of $Q$ sums to zero. These diagonal entries are not arrows or negative event rates. For example the direct $1\to 5$ rate is $0.004$ per year, while death after reaching another state can occur at a much larger rate. The printed fit has no covariates on the intensities, so it estimates one pooled rate system and contains no treatment-effect comparison, despite the trial's treatment randomization. The reported $-2\ell$ describes the fitted likelihood level; by itself it is not a goodness-of-fit test without a reference or competing model.

For a homogeneous continuous-time Markov chain, the holding time in a living state $r$ before the next transition has survival probability $e^{-{\lambda }_{r}t}$, where ${\lambda }_r={\sum }_{s\ne r}{q}_{rs}=-q_{rr}$. Indeed, conditional on still occupying that state, the probability of leaving in the next $dt$ is ${\lambda }_{r}dt+o(dt)$, so its survival function satisfies $S^{\prime }(t)=-{\lambda }_{r}S(t)$ and $S(0)=1$. Integrating gives the mean holding time from a transition intensity matrix:This is the mean time until any allowed exit from that state, not necessarily the patient's remaining lifetime.

For confidence limits, use the supplied standard error $s_r$ of the corresponding diagonal generator estimate. Changing its sign does not change its standard error, so it is also the standard error of ${\hat{\lambda }}_r$. The delta method, with derivative $d(1/\lambda )/d\lambda =-1/{\lambda }^2$, givesThe resulting approximate 95% Wald confidence intervals areThe diagonal standard errors already incorporate the covariance among outgoing rates. They should not be replaced by the square root of a sum of squared off-diagonal standard errors, which would silently assume those rate estimates independent.

Alternatively construct a positive rate confidence interval and apply the confidence interval for an inverse rate transformation. Here every normal rate interval has a positive lower endpoint, givingThese approximate intervals are, respectively, $(10.863,12.830)$, $(4.497,7.161)$, $(2.397,3.117)$ and $(1.548,2.631)$ years. They are asymmetric because inversion is nonlinear; positive log-rate or likelihood-based intervals could also be inverted. The symmetric delta intervals and the inverted-rate intervals are two legitimate first-order approaches and need not have numerically identical endpoints.

The death state has zero exit rate, so its holding time is infinite. There is no finite mean-sojourn estimate or ordinary finite confidence interval for this structurally absorbing state. All intervals above concern the mean parameter, not a prediction interval for one patient's realized holding time.

The transition matrix over five years is $P(5)=e^{5Q}$. Its first row conditions on starting in state 1. The displayed approximate probabilities mean that, five years later, about 65% remain disease-free, 6% have local recurrence only, 10% have distant metastasis only, 2% have both, and 16% have died. They are probabilities of the states occupied at five years; they are not the probabilities of the next jump destinations or annual transition intensities. In particular the positive probability of state 4 does not require a direct $1\to 4$ arrow: the paths $1\to 2\to 4$ and $1\to 3\to 4$ contribute to it. The probability of death includes deaths after any allowed sequence of intermediate states.

The printed row sums to $0.99$ rather than one because the entries are approximate; an exact stochastic transition row sums to one. Exponentiating the displayed three-decimal generator gives approximately $(0.65377,0.06490,0.09775,0.02532,0.15826)$, illustrating that the edited, low-precision table should not be treated as an exact matrix exponential of its separately rounded rates.

State 5 is an absorbing state, so once entered it can never be left. Its generator row is zero, and the last row of every positive-time transition matrix is thereforeThis explains all four leading zeroes in the printed last row.