Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 207 / 1

Write $Z(u)$ for the continuous-time Markov chain and ${\mathbb{E}}_r$ for expected value conditional on $Z(0)=r$. The occupation time of a continuous-time Markov chain in state $s$ is the integral of its indicator random variable. By Tonelli theorem,The interchange is valid because the integrand is nonnegative; indeed the random integral is bounded by $t$.

Let $N(u)$ count all jumps of the continuous-time Markov chain. Its Markov-chain entry count into state $s$ isThusThis is a Lebesgue-Stieltjes integral against a counting process, so each jump contributes one. The initial occupation of state $s$ is not counted as an entry. An ordinary time integral of an indicator random variable asserting a jump at exactly time $u$ would be zero, since individual jump times have zero Lebesgue measure. The requested integral must therefore use this counting-measure interpretation. Assume non-explosion and finite expected jump counts on bounded intervals, as holds for a finite-state continuous-time Markov chain.

For $i\ne s$, the Markov property and the definition of transition intensity giveSumming over the possible pre-jump states gives the density of the expected Markov-chain entry count:For a finite-state continuous-time Markov chain, the chance of two or more jumps in this short interval is $o(du)$. There is no $q_{ss}$ contribution: diagonal entries of the infinitesimal generator encode exit rates, not jumps into the same state. With repeated entries possible, this density is an expected event rate rather than the probability density function of a single normalized random time.

Integrate the expected Markov-chain entry count density and use the occupation time of a continuous-time Markov chain formula:Each term is a transition intensity multiplied by the expected exposure time to that transition. This counts every entry, including returns, rather than only the first visit. For countably many states the same nonnegative interchange follows from Tonelli theorem, provided the resulting expected count is finite. The sum excludes $i=s$.

Use an irreversible continuous-time multi-state model $1\to 2\to 3$, with state 3 absorbing. Let $a=q_{12}(50)$ and $b=q_{23}(50)$, and define log-linear age effects byFor a closed-form panel-observed multi-state likelihood, freeze the age covariate at the start of each observation interval. The transition intensities are then $(a,b)$ over the first interval and $(\lambda ,\mu )=(a{e}^{5{\beta }_{12}},b{e}^{5{\beta }_{23}})$ over the second. Conditional on the initial state, the likelihood contribution isIndeed, $p_{11}(5)=e^{-5a}$ and $p_{13}(1)=1-p_{11}(1)-p_{12}(1)$. In this irreversible continuous-time Markov chain, the intermediate state has probabilityThe second expression is the continuous limit of the first and handles equal rates. The observed two-state jump across an interval is compatible with adjacent instantaneous transitions: an unobserved visit to state 2 occurs between the examinations.

The four-parameter panel-observed multi-state likelihood requires the following additional assumptions.

If age is instead updated continuously, the model is time-inhomogeneous and the displayed homogeneous transition formula is not exact. For example, setThen the exact irreversible likelihood contribution isThus the simple closed form uses an explicit frozen-age approximation; adjacency alone and continuously evolving age do not determine that form.

Under the frozen-age exponential distribution convention, the mean waiting time is the reciprocal of the outgoing transition intensity. The beliefs therefore specifyBoth rates double over ten years. ConsequentlyThese calibrate the age-specific constant-rate means; they are not exact mean ages of future events when age-dependent rates continue increasing throughout the waiting time.

For a literal continuously ageing interpretation, a rate $q{e}^{\beta u}$ gives a Gompertz distribution for the waiting time. For $\beta >0$, its mean isHere $E_1$ is the exponential integral. Solving $m(a,{\beta }_{12})=10$ and $m(a{e}^{10{\beta }_{12}},{\beta }_{12})=5$, and the analogous pair with means 2 and 1, instead giveswith rates and slopes in inverse years. These are an alternative continuous-age calibration, not the frozen-age answer boxed above.