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

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.