A frozen-age approximation holds an age covariate constant at the start of each observation interval, allowing homogeneous <transition probabilities> to enter a <panel-observed multi-state likelihood>. It approximates a model with continuously changing age. Reciprocal rate calibration of mean waiting times is exact for a fixed-age <exponential distribution>, but not for a continuously ageing <Gompertz distribution>.

Frozen-age approximation in a multi-state model

Under the frozen-age <exponential distribution> convention, the mean waiting time is the reciprocal of the outgoing <transition intensity>. The beliefs therefore specify
$$
q_{12}(50)=\frac1{10},\quad q_{12}(60)=\frac15,\qquad
q_{23}(50)=\frac12,\quad q_{23}(60)=1.
$$
Both rates double over ten years. Consequently
$$
\boxed{a=0.1\ \text{year}^{-1},\qquad b=0.5\ \text{year}^{-1},\qquad
\beta_{12}=\beta_{23}=\frac{\log2}{10}\ \text{year}^{-1}.}
$$
These 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 is
$$
m(q,\beta)=\int_0^\infty\exp\!\left[-\frac q\beta(e^{\beta u}-1)\right]du
=\frac{e^{q/\beta}}\beta E_1(q/\beta),\qquad
E_1(z)=\int_z^\infty\frac{e^{-v}}v\,dv.
$$
Here $E_1$ is the <exponential integral>. Solving $m(a,\beta_{12})=10$ and $m(ae^{10\beta_{12}},\beta_{12})=5$, and the analogous pair with means 2 and 1, instead gives
$$
(a,\beta_{12})\approx(0.0391767,0.1136031),\qquad
(b,\beta_{23})\approx(0.4328981,0.0763367),
$$
with rates and slopes in inverse years. These are an alternative continuous-age calibration, not the frozen-age answer boxed above.

Solution

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

{c}
{title2=$h(t)=qe^{\beta t}$}
{wiki}

A positive, exponentially increasing <hazard function> $h(t)=qe^{\beta t}$ gives the <survivor function>
$$
S(t)=\exp\!\left[-\frac q\beta(e^{\beta t}-1)\right],\qquad q,\beta>0.
$$
Its mean is $\beta^{-1}e^{q/\beta}E_1(q/\beta)$, where $E_1$ is the <exponential integral>. Unlike the <exponential distribution>, its future mean waiting time is not the reciprocal of its current <hazard function>.

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 Incoming links: Gompertz distribution