Calculate state-dependent probabilities for continuous time Markov models.

A disability income contract pays a benefit only while the insured occupies the "disabled" state. To price it you need not just survival, but state-occupancy probabilities at every future moment. Continuous-time Markov machinery delivers them.

Label states $0,1,\dots ,n$. Transition intensities ${\mu }_x^{ij}$ are instantaneous rates from $i$ to $j$. Two probability objects matter:

• ${}_t{p}_x^{ij}$: occupy $j$ at time $t$, starting in $i$, with any path allowed.
• ${}_t{p}_x^{\overline{ii}}$: stay continuously in $i$ from time 0 to $t$ (no exits, no returns).

These coincide only when $i$ cannot be re-entered from any other state.

The chance of leaving state $i$ over $(s,s+ds)$ is the sum of all exit intensities. Therefore:

$${}_t{p}_x^{\overline{ii}}=\exp \left(-{\int }_0^t{\mu }_{x+s}^{i\bullet }ds\right),{\mu }_x^{i\bullet }=\sum _{k\ne i}{\mu }_x^{ik}$$

When intensities are constant, the integral collapses to ${\mu }^{i\bullet }\cdot t$.

KEY: The overline distinguishes sojourn from occupancy. ${}_t{p}_x^{11}$ lets the life leave state 1 and return; ${}_t{p}_x^{\overline{11}}$ does not.

Differentiate by conditioning on the state just before $t+dt$:

$$\frac{d}{dt}{}_t{p}_x^{ij}=\sum _{k\ne j}({}_t{p}_x^{ik}{\mu }_{x+t}^{kj}-{}_t{p}_x^{ij}{\mu }_{x+t}^{jk})$$

Common mistakes

• Confusing ${}_t{p}_x^{ii}$ with ${}_t{p}_x^{\overline{ii}}$. The overline forbids returns. Whenever $i$ is reachable from another state, occupancy strictly exceeds sojourn.
• Forgetting the negative outflow term. Writing only the inflow into $j$ and missing $-{}_t{p}_x^{ij}{\sum }_k{\mu }^{jk}$ produces probabilities that drift above 1.
• Sojourn with the wrong total intensity. Use the sum of every exit from $i$, not just transitions to death. A common slip uses ${\mu }^{02}$ alone instead of ${\mu }^{01}+{\mu }^{02}$.

Bottom line

• Notation: ${}_t{p}_x^{ij}$ is the probability a life in state $i$ at age $x$ is in state $j$ at age $x+t$. The overline ${}_t{p}_x^{\overline{ii}}$ requires staying in $i$ throughout.
• Boundary/identity: ${}_0{p}_x^{ii}=1$ with zero off-diagonal, and ${\sum }_j{}_t{p}_x^{ij}=1$ at every $t$; absorbing states accumulate the departing mass.
• Sojourn formula: ${}_t{p}_x^{\overline{ii}}=\exp \left(-{\int }_0^t{\sum }_{k\ne i}{\mu }_{x+s}^{ik}ds\right)$, summing over every exit from $i$.
• First-transition integral: for simple no-return path structures one integral over the first-transition time gives ${}_t{p}_x^{ij}$ directly, as with permanent disability and ${}_t{p}_x^{01}$.

Exam shortcut

If the disabled state has no recovery arrow, ${}_t{p}_x^{11}={}_t{p}_x^{\overline{11}}$: use the simple exponential and skip the ODE. DECISION: Constant intensities and at most one returnable state → one integral. Time-varying intensities or two-way transitions → Euler march. Makeham/Gompertz force on a sojourn → closed-form using $(c^t-1)/\ln c$. Run the row-sum audit after each Euler step: confirm ${\sum }_j{}_t{p}_x^{ij}=1$. Drift above 1.005 means $h$ is too large.

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Learning objectives

• 1e

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