Risk Modelling and Survival Analysis · Transition intensities dependent on age (exact or census)

Multiple State Models and Transition Intensities Explained

Updated 11 October 2026 · Fact-checked

A multiple state model describes a life moving between states, such as healthy, sick and dead, as a Markov jump process. Transition intensities μ_ij(x) give the instantaneous rate of moving from state i to state j. Kolmogorov forward equations turn these intensities into transition probabilities. Solve them by writing one equation per state.

Understand Multiple State Models and Transition Intensities

A multiple state model has a finite set of states, for example Healthy, Sick and Dead. At any time the life is in exactly one state. It can jump from one state to another at any moment. This is a continuous-time, discrete-state process.

The model is Markov. Given the current state, the future does not depend on how the life got there. In the time-inhomogeneous case, the future depends on the current state and the current age or time, not on the earlier path. This is what makes the maths tractable.

A transition intensity μ_ij(x) is the instantaneous rate of moving from state i to state j at age x, for i ≠ j. Formally, μ_ij(x) = lim (h→0+) of h⁻¹ × P(life in j at x+h | life in i at x). For small h, the probability of moving from i to j in the next h years is about h × μ_ij(x), with an error of o(h). The two-state alive-dead model is the special case where μ_01(x) is the force of mortality.

The transition probability _tp_x^ij is the probability of being in state j at age x+t, given state i at age x. It includes paths with several jumps. The Kolmogorov forward equations link these probabilities to the intensities. They look at the very last small interval before time x+t. Probability flows into state j from other states and flows out of j to other states.

In the Healthy-Sick-Dead model, the intensities are usually given as functions of age or as constants. Constant intensities make the model time-homogeneous and the equations easier to solve. In exam questions you often set up the equations, solve one simple case, or use them numerically with a small step.

Key rules to remember

Definition of transition intensity: μ_ij(x) = lim (h→0+) [ _hp_x^ij ÷ h ], for i ≠ j; Equivalent to _hp_x^ij = h × μ_ij(x) + o(h). It is a rate, not a probability, and can exceed 1.
Kolmogorov forward equation (general): d/dt _tp_x^ij = Σ over k ≠ j of [ _tp_x^ik × μ_kj(x+t) − _tp_x^ij × μ_jk(x+t) ]; Inflow to j from every other state k, minus outflow from j to every other state k. Start in i at age x, end in j at age x+t.
Initial conditions: _0p_x^ii = 1 and _0p_x^ij = 0 for i ≠ j; Needed to solve the differential equations.
Total probability: Σ over j of _tp_x^ij = 1; Use it as a check, and to get one probability from the others.
Probability of staying in a state throughout: _tp_x^ii(bar) = exp( −∫₀ᵗ Σ over j ≠ i of μ_ij(x+s) ds ); This is the probability of remaining continuously in state i, not just being in i at time t. Often written with a bar over the p.
Constant total exit intensity: If μ_ij(x) = μ_ij is constant, holding time in i ~ Exponential(Σ_j μ_ij); The mean time spent in i per visit is 1 ÷ Σ_j μ_ij.
Small-interval approximation (Euler step): _{t+h}p_x^ij ≈ _tp_x^ij + h × d/dt _tp_x^ij; Used for numerical solutions when no closed form is practical. Use small h.

How to solve Multiple State Models and Transition Intensities questions

Use this method for any question that gives a multi-state model and asks for a probability, an equation or an expected time.

1Draw the state diagram. Label each state and put each intensity on its arrow. Check whether the intensities depend on age.
2Identify the start state i and the target state j. Note the age x and the time t.
3Decide if the question wants the probability of being in j at time t, or of staying in i throughout. The second has a closed form using only the exit intensities.
4For the forward equation, write for state j: inflow terms (probability in k times μ_kj) minus outflow terms (probability in j times the sum of intensities leaving j).
5State the initial conditions at t = 0.
6Solve. If a state has no way back, solve it first, often with an integrating factor. Then substitute into the next equation.
7Use Σ_j _tp_x^ij = 1 to find any remaining probability, and check every answer lies between 0 and 1.
8Write the answer with notation and units. State any assumption such as constant intensities.

Quickest way: Draw, read off, subtract

When to use it: Use when you need a forward equation or a simple probability and time is short.

Draw the diagram first. Most marks are for correct arrows and intensities.
For each state, read the arrows in and out. Inflow minus outflow gives the equation directly.
For a state you cannot leave, such as Dead, you can skip its equation and use the total probability rule.
For staying in a state, use exp(−∫ total exit intensity). No equation needed.
If a state can only be left, not entered, the probability is just the exponential of minus the exit intensity integral.

Common mistakes in Multiple State Models and Transition Intensities

Treating μ_ij(x) as a probability.
The notation looks like a probability and the model is built from small-interval probabilities.
Fix: Remember it is a rate per year. Only h × μ_ij(x) is approximately a probability, for small h.
Using the wrong age in the intensity, writing μ(x) instead of μ(x+t).
The equation is about the process at time t after starting age x, which is easy to forget.
Fix: In the forward equation, always evaluate intensities at x+t. Constant intensities avoid the issue.
Mixing up _tp_x^ii and the probability of staying in i continuously.
If the life can leave and come back, being in i at time t does not mean it never left.
Fix: The exponential formula gives only the continuous-stay probability. Use it only for that. With no return paths the two are equal.
Leaving out an outflow or inflow term in the forward equation.
Students rush and read only some arrows on the diagram.
Fix: Go through every arrow touching the state. Tick each off. Check that the equations sum to zero across all states.
Forgetting initial conditions or applying them to the wrong probability.
The differential equation feels like the whole answer.
Fix: Write _0p_x^ii = 1 and the others 0 before solving. Use them to fix the constant of integration.
Assuming the process remembers past states, for example that sickness duration affects recovery.
Real life suggests duration matters.
Fix: Under the Markov assumption only the current state and age matter. Duration dependence needs a different, semi-Markov model, and the question will say so.

Worked examples

Example 1

A healthy-sick-dead model has constant intensities per year: healthy to sick 0.10, healthy to dead 0.02, sick to healthy 0.30, sick to dead 0.08. A life is healthy at age 50. Write the Kolmogorov forward equations for _tp_50^HH and _tp_50^HS, and find the probability the life stays healthy throughout the next 2 years.

Show the solution

Label states H, S, D. Intensities: μ_HS = 0.10, μ_HD = 0.02, μ_SH = 0.30, μ_SD = 0.08.
For H: inflow from S is _tp^HS × μ_SH = 0.30 × _tp^HS. Outflow is _tp^HH × (μ_HS + μ_HD) = 0.12 × _tp^HH.
So d/dt _tp_50^HH = 0.30 × _tp_50^HS − 0.12 × _tp_50^HH.
For S: inflow from H is 0.10 × _tp^HH. Outflow is (μ_SH + μ_SD) × _tp^HS = 0.38 × _tp^HS.
So d/dt _tp_50^HS = 0.10 × _tp_50^HH − 0.38 × _tp_50^HS.
Initial conditions: _0p^HH = 1, _0p^HS = 0.
Staying healthy throughout uses only the exit intensity from H, which is 0.10 + 0.02 = 0.12.
Probability = exp(−0.12 × 2) = exp(−0.24) = 0.7866.

Answer: d/dt _tp^HH = 0.30 _tp^HS − 0.12 _tp^HH; d/dt _tp^HS = 0.10 _tp^HH − 0.38 _tp^HS; with _0p^HH = 1 and _0p^HS = 0. The probability of staying healthy throughout 2 years is exp(−0.24) ≈ 0.787.

Example 2

A two-state-exit model has states A (active), R (retired) and D (dead). Constant intensities: A to R 0.04, A to D 0.01, R to D 0.05. R has no exit except D. A life is in state A at time 0. Find the probability of being in state R at time 1 year, using the forward equation.

Show the solution

Exit intensity from A is 0.04 + 0.01 = 0.05. There is no return to A.
So _tp^AA = exp(−0.05t).
Forward equation for R: d/dt _tp^AR = 0.04 × _tp^AA − 0.05 × _tp^AR.
Substitute: d/dt _tp^AR + 0.05 _tp^AR = 0.04 exp(−0.05t).
The integrating factor is exp(0.05t). Then d/dt [ exp(0.05t) × _tp^AR ] = 0.04.
Integrate: exp(0.05t) × _tp^AR = 0.04t + c. At t = 0, _tp^AR = 0, so c = 0.
So _tp^AR = 0.04t × exp(−0.05t).
At t = 1: 0.04 × exp(−0.05) = 0.04 × 0.951229 = 0.03805.

Answer: _1p^AR = 0.04 × exp(−0.05) ≈ 0.0380.

Exam tips

Draw the diagram before anything else, even if the question gives one in words. It earns marks and prevents missing terms.
In written answers, state the Markov assumption and whether intensities are constant. Examiners reward stated assumptions.
If a state has no return paths, solve its probability first, then substitute into the next equation using an integrating factor.
In MCQs, check the answer lies between 0 and 1 and that probabilities across states sum to 1.
Show the equation, the initial condition and the solution. Marks are split across these steps, so a final number alone loses marks.

Practice questions from Transition intensities dependent on age (exact or census)

Multiple State Models and Transition Intensities: frequently asked questions

What is the difference between a transition intensity and a transition probability?

A transition intensity μ_ij(x) is an instantaneous rate at age x. A transition probability _tp_x^ij is the chance of being in state j at age x+t, given state i at age x. Intensities are inputs. Probabilities are what you derive from them.

How do I derive the Kolmogorov forward equations?

Consider the process at time t+h. The life is in j if it was in j and did not leave, or was in another state k and moved to j. Write this using h × μ terms, subtract _tp^ij, divide by h and let h tend to 0. This gives inflow minus outflow.

When can I use the exponential formula for staying in a state?

Use it for the probability of remaining in a state continuously for time t. It equals exp of minus the integral of the total exit intensity. It does not give the probability of being in that state at time t if the life can leave and return.

Do I need to solve the forward equations by hand in the exam?

Often you only set them up, or solve a simple case with one-way movement. For more complex models, you may be asked to use a small time step numerically. Practise both.