Risk Modelling and Survival Analysis · Maximum likelihood estimators for transition intensities

Multiple-State Models: Estimating Several Transition Intensities

Updated 11 October 2026 · Fact-checked

In a Markov multiple-state model with constant intensities, the maximum likelihood estimate of the intensity from state i to state j is μ̂ij = Nij ÷ Ti. Nij is the number of observed i→j transitions. Ti is the total time all lives spent in state i. Each intensity is estimated separately.

Understand Multiple-State Models and Estimating Several Intensities

A multiple-state model lets a life move between states, for example Healthy, Sick and Dead. Each possible move i→j has a transition intensity μij. Over a short time dt, the probability of moving from i to j is about μij dt. In this topic we assume the intensities are constant over the period studied and that the process is Markov, so the future depends only on the current state.

The data for each life is a path: how long it stayed in each state, and which state it moved to when it left. Some lives are still in a state when observation stops. These are censored. They contribute time spent but no transition.

The likelihood for one life is a product of terms. For each stay in state i, you get a factor for surviving there, which is exp(-μi· × time), where μi· is the sum of all intensities out of state i. If the life then moves to j, you add a factor μij. Multiply over all lives and all stays.

This product separates into pieces, one per transition type. For each i→j, the log-likelihood contribution is Nij ln μij - μij Ti. Setting the derivative to zero gives μ̂ij = Nij ÷ Ti. Because the likelihood factorises, you can estimate each intensity on its own.

The key point: the time Ti counts all time in state i, not just time before a move to j. A life that leaves i for k still spent time in i. That time belongs in the denominator of every intensity out of i.

Key rules to remember

Likelihood for constant intensities: L ∝ Π over i≠j of exp(-μij Ti) × μij^Nij; Ti is total time in state i. Nij is the number of i→j transitions. Constants and the Markov assumption are required.
MLE of a transition intensity: μ̂ij = Nij ÷ Ti; Use the same Ti for every intensity leaving state i.
Log-likelihood: ln L = Σ (Nij ln μij - μij Ti) + constant; Differentiate with respect to each μij separately.
Variance of the estimator (asymptotic): Var(μ̂ij) ≈ μij ÷ E[Ti], estimated by Nij ÷ Ti²; Based on the second derivative: -d²lnL/dμij² = Nij ÷ μij². It is an approximation for large samples.
Distribution of the estimator: μ̂ij is approximately Normal(μij, μij ÷ E[Ti]); Use for confidence intervals: μ̂ij ± 1.96 × √(Nij) ÷ Ti for 95%.
Total exit intensity: μi· = Σ over j≠i of μij, estimated by (Σj Nij) ÷ Ti; Waiting time in state i is exponential with this rate.

How to solve Multiple-State Models and Estimating Several Intensities questions

Use this method for any question that gives observed transitions and time in states and asks for estimated intensities or their likelihood.

1Draw the model. List every state and every allowed transition i→j.
2From the data, count Nij for each allowed transition. Do not count censored exits.
3Work out Ti for each state: add up the time all lives spent in it, including time of lives still there at the end.
4Write the likelihood as a product over transitions of μij^Nij × exp(-μij Ti), if the question asks for it. Take logs.
5Differentiate ln L with respect to each μij. Set to zero to get μ̂ij = Nij ÷ Ti.
6Calculate each estimate. Check units, such as per year.
7If asked for variance or a confidence interval, use Var ≈ Nij ÷ Ti² and the Normal approximation.
8If asked for a probability or expected value, plug the estimates into the model, for example exp(-μ̂i· t) for staying in state i for time t.

Quickest way: Tally and divide

When to use it: Use when the question gives a table of lives or transitions and asks only for the estimates.

Make a small table with a row per state.
In each row, write total time Ti and the counts Nij for each exit.
Divide each count by the row's Ti.
Sanity check: intensities out of one state share the same denominator, and the total exit rate is total exits ÷ Ti.
Only derive the MLE from scratch if the question says 'show that' or asks for the likelihood.

Common mistakes in Multiple-State Models and Estimating Several Intensities

Using only the time before a transition to j as the denominator for μij.
Students think each intensity needs its own waiting time.
Fix: Use total time in state i for every intensity out of i. Time in i before a move to k still counts.
Counting censored lives as transitions.
A life leaving the study looks like it left the state.
Fix: Add the censored life's time to Ti but leave Nij unchanged.
Leaving out time spent by lives still in the state at the end of the study.
Focus is on lives that moved.
Fix: Sum time for every life, whether it moved or not, up to the end of observation.
Counting time in the Dead state, or using it as a denominator.
All states are treated alike.
Fix: Dead is absorbing with no exit intensities. It has no Ti needed.
Mixing up N and T when writing the variance, e.g. Nij ÷ Ti instead of Nij ÷ Ti².
Memorising the estimate and the variance together.
Fix: Derive it: Var ≈ μ̂ ÷ Ti = Nij ÷ Ti². Check that the units are per-year squared.
Forgetting that the method needs constant intensities.
The formula is applied automatically.
Fix: State the assumption in written answers. If intensities vary by age, split data into age bands and estimate each band separately.

Worked examples

Example 1

A study follows 5 lives in a Healthy-Sick-Dead model. Time in Healthy totals 40 years. Time in Sick totals 10 years. Observed transitions: Healthy→Sick 6, Healthy→Dead 2, Sick→Healthy 4, Sick→Dead 3. Estimate all four intensities.

Show the solution

Healthy exits use T_H = 40. Sick exits use T_S = 10.
μ̂HS = 6 ÷ 40 = 0.15.
μ̂HD = 2 ÷ 40 = 0.05.
μ̂SH = 4 ÷ 10 = 0.4.
μ̂SD = 3 ÷ 10 = 0.3.

Answer: μ̂HS = 0.15, μ̂HD = 0.05, μ̂SH = 0.4, μ̂SD = 0.3, all per year.

Example 2

Using the data above, (a) estimate the standard deviation of μ̂HS, and (b) estimate the probability that a Healthy life stays Healthy for at least 2 years.

Show the solution

(a) Var(μ̂HS) ≈ N ÷ T² = 6 ÷ 1,600 = 0.00375.
Standard deviation = √0.00375 = 0.0612.
(b) Total exit rate from Healthy is 0.15 + 0.05 = 0.20.
Probability of staying at least 2 years = exp(-0.20 × 2) = exp(-0.4).
exp(-0.4) = 0.6703.

Answer: (a) Standard deviation ≈ 0.0612. (b) Probability ≈ 0.670.

Exam tips

Write the allowed transitions first. It prevents missing a state or counting a transition that cannot happen.
Show the likelihood and the derivative when the question says 'derive'. Marks are for the form Nij ln μij - μij Ti.
State assumptions: constant intensities and the Markov property.
Questions often add a follow-on, such as a confidence interval or a probability. Keep the estimates unrounded until the end.
In computer-based questions, tabulate the counts and times in R before dividing, so you can check them.

Practice questions from Maximum likelihood estimators for transition intensities

Multiple-State Models and Estimating Several Intensities: frequently asked questions

Why can I estimate each intensity separately?

The likelihood is a product of terms, each involving only one μij. Maximising a product of separate terms means maximising each one. So each intensity is estimated on its own.

What is Ti in the Healthy-Sick-Dead model?

Ti is the total time all lives spent in state i during the observation. It includes time of lives that moved on and of lives that stayed in the state until the end.

How do I handle a life that is still alive at the end of the study?

Add its time in the current state to Ti. Do not add a transition. The life is censored, so it gives time but no event.

How is this different from the two-state model?

The two-state model has one intensity, with deaths divided by time lived. The multiple-state model repeats the same idea for every transition, using the time in the state the life is leaving.