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

Maximum Likelihood Estimator of a Transition Intensity

Q: Why is the MLE of μ equal to deaths divided by total time lived?

The likelihood is μ^d e^(−μv). Maximising its log, d ln μ − μv, gives d/μ = v. So μ̂ = d ÷ v, the observed rate of events per unit of time at risk.

Q: How do I prove the estimate is a maximum and not a minimum?

Take the second derivative of the log-likelihood, which is −d/μ². For d > 0 this is negative for all μ > 0, so the stationary point is a maximum.

Updated 11 October 2026 · Fact-checked

For a constant transition intensity μ, the maximum likelihood estimator is μ̂ = d ÷ v, where d is the number of observed transitions and v is the total time spent in the starting state (waiting time). You derive it by maximising the log-likelihood d ln μ − μv and checking that the second derivative is negative.

Understand Maximum Likelihood Estimator of Transition Intensity

A transition intensity is the instantaneous rate of moving from one state to another. In a two-state alive-dead model it is the force of mortality μ. If μ is assumed constant over the period studied, the time a life spends in the alive state before moving is exponentially distributed with rate μ.

Now think of the data. Each life i is observed for a time v_i. Either the life dies at the end of that time, or observation stops while the life is still alive (censored). A life that dies contributes the density μe^(−μv_i) to the likelihood. A censored life contributes the survival probability e^(−μv_i). Every life contributes the factor e^(−μv_i). Each death adds one extra factor of μ.

Multiply across lives. With d deaths and total observed time v = Σ v_i, the likelihood is L(μ) = μ^d e^(−μv). Taking logs gives ln L = d ln μ − μv. The data enter only through d and v, so these are the key quantities.

Differentiate and set to zero: d/μ − v = 0, so μ̂ = d ÷ v. The second derivative is −d/μ², which is negative whenever d > 0. So this is a maximum. The result matches intuition: deaths per unit of time lived. The same logic gives, for a multi-state model, μ̂_ij = d_ij ÷ v_i, where v_i is the total time spent in state i.

The estimator is for a constant intensity over the observation period. If intensity varies with age, you split the data into age groups and estimate separately in each.

Key rules to remember

Likelihood (constant μ, two-state): L(μ) = μ^d × e^(−μv); d = number of deaths, v = total waiting time (sum of all observed times). Constants not involving μ are dropped.
Log-likelihood: ln L = d ln μ − μv; Add a constant if you keep the full likelihood; it does not affect the maximum.
Maximum likelihood estimator: μ̂ = d ÷ v; Number of transitions divided by total time spent in the starting state.
Second-derivative check: d²(ln L)/dμ² = −d ÷ μ² < 0 (for d > 0); Confirms a maximum. If d = 0, the likelihood is decreasing and the formula gives μ̂ = 0 at the boundary.
Multi-state version: μ̂_ij = d_ij ÷ v_i; d_ij = observed transitions from i to j; v_i = total time spent in state i by all lives.
Estimator as a random variable: μ̃ = D ÷ V; Use capital letters when treating the estimator as random; lower case for the estimate from observed data.

How to solve Maximum Likelihood Estimator of Transition Intensity questions

Use this method whether the question asks you to derive the estimator, calculate an estimate, or check it is a maximum.

1Define the model: states, the transition of interest, and the assumption that the intensity is constant.
2Write each life's contribution: μe^(−μv_i) if the transition occurs, e^(−μv_i) if observation ends first.
3Multiply to get L(μ) = μ^d e^(−μv), with d the number of transitions and v the total time in the starting state.
4Take logs: ln L = d ln μ − μv (plus constants).
5Differentiate with respect to μ and set equal to zero: d/μ − v = 0, so μ̂ = d ÷ v.
6Take the second derivative, −d/μ², and state it is negative, so this is a maximum.
7For numerical questions, compute d and v carefully, using consistent time units, and give μ̂ with units (per year).
8For multi-state models, repeat for each transition, using time in the starting state as the denominator.

Quickest way: Count and divide

When to use it: Numerical questions that give you data and ask for the estimate, when no derivation is requested.

Count the transitions of the type asked for. Ignore other transitions.
Add up the time every life spent in the starting state, including those who left or were censored.
Divide: μ̂ = d ÷ v.
Check units and sanity: the answer should be near the plausible rate, for example 0.01 to 0.05 for adult mortality per year.
If asked for a probability, use p = e^(−μ̂t) afterwards.

Common mistakes in Maximum Likelihood Estimator of Transition Intensity

Using the number of lives at the start as the denominator instead of total time lived.
Students mix this up with a simple proportion of deaths in the initial group.
Fix: Use total waiting time v. Add the time each life actually spent in the state, including time of survivors and those who died early.
Leaving out the censored lives or their time.
Students think only the deaths provide information.
Fix: Censored lives contribute e^(−μv_i) to the likelihood. Their time counts in v, though they add nothing to d.
Skipping the check that the stationary point is a maximum.
Students stop once they get d ÷ v.
Fix: Always differentiate again. State that −d/μ² < 0 for d > 0, so it is a maximum.
Including time spent in other states in the denominator.
Students use total observation time in multi-state models.
Fix: The denominator is time in the state the transition leaves. For μ̂_ij use v_i only.
Mixing up units of time, such as months and years.
Data are given in mixed units.
Fix: Convert everything to years before summing. The answer is then a rate per year.
Writing the likelihood with the wrong exponent on μ.
Students put the number of lives, not the number of transitions, as the power.
Fix: The power of μ is d, the number of transitions that occurred.

Worked examples

Example 1

A study follows 5 lives. Two die, after 1.5 years and 2.5 years. The other three are alive when the study ends, having been observed for 3 years, 4 years and 2 years. Assuming a constant force of mortality μ, find the maximum likelihood estimate of μ.

Show the solution

Number of deaths: d = 2.
Total time lived: v = 1.5 + 2.5 + 3 + 4 + 2 = 13 years.
Likelihood: L(μ) = μ² e^(−13μ), so ln L = 2 ln μ − 13μ + constant.
Set the derivative to zero: 2/μ − 13 = 0, so μ̂ = 2 ÷ 13 = 0.1538.
Second derivative: −2/μ² < 0, so this is a maximum.

Answer: μ̂ = 2/13 ≈ 0.154 per year.

Example 2

In a three-state model (healthy H, sick S, dead D), 40 lives are observed. Total time spent healthy is 120 years. During this time, 9 moved to S and 3 moved to D. Assuming constant intensities, estimate μ_HS and μ_HD and say what the total exit intensity from H is.

Show the solution

Each intensity uses the time in the starting state, v_H = 120 years.
μ̂_HS = 9 ÷ 120 = 0.075.
μ̂_HD = 3 ÷ 120 = 0.025.
Total exit intensity from H is the sum: 0.075 + 0.025 = 0.100, which equals 12 ÷ 120.
This agrees with treating all exits as one transition type.

Answer: μ̂_HS = 0.075 per year, μ̂_HD = 0.025 per year, and the total exit intensity from H is 0.100 per year.

Exam tips

Show the likelihood, the log-likelihood, the derivative and the second derivative in order. Each earns method marks in written questions.
State the constant intensity assumption clearly at the start of the answer.
In numerical questions, list d and v separately before dividing, so partial credit is possible if one is wrong.
Be ready to extend the derivation to a multi-state model, where v is time in the starting state only.
In MCQs, check units and which state's waiting time the question wants before computing.

Practice questions from Maximum likelihood estimators for transition intensities

Maximum Likelihood Estimator of Transition Intensity: frequently asked questions

Why is the MLE of μ equal to deaths divided by total time lived?

The likelihood is μ^d e^(−μv). Maximising its log, d ln μ − μv, gives d/μ = v. So μ̂ = d ÷ v, the observed rate of events per unit of time at risk.

How do I prove the estimate is a maximum and not a minimum?

Take the second derivative of the log-likelihood, which is −d/μ². For d > 0 this is negative for all μ > 0, so the stationary point is a maximum.

What happens to censored lives in the likelihood?

A censored life contributes the probability of surviving its observed time, e^(−μv_i). It adds to the total time v but not to the number of deaths d.

Does the formula work if there are no deaths?

If d = 0, the likelihood e^(−μv) decreases as μ increases, so the maximum is at μ = 0. The formula gives 0, but it is a boundary value, not a stationary point found by calculus.