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

Likelihood Estimation of Constant Transition Intensities

Q: What is the MLE of the force of mortality in a two-state model?

If the force is constant, the MLE is μ̂ = d ÷ v. Here d is the number of deaths and v is the total time lived by all lives. It is a rate per unit time.

Q: How do I derive the variance of the estimator of μ?

Find the second derivative of the log-likelihood, which is −D ÷ μ². Take the negative expectation to get the information E[D] ÷ μ². Its inverse is the asymptotic variance, μ² ÷ E[D], which equals μ ÷ E[V] because E[D] = μE[V].

Q: What is central exposed to risk?

It is the total time that lives are observed in the state, summed across all lives. It is the denominator of the estimator μ̂ = d ÷ v. Lives that die or leave part-way contribute only the time they were observed.

Q: Is the estimator exactly unbiased?

No. The variance result is asymptotic, and for small samples the estimator can be biased. The results hold approximately for large samples.

Updated 11 October 2026 · Fact-checked

For a constant force μ in a two-state model, the likelihood is L = μ^d × e^(−μv), where d is the number of transitions and v is the total waiting time (central exposed to risk). Maximising gives μ̂ = d ÷ v. Its asymptotic variance is μ ÷ E[V], estimated by μ̂² ÷ d.

Understand Likelihood Estimation of Constant Transition Intensities

Suppose you watch a group of lives over a period. Each life is in the alive state until it moves to the dead state, or until you stop observing it. If the force of mortality is a constant μ, the time a life spends alive is exponential with parameter μ. This is the simplest model, and it is the base for more general multi-state work.

For each life you record two things. One is the waiting time spent in the state, the time observed from entry to exit or to the end of observation. The other is whether a transition happened (death) or the observation was censored. Add the waiting times of all lives to get the total waiting time v. This is the central exposed to risk. Add the transitions to get d, the number of deaths.

A life that dies after time t contributes the density μe^(−μt). A life that is censored at time t contributes the survival probability e^(−μt). Multiply over all lives. Each death contributes one factor of μ, and every life contributes e^(−μ × its time observed). So the likelihood depends on the data only through d and v: L(μ) = μ^d e^(−μv).

Take logs, differentiate, set to zero and you get μ̂ = d ÷ v. This is the number of deaths per unit of time lived. The estimator is random because d and V are random, so you study its variance. For large samples, μ̂ is approximately normal with mean μ and variance equal to the Cramér-Rao lower bound, the inverse of the Fisher information, which is the negative expected second derivative of the log-likelihood. Under the exponential model this gives variance ≈ μ ÷ E[V], which you estimate by μ̂² ÷ d.

The same idea extends to several states. Each transition type i to j has its own intensity, estimated by the number of i to j transitions divided by the total time spent in state i. The likelihood factorises across intensities.

Key rules to remember

Likelihood (constant μ): L(μ) = μ^d × e^(−μv); d = number of deaths (transitions); v = total waiting time observed. Valid for exact observation with censoring that is non-informative.
Log-likelihood: ln L = d ln μ − μv; Differentiate with respect to μ to find the maximum.
Maximum likelihood estimator: μ̂ = d ÷ v; As a random variable, μ̂ = D ÷ V. Check the second derivative −d ÷ μ² is negative, so this is a maximum.
Asymptotic variance: Var(μ̂) ≈ μ ÷ E[V]; Approximate and valid for large samples. Equivalent form: μ² ÷ E[D].
Estimated variance: Var(μ̂) ≈ μ̂² ÷ d = d ÷ v²; Use the observed values when μ and E[V] are unknown.
Approximate 95% confidence interval: μ̂ ± 1.96 × √(μ̂² ÷ d) = μ̂ (1 ± 1.96 ÷ √d); Based on the asymptotic normality of μ̂.
Multi-state estimator: μ̂ij = dij ÷ vi; dij = observed transitions from state i to j; vi = total time spent in state i.

How to solve Likelihood Estimation of Constant Transition Intensities questions

Use this method for any question that asks you to derive or apply the MLE of a constant intensity.

1Define the model: states, the transition intensity μ (or μij), and the assumption that it is constant over the period.
2Identify the data: the number of transitions d and the total waiting time v in the relevant state. Check whether lives are censored.
3Write each life's contribution: μe^(−μt) for a transition at time t, e^(−μt) for a censored life. Multiply to get L = μ^d e^(−μv).
4Take the log: ln L = d ln μ − μv. Differentiate: d ÷ μ − v. Set it to zero to get μ̂ = d ÷ v.
5Confirm it is a maximum with the second derivative: −d ÷ μ² < 0.
6For the variance, use the information: E[−d²lnL/dμ²] = E[D] ÷ μ² = E[V] ÷ μ. The asymptotic variance is its inverse, μ ÷ E[V].
7Substitute the observed values to get an estimate and, if asked, a confidence interval. State that the result is approximate for large samples.

Quickest way: Deaths divided by time lived

When to use it: Use when the question gives you d and v (or individual times) and asks for an estimate, a standard error or a confidence interval.

Add up the times lived by all lives to get v. Count deaths to get d.
Compute μ̂ = d ÷ v.
Compute the standard error as μ̂ ÷ √d.
Form the interval μ̂ ± 1.96 × μ̂ ÷ √d for 95%.
Check the units: the answer is per year if v is in years.

Common mistakes in Likelihood Estimation of Constant Transition Intensities

Using the number of lives at the start instead of total waiting time as the denominator.
It looks like the usual death probability d ÷ n.
Fix: The intensity is a rate per unit time. Divide by the total time lived v, including the part-year of those who die or leave.
Leaving out the contribution of censored lives to the likelihood.
Students think only deaths carry information.
Fix: Censored lives contribute e^(−μt). Their time adds to v even though they add nothing to d.
Writing the variance as μ̂ ÷ d, or giving d ÷ v (which is just μ̂ itself) as the variance.
Confusion between variance and the estimator itself, or mixing up D and V.
Fix: Var(μ̂) ≈ μ̂² ÷ d = d ÷ v². Check dimensions: d ÷ v is a rate, but a variance has units of rate squared.
Forgetting to check the second derivative or to say the variance is asymptotic.
Students rush to the answer.
Fix: Show that the second derivative is negative. State that the variance result is approximate and valid for large samples.
Treating D and V as fixed when deriving the variance.
The estimate d ÷ v is computed from fixed data values.
Fix: In the derivation, D and V are random. The information uses E[D] = μE[V].
Mixing up the central exposed to risk with the initial exposed to risk.
Both are called exposed to risk.
Fix: Central exposed to risk is the total time lived and pairs with μ̂ = d ÷ v. Initial exposed to risk counts lives at the start and pairs with q̂ = d ÷ E.

Worked examples

Example 1

In a study of 200 policyholders, 8 deaths occurred and the total time observed was 1,600 years. Assuming a constant force of mortality μ, find the MLE of μ, an estimate of its variance, and an approximate 95% confidence interval.

Show the solution

The number of lives (200) is not needed. Only d and v matter. Time is measured in years, so rates are per year.
d = 8 and v = 1,600.
μ̂ = 8 ÷ 1,600 = 0.005.
Estimated variance = μ̂² ÷ d = 0.000025 ÷ 8 = 0.000003125.
Standard error = 0.005 ÷ √8 = 0.005 ÷ 2.8284 = 0.001768.
95% interval = 0.005 ± 1.96 × 0.001768 = 0.005 ± 0.003465.
Interval = (0.00154, 0.00846).

Answer: μ̂ = 0.005 per year; estimated variance ≈ 3.125 × 10⁻⁶; approximate 95% confidence interval (0.0015, 0.0085) per year.

Example 2

Lives are observed with a constant force μ. Out of n lives, d die and the rest are censored. Derive the MLE of μ and its asymptotic variance.

Show the solution

Let the time observed for life i be tᵢ, and v = Σtᵢ.
A death at tᵢ contributes μe^(−μtᵢ). A censored life contributes e^(−μtᵢ).
L(μ) = μ^d e^(−μv).
ln L = d ln μ − μv.
d(ln L)/dμ = d ÷ μ − v = 0, so μ̂ = d ÷ v.
d²(ln L)/dμ² = −d ÷ μ² < 0, so this is a maximum.
Treat D as random. The information is E[D] ÷ μ² = μE[V] ÷ μ² = E[V] ÷ μ.
Asymptotic variance is the inverse: μ ÷ E[V].

Answer: μ̂ = d ÷ v, with asymptotic variance μ ÷ E[V] (equivalently μ² ÷ E[D]), estimated by μ̂² ÷ d.

Exam tips

In a derivation question, show every step: likelihood, log-likelihood, first derivative, second derivative and the variance. Marks are for each step.
State the assumptions: constant intensity, observation of exact times, and non-informative censoring.
Quote the variance as asymptotic or approximate. Examiners look for this wording.
In MCQs, check whether the question gives central exposed to risk (use d ÷ v) or initial exposed to risk (use d ÷ E).
In the computer-based paper, compute d and v from the data first, then apply the formula and report the standard error with the estimate.

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