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

Two-State Model and Observed Data Setup for Survival Analysis

Updated 11 October 2026 · Fact-checked

The two-state model has states alive and dead, with a constant force of mortality μ. For each life you record the waiting time in the alive state and whether death occurred. The likelihood is L(μ) = μ^d × e^(−μv), where d is the number of deaths and v is the total waiting time.

Understand Two-State Model and Observed Data Setup

The two-state model is the simplest survival model. A life is either alive or dead. Death is a one-way move, so there is no return. The only thing that governs the move is the transition intensity from alive to dead, which is the force of mortality μ.

We assume μ is constant. That means the future lifetime T of a life in the alive state is exponential with parameter μ. The survival probability over time t is e^(−μt). The density of death at time t is μe^(−μt). Constant μ is an assumption. It is reasonable over a short age range, so say it in your answer.

Now think about the data. You observe a group of lives for a period. For life i you record the waiting time v_i, which is the time spent in the alive state while under observation. You also record an indicator d_i, which is 1 if the life died during observation and 0 if it did not. A life that is still alive when observation stops, or that leaves the study for another reason, is censored.

The likelihood contribution of each life depends on what you saw. A death at time v_i contributes the density μe^(−μv_i). A censored life contributes only the survival probability e^(−μv_i), because all you know is that it survived to v_i. Combining the two, each life contributes μ^(d_i) × e^(−μv_i).

Multiply across independent lives. The likelihood is L(μ) = μ^d × e^(−μv). Here d = Σd_i is the total deaths and v = Σv_i is the total waiting time. Only d and v matter, not the individual times. This is the setup you carry into the maximum likelihood step.

Key rules to remember

Survival probability, constant force: tp_x = e^(−μt); Holds when μ is constant over the period of observation.
Density of death at time t: f(t) = μ × e^(−μt); Used for a life observed to die at waiting time t.
Contribution of one life: L_i(μ) = μ^(d_i) × e^(−μv_i); d_i = 1 if death observed, 0 if censored. v_i is the time observed in the alive state.
Full likelihood: L(μ) = μ^d × e^(−μv); d = Σd_i is total deaths. v = Σv_i is total waiting time. Assumes lives are independent.
Log-likelihood: ln L(μ) = d ln μ − μv; Differentiate this to find the MLE in the next topic.
Maximum likelihood estimate: μ̂ = d ÷ v; Obtained by setting d ÷ μ − v = 0. Deaths divided by total time at risk.

How to solve Two-State Model and Observed Data Setup questions

Use this method for any question that asks you to set up or write the likelihood in the two-state model.

1State the model: two states, alive and dead, with constant force of mortality μ. Note that lives are assumed independent.
2Define the data for each life: waiting time v_i and death indicator d_i.
3Identify which lives died and which were censored. Censoring includes survivors at the end of the study and withdrawals.
4Write each life's contribution: μe^(−μv_i) for a death, e^(−μv_i) for a censored life.
5Multiply the contributions to get L(μ) = μ^d × e^(−μv), where d is total deaths and v is total waiting time.
6Take logs if asked: ln L = d ln μ − μv.
7If asked for the estimate, differentiate, set to zero and solve to get μ̂ = d ÷ v. Check the second derivative is negative.
8 Write the answer with units, such as per year.

Quickest way: Count deaths, add up time

When to use it: Use when the question gives you raw data and asks for the likelihood or the estimate under time pressure.

Count the deaths. This is d.
Add up every life's time under observation, deaths and survivors alike. This is v.
Write L(μ) = μ^d × e^(−μv).
If an estimate is needed, write μ̂ = d ÷ v.
Check that v includes censored lives and that the units of v match the units of μ.

Common mistakes in Two-State Model and Observed Data Setup

Leaving censored lives out of the total waiting time.
Students think only deaths carry information.
Fix: Include every life's time in v. Censored lives add to v but not to d.
Using μe^(−μv) for every life, including censored ones.
Students forget that a censored life has no observed death.
Fix: Use e^(−μv_i) for censored lives. The factor μ appears only for deaths.
Putting the number of lives in place of the number of deaths as the power of μ.
Students count everyone in the study instead of those who died.
Fix: The power of μ is d, the deaths. Count the indicator d_i = 1 only.
Mixing time units, such as months for some lives and years for others.
Data comes in different forms and is added without converting.
Fix: Convert everything to one unit before summing v. Then μ is in that unit.
Not stating the constant force assumption or independence of lives.
Students treat them as obvious and skip them.
Fix: Write both assumptions in one line at the start. They justify the product form of the likelihood.
Treating d and v as random in the setup when the data are fixed numbers.
The same symbols D and V are used later for the estimator's distribution.
Fix: For a given data set, use lower-case d and v. Use capital D and V only when discussing the estimator as a random variable.

Worked examples

Example 1

Five lives are observed under a two-state model with constant force of mortality μ. Two die, at times 1.5 and 2.5 years. The other three are still alive when observation ends, at times 3, 3 and 2 years. Write down the likelihood and the maximum likelihood estimate of μ.

Show the solution

Deaths: d = 2.
Total waiting time: v = 1.5 + 2.5 + 3 + 3 + 2 = 12 years.
Likelihood: L(μ) = μ² × e^(−12μ).
Log-likelihood: ln L = 2 ln μ − 12μ.
Differentiate: 2 ÷ μ − 12 = 0, so μ = 2 ÷ 12 = 1/6.
Second derivative is −2 ÷ μ², which is negative, so this is a maximum.

Answer: L(μ) = μ² e^(−12μ) and μ̂ = 1/6 ≈ 0.1667 per year.

Example 2

A study follows 100 lives for one year. Eight die, each after an average of 0.4 years. The 92 survivors are observed for the full year. Assuming constant force μ, write the likelihood and find μ̂.

Show the solution

Deaths: d = 8.
Time from deaths: 8 × 0.4 = 3.2 years.
Time from survivors: 92 × 1 = 92 years.
Total waiting time: v = 3.2 + 92 = 95.2 years.
Likelihood: L(μ) = μ⁸ × e^(−95.2μ).
Estimate: μ̂ = d ÷ v = 8 ÷ 95.2 = 0.08403.

Answer: L(μ) = μ⁸ e^(−95.2μ) and μ̂ ≈ 0.0840 per year.

Exam tips

Write the likelihood in terms of d and v first. Marks are usually given for this form before any calculus.
Always show how each life contributes, one line for deaths and one for censored lives.
Define your symbols: say what d, v and μ mean and in what units.
Check the total waiting time by adding the deaths' times and the survivors' times separately.
If the question says 'derive', show the log-likelihood, the derivative and the second-derivative check.

Practice questions from Maximum likelihood estimators for transition intensities

Two-State Model and Observed Data Setup: frequently asked questions

Why does a censored life contribute only e^(−μv)?

You know only that the life survived to time v. The probability of that is e^(−μv). You do not know when death will occur, so no density term appears.

What is the waiting time in the two-state model?

It is the time a life spends in the alive state while under observation. For a life that dies, it runs from the start of observation to death. For a censored life, it runs to the end of observation or withdrawal.

Is constant force of mortality realistic?

Not over a long age range, since mortality changes with age. It is a modelling assumption that works over short periods or narrow age bands. State it clearly in your answer.

Does the likelihood depend on individual lifetimes?

No. It depends only on the total deaths d and total waiting time v. These two numbers are enough to find the maximum likelihood estimate.