Risk Modelling and Survival Analysis · Estimation procedures for lifetime distributions

Likelihood Functions for Censored Survival Data

Q: Why do censored lives contribute S(c) to the likelihood?

You only know the life survived to c, so the information is the probability of surviving past c. That probability is S(c). The life adds exposure but no death.

Q: What is the MLE of a constant force of mortality?

It is the number of deaths divided by the total time lived by all lives in the study. In symbols, μ̂ = d ÷ v. Censored lives add to v but not to d.

Q: What is the variance of the MLE for a constant hazard?

Asymptotically, Var(μ̂) ≈ μ² ÷ d. In practice you estimate it by μ̂² ÷ d. The standard error is μ̂ ÷ √d.

Q: How does left truncation change the likelihood?

Each life's contribution is divided by the survival probability at its entry time. This conditions on the life being alive when it entered. For a constant hazard you simply count time from entry.

Updated 11 October 2026 · Fact-checked

The likelihood is a product of one term per life. An observed death at time t contributes the density f(t). A life censored at c contributes the survival probability S(c). Truncation divides by survival at the entry time. For a constant hazard μ, the MLE is the number of deaths divided by total time observed.

Understand Likelihood Functions for Censored Survival Data

A lifetime study rarely watches every life until death. Some people leave the study, or the study ends while they are alive. You then know only that their lifetime exceeds some time. This is right censoring. Throwing these lives away would bias your estimate, so you keep them and use what you know.

The likelihood is the joint probability (or density) of the data you saw, viewed as a function of the parameters. You assume lives are independent, so you multiply one factor per life. Each factor depends on what you observed for that life.

If you saw a death at time t, you know the lifetime exactly. The factor is the density f(t) = S(t)μ(t). If you saw the life survive to time c and then be censored, you only know T > c. The factor is S(c). Both factors contain S, so it helps to write them with a indicator: L = Π [μ(tᵢ)]^δᵢ × S(tᵢ), where δᵢ = 1 for a death and 0 for a censored life.

Left truncation arises when a life enters observation at time x₀ only because it was alive then (for example, a policyholder who joins at age 40). You then condition on survival to the entry time. Each factor is divided by S(entry time). Ignoring this overstates survival.

For a constant hazard μ, S(t) = e^(−μt). The likelihood becomes L = μ^d × e^(−μv), where d is the number of deaths and v is the total time lived by all lives (deaths and censored). Maximising gives μ̂ = d ÷ v. Its variance comes from the second derivative of the log-likelihood and is approximately μ² ÷ d.

Key rules to remember

Exact death contribution: f(t) = S(t) × μ(t); Use when the death time is observed exactly.
Right-censored contribution: P(T > c) = S(c); Use when the life is known only to survive beyond c.
Left-truncated contribution: f(t) ÷ S(x₀) for a death; S(c) ÷ S(x₀) for a censored life; x₀ is the time the life enters observation. Condition on survival to x₀.
General likelihood: L = Π [μ(tᵢ)]^δᵢ × exp(−∫ μ(s) ds from entry to tᵢ); δᵢ = 1 for death, 0 for censored. Entry is 0 if there is no truncation.
Constant hazard likelihood: L(μ) = μ^d × e^(−μv); d = number of deaths; v = total observed time for all lives.
Constant hazard MLE: μ̂ = d ÷ v; Check it is a maximum: d²ln L/dμ² = −d ÷ μ² < 0.
Variance of MLE: Var(μ̂) ≈ μ² ÷ d, estimated by μ̂² ÷ d = d ÷ v²; Asymptotic result from the Cramér-Rao lower bound, using the information −E[d²ln L/dμ²].

How to solve Likelihood Functions for Censored Survival Data questions

Use this method for any likelihood or MLE question on lifetime data.

1List each life: entry time, exit time and whether exit was a death or censoring.
2Write each life's factor: μ(t)S(t) for a death, S(c) for censoring, each divided by S(entry) if left-truncated.
3Multiply the factors to get L, using the given hazard or distribution for S and μ.
4Take the log to get ln L. Simplify so d, v or other sums appear.
5Differentiate with respect to the parameter and set equal to zero. Solve for the estimate.
6Check the second derivative is negative to confirm a maximum.
7If asked for variance, use −1 ÷ (d²ln L/dμ²) evaluated at the MLE, or the information formula.
8Substitute the data and give the answer with units (per year).

Quickest way: Constant hazard shortcut

When to use it: Use when the hazard is constant and the question gives deaths and total exposure, or lets you add them up.

Count deaths d.
Add up total time observed v for every life, including censored and dead lives (exit time minus entry time).
Compute μ̂ = d ÷ v.
For the standard error, compute μ̂ ÷ √d.
For a confidence interval, use μ̂ ± 1.96 × μ̂ ÷ √d.

Common mistakes in Likelihood Functions for Censored Survival Data

Leaving out censored lives' time from v.
Students think only deaths matter.
Fix: Add every life's observed time. Censored lives add to v but not to d.
Using f(c) for a censored life instead of S(c).
Mixing up exact and censored information.
Fix: Ask: did I see the death? If not, use S.
Forgetting to divide by S(entry) for left-truncated lives.
Students treat entry as time zero.
Fix: Measure time from entry, or divide by S at entry. For a constant hazard, count only time after entry.
Giving Var(μ̂) = μ ÷ d or d ÷ v instead of μ² ÷ d.
Confusion with the Poisson count variance.
Fix: Derive it: −d²lnL/dμ² = d ÷ μ², so variance is μ² ÷ d.
Not checking that the stationary point is a maximum.
Students rush the calculus.
Fix: Write the second derivative and note it is negative.
Using the wrong time unit.
Mixing months and years.
Fix: Convert all times to one unit before summing v.

Worked examples

Example 1

Five lives are observed from time 0. Two die at times 3 and 5 years. Three are censored at times 4, 6 and 10 years. Assuming a constant force of mortality μ, find the MLE of μ and its estimated standard error.

Show the solution

Deaths: d = 2.
Total time: v = 3 + 5 + 4 + 6 + 10 = 28 years.
Likelihood: L = μ² × e^(−28μ).
ln L = 2 ln μ − 28μ. Differentiate: 2 ÷ μ − 28 = 0, so μ̂ = 2 ÷ 28 = 0.07143.
Second derivative: −2 ÷ μ², which is negative, so this is a maximum.
Variance ≈ μ̂² ÷ d = 0.07143² ÷ 2 = 0.002551.
Standard error = √0.002551 = 0.0505.

Answer: μ̂ = 1/14 ≈ 0.0714 per year, with estimated standard error ≈ 0.0505.

Example 2

Three lives enter a study at age 60 having survived to that age, and follow the exponential model S(t) = e^(−λt) measured from entry. Life A dies after 2 years. Life B is censored after 5 years. Life C dies after 3 years. Write the likelihood, find λ̂, and the variance of λ̂.

Show the solution

Life A contributes f(2) = λe^(−2λ).
Life B contributes S(5) = e^(−5λ).
Life C contributes f(3) = λe^(−3λ).
Because time is measured from entry and the lives are alive at entry, the conditioning factor S(0) = 1.
L = λ² × e^(−λ(2+5+3)) = λ² e^(−10λ).
ln L = 2 ln λ − 10λ. Setting 2 ÷ λ − 10 = 0 gives λ̂ = 0.2.
Second derivative = −2 ÷ λ² < 0, a maximum.
Information = 2 ÷ λ², so Var(λ̂) ≈ λ² ÷ 2. At λ̂ = 0.2 this is 0.04 ÷ 2 = 0.02.

Answer: L = λ² e^(−10λ); λ̂ = 0.2 per year; estimated variance = 0.02.

Exam tips

Write the contribution of each life in a short table (entry, exit, death or censored) before writing L. It earns method marks.
State assumptions clearly: independent lives, non-informative censoring, and constant hazard if used.
In MCQs, compute d and v first. The answer is usually d ÷ v and the trap is leaving out censored time.
Always show the second-derivative check and the variance derivation in written answers.
For Paper B in R, use the same logic: sum the time, count the deaths, and compare with a survival package fit if asked.

Practice questions from Estimation procedures for lifetime distributions