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Competing risks and cumulative incidence — Examples

1 · Prerequisites

2 · Summary

The finite counterexample exhibits distinct latent joint failure-time laws with the same observed first-time/cause law. The two-group mortality example computes all-cause risk-set updates and cause-specific product limits exactly; it shows how higher competing mortality lowers observed cancer incidence while the competing-cause-as-censoring complement remains a different quantity.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The observed first time and cause do not identify the latent joint failure-time law

Statement refuted

Without restrictions on the joint law of the latent cause-specific times, the law of the observed first time and cause uniquely determines that latent joint law.

Facts & Assumptions

Given: A parametric model whose parameter is the full joint law of two latent cause-specific event times and whose observation is only the first time and its cause. Refute identifiability of the latent joint law.

[F1]

Competing-risk data, cause-specific hazard measures, and cumulative incidence: $(T, D)$ is obtained by mapping latent cause times to their unique minimum and its cause.

[F2]

Parametric statistical models: a model is a parameter map $θ \mapsto P_{θ}$ of sampling laws.

[F3]

Observational equivalence: $θ$ and $η$ are observationally equivalent exactly when $P_{θ} = P_{η}$ .

[F4]

Estimands on a statistical model: an estimand is a specified function of the model parameter.

[F5]

Identifiability of an estimand: an estimand is identifiable only if equal sampling laws imply equal target values.

Counterexample

Use two causes and parameter set $Θ = {0, 1}$ . Under $θ = 0$ , give $(T_{1}, T_{2})$ mass $1 / 2$ at $(1, 2)$ and mass $1 / 2$ at $(2, 1)$ . Under $θ = 1$ , give it mass $1 / 2$ at $(1, 3)$ and mass $1 / 2$ at $(3, 1)$ . Each law has finite positive times and no simultaneous first causes. The sampling law is the pushforward under $(T_{1}, T_{2}) \mapsto (T, D)$ , where $T = \min (T_{1}, T_{2})$ and $D$ is the unique minimizing cause.

1.1givenF4algebra

Let $ψ (θ)$ be the full joint law of $(T_{1}, T_{2})$ under parameter $θ$ . This is an estimand by [F4]. The two laws specified above differ: under $θ = 0$ , $P (T_{2} = 2) = 1 / 2$ , while under $θ = 1$ , $P (T_{2} = 2) = 0$ .

1.2givenF1F2F3algebra

Under either parameter, the first event is at time $1$ . The outcome $(T, D) = (1, 1)$ has probability $1 / 2$ from the atom with $T_{1} = 1$ , and $(T, D) = (1, 2)$ has probability $1 / 2$ from the atom with $T_{2} = 1$ ; no other observed pair occurs. Thus the two pushforward sampling laws are equal, so the parameters are observationally equivalent by [F2]–[F3].

2.1step 1.1step 1.2F5∎

Step 1.1 gives $ψ (0) \neq ψ (1)$ , while step 1.2 gives equal sampling laws. This violates the defining implication in [F5], so the latent joint law is not identifiable from the observed first time and cause.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Higher competing mortality lowers incidence in a two-time table

Example

Cause 1 is cancer death and cause 2 is cardiovascular death. Consider two exact groups of 16 right-censored records, with follow-up through time $2$ . Remaining event-free subjects are censored at $2$ . At time $1$ , the low-CV group has 4 cancer deaths and 4 cardiovascular deaths; the high-CV group has 4 cancer deaths and 8 cardiovascular deaths. At time $2$ , the corresponding risk sets and cancer deaths are 8 and 2 in the low-CV group, and 4 and 1 in the high-CV group. Events from different causes at the same time are grouped in the common risk-set update.

Facts & Assumptions

Given: Both groups have 16 subjects; events are observed at times $1$ and $2$ with the risk and cause counts stated above. Compute the cause-1 Aalen–Johansen incidence and compare it with the complement of the cause-specific Kaplan–Meier product that treats cause 2 as censored.

[F1]

Competing-risk data, cause-specific hazard measures, and cumulative incidence: the observed cause- $k$ cumulative incidence is $F_{k} (t) = P (T \leq t, D = k)$ .

[F2]

Aalen–Johansen cumulative-incidence estimator: at an event time $u$ , the cause- $k$ increment is $\hat{S} (u -) Δ N_{k} (u) / Y (u)$ and all-cause survival is multiplied by $1 - Δ N (u) / Y (u)$ .

[F3]

Event-time observation, censoring, and truncation schemes: under right censoring, $U = \min (T, C)$ and $δ = 1 {T \leq C}$ ; an event tied with censoring is counted as an event.

[F4]

Nelson–Aalen, Kaplan–Meier, and Greenwood estimators: the single-event Kaplan–Meier product multiplies $1 - Δ N (u) / Y (u)$ over event times, with tied event counts grouped in the risk set.

Verification

1.1F2givenalgebra

At time $1$ , both groups have $Y (1) = 16$ and $Δ N_{1} (1) = 4$ . The low-CV group has $Δ N (1) = 8$ , so $\hat{S} (1) = 1 - 8 / 16 = 1 / 2$ and ${\hat{F}}_{1} (1) = 4 / 16 = 1 / 4$ . The high-CV group has $Δ N (1) = 12$ , so $\hat{S} (1) = 1 - 12 / 16 = 1 / 4$ and ${\hat{F}}_{1} (1) = 4 / 16 = 1 / 4$ .

2.1F1step 1.1F2givenalgebra

At time $2$ , the low-CV risk count is $8$ and $Δ N_{1} (2) = 2$ , giving increment $(1 / 2) (2 / 8) = 1 / 8$ and ${\hat{F}}_{1} (2) = 1 / 4 + 1 / 8 = 3 / 8$ . The high-CV risk count is $4$ and $Δ N_{1} (2) = 1$ , giving increment $(1 / 4) (1 / 4) = 1 / 16$ and ${\hat{F}}_{1} (2) = 1 / 4 + 1 / 16 = 5 / 16$ . Thus higher cardiovascular mortality lowers the observed cancer cumulative incidence even though both groups have cause-1 event fractions $1 / 4$ at each event time.

3.1F1F3F4step 2.1givenalgebra∎

If cause-2 deaths are instead recoded as censorings for the cause-1 Kaplan–Meier calculation, the low-CV product is $(1 - 4 / 16) (1 - 2 / 8) = (3 / 4)^{2} = 9 / 16$ , and the high-CV product is $(1 - 4 / 16) (1 - 1 / 4) = (3 / 4)^{2} = 9 / 16$ . Both complements are $7 / 16$ , which differs from both observed cause-1 incidences, $3 / 8$ and $5 / 16$ . The cause-specific product-limit complement therefore does not equal the observed cause-specific cumulative incidence in these records.

Competing risks and cumulative incidence — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The observed first time and cause do not identify the latent joint failure-time law

Statement refuted

Facts & Assumptions

Counterexample

Higher competing mortality lowers incidence in a two-time table

Example

Facts & Assumptions

Verification

Sources