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Cause-specific cumulative incidence as an integral against its hazard measure

Statement

For every cause k∈{1,…,K} and every t≥0, the cumulative incidence decomposes as Fk(t)=∫[0,t]S(u−) dΛk(u). If νk has a Lebesgue density fk and S(u)>0 for every u∈(0,t], then dΛk(u)=λk(u) du for Lebesgue-a.e. u∈(0,t] with λk(u):=fk(u)/S(u), and Fk(t)=∫0tS(u)λk(u) du.

Facts & Assumptions

Given: A finite-cause competing-risk model as in Competing-risk data, cause-specific hazard measures, and cumulative incidence, its event subdistribution νk, survival function S, and cause-specific hazard measure Λk.

[F1]

Competing-risk data, cause-specific hazard measures, and cumulative incidence: Fk(t)=νk([0,t]), S(u−)=P(T≥u), and Λk=w νk with w(u)=1/S(u−) when S(u−)>0 and w(u)=0 otherwise.

[F2]

The measure with density f relative to μ: if w is nonnegative and measurable, then w νk is the measure defined by (w νk)(B)=∫Bw dνk.

[F3]

Integrating against a density agrees with integrating the product: for nonnegative measurable f,g, ∫g d(f dμ)=∫gf dμ.

[F4]

Every monotone real function is Borel measurable: every monotone real function is Borel measurable.

Proof

1.1F1givenalgebra

The set function νk(B)=P(T∈B,D=k) is a finite measure: it is the restriction of the law of (T,D) to the measurable cause slice D=k, so disjoint Borel sets have disjoint preimages and probability countable additivity applies.

2.1F1F4step 1.1

The function s(u):=S(u−)=P(T≥u) is nonincreasing, hence Borel by [F4]. Thus R:={u:s(u)>0} is Borel and w=s−11R is nonnegative Borel.

3.1F1step 2.1given

The set Rc is an upper interval. If Rc=[a,∞), then νk(Rc)≤P(T≥a)=s(a)=0. If Rc=(a,∞), then {T>a}=⋃m≥1{T≥a+1/m} and each event on the right has probability s(a+1/m)=0; hence again νk(Rc)=0. The empty case is immediate.

3.2F2step 2.1

By [F2], Λk=w νk is a measure.

4.1F1F3step 3.1step 3.2algebra

Applying [F3] with g(u)=1[0,t](u)s(u) gives ∫[0,t]s(u) dΛk(u)=∫[0,t]s(u)w(u) dνk(u)=∫[0,t]1R(u) dνk(u)=νk([0,t])=Fk(t). The penultimate equality uses step 3.1, so this proves the measure-valued identity, including atoms and t=0.

5.1F1F3givenstep 4.1algebra∎

Suppose νk(du)=fk(u) du and S(u)>0 for every u∈(0,t]. For each integer m≥1, the set Am:={u∈(0,t]:P(T=u)≥1/m} has at most m points, since distinct singleton events are disjoint and their probabilities sum to at most 1. Every positive atom belongs to some Am, so A:=⋃m≥1Am is countable and therefore Lebesgue-null. For 0<u≤t, the definitions s(u)=P(T≥u) and S(u)=P(T>u) give s(u)−S(u)=P(T=u); hence s(u)=S(u) outside A. Also s(u)≥S(u)>0, so w(u)=1/s(u) on (0,t]. Since dνk=fk du, [F1, F3] give dΛk=wfk du, whose density agrees with fk/S=λk Lebesgue-a.e. on (0,t]. Finally, νk({0})=0 because νk has a Lebesgue density, so the point 0 contributes nothing to either side of step 4.1; substituting the density on (0,t] into that identity gives the stated formula, including t=0.

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