Heckman and Honoré (1989)

The Identifiability of the Competing Risks Model

These notes are based on the following article:

Heckman, James J. and Bo E. Honoré (1989). The identifiability of the competing risks model. Biometrika, 76: 325–330.

The Classical Competing Risks Model

Cox and Tsiatis Nonidentification Theorem

Importance of Dependence

Overview

Proportional Hazards Model

Proportional Hazards and Competing Risks

Introducing Dependence

S j(t|x)=exp{Z j(t)ϕ j(x)}.

S(t 1,t 2|x)=K[exp{Z 1(t)ϕ 1(x)},exp{Z 2(t)ϕ 2(x)}].

(1)S(t 1,t 2|x)=K(exp[Z 1(t 1)ϕ 1(x)],exp[Z 2(t 2)ϕ 2(x)]).

Generalization: Mixed Proportional Hazards

S(t 1,t 2|x)= Ωexp[Z 1(t 1)e xβ 1+c 1ω]exp[Z 2(t 2)e xβ 2+c 2ω]dG(ω).

K(η 1,η 2)= Ωη 1 exp(c 1ω)η 2 exp(c 2ω)dG(ω).

Generalization: Accelerated hazards

S(t|x)=exp[Z{tϕ(x)}]

S(t 1,t 2|x)=K(exp[Z 1{t 1ϕ 1(x)}],exp[Z 2{t 2ϕ 2(x)}]).

Identification Theorem

Assume that (T 1,T 2) has joint distribution (1). Then Z 1, Z 2, ϕ 1, ϕ 2, and K are identified from the minimum of (T 1,T 2) under the following assumptions:

  1. K is continuously differentiable with partial derivatives K 1 and K 2 and for i=1,2, lim nK i(η 1n,η 2n) is finite for all sequences η 1n, η 2n for which η 1n1 and η 2n1 for n. We also assume that K is strictly increasing in each of its arguments.

  2. Z 1(1)=Z 2(1)=1 and ϕ 1(x 0)=ϕ 2(x 0)=1 for some x 0.

  3. The support of {ϕ 1(x),ϕ 2(x)} is (0,)×(0,).

  4. Z 1 and Z 2 are nonnegative, differentiable, strictly increasing functions, except that we allow them to be infinite for finite t.

Notes about these assumptions:

  1. K is already weakly increasing.

  2. This is an innocuous normalization since ϕ j(x) and Z j(t) are not jointly identified to scale.

  3. This is satisfied, for example, when ϕ j(x)=exp(xβ j) and there is a common covariate with support and different coefficients.

Mapping Observables to Unobservables

Observed distributions:

Q 1(t|x)=Pr(T 1t,T 2T 1|x)Q 2(t|x)=Pr(T 2t,T 1T 2|x).

Tsiatis (1975) establishes the following mappings:

Q 1t(t|x)=[St 1] t 1=t 2=tQ 2t(t|x)=[St 2] t 1=t 2=t.

We have Q 1t(t|x)=K 1[exp{Z 1(t)ϕ 1(x)},exp{Z 2(t)ϕ 2(x)}]exp{Z 1(t)ϕ 1(x)}Z 1(t)ϕ 1(x).

Identification of ϕ j

Taking the ratio of Q 1(t|x)t at x and x 0 yields

K 1[exp{Z 1(t)ϕ 1(x)},exp{Z 2(t)ϕ 2(x)}]exp{Z 1(t)ϕ 1(x)}Z 1(t)ϕ 1(x)K 1[exp{Z 1(t)ϕ 1(x 0)},exp{Z 2(t)ϕ 2(x 0)}]exp{Z 1(t)ϕ 1(x 0)}Z 1(t)ϕ 1(x 0).

Taking t0 and using the normalization yields ϕ 1(x). Our choice of x was arbitrary so ϕ 1(x) is identified on the entire support of X. Similarly for ϕ 2(x).

Identification of K

We know S(t,t|x) since S(t,t|x)=Q 1(t|x)+Q 2(t|x). Furthermore, S(t,t|x)=K(exp[Z 1(t)ϕ 1(x)],exp[Z 2(t)ϕ 2(x)]).

Setting t=1 gives S(1,1|x)=K(exp[ϕ 1(x)],exp[ϕ 2(x)]). and letting ϕ 1(x) and ϕ 2(x) vary over (0,) 2 (by Assumption 3) yields K.

Identification of Z j

S(t,t|x n)=K(exp[Z 1(t)ϕ 1(x n)],exp[Z 2(t)ϕ 2(x n)])

Conclusion

Identification argument:

Implications of Nonparametric Identification: