Jason Blevins’s Notebook
Heckman and Honoré (1989)

The Identifiability of the Competing Risks Model

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

Generalization: Mixed Proportional Hazards

Generalization: Accelerated hazards

S(tx)=exp[Z{tϕ(x)}]
S(t | x) = \exp\left[ -Z\lbrace t \phi(x) \rbrace \right]

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 1 n,η 2 n) is finite for all sequences η 1 n, η 2 n for which η 1 n1 and η 2 n1 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 (tx)=Pr(T 1 t,T 2 T 1 x)Q 2 (tx)=Pr(T 2 t,T 1 T 2 x).
Q_1(t|x) = \Pr( T_1 \ge t, T_2 \ge T_1 |x) \quad Q_2(t|x) = \Pr( T_2 \ge t, T_1 \ge T_2 |x).

Tsiatis (1975) establishes the following mappings:

Q 1 t(tx)=[St 1 ] t 1 =t 2 =tQ 2 t(tx)=[St 2 ] t 1 =t 2 =t.
\frac{\partial Q_1}{\partial t}(t|x) = \left[ \frac{\partial S}{\partial t_1} \right]_{t_1 = t_2 = t}\quad \frac{\partial Q_2}{\partial t}(t|x) = \left[ \frac{\partial S}{\partial t_2} \right]_{t_1 = t_2 = t}.

We have

Q 1 t(tx)=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).
\frac{\partial Q_1}{\partial t}(t | x) = -K_1\left[ \exp\left\lbrace -Z_1(t)\phi_1(x) \right\rbrace, \exp\left\lbrace -Z_2(t)\phi_2(x) \right\rbrace \right] \exp\left\lbrace -Z_1(t)\phi_1(x) \right\rbrace Z'_1(t)\phi_1(x).

Identification of ϕ j

Taking the ratio of Q 1 (tx)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 ).
\frac{ K_1\left[ \exp\left\lbrace -Z_1(t)\phi_1(x) \right\rbrace, \exp\left\lbrace -Z_2(t)\phi_2(x) \right\rbrace \right] \exp\left\lbrace -Z_1(t)\phi_1(x) \right\rbrace Z'_1(t)\phi_1(x) }{ K_1\left[ \exp\left\lbrace -Z_1(t)\phi_1(x_0) \right\rbrace, \exp\left\lbrace -Z_2(t)\phi_2(x_0) \right\rbrace \right] \exp\left\lbrace -Z_1(t)\phi_1(x_0) \right\rbrace Z'_1(t)\phi_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,tx) since S(t,tx)=Q 1 (tx)+Q 2 (tx). Furthermore,

S(t,tx)=K(exp[Z 1 (t)ϕ 1 (x)],exp[Z 2 (t)ϕ 2 (x)]).
S(t, t | x) = K\left( \exp\left[ -Z_1(t) \phi_1(x) \right], \exp\left[ -Z_2(t) \phi_2(x) \right] \right).

Setting t=1 gives

S(1 ,1 x)=K(exp[ϕ 1 (x)],exp[ϕ 2 (x)]).
S(1, 1 | x) = K\left( \exp[-\phi_1(x)], \exp[-\phi_2(x)] \right).

and letting ϕ 1 (x) and ϕ 2 (x) vary over (0 ,) 2 (by Assumption 3) yields K.

Identification of Z j

S(t,tx n)=K(exp[Z 1 (t)ϕ 1 (x n)],exp[Z 2 (t)ϕ 2 (x n)])
S(t,t|x_n) = K\left( \exp[-Z_1(t)\phi_1(x_n)], \exp[-Z_2(t)\phi_2(x_n)] \right)

Conclusion

Identification argument:

Implications of Nonparametric Identification: