Accelerated failure time vs Cox proportional hazards mixture cure models: David vs Goliath?

Abstract

A mixture cure model relies on a model for the cure probability and a model for the survival function of the uncured subjects. For the latter, one often uses a Cox proportional hazards model. We show the identifiability of this model under weak assumptions. The model assumes that the cure threshold is the same for all values of the covariates, which might be unrealistic in certain situations. An alternative mixture cure model is the accelerated failure time (AFT) model. We also show the identifiability of this model under minimal assumptions. The cure threshold in this model depends on the covariates, which often leads to a better fit of the data. This is especially true when the follow-up period is insufficient for certain values of the covariates. We study these two models via simulations both when the follow-up is sufficient and when it is insufficient. Moreover, the two models are applied to data coming from a breast cancer clinical trial. We show that the AFT and the Cox model both fit the data well in the region of sufficient follow-up, but differ drastically outside that region.

Appendix: Proofs of main results

In this Appendix, we provide the proofs of Propositions 1 and 2.

Proof of Proposition 1

We need to show that if

$$\begin{aligned}&\Big \{p(z) f_u(y| x)\Big \}^{\delta } \ \Big \{1 - p(z) + p(z)S_u(y| x)\Big \}^{1 - \delta } \nonumber \\&\quad = \Big \{{\tilde{p}}(z) {\tilde{f}}_u(y|x) \Big \}^{\delta } \ \Big \{1 - {\tilde{p}}(z) + {\tilde{p}}(z) \tilde{S}_u(y|x)\Big \}^{1 - \delta } \end{aligned}$$

(7.1)

for all realizations $(y, \delta , x, z)$ of $(Y, \Delta , X, Z)$, then ${\gamma }=\tilde{{\gamma }}, {\beta }=\tilde{{\beta }}$ and $S_0 \equiv {\tilde{S}}_0$. Here, $p(z) = \exp (\gamma ^t z) / [1+\exp (\gamma ^t z)]$, ${\tilde{p}}(z) = \exp ({\tilde{\gamma }}^t z) / [1+\exp ({\tilde{\gamma }}^t z)]$, $S_u(y| x) = S_0(y)^{\exp (\beta ^t x)}$, ${\tilde{S}}_u(y| x) = {\tilde{S}}_0(y)^{\exp ({\tilde{\beta }}^t x)}$, and $f_u(y|x)$ and ${\tilde{f}}_u(y|x)$ are the corresponding probability density functions.

First, consider $y>\tau$ and $\delta =0$. Note that

$$\begin{aligned} P(Y> \tau , \Delta =0 \ | \ X, Z)= & {} P(C>\tau , C\le T \ | \ X, Z) \\= & {} P(C>\tau , T=\infty \ | \ X, Z) \\= & {} P(C>\tau \ | \ X, Z) \ (1-p(Z)) >0, \end{aligned}$$

thanks to (A)(i), (B)(i) and (B)(ii). Hence, $(y>\tau , \delta =0, x, z)$ is a possible realization of $(Y,\Delta , X, Z)$. Equation (7.1) reduces in this case to $1 - p(z) = 1 - \tilde{p}(z)$. It follows that $\gamma \equiv {\tilde{\gamma }}$, since $\hbox {Var}(Z)$ is positive definite by assumption (A)(ii).

Next, consider any $0 < y \le \tau$. Then,

$$\begin{aligned} P(Y \le y, \Delta =1 \ | \ X, Z)= & {} P(T\le y, T \le C \ | \ X, Z) \nonumber \\= & {} \int _0^y P(C > t \ | \ X,Z) \, d P(T \le t \ | \ X,Z). \end{aligned}$$

(7.2)

Hence, the corresponding sub-density equals $p(Z) P(C > y \ | \ X,Z) f_u(y|X)$. Since the support of $f_u(\cdot |X)$ is $[0,\tau ]$ and since $p(Z) P(C> \tau \ | \ X,Z) >0$ thanks to assumptions (A)(i), (B)(i) and (B)(ii), it follows that the support of Y when $\Delta =1$ and given X and Z is also $[0,\tau ]$. For any $0 < y \le \tau$ and $\delta =1$, the likelihood contribution in (7.1) is such that $p(Z) f_u(y| X) = {\tilde{p}}(Z) \tilde{f}_u(y| X)$. Since $p \equiv {\tilde{p}} >0$, it follows that $f_u(y| X)=\tilde{f}_u(y| X)$. Since this is true for all $0 < y \le \tau$, it follows that $S_u(y|X)={\tilde{S}}_u(y|X)$. Hence,

$$\begin{aligned} S_0(y)^{\exp (\beta ^t X)} = {\tilde{S}}_0(y)^{\exp ({\tilde{\beta }}^t X)}. \end{aligned}$$

Taking a logarithmic transformation at both sides we get

$$\begin{aligned} \exp (\beta ^t X) \log S_0(y) = \exp ({\tilde{\beta }}^t X) \log {\tilde{S}}_0(y), \end{aligned}$$

or equivalently, $(\beta -{\tilde{\beta }})^t X = \log \big [\log \tilde{S}_0(y) / \log S_0(y)\big ]$ for all $0 < y \le \tau$. It follows that

$$\begin{aligned} (\beta -{\tilde{\beta }})^t \text{ Var }(X) (\beta -{\tilde{\beta }}) = 0. \end{aligned}$$

Since Var(X) is positive definite by assumption (A)(ii), this is only possible if $\beta ={\tilde{\beta }}$, which implies that $S_0 \equiv {\tilde{S}}_0$. $\square$

Proof of Proposition 2

We need to show that if

$$\begin{aligned}&\Big \{p(z) f_u(y|x)\Big \}^{\delta } \ \Big \{1 - p(z) + p(z)S_u(y| x)\Big \}^{1 - \delta } \nonumber \\&\quad = \Big \{{\tilde{p}}(z) {\tilde{f}}_u(y|x) \Big \}^{\delta } \ \Big \{1 - {\tilde{p}}(z) + {\tilde{p}}(z) \tilde{S}_u(y|x)\Big \}^{1 - \delta } \end{aligned}$$

(7.3)

for all realizations $(y, \delta , x, z)$ of $(Y, \Delta , X, Z)$, then ${\gamma }=\tilde{{\gamma }}, {\beta }=\tilde{{\beta }}$ and $S_0 \equiv {\tilde{S}}_0$. Here, $p(z) = \exp (\gamma ^t z) / [1+\exp (\gamma ^t z)]$, ${\tilde{p}}(z) = \exp ({\tilde{\gamma }}^t z) / [1+\exp ({\tilde{\gamma }}^t z)]$, $S_u(y| x) = S_0(y \exp (-\beta ^t x))$, ${\tilde{S}}_u(y| x) = {\tilde{S}}_0(y \exp (-{\tilde{\beta }}^t x))$, and $f_u(y|x)$ and ${\tilde{f}}_u(y|x)$ are the corresponding probability density functions.

First, consider any $0 < y \le \tau _0 \exp (\beta ^t x)$ with $(x,z) \in S$. Then, we know from (7.2) that the conditional sub-density of the uncensored Y-values equals $p(z) P(C> y \ | \ X=x,Z=z) f_u(y|x) = p(z) P(C > y \ | \ X=x,Z=z) f_0(y \exp (-\beta ^t x)) \exp (-\beta ^t x)$. Since the support of $f_0(\cdot )$ is $[0,\tau _0]$ and since $p(z) P(C> \tau _0 \exp (\beta ^t x) \ | \ X=x,Z=z) >0$ thanks to assumptions (A)(i), (C)(i) and (C)(ii), it follows that the support of Y when $\Delta =1$, $X=x$ and $Z=z$ is $[0,\tau _0 \exp (\beta ^t x)]$. Hence, $\beta ^t x = {\tilde{\beta }}^t x$, which is only possible if $\beta ={\tilde{\beta }}$ since Var$(X|X \in S_X)>0$.

Next, consider $y> \tau _0 \exp (\beta ^t x)$ and $\delta =0$ with $(x,z) \in S$. Note that

$$\begin{aligned}&P(Y> \tau _0\exp (\beta ^t x), \Delta =0 \ | \ X=x, Z=z) \\&\quad = P(C> \tau _0\exp (\beta ^t x), C\le T \ | \ X=x, Z=z) \\&\quad = P(C> \tau _0\exp (\beta ^t x), T=\infty \ | \ X=x, Z=z) \\&\quad = P(C> \tau _0\exp (\beta ^t x) \ | \ X=x, Z=z) \ (1-p(z)) >0, \end{aligned}$$

thanks to (A)(i), (C)(i) and (C)(ii). Hence, $(y> \tau _0\exp (\beta ^t x), \delta =0, x, z)$ is a possible realization of $(Y,\Delta , X, Z)$. Equation (7.3) reduces in this case to $1 - p(z) = 1 - {\tilde{p}}(z)$. It follows that $\gamma = {\tilde{\gamma }}$ since $\hbox {Var}(Z|Z \in S_Z)>0$.

Finally, for any $0 < y \le \tau _0 \exp (\beta ^t x)$ and $\delta =1$, the likelihood contribution in (7.3) is such that $p(z) f_u(y|x) = {\tilde{p}}(z) \tilde{f}_u(y| x)$. Since $p(z) = {\tilde{p}}(z) >0$, it follows that $f_u(y|x)={\tilde{f}}_u(y| x)$. Since this is true for all $0 < y \le \tau _0 \exp (\beta ^t x)$, it follows that $S_u(y| x)={\tilde{S}}_u(y| x)$ for all $0 < y \le \tau _0 \exp (\beta ^t x)$, and hence $S_0 \equiv {\tilde{S}}_0$. $\square$