On logistic regression with right censored data, with or without competing risks, and its use for estimating treatment effects

Abstract

Simple logistic regression can be adapted to deal with right-censoring by inverse probability of censoring weighting (IPCW). We here compare two such IPCW approaches, one based on weighting the outcome, the other based on weighting the estimating equations. We study the large sample properties of the two approaches and show that which of the two weighting methods is the most efficient depends on the censoring distribution. We show by theoretical computations that the methods can be surprisingly different in realistic settings. We further show how to use the two weighting approaches for logistic regression to estimate causal treatment effects, for both observational studies and randomized clinical trials (RCT). Several estimators for observational studies are compared and we present an application to registry data. We also revisit interesting robustness properties of logistic regression in the context of RCTs, with a particular focus on the IPCW weighting. We find that these robustness properties still hold when the censoring weights are correctly specified, but not necessarily otherwise.

Appendix A

1.1 Appendix A.1: Proof of Theorem 1

Overall, we follow similar lines as those of Bang and Tsiatis (2000). First, note that for all $s\in [0,t]$,

$$\begin{aligned} W(s) =\frac{\Delta (s)}{G_c(s \wedge {{\widetilde{T}}})} = 1 - \int _0^s \frac{dM^c(u)}{G_c(u)} \ , \end{aligned}$$

(15)

with $\Delta (s) = 1\!\! 1 \{ s \wedge T \le C \}$, $Y(u)=1\!\! 1 \{ {\widetilde{T}}\ge u \}$ and $M^c(u)=1\!\! 1 \{ {\widetilde{T}}\le u, \Delta =0 \} - \int _0^u Y(v)\lambda _c(v)dv$. The second equality in (15) has been pointed out by Robins and Rotnitzky (1992) for $s=\infty$ and it can be shown as follows for any $s\in [0,t]$ as follows.

$$\begin{aligned} 1 - \int _0^s \frac{dM^c(u)}{G_c(u)}&= 1 - \left\{ \frac{(1-\Delta )1\!\! 1 \{ {\widetilde{T}} \le s \}}{G_c({\widetilde{T}})} - \int _0^{{\widetilde{T}} \wedge s} \frac{\lambda _c(u)}{G_c(u)}du \right\} \\&= 1 - \left\{ \frac{(1-\Delta )1\!\! 1 \{ {\widetilde{T}} \le s \}}{G_c({\widetilde{T}})} - \left[ \frac{1}{G_c({\widetilde{T}} \wedge s)} - \frac{1}{G_c(0)} \right] \right\} \ , \end{aligned}$$

because

$$\begin{aligned} \frac{\lambda _c(u)}{G_c(u)}du= -\frac{dG_c(u)}{\{G_c(u)\}^2} \quad \text{ and } \quad \frac{d}{du}\left( \frac{1}{G_c(u)} \right) = -\frac{dG_c(u)}{\{G_c(u)\}^2} \ . \end{aligned}$$

Since $G_c(0)=1$, it follows

$$\begin{aligned} 1 - \int _0^s \frac{dM^c(u)}{G_c(u)} = \left\{ \begin{array}{ll} \Delta / G_c({\widetilde{T}}) &{} \text{ if } \quad {\widetilde{T}} \le s \\ 1 / G_c(s) &{} \text{ if } \quad {\widetilde{T}} > s\end{array} \right\} = W(s) \ . \end{aligned}$$

Second, recall a well-known martingale integral representation for the Kaplan-Meier estimator, see e.g. Andersen et al. (1993, Sec. IV.3), for all $s\in [0,t]$,

$$\begin{aligned} \frac{{\widehat{G}}_c(s) - G_c(s) }{ G_c(s)}&= - \sum _{i=1}^n \int _0^s \frac{{\widehat{G}}_c(u-)}{Y_\bullet (u)} \frac{dM^c_i(u)}{G_c(u)}, \end{aligned}$$

(16)

where $Y_\bullet (u)=\sum _{i=1}^n 1\!\! 1 \{ {\widetilde{T}}_i\ge u \}$. Third, note that

$$\begin{aligned} n^{-1}Y_\bullet (u)= {\widehat{G}}_c(u-) {\widehat{S}} (u-) \ , \end{aligned}$$

(17)

where ${\widehat{S}} (u-)$ is the Kaplan-Meier estimator for $S(u-)=P(T>u-)$. With the notations

$$\begin{aligned} {\widehat{U}}_{oipcw}(\varvec{\beta })&= n^{-1} \sum _{i=1}^n {\varvec{X}}_i \Big \{{\widehat{W}}_i(t) D_i(t) - Q(t,{\varvec{X}}_i,\varvec{\beta })\Big \} \\ \quad U_{oipcw}(\varvec{\beta })&= n^{-1} \sum _{i=1}^n {\varvec{X}}_i \Big \{ W_i(t) D_i(t) - Q(t,{\varvec{X}}_i,\varvec{\beta })\Big \} \\ \text{ and } \quad U_{glm}(\varvec{\beta })&= n^{-1} \sum _{i=1}^n {\varvec{X}}_i \Big \{ D_i(t) - Q(t,{\varvec{X}}_i,\varvec{\beta })\Big \} \ , \end{aligned}$$

we first note that

$$\begin{aligned} n^{1/2} {\widehat{U}}_{oipcw}(\varvec{\beta })&= \underbrace{n^{1/2} U_{oipcw}(\varvec{\beta })}_{(*)} \\&\quad + \underbrace{n^{-1/2} \sum _{j=1}^n {\varvec{X}}_j D_j(t) \frac{ \Delta _j(t) }{{\widehat{G}}_c(t \wedge \widetilde{T}_j)}\left\{ \frac{G_c(t \wedge {{\widetilde{T}}}_j) - {\widehat{G}}_c(t \wedge {{\widetilde{T}}}_j) }{ G_c(t \wedge {{\widetilde{T}}}_j)} \right\} }_{(**)} \ . \end{aligned}$$

From (15), we get

$$\begin{aligned} (*) =&\ n^{1/2} U_{glm}(\varvec{\beta }) - n^{-1/2} \sum _{i=1}^n \int _0^t {\varvec{X}}_i D_i(t) \frac{dM^c_i(u)}{G_c(u)} \ , \end{aligned}$$

and from (16), we get

$$\begin{aligned} (**)&= n^{-1/2} \sum _{j=1}^n {\varvec{X}}_j D_j(t) \frac{ \Delta _j(t) }{{\widehat{G}}_c(t \wedge {{\widetilde{T}}}_j)}\left\{ - \sum _{i=1}^n \int _0^{t \wedge {{\widetilde{T}}}_j} \frac{\widehat{G}_c(u-)}{Y_\bullet (u)} \frac{dM^c_i(u)}{G_c(u)} \right\} \\&= - n^{-1/2} \sum _{i=1}^n \int _0^t\left[ \frac{\widehat{G}_c(u-)}{Y_\bullet (u)} \sum _{j=1}^n {\varvec{X}}_j D_j(t) \frac{ \Delta _j(t) }{{\widehat{G}}_c(t \wedge {{\widetilde{T}}}_j)}1\!\! 1 \{ {{\widetilde{T}}}_j \ge u \} \right] \frac{ dM^c_i(u)}{G_c(u)} \end{aligned}$$

and from (17), we get

$$\begin{aligned} (**)&= - n^{-1/2} \sum _{i=1}^n \int _0^t {\widehat{E}}[{\varvec{X}}D(t)|T\ge u] \frac{ dM^c_i(u) }{G_c(u)} \end{aligned}$$

where

$$\begin{aligned} {\widehat{E}}[{\varvec{X}}D(t)|T>u]&= \frac{1}{n{\widehat{S}}(u-)}\sum _{i=1}^n \frac{ \Delta _i(t) 1\!\! 1 \{ {{\widetilde{T}}}_i \ge u \} {\varvec{X}}_i D_i(t) }{ {\widehat{G}}_c(t \wedge {{\widetilde{T}}}_i)}\\&= E\big [ {\varvec{X}}D(t) \, \big | \, T \ge u \big ] + o_p(1) \ , \end{aligned}$$

because of independent censoring and the uniform convergence of the Kaplan-Meier estimator ${\widehat{G}}_c(\cdot )$ in [0, t]. Consequently,

$$\begin{aligned} n^{1/2} {\widehat{U}}_{oipcw}(\varvec{\beta })&= \ n^{1/2} U_{glm}(\varvec{\beta }) - n^{-1/2} \\&\quad \sum _{i=1}^n \int _0^t \Big \{ {\varvec{X}}_i D_i(t) - {\widehat{E}}[{\varvec{X}}D(t)\, | \,T\ge u] \Big \} \frac{dM^c_i(u)}{G_c(u)} \\&= \ n^{1/2} U_{glm}(\varvec{\beta }) - n^{-1/2} \\&\quad \sum _{i=1}^n \int _0^t \Big \{ {\varvec{X}}_i D_i(t) - E[{\varvec{X}}D(t) \, | \, T\ge u] \Big \} \frac{dM^c_i(u)}{G_c(u)} \\&\quad + o_p(1) \end{aligned}$$

For $n^{1/2} {\widehat{U}}_{ipcw-glm}(\varvec{\beta })$, the result follows similarly, since the calculation is similar except from ${\varvec{X}}_i D_i(t)$ being replaced with ${\varvec{X}}_i \{ D_i(t) - Q(t,{\varvec{X}}_i)\}$ in $(*)$ and $(**)$.

1.2 Appendix A.2: variance estimator ${\widehat{\varvec{\Sigma }}}_m$

From the above derivations, the same calculations without using

$$\begin{aligned} n^{1/2} U_{oipcw}(\varvec{\beta }) =&\ n^{1/2} U_{glm}(\varvec{\beta }) - n^{-1/2} \sum _{i=1}^n \int _0^t {\varvec{X}}_i D_i(t) \frac{dM^c_i(u)}{G_c(u)} \end{aligned}$$

lead to $n^{1/2} {\widehat{U}}_{oipcw}(\varvec{\beta }) = n^{-1/2} \sum _{i=1}^n \varvec{\epsilon }_i+ o_p(1)$ where

$$\begin{aligned} \varvec{\epsilon }_i= {\varvec{X}}_i \Big \{ W_i(t) D_i(t) - Q(t,{\varvec{X}}_i,\varvec{\beta })\Big \} + \int _0^t E[{\varvec{X}}D(t) \, | \, T\ge u] \frac{dM^c_i(u)}{G_c(u)} \ . \end{aligned}$$

One can consistently estimate $\varvec{\epsilon }_i$ by

$$\begin{aligned} \widehat{\varvec{\epsilon }}_i= {\varvec{X}}_i \Big \{ {\widehat{W}}_i(t) {{\widetilde{D}}}_i(t)- Q(t,{\varvec{X}}_i,\widehat{\varvec{\beta }}_{oipcw})\Big \} + \int _0^t {\widehat{E}}[{\varvec{X}}D(t) \, | \, T\ge u] \frac{d\widehat{M}^c_i(u)}{{\widehat{G}}_c(u)} \ , \end{aligned}$$

where ${\widehat{M}}_i^c(s)=1\!\! 1 \{ {\widetilde{T}}_i \le s, \Delta _i =0 \} - \int _0^sY_i(v) d \widehat{\Lambda }_c(v)$, where ${\widehat{\Lambda }}_c(s)$ is the Nelson-Aalen estimator of the cumulative hazard of C at time s and

$$\begin{aligned} {\widehat{E}}[{\varvec{X}}D(t)|T>u]&= \frac{1}{n{\widehat{S}}(u-)}\sum _{i=1}^n \frac{ \Delta _i(t) 1\!\! 1 \{ {{\widetilde{T}}}_i \ge u \} {\varvec{X}}_i {\widetilde{D}}_i(t) }{ {\widehat{G}}_c(t \wedge {{\widetilde{T}}}_i)} \ . \end{aligned}$$

Consequently, one can consistently estimate $\varvec{\Omega }_{oipcw}$ by ${\widehat{\varvec{\Omega }}}_{oipcw}= (1/n) \sum _{i=1}^n\widehat{\varvec{\epsilon }}_i\widehat{\varvec{\epsilon }}_i^T$. Similarly, to estimate $\varvec{\Omega }_{ipcw-glm}$ we can use ${\widehat{\varvec{\Omega }}}_{ipcw-glm}= (1/n) \sum _{i=1}^n \widehat{\varvec{\omega }}_i \widehat{\varvec{\omega }}_i^T$ with

$$\begin{aligned} \widehat{\varvec{\omega }}_i&= \widehat{W}_i(t) {\varvec{X}}_i \Big \{ {{\widetilde{D}}}_i(t) - Q(t,{\varvec{X}}_i,\widehat{\varvec{\beta }}_{ipcw-glm})\Big \} \\&\quad + \int _0^t \widehat{E}\big [{\varvec{X}}\{D(t) - Q(t,{\varvec{X}}_i,\varvec{\beta })\} \, \big | \, T\ge u\big ] \frac{d\widehat{M}^c_i(u)}{{\widehat{G}}_c(u)} \end{aligned}$$

where

$$\begin{aligned}&{\widehat{E}}\big [{\varvec{X}}\{D(t) - Q(t,{\varvec{X}}_i,\varvec{\beta })\} \, \big | \, T\ge u\big ] \\&\quad = \frac{1}{n{\widehat{S}}(u-)}\sum _{i=1}^n \frac{ \Delta _i(t) 1\!\! 1 \{ {{\widetilde{T}}}_i \ge u \} {\varvec{X}}_i\big \{ {\widetilde{D}}_i(t) - Q(t,{\varvec{X}}_i,\widehat{\varvec{\beta }}_{ipcw-glm}) \big \} }{ {\widehat{G}}_c(t \wedge {{\widetilde{T}}}_i)} \ . \end{aligned}$$

Consistent estimators of $\varvec{{{\mathcal {I}}}}$ are ${\widehat{\varvec{{{\mathcal {I}}}}}}_m \!=\! n^{-1} \sum _{i=1}^n \left[ {\varvec{X}}_i^{2} Q(t,{\varvec{X}}_i,\widehat{\varvec{\beta }}_m) \left\{ 1 \!-\! Q(t,{\varvec{X}}_i,\widehat{\varvec{\beta }}_m) \right\} \right]$, for $m=oipcw$ or $ipcw-glm$. Finally, $\varvec{\Sigma }_m= \varvec{{{\mathcal {I}}}}^{-1}\varvec{\Omega }_m\varvec{{{\mathcal {I}}}}^{-1}$ can be estimated by ${\widehat{\varvec{\Sigma }}}_m= {\widehat{\varvec{{{\mathcal {I}}}}}}_m^{-1} \widehat{\varvec{\Omega }}_{m} {\widehat{\varvec{{{\mathcal {I}}}}}}_m^{-1}$.

1.3 Appendix A.3: Proof of Proposition 4

A proof was provided by Bartlett (2018, Appendix A.2), in a slightly different context. We repeat the main arguments here for completeness. First, we note that a Taylor-expansion and the results from Sects. 2.5 and 2.7 imply that

$$\begin{aligned} \sqrt{n}\left\{ \widehat{F_1}^g(t,a) - F_1(t,a)\right\} = \frac{1}{\sqrt{n}}\sum _{i=1}^n\Big \{ {Q}(t,a,{\varvec{L}}_i) - F_1(t,a) + {\varvec{B}}_{\varvec{\beta }}(a) \varvec{\psi }_i \Big \} + o_p(1) \end{aligned}$$

with $\varvec{\psi }_i=\varvec{{{\mathcal {I}}}}^{-1} \varvec{\Phi }\big \{{\varvec{X}}_i, D_i(t), {{\widetilde{T}}}_i, {\widetilde{\eta }}_i , t \big \}$. Here, $\varvec{\Phi }$ denotes $\varvec{\Phi }_m$ or $\varvec{\Phi }_m^{Aug}$, for $m=opicw$ or $ipcw-glm$, depending on which estimator ${\widehat{\varvec{\beta }}}$ we plugged-in to define ${\widehat{F_1}}^g(t,a)$. Formulas for the different versions of $\varvec{\Phi }$ can be found in Theorem 1 or in the proof of Proposition 2. Hence, it only remains to prove that $\text{ Cov }\Big \{ {Q}(t,a,{\varvec{L}}) \, , \, {\varvec{B}}_{\varvec{\beta }}(a) \varvec{\psi }\Big \}= 0$. This can be done by expanding the covariance, using the conditional covariance formula, as

$$\begin{aligned}&\text{ Cov }\Big \{ E\big [ {Q}(t,a,{\varvec{L}}) \, \big \vert \, {\varvec{X}}\big ] \, , \, E\big [ {\varvec{B}}_{\varvec{\beta }}(a) \varvec{\psi }\, \big \vert \, {\varvec{X}}\big ] \Big \} + E\Big [ \text{ Cov }\big \{ {Q}(t,a,{\varvec{L}}) \, , \, {\varvec{B}}_{\varvec{\beta }}(a) \varvec{\psi }\, \big \vert \, {\varvec{X}}\big \} \Big ] \end{aligned}$$

The second term is 0, because the first component is constant, conditional on ${\varvec{X}}$. We now explain why the first term is also 0. First note, that

$$\begin{aligned} E\Big [ {\varvec{B}}_{\varvec{\beta }}(a) \varvec{\psi }\, \Big \vert \, {\varvec{X}}\Big ]&={\varvec{B}}_{\varvec{\beta }}(a)\varvec{{{\mathcal {I}}}}^{-1} E\Big [ \varvec{\Phi }\{\dots \} \, \Big \vert \, {\varvec{X}}\Big ] \end{aligned}$$

and

$$\begin{aligned} E\Big [ \varvec{\Phi }\{\dots \} \, \Big \vert \, {\varvec{X}}\Big ]&= E\Big [ {\varvec{X}}\big \{ D(t) - Q(t,{\varvec{X}})\big \} - \int _0^t \varvec{\varphi }\big ({\varvec{X}}, D(t),s\big )\frac{dM^c(s)}{G_c(s)} \, \Big \vert \, {\varvec{X}}\Big ] \ . \end{aligned}$$

Second, note that $E\big [ {\varvec{X}}\big \{ D(t) - Q(t,{\varvec{X}})\big \} \, \big \vert \, {\varvec{X}}\big ]= {\varvec{X}}\big \{ F_1(t,{\varvec{X}}) - Q(t,{\varvec{X}})\big \} ={\varvec{0}}$ because the model is assumed to be well specified. Third, $E\big [ \int _0^t \varvec{\varphi }\big ({\varvec{X}}, D(t),s\big ) \frac{dM^c(s)}{G_c(s)} \, \big \vert \, {\varvec{X}}\big ]=0$ follows by independent censoring and standard martingale theory (Aalen et al. 2008, Sec. 2.2).

1.4 Appendix B

1.5 Sketch of Proof for Theorem 2

A proof was provided by Rosenblum and Steingrimsson (2016) for the uncensored case. We here essentially repeat their main arguments, which also apply in our case, up to minor differences introduced by the IPCW weights. First, note that ${\widehat{\varvec{\beta }}}_{oipcw}$ is consistent for $\varvec{\beta }$, where $\varvec{\beta }$ is the solution to

$$\begin{aligned} E\left[ {\varvec{X}}\Big \{ D(t) \cdot W(t,{\varvec{X}}) - Q(t,{\varvec{X}},\varvec{\beta })\Big \} \right] = {\varvec{0}} \ , \end{aligned}$$

(18)

with ${\varvec{X}}=(1,A,{\varvec{L}})^T$ and $W(t,{\varvec{X}})=\Delta (t)/G_C(t \wedge {{\widetilde{T}}},{\varvec{X}})$, using a notation that emphasizes the (potential) dependence of the IPCW weight on ${\varvec{X}}$. Second, note that the first equation in (18), which corresponds to the first component of ${\varvec{X}}=(1,A,{\varvec{L}})^T$, is

$$\begin{aligned} E\big [ D(t) W(t,{\varvec{X}}) \big ]&= E\big [ Q(t,{\varvec{X}},\varvec{\beta }) \big ] \nonumber \\&= \sum _{a \in \{0,1\}} E\Big [ Q\big (t,(1,A,{\varvec{L}})^T,\varvec{\beta }\big ) \, \big \vert \, A=a \Big ] \cdot \pi (a) \nonumber \\&= E\big [ Q\big (t,(1,1,{\varvec{L}})^T,\varvec{\beta }\big ) \big ]\cdot \pi (1) + E\left[ Q\big (t,(1,0,{\varvec{L}})^T,\varvec{\beta }\big ) \right] \cdot \pi (0) \end{aligned}$$

(19)

where $\pi (a)=P(A=a)$, for $a=0,1$, and where the last equality follows because A is assumed independent of ${\varvec{L}}$ (randomization), hence the conditioning disappears in the expectations. Similarly, the second equation in (18), which corresponds to the second component of ${\varvec{X}}=(1,A,{\varvec{L}})^T$, is

$$\begin{aligned} E\big [ A D(t) W(t,{\varvec{X}}) \big ]&= E\big [ A Q(t,{\varvec{X}},\varvec{\beta }) \big ] \nonumber \\&= E\big [ Q(t,(1,1,{\varvec{L}})^T,\varvec{\beta }) \big ] \cdot \pi (1) \ . \end{aligned}$$

(20)

Furthermore,

$$\begin{aligned} E\big [ D(t) W(t,{\varvec{X}}) \big ]=E\big [A D(t) W(t,{\varvec{X}}) \big ] + E\big [(1-A) D(t) W(t,{\varvec{X}}) \big ] \ . \end{aligned}$$

(21)

By plugging-in (19) and (20) into (21), we find

$$\begin{aligned} E\big [(1-A) D(t) W(t,{\varvec{X}}) \big ]= E\left[ Q\big (t,(1,0,{\varvec{L}})^T,\varvec{\beta }\big ) \right] \cdot \pi (0) \ . \end{aligned}$$

(22)

Further note that we also have the identities

$$\begin{aligned} E\big [A D(t) W(t,{\varvec{X}})\big ]&= E\big [ D(t) W(t,{\varvec{X}})| A=1 \big ] \cdot \pi (1) \end{aligned}$$

(23)

$$\begin{aligned} \text{ and } \quad E\big [(1-A) D(t) W(t,{\varvec{X}})\big ]&= E\big [ D(t) W(t,{\varvec{X}})| A=0 \big ] \cdot \pi (0) \ . \end{aligned}$$

(24)

Hence, Eqs. (23) and (20) give

$$\begin{aligned} E\big [ D(t) W(t,{\varvec{X}}) \, \big | \, A=1 \big ]=E\big [ Q\big (t,(1,1,{\varvec{L}})^T,\varvec{\beta }\big ) \big ] \ , \end{aligned}$$

(25)

and Eqs. (24) and (22) give

$$\begin{aligned} E\big [ D(t) W(t,{\varvec{X}}) \, \big | \, A=0 \big ]=E\big [ Q(t,(1,0,{\varvec{L}})^T,\varvec{\beta }) \big ] \ . \end{aligned}$$

(26)

The robustness result for the G-computation estimator, that is, $E\big [ Q(t,(1,a,{\varvec{L}}),\varvec{\beta }) \big ] = F_1(t,a)$, for $a=0,1$, therefore follows for OIPCW because $E\big [ D(t) W(t,{\varvec{X}})| A=a \big ]=F_1(t,a)$, when the censoring adjustment is correct.

To show that ${\widehat{\beta }}_A$ converges in probability towards zero if and only if $F_1(t,0) = F_1(t,1)$, we here repeat the argument of Rosenblum and Steingrimsson (2016). As, we have just shown that

$$\begin{aligned} F_1(t,0) - F_1(t,1)&= E\big [ \text{ expit }(\beta _0 + \varvec{\beta }_{{\varvec{L}}} {\varvec{L}}) -\text{ expit }(\beta _0 + \beta _A + \varvec{\beta }_{{\varvec{L}}} {\varvec{L}})\big ] \end{aligned}$$

and since $x\mapsto \text{ expit }(x)$ is monotonically increasing, then it follows that $F_1(t,0) - F_1(t,1)=0$ if and only if $\beta _A=0$. This completes the proof for the OIPCW approach.

The same arguments apply for IPCW-GLM too. Briefly, instead of the solution to (18), we consider the solution to

$$\begin{aligned} E\Big [ {\varvec{X}}W(t,{\varvec{X}}) \Big \{ D(t) - Q(t,\varvec{{\varvec{X}}},\varvec{\beta }) \Big \} \Big ] = {\varvec{0}} \ . \end{aligned}$$

Instead of (25) and (26), this leads to

$$\begin{aligned} E\big [D(t) W(t,{\varvec{X}}) \, \big | \, A=0 \big ]&= E\big [ W(t,{\varvec{X}}) Q\big (t,(1,0,{\varvec{L}})^T,\varvec{\beta }\big )\, \big | \, A=0 \big ] \end{aligned}$$

(27)

$$\begin{aligned} E\big [ D(t) W(t,{\varvec{X}}) \, \big | \, A=1 \big ]&= E\big [ W(t,{\varvec{X}}) Q(t,(1,1,{\varvec{L}})^T,\varvec{\beta })\, \big | \, A=1 \big ] \ . \end{aligned}$$

(28)

Note that in the expectations in (27) and (28) the conditioning does not vanish as in (26) and (25), because of the weight $W(t,{\varvec{X}})$. If the censoring adjustment is correct, then $E\left[ W(t,{\varvec{X}}) \, \big | \, A=a,{\varvec{X}}\right] =1$ for $a=0,1$, which together with the law of iterated expectations implies that the right-hand side of (27) and (28) are the same as that of (25) and (26). Hence (27) and (28) are equivalent to (26) and (25) and the rest of the proof is identical to that of OIPCW.

1.6 Sketch of Proof for Theorem 3

The proof exploits the three key equalities (29), (30) and (31) below. They are proven at the end of this section, after the proof of the theorem which exploits them. First, the stratified Kaplan-Meier estimator of $P(C>s|A=a)$, for $s\in [0,t]$, will converge to ${{\widetilde{G}}}_c(s,A)=\exp \left\{ -\int _0^s {\widetilde{\lambda }}_c(u,A) du\right\}$, where

$$\begin{aligned} {\widetilde{\lambda }}_c(u,A)=\lim _{h \rightarrow 0} \frac{1}{h} P\big ( C \in [u,u+h] \, \big \vert \, C> u, T>u, A\big ) = \widetilde{\lambda }_0(u) + A \lambda _1(u) \ , \end{aligned}$$

(29)

where ${\tilde{\lambda }}_0(u)$ does not depends on $(A,{\varvec{L}})$. Second,

$$\begin{aligned} E\Big [ {{\widetilde{W}}}(t,A) D(t) \Big | A=0\Big ] = E\Big [{{\widetilde{W}}}(t,A) D(t) \Big | A=1\Big ] \ , \end{aligned}$$

(30)

where ${{\widetilde{W}}}(t,A)=\Delta (t)/{{\widetilde{G}}}_c(t\wedge T,A)$, where the notation ${{\widetilde{W}}}(t,A)$ emphasizes both the mispecification of the censoring adjustment and the dependence on A. Third, for $a=0,1$,

$$\begin{aligned} E\left[ {{\widetilde{W}}}(t,A) Q\big (t,(1,a,{\varvec{L}})^T,\varvec{\beta }\big )\, \Big \vert \, A=a \right]&= E\Big [ r_c(T \wedge t,{\varvec{L}}) Q\big (t,(1,a,{\varvec{L}})^T,\varvec{\beta }\big ) \Big ] \ , \end{aligned}$$

(31)

where $r_c(T \wedge t,{\varvec{L}})\ge 0$ depends on $({\varvec{L}},T)$ but not on A (see Sect. 7.2.3 for details).

Other useful results are “new versions” of (25) and (26) (for OIPCW) and (27) and (28) (for IPCW-GLM). The equations of (25), (26), (27) and (28) still hold under the weaker assumptions of Theorem 2, up to the minor difference that $W(t,{\varvec{X}})$ should now be replaced by ${{\widetilde{W}}}(t,A)$. Indeed, (25), (26), (27) and (28) were derived without making any assumptions on the censoring adjustment. They were only consequences of the estimating equations. Below we refer to these “new versions” when we cite (25), (26), (27) or (28).

Let us now consider the OIPCW case. Equations (30), (25) and (26), imply

$$\begin{aligned} E\big [ Q(t,(1,1,{\varvec{L}})^T,\varvec{\beta }) \big ]=E\big [ Q(t,(1,0,{\varvec{L}})^T,\varvec{\beta }) \big ] \end{aligned}$$

which proves that ${\widehat{F_1}}^g(t,1)-\widehat{F_1}^g(t,0)$ convergences to 0, for OIPCW. Using the same monotonicity argument as in the Proof of Theorem 2, this implies that ${\widehat{\beta }}_A$ also convergences to 0.

Let us now consider the IPCW-GLM case. Equations (30), (27) and (28), imply

$$\begin{aligned} E\Big [ {{\widetilde{W}}}(t,A) Q\big (t,(1,1,{\varvec{L}})^T,\varvec{\beta }\big ) \, \Big \vert \, A=1 \Big ]=E\Big [ {{\widetilde{W}}}(t,A) Q\big (t,(1,0,{\varvec{L}})^T,\varvec{\beta }\big ) \, \Big \vert \, A=0 \Big ] \end{aligned}$$

which, using (31), further leads to

$$\begin{aligned} E\Big [ r_c(T \wedge t,{\varvec{L}}) \Big \{ \text{ expit }(\beta _0 + \varvec{\beta }_{{\varvec{L}}} {\varvec{L}}) -\text{ expit }(\beta _0 + \beta _A + \varvec{\beta }_{{\varvec{L}}} {\varvec{L}}) \Big \} \Big ]=0 \ . \end{aligned}$$

Because $r_c(T \wedge t,{\varvec{L}})> 0$ and $x\mapsto \text{ expit }(x)$ is monotonically increasing, then it follows that $\beta _A=0$, which proves that ${\widehat{\beta }}_A$ convergences to 0. This further implies that $E\big [ Q(t,(1,1,{\varvec{L}})^T,\varvec{\beta }) \big ]=E\big [ Q(t,(1,0,{\varvec{L}})^T,\varvec{\beta }) \big ]$, which proves that ${\widehat{F_1}}^g(t,1)-\widehat{F_1}^g(t,0)$ convergences to 0.

1.6.1 Proof of (29)

By construction, the stratified Kaplan-Meier estimator of $P(C>s|A=a)$, for $s\in [0,t]$, will converge to $\widetilde{G}_c(s,A)=\exp \left\{ - \int _0^s {\widetilde{\lambda }}_c(u,A) du\right\}$, where

$$\begin{aligned} {\widetilde{\lambda }}_c(u,A)&= \lim _{h \rightarrow 0} \frac{1}{h} P\big ( C \in [u,u+h] \, \big \vert \, C> u, T>u, A\big ) \\&= \frac{ \lim _{h \rightarrow 0} \frac{1}{h} P\big ( C \in [u,u+h], T>u \, \vert \, A\big ) }{ P( C>u, T>u \, \vert \, A)} \end{aligned}$$

First, let’s look at the numerator,

$$\begin{aligned}&\lim _{h \rightarrow 0} \frac{1}{h} P\big ( C \in [u,u+h], T>u \, \vert \, A\big )\\&\quad = \lim _{h \rightarrow 0} \frac{1}{h} E \Big [ 1\!\! 1 \{ C \in [u,u+h] \}\cdot 1\!\! 1 \{ T>u \} \, \Big \vert \, A\Big ]\\&\quad = \lim _{h \rightarrow 0} \frac{1}{h} E \Big [ E \Big ( 1\!\! 1 \{ C \in [u,u+h] \}\cdot 1\!\! 1 \{ T>u \} \, \Big \vert \, A, {\varvec{L}}\Big ) \, \Big \vert \, A \Big ]\\&\quad = E \Big [ \lambda _c(u,A,{\varvec{L}}) G_c(u,A,{\varvec{L}}) S(u,{\varvec{L}}) \, \Big \vert \, A \Big ] \quad \text{ as } \quad C \perp \!\!\! \perp T \, \vert \, (A,{\varvec{L}}) \ , \end{aligned}$$

where $G_c(u,A,{\varvec{L}})=P(C>u \vert A,{\varvec{L}})$ and $S(u,{\varvec{L}})= 1- F(u,A,{\varvec{L}}) - F_2(u,A,{\varvec{L}})$, which depends on ${\varvec{L}}$ but not on A, by the assumption of no conditional treatment effects. This further leads to

$$\begin{aligned}&\lim _{h \rightarrow 0} \frac{1}{h} P\big ( C \in [u,u+h], T>u \, \vert \, A\big )\\&\quad = E \Big [ \big \{ \lambda _0(t)+A \lambda _1(t) + \lambda _2(t,{\varvec{L}})\big \} e^{-\int _0^u \big \{\lambda _0(s) + A\lambda _1(s)+\lambda _2(s,{\varvec{L}})\big \}ds} S(u,{\varvec{L}}) \, \Big \vert \, A \Big ] \\&\quad = \tau (A,u)\Big [ \ \big \{ \lambda _0(t)+A \lambda _1(t) \big \} \gamma (u) \ + \ \xi (u) \ \Big ] \end{aligned}$$

with

$$\begin{aligned} \tau (A,u)&= e^{-\int _0^u \big \{\lambda _0(s) + A\lambda _1(s)\big \}ds} \\ \gamma (u)&= E \Big [ e^{-\int _0^u \lambda _2(s,{\varvec{L}})ds} S(u,{\varvec{L}}) \Big ] \\ \xi (u)&= E \Big [e^{-\int _0^u \lambda _2(s,{\varvec{L}})ds} S(u,{\varvec{L}}) \lambda _2(t,{\varvec{L}}) \Big ] \end{aligned}$$

where the conditioning on A in the expectations vanishes because we assumed $A\perp \!\!\! \perp L$ (randomization). Similarly, it follows $P( C>u, T>u \, \vert \, A)= \tau (A,u) \gamma (u)$. Hence, $\widetilde{\lambda }_c(u,A) = {\widetilde{\lambda }}_0(u) + A \lambda _1(u)$ with ${\widetilde{\lambda }}_0(u)= \lambda _0(u) + \xi (u)/\gamma (u)$.

1.6.2 Proof of (30)

Using result (29) and the above definition of $\widetilde{G}_c(t,A)$, we now show (30). First, we note that result result (29) implies that the ratio $r_c(t,{\varvec{L}}) = G_c(t,A,{\varvec{L}})/{{\widetilde{G}}}_C(t,A) = \exp [ - \int _0^t \{\lambda _0(s) + \lambda _2(s,{\varvec{L}}) - {\widetilde{\lambda }}_0(s)\}ds ]$ does not depend on A, but on ${\varvec{L}}$ and t (hence the notation $r_c(t,{\varvec{L}})$). Therefore, we have

$$\begin{aligned} E\Big [ {{\widetilde{W}}}(t,A) D(t) \Big | A=0\Big ]&= E\Big [ E \Big \{ \frac{1\!\! 1 \{ T\wedge t \le C \}}{{{\widetilde{G}}}_c(t\wedge T,A)} D(t) \Big | T, \eta ,{\varvec{L}}, A=0 \Big \}\Big | A=0\Big ] \\&= E\left[ \frac{E \Big \{ 1\!\! 1 \{ t\wedge T \le C \} \Big | T, \eta ,{\varvec{L}}, A=0 \Big \}}{{{\widetilde{G}}}_C(T\wedge t,0)} D(t) \Big | A=0\right] \\&= E\left[ r_c(T \wedge t,{\varvec{L}}) D(t) \, \Big | \, A=0\right] \quad \text{ as }\quad C \perp \!\!\! \perp (T,\eta ) \vert (A,{\varvec{L}}) \\&= E\left[ E\Big \{ \int _0^t r_c(s,{\varvec{L}}) dD(s)\, \Big | \, {\varvec{L}}, A=0\Big \} \Big | A=0\right] \\&= E\left[ \int _0^t r_c(s,{\varvec{L}}) E\left\{ dD(s) \, \Big | \, {\varvec{L}}, A=0\right\} \Big | A=0\right] \\&= E\left[ \int _0^t r_c(s,{\varvec{L}}) dF_1(t,0,{\varvec{L}}) \, \Big | \, A=0\right] \\&= E\left[ \int _0^t r_c(s,{\varvec{L}}) dF_1(t,1,{\varvec{L}}) \, \Big | \, A=0\right] \\&\qquad \quad \text{(no } \text{ conditional } \text{ treatment } \text{ effect) }\\&= E\left[ \int _0^t r_c(s,{\varvec{L}}) dF_1(t,1,{\varvec{L}}) \, \Big | \, A=1\right] \\&\quad \qquad \text{(randomization, } \text{ i.e., } A \perp \!\!\! \perp {\varvec{L}})\\&= E\Big [ {{\widetilde{W}}}(t,A) D(t) \, \Big | \, A=1\Big ] \end{aligned}$$

1.6.3 Proof of (31)

As in the Proof of (30), here again we use that (29) implies that $r_c(t,{\varvec{L}}) = G_c(t,A,{\varvec{L}})/{{\widetilde{G}}}_C(t,A)$ does not depend on A, but on ${\varvec{L}}$ and t. Therefore,

$$\begin{aligned}&E\left[ {{\widetilde{W}}}(t,A) Q(t,(1,a,{\varvec{L}})^T,\varvec{\beta })\, \vert \, A=a \right] \\&\quad = E\Big [ E \Big \{ \frac{1\!\! 1 \{ T\wedge t \le C \}}{{{\widetilde{G}}}_c(t\wedge T,A)} Q(t,(1,a,{\varvec{L}})^T,\varvec{\beta })\Big | T, {\varvec{L}},A \Big \}\Big | A=a\Big ] \\&\quad = E\left[ \frac{E \Big \{ 1\!\! 1 \{ t\wedge T \le C \} \Big | T, {\varvec{L}},A \Big \}}{{{\widetilde{G}}}_C(T\wedge t,A)} Q(t,(1,a,{\varvec{L}})^T,\varvec{\beta }) \Big | A=a\right] \\&\quad = E\left[ r_c(T \wedge t,{\varvec{L}}) Q(t,(1,a,{\varvec{L}})^T,\varvec{\beta }) \Big | A=a\right] \qquad \text{ as }\quad C \perp \!\!\! \perp T \vert (A,{\varvec{L}}) \\&\quad = E\big [ r_c(T \wedge t,{\varvec{L}}) Q(t,(1,a,{\varvec{L}})^T,\varvec{\beta }) \big ] \ , \end{aligned}$$

where the last equality follows from the randomization assumption $A \perp \!\!\! \perp {\varvec{L}}$ and the assumption of no conditional treatment effect, which further implies $A \perp \!\!\! \perp T$.