From a Blipped-Down Time to an Estimating Equation
In a structural nested failure-time model, a candidate blip parameter tells us how to remove the treatment effect from an observed failure time. G-estimation searches for the candidate under which treatment no longer predicts the resulting transformed time, after accounting for treatment assignment given baseline covariates. Here the focus is the estimating equation created by that test, not the mechanics of constructing the failure-time blip.
Consider a single baseline binary treatment \(A\in\{0,1\}\), baseline covariates \(H\), and an uncensored observed failure time \(T\). Assume a rank-preserving accelerated failure-time model \(T=T^0\exp(\psi_0 A)\), where \(T^0\) is the failure time under no treatment and \(\psi_0\) is the blip parameter. For a candidate value \(\psi\), the transformed time is \(U(\psi)=T\exp(-\psi A)\). At the true value, \(U(\psi_0)=T^0\).
The indicator is candidate-dependent: changing \(\psi\) changes both the transformed time and who is retained. The equation uses the original sample size \(n\) as its denominator. Dividing by the number retained at a candidate value rescales the score by a positive factor whenever anyone is retained, so it does not change that candidate's zero; it does change the numerical score value.
Why the True Parameter Solves the Population Equation
The central identifying condition is conditional exchangeability: \(A\) is independent of \(T^0\) given \(H\). This is the baseline-treatment version of no unmeasured confounding. Assume consistency, the stated rank-preserving model, \(0<e(H)<1\) almost surely, and finite expectations for the score. At the true parameter, retention depends on \(T^0\), not on treatment once \(H\) and \(T^0\) are fixed. Consequently, treatment residuals have mean zero even among the retained individuals.
Proof. At \(\psi_0\), the model gives \(U(\psi_0)=T^0\), and hence \(R_{\psi_0}=I(T^0\leq\tau)\). Conditional on \(H,T^0\), exchangeability gives \(E[A\mid H,T^0]=E[A\mid H]=e(H)\). Therefore $$ \begin{aligned} &E\left[R_{\psi_0}\{A-e(H)\}g\{U(\psi_0),H\}\mid H,T^0\right]\\ &\quad=I(T^0\leq\tau)g(T^0,H) \left\{E[A\mid H,T^0]-e(H)\right\}=0. \end{aligned} $$ Taking expectations of both sides proves the result. Integrability ensures the expectation is well-defined. \(\square\)
This establishes that the true parameter is a population root; it does not establish that it is the only root. Identification also requires that other candidate values produce a nonzero population score, for the selected \(g\) and \(\tau\). A sample root is therefore meaningful only when the treatment model, blip model, and identifying moment together distinguish the target value.
Worked Examples
Worked Example: Solving a Score for an Acceleration Parameter
Suppose treatment is randomized with known \(e=0.5\), there is no right-censoring, and the observed data are:
| Individual | \(A\) | Observed \(T\) |
|---|---|---|
| 1 | 0 | 4 |
| 2 | 0 | 6 |
| 3 | 1 | 8 |
| 4 | 1 | 12 |
Use \(g(U,H)=U\), with no covariates, and \(\tau=100\). On the candidate range \(-1\leq\psi\leq1\), all four observations are retained: the largest transformed treated time is \(12\exp(1)<100\). The score is $$ \widehat S_4(\psi)=\frac{1}{4} \left[-\frac12(4)-\frac12(6)+\frac12(8e^{-\psi})+\frac12(12e^{-\psi})\right] =\frac{-5+10e^{-\psi}}{4}. $$ Setting the numerator to zero gives \(10e^{-\psi}=5\), so \(e^{-\psi}=1/2\) and \(\hat\psi=\log 2\). At this root, treated transformed times are \(8/2=4\) and \(12/2=6\), matching the two untreated times. This is a sample illustration of the moment equation, not a claim that four observations provide precise inference.
Worked Example: Checking Contributions Under Artificial Retention
Let \(e=0.5\), \(g(U)=U\), and suppose that at a particular candidate value the retained observations are as follows. The third individual is artificially censored before the chosen transformed-time horizon, so \(R_\psi=0\) for that individual.
| Individual | \(R_\psi\) | \(A\) | \(U(\psi)\) | \(R_\psi(A-0.5)U(\psi)\) |
|---|---|---|---|---|
| 1 | 1 | 0 | 2 | \(-1\) |
| 2 | 1 | 1 | 2 | \(1\) |
| 3 | 0 | 1 | 3 | \(0\) |
| 4 | 1 | 0 | 4 | \(-2\) |
The original-sample score is $$ \widehat S_4(\psi)=\frac{-1+1+0-2}{4}=-0.5. $$ Thus this candidate is not a root. If the average is instead taken over the three retained individuals, it is \((-1+1-2)/3=-2/3\), also not zero. Individual 4 contributes \((0-0.5)4=-2\); being retained does not make its contribution vanish. This check illustrates why each retained observation's treatment residual must be multiplied by its transformed time before summing.
Worked Example: Candidate-Dependent Retention at a Fixed Horizon
Suppose the true parameter is \(\psi_0=\log 2\), treatment is randomized with \(e=0.5\), and the untreated failure times are 3 and 8 in each treatment group. Under the model, untreated individuals have observed times 3 and 8, while treated individuals have observed times 6 and 16. Choose \(\tau=5\) and \(g(U)=U\).
At the true candidate, the untreated observed-time cutoff is 5 and the treated cutoff is \(5\exp(\log 2)=10\). Thus the observed failure at 3 is retained in each group, while the failure at 8 among untreated individuals and the failure at 16 among treated individuals are excluded. The retained transformed times are both 3: the untreated value is \(3\), and the treated value is \(6\exp(-\log 2)=3\). Their score contributions are \((0-0.5)3=-1.5\) and \((1-0.5)3=1.5\), whose sum is zero. Retention has selected failures by the same transformed-time horizon in both groups, as required at the true candidate.
Estimation, Diagnostics, and Failure Modes
In practice, compute the score over a scientifically plausible parameter range, updating \(U_i(\psi)\) and \(R_{\psi,i}\) at every candidate. For a scalar parameter, a sign change can help bracket a root, but a grid alone can miss a narrow or tangential root. A numerical root finder should be checked by substituting its answer back into the original score and reporting the score residual and the number retained. With a vector blip parameter or multiple instruments, solve the vector moment equations jointly or minimize a specified quadratic criterion.
A standard consistency argument needs more than the zero-mean theorem. Let \(S(\psi)\) denote the population score and suppose the parameter space is compact, \(S\) is continuous on the parameter space and has a unique zero \(\psi_0\), and the empirical score converges uniformly in probability to \(S\). Also assume a sample root exists with probability tending to one and that its score residual tends to zero. Then the sample root is consistent:
Proof. Fix any \(\epsilon>0\). By compactness and continuity, uniqueness of the zero implies that the minimum of \(|S(\psi)|\) over the compact set \(\{ \psi:|\psi-\psi_0|\geq\epsilon\}\) is some \(c_\epsilon>0\). Uniform convergence gives \(P(\sup_\psi|\widehat S_n(\psi)-S(\psi)|<c_\epsilon/3)\to1\). The approximate-root condition gives \(P(|\widehat S_n(\hat\psi)|<c_\epsilon/3)\to1\). On the intersection of these events, a value with \(|\hat\psi-\psi_0|\geq\epsilon\) would satisfy \(|S(\hat\psi)|\leq|\widehat S_n(\hat\psi)|+|\widehat S_n(\hat\psi)-S(\hat\psi)|<2c_\epsilon/3\), contradicting the definition of \(c_\epsilon\). Hence \(P(|\hat\psi-\psi_0|\geq\epsilon)\to0\), which is consistency. \(\square\)
This argument makes the main failure modes explicit. If \(S\) has several roots, the data and moment do not uniquely identify the parameter. If the score is nearly flat near its root, the estimate can be highly sensitive to small data or nuisance-model changes. If treatment propensities are near zero or one, the treatment residual has little variation in some covariate strata, weakening the comparison. Sparse retention at a candidate can also make the empirical score erratic; no retained observations means the score contains no information there.
The propensity must be appropriate for the conditioning set \(H\). Replacing \(e(H)\) with a constant when treatment depends on \(H\) can leave a treatment-covariate association in the residual score, even if the blip function is correct. When the propensity is estimated, its estimation method and uncertainty matter; the known-propensity equation above does not by itself justify treating \(\hat e\) as fixed. Likewise, a misspecified blip transformation may fail to remove the treatment association, and a poorly chosen \(g\) may have little power to detect the remaining association.
The method relies on a particular model and moment, rather than comparing treatment groups with an arbitrary survival summary. Alternative choices include using several functions \(g\) to impose more than one moment, or using other valid instruments based on treatment assignment. More moments can improve discrimination but also make the procedure sensitive to misspecification and require a rule for combining equations. The single-moment approach is simple, but it can have multiple roots or fail to detect dependence that the chosen \(g\) does not capture. The present derivation assumes uncensored failure times; ordinary right-censoring requires additional adjustment and cannot be handled merely by substituting a censored time for \(T\).
Common Pitfalls
- Keeping only the treatment residual in mind. The contribution is \(R_\psi(A-e(H))g(U(\psi),H)\), not just \(R_\psi(A-e(H))\). A retained individual with \(A=0\) and positive \(U\) generally contributes a negative value when \(e(H)>0\).
- Holding the retained set fixed while varying \(\psi\). Both the transformed time and the artificial retention indicator depend on the candidate. Reusing one retained set tests a different equation.
- Calling every zero the estimate. A root is interpretable as the target only with identification, correct treatment adjustment, and a suitable blip model. Check for additional roots and weakly varying scores.
- Ignoring the treatment model. The residual \(A-e(H)\) is designed to remove treatment association explained by \(H\). An incorrect propensity can undermine that interpretation.
Check Your Understanding
Try each question, then check the brief solution immediately below it.
- Under the rank-preserving model, what does \(U(\psi_0)\) equal, and why does that matter for the score?
- With \(e=0.4\), \(R_\psi=1\), \(A=0\), and \(U(\psi)=5\), calculate the contribution when \(g(U)=U\).
- Why does dividing a candidate's score by the number retained not change its zero, provided at least one observation is retained?
- Name two reasons that a population score can have a zero other than the true blip parameter.
Solutions. (1) It equals \(T^0\), so at the true candidate the transformed time is the untreated failure time; conditional exchangeability then gives the treatment-residual moment its zero expectation. (2) The contribution is \((0-0.4)5=-2\). (3) The retained-sample average is the original-sample score multiplied by \(n\) divided by the positive retained count, so it is zero exactly when the original score is zero. (4) The moment may not uniquely identify the parameter, or the chosen \(g\) may yield a zero despite remaining treatment dependence; model or propensity misspecification can also distort the score.