Deriving the Failure-Time Blip Transformation

Why transform failure time?

A failure-time structural nested model describes how a treatment decision changes the remaining time to failure. Its blip transformation asks a focused question: if a proposed treatment effect were removed from the observed failure time, what failure time would remain? That transformed time is useful because a correct candidate blip should leave a time corresponding to the outcome with the relevant treatment effect removed.

The direction of the transformation matters. If treatment multiplies remaining survival time by 1.5, then removing its effect means dividing the observed remaining time by 1.5, not multiplying by 1.5. With repeated decisions, each treatment changes the remaining time from its own decision point onward. The transformations must therefore be applied backward, beginning with the last treatment.

One decision: deriving the inverse blip

Suppose treatment is assigned at time \(t\), while the individual is still alive. Let \(T\) denote the observed failure time. Write \(T^0\) for the failure time under no treatment at this decision, holding the relevant past history fixed. Let \(A\) be the treatment level, and let \(b(A,H;\psi)\) be its log acceleration, possibly depending on the history \(H\) available at the decision and parameter \(\psi\). The structural model for the remaining survival time is

T=t+\exp\{b(A,H;\psi)\}(T^0-t).

This is a model for time remaining after \(t\), rather than a multiplier applied to the entire calendar failure time. A positive value of \(b\) accelerates the remaining time by a factor greater than one and therefore lengthens survival; a negative value shortens it. Solving the equation for \(T^0\) gives the inverse, or blipped-down, time:

T^0=t+\{T-t\}\exp\{-b(A,H;\psi)\}.

At baseline, \(t=0\), this reduces to \(T^0=T\exp\{-b(A,H;\psi)\}\). If no treatment is given and the model has \(b(0,H;\psi)=0\), the transformation leaves the observed time unchanged.

Definition: The failure-time blip transformation for one treatment decision at time \(t\) is \(t+(T-t)\exp\{-b(A,H;\psi)\}\). It removes the candidate treatment acceleration from the observed remaining survival time. The resulting value is a candidate untreated failure time for that decision, not automatically an identified or observed counterfactual outcome.

The model requires a positive acceleration factor, ensured by the exponential form. It also requires that the treatment decision occurs before failure, so the observed remaining time \(T-t\) is positive. To interpret the transformed value as a failure time on the relevant post-decision scale, it must be at least \(t\). That interpretation can fail for some candidate parameters and data values; an algebraic transformation can still be computed, but it does not then describe a feasible post-decision failure time.

Worked Example: Removing a Baseline Acceleration

Suppose treatment is assigned at baseline, the observed failure time is \(T=12\) years, and the proposed log acceleration is \(b= \log(1.5)\). Under this candidate, treatment multiplies survival time by \(\exp(b)=1.5\). The blipped-down time is

T^0=12\exp\{-\log(1.5)\} =\frac{12}{1.5}=8\text{ years}.

Checking in the forward direction, \(8\times1.5=12\) years, which recovers the observed time. Multiplying 12 by 1.5 would instead give 18 years and would apply the acceleration again rather than remove it.

Repeated decisions: backward recursion

Now suppose there are \(K\) treatment decisions at ordered times \(t_1<\cdots<t_K\). Let \(A_j\) be the treatment at time \(t_j\), and let \(H_j\) be the history available just before that decision. The candidate log acceleration at decision \(j\) is \(b_j=b(A_j,H_j;\psi)\). Assume each decision is made before failure and that the structural model applies to the remaining time after that decision.

Start with \(U_K=T\), the observed failure time after all decisions. Undo the last treatment first. This yields a time \(U_{K-1}\) with the last blip removed. Then undo the treatment at \(t_{K-1}\), using \(U_{K-1}\) as the failure time entering that step. Continue backward until all specified blips have been removed.

U_K=T,\qquad U_{j-1}=t_j+(U_j-t_j)\exp(-b_j), \quad j=K,K-1,\ldots,1.

Here \(U_0\) is the fully blipped-down time. The recursion is not obtained by multiplying the observed \(T\) by every inverse acceleration: each decision acts on a remaining interval whose starting point is that decision time.

Result: Under the stated structural model, backward recursion removes the candidate blips one at a time. At step \(j\), \(U_j\) is the time entering that step and \(U_{j-1}\) is the time after removing the blip at \(t_j\). The recursion gives the candidate failure time with all \(K\) modeled treatment effects removed.

Worked Example: Two Treatment Decisions

Suppose decisions occur at \(t_1=2\) and \(t_2=5\) years. The observed failure time is \(T=11\) years. The candidate acceleration factors are 2 at the first decision and 1.5 at the second, so \(b_1=\log 2\) and \(b_2=\log 1.5\). Start at the final observed time and remove the later blip:

U_2=11,\qquad U_1=5+(11-5)\exp\{-\log(1.5)\} =5+\frac{6}{1.5}=9.

Now remove the earlier blip using the transformed time \(U_1=9\):

U_0=2+(9-2)\exp\{-\log 2\} =2+\frac{7}{2}=5.5.

Check by applying the forward accelerations in chronological order. Starting from \(U_0=5.5\), the time after the first treatment is \(2+2(5.5-2)=9\). The time after the second is \(5+1.5(9-5)=11\), the observed value. The calculation confirms both the order and the signs of the inverse transformations.

Expanded form and its derivation

The recursion can also be written as a sum of the time intervals between decisions, each multiplied by the inverse accelerations accumulated before that interval. This makes clear why the calendar time itself is not multiplied by one overall factor.

U_0 =t_1 +\exp(-b_1)(t_2-t_1) +\exp\{-(b_1+b_2)\}(t_3-t_2) +\cdots +\exp\left\{-\sum_{j=1}^{K}b_j\right\}(T-t_K).

For \(K=1\), the formula is \(U_0=t_1+\exp(-b_1)(T-t_1)\), exactly the one-decision transformation. For \(K=2\), substituting \(U_2=T\) into the recursion gives

\begin{aligned} U_0 &=t_1+\exp(-b_1)(U_1-t_1)\\ &=t_1+\exp(-b_1)(t_2-t_1) +\exp\{-(b_1+b_2)\}(T-t_2). \end{aligned}

For the general case, expand the first recursion step: \(U_0=t_1+\exp(-b_1)(U_1-t_1)\). The quantity \(U_1\) is the same backward recursion beginning at \(t_2\). Substituting its expanded expression multiplies every interval after \(t_2\) by the additional factor \(\exp(-b_1)\), while the first interval receives only \(\exp(-b_1)\). Repeating this substitution adds one inverse factor for each earlier decision to every later interval, producing the displayed sum. This proves the expanded form by repeated substitution. \(\square\)

The same recursion also follows directly from the forward model. At decision \(j\), the post-decision time is \(U_j\), and the structural model says \(U_j=t_j+\exp(b_j)(U_{j-1}-t_j)\). Rearranging gives \(U_{j-1}=t_j+\exp(-b_j)(U_j-t_j)\). Applying this identity from \(j=K\) down to \(1\) proves that the recursion removes each modeled treatment effect in turn. \(\square\)

Worked Example: Checking the Expanded Formula

Use the two-decision example with \(t_1=2\), \(t_2=5\), \(T=11\), and acceleration factors 2 and 1.5. The expanded expression has three pieces: time before the first decision, the interval between decisions, and time after the second decision.

\begin{aligned} U_0 &=2+\frac{1}{2}(5-2) +\frac{1}{2(1.5)}(11-5)\\ &=2+1.5+2\\ &=5.5. \end{aligned}

This agrees with the recursive calculation. The interval from year 2 to year 5 is divided by the first acceleration only; the interval after year 5 is divided by both accelerations. That is why the final interval has denominator \(2(1.5)=3\).

Assumptions, interpretation, and alternatives

The algebra above is exact once the structural model is specified. Its causal interpretation requires more. Consistency links the observed failure time to the counterfactual failure time under the treatment actually received. The model must correctly describe the effect of each treatment on remaining survival time, including any modeled dependence on the history \(H_j\). A treatment decision must occur while the person is at risk, and the candidate transformations must remain meaningful on the post-decision time scale.

In a rank-preserving model, the transformed value can be interpreted at the individual level as the failure time under the blipped-down treatment history. In a distributional failure-time model, the structural claim concerns the distribution of the transformed time rather than requiring each individual's potential failure times to obey a deterministic rank-preserving relation. The inverse algebra is the same, but those interpretations are not interchangeable.

A whole-trajectory accelerated failure-time model typically specifies how a treatment history scales a failure time under a particular parametric model. A structural nested failure-time model instead blips down the effect of one decision at a time, conditional on the history at that decision. The distinction is especially useful when treatment is assigned repeatedly. A hazard-ratio model, such as a proportional hazards model, parameterizes a different feature of survival distributions; its coefficient is not generally the log acceleration \(b_j\).

The transformation alone does not establish that a causal effect is identified, select the correct candidate parameter, or adjust for censoring. Those require additional assumptions and methods. In particular, a transformed value that falls before a later treatment decision cannot be interpreted as survival to that decision. Such values signal that the candidate blips and observed treatment schedule are incompatible for that individual under the simple post-decision interpretation. The transformation should not be silently repaired by truncating the value.

Diagnostics and common pitfalls

Before interpreting a transformed time, check that the decision times are ordered, each decision precedes observed failure, and each recursive remaining interval is nonnegative. Confirm that treatment coded as zero has zero blip when that is the model convention. A direct forward reconstruction is a useful arithmetic check: begin with \(U_0\), reapply the accelerations in chronological order, and verify that the result is \(T\).

Wrong sign: multiplying by \(\exp(b_j)\) while blipping down applies the forward effect. The inverse uses \(\exp(-b_j)\).
Wrong time interval: scaling \(T\) rather than \(T-t_j\) incorrectly scales survival time before the decision. The blip acts on time remaining after \(t_j\).
Wrong order: with repeated decisions, starting at the earliest treatment ignores the fact that later treatments act on a time already affected by earlier treatments. Apply inverse blips from last to first.
Overstated interpretation: a computed transformed value is not by itself proof that it is a counterfactual outcome. That interpretation depends on consistency and the structural model; identification needs further assumptions.

The transformation is therefore a model-based calculation, not a standalone causal estimator. Its value is that it makes the proposed failure-time effects explicit and gives a precise object that later analyses can use.

Check Your Understanding

Work through each question using the inverse-blip definition and the recursive order. Short solutions follow.

At time \(t=4\), the observed failure time is 10 and the candidate acceleration factor is 2. What is the blipped-down time?
Why does the transformation at time \(t\) act on \(T-t\), rather than on \(T\)?
For two decisions at \(t_1<t_2\), which blip is removed first, and why?
With decisions at times 1 and 3, observed failure at 7, and acceleration factors 2 and 2, calculate the fully blipped-down time.

Solutions. (1) \(4+(10-4)/2=7\). (2) Treatment at \(t\) changes remaining survival after the decision, not time already elapsed. (3) Remove the decision at \(t_2\) first; it acts on the latest remaining-time segment, so the earlier transformation must use the resulting time. (4) First \(U_1=3+(7-3)/2=5\); then \(U_0=1+(5-1)/2=3\). Reapplying the first acceleration gives \(1+2(3-1)=5\), and the second gives \(3+2(5-3)=7\), confirming the calculation.