Tutorials › Multiple Imputation by Chained Equations (MICE) › Conditional Draws for Right-Censored Event Times

Survival and Time-Varying Data · Tutorial 409 of 497

Conditional Draws for Right-Censored Event Times

Learn how to draw an event time beyond its censoring time without assigning an impossible value or ignoring the information in the observed follow-up.

Advanced 10 min read

What You'll Learn

  • Define the conditional event-time distribution for a right-censored observation.
  • Derive its conditional survival function and inverse-survival draw.
  • Calculate conditional survival probabilities and a censored-time draw.
  • Identify the assumptions that make conditioning on survival through follow-up valid.
  • Diagnose implausible draws and distinguish conditional draws from common shortcuts.

Why a censored time needs a conditional draw

A right-censored observation tells us more than “the event time is missing.” If a person is followed for \(c\) months without the event, their event time \(T\) is known to exceed \(c\). An imputation that assigns a value at or before \(c\) contradicts what was observed; a draw from the event-time distribution without conditioning on survival through \(c\) can make exactly that error.

The relevant draw is from the distribution of \(T\) among individuals who have survived event-free beyond their own censoring time, conditional on the predictors and history used by the imputation model. It represents an unobserved event time, not an event newly observed during the original follow-up.

Definition: Let \(T\) be an event time and \(Z\) the covariates or other conditioning information in the event-time model. Define the conditional survival function as \(S(t\mid Z)=P(T>t\mid Z)\). For an observation known to be event-free at censoring time \(c\), the conditional distribution of \(T\) is the distribution given \(T>c\) and \(Z\). For \(t\ge c\), its survival function is \(P(T>t\mid T>c,Z)=S(t\mid Z)/S(c\mid Z)\).

Write the observed follow-up time as \(Y=\min(T,C)\), where \(C\) is the censoring time, and let \(\Delta=1\) indicate an observed event and \(\Delta=0\) indicate right censoring. Under the usual exact-time convention, \(\Delta=0\) at \(Y=c\) means \(T>c\). If an event is observed, its event time is not drawn from this conditional distribution. This tutorial concerns the censored case.

Deriving the conditional distribution

For any \(t\ge c\), the event \(\{T>t\}\) is contained in \(\{T>c\}\). The definition of conditional probability therefore gives

$$ P(T>t\mid T>c,Z) =\frac{P(T>t,\ T>c\mid Z)}{P(T>c\mid Z)} =\frac{S(t\mid Z)}{S(c\mid Z)}. $$

This ratio is the conditional survival function: it starts at 1 when \(t=c\) and, as \(t\) increases, gives the probability of remaining event-free beyond \(t\), among those already known to have survived through \(c\). Subtracting it from 1 gives the conditional cumulative distribution function:

$$ F_c(t\mid Z) =P(T\le t\mid T>c,Z) =1-\frac{S(t\mid Z)}{S(c\mid Z)},\qquad t\ge c. $$

For a continuous event-time distribution, draw \(V\sim\mathrm{Uniform}(0,1)\) and solve \(F_c(T^*\mid Z)=V\). Equivalently, since \(U=1-V\) is also uniform on \((0,1)\), solve \(S(T^*\mid Z)=U S(c\mid Z)\). Thus the draw is above \(c\) by construction. If the conditional survival curve is available only numerically, find the time at which it reaches the target \(U S(c\mid Z)\), using the curve’s specified interpolation or inverse method.

$$ S(T^*\mid Z)=U S(c\mid Z),\qquad U\sim\mathrm{Uniform}(0,1). $$

If the conditional cumulative hazard \(H(t\mid Z)=-\log S(t\mid Z)\) is available and the survival function is positive, taking negative logarithms yields an equivalent equation:

$$ H(T^*\mid Z)=H(c\mid Z)-\log U. $$

This follows because \(-\log\{U S(c\mid Z)\}=-\log U-\log S(c\mid Z)\). It is a convenient transformation, not a separate imputation rule. The required hazard or survival curve must describe the same event-time distribution and covariate history used to condition the draw.

Worked Example: Conditional Survival Beyond Follow-Up

Suppose a hypothetical fitted event-time distribution, for a specified covariate profile, gives \(S(8)=0.70\) and \(S(12)=0.35\), with times measured in months. An individual is right-censored at \(c=8\) months. Find the probability that the event time exceeds 12 months, conditional on surviving through censoring.

The required probability is the conditional survival ratio:

$$ P(T>12\mid T>8,Z)=\frac{S(12\mid Z)}{S(8\mid Z)} =\frac{0.35}{0.70}=0.500. $$

Thus, under this specified distribution, the conditional probability that the event occurs after 12 months is 0.500. The conditional probability of an event by 12 months, but after censoring at 8 months, is \(1-0.500=0.500\). This calculation assumes that both survival values refer to the same covariate profile and the same event definition.

Conditions for a valid conditional draw

The conditioning calculation is algebraic, but using it to impute a censored event time depends on the event-time model being appropriate. The key requirements are:

  • The censoring information means survival beyond \(c\). The observation must genuinely be right-censored at \(c\), rather than having an unknown event time with no known lower bound.
  • The survival model describes the relevant conditional distribution. It must include the predictors needed to represent differences in event-time distributions. As in the earlier material on Cause-Specific Hazards as Imputation Predictors, event type and predictors must be aligned with the target event process.
  • Censoring is conditionally independent of the event time, or its dependence is modeled. Informally, after conditioning on the information in the model, the censoring mechanism must not reveal additional information about the unobserved event time. If people with shorter latent event times are more likely to be censored for reasons not captured by the model, conditioning only on \(T>c\) may not give the right distribution.
  • The conditioning information is available at the relevant times. With time-varying predictors, do not use future values as though they were known when predicting the event time. If the future predictor path is itself unknown, a model for that path or a clearly defined prediction strategy is needed.
  • The model supports the requested tail draw. If \(S(c\mid Z)=0\), survival to \(c\) has zero probability under the model, so the conditional distribution is undefined. More generally, a model that assigns negligible survival beyond \(c\) may be incompatible with the observed censoring pattern.

In multiple imputation, uncertainty in estimated model parameters also matters. The inverse-survival calculation describes a draw when the relevant survival curve is fixed. A proper imputation procedure should account for uncertainty in the event-time model, typically by drawing model parameters as well as drawing a uniform random value. Repeating only the uniform draw from one fixed fitted curve can understate imputation uncertainty.

Worked Example: Drawing a Conditional Event Time

Consider a hypothetical survival curve for one covariate profile. At a censoring time of 6 months, \(S(6)=0.80\). The curve has \(S(10)=0.50\) and \(S(15)=0.25\). For this illustration, suppose the curve is interpolated linearly on the cumulative-hazard scale between 10 and 15 months. Draw a conditional event time using \(U=0.40\).

First calculate the survival target, which must be lower than \(S(6)\):

$$ U S(6)=0.40(0.80)=0.320. $$

At 10 months survival is 0.50, and at 15 months it is 0.25, so the target 0.320 lies between these values. To apply the stated interpolation, calculate the cumulative hazards at the endpoints and at the target:

$$ H(10)=-\log(0.50)=0.6931,\qquad H(15)=-\log(0.25)=1.3863, $$ $$ H(T^*)=-\log(0.320)=1.1394. $$

Linear interpolation in cumulative hazard places the draw at the fraction

$$ \frac{1.1394-0.6931}{1.3863-0.6931}=0.644. $$

The interval from 10 to 15 months is 5 months, so

$$ T^*=10+0.644(5)=13.22\text{ months}. $$

The draw exceeds the censoring time of 6 months and corresponds to the requested conditional probability level. The interpolation assumption is part of this numerical illustration; another survival-curve representation can give a different time at the same target survival. In an application, use the fitted curve’s defined inverse rather than silently choosing an interpolation rule.

Practical checks and common failure modes

A conditional draw should be checked against the information that generated it. Every imputed time for a right-censored record must exceed that record’s censoring time. Also check that the target survival value is within the range supported by the fitted curve, that the inverse is numerically stable, and that draws do not pile up at an artificial maximum follow-up time. Compare imputed times across censoring times and important predictors: systematic patterns that conflict with the fitted survival distributions can indicate a model or implementation problem.

A curve that does not approach zero requires particular care. For example, a model with a nonzero probability of never experiencing the event has a probability mass at an infinite event time. Its conditional distribution includes that mass; a finite-time inverse alone cannot represent it. Likewise, a discrete-time event model can have jumps rather than a continuous distribution. The conditional probability formula still applies, but a continuous inverse calculation should not be used as if the distribution had no jumps.

Competing events also require a clearly defined target. If the goal is to impute time to a particular cause, the probability of remaining free of that cause is not automatically the same as the probability of remaining free of every event. The event type and time must be treated consistently with the estimand and the model; a cause-specific event-time draw cannot be interpreted as an all-cause event-time draw without justification.

Worked Example: Why an Unconditional Draw Can Be Impossible

Suppose an individual is known to be event-free through 9 months. A draw from the unconditional event-time distribution gives 7 months. This draw is incompatible with the observation because \(7<9\), so it assigns an event before the individual was known to be event-free.

Assume the fitted curve gives \(S(9)=0.60\). For a conditional draw with \(U=0.30\), the target survival is

$$ U S(9)=0.30(0.60)=0.180. $$

The conditional draw is the time at which the fitted curve reaches survival 0.180. Because this target is below \(S(9)=0.60\), the resulting time is later than 9 months, provided the curve is nonincreasing and the target is reached. This is the essential correction: the draw must be from the distribution truncated to \(T>9\), not from the unconditional event-time distribution.

A frequent mistake is to replace a censored event time by \(c\), or by \(c\) plus a small constant. That value is not a random draw from the event-time distribution; it fabricates a precise event time at follow-up and discards uncertainty about how long after censoring the event would occur. Another mistake is to draw unconditionally and discard values below \(c\) without checking the implied distribution or implementation. Rejection sampling can reproduce the conditional distribution in principle, but directly using the conditional survival function makes the target explicit and avoids inefficient or error-prone rejection.

There are settings where no event-time imputation is needed: a survival analysis can retain right-censored observations as censored rather than pretending their event times were observed. Conditional imputation is relevant when the analysis or imputation procedure requires completed event-time values. It does not turn follow-up after \(c\) into observed data, and it cannot correct a misspecified event-time or censoring model.

Key takeaway: For a right-censored event time at \(c\), draw from \(T\mid T>c,Z\), not from the unconditional distribution. Its survival function is \(S(t\mid Z)/S(c\mid Z)\), so an inverse-survival draw solves \(S(T^*\mid Z)=U S(c\mid Z)\).

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. A hypothetical model gives \(S(5\mid Z)=0.75\) and \(S(9\mid Z)=0.30\). For a person censored at 5 months, calculate the conditional probability of surviving beyond 9 months.
  2. For a person censored at \(c\), explain why drawing from the unconditional distribution can produce an invalid imputed time.
  3. If \(S(c\mid Z)=0.50\) and the uniform draw is \(U=0.20\), what survival value should the inverse-survival calculation target?
  4. Name one assumption needed for conditioning only on \(T>c\) to describe the event time appropriately, and give one diagnostic for an implemented draw.

Solutions. (1) \(S(9\mid Z)/S(5\mid Z)=0.30/0.75=0.400\), so the conditional probability is 0.400. (2) The unconditional draw is not restricted to times greater than \(c\), so it may place the event before the observed event-free follow-up. (3) The target is \(U S(c\mid Z)=0.20(0.50)=0.100\). (4) One assumption is conditional independence of censoring and event time given the modeled information; one diagnostic is to verify that every imputed time for a censored record exceeds its censoring time.

Clinical Biostats