Why the Censoring Threshold Changes the Draw
When an event time is right-censored at time \(c\), the event is known not to have occurred by \(c\). If an imputation procedure calls for a latent event-time draw, that draw must therefore be greater than \(c\). Drawing an event time from the ordinary, unconditional model does not enforce this information: it can produce values before the censoring threshold.
The central operation is conditioning the event-time distribution on \(T>c\). Under parametric models, this conditional distribution can be sampled directly using its survival function. Direct inversion is efficient and, when the model and assumptions are appropriate, produces draws with exactly the required conditional distribution.
For a censored person, the observed information is that their event time exceeds their censoring time. The conditional distribution uses that information rather than treating the censoring time as an event time. The formulas below first treat the model parameters as fixed. In a multiple-imputation procedure, parameter uncertainty can also be represented by drawing parameters for each imputation and then drawing the event time conditionally using those parameter values.
Assumptions and Conditional Distribution
Suppose, conditional on the covariates used in the event-time model, \(T\) follows the specified parametric distribution. For the model to describe censored observations using their event-time information, censoring must be non-informative given those covariates: among people with the same covariate values, censoring must not provide additional information about their event time beyond the fact that it exceeds \(c\). The parametric survival model must also be an adequate description of the event-time distribution in the relevant range, including times beyond observed censoring thresholds.
For \(t\ge c\), the conditional cumulative distribution function follows by the definition of conditional probability:
The equality assumes a continuous event time, so the probability of an event exactly at \(c\) is zero. A draw can be generated by setting \(F_c(T^*)=U\), where \(U\) is uniform on \((0,1)\), and solving for \(T^*\). The result must be at least \(c\); a formula that can return a value below \(c\) has not sampled from the desired conditional distribution.
Exponential Model: A Memoryless Residual Time
Under an exponential model with rate \(\lambda>0\), measured in inverse units of time, \(S(t)=\exp(-\lambda t)\) for \(t\ge0\). Substitution into the conditional survival function gives:
This is the survival function of \(c+E\), where \(E\) is exponential with rate \(\lambda\). Thus an exponential event time that has already survived to \(c\) has a remaining wait with the same exponential distribution as at time zero. This is the memoryless property. Inverse transformation gives the following draw, using a uniform \(U\) on \((0,1)\):
Worked Example: An Exponential Draw Beyond 8 Months
Suppose a hypothetical event-time model has rate \(\lambda=0.10\) per month, and a person's event time is right-censored at \(c=8\) months. For a direct conditional draw, take \(U=0.60\).
The exponential inverse-transform formula gives \(T^*=c-\log(1-U)/\lambda\). Substituting the values, \(T^*=8-\log(0.40)/0.10=8+9.163=17.16\) months.
The draw exceeds 8 months, as required. As a further check, the conditional probability that this event time exceeds 20 months is \(P(T>20\mid T>8)=\exp[-0.10(20-8)]=\exp(-1.2)=0.3010\). This checks the model-based conditional tail probability; it is not a probability that the particular imputed event time exceeds 20 months after the draw has been observed.
Weibull Model: Conditional Residual Time Depends on \(c\)
Use the Weibull convention \(S(t)=\exp[-(t/\eta)^k]\), where \(k>0\) is the shape and \(\eta>0\) is the scale, in the same time units as \(t\). The conditional survival function and direct draw are:
Unlike the exponential case, this generally cannot be written as \(c\) plus an independent Weibull residual with the same parameters. The Weibull hazard is \(h(t)=(k/\eta)(t/\eta)^{k-1}\). It increases with time when \(k>1\), decreases when \(k<1\), and is constant when \(k=1\). Consequently, the distribution of remaining time depends on the threshold \(c\), except in the exponential case \(k=1\).
Worked Example: A Weibull Draw Beyond 8 Months
Consider a hypothetical Weibull model with shape \(k=2\), scale \(\eta=10\) months, and censoring threshold \(c=8\) months. Use \(U=0.50\), so \(-\log(1-U)=-\log(0.50)=0.6931\).
The conditional draw is \(T^*=10[(8/10)^2+0.6931]^{1/2} =10[0.6400+0.6931]^{1/2} =10(1.3331)^{1/2} =11.55\) months.
The value is above the censoring threshold. Since \(U=0.50\), it is the conditional median: the conditional probability of exceeding it is \(\exp[-\{(11.55/10)^2-(8/10)^2\}]\), approximately \(0.5000\). For comparison, the unconditional Weibull median is \(10\sqrt{-\log(0.50)}=8.326\) months, which is below the threshold. An unconditional draw at this quantile would contradict the observed survival to month 8.
Why Direct Inversion Works
The inverse-transform argument establishes that these formulas generate the target conditional distribution, rather than just values that happen to exceed \(c\). For any continuous, strictly increasing conditional distribution function \(F_c\), set \(T^*=F_c^{-1}(U)\). Then for \(t\ge c\):
because \(P(U\le q)=q\) for \(0\le q\le1\). Substituting the conditional distribution functions derived above and solving \(F_c(T^*)=U\) gives the exponential and Weibull formulas. Their outputs are at or above \(c\): for the Weibull formula, \(-\log(1-U)>0\), so the quantity inside the brackets exceeds \((c/\eta)^k\); taking the positive \(k\)th root and multiplying by \(\eta\) yields \(T^*>c\). The same conclusion follows directly from the positive exponential residual.
Direct Draws, Unconditional Draws, and Rejection Sampling
An ordinary unconditional draw is not a substitute for a conditional draw. It may fall below the known threshold, and even keeping only the unconditional draws that exceed \(c\) is a different computational route with an important distinction: retaining only those draws is valid rejection sampling, while using the first unconditional draw regardless of its value is not.
Worked Example: Rejection Sampling Under an Exponential Model
Suppose a hypothetical event time follows an exponential model with rate \(\lambda=0.20\) per month and is censored at \(c=5\) months. The unconditional probability of a candidate exceeding the threshold is \(P(T>5)=\exp(-0.20\times5)=\exp(-1)=0.3679\).
For illustration, generate unconditional candidates by \(T=-\log(1-U)/0.20\). With \(U=0.20\), the candidate is \(-\log(0.80)/0.20=1.116\) months, so reject it because it is below 5 months. With a new independent \(U=0.70\), the candidate is \(-\log(0.30)/0.20=6.020\) months, so retain it. Repeating this procedure until a candidate exceeds 5 months produces draws from the exponential distribution conditional on \(T>5\).
The expected number of unconditional candidates per retained draw is \(1/0.3679=2.718\). Direct conditional inversion avoids rejected candidates. Rejection sampling is nevertheless valid when only candidates exceeding the threshold are retained; drawing once unconditionally and retaining even a candidate below 5 months is not.
A useful way to compare the approaches is to ask whether they produce the conditional law and whether they enforce the observed survival information:
| Approach | Does it target \(T\mid T>c\)? | Main consideration |
|---|---|---|
| Direct conditional inversion | Yes, if the conditional survival model is correct | Efficient; formulas are available here for exponential and Weibull models |
| Unconditional draw used as-is | No | Can fall below \(c\) and ignores known survival to \(c\) |
| Unconditional rejection sampling, retaining only \(T>c\) | Yes | Simple but potentially inefficient when \(P(T>c)\) is small |
Diagnostics, Failure Modes, and Limits
Check the draw mechanism as well as the event-time model. Every draw for a record censored at \(c\) should be at least \(c\). For an exponential model, verify the residual \(T^*-c\) is positive; for a Weibull model, verify that the result is positive and satisfies the conditional inverse relation. Across many simulated draws at fixed parameters and \(c\), the fraction exceeding a later time \(t\) should be close to \(S(t)/S(c)\). For example, under the exponential model that ratio is \(\exp[-\lambda(t-c)]\).
- Draws below the threshold: This usually indicates the unconditional inverse formula was used instead of the conditional formula, or the threshold was entered in inconsistent units.
- Implausible upper tails: A correct conditional calculation can still produce extreme values. Check model fit and the assumed tail form; conditional inversion does not make an unsuitable parametric model appropriate.
- Incorrect parameter convention: Here \(\eta\) is the Weibull scale, not a rate, and \(k\) is the shape. Confirm software uses the same parameterization before translating a formula.
- Informative censoring: If censoring conveys information about the event time beyond \(T>c\), conditioning only on survival to \(c\) may not describe the needed distribution.
These formulas concern right-censoring at a known threshold under a continuous parametric event-time model. They do not directly handle interval censoring, where the event is known to fall between two times, or competing events that change the meaning of the event being imputed. They also do not establish that an event time should always be imputed as a complete value for subsequent analysis; the analysis must respect the design and censoring structure. Finally, treating fitted parameters as fixed in every imputation omits parameter uncertainty. When that uncertainty is relevant, draw model parameters for each imputation and then apply the conditional event-time draw using that parameter draw.
Common Mistake
Wrong reasoning: “The model says an event time can be sampled from its unconditional distribution, so one ordinary draw is a reasonable imputation for a censored person.” This fails because the observed record already establishes \(T>c\); the unconditional distribution assigns positive probability to values that contradict that fact. Correction: use \(S(t)/S(c)\) to define the conditional distribution and draw from it, or use rejection sampling that discards every candidate at or below \(c\). A draw from the conditional distribution respects the censoring information, although the parametric model and censoring assumptions still need to be credible.
Check Your Understanding
Answer each question, then compare with the solutions.
- For a right-censored observation at \(c\), write the conditional survival function of \(T\) beyond \(c\) in terms of \(S(t)\) and \(S(c)\).
- Under an exponential model with rate \(0.25\) per day and censoring at \(c=4\) days, calculate the conditional draw for \(U=0.50\).
- For a Weibull model with \(k=1\), what familiar model does the conditional distribution reduce to, and what is its rate?
- Explain why drawing unconditionally and retaining only candidates greater than \(c\) differs from drawing unconditionally once and keeping that value regardless.
Solutions. (1) For \(t\ge c\), \(P(T>t\mid T>c)=S(t)/S(c)\). (2) The draw is \(4-\log(1-0.50)/0.25=4+0.6931/0.25=6.772\) days, which exceeds 4 days. (3) When \(k=1\), \(S(t)=\exp(-t/\eta)\), so the model is exponential with rate \(1/\eta\); its remaining time beyond \(c\) is exponential with that same rate. (4) Rejection sampling conditions on exceeding \(c\) by discarding all candidates that fail that condition, so the retained values have the conditional distribution. Keeping a single unconditional draw regardless of its value does not condition on \(T>c\) and can produce a value that contradicts the censoring information.