Why is large amounts of censoring bad in survival analysis?

Question

I am trying to understand why large amounts of censoring (ie. many patients are censored) is undesirable in Survival Analysis.

As a proof of concept, suppose there are 5 patients - all patients enter the study at the same time:

• patient1 has event at t1
• patient2 has event at t2
• patient3 drops out of the study at t3
• patient4 has event at t4
• and when the study is over at t5, patient 5 has not had the event
• t5 > t4 > t3 > t2 > t1

(Semi Parametric Approach) Here is my attempt to write the model and likelihood for a Cox-PH regression in this situation:

$$ h(t|X) = h_0(t) \exp(\beta^T X) $$ $$ L(\beta) = \prod_{i: \delta_i = 1} \frac{h(t_i|X_i)}{\sum_{j: t_j \geq t_i} \exp(\beta^T X_j)} $$ $$ L(\beta) = \frac{h(t_1|X_1)}{\sum_{j: t_j \geq t_1} \exp(\beta^T X_j)} \times \frac{h(t_2|X_2)}{\sum_{j: t_j \geq t_2} \exp(\beta^T X_j)} \times \frac{h(t_4|X_4)}{\sum_{j: t_j \geq t_4} \exp(\beta^T X_j)} $$

$$ L(\beta) = \frac{h(t_1|X_1)}{\exp(\beta^T X_1) + \exp(\beta^T X_2) + \exp(\beta^T X_3) + \exp(\beta^T X_4) + \exp(\beta^T X_5)} \times \frac{h(t_2|X_2)}{\exp(\beta^T X_2) + \exp(\beta^T X_3) + \exp(\beta^T X_4) + \exp(\beta^T X_5)} \times \frac{h(t_4|X_4)}{\exp(\beta^T X_3) + \exp(\beta^T X_5)} $$

(Parametric Approach) Here is my attempt to write the model and likelihood for a AFT model in this situation (note that the likelihood is based on the distribution of $\epsilon$ and not the survival times $T$ . I have heard that if we pick $T$ to have distributions such as Exponential or Weibull, then $\epsilon$ results in Extreme Value Distribution such as the Gumbel Distribution ):

$$ \log(T) = \mu + \beta^T X + \sigma \epsilon $$

$$ L(\mu, \sigma, \beta) = \prod_{i=1}^{n} \left[ f\left( \frac{\log(t_i) - \mu - \beta^T X_i}{\sigma} \right) \right]^{\delta_i} \left[ 1 - F\left( \frac{\log(t_i) - \mu - \beta^T X_i}{\sigma} \right) \right]^{1-\delta_i} $$

$$ L(\mu, \sigma, \beta) = \left[ f\left( \frac{\log(t_1) - \mu - \beta^T X_1}{\sigma} \right) \right] \times \left[ f\left( \frac{\log(t_2) - \mu - \beta^T X_2}{\sigma} \right) \right] \times \left[ 1 - F\left( \frac{\log(t_3) - \mu - \beta^T X_3}{\sigma} \right) \right] \times \left[ f\left( \frac{\log(t_4) - \mu - \beta^T X_4}{\sigma} \right) \right] \times \left[ 1 - F\left( \frac{\log(t_5) - \mu - \beta^T X_5}{\sigma} \right) \right] $$

So in the Cox-Ph and AFT model, how is inference (e.g. perhaps it results in high variance, high bias, non-consistency, larger sample sizes to achieve compared results vs smaller sample sizes with lesser censoring) and parameter estimation negatively affected when large numbers of the patients are censored? Does the mathematical optimization become difficult (e.g. incomplete matrix rank, matrix inverses not defined, non-identifiable model)?

Answer 1

From a purely intuitive point of view: Imagine the extreme case where all the cases are censored. Then you couldn't say anything. Now imagine the other extreme, where no one is censored. Then you wouldn't need survival analysis.

Each observation that is censored is one more case where we don't know what will happen, but have to "guess". The different survival analysis methods are different ways of guessing, with different properties and ways they can go wrong, but all are guesses as opposed to knowledge.

"The patient died on Dec 12, 2023" tells us a lot more than "the patient will die some time after this study ends."

Answer 2

Keep it simple. First think of the purpose of estimation. One of the purposes is to estimate absolute survival probabilities. If you had 10,000 patients and no events you estimate S(t)=1.0 with high precision, so no problem even with all observations censored. For estimation of relative effects, consider the exponential distribution which is a special case of Weibull. The variance of the log hazard rate estimate in an exponential distribution is the reciprocal of the number of uncensored observations. So for estimating relative effects censoring is "bad". We need a similar result for accelerated failure time distributions such as what the log-normal survival model uses. The result will be similar to the exponential case, but not quite the same.

Answer 3

I'm going to give a slightly different answer from the previous responses, that also has a more visual tint. For simplicity, suppose we want to estimate the survival function up to 1 year. To show how censoring impacts the analysis, I am going to use nonparametric bounds.

The bounds are the best/worst cases on the survival function given the observed data that are going to place no parametric assumptions. The bounds represent the extremes of what censoring is capable of doing to our estimates given the observed data. For the upper bound on the survival function, we will assume that all censored individuals do not ever have the event (up to 1 year). For the lower bound, we will assume that all censored individuals immediately had the event after they were censored.

The following is some code to generate survival times for $n=10000$ observations with different extents of censoring. In the zero case, there is no censoring. In the first and second cases, more censoring occurs.

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from lifelines import KaplanMeierFitter
n = 10000
t_max = 1
t = 0.7 * np.random.weibull(1.5, size=n)   # Event times
t = np.where(t >= t_max, t_max, t)         # Admin censoring at 1 year
c0 = 2                                     # Censoring (none)
c1 = 1.5 * np.random.weibull(2., size=n)   # Censoring (some)
c2 = 0.9 * np.random.weibull(0.9, size=n)  # Censoring (more)

For c0, there is no censoring. The following code is for the bounds (since no censoring, the bounds are the same as the point estimate of the Kaplan-Meier)

t0_star = t                           # New variable for t
delta0 = np.where(t >= t_max, 0, 1)   # Event indicator (all events expect admin censor)
Using Kaplan-Meier, since equivalent to CDF here
km0 = KaplanMeierFitter()
km0.fit(t0_star, delta0)
km0_St = km0.survival_function_  # Survival function

The following is code to get the bounds for censoring in case 1 (c1).

# Setting up data we get to observed in this case
t1_star = np.min([t, c1], axis=0)
delta1 = np.where((t <= c1) & (t < t_max), 1, 0)
Upper bound computation
t1_staru = np.where(delta1 == 0, t_max, t1_star)  # All never events
km1u = KaplanMeierFitter()
km1u.fit(t1_staru, delta1)
km1u_St = km1u.survival_function_
Lower bound computation
delta1l = np.where(t1_star < t_max, 1, delta1)  # All events at censoring times
km1l = KaplanMeierFitter()
km1l.fit(t1_star, delta1l)
km1l_St = km1l.survival_function_
Merging bounds into single data object for plotting
bounds = pd.merge(km1l_St, km1u_St, left_index=True, right_index=True, how='outer')
bounds.ffill(inplace=True)
Plotting the bounds
plt.fill_between(bounds.index, bounds.KM_estimate_x, bounds.KM_estimate_y, step='post')
plt.show()

When we apply this for each of the scenarios, we can get the following plot

So, as you can see, the bounds get wider the more censoring that occurs. Large amounts of censoring is bad in survival analysis because we have less information in the data. To compute point estimates for the survival function (with Cox, AFT, Kaplan-Meier, etc. models), we need to rely on assumptions regarding how censoring occurs. The more censoring that occurs, the more we have to 'lean' on those assumptions. So, that is why less censoring is generally better.

Answer 4

Imagine a two-armed clinical trial intended to study a treatment for some long-term condition, but which only follows-up patients for a single day. This is obviously a rather stupid design, as very likely none of the patients will have an event in just 1 day, and you'll learn absolutely nothing about which treatment would have been more effective over a longer term. The follow-up times are just too short, resulting in everybody being censored.

Note that this scenario is really no different from having a longer follow-up period but still seeing mostly censored individuals. In a survival analysis where most individuals are censored, it suggests that your follow-up period may be inappropriately short to effectively study longer-term effects. It would be no different having a 1-year follow-up time for events that occur on a 100-year time scale. The absolute length of follow-up time or number of individuals in the survival analysis isn't what gives you power, it's the number of events, which is related to how long the follow-up is relative to the event timing. Some amount of censored data is expected, but if a very large proportion of the cohort is censored, intuitively you haven't followed them long enough to study your event of interest, and your power will suffer.

That said, since power is determined by the number of events, you can mitigate a high censoring rate somewhat simply by having a larger cohort with a short follow-up time, allowing you to collect more early events by chance alone. This does, however, run the risk of finding effects unique to the early time points, rather than actually looking at the time scale relevant to most individuals - even if the N were so large as to have proper power with a 99% censor rate, I'd be cautious.

Answer 5

Other answers correctly explain the main problem with censoring, that is the loss of power because the sample is actually becoming smaller as more patients leave.

However I think there may be another problem even if the sample is large enough to keep the analysis reasonably powerful after a lot of censoring: censoring may not be completely at random. We assume that the remaining patients are a representative sample of the population, but if the probability of patients leaving is somehow related with the condition we are studying, the remaining sample could be biased. And of course, the more censored it became, the more biased it can be.

In other words, for example, if the healthiest patients are more likely to leave the study, and a lot of patients leave, we might end seriously underestimating survival time. Or if the patients in worse condition are more likely to abandon, and a lot of patients abandon, we may end overestimating survival time.

If there is little censoring (very few people leave) our end sample will be very similar to the original sample, and therefore the potential bias will be less than when there is a large amount of censoring.

Answer 6

Nitpicking point: there are some answers something along the lines of "if all the data was censored, you would not be able to estimate the survival curve".

To be slightly more precise, if all the data were right censored (which is the most common type of censoring), then at best you could estimate an lower bounds of the survival probability at the observed censoring points.

However, there are some cases in which we can get reasonable estimates when 100% of the data is censored. For example, current status data occurs when we inspect each item at one time and observe that the event has not occurred (right censored) or already occurred at the inspection time (left censored). In this case, 100% of the data is censored but we still can get consistent estimates of the survival distribution!

But to drive the point home that uncensored data is more informative, the convergence rate of the Non-Parametric Maximum Likelihood Estimate (NPMLE, generalization of the Kaplan Meier curves) with current status data is $n^{-1/3}$ compared to the more standard convergence rate of $n^{-1/2}$. So this convergence rate reinforces our heuristic understanding that knowing the exact event time should be more informative than knowing that the event occurred before or after a given time.

Answer 7

As an illustrative example, consider estimation of the failure survival curve under independent censoring. Let $T,C>0$ be the potential failure and censoring times, respectively. Define $F(u) = P(T>u)$ and $G(u) = P(C \geq u)$. The Kaplan Meier estimator is an asymptotically efficient nonparametric estimator of $P(T>u)$ and asymptotically follows a Gaussian process with covariance function $$\sigma^2(s, t) = F(s) F(t) \int_{(0,s \wedge t]} \frac{1}{G(u)} \,\mathrm{d} \left\{ \frac{1}{F(u)} \right\}.$$ Notice the $G$ in the denominator: if censoring is more aggressive, then $G$ is smaller, and $\sigma^2$ is larger. In fact, this argument establishes the following result.

Theorem: Let $T>0$ and $C_{\text{hi}},C_{\text{lo}}>0$. Consider observed data under $C_{\text{hi}}$ and under $C_{\text{lo}}$. If $C_{\text{hi}}$ stochastically dominates $C_{\text{lo}}$, then the Kaplan Meier estimator calculated under $C_{\text{hi}}$ has lower variance than the Kaplan Meier estimator calculated under $C_{\text{lo}}$.