1
$\begingroup$

Hello I am learning about survival analysis and am getting exposed to interval censoring.

I was curious why when using the survfit function from survival package that the number at risk at time 0 is 1 greater then the number of observations? Is there a theoretical reason or it it a problem with my R code?

Continuous Time

set.seed(123)
library(survival)

# Continuous Time ----
size = 10
deathtime <- seq(1, size)
death = rep(1, size)
df <- data.frame(deathtime, death)

surv.obj<- Surv(df$deathtime, df$death)

summary(survfit(surv.obj ~ 1))

enter image description here

Makes sense at time 1 there is 10 at risk since there is 10 observations.

Interval Time


# Interval Time ----
start <- seq(0, size - 1)
end <- seq(1,size)
death <- rep(1, size)

df.i <- data.frame(start, end, death)

surv.obj.i <- Surv(time = df.i$start,
                   time2 = df.i$end,
                   event = df.i$death,
                   type = "interval")

summary(survfit(surv.obj.i ~ 1))

enter image description here

Why does with interval censoring at time 0 there is a number of risk of 11? There is only 10 observations.

$\endgroup$

1 Answer 1

3
$\begingroup$

That has to do with the Turnbull estimate of the survival curve based on the interval-censored data. Simply printing the survfit object provides the needed hint:

> survfit(surv.obj.i ~ 1)
Call: survfit(formula = surv.obj.i ~ 1)

     records  n events median 0.95LCL 0.95UCL
[1,]      10 11     11    4.5       2      NA

The function knows that there were only 10 records. But the Turnbull handling of interval-censored data must allow for the possibility of an event at either end of each of your 10 time intervals. With potential events thus starting at time = 0 through time = 10, that's counted by the software as 11 potential (limiting) event times.

For more general ways of handling interval-censored data, you might examine the documentation of the icenReg package. For a small number of time intervals shared among all individuals, discrete-time survival analysis might be preferable. See this answer to your related question for some details on discrete-time models and links to more resources.

$\endgroup$
1
  • $\begingroup$ And the Turnbull estimator also estimates the number of risk and that is why the number at risk is not a round number. Turnbull I see. Thank you for sharing. $\endgroup$
    – Vefeagins
    Oct 5, 2021 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.