Questions tagged [kaplan-meier]

The Kaplan-Meier estimator is a common non-parametric method for survival analysis and for plotting survival graphs. The survival function $S(t)$ calculates the probability of survival past time $t$. It is most useful in comparing the survival of different groups while properly handling censored data.

The Kaplan-Meier estimator is a common non-parametric method for survival analysis and for plotting survival graphs. The survival function $S(t)$ calculates the probability of survival past time $t$. It is most useful in comparing the survival of different groups while properly handling censored data. It is not well suited for estimating survival adjusted for covariates. In such a case, it is better to use other parametric or semi-parametric models, such as the Cox Proportional Hazards model.

269 questions
23
votes
2 answers

Bayesian Survival Analysis: please, write me a prior for Kaplan Meier!

Consider right-censored observations, with events at times $t_1, t_2, \dots$. The number of susceptible individuals at time $i$ is $n_i$, and the number of events at time $i$ is $d_i$. The Kaplan-Meier or product estimator arises naturally as a MLE…
Elvis
  • 11,870
  • 36
  • 56
20
votes
1 answer

Basic questions about discrete time survival analysis

I am attempting to carry out a discrete time survival analysis using a logistic regression model, and I'm not sure I completely understand the process. I would greatly appreciate assistance with a few basic questions. Here is the set up: I'm…
Talbot Katz
  • 361
  • 2
  • 5
17
votes
3 answers

How can machine learning models (GBM, NN etc.) be used for survival analysis?

I know that traditional statistical models like Cox Proportional Hazards regression & some Kaplan-Meier models can be used to predict days till next occurrence of an event say failure etc. i.e Survival analysis Questions How can regression version…
15
votes
2 answers

Estimating a survival probability in R

Based on a sample of $n$ survival times, I would like to estimate the probability of surviving time $t$, for some specific $t$, using the Kaplan-Meier estimator. Is it possible to do this in R? Please, note that $t$ is not necessarily an event time.…
user7064
  • 1,685
  • 5
  • 23
  • 39
13
votes
2 answers

Kaplan-Meier curves seem to say otherwise than the Cox regression

In R, I am doing survival data analysis of cancer patients. I have been reading very helpful stuff about survival analysis in CrossValidated and other places and think I understood how to interpret the Cox regression results. However, one result…
francoiskroll
  • 427
  • 1
  • 4
  • 12
12
votes
1 answer

Kaplan-Meier(K-M) vs. Cox Regression

I am very new to survival analysis. I am looking for differences between these two methods - Kaplan-Meier(K-M) vs. Cox Regression. KM Survival Analysis cannot use multiple predictors, whereas Cox Regression can. KM Survival Analysis can run only on…
Riya
  • 589
  • 2
  • 7
  • 15
10
votes
2 answers

Kaplan-Meier multiple group comparisons

Let's say I have the following data frame library(survival) library(multcomp) data(cml) cml$group<-sample(1:5, 507, replace=T) plot(survfit(Surv(time=cml$time, cml$status)~factor(cml$group))) (survdiff(Surv(time=cml$time,…
ECII
  • 1,791
  • 2
  • 17
  • 25
10
votes
1 answer

Variance of the Kaplan-Meier estimate for dependent observations

Can someone help me find a way to estimate the variance of the Kaplan-Meier estimate with dependent observations? Specifically, I have failure time data from patients with several different observations for each patient (and different patients may…
Anton
  • 243
  • 1
  • 8
7
votes
1 answer

Comparison of Kaplan-Meier curves across ordered groups

I am familiar with the log-rank test for comparing multiple Kaplan-Meier curves, but I am looking for a test that will compare across ordered groups (an ordinal variable). A significant result from the log rank test indicates that at least one of…
Brian Diggs
  • 1,021
  • 8
  • 17
7
votes
3 answers

R packages (or SAS code) to produce two simultaneous Kaplan-Meier curves?

There's a way to do survival analysis of two (or more I suppose) mutually exclusive competing risks as a mixture of two different survival curves. Something like what you see in A.C. Ghani et al. Methods for Estimating the Case Fatality Ratio for a…
Fomite
  • 21,264
  • 10
  • 78
  • 137
7
votes
1 answer

projecting survival curves estimates into the future

Lets suppose I have a have a survival curve from 0 to 6000 days using Kaplan -Meier curves. How would I be able to project future survival rates from 6001 and forward ? Is there a function or extrapolation method I can use ? Below is an example,…
forecaster
  • 7,349
  • 9
  • 43
  • 81
7
votes
1 answer

Difference between Kaplan Meier Estimator and the Empirical CDF

In survival analysis, you often use the nonparametric maximum likelihood estimator (i.e. Kaplan-Meier estimator) of the survival function $S(t)$. Since $S(t) = 1 - F(t)$, shouldn't we also be able to estimate the empirical cdf and use the…
Good Guy Mike
  • 611
  • 9
  • 22
7
votes
2 answers

Confidence interval before the first event in a Kaplan–Meier curve

The confidence intervals for Kaplan–Meier curves in survival analysis only exist for times after the first (non-censored) event. Example R code: set.seed(1) library(survival) n = 30 x = 10 + sort(10*rexp(n)) u = rep(0, n) u[15] = 1 l =…
Karl Ove Hufthammer
  • 5,202
  • 2
  • 23
  • 32
6
votes
1 answer

Why does my proof for showing that the Kaplan-Meier estimate is unbiased not work?

I know that the Kaplan-Meier estimator is biased because my textbook says so. However, I don't understand why the following proof doesn't work: Let $\hat{S}(t)$ be the Kaplan-Meier estimate for the survival function $S(t)\equiv P(T_i > t)$ where…
TrynnaDoStat
  • 7,414
  • 3
  • 23
  • 39
6
votes
2 answers

Kaplan-Meier p-values

I have breast cancer trial data where, for subgroups 1 and 2, two treatment arms are being compared. The results are: For subgroup 1, N=33, median survival in trt1=10mo, median survival in trt2=5.7mo, p-value for difference=0.18. For subgroup 2,…
vchhaya
  • 61
  • 1
  • 2
1
2 3
17 18