1 / 70

Kaplan-Meier methods and Parametric Regression methods

Kaplan-Meier methods and Parametric Regression methods. Kristin Sainani Ph.D. http://www.stanford.edu/~kcobb Stanford University Department of Health Research and Policy. More on Kaplan-Meier estimator of S(t) (“product-limit estimator” or “KM estimator”).

erik
Télécharger la présentation

Kaplan-Meier methods and Parametric Regression methods

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Kaplan-Meier methods and Parametric Regression methods Kristin Sainani Ph.D.http://www.stanford.edu/~kcobbStanford UniversityDepartment of Health Research and Policy

  2. More on Kaplan-Meier estimator of S(t)(“product-limit estimator” or “KM estimator”) • When there are no censored data, the KM estimator is simple and intuitive: • Estimated S(t)= proportion of observations with failure times > t. • For example, if you are following 10 patients, and 3 of them die by the end of the first year, then your best estimate of S(1 year) = 70%. • When there are censored data, KM provides estimate of S(t) that takes censoring into account (see last week’s lecture). • If the censored observation had actually been a failure: S(1 year)=4/5*3/4*2/3=2/5=40% • KM estimator is defined only at times when events occur! (empirically defined)

  3. KM (product-limit) estimator, formally

  4. Observed event times The risk set nj at time tjconsists of the original sample minus all those who have been censored or had the event before tj Typically dj= 1 person, unless data are grouped in time intervals (e.g., everyone who had the event in the 3rd month). dj/nj=proportion that failed at the event time tj 1-dj/nj=proportion surviving the event time S(t) represents estimated survival probability at time t: P(T>t) Multiply the probability of surviving event time t with the probabilities of surviving all the previous event times. KM (product-limit) estimator, formally This formula gives the product-limit estimate of survival at each time an event happens.

  5. Example 1: time-to-conception for subfertile women “Failure” here is a good thing. 38 women (in 1982) were treated for infertility with laparoscopy and hydrotubation. All women were followed for up to 2-years to describe time-to-conception. The event is conception, and women "survived" untilthey conceived. Example from: BMJ, Dec 1998; 317: 1572 - 1580.

  6. Raw data: Time (months) to conception or censoring in 38 sub-fertile women after laparoscopy and hydrotubation (1982 study) Conceived (event) Did not conceive (censored) 1 2 1 3 1 4 1 7 1 7 1 8 2 8 2 9 2 9 2 9 2 11 3 24 3 24 3 4 4 4 6 6 9 9 9 10 13 16 Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy and hydrotubation. BMJ 1982; 284: 1013-1014

  7. Corresponding Kaplan-Meier Curve S(t) is estimated at 9 event times. (step-wise function)

  8. Raw data: Time (months) to conception or censoring in 38 sub-fertile women after laparoscopy and hydrotubation (1982 study) Conceived (event) Did not conceive (censored) 1 2 1 3 1 4 1 7 1 7 1 8 2 8 2 9 2 9 2 9 2 11 3 24 3 24 3 4 4 4 6 6 9 9 9 10 13 16 Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy and hydrotubation. BMJ 1982; 284: 1013-1014

  9. Raw data: Time (months) to conception or censoring in 38 sub-fertile women after laparoscopy and hydrotubation (1982 study) Conceived (event) Did not conceive (censored) 1 2 1 3 1 4 1 7 1 7 1 8 2 8 2 9 2 9 2 9 2 11 3 24 3 24 3 4 4 4 6 6 9 9 9 10 13 16 Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy and hydrotubation. BMJ 1982; 284: 1013-1014

  10. Corresponding Kaplan-Meier Curve 6 women conceived in 1st month (1st menstrual cycle). Therefore, 32/38 “survived” pregnancy-free past 1 month.

  11. S(t=1) = 32/38 = 84.2% S(t) represents estimated survival probability: P(T>t) Here P(T>1). Corresponding Kaplan-Meier Curve

  12. Raw data: Time (months) to conception or censoring in 38 sub-fertile women after laparoscopy and hydrotubation (1982 study) Conceived (event) Did not conceive (censored) 1 2.1 1 3 1 4 1 7 1 7 1 8 2 8 2 9 2 9 2 9 2 11 3 24 3 24 3 4 4 4 6 6 9 9 9 10 13 16 Important detail of how the data were coded:Censoring at t=2 indicates survival PAST the 2nd cycle (i.e., we know the woman “survived” her 2nd cycle pregnancy-free). Thus, for calculating KM estimator at 2 months, this person should still be included in the risk set. Think of it as 2+ months, e.g., 2.1 months. Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy and hydrotubation. BMJ 1982; 284: 1013-1014

  13. Corresponding Kaplan-Meier Curve

  14. S(t=2) = ( 84.2%)*(84.4%)=71.1% Corresponding Kaplan-Meier Curve 5 women conceive in 2nd month. The risk set at event time 2 included 32 women. Therefore, 27/32=84.4% “survived” event time 2 pregnancy-free. Can get an estimate of the hazard rate here, h(t=2)= 5/32=15.6%. Given that you didn’t get pregnant in month 1, you have an estimated 5/32 chance of conceiving in the 2nd month. And estimate of density (marginal probability of conceiving in month 2): f(t)=h(t)*S(t)=(.711)*(.156)=11%

  15. Raw data: Time (months) to conception or censoring in 38 sub-fertile women after laparoscopy and hydrotubation (1982 study) Conceived (event) Did not conceive (censored) 1 2.1 1 3.1 1 4 1 7 1 7 Risk set at 3 months includes 26 women 1 8 2 8 2 9 2 9 2 9 2 11 3 24 3 24 3 4 4 4 6 6 9 9 9 10 13 16 Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy and hydrotubation. BMJ 1982; 284: 1013-1014

  16. Corresponding Kaplan-Meier Curve

  17. S(t=3) = ( 84.2%)*(84.4%)*(88.5%)=62.8% Corresponding Kaplan-Meier Curve 3 women conceive in the 3rd month. The risk set at event time 3 included 26 women. 23/26=88.5% “survived” event time 3 pregnancy-free.

  18. Raw data: Time (months) to conception or censoring in 38 sub-fertile women after laparoscopy and hydrotubation (1982 study) Conceived (event) Did not conceive (censored) 1 2 1 3.1 1 4 1 7 1 7 1 8 2 8 2 9 2 9 2 9 2 11 3 24 3 24 3 4 4 4 6 6 9 9 9 10 13 16 Risk set at 4 months includes 22 women Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy and hydrotubation. BMJ 1982; 284: 1013-1014

  19. Corresponding Kaplan-Meier Curve

  20. S(t=4) = ( 84.2%)*(84.4%)*(88.5%)*(86.4%)=54.2% Hazard rates (conditional chances of conceiving, e.g. 100%-84%) look similar over time. Corresponding Kaplan-Meier Curve 3 women conceive in the 4th month, and 1 was censored between months 3 and 4. The risk set at event time 4 included 22 women. 19/22=86.4% “survived” event time 4 pregnancy-free. And estimate of density (marginal probability of conceiving in month 4): f(t)=h(t)*S(t)=(.136)* (.542)=7.4%

  21. Raw data: Time (months) to conception or censoring in 38 sub-fertile women after laparoscopy and hydrotubation (1982 study) Conceived (event) Did not conceive (censored) 1 2 1 3 1 4.1 1 7 1 7 1 8 2 8 2 9 2 9 2 9 2 11 3 24 3 24 3 4 4 4 6 6 9 9 9 10 13 16 Risk set at 6 months includes 18 women Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy and hydrotubation. BMJ 1982; 284: 1013-1014

  22. Corresponding Kaplan-Meier Curve

  23. S(t=6) = (54.2%)*(88.8%)=42.9% Corresponding Kaplan-Meier Curve 2 women conceive in the 6th month of the study, and one was censored between months 4 and 6. The risk set at event time 5 included 18 women. 16/18=88.8% “survived” event time 5 pregnancy-free.

  24. S(t=13)  22% (“eyeball” approximation) Skipping ahead to the 9th and final event time (months=16)…

  25. Raw data: Time (months) to conception or censoring in 38 sub-fertile women after laparoscopy and hydrotubation (1982 study) Conceived (event) Did not conceive (censored) 1 2 1 3 1 4 1 7 1 7 1 8 2 8 2 9 2 9 2 9 2 11 3 24 3 24 3 4 4 4 6 6 9 9 9 10 13 16 2 remaining at 16 months (9th event time) Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy and hydrotubation. BMJ 1982; 284: 1013-1014

  26. S(t=16) =( 22%)*(2/3)=15% Skipping ahead to the 9th and final event time (months=16)… Tail here just represents that the final 2 women did not conceive (cannot make many inferences from the end of a KM curve)!

  27. Kaplan-Meier: SAS output The LIFETEST Procedure Product-Limit Survival Estimates Survival Standard Number Number time Survival Failure Error Failed Left 0.0000 1.0000 0 0 0 38 1.0000 . . . 1 37 1.0000 . . . 2 36 1.0000 . . . 3 35 1.0000 . . . 4 34 1.0000 . . . 5 33 1.0000 0.8421 0.1579 0.0592 6 32 2.0000 . . . 7 31 2.0000 . . . 8 30 2.0000 . . . 9 29 2.0000 . . . 10 28 2.0000 0.7105 0.2895 0.0736 1127 2.0000* . . . 11 26 3.0000 . . . 12 25 3.0000 . . . 13 24 3.0000 0.6285 0.3715 0.0789 14 23 3.0000* . . . 14 22 4.0000 . . . 15 21 4.0000 . . . 16 20 4.0000 0.5428 0.4572 0.0822 17 19 4.0000* . . . 17 18

  28. Kaplan-Meier: SAS output Survival Standard Number Number time Survival Failure Error Failed Left 6.0000 . . . 18 17 6.0000 0.4825 0.5175 0.0834 19 16 7.0000* . . . 19 15 7.0000* . . . 19 14 8.0000* . . . 19 13 8.0000* . . . 19 12 9.0000 . . . 20 11 9.0000 . . . 21 10 9.0000 0.3619 0.6381 0.0869 22 9 9.0000* . . . 22 8 9.0000* . . . 22 7 9.0000* . . . 22 6 10.0000 0.3016 0.6984 0.0910 23 5 11.0000* . . . 23 4 13.0000 0.2262 0.7738 0.0944 24 3 16.0000 0.1508 0.8492 0.0880 25 2 24.0000* . . . 25 1 24.0000* . . . 25 0 NOTE: The marked survival times are censored observations.

  29. Not so easy to get a plot of the actual hazard function! In SAS, need a complicated MACRO, and depends on assumptions…here’s what I get from Paul Allison’s macro for these data…

  30. Linear cumulative hazard function indicates a constant hazard. See lecture 1 if you want more math! At best, you can get the cumulative hazard function…

  31. Cumulative Hazard Function • If the hazard function is constant, e.g. h(t)=k, then the cumulative hazard function will be linear (and higher hazards will have steeper slopes): • If the hazard function is increasing with time, e.g. h(t)=kt, then the cumulative hazard function will be curved up, for example h(t)=kt gives a quadratic: • If the hazard function is decreasing over time, e.g. h(t)=k/t, then the cumulative hazard function should be curved down, for example:

  32. Kaplan-Meier: example 2 Researchers randomized44 patients with chronic active hepatitis were to receive prednisolone or no treatment (control), then compared survival curves. Example from: BMJ 1998;317:468-469 ( 15 August )

  33. Survival times (months) of 44 patients with chronic active hepatitis randomised to receive prednisolone or no treatment. Prednisolone (n=22) Control (n=22) 2 2 6 3 12 4 54 7 56 * 10 68 22 89 28 96 29 96 32 125* 37 128* 40 131* 41 140* 54 141* 61 143 63 145* 71 146 127* 148* 140* 162* 146* 168 158* 173* 167* 181* 182* Data from: BMJ 1998;317:468-469 ( 15 August ) *=censored

  34. Big drops at the end of the curve indicate few patients left. E.g., only 2/3 (66%) survived this drop. Kaplan-Meier: example 2 Are these two curves different? Misleading to the eye—apparent convergence by end of study. But this is due to 6 controls who survived fairly long, and 3 events in the treatment group when the sample size was small.

  35. Control group: Survival Standard Number Number time Survival Failure Error Failed Left 0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 3.000 0.9091 0.0909 0.0613 2 20 4.000 0.8636 0.1364 0.0732 3 19 7.000 0.8182 0.1818 0.0822 4 18 10.000 0.7727 0.2273 0.0893 5 17 22.000 0.7273 0.2727 0.0950 6 16 28.000 0.6818 0.3182 0.0993 7 15 29.000 0.6364 0.3636 0.1026 8 14 32.000 0.5909 0.4091 0.1048 9 13 37.000 0.5455 0.4545 0.1062 10 12 40.000 0.5000 0.5000 0.1066 11 11 41.000 0.4545 0.5455 0.1062 12 10 54.000 0.4091 0.5909 0.1048 13 9 61.000 0.3636 0.6364 0.1026 14 8 63.000 0.3182 0.6818 0.0993 15 7 71.000 0.2727 0.7273 0.0950 16 6 127.000* . . . 16 5 140.000* . . . 16 4 146.000* . . . 16 3 158.000* . . . 16 2 167.000* . . . 16 1 182.000* . . . 16 0 6 controls made it past 100 months.

  36. 5/6 of 54% rapidly drops the curve to 45%. 2/3 of 45% rapidly drops the curve to 30%. treated group: Survival Standard Number Number time Survival Failure Error Failed Left 0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 6.000 0.9091 0.0909 0.0613 2 20 12.000 0.8636 0.1364 0.0732 3 19 54.000 0.8182 0.1818 0.0822 4 18 56.000* . . . 4 17 68.000 0.7701 0.2299 0.0904 5 16 89.000 0.7219 0.2781 0.0967 6 15 96.000 . . . 7 14 96.000 0.6257 0.3743 0.1051 8 13 125.000* . . . 8 12 128.000* . . . 8 11 131.000* . . . 8 10 140.000* . . . 8 9 141.000* . . . 8 8 143.000 0.5475 0.4525 0.1175 9 7 145.000* . . . 9 6 146.000 0.4562 0.5438 0.1285 10 5 148.000* . . . 10 4 162.000* . . . 10 3 168.000 0.3041 0.6959 0.1509 11 2 173.000* . . . 11 1 181.000* . . . 11 0

  37. Point-wise confidence intervals We will not worry about mathematical formula for confidence bands. The important point is that there is a confidence interval for each estimate of S(t). (SAS uses Greenwood’s formula.)

  38. Log-rank test Test of Equality over Strata Pr > Test Chi-Square DF Chi-Square Log-Rank 4.6599 1 0.0309 Wilcoxon 6.5435 1 0.0105 -2Log(LR) 5.4096 1 0.0200 Chi-square test (with 1 df) of the (overall) difference between the two groups. Groups appear significantly different.

  39. Log-rank test Log-rank test is just a Cochran-Mantel-Haenszel chi-square test! Anyone remember (know) what this is?

  40. Event No Event Group 1 a b Group 2 c d CMH test of conditional independence K Strata = unique event times Nk

  41. Event No Event Group 1 a b Group 2 c d CMH test of conditional independence K Strata = unique event times Nk

  42. Event No Event Group 1 a b Group 2 c d Why is this the expected value in each stratum? Variance is the variance of a hypergeometric distribution CMH test of conditional independence How do you know that this is a chi-square with 1 df?

  43. At risk=22 1st event at month 2. Event time 1 (2 months), control group: Survival Standard Number Number time Survival Failure Error Failed Left 0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 3.000 0.9091 0.0909 0.0613 2 20 4.000 0.8636 0.1364 0.0732 3 19 7.000 0.8182 0.1818 0.0822 4 18 10.000 0.7727 0.2273 0.0893 5 17 22.000 0.7273 0.2727 0.0950 6 16 28.000 0.6818 0.3182 0.0993 7 15 29.000 0.6364 0.3636 0.1026 8 14 32.000 0.5909 0.4091 0.1048 9 13 37.000 0.5455 0.4545 0.1062 10 12 40.000 0.5000 0.5000 0.1066 11 11 41.000 0.4545 0.5455 0.1062 12 10 54.000 0.4091 0.5909 0.1048 13 9 61.000 0.3636 0.6364 0.1026 14 8 63.000 0.3182 0.6818 0.0993 15 7 71.000 0.2727 0.7273 0.0950 16 6 127.000* . . . 16 5 140.000* . . . 16 4 146.000* . . . 16 3 158.000* . . . 16 2 167.000* . . . 16 1 182.000* . . . 16 0

  44. At risk=22 1st event at month 2. Event time 1 (2 months), treated group: Survival Standard Number Number time Survival Failure Error Failed Left 0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 6.000 0.9091 0.0909 0.0613 2 20 12.000 0.8636 0.1364 0.0732 3 19 54.000 0.8182 0.1818 0.0822 4 18 56.000* . . . 4 17 68.000 0.7701 0.2299 0.0904 5 16 89.000 0.7219 0.2781 0.0967 6 15 96.000 . . . 7 14 96.000 0.6257 0.3743 0.1051 8 13 125.000* . . . 8 12 128.000* . . . 8 11 131.000* . . . 8 10 140.000* . . . 8 9 141.000* . . . 8 8 143.000 0.5475 0.4525 0.1175 9 7 145.000* . . . 9 6 146.000 0.4562 0.5438 0.1285 10 5 148.000* . . . 10 4 162.000* . . . 10 3 168.000 0.3041 0.6959 0.1509 11 2 173.000* . . . 11 1 181.000* . . . 11 0

  45. Event No Event treated 1 21 control 1 21 Stratum 1= event time 1 Event time 1: 1 died from each group. (22 at risk in each group) 44

  46. At risk=21 Next event at month 3. Event time 2 (3 months), control group: Survival Standard Number Number time Survival Failure Error Failed Left 0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 3.000 0.9091 0.0909 0.0613 2 20 4.000 0.8636 0.1364 0.0732 3 19 7.000 0.8182 0.1818 0.0822 4 18 10.000 0.7727 0.2273 0.0893 5 17 22.000 0.7273 0.2727 0.0950 6 16 28.000 0.6818 0.3182 0.0993 7 15 29.000 0.6364 0.3636 0.1026 8 14 32.000 0.5909 0.4091 0.1048 9 13 37.000 0.5455 0.4545 0.1062 10 12 40.000 0.5000 0.5000 0.1066 11 11 41.000 0.4545 0.5455 0.1062 12 10 54.000 0.4091 0.5909 0.1048 13 9 61.000 0.3636 0.6364 0.1026 14 8 63.000 0.3182 0.6818 0.0993 15 7 71.000 0.2727 0.7273 0.0950 16 6 127.000* . . . 16 5 140.000* . . . 16 4 146.000* . . . 16 3 158.000* . . . 16 2 167.000* . . . 16 1 182.000* . . . 16 0

  47. At risk=21 No events at 3 months Event time 2 (3 months), treated group: Survival Standard Number Number time Survival Failure Error Failed Left 0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 6.000 0.9091 0.0909 0.0613 2 20 12.000 0.8636 0.1364 0.0732 3 19 54.000 0.8182 0.1818 0.0822 4 18 56.000* . . . 4 17 68.000 0.7701 0.2299 0.0904 5 16 89.000 0.7219 0.2781 0.0967 6 15 96.000 . . . 7 14 96.000 0.6257 0.3743 0.1051 8 13 125.000* . . . 8 12 128.000* . . . 8 11 131.000* . . . 8 10 140.000* . . . 8 9 141.000* . . . 8 8 143.000 0.5475 0.4525 0.1175 9 7 145.000* . . . 9 6 146.000 0.4562 0.5438 0.1285 10 5 148.000* . . . 10 4 162.000* . . . 10 3 168.000 0.3041 0.6959 0.1509 11 2 173.000* . . . 11 1 181.000* . . . 11 0

  48. Event No Event treated 0 21 control 1 20 Stratum 2= event time 2 Event time 2: At 3 months, 1 died in the control group. At that time 21 from each group were at risk 42

  49. At risk=20 1 event at month 4. Event time 3 (4 months), control group: Survival Standard Number Number time Survival Failure Error Failed Left 0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 3.000 0.9091 0.0909 0.0613 2 20 4.000 0.8636 0.1364 0.0732 3 19 7.000 0.8182 0.1818 0.0822 4 18 10.000 0.7727 0.2273 0.0893 5 17 22.000 0.7273 0.2727 0.0950 6 16 28.000 0.6818 0.3182 0.0993 7 15 29.000 0.6364 0.3636 0.1026 8 14 32.000 0.5909 0.4091 0.1048 9 13 37.000 0.5455 0.4545 0.1062 10 12 40.000 0.5000 0.5000 0.1066 11 11 41.000 0.4545 0.5455 0.1062 12 10 54.000 0.4091 0.5909 0.1048 13 9 61.000 0.3636 0.6364 0.1026 14 8 63.000 0.3182 0.6818 0.0993 15 7 71.000 0.2727 0.7273 0.0950 16 6 127.000* . . . 16 5 140.000* . . . 16 4 146.000* . . . 16 3 158.000* . . . 16 2 167.000* . . . 16 1 182.000* . . . 16 0

  50. At risk=21 Event time 3 (4 months), treated group: Survival Standard Number Number time Survival Failure Error Failed Left 0.000 1.0000 0 0 0 22 2.000 0.9545 0.0455 0.0444 1 21 6.000 0.9091 0.0909 0.0613 2 20 12.000 0.8636 0.1364 0.0732 3 19 54.000 0.8182 0.1818 0.0822 4 18 56.000* . . . 4 17 68.000 0.7701 0.2299 0.0904 5 16 89.000 0.7219 0.2781 0.0967 6 15 96.000 . . . 7 14 96.000 0.6257 0.3743 0.1051 8 13 125.000* . . . 8 12 128.000* . . . 8 11 131.000* . . . 8 10 140.000* . . . 8 9 141.000* . . . 8 8 143.000 0.5475 0.4525 0.1175 9 7 145.000* . . . 9 6 146.000 0.4562 0.5438 0.1285 10 5 148.000* . . . 10 4 162.000* . . . 10 3 168.000 0.3041 0.6959 0.1509 11 2 173.000* . . . 11 1 181.000* . . . 11 0

More Related