1. 01/20151 EPI 5344: Survival Analysis in Epidemiology Hazard March 3, 2015 Dr. N. Birkett, School of Epidemiology, Public Health & Preventive Medicine, University of Ottawa

2. 01/20152 Hazard (1) h(t) –Instantaneous hazard –Rate of event occurring at time ‘t’, conditional having survived event-free until time ‘t’ H(t) –Cumulative hazard –the ‘sum’ of all hazards from time ‘0’ to time ‘t’ –Area under the h(t) curve from ‘0’ to ‘t’

3. 01/20153 Hazard (2) Simplest survival model assumes a constant hazard –Yields an exponential survival curve –Leads to basic epidemiology formulae for incidence, etc. –More next week Can extend it using the piecewise model –Fits a different constant hazard for given follow- up time intervals.

4. 01/20154 Hazard estimation (1) If hazard is not constant, how does it vary over time?

5. 01/20155 Hazard estimation (2) How can we estimate the hazard? –Parametric methods (not discussed today) –Non-parametric methods We can estimate: –h(t) –H(t)

6. 01/20156 Hazard estimation (3) Preference is to estimate H(t) –Nelson-Aaalen method is main approach. Let’s look at direct estimation of h(t) –Works from a piece-wise constant hazard model Start by dividing follow-up time into intervals –Actuarial has pre-defined intervals –KM uses time between events as intervals.

7. 01/20157 Hazard estimation (4) Direct hazard estimation has issues –h(t) shows much random variation –Unstable estimates due to small event numbers in time intervals –Works ‘best’ for actuarial method since intervals are pre-defined –Length is generally the same for each interval (u i ).

8. 01/20158 h(t) direct estimation

9. 01/20159 Hazard estimation (5) Actuarial method to estimate h(t) –Length is generally the same for each interval (u i ). Standard ID formula from Epi

10. 01/201510 ABCDEFGHI Year# people still alive # lost# people dying in this year Effective # at risk Prob die in year Prob survive this year Cum. Prob of surviving to this year Cum. Prob of dying by this year 199010,0005,0001,5007,5000.20.8 0.2 1991 3,5001,750 5252,6250.20.80.640.36 1992 1,225 612 184 9190.20.80.510.49 1993 429 215 64 3220.20.80.410.59 1994 150 75 23 1130.20.80.330.67 Last week, we used this data to illustrate actuarial method Let’s use it to estimate h(t)

11. 01/201511 NtNt WtWt ItIt h(t) Year# people still alive # lost# people dying in this year Effective # at risk Prob die in year 199010,0005,0001,5007,5000.2 1991 3,5001,750 5252,6250.2 1992 1,225 612 184 9190.2 1993 429 215 64 3220.2 1994 150 75 23 1130.2 1990: NtNt WtWt ItIt h(t) Year# people still alive # lost# people dying in this year Effective # at risk Prob die in year 199010,0005,0001,5007,5000.20.222 1991 3,5001,750 5252,6250.2 1992 1,225 612 184 9190.2 1993 429 215 64 3220.2 1994 150 75 23 1130.2 1991: NtNt WtWt ItIt h(t) Year# people still alive # lost# people dying in this year Effective # at risk Prob die in year 199010,0005,0001,5007,5000.20.222 1991 3,5001,750 5252,6250.20.222 1992 1,225 612 184 9190.2 1993 429 215 64 3220.2 1994 150 75 23 1130.2 NtNt WtWt ItIt h(t) Year# people still alive # lost# people dying in this year Effective # at risk Prob die in year 199010,0005,0001,5007,5000.20.222 1991 3,5001,750 5252,6250.20.222 1992 1,225 612 184 9190.20.222 1993 429 215 64 3220.20.222 1994 150 75 23 1130.20.222

12. 01/201512 Hazard estimation (6) Person-time variant –Divide follow-up time into fixed intervals –Compute actual person-time in each interval (rather than using approximation). –Gives a slightly smoother curve

13. 01/201513 Hazard estimation (7) Kaplan-Meier method to estimate h(t) –‘interval’ is time between death events Varies irregularly –Formula has same structure as above person- time estimate given above d i = # with event u i = t i+1 – t i n i = size of risk set at ‘t’

14. 01/201514 Hazard estimation (8) Issues with using KM method to estimate h(t) –Normally, only have 1 or 2 in numerator –Makes estimates ‘unstable’ liable to considerable random variation and noise –Do not usually estimate h(t) from KM methods –Use a Kernel Smoothing approach to improve estimates

15. 01/201515 Estimating Cumulative hazard: H(t) –Measures the area under the h(t) curve. Tends to be more stable since it is based on number of events from ‘0’ to ‘t’ rather than number in the last interval Hazard estimation (9)

16. 01/201516 Simple approach –Estimate h(t) assuming a piece-wise constant model –H(t) is the sum of the pieces. –For each ‘piece’ before time ‘t’, compute product of the estimated ‘h i ’ for the interval multiplied by the length of the interval it is based on. –Add these up across all ‘pieces’ before time ‘t’. width of last ‘piece’ is up to ‘t’ only –Relates to the density method from epi Hazard estimation (10)

17. 01/201517 H(t) estimation based on piecewise estimation of h(t)

18. 01/201518 Four ways you can do this: Actuarial using ‘epi’ formula Actuarial using Person-time method Kaplan-Meier approach using Nelson-Aalen estimator Kaplan-Meier approach using –log(S(t)) We’ve discussed methods 1 and 2. Generate h(t) Just add things up Let’s talk about 3 and 4 Hazard estimation (11)

19. 01/201519 Nelson-Aalen estimator for H(t) Apply above approach defining intervals by using the time points for events Most commonly used approach to estimate H(t) Related to Kaplan-Meier method Compute H(t) at each time when event happens: Hazard estimation (12) d i = # with event at ‘t i ’ n i = size of risk set at ‘t i ’

20. 01/201520 Another approach to estimate H(t) Use -log(S(t)) –from our basic formulae, we have: –Estimate S(t) and convert using this formula Hazard estimation (13)

21. 01/201521 For those who care, methods 3 and 4 are very similar From KM, the estimate of S(t) is: Hazard estimation (14)

22. 01/201522 Hence, we have: But, for small values, we have: So, we get: Hazard estimation (15)

23. 01/201523 Numerical example IDTime(mons)Censored 114XXXXX 222 329 437XXXXX 545XXXXX 646 761 876XXXXX 992XXXXX 10111XXXXX Very coarse: 10 events in 10 years

24. Year# people under follow- up # lost# people dying in this year h(t)H(t) 0-1100000 1-21011 2-3 3-4 4-5 5-6 6-7 7-8 8-9 01/201524 Actuarial Method for h(t) Year# people under follow- up # lost# people dying in this year h(t)H(t) 0-1100000 1-210110.111 2-3801 3-4 4-5 5-6 6-7 7-8 8-9 Year# people under follow- up # lost# people dying in this year h(t)H(t) 0-1100000 1-210110.111 2-38010.1330.244 3-4721 4-5 5-6 6-7 7-8 8-9 Year# people under follow- up # lost# people dying in this year h(t)H(t) 0-1100000 1-210110.111 2-38010.1330.244 3-47210.1820.426 4-540000.426 5-64010.2860.712 6-731000.712 7-821000.712 8-911000.712

25. Nelson-Aalen estimate of H(t) IntervalComputationH(t) from actuarial method 0-22H(t) = 00.111 22 + -29H(t) =0.111 29 + -46H(t) =0.426 46 + -51H(t) =0.712 51 + H(t) =0.712 01/201525

26. 01/201526 A new example

27. 01/201527 H(t) has many uses, largely based on: Hazard estimation (10)

28. Nelson-Aalen (2) Estimating H(t) gives another way to estimate S(t). Uses formula: 01/201528 IntervalH(t)S(t)Cum Incid(t) 0-2201.00.0 22 + -290.1110.8950.105 29 + -460.2360.7900.210 46 + -510.4360.6470.353 51 + 0.6860.5040.496

29. 01/201529

30. 01/201530 Key for testing proportional hazards assumption (later)

31. 01/201531 Suppose the hazard is a constant (λ), then we have: Plot ‘ln(S(t))’ against ‘t’. A straight line indicates a constant hazard. Approach can be used to test other models (e.g. Weibull).

32. Smoothing & hazard estimation Using KM to give a direct estimate of h(t) is very unstable –Only 1 event per time point Instead, apply a smoothing method to generate an estimate of h(t) 01/201532

33. Example (from Allison) Recidivism data set –432 male inmates released from prison –Followed for 52 weeks –Dates of re-arrests were recorded –Study designed to examine the impact of a financial support programme on reducing re- arrest 01/201533

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35. 01/201535 Simple hazard estimates using actuarial method Adjusted hazard estimates using actuarial method: last interval ends at 53 weeks, not 60 weeks

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38. Proportional Hazards 01/201538

39. Proportional Hazards (1) Suppose we have two groups followed over time (say treatment groups in an RCT). How will the hazards in the two groups relate? –There need be no specific relationship –They could even go in opposite directions 01/201539

40. 01/201540 Hazard functions for 2 hypothetical groups in a RCT

41. Proportional Hazards (2) Often, it is reasonable to place restrictions on how the hazards relate Consider a situation where the hazard is constant over time: –Experimental:λ e –Control: λ c 01/201541

42. 01/201542 λcλc λeλe The ratio of the hazard in one group to the other is constant for all follow-up time. A simple example of Proportional Hazards

43. Proportional Hazards (3) What if the hazard is not constant over time? –Relationship between curves can be complex –It is common to make the assumption that the hazard curves are proportional over all follow- up time 01/201543

44. 01/201544 λcλc λeλe The hazard in the experimental group is a constant multiple of that in the control group for all follow-up time. Proportional Hazards (PH)

45. Proportional Hazards (4) PH is easier to see if we look at the logarithm of the hazards. 01/201545  The difference in the log-hazards is constant over time. – Means that the curves are a fixed distance apart

46. 01/201546 λcλc λeλe

47. Proportional Hazards (5) If PH is true, then we frequently designate one group as the reference group (0). 01/201547 Re-write this to get:

48. Proportional Hazards (6) In above equation, HR can be affected by patient characteristics –Age –Sex –Residence –Baseline disease severity Can model this as: 01/201548

49. Proportional Hazards (7) Most common form for this model is: 01/201549 Model underlies the Cox regression approach.

50. Reminder & Warning Proportional Hazards is an ASSUMPTION It need not be true Not all probability models for survival curves leads to PH PH is less likely to be true when the follow- up time gets very long 01/201550

51. 01/201551