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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.
Citation preview
101/2015
EPI 5344:Survival Analysis in
EpidemiologyHazard
March 3, 2015
Dr. N. Birkett,School of Epidemiology, Public Health &
Preventive Medicine,University of Ottawa
201/2015
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’
301/2015
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.
401/2015
Hazard estimation (1)• If hazard is not constant, how does it vary
over time?
501/2015
Hazard estimation (2)
• How can we estimate the hazard? – Parametric methods (not discussed today)– Non-parametric methods
• We can estimate:– h(t)– H(t)
601/2015
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.
701/2015
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
(ui).
01/2015 8
h(t) direct estimation
901/2015
Hazard estimation (5)• Actuarial method to estimate h(t)
– Length is generally the same for each interval (ui).
Standard ID formula from Epi
01/2015 10
A B C D E F G H IYear # 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
1990 10,000 5,000 1,500 7,500 0.2 0.8 0.8 0.2
1991 3,500 1,750 525 2,625 0.2 0.8 0.64 0.36
1992 1,225 612 184 919 0.2 0.8 0.51 0.49
1993 429 215 64 322 0.2 0.8 0.41 0.59
1994 150 75 23 113 0.2 0.8 0.33 0.67
• Last week, we used this data to illustrate actuarial method• Let’s use it to estimate h(t)
01/2015 11
Nt Wt It h(t)Year # people
still alive# lost # people
dying in this year
Effective # at risk
Prob die in year
1990 10,000 5,000 1,500 7,500 0.2
1991 3,500 1,750 525 2,625 0.2
1992 1,225 612 184 919 0.2
1993 429 215 64 322 0.2
1994 150 75 23 113 0.2
1990:
Nt Wt It h(t)Year # people
still alive# lost # people
dying in this year
Effective # at risk
Prob die in year
1990 10,000 5,000 1,500 7,500 0.2 0.222
1991 3,500 1,750 525 2,625 0.2
1992 1,225 612 184 919 0.2
1993 429 215 64 322 0.2
1994 150 75 23 113 0.2
1991:
Nt Wt It h(t)Year # people
still alive# lost # people
dying in this year
Effective # at risk
Prob die in year
1990 10,000 5,000 1,500 7,500 0.2 0.222
1991 3,500 1,750 525 2,625 0.2 0.222
1992 1,225 612 184 919 0.2
1993 429 215 64 322 0.2
1994 150 75 23 113 0.2
Nt Wt It h(t)Year # people
still alive# lost # people
dying in this year
Effective # at risk
Prob die in year
1990 10,000 5,000 1,500 7,500 0.2 0.222
1991 3,500 1,750 525 2,625 0.2 0.222
1992 1,225 612 184 919 0.2 0.222
1993 429 215 64 322 0.2 0.222
1994 150 75 23 113 0.2 0.222
1201/2015
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
1301/2015
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
di = # with eventui = ti+1 – ti
ni = size of risk set at ‘t’
1401/2015
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
1501/2015
• 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)
1601/2015
• 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 ‘hi’ 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)
01/2015 17
H(t) estimation based on piecewise estimation of h(t)
1801/2015
• 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)
1901/2015
• 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)
di = # with event at ‘ti’ni = size of risk set at ‘ti’
2001/2015
• 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)
2101/2015
• For those who care, methods 3 and 4 are very similar
• From KM, the estimate of S(t) is:
Hazard estimation (14)
2201/2015
• Hence, we have:
• But, for small values, we have:• So, we get:
Hazard estimation (15)
2301/2015
Numerical exampleID Time(mons) Censored
1 14 XXXXX
2 22
3 29
4 37 XXXXX
5 45 XXXXX
6 46
7 61
8 76 XXXXX
9 92 XXXXX
10 111 XXXXX
Very coarse:10 events in 10 years
Year # people under follow-up
# lost # people dying in this year
h(t) H(t)
0-1 10 0 0 0 0
1-2 10 1 1
2-3
3-4
4-5
5-6
6-7
7-8
8-9
01/2015 24
Actuarial Method for h(t)
Year # people under follow-up
# lost # people dying in this year
h(t) H(t)
0-1 10 0 0 0 0
1-2 10 1 1 0.111 0.111
2-3 8 0 1
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)
0-1 10 0 0 0 0
1-2 10 1 1 0.111 0.111
2-3 8 0 1 0.133 0.244
3-4 7 2 1
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)
0-1 10 0 0 0 0
1-2 10 1 1 0.111 0.111
2-3 8 0 1 0.133 0.244
3-4 7 2 1 0.182 0.426
4-5 4 0 0 0 0.426
5-6 4 0 1 0.286 0.712
6-7 3 1 0 0 0.712
7-8 2 1 0 0 0.712
8-9 1 1 0 0 0.712
25
Nelson-Aalen estimate of H(t)Interval Computation H(t) from actuarial method
0-22 H(t) = 0 0.111
22+-29 H(t) = 0.111
29+-46 H(t) = 0.426
46+-51 H(t) = 0.712
51+ H(t) = 0.712
01/2015
2601/2015
A new example
2701/2015
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:
01/2015
Interval H(t) S(t) Cum Incid(t)0-22 0 1.0 0.0
22+-29 0.111 0.895 0.105
29+-46 0.236 0.790 0.210
46+-51 0.436 0.647 0.353
51+ 0.686 0.504 0.496
2901/2015
3001/2015
Key for testing proportional hazards assumption (later)
3101/2015
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)
01/2015
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
01/2015
3401/2015
3501/2015
Simple hazard estimates using actuarial method
Adjusted hazard estimates using actuarial method: last interval ends at 53 weeks, not 60 weeks
3601/2015
3701/2015
38
Proportional Hazards
01/2015
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
01/2015
4001/2015
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: λe
– Control: λc
01/2015
4201/2015
λc
λe
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
01/2015
4401/2015
λc
λe
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.
01/2015
• The difference in the log-hazards is constant over time.– Means that the curves are a fixed distance apart
4601/2015
λc
λe
47
Proportional Hazards (5)• If PH is true, then we frequently designate
one group as the reference group (0).
01/2015
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:
01/2015
49
Proportional Hazards (7)
• Most common form for this model is:
01/2015
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
01/2015
5101/2015
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