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

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EPI 5344:Survival Analysis in

EpidemiologyHazard

March 3, 2015

Dr. N. Birkett,School of Epidemiology, Public Health &

Preventive Medicine,University of Ottawa

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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’

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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.

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Hazard estimation (1)• If hazard is not constant, how does it vary

over time?

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Hazard estimation (2)

• How can we estimate the hazard? – Parametric methods (not discussed today)– Non-parametric methods

• We can estimate:– h(t)– H(t)

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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.

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• 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).

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h(t) direct estimation

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Hazard estimation (5)• Actuarial method to estimate h(t)

– Length is generally the same for each interval (ui).

Standard ID formula from Epi

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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)

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Nt Wt It h(t)Year # 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:



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:



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



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

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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

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• 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’

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• 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

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• 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


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• 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


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H(t) estimation based on piecewise estimation of h(t)

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• 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


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• 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:


di = # with event at ‘ti’ni = size of risk set at ‘ti’

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• Another approach to estimate H(t)• Use -log(S(t))

– from our basic formulae, we have:

– Estimate S(t) and convert using this formula


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• For those who care, methods 3 and 4 are very similar

• From KM, the estimate of S(t) is:


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• Hence, we have:

• But, for small values, we have:• So, we get:


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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

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Actuarial Method for h(t)



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



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



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

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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

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A new example

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H(t) has many uses, largely based on:


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Nelson-Aalen (2)

Estimating H(t) gives another way to estimate S(t). Uses formula:

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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

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Key for testing proportional hazards assumption (later)

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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).

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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)

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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

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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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Proportional Hazards

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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

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Hazard functions for 2 hypothetical groups in a RCT

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• 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

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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

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• 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

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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)

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Proportional Hazards (4)• PH is easier to see if we look at the

logarithm of the hazards.

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• The difference in the log-hazards is constant over time.– Means that the curves are a fixed distance apart

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λc

λe

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Proportional Hazards (5)• If PH is true, then we frequently designate

one group as the reference group (0).

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Re-write this to get:

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Proportional Hazards (6)• In above equation, HR can be affected by

patient characteristics– Age– Sex– Residence– Baseline disease severity

• Can model this as:

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• Most common form for this model is:

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Model underlies the Cox regression approach.

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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

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Documents

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