Upload
rolf-miller
View
218
Download
1
Tags:
Embed Size (px)
Citation preview
Types of outcome
Continuous OLS
Linear regression
Binary Binary regression
Logistic or probit regression
Time to event data Survival or event history analysis
Examples of time to event data
Time to death Time to incidence of disease Unemployed - time till find job Time to birth of first child Smokers – time till quit smoking
Time to event data
Analyse durations or length of time to reach endpoint
Data are usually censored Don’t follow sample long enough for everyone to get to the
endpoint (e.g. death)
States
States are categories of the outcome variable of interest Each person occupies exactly one state at any moment in
time Examples
alive, dead single, married, divorced, widowed never smoker, smoker, ex-smoker
Set of possible states called the state space
Events
A transition from one state to another From an origin state to a destination state Possible events depend on the state space Examples
From smoker to ex-smoker From married to widowed
Not all transitions can be events E.g. from smoker to never smoker
Risk period
Not all people can experience each state throughout the study period
To be able to have a particular event, one must be in the origin state at some stage
Example can only experience divorce if married
The period of time that someone is at risk of a particular event is called the risk period
All subjects at risk of an event at a point in time called the risk set
Duration
Event history analysis is to do with the analysis of the duration of a nonoccurrence of an event or the length of time during the risk period
Examples Duration of marriage Length of life
In practice we model the probability of a transition conditional on being in the risk set
Example data
ID Entry date Died End date
1 01/01/1991 01/01/2008
2 01/01/1991 01/01/2000 01/01/2000
3 01/01/1995 01/01/2005
4 01/01/1994 01/07/2004 01/07/2004
Censoring
An observation is censored if it has incomplete information
We will only consider right censoring
That is, the person did not have an event during the time that they were studied
Common reasons for right censoring the study ends
the person drops-out of the study
the person has to be taken off a drug
Data
Survival or event history data characterised by 2 variables Time or duration of risk period
Failure (event)
• 1 if not survived or event observed
• 0 if censored or event not yet occurred
What is the data structure?
ID Entry date Died End date Duration Event
1 01/01/1991 01/01/2008 17.0 0
2 01/01/1991 01/01/2000 01/01/2000 9.0 1
3 01/01/1995 01/01/2005 10.0 0
4 01/01/1994 01/07/2004 01/07/2004 10.5 1
The row is a personThe tricky part is often calculating the durationRemember we need an indicator for observed events/ censored cases
Worked example
Random 20% sample from BHPS
Waves 1 – 15
One record per person/wave
Outcome: Duration of cohabitation
Conditions on cohabiting in first wave
Survival time: years from entry to the study in 1991 till year living without a partner
The data
+----------------------------+ | pid wave mastat | |----------------------------| | 10081798 1 married | | 10081798 2 married | | 10081798 3 married | | 10081798 4 married | | 10081798 5 married | | 10081798 6 married | | 10081798 7 widowed | | 10081798 8 widowed | | 10081798 9 widowed | | 10081798 10 widowed | | 10081798 11 widowed | | 10081798 12 widowed | | 10081798 13 widowed | | 10081798 14 widowed | | 10081798 15 widowed | |----------------------------|
Duration = 6 years
Event = 1
Ignore data after event = 1
The data (continued)
+----------------------------+ | pid wave mastat | |----------------------------| | 10162747 1 living a | | 10162747 2 living a | | 10162747 3 living a | | 10162747 4 living a | | 10162747 5 living a | | 10162747 6 living a | | 10162747 10 separate | | 10162747 11 . | | 10162747 12 . | | 10162747 13 . | | 10162747 14 never ma | | 10162747 15 never ma | +----------------------------+
Note missing waves before event
Preparing the data
. sort pid wave . generate skey=1 if wave==1&(mastat==1|mastat==2) . by pid: replace skey=skey[_n-1] if wave~=1 . keep if skey==1 . drop skey . . stset wave,id(pid) failure(mastat==3/6) id: pid failure event: mastat == 3 4 5 6 obs. time interval: (wave[_n-1], wave] exit on or before: failure ------------------------------------------------------------------------------ 15058 total obs. 1628 obs. begin on or after (first) failure ------------------------------------------------------------------------------ 13430 obs. remaining, representing 1357 subjects 270 failures in single failure-per-subject data 13612 total analysis time at risk, at risk from t = 0 earliest observed entry t = 0 last observed exit t = 15
Select records for respondents who were cohabiting in 1991
Declare that you want to set the data to survival time
Important to check that you have set data as intended
Checking the data setup
. list pid wave mastat _st _d _t _t0 if pid==10081798,sepby(pid) noobs +-------------------------------------------------+ | pid wave mastat _st _d _t _t0 | |-------------------------------------------------| | 10081798 1 married 1 0 1 0 | | 10081798 2 married 1 0 2 1 | | 10081798 3 married 1 0 3 2 | | 10081798 4 married 1 0 4 3 | | 10081798 5 married 1 0 5 4 | | 10081798 6 married 1 0 6 5 | | 10081798 7 widowed 1 1 7 6 | | 10081798 8 widowed 0 . . . | | 10081798 9 widowed 0 . . . | | 10081798 10 widowed 0 . . . | | 10081798 11 widowed 0 . . . | | 10081798 12 widowed 0 . . . | | 10081798 13 widowed 0 . . . | | 10081798 14 widowed 0 . . . | | 10081798 15 widowed 0 . . . | +-------------------------------------------------+ 1 if observation is to be used
and 0 otherwise
1 if event, 0 if censoring orevent not yet occurred
time of exit
time of entry
Checking the data setup
. list pid wave mastat _st _d _t _t0 if pid==10162747,sepby(pid) noobs +--------------------------------------------------+ | pid wave mastat _st _d _t _t0 | |--------------------------------------------------| | 10162747 1 living a 1 0 1 0 | | 10162747 2 living a 1 0 2 1 | | 10162747 3 living a 1 0 3 2 | | 10162747 4 living a 1 0 4 3 | | 10162747 5 living a 1 0 5 4 | | 10162747 6 living a 1 0 6 5 | | 10162747 10 separate 1 1 10 6 | | 10162747 11 . 0 . . . | | 10162747 12 . 0 . . . | | 10162747 13 . 0 . . . | | 10162747 14 never ma 0 . . . | | 10162747 15 never ma 0 . . . | +--------------------------------------------------+ How do we know when
this person separated?
Trying again!
. fillin pid wave . stset wave,id(pid) failure(mastat==3/6) exit(mastat==3/6 .) id: pid failure event: mastat == 3 4 5 6 obs. time interval: (wave[_n-1], wave] exit on or before: mastat==3 4 5 6 . ------------------------------------------------------------------------------ 20355 total obs. 7524 obs. begin on or after exit ------------------------------------------------------------------------------ 12831 obs. remaining, representing 1357 subjects 234 failures in single failure-per-subject data 12831 total analysis time at risk, at risk from t = 0 earliest observed entry t = 0 last observed exit t = 15
. list pid wave mastat _st _d _t _t0 if pid==10162747,sepby(pid) noobs +--------------------------------------------------+ | pid wave mastat _st _d _t _t0 | |--------------------------------------------------| | 10162747 1 living a 1 0 1 0 | | 10162747 2 living a 1 0 2 1 | | 10162747 3 living a 1 0 3 2 | | 10162747 4 living a 1 0 4 3 | | 10162747 5 living a 1 0 5 4 | | 10162747 6 living a 1 0 6 5 | | 10162747 7 . 1 0 7 6 | | 10162747 8 . 0 . . . | | 10162747 9 . 0 . . . | | 10162747 10 separate 0 . . . | | 10162747 11 . 0 . . . | | 10162747 12 . 0 . . . | | 10162747 13 . 0 . . . | | 10162747 14 never ma 0 . . . | | 10162747 15 never ma 0 . . . | +--------------------------------------------------+
Checking the new data setup
Now censored instead of an event
Summarising time to event data
Individuals followed up for different lengths of time
So can’t use prevalence rates (% people who have an event)
Use rates instead that take account of person years at risk Incidence rate per year
Death rate per 1000 person years
Summarising time to event data
Number of observationsPerson-years Rate per year
<25% of sample had event by 15 elapsed years
. stsum failure _d: mastat == 3 4 5 6 analysis time _t: wave exit on or before: mastat==3 4 5 6 . id: pid | incidence no. of |------ Survival time -----| | time at risk rate subjects 25% 50% 75% ---------+--------------------------------------------------------------------- total | 12831 .0182371 1357 . . .
List the cumulative hazard function
Default is the survivor function. sts list, failure failure _d: mastat == 3 4 5 6 analysis time _t: wave exit on or before: mastat==3 4 5 6 . id: pid Beg. Net Failure Std. Time Total Fail Lost Function Error [95% Conf. Int.] ------------------------------------------------------------------------------- 2 1357 29 162 0.0214 0.0039 0.0149 0.0306 3 1166 33 89 0.0491 0.0061 0.0384 0.0625 4 1044 16 64 0.0636 0.0070 0.0513 0.0789 5 964 35 58 0.0976 0.0088 0.0818 0.1164 6 871 12 34 0.1101 0.0094 0.0931 0.1300 7 825 20 24 0.1316 0.0103 0.1128 0.1534 8 781 14 17 0.1472 0.0109 0.1271 0.1701 9 750 12 30 0.1609 0.0115 0.1398 0.1848 10 708 15 23 0.1786 0.0121 0.1563 0.2038 11 670 9 32 0.1897 0.0125 0.1666 0.2155 12 629 8 16 0.2000 0.0128 0.1762 0.2266 13 605 13 24 0.2172 0.0134 0.1922 0.2449 14 568 8 24 0.2282 0.0138 0.2025 0.2566 15 536 10 526 0.2426 0.0143 0.2160 0.2719 -------------------------------------------------------------------------------
Graphs of survival time
Kaplan-Meier estimate of survival curve
The Kaplan-Meier method estimates the cumulative probability of an individual surviving after baseline to any time, t
Kaplan-Meier graphs
Can read off the estimated probability of surviving a relationship at any time point on the graph E.g. at 5 years 88% are still cohabiting
The survival probability only changes when an event occurs
So the graph is stepped and not a smooth curve
0.0
00
.25
0.5
00
.75
1.0
0
0 5 10 15analysis time
sex = male sex = female
Comparing survival by group using Kaplan-Meier graphs
Testing equality of survival curves among groups
The log-rank test
A non –parametric test that assesses the null hypothesis that there are no differences in survival times between groups
. sts test sex, logrank failure _d: mastat == 3 4 5 6 analysis time _t: wave exit on or before: mastat==3 4 5 6 . id: pid Log-rank test for equality of survivor functions | Events Events sex | observed expected -------+------------------------- male | 98 113.59 female | 136 120.41 -------+------------------------- Total | 234 234.00 chi2(1) = 4.25 Pr>chi2 = 0.0392
Log-rank test example
Significant difference between men and women
Event History with Cox ModelEvent History with Cox regression model
No longer modelling the duration
Modelling the hazard
Hazard: measure of the probability that an event occurs at time t conditional on it not having occurred up until t
Also known as the Cox proportional hazard model
Some hazard shapes
Increasing Onset of Alzheimer's
Decreasing Survival after surgery
U-shaped Age specific mortality
Constant Time till next email arrives
Cox regression model
Regression model for survival analysis Can model time invariant and time varying
explanatory variables Produces estimated hazard ratios (sometimes
called rate ratios or risk ratios) Regression coefficients are on a log scale
Exponentiate to get hazard ratio Similar to odds ratios from logistic models
Cox regression equation
).......exp()()( 22110 inniii xxxthth
)(0 th
)(thi
is the baseline hazard function and can take any formIt is estimated from the data (non parametric)
is the hazard function for individual i
inii xxx ,....,, 21
n ,....,, 21
are the covariates
are the regression coefficients estimated from the data
Effect of covariates is constant over time (parameterised)This is the proportional hazards assumption
Therefore, Cox regression referred to as a semi-parametric model
Cox regression in Stata
Will first model a time invariant covariate (sex) on risk of partnership ending
Then will add a time dependent covariate (age) to the model
Cox regression in Stata
. stcox female failure _d: mastat == 3 4 5 6 analysis time _t: wave exit on or before: mastat==3 4 5 6 . id: pid Cox regression -- Breslow method for ties No. of subjects = 1357 Number of obs = 12337 No. of failures = 234 Time at risk = 12337 LR chi2(1) = 4.18 Log likelihood = -1574.5782 Prob > chi2 = 0.0409 ------------------------------------------------------------------------------ _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- female | 1.30913 .1734699 2.03 0.042 1.009699 1.697358 ------------------------------------------------------------------------------
Interpreting output from Cox regression
Cox model has no intercept It is included in the baseline hazard
In our example, the baseline hazard is when sex=1 (male)
The hazard ratio is the ratio of the hazard for a unit change in the covariate HR = 1.3 for women vs. men The risk of partnership breakdown is increased by 30% for women
compared with men
Hazard ratio assumed constant over time At any time point, the hazard of partnership breakdown for a woman
is 1.3 times the hazard for a man
Interpreting output from Cox regression (ii)
The hazard ratio is equivalent to the odds that a female has a partnership breakdown before a man
The probability of having a partnership breakdown first is = (hazard ratio) / (1 + hazard ratio)
So in our example, a HR of 1.30 corresponds to a
probability of 0.57 that a woman will experience a partnership breakdown first
The probability or risk of partnership breakdown can be different each year but the relative risk is constant
So if we know that the probability of a man having a partnership breakdown in the following year is 1.5% then the probability of a woman having a partnership breakdown in the following year is
0.015*1.30 = 1.95%
.01
2.0
14
.01
6.0
18
.02
Sm
oot
hed
haza
rd fu
nct
ion
4 6 8 10 12analysis time
hazard function varying over time
Cox proportional hazards regression:
Time dependent covariates
Examples Current age group rather than age at baseline
GHQ score may change over time and predict break-ups
Will use age to predict duration of cohabitation Nonlinear relationship hypothesised
Recode age into 8 equally spaced age groups
Cox regression with time dependent covariates
. xi: stcox female i.agecat i.agecat _Iagecat_0-7 (naturally coded; _Iagecat_0 omitted) failure _d: mastat == 3 4 5 6 analysis time _t: wave exit on or before: mastat==3 4 5 6 . id: pid Cox regression -- Breslow method for ties No. of subjects = 1357 Number of obs = 12337 No. of failures = 234 Time at risk = 12337 LR chi2(8) = 78.44 Log likelihood = -1537.4472 Prob > chi2 = 0.0000 ------------------------------------------------------------------------------ _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- female | 1.3705 .1842481 2.34 0.019 1.05304 1.783666 _Iagecat_1 | .5838602 .1883578 -1.67 0.095 .3102449 1.098786 _Iagecat_2 | .311325 .1039311 -3.50 0.000 .1618279 .5989281 _Iagecat_3 | .2136714 .0737986 -4.47 0.000 .1085813 .4204725 _Iagecat_4 | .2225187 .0811395 -4.12 0.000 .1088888 .4547261 _Iagecat_5 | .4770023 .1691695 -2.09 0.037 .238035 .9558732 _Iagecat_6 | 1.203702 .4306775 0.52 0.604 .5969856 2.427023 _Iagecat_7 | 1.644141 .9677715 0.84 0.398 .518688 5.21161 ------------------------------------------------------------------------------
Cox regression assumptions
Assumption of proportional hazards
No censoring patterns
True starting time
Plus assumptions for all modelling Sufficient sample size, proper model specification, independent
observations, exogenous covariates, no high multicollinearity, random sampling, and so on
Proportional hazards assumption
Cox regression with time-invariant covariates assumes that the ratio of hazards for any two observations is the same across time periods
This can be a false assumption, for example using age at baseline as a covariate
If a covariate fails this assumption for hazard ratios that increase over time for that covariate,
relative risk is overestimated for ratios that decrease over time, relative risk is
underestimated standard errors are incorrect and significance tests are
decreased in power
Testing the proportional hazards assumption
Graphical methods Comparison of Kaplan-Meier observed & predicted curves
by group. Observed lines should be close to predicted Survival probability plots (cumulative survival against time
for each group). Lines should not cross Log minus log plots (minus log cumulative hazard against
log survival time). Lines should be parallel
Testing the proportional hazards assumption
Formal tests of proportional hazard assumption
Include an interaction between the covariate and a function of time. Log time often used but could be any function. If significant then assumption violated
Test the proportional hazards assumption on the basis of partial residuals. Type of residual known as Schoenfeld residuals.
When assumptions are not met
If categorical covariate, include the variable as a strata variable
Allows underlying hazard function to differ between categories and be non proportional
Estimates separate underlying baseline hazard for each stratum
When assumptions are not met
If a continuous covariate
Consider splitting the follow-up time. For example, hazard may be proportional within first 5 years, next 5-10 years and so on
Could covariate be included as time dependent covariate?
There are different survival regression methods (e.g. parametric model)
Censoring assumptions
Censored cases must be independent of the survival distribution. There should be no pattern to these cases, which instead should be missing at random.
If censoring is not independent, then censoring is said to be informative
You have to judge this for yourself Usually don’t have any data that can be used to test the
assumption Think carefully about start and end dates Always check a sample of records
True starting time
The ideal model for survival analysis would be where there is a true zero time
If the zero point is arbitrary or ambiguous, the data series will be different depending on starting point. The computed hazard rate coefficients could differ, sometimes markedly
Conduct a sensitivity analysis to see how coefficients may change according to different starting points
Other extensions to survival analysis
Discrete (interval-censored) survival times
Repeated events
Multi-state models (more than 1 event type) Transition from employment to unemployment or leaving
labour market
Modelling type of exit from cohabiting relationship- separation/divorce/widowhood
Could you use logistic regression instead?
May produce similar results for short or fixed follow-up periods Examples
• everyone followed-up for 7 years
• maximum follow-up 5 years
Results may differ if there are varying follow-up times
If dates of entry and dates of events are available then better to use Cox regression
Finally….
This is just an introduction to survival/ event history analysis
Only reviewed the Cox regression model Also parametric survival methods
But Cox regression likely to suit type of analyses of interest to sociologists
Consider an intensive course if you want to use survival analysis in your own work