Kristin Sainani Ph.D.http://www.stanford.edu/~kcobbStanford UniversityDepartment of Health Research and Policy
Note: also the recent diabetes study in a Swedish population, where they reduced heart disease 50% among type II diabetics. Very tough intervention. Multiple drugs. Attempts to quit smoking. Diet and exercise. Not everyone participated fully; for example, smoking cessation was utter failure. If everyone had participated more fully in the regime, there might have even been a stronger effect.
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33111
1.
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Cox Regression II
Kristin Sainani Ph.D.
http://www.stanford.edu/~kcobb
Stanford University
Department of Health Research and Policy
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Topics
Stratification
Age as time scale
Residuals
Repeated events
Intention-to-treat analysis for RCTs
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1. Stratification
Violations of PH assumption can be resolved by:
•Adding time*covariate interaction
•Adding other time-dependent version of the covariate
•Stratification
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Stratification
•Different stratum are allowed to have different baseline
hazard functions.
•Hazard functions do not need to be parallel between
different stratum.
•Essentially results in a “weighted” hazard ratio being
estimated: weighted over the different strata.
•Useful for “nuisance” confounders (where you do not care
to estimate the effect).
•Does not allow you to evaluate interaction or confounding
of stratification variable (will miss possible interactions).
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Males: 1, 3, 4, 10+, 12, 18 (subjects 1-6)
Females: 1, 4, 5, 9+ (subjects 7-10)
Example: stratify on gender
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Age is a common confounder in Cox
Regression, since age is strongly related to
death and disease.
You may control for age by adding baseline
age as a covariate to the Cox model.
A better strategy for large-scale longitudinal
surveys, such as NHANES, is to use age as
your time-scale (rather than time-in-study).
You may additionally stratify on birth cohort to
control for cohort effects.
2. Using age as the time-scale
in Cox Regression
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Age as time-scale
The risk set becomes everyone who was at
risk at a certain age rather than at a certain
event time.
The risk set contains everyone who was still
event-free at the age of the person who had
the event.
Requires enough people at risk at all ages
(such as in a large-scale, longitudinal survey).
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The likelihood with age as time
Event times: 3, 5, 7+, 12, 13+ (years-in-study)
Baseline ages: 28, 25, 40, 29, 30 (years)
Age at event or censoring: 31, 30, 47+, 41, 43+
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3. Residuals
Residuals are used to investigate the
lack of fit of a model to a given subject.
For Cox regression, there’s no easy
analog to the usual “observed minus
predicted” residual of linear regression
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Martingale residual
ci (1 if event, 0 if censored) minus the estimated
cumulative hazard to ti (as a function of fitted model) for
individual i:
ci-H(ti,Xi, ßi)
E.g., for a subject who was censored at 2 months, and whose predicted
cumulative hazard to 2 months was 20%
Martingale=0-.20 = -.20
E.g., for a subject who had an event at 13 months, and whose predicted
cumulative hazard to 13 months was 50%:
Martingale=1-.50 = +.50
Gives excess failures.
Martingale residuals are not symmetrically distributed,
even when the fitted model is correctly, so transform to
deviance residuals...
12.
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Deviance Residuals
The deviance residual is a normalized
transform of the martingale residual.
These residuals are much more
symmetrically distributed about zero.
Observations with large deviance
residuals are poorly predicted by the
model.
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Deviance Residuals
Behave like residuals from ordinary linear
regression
Should be symmetrically distributed around 0
and have standard deviation of 1.0.
Negative for observations with longer than
expected observed survival times.
Plot deviance residuals against covariates to
look for unusual patterns.
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Deviance Residuals
In SAS, option on the output statement:
Output out=outdata resdev=Varname
**Cannot get diagnostics in SAS if time-
dependent covariate in the model
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Example: uis data
Out of 628
observations,
a few in the
range of 3-SD
is not
unexpected
Pattern looks
fairly symmetric
around 0.
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Example: uis data
What do you
think this
cluster
represents?
19.
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Schoenfeld residuals
Schoenfeld (1982) proposed the first set of
residuals for use with Cox regression
packages
Schoenfeld D. Residuals for the proportional
hazards regresssion model. Biometrika, 1982,
69(1):239-241.
Instead of a single residual for each
individual, there is a separate residual for
each individual for each covariate
Note: Schoenfeld residuals are not defined for
censored individuals.
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Schoenfeld residuals
The Schoenfeld residual is defined as the covariate
value for the individual that failed minus its expected
value. (Yields residuals for each individual who failed,
for each covariate).
Expected value of the covariate at time ti = a weighted-
average of the covariate, weighted by the likelihood of
failure for each individual in the risk set at ti.
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was 56 rather than older?
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Example
5 people left in our risk set at event
time=7 months:
Female 55-year old smoker
Male 45-year old non-smoker
Female 67-year old smoker
Male 58-year old smoker
Male 70-year old non-smoker
The 55-year old female smoker is the one who
has the event…
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Example
Based on our model, we can calculate a
predicted probability of death by time 7 for
each person (call it “p-hat”):
Female 55-year old smoker: p-hat=.10
Male 45-year old non-smoker : p-hat=.05
Female 67-year old smoker : p-hat=.30
Male 58-year old smoker : p-hat=.20
Male 70-year old non-smoker : p-hat=.30
Thus, the expected value for the AGE of the person who
failed is:
55(.10) + 45 (.05) + 67(.30) + 58 (.20) + 70 (.30)= 60
And, the Schoenfeld residual is: 55-60 = -5
23.
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Example
Based on our model, we can calculate a
predicted probability of death by time 7 for
each person (call it “p-hat”):
Female 55-year old smoker: p-hat=.10
Male 45-year old non-smoker : p-hat=.05
Female 67-year old smoker : p-hat=.30
Male 58-year old smoker : p-hat=.20
Male 70-year old non-smoker : p-hat=.30
The expected value for the GENDER of the person who
failed is:
0(.10) + 1(.05) + 0(.30) + 1 (.20) + 1 (.30)= .55
And, the Schoenfeld residual is: 0-.55 = -.55
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Schoenfeld residuals
Since the Schoenfeld residuals are, in
principle, independent of time, a plot that
shows a non-random pattern against time is
evidence of violation of the PH assumption.
Plot Schoenfeld residuals against time to evaluate
PH assumption
Regress Schoenfeld residuals against time to test
for independence between residuals and time.
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Schoenfeld residuals
In SAS:
option on the output statement:
Output out=outdata ressch= Covariate1
Covariate2 Covariate3
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Summary of the many ways to
evaluate PH assumption…
1. Examine log(-log(S(t)) plots
PH assumption is supported by parallel lines and refuted by lines that cross
or nearly cross
Must use categorical predictors or categories of a continuous predictor
2. Include interaction with time in the model
PH assumption is supported by non-significant interaction coefficient and
refuted by significant interaction coefficient
Retaining the interaction term in the model corrects for the violation of PH
Don’t complicate your model in this way unless it’s absolutely necessary!
3. Plot Schoenfeld residuals
PH assumption is supported by a random pattern with time and refuted by a
non-random pattern
4. Regress Schoenfeld residuals against time to test for
independence between residuals and time.
PH assumption is supported by a non-significant relationship between
residuals and time, and refuted by a significant relationship
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Death (presumably) can only happen
once, but many outcomes could happen
twice…
Fractures
Heart attacks
Pregnancy
Etc…
4. Repeated events
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Strategy 1: run a second Cox
regression (among those who had a
first event) starting with first event time
as the origin
Repeat for third, fourth, fifth, events,
etc.
Problems: increasingly smaller and smaller
sample sizes.
Repeated events: 1
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Treat each interval as a distinct
observation, such that someone who
had 3 events, for example, gives 3
observations to the dataset
Major problem: dependence between the
same individual
Repeated events: Strategy 2
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Stratify by individual (“fixed effects partial
likelihood”)
In PROC PHREG: strata id;
Problems:
does not work well with RCT data
requires that most individuals have at least 2
events
Can only estimate coefficients for those covariates
that vary across successive spells for each
individual; this excludes constant personal
characteristics such as age, education, gender,
ethnicity, genotype
Strategy 3
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5. Considerations when
analyzing data from an RCT…
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Intention-to-Treat Analysis
Intention-to-treat analysis: compare
outcomes according to the groups to
which subjects were initially assigned,
regardless of which intervention they
actually received.
Evaluates treatment effectiveness
rather than treatment efficacy
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Why intention to treat?
Non-intention-to-treat analyses lose the
benefits of randomization, as the groups may
no longer be balanced with regards to factors
that influence the outcome.
Intention-to-treat analysis simulates “real life,”
where patients often don’t adhere perfectly to
treatment or may discontinue treatment
altogether.
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Drop-ins and Drop-outs:
example, WHI
Both women on
placebo and women
on active treatment
discontinued study
medications.
Women on placebo “dropped
in” to treatment because
their regular doctors put
them on hormones (dogma=
“hormones are good”).
Women on treatment
“dropped in” to treatment
because their doctors took
them off study drugs and
put them on hormones to
insure they were on
hormones and not placebo.
Women’s Health Initiative Writing
Group. JAMA. 2002;288:321-333.
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Effect of Intention to treat on
the statistical analysis
Intention-to-treat analyses tend to
underestimate treatment effects;
increased variability “waters down”
results.
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Example
Take the following hypothetical RCT:
Treated subjects have a 25% chance of dying during the 2-year
study vs. placebo subjects have a 50% chance of dying.
TRUE RR= 25%/50% = .50 (treated have 50% less chance of
dying)
You do a 2-yr RCT of 100 treated and 100 placebo subjects.
If nobody switched, you would see about 25 deaths in the treated
group and about 50 deaths in the placebo group (give or take a
few due to random chance).
∴Observed RR≅ .50
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Example, continued
BUT, if early in the study, 25 treated subjects switch
to placebo and 25 placebo subjects switch to
control.
You would see about
25*.25 + 75*.50 = 43-44 deaths in the placebo group
And about
25*.50 + 75*.25 = 31 deaths in the treated group
Observed RR = 31/44 ≅ .70
Diluted effect!
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References
Paul Allison. Survival Analysis Using SAS. SAS Institute
Inc., Cary, NC: 2003.
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