Lecture 17: Survival Analysis -- Cox proportional Hazards Ani - - PowerPoint PPT Presentation

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Lecture 17: Survival Analysis

• - Cox proportional Hazards

Ani Manichaikul amanicha@jhsph.edu 14 May 2007

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Survival Analysis

n Suppose we have designed a study to

estimate survival after chemotherapy treatment for patients with a certain cancer

n Patients received chemotherapy

between 1990 and 1994 and were followed until death or the year 2000, whichever occurred first

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Survival Analysis

n In this study the event of interest is

death

n The time clock starts as soon as the

subject finishes his/her chemotherapy treatments

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Dies 1990 1995 2000

Survival Analysis

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Dies 1990 1995 2000

Patient one enters in 1990, dies in 1995: Patient one survives five years

Survival Analysis

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1990 1995 2000

Lost to Follow-up

Survival Analysis

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1990 1995 2000

Patient two enters in 1991, drops out in 1997: Patient two is lost to follow-up after six years

Survival Analysis

Lost to Follow-up

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1990 1995 2000

Withdrawn Alive (Administratively Censored)

Survival Analysis

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1990 1995 2000

Patient three enters in 1993, is still alive at end of study: Patient three is still alive after seven years

Survival Analysis

Withdrawn Alive

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Survival Analysis

n Patient:

n 1:

1990

→ 1995 5 years

n 2:

1991

→ 1997 6+ years

n 3:

1993

→ 2000 7+ years

n Patients two and three are called

censored observations

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Central Problem

n Estimation of the survival curve n S(t) = Proportion surviving at least to

time t or beyond

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Approaches

n Life table method

n Grouped in intervals

n Kaplan-Meier (1958)

n Ungrouped data n Small samples

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) _ _ (Pr ) ( ) ( ) ( ) ( Time Event evious t n t y t n t S S ×         − =

Kaplan-Meier Estimate

n Curve can be estimated at each event,

but not at censoring times

n y(t) = # events at time t n n(t) = # subjects at risk for event at

time t

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) _ _ (Pr ) ( ) ( ) ( ) ( Time Event evious t n t y t n t S S ×         − =

Proportion of original sample making it to time t

Kaplan-Meier Estimate

n Curve can be estimated at each event,

but not at censoring times

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Proportion surviving to time t who survive beyond time t

Kaplan-Meier Estimate

n Curve can be estimated at each event,

but not at censoring times

) _ _ (Pr ) ( ) ( ) ( ) ( Time Event evious t n t y t n t S S ×         − =

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n Start estimate at first event time

n No Chemotherapy Group: Time = 5

833 . 12 10 12 2 12 ) 5 ( ) 5 ( ) 5 ( ) 5 ( = = − =         − = n y n S

Kaplan-Meier Estimate

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n No Chemotherapy group: Time= 8

n 2nd event time

666 . 833 . 10 8 ) 833 (. 10 2 10 ) 5 ( ) 8 ( ) 8 ( ) 8 ( ) 8 ( = × = ×       − = ×         − = S n y n S

Kaplan-Meier Estimate

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Kaplan-Meier Estimate

n Skip over censoring times: Remove

from number at risk for next event time

n Continue through final event time

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Kaplan-Meier Estimate

n Graph is a step function n “Jumps” at each observed event time n Nothing is assumed about curved shape

between each observed event time

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Kaplan-Meier Estimate

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Confidence Interval for S(t)

Greenwood’s Formula Complementary log-log transformation

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Greenwood’s Formula

n Variance of S(t) n Standard Error

( )

∑

≤

− =

t t j j j j j

j

y n n y t S t S Var

: 2

) ( ˆ )] ( ˆ [

)] ( ˆ [ ) ( SEGW t S Var t =

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95% Confidence Interval

n Using Greenwood’s formula, and

approximate 95% CI for S(t) is

n There is a “problem”: the 95%

Confidence Interval is not constrained to lie within the interval (0,1)

) ( SE * 96 . 1 ) ( ˆ

GW t

t S ±

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Alternative Confidence Interval

n Complementary log-log transformation n Variance of CLL:

)] ( ˆ log log[ ) ( ˆ t S t − = υ

( ) ( )

2 : :

log )] ( ˆ ( Var[                 − − =

∑ ∑

≤ ≤ t t j j j j j t t j j j j j

j j

y n n y y n n y t υ

)] ( ˆ ( Var[ (t) SECLL t υ =

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95% CI based on complementary log-log transformation

n Use CLL to obtain 95% confidence

interval on S(t)

n Get 95% CI for :

) (t ν

) ( SE * 96 . 1 ) ( ˆ

CLL t

t ± υ

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n Transform back to get 95% for S(t):

Use the inverse transformation to get the 95% CI for S(t):

( )

(t)

• e

e ) (

ν

= t S

( ) ( )

( )

) ( SE * 96 . 1 ) ( CLL SE * 96 . 1 ) ( ˆ ) ( CLL SE * 96 . 1 ) ( ˆ

)] ( ˆ [ ] e , e [

• e
• e

t CLL t t t t

e

t S

± − +

=

υ υ

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Back to the AML Data

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Kaplan-Meier Estimates

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95% CI: Greenwood

n VarGreenwood(13)] = 0.8182

= (0.116)2

n 95% CI Greenwood

= .818 1.96* (.116) = (.586, 1.05) 1.05 is out of Range!

      + 9 * 10 1 10 * 11 1

±

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Better 95% CI CLL transformation

n n

605 . 1 )] ( ˆ log log[ ) ( ˆ − = − = t S t υ

708 . ) 13 ( 502 . 04027 . 0202 . 10 9 log 11 10 log 90 1 110 1 )] 13 ( ˆ ( Var[

2

= = =       +       + =

CLL

SE υ

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Better 95% CI CLL transformation

n 95% CLL for S(13)

Does not contain 1!

[ ]

) 952 ,. 437 (. 818 .

) 708 (. * 96 . 1

= =

±

e

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95% CI for S(t) in the maintained

• n chemotherapy group

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95% CI for S(t) in the not maintained on chemo group

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Regression in Survival Analysis

n The Kaplan-Meier estimate and log-rank

tests are great ways to compare survival between groups without making too many assumptions.

n But…we also want a simple summary

measure that compares groups Solution: Regression Analysis

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Regression in Survival Analysis

n The regression model for the hazard

function (instantaneous incidence rate) as a function of p explanatory X variables is specified as:

n log hazard: n hazard:

p p X

X X t X t β β β λ λ + + + + = ... ) ( log ) ; ( log

2 2 1 1

( )( ) (

)

( )

β β β β

λ λ λ

X X X X

e t e e e t X t

p p

) ( ... ) ( ) ; (

2 2 1 1

= =

(Vector of X’s)

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Interpretations

n 0(t): Hazard (incidence) rate as a

function of time when all X’s are zero

n

• ften must center Xs to make 0(t)

interpretable

n exp{ 1} : the relative hazard associated

with a 1 unit change in X1 (i.e., X1+ 1 - vs- X1), holding other Xs constant, independent of time

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Interpretations: Relative Risk

n exp{ 1} : the relative risk for X1+ 1 -vs-

X1, holding other Xs constant, independent of time

n Other s have similar interpretations

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Interpretations

n e: “multiplies” the baseline hazard

0(t) by the same amount regardless of

the time t.

n This is therefore a “proportional

hazards” model

n the effect of any (fixed) X is the same at

any time during follow-up

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Note

n is the focus whereas 0(t) is a nuisance

variable

n David Cox (1972) showed how to estimate

without having to assume a model for 0(t)

n “Semi-parametric”

n 0(t) is the baseline hazard

n “non-parametric part of the model

n are the regression coefficients

n “parametric” part of the model

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Why Cox Proportional Hazards Model is Different?

n It uses the partial likelihood, not the

likelihood

n We do not assume a particular

distribution for the failure time; we only assume proportional hazards

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Results from AML Data

n Semi-parametric model for the hazard

(incidence) rate for the AML data

n Where

n i(t) is the hazard for person i at week t n 0(t) is the hazard if Xi = 0 (not maintained

group), and is the multiplicative effect of Xi= 1 (maintained group)

β

λ λ

i

X i

e t t ) ( ) ( =

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Results from AML Data:

n e-0.812 = 0.44: relative rate of AML

relapse maintained vs not maintained

n 1/.44 = 2.25 relative rate of AML relapse

not-maintained vs maintained

n 95% CI: [e-.812-1.96* .521, e-.812+ 1.96* .521]

n (1.22, 6.25)

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Example: CABG surgery

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Cox model to compare two treatments, controlling for several predictors (Fisher and Van Belle, 1993)

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Compare surgical (CABG) with medical treatment for left main coronary heart disease

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Use mortality (time to death) as the response variable

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Control for 7 risk factors (age at baseline and 6 coronary status measures) in making the comparison

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Time variable is time from treatment initiation to death or censoring due to the end of the study or lost to follow-up

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Variables

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Cox PH: CABG Surgery

n Model for the log hazard rate (incidence

• f death):

n Model for the hazard rate

8 8 2 2 1

... ) ( log ) ; ( log X X X t X t β β β λ λ + + + + =

( )

8 8 2 2 1 1

... 0 )

( ) ; (

X X X

e t X t

β β β

λ λ

+ + +

=

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Cox Model Results

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Question:

n What is the relative risk of death for the

CABG group compared to the medical group, adjusting for age and other risk factors?

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Answer (Interpretation):

n e-1.0777 = .34 n 66% reduction in the risk of death for

• therwise comparable patients treated

with CABG compared with patients treated medically

n note the coding 2= CABG, 1= Medical

gives the same results as 1= CABG and 0= Medical

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What is the interpretation of each coefficient?

n CHFSCR: Controlling for type of treatment

and other risk factors, the risk of death, as estimated from a Cox model, is e.2985 = 1.35 times higher per unit difference in CHF score

n AGE: Controlling for type of treatment and

• ther risk factors, the risk of death, as

estimated from a Cox model, is e.0423 = 1.04 times higher per year of age

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Interpretation

n HYPTEN: Controlling for type of treatment

and other risk factors, the risk of death, as estimated from a Cox model, is e-.5428 = 0.58 times lower for patients who have a history of hypertension compared with those who do not

n Anyone know why a history of hypertension

should lower risk following treatment?

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Question:

n What is the relative risk death for

(A) a medically treated 45-year old

• vs-

(B) a surgically treated 75 year old who otherwise have comparable risk factors?

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Answer:

n log hazard for (A) =

const + 1* (-1.0777) + 45* (.0423) = const + .8258

n log hazard for (B) =

const + 2* (-1.0777) + 75* (.0423) = const + 1.017

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Interpretation:

n Difference in log hazards, (B) -vs- (A):

(const+ 1.017) - (const+ .8258) = .1913

n Relative Risk, (B) -vs- (A):

e.1913 = 1.21 higher risk for older, surgically treated patient than for younger, medically treated patient

n Is the assumption of “otherwise comparable risk

factors” reasonable?

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Question:

n How much higher is the risk of a 70

year old patient compared with a 60 year old patient, assuming treatment and other coronary risk factors are the same?

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Answer:

n The estimated difference in log hazards for

two patients whose ages differ by 10 years, holding other predictors fixed is 10×AGE= 10 ×.0423 = .423 RR = e.423 = 1.53

n A ten year difference in the age at initiation

• f treatment increases the risk of subsequent

mortality by 50%

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Another Question:

n How would you determine whether the

mortality advantage of CABG over medical treatment was greater for younger patients than for older patients?