Post on 21-Sep-2020
Survival Analysis for Randomized Clinical Trials
II Cox Regression
Ziad Taib
Biostatistics
AstraZeneca
February 26, 2008
Plan• Introduction to the proportional hazard
model (PH)
• Partial likelihood
• Comparing two groups
• A numerical example
• Comparison with the log-rank test
The model
)exp()()( 110 ikkii zzthth ββ ++=
Understanding “baseline hazard” h0(t)
)]()(exp[)(
)(111 jkikkji
j
i zzzzth
th −++−= ββ
Cox Regression Model• Proportional hazard. • No specific distributional assumptions (but
includes several important parametric models as special cases).
• Partial likelihood estimation (Semi-parametric in nature).
• Easy implementation (SAS procedure PHREG).• Parametric approaches are an alternative, but they
require stronger assumptions about h(t).
Cox proportional hazards model, continued
• Can handle both continuous and categorical predictor variables (think: logistic, linear regression)
• Without knowing baseline hazard ho(t), can still calculate coefficients for each covariate, and therefore hazard ratio
• Assumes multiplicative risk—this is the proportional hazard assumption– Can be compensated in part with interaction terms
Cox Regression
• In 1972 Cox suggested amodel for survival data that would make it possible to take covariates into account. Up to then it was customary to discretise continuos variables and build subgroups.
• Cox idea was to model the hazard rate function
tthtTttTtP ∆=≥∆+<≤ )()|(
where h(t) is to be understood as an intensity i.e. a probability by time unit. Multiplied by time we get a probability. Think of the analogy with speed as distance by time unit. Multiplied by time we get distance.
Cox’ suggestion is to model h using
( )( ) vector.covariate the,...,
vector;parameter the,...,
;)()(
1
1
0
iki
kT
i
ZZ
eththT
==
=
i
Zβ
Z
β
i
ββ
where each parameter is a measure of the importance of the corresponding variable.
A consequence of this is that two individuals with different covariate values will have hazard rate functions which differ by a multiplicative term which is the same for all values of t.
The covariates can be discrete, continuous or even categorical. It is also possible to generalise the model to allow time varying covariates.
Cthth
Ceeth
eth
th
th
iethth
T
T
T
T
i
).()(
;)(
)(
)(
)(
;2,1 ;)()(
21
)
0
0
2
1
0
=
===
==
21
2
1
i
Z-(Zβ
Zβ
Zβ
Zβ
Example
Assume we have a situation with one covariate that takes two different values 0 and 1. This is the case when we wish to compare two treatments
β
ββ
ethth
thth
k
).()(
);()(
;1
;;1
02
01
1
=
===
===
21 Z 0;Z
β
h1(t)
h2(t)
t
From the definition we
see immediately that
( )
[ ] iT
e
iT
t
t
iT
t
i
tS
e
e
e
tS
e
duuh
eduuh
duuh
i
Zβ
Zβ
Zβ
)(0
)(
)(
)(
0
0
0
0
0
=
∫=
∫=
∫=
−
−
−
• One of the advantages of the PH model is that we can estimate the parameters and make comparisons between treatments without having to estimate the baseline.
• The baseline can be interpreted as the case corresponding to an individual with covariate values zero.
• The epitate semi-parametric is due to the fact that we do not model h0 explicitley. Usual likelihood theory does not apply.
Comments
Partial likelihood• The key to understanding partial likelihood is the following
conditional probability:P(patient j dies under ∆t | one patient dies during ∆t)
=P(patient j dies under ∆t)/P(one patient dies during ∆t)
=P(patient j dies under ∆t) / Σi P( one patient dies under ∆t | patient i dies during ∆t) P(patient i dies during ∆t)
=P(patient j dies under ∆t) / Σi P(patient i dies during ∆t)
∑∑===
∩=
ii
j
iii
jjjj AP
AP
APABP
AP
BP
AP
BP
BAPBAP
)(
)(
)()|(
)(
)(
)(
)(
)()|(
)|( BAP j
)(
)(
)(
)(
BP
AP
BP
BAP jj =∩
∑i
ii
j
APABP
AP
)()|(
)(
∑i
i
j
AP
AP
)(
)(
Therefore P(patient j dies under ∆t | one patient dies during ∆t)
(*))(
)(
)(
)(
0
0 ==∆
∆=∆
∆=
∑∑∑iii
i
j
iT
T
iT
jT
e
e
teth
teth
tth
tthZβ
Zβ
Zβ
Zβ j
• Let R(t)= # of patients at risk at time t (i.e alive and under observation at t).
• Let further Rj=R(tj).
tthtTttTtP iii ∆=≥∆+<≤ )()|(P(patient i dies under ∆t) =
From this it follows that we can base our inference on the likelihood function
( )
( ) ( ) (**)
)(
)(),(
11
10
====
∆∆
=
∏∏ ∑
∏ ∑
==∈
=∈
ββ
β
Zβ
Zβ
LLe
e
tth
tththL
J
jj
J
j
Ri
J
jRi
i
j
j
iT
jT
j
∑∈
∆∆
=
jRii
j
tth
tth
)(
)(
P(patient j dies under ∆t | one patient dies during ∆t)
Then
Comments• If we treat this expression as an ordinary likelihood function we
can get parameter estimates.• Due to non-linearity we cannot hope finding explicit formulas
for the estiamtes. We have to rely on numerical maximisation methods such as Newton-Raphson.
• On the top of the usual score functions, we can consider the second derivative to compute the information and use this to estimate the covariance matrix of the parameter vector .
• As for the ordinary ML method, the estimates are asymptotically unbiased and and multivariate normal with mean and covariance matrix that can be estimated using and the inverse of the covariance matrix.
β̂
Let L(β) stand for the likelihood function and β for some parameter (vector) of interest. Then, the maximum likleihood estimates are found by solving the set of equations
It can be shown that the estimate to have an asymptotically normal distribution with means βi and variance/covariance matrix
Maximum Likelihood estimates
Define the information matrix, I(β) to have elements
,...,2,1 ,0)(log ==
∂∂
iL
iββ
.)(log
)(2
∂∂
∂−=ji
ij
LEI
ββββ
[ ] 1)( −βI
• We take logLj = lj
( )
( )
( )( )∑
∑ ∑
∑
=
= ∈
=
=
−=
=
J
jj
J
j Rij
T
J
jj
l
eLog
LogL
LogL
j
iT
1
1
1
β
Zβ
β
β
Zβ
Score and information( )
−−=∂∂
−=∂∂
=
∑
∑
∑
∑
∑
∑
∈
∈
∈
∈
∈
∈
22
2
2
j
iT
j
iT
j
iT
j
iT
ki
k
j
iT
j
iT
Ri
Riki
Ri
Rij
Ri
Riki
kjk
jk
e
eZ
e
eZl
e
eZ
Zl
U
Zβ
Zβ
Zβ
Zβ
Zβ
Zβ
β
β
β
• The two derivatives can be used to get parameter estimates and information .
• One difficulty we avoided so far is the occurence of ties.
Ties• dj = the # of failures at tj.• sj = the sum of the covariate values for patients who die at time tj.• Notice that sum (arbitrary) order must be imposed on the ties.
( ) ( )∑∑ ∑== ∈
=
−=
J
jj
J
j Rijj
T leLogdLogLj
iT
11
Zβsββ
−−=∂∂
−=∂∂
∑
∑
∑
∑
∑
∑
∈
∈
∈
∈
∈
∈
22
2
2
j
iT
j
iT
j
iT
j
iT
ki
k
j
iT
j
iT
Ri
Riki
Ri
Ri
jj
Ri
Riki
jkjk
j
e
eZ
e
eZ
dl
e
eZ
dsl
Zβ
Zβ
Zβ
Zβ
Zβ
Zβ
β
β
A numerical ExampleTIME TO RELIEF OF ITCH SYMPTOMS FOR PATIENTS USING A STANDARD AND EXPERIMENTAL CREAM
What about a t-test?
The mean difference in the time to “cure” of 1.2 days is not statistically significant between the two groups.
Using SASPROC PHREG ; MODEL RELIEFTIME * STATUS(0) = DRUG ;
Note that the estimate of the DRUG variable is -1.3396 with a p value of 0.0346. The negative sign indicates a negative association between the hazard of being cured and the DRUG variable. But the variable DRUG is coded 1 for the new drug and coded 2 for the standard drug. Therefore the hazard of being cured is lower in the group given the standard drug. This is an awkward but accurate way of saying that the new drug tends to produce a cure more quickly than the standard drug. The mean time to cure is lower in the group given the new drug. There is an inverse relationship between the average time to an event and the hazard of that event.
• At each time point the cure rate of the standard drug is about 25% of that of the new drug. Put more positively we might state that the cure rate is 1.8 times higher in the group given the experimental cream compared to the group given the standard cream.
ˆ 2ˆ(2 1) 1.33962
ˆ 11
( )0.262
( )
h t ee e
h t e
ββ
β
β− −
−= = = =
The ratio of the hazards is given by
Comparing two groups in general
Let • nAj be the number at risk in group 1 at time tj• nBj be the number at risk in group 2 at time tjthen
( )
( ) ∑
∑
=
=
+−
+=
+−=
J
j BA
A
BA
A
j
J
j BA
A
jj
jj
j
jj
j
jj
j
nen
en
nen
endI
nen
endsU
1
2
1
β
β
β
β
β
β
β
β ( )( ) ;0
0 22
χ≅=I
UQ
The score test
A numerical example• Assume that a study aiming at comparing two
groups resulted in the following data
• U(0)=-10.2505
• I(0)=6.5957
• ;509192.1ˆ −=β 2211.0
;)(
)(
)(
)(
;2,1 ;)()(
150919ˆ
0.0
0
2
1
0
==
==
==
−ee
eeth
eth
th
th
ieththT
i
β
ββ
β .1
Zβ i
The treatment reduces the hazard rate to almost ¼ compared with placebo.
Log-rank vs the score test (PHREG)
• The two methods give very similar answers, why?• On one hand the score test can be based on the
statistic( )( )
;
0
0
2
12
2
1
2
χ≅
−
=
=
∑
∑
=
=
J
j
BAj
J
j j
Aj
A
j
jj
j
j
n
nnd
n
ndd
I
UQ
( )
( )
;
groups;both in risk at #
groups;both in events #
;0
;0
12
1
j
jj
jj
j
jj
j
j
Aj
BAj
BAj
J
j
BAj
J
j j
Aj
A
ds
nnn
ddd
n
nndI
n
nddU
=
=+=
=+=
=
−=
∑
∑
=
=
nAjnAj-dAjdAjDRUG
njnj-djdjTOTAL
nBjnBj-dBjdBjCONTROL
NO CURECURE
• Compare this to the Log-rank statistic
[ ]21
2
χ≅=− UVar
UQ HM
[ ] [ ] ( )( )
1
D-Tar ;
2tt2
ttt
tttt
t
tttt TT
MNDAV
T
NDAE
−==== σµ
( )∑=
−=k
tttAU
1
µ
[ ] [ ] ( )( )
1ar ;
22
nn
nn-dnddVσ
n
nddE
j
jj
j
j
j
j
BAjjj
Ajj
Aj
Aj −====µ
( )( )
;
1
2
12
2
1χ≅
−
−
=
∑
∑
=
=
− J
j j
BAjjj
J
j j
Aj
A
HM
j
jj
j
j
nn
nn-dnd
n
ndd
Q
;2
12
2
1χ≅
−
=
∑
∑
=
=
J
j
BAj
J
j j
Aj
A
j
jj
j
j
n
nnd
n
ndd
Q ;
)1(
)(2
12
2
1χ≅
−
−
=
∑
∑
=
=
− J
j j
BAjjj
J
j j
Aj
A
HM
j
jj
j
j
nn
nn-dnd
n
ndd
Q
LOG-RANK TESTSCORE TEST
• dj = 1 i.e. no ties and the two formulas are identical. In general the two formulas are approximately the same when nj is large.
• For the example in a case with ties there is a slight difference
- Q=15.930
- QM-H=16.793
Generalizations of Cox regression
1. Time dependent covariates
2. Stratification
3. General link function
4. Likelihood ratio tests
5. Sample size determination
6. Goodness of fit
7. SAS
References
• Cox & Oakes (1984) “Analysis of survival data”. Chapman & Hall.
• Fleming & Harrington (1991) “Counting processes and survival analysis”. Wiley & Sons.
• Allison (1995). “Survival analysis using the SAS System”. SAS Institute.
• Therneau & Grambsch (2000) “Modeling Survival Data”. Springer.
• Hougaard (2000) “Analysis of Multivariate survival data”. Springer.
Questions or Comments?