7.2 Estimation of Survival Function
(I) Parametric approach
Suppose
t1 , t 2 ,, t n are failure times corresponding to censor
indicators
w1 , w2 ,, wn ( wi  1, death; wi  0,
censoring). Then the likelihood function is
n
 f t  
   i  S ti 
i 1  S t i  
n
L     f ti  S ti 
1 wi
wi
i 1
n
wi
   ti  i S ti 
w
i 1
(as
wi  1   f t i 
S ti 1 w
w
1 w
 0   f t i  S t i 
wi
wi
i
i
i
 f t i 
 S t i   P T  t i 
)
 t , S t 
where
depends on some parameter

. Then, the

L ˆ
 0.
parameter estimate ˆ can be obtained by solving

Then,
̂ t 
and Ŝ t  can be obtained by evaluating

at
ˆ .
Example:
Let T have exponential density. Then,
and
 t    . Then,
1
f t   e t , S t   e t ,
n
L    wi e  ti
i 1
and further
l    log L  
n
 w
i 1
i
log    ti 
 n

 n

   wi  log       ti 
 i 1

 i 1 
Thus,
n
l  


w
i
i 1

n
n
  t i  0  ˆ 
i 1
w
i
i 1
n
t
i 1
.
i
1


E
T

 the estimate for mean survival
(Intuitively,

n
1

ˆ
time is 
t
i 1
n
i
 wi
) Then,
Sˆ t   e̂t .
For example, in
i 1
the motivating example,
n
ˆ 
w
t
i 1
Then,
i
i 1
n

1111
4

6  7  9.5  7  10  7  6  11 63.5 .
i
Sˆ t   e

4t
63.5
.
2
(II) Nonparametric approach
Let t (1)  t ( 2 )    t ( m ) be death times. The number of
individuals who alive just before time t ( j ) , including those who are
about to die at this time, will be denoted n j , for j  1,2,, m ,
and d
j
will denote the number who die at this time. Thus, we have
the following table:
t (1)
t( 2)
n1
n2
d1
d2
…
t( m)
…
nm
…
dm
t
t (1)
t (2)
t(k )
t ( k 1)
t (m )
Then, for t ( k )  t  t ( k 1) ,
 nj  d j
ˆ
S t    

nj
j 1 
k




 nk  d k






nk




dk 
d1 
d2 






1

1


1






n1 
n2 
nk 



 1  ˆ t1  1  ˆ t 2   1  ˆ t k 
 n1  d1


n1


 n2  d 2



n2


 
3





Ŝ t  is referred to as Kaplan-Meier estimate.
Note:
Intuitively, if T is a discrete random variable taking values
t (1)  t (2)  
with associated probability function
PT  t(i ) , i  1, 2,  ,
then
 t(i )   PT  t(i ) | T  t(i )  
f t(i ) 
S t(i )  .
Then, for t ( k )  t  t ( k 1) ,
t
t (1)
t (2)
t(k )
t ( k 1)
t (m )
S t   PT  t   P T  t( k 1) 

P T  t( 2 )  P T  t( 3)  P T  t( k 1) 


P T  t(1)  P T  t( 2 )  P T  t( k ) 
 P T  t(1)   P T  t(1)   P T  t( k )   P T  t( k ) 







P
T

t
P
T

t
(1)
(k )

 

 P T  t(1)   P T  t( 2 )   P T  t( k ) 
 1 
 1 
  1 







P
T

t
P
T

t
P
T

t





(1) 
(
2
)
(
k
)

 
 1   t(1)  1   t( 2 )   1   t( k ) 


 
4



since P T  t (1)  1 .
Example (continue):
In the motivating example, we have
t(i )
6
ni
8
di
1
7
6
2
11
1
1
Thus,
1, 0  t  6

 n1  d1 8  1

 0.875, 6  t  7

n
8
 1

Sˆ t    n1  d1  n2  d 2   8  1  6  2  7 , 7  t  11
 n1  n2 
8
6
12

 n1  d1  n2  d 2  n3  d 3   8  1  6  2  1  1  0, 11  t
 n1  n2  n3 
8
6
1
0.6
0.4
0.2
0.0
Survival function
0.8
1.0
The plot of the survival function is
0
2
4
6
t
5
8
10