Id |
Time |
M(0) |
M(βˆ) |
A |
1 |
1 − 1/19 = 18/19 |
0.85531 |
B |
1 |
0 − 1/19 = −1/19 |
-0.02593 |
C |
2 |
1 − (1/19 + 5/8) = 49/152 |
0.17636 |
D |
2 |
1 − (1/19 + 5/8) = 49/152 |
0.17636 |
E |
2 |
1 − (1/19 + 5/8) = 49/152 |
0.65131 |
F |
2 |
0 − (1/19 + 5/8) = −103/152 |
-0.82364 |
G |
3 |
0 − (1/19 + 5/8) = −103/152 |
-0.34869 |
H |
4 |
1 − (1/19 + 5/8 + 2/3) = −157/456 |
-0.64894 |
I |
5 |
0 − (1/19 + 5/8 + 2/3) = −613/456 |
-0.69808 |
Score residuals at β = 0 are
Id |
Time |
Score |
A |
1 |
(2 − 13/19)(1 − 1/19) |
B |
1 |
(0 − 13/19)(0 − 1/19) |
C |
2 |
(1 − 13/19)(0 − 1/19) + (1 − 11/16)(1 − 5/8) |
D |
2 |
(1 − 13/19)(0 − 1/19) + (1 − 11/16)(1 − 5/8) |
E |
2 |
(0 − 13/19)(0 − 1/19) + (0 − 11/16)(1 − 5/8) |
F |
2 |
(1 − 13/19)(0 − 1/19) + (1 − 11/16)(0 − 5/8) |
G |
3 |
(1 − 13/19)(0 − 1/19) + (0 − 11/16)(0 − 5/8) |
H |
4 |
(1 − 13/19)(0 − 1/19) + (1 − 11/16)(0 − 5/8) +(1 − 2/3)(1 − 2/3) |
I |
5 |
(1 − 13/19)(0 − 1/19) + (1 − 11/16)(0 − 5/8) +(0 − 2/3)(0 − 2/3) |
SAS returns the unweighted residuals as given above; it is the weighted sum of residuals that totals zero, = 0, likewise for the score and Schoenfeld residuals evaluated at βˆ. S also returns unweighted residuals by default, with an option to return the weighted version. Whether the weighted or the unweighted form is more useful depends on the intended application, neither is more “correct” than the other. S does differ for the dfbeta residuals, for which the default is to return weighted values. For the third observation in this data set, for instance, the unweighted dfbeta is an approximation to the change in βˆ that will occur if the case weight is changed from 2 to 3, corresponding to deletion of one of the three “subjects” that this observation represents, and the weighted form approximates a change in the case weight from 0 to 3, i.e., deletion of the entire observation.
The increments of the Nelson-Aalen estimate of the hazard are shown in the rightmost column of table 3. The hazard estimate for a hypothetical subject with covariate X† is Λi(t) = exp(X†β)Λ0(t) and the survival estimate is Si(t) = exp(−Λi(t)). The two term of the variance, for X† = 0, are Term1 + d0V d:
Time |
Term 1 |
1 |
1/(r2 + 11r + 7)2 |
2 |
1/(r2 + 11r + 7)2 + 10/(11r + 5)2 |
4 |
1/(r2 + 11r + 7)2 + 10/(11r + 5)2 + 2/(2r + 1)2 |
Time |
d |
1 |
(2r2 + 11r)/(r2 + 11r + 7)2 |
2 |
(2r2 + 11r)/(r2 + 11r + 7)2 + 110r/(11r + 5)2 |
4 |
(2r2 + 11r)/(r2 + 11r + 7)2 + 110r/(11r + 5)2 + 4r/(2r + 1)2 |
For β = log(2) and X† = 0, where k ≡ the variance of βˆ = 1/2.153895 this reduces to
Time |
Variance |
|
1 |
1/1089 |
+ k(30/1089)2 |
2 |
(1/1089+ 10/729) |
+ k(30/1089 + 220/729)2 |
4 |
(1/1089+ 10/729 + 2/25) |
+ k(30/1089 + 220/729 + 8/25)2 |
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