Page 261 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

0.45 r = 0.5
0.4
r = 1.0
0.35 r = 1.5
Hazard function 0.25
0.3
0.2
0.15
0.1
0.05
0
1 2 3 4 5 6 7 8 9 10
Survival time (years)
Fig. 14.2. The Weibull hazard function with a λ = 0.20 and with various ρ values.

Table 14.2. Some commonly used survival distributions
and their parameters.

Distributions h(t) S(t) f(t)
Exponential l exp−(lt) lexp−(lt)r
Weibull rl(lt) ρ−1 exp−(lt)r rl(lt) ρ−1 exp−(lt)r
Log-logistic lrt r-1 1 lrt r-1
r 2
1 + lt r 1+ lt r 1 ( + lt )

14.4.4 Non-parametric estimation of the survival function

The survival function, S(t), can be estimated from the parametric functions mentioned above.
A non-parametric estimation of the survival function can be obtained using the Kaplan–
Meier estimator (Kaplan and Meier, 1958). Let T represent failure times ordered from the
i
first occurrence to the last. At T, let the number of animals that could have died (at risk) be
i
denoted by n and the number that actually died as d. The Kaplan–Meier estimator then is:
i i
() =
ˆ
i
i
St ∏ ⎛ ⎜ n − d ⎞ ⎟
| iT i < t ⎝ n i ⎠
The usefulness of the Kaplan–Meier estimate of the survival function is that it could
be used to check if the survival trait follows a particular parametric distribution.
For instance, the appropriateness of a Weibull model can be evaluated by plotting
log(−log(S ˆ (t))) versus log(t), where S ˆ (t) is the Kaplan–Meier estimate. This should result
in a straight line with intercept rlog(t) and slope r, given that:
S(t) = exp(−(lt)r) ® −log(S(t)) = lt ® log(−log(S(t))) = log(l) + rlog(t)
r
Similarly for the exponential distribution:
S(t) = exp(−lt) ® −log(S(t)) = lt

Therefore, the test for an exponential model will involve the plot of −log(S ˆ (t)) versus t,
which should give straight line passing through the origin with slope l.

Survival Analysis 245