Nonparametric Methods

In the previous sections, the focus was on developing parametric models for lifetimes and fitting them to a data set. The emphasis switches here to letting the data speak for itself, rather than approximating the lifetime distribution by one of the parametric models. There are several reasons to take this approach. First, it is not always possible to find a parametric model that adequately describes the lifetime distribution. This is particularly true of data arising from populations with nonmonotonic hazard functions. The most popular parametric models, such as the Weibull distribution, have monotonic hazard functions. A nonparametric analysis might provide more accurate estimates. Second, data sets are often so small that fitting a parametric model results in parameter estimators with confidence intervals that are so wide that the models are of little practical use. To cover all nonparametric methods used

Reliability 3-33 in reliability is not possible due to space constraints. We focus on just one such method here:

the Kaplan–Meier product-limit survivor function estimate and the associated Greenwood’s formula for assessing the precision of the estimate.

Let y1< y2< · · · < y_k be the k distinct failure times, and let d_j denote the number of observed failures atyj, j = 1, 2, . . ., k. Let nj=n(yj) denote the number of items on test just before timeyj, j = 1, 2, . . ., k, and it is customary to include any values that are censored at yj in this count. Also, letR(yj) be the set of all indexes of items that are at risk just before timeyj, j = 1, 2, . . ., k.

The search for a nonparametric survivor function estimator begins by assuming that the data were drawn from a discrete distribution with mass values aty1, y2, . . ., yk. For a discrete distribution,h(yj) is a conditional probability with interpretationh(yj) =P [T = yj|T ≥ yj].

For a discrete distribution, the survivor function can be written in terms of the hazard function at the mass values:

S(t) = 

j∈R(t)^

[1− h(yj)] t ≥ 0

where R(t)^ is the complement of the risk set at time t. Thus a reasonable estimator for S(t) is

j∈R(t)^[1− ˆh(y_j)], which reduces the problem of estimating the survivor function to that of estimating the hazard function at each mass value. An appropriate element for the likelihood function at mass valuey_j is

h(yj)^dj[1− h(yj)]^nj−dj

forj = 1, 2, . . ., k. This expression is correct because d_jis the number of failures aty_j,h(y_j) is the conditional probability of failure at y_j, n_j− d_j is the number of items on test not failing aty_j, and 1− h(y_j) is the probability of failing after timey_j conditioned on survival to timey_j. Thus the likelihood function forh(y1), h(y2), . . ., h(y_k) is

L(h(y1), h(y2), . . ., h(yk)) =

k j=1

h(yj)^dj[1− h(yj)]^nj−dj

and the log likelihood function is logL(h(y1), h(y2), . . ., h(yk)) =

k j=1

{djlogh(yj) + (nj− dj) log [1− h(yj)]}

Theith element of the score vector is

∂ log L(h(y1), h(y2), . . ., h(yk)) of d_i to n_i is an appropriate estimate of the conditional probability of failure at time y_i. This derivation may strike a familiar chord since, at each timey_i, estimatingh(y_i) withd_i divided byn_iis equivalent to estimating the probability of success, that is, failing at timey_i, for each of theni items on test. Thus, this derivation is equivalent to finding the maximum likelihood estimators for the probability of success fork binomial random variables.

Using this particular estimate for the hazard function aty_i, the survivor function estimate

commonly known as the Kaplan–Meier or product-limit estimate. One problem that arises with the product-limit estimate is that it is not defined past the last observed failure time.

The usual way to handle this problem is to cut the estimator off at the last observed failure timeyk. The following example illustrates the product-limit estimate.

Example 3.16

An experiment is conducted to determine the effect of the drug 6-mercaptopurine (6-MP) on leukemia remission times (Lawless, 2003, page 5). A sample ofn = 21 leukemia patients is treated with 6-MP, and the remission times are recorded. There are r = 9 individuals for whom the remission time is observed, and the remission times for the remaining 12 individuals are randomly censored on the right. There are k = 7 distinct observed failure times. Letting an asterisk denote a censored observation, the remission times (in weeks) are

6 6 6 6^∗ 7 9^∗ 10 10^∗ 11^∗ 13 16 17^∗ 19^∗ 20^∗ 22 23 25^∗ 32^∗ 32^∗ 34^∗ 35^∗ Find an estimate forS(14).

Table 3.1 gives the values ofyj,dj,nj, and 1− dj/nj forj = 1, 2, . . ., 7. In particular, the product-limit survivor function estimate att = 14 weeks is

S(14) =ˆ 

TABLE 3.1 Product-Limit Calculations for the 6-MP Data

j yj dj nj 1−dj

Reliability 3-35

0 0.0 0.2 0.4 0.6 0.8 1.0

t

S(t)

5 10 15 20

FIGURE 3.16 Product-limit survivor function estimate for the 6-MP data.

The product-limit survivor function estimate for allt values is plotted in Figure 3.16. Down-ward steps occur only at observed failure times. The effect of censored observations in the survivor function estimate is a larger downward step at the next subsequent failure time.

If there are ties between the observations and a censoring time, as there is at time 6, our convention of including the censored values in the risk set means that there will be a larger downward step following this tied value. Note that the estimate is truncated at time 23, the

last observed failure time. 

To find an estimate for the variance of the product-limit estimate is significantly more difficult than for the uncensored case. The Fisher and observed information matrices require a derivative of the score vector:

−∂²logL(h(y1), h(y2), . . ., h(y_k))

∂h(yi)∂h(yj) = d_i

h(yi)² + n_i− d_i (1− h(yi))²

wheni = j and 0 otherwise, for i = 1, 2, . . ., k, j = 1, 2, . . ., k. Both the Fisher and observed information matrices are diagonal. Replacingh(yi) by its maximum likelihood estimate, the diagonal elements of the observed information matrix are



−∂²logL(h(y1), h(y2), . . ., h(y_k))

∂h(y_i)²



h(yi)=di/ni

= n³_i d_i(n_i− d_i)

fori = 1, 2, . . ., k. Using this fact and some additional approximations, an estimate for the variance of the survivor function is

V [ ˆˆ S(t)] = [ ˆS(t)]²

j∈R(t)^

d_j n_j(n_j− d_j)

commonly known as “Greenwood’s formula.” The formula can be used to find asymptotically valid confidence intervals forS(t) by using the normal critical values as in the uncensored case:

wherezα/2 is the 1− α/2 fractile of the standard normal distribution.

Example 3.17

For the 6-MP treatment group in the previous example, give a 95% confidence interval for the probability of survival to time 14.

The point estimator for the probability of survival to time 14 from the previous example is S(14) ∼ˆ = 0.69. Greenwood’s formula is used to estimate the variance of the survivor function estimator at time 14:

Thus an estimate for the standard deviation of the survivor function estimate att = 14 is

√0.011 = 0.11. A 95% confidence interval for S(14) is

Figure 3.17 shows the 95% confidence bands for the survivor function for all t values.

These have also been cut off at the last observed failure time, t = 23. The bounds are particularly wide as there are onlyr = 9 observed failure times.