If we give the survival function a specific functional form, we can estimate the survival curve based on just a few parameter estimates.
Parametric Survival Curves
If we give the survival function a specific
functional form, we can estimate the survival curve based on just a few
parameter estimates. We will illustrate this proce-dure with the negative
exponential and Weibull distributions.
The negative exponential, a simple one-parameter
family of probability distribu-tions, models well the lifetime distributions
for some products, such as electric light bulbs; i.e., it is useful in
describing their time to failure.
The Weibull distribution is a two-parameter family
of distributions that has been used even more widely than the negative
exponential to model time to failure for manufactured products. The Weibull
distribution shares one major characteristic with the normal distribution
model; i.e., it is a limiting distribution. Each distribu-tion is successful
under certain circumstances.
Whereas the normal distribution is a limiting
distribution for sums or averages of independent observations with the same
distribution, the Weibull is a limiting dis-tribution for the smallest value in
a sample of independent observations with the same distribution.
Recall that in Chapter 7 we saw that as the sample
size (n) increases, the sam-pling
distribution of means becomes more and more similar to a normal distribu-tion.
Because the distribution continues to become close to the normal distribution
as the sample size increases, we call the normal distribution a limiting
distribution. Similarly, if we have a sample of size n, the probability distribution for the smallest value among the n observations approaches the Weibull
distribution more closely as the sample size n increases. To obtain standard forms for the Weibull as we did
with the normal distribution, we subtract a constant from the original
statistic (e.g., minimum value in the sample) and then divide the result by
another constant.
This procedure is analogous to Z = (X – μ)/(σ/√n) for
the standard normal dis-tribution. The normal distribution works well when the
variable of interest can be viewed as a sum. The Weibull works well when the
variable of interest can be viewed as the smallest value.
For mortality, we can think of time to death as the
time when an illness, exposure factor, or other occurrence causes a person to
die. Mortality can be modeled in terms of many competing causes. For example, a
person who dies in an automobile accident is no longer at risk of dying from
coronary heart disease. A mortality mod-el can sort these competing causes in
order to determine which one occurs first. Suppose we specify the observed time
of death that occurs for the first of these competing causes. We denote this
time as the minimum of random times to death. In this particular situation, the
Weibull model should fit well.
For the negative exponential distribution, the
survival function S(t) = e–λt for all t ≥ 0. The single parameter λ is called the rate parameter, which is also equal to the so-called
hazard function or instantaneous death rate. The term λ represents the limit of the
probability of death in the next instant of time given survival up to time t. Its mathematical definition is given
in the next paragraph.
In survival analysis, the distribution function F(t)
is defined as F(t) = P(X ≤ t) = 1 – S(t). For those who have
studied differential equations, we note that the density function for
continuous functions F(t) is the first derivative of f and is denoted as f (t). The hazard
function h(t) is defined as h(t) =
f (t)/S(t). We interpret h(t)
as the rate of occurrence of an event
that happens in a small interval beyond t,
given that it has not occurred by t.
For the negative exponential model, F(t)
= 1 – e– λ t and f (t) = λe– λt. So h(t) = λe– λt/e– λt = λ. The exponential model has the property of a constant hazard rate. This is sometimes called the lack of memory property because the rate
does not de-pend on t. Note that
hazard rates usually depend on the time t.
The negative exponential model can be used for
studying light bulbs, which are no more likely to fail in the next five minutes
when they have been on for one hun-dred hours than they are in the first five
minutes after being installed. This unusual property is one of the reasons why,
although good for modeling the life of light bulbs, the exponential is not a
good model in general. For many products we expect the hazard rate to increase
with age. Display 15.3, which is based on the survival function, defines the
negative exponential model.
A common model for mortality is the so-called
bathtub-shaped hazard rate func-tion. At or near birth, the hazard rate is
high, but once the baby survives for a few days the hazard rate drops
significantly. For many years, the hazard rate stays flat (constant). But as
the person ages, the hazard rate starts to increase sharply. This function
would have the shape of a bathtub.
The Weibull model can be viewed also as a
generalization of the negative expo-nential. It is determined by two parameters,
λ and β, where λ refers to a rate para-meter and β refers to the shape of the
parameter distribution. The case β = 1 is the negative exponential (for reasons explained in the next
paragraph). The model can be defined by its distribution F(t), survival function S(t),
density function f (t), or hazard function h(t).
The latter, h(t), can be used to derive mathematically each of the other three
functions: F(t), S(t), and f (t). So we can describe
the Weibull by its hazard function h(t). (Refer to Display 15.4 for the
Weibull model.)
The Weibull model can have an increasing hazard
rate, a decreasing hazard rate, or in the special case of the negative
exponential, a constant hazard rate. The Weibull does not exhibit a bathtub
shape. To obtain the bathtub shape, we need a more complex parametric model.
Such models are beyond the scope of this course.
We note that for β > 1, the hazard function is increasing in t; for β = 1, it is a constant function of t;
and for β < 1
it is decreasing in t.
For complete data, likelihood methods are used to find the estimates of the para-meters for survival distributions. Sometimes survival times are right-censored; the estimation problem becomes more complicated. Many fine texts, including Lawless (1982), provide methods for estimation (point estimates and confidence intervals) and testing model parameters.
S(t) = exp(–λt)
where t ≥ 0, and λ > 0 is the rate parameter. F(t) = 1 – exp(–λt), f (t) = λ exp(–λt), and h(t) = λ.
h(t) =
λb(λt)β–1
where t ≥ 0, λ > 0 is the rate parameter, and β > 0 is the shape parameter. S(t) = exp[–(λt)β] and f (t) = λβ(λt)β–1 exp[–(λt)β].
For the negative exponential, the point estimate of λ is simply the number of events divided by the total time on test, where the
total time on test is defined as the sum of the survival times for all the
patients (time to censoring is used for the right-censored cases). Once the
parameter λ has been estimated, the survival curve esti-mate is determined by
plugging the estimate for λ into the formula. So if the estimate for λ is denoted λh and the
estimate for the survival curve is Sh(t), then Sh(t) = e–λht.
Let us consider the data in Table 15.1 again. There
are four events (deaths) at 11.8, 5.4, 1.5, and 4.3 months into the trial and
six censored times at 3.2, 12.5, 17.6, 13.3, 15.0, and 13.0 months. The
estimate λh is just the number of events/total time on test = 4/(11.8 + 5.4 + 1.5 +
4.3 + 3.2 + 12.5 + 17.6 + 13.3 + 15.0 + 13.0) = 4/97.6 = 0.041. So Sh(t) = exp(–0.041t).
Refer to Table 15.4. The column labeled “Estimated
Cumulative Survival” com-pares the survival estimates at the event time points,
Sh(tj), for the negative exponential with the results for
the Kaplan–Meier (KM) estimates (KM given in parentheses). The discrepancies
between the negative exponential and the Kaplan–Meier estimates indicate that
the exponential does not fit this model well. The discrepancy is particu-larly
noticeable at time 5.4 months, when the parametric estimate is 0.801 and the
Kaplan–Meier is 0.675. However, the sample size is small, and this discrepancy
may not be statistically significant. Note that for the exponential model the
estimates Sh(tj) = e–λhtj. So, since λh = 0.041 at t1 =
1.5, Sh(t1) = exp[–0.041 (1.5)] =
exp(–0.0615) = 0.940. At t2
= 4.3, Sh(t2) = exp[–0.041 (4.3)] =
exp(–0.1763) = 0.838.
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