Parametric Survival Distributions

Why Specify an Event-Time Distribution?

A parametric survival model describes the distribution of a positive event time using a small number of parameters. Its advantage is a compact representation: once the distribution is specified, survival probabilities, event-time densities, and hazards follow from the same model. Its risk is equally important: a chosen family restricts the possible hazard shapes. A model with a constant hazard cannot represent a sharply increasing one, for example.

The focus here is the distribution of an event time \(T\), not a regression model for covariates. Assume \(T>0\) is a continuous event time measured from a clearly defined origin. Right-censored observations do not change the definition of the latent event-time distribution; they provide incomplete information about \(T\). The formulas below describe the distribution one posits for \(T\).

Definition: The survival function is \(S(t)=P(T>t)\), the density is \(f(t)\), and the cumulative hazard is \(H(t)=-\log S(t)\). Where \(S(t)>0\), the hazard is \(h(t)=f(t)/S(t)\). It is the instantaneous event rate at time \(t\), conditional on having remained event-free up to \(t\), and has units of inverse time.

For a continuous event time, \(f(t)=-S'(t)\), so \(h(t)=-\frac{d}{dt}\log S(t)=H'(t)\). Thus a proposed survival function immediately gives both cumulative hazard and hazard. These relationships are also useful checks: \(S(0)=1\) for a model with no probability mass at time zero, \(S(t)\) must not increase, and a hazard must be nonnegative.

H(t)=-\log S(t), \qquad h(t)=H'(t)=-\frac{d}{dt}\log S(t)=\frac{f(t)}{S(t)}.

Five Common Parametric Families

The parameterizations below use time \(t\geq 0\). Parameters are chosen so that their units and roles are explicit. In particular, a scale parameter has units of time, while a rate or hazard parameter has units of inverse time.

Family	Survival and hazard	Typical hazard shape
Exponential, rate \(\lambda>0\)	\(S(t)=e^{-\lambda t}\); \(h(t)=\lambda\)	Constant
Weibull, shape \(k>0\), scale \(\eta>0\)	\(S(t)=e^{-(t/\eta)^k}\); \(h(t)=\frac{k}{\eta}(t/\eta)^{k-1}\)	Decreasing if \(k<1\), constant if \(k=1\), increasing if \(k>1\)
Gompertz, initial rate \(\lambda>0\), slope \(\gamma\)	\(h(t)=\lambda e^{\gamma t}\); \(H(t)=\frac{\lambda}{\gamma}(e^{\gamma t}-1)\) for \(\gamma\ne0\)	Increasing if \(\gamma>0\); decreasing if \(\gamma<0\)
Log-normal, \(\log T\sim N(\mu,\sigma^2)\)	\(S(t)=1-\Phi(z)\); \(h(t)=\frac{\phi(z)}{t\sigma[1-\Phi(z)]}\), where \(z=(\log t-\mu)/\sigma\)	Typically rises and later declines
Log-logistic, shape \(a>0\), scale \(b>0\)	\(S(t)=\frac{1}{1+(t/b)^a}\); \(h(t)=\frac{(a/b)(t/b)^{a-1}}{1+(t/b)^a}\)	Can rise and then decline if \(a>1\)

Here \(\phi\) and \(\Phi\) are the standard normal density and cumulative distribution function. In the log-normal model, \(\mu\) is the mean of \(\log T\) and \(\sigma\) is its standard deviation; they are not the mean and standard deviation of \(T\). The log-logistic formula follows by differentiating its survival function and dividing the resulting density by survival.

Gompertz qualification: For \(\gamma=0\), take the continuous limit \(H(t)=\lambda t\), giving the exponential model. If \(\gamma\geq0\), \(S(t)\) tends to zero as \(t\) tends to infinity, so the model assigns probability one to an eventual event. If \(\gamma<0\), \(H(t)\) tends to the finite value \(\lambda/(-\gamma)\), and \(S(t)\) tends to a positive limit. This specification therefore includes a probability of no event ever; use it only if that feature is intended.

Worked Examples: Reading Hazard Shapes From Parameters

Worked Example: Exponential and Weibull Hazards

Suppose time is measured in months. An exponential model has rate \(\lambda=0.10\) per month. At \(t=10\) months, its survival probability and hazard are

S(10)=e^{-0.10(10)}=e^{-1}=0.3679,\qquad h(10)=0.10\text{ per month}.

The constant hazard says that, at any time \(t\), the instantaneous event rate among those still event-free is \(0.10\) per month. It does not say that each individual has a 10% probability of an event in every month; a hazard is a rate, not a discrete-time probability.

Now use a Weibull model with shape \(k=2\) and scale \(\eta=10\) months. At 10 months,

S(10)=e^{-(10/10)^2}=e^{-1}=0.3679,\qquad h(10)=\frac{2}{10}(10/10)^{2-1}=0.2000\text{ per month}.

The two models have the same survival at this one time, but different hazards there. The Weibull hazard increases linearly with time for \(k=2\); the exponential hazard remains at \(0.10\) per month. Matching a single survival probability does not make two distributions equivalent.

Worked Example: An Increasing Gompertz Hazard

Let the hypothetical Gompertz hazard be \(h(t)=\lambda e^{\gamma t}\), with \(\lambda=0.020\) per month and \(\gamma=0.030\) per month. The cumulative hazard at \(t=24\) months is

H(24)=\frac{0.020}{0.030}\left(e^{0.030(24)}-1\right) =\frac{2}{3}(e^{0.72}-1) \approx \frac{2}{3}(2.05443-1)=0.70296.

Therefore \(S(24)=e^{-0.70296}\approx0.4951\): under this specified distribution, about 49.51% of event times exceed 24 months. The instantaneous hazard at that time is

h(24)=0.020e^{0.72}=0.04109\text{ per month}.

Because \(\gamma>0\), the hazard increases with time. The initial hazard at time zero is \(0.020\) per month, and by 24 months it is about \(0.04109\) per month. This is a change in the conditional instantaneous rate, not a direct statement about the probability of an event during a particular 24-month interval.

Worked Example: Computing a Log-Normal Hazard

Suppose a hypothetical event time in months follows \(\log T\sim N(\log 12,0.5^2)\). The median event time is 12 months, since \(\log 12\) is the median of \(\log T\). At \(t=60\) months, the standardized log time is

z=\frac{\log(60)-\log(12)}{0.5} =\frac{\log(5)}{0.5} \approx 3.218876.

For this value, \(\phi(z)\approx0.0022440\) and \(1-\Phi(z)\approx0.00064347\). Substituting into the hazard formula gives

h(60)=\frac{\phi(z)}{60(0.5)[1-\Phi(z)]} \approx\frac{0.0022440}{30(0.00064347)} \approx0.1162\text{ per month}.

The small survival probability at 60 months is not itself a hazard. The hazard compares density at that time with the probability of surviving beyond it. This example also illustrates why the log-normal hazard should be evaluated from its formula rather than inferred from the fact that the distribution is right-skewed: its hazard can rise and later decline.

Worked Example: A Log-Logistic Hazard That Peaks

For a log-logistic model with shape \(a=2\) and scale \(b=10\) months, the survival probability at \(t=10\) months is

S(10)=\frac{1}{1+(10/10)^2}=\frac{1}{2}=0.5.

Its hazard at that time is

h(10)=\frac{(2/10)(10/10)^{2-1}}{1+(10/10)^2} =\frac{0.2}{2} =0.1000\text{ per month}.

For this particular shape, the hazard increases before \(t=10\) and decreases afterward. For example, \(h(5)=[0.2(0.5)]/[1+0.5^2]=0.0800\) per month, while \(h(20)=[0.2(2)]/[1+2^2]=0.0800\) per month. The scale \(b\) is also the median, because \(S(b)=1/2\), but it is not generally the mean.

Why the Hazard Shapes Differ

The hazard is a compact way to compare what the families imply about how event risk changes among those still at risk. The exponential model is the special constant-hazard benchmark. The Weibull family adds a monotone power-shaped hazard: its shape parameter \(k\) controls whether the hazard falls, stays level, or rises. The Gompertz family instead makes the log hazard linear in time, since \(\log h(t)=\log\lambda+\gamma t\). The log-normal and log-logistic families can accommodate hazards that rise and later fall.

For the Weibull formula, the shape follows directly from the power of time:

h(t)=\frac{k}{\eta^k}t^{k-1}.

When \(k>1\), the exponent \(k-1\) is positive and the hazard increases with \(t\); when \(0<k<1\), it decreases; and when \(k=1\), it is constant at \(1/\eta\). The exponential model is thus the Weibull model with \(k=1\) and \(\lambda=1/\eta\).

For the log-logistic family with \(a>1\), the hazard reaches its maximum at \(t=b(a-1)^{1/a}\). This follows by setting the derivative of its log hazard to zero:

\frac{d}{dt}\log h(t) =\frac{a-1}{t}-\frac{a t^{a-1}}{b^a+t^a}=0 \quad\Longleftrightarrow\quad (a-1)b^a=t^a.

Solving gives the stated peak time. For \(a=2,b=10\), it is \(10\) months, as in the worked example. When \(a\leq1\), this interior peak does not occur; the hazard is decreasing (with the \(a=1\) case equal to \(1/(b+t)\)).

State the time scale and origin.
A rate of \(0.1\) per month and a scale of 10 months describe the same time unit, but their numbers cannot be interpreted without that unit.

Translate the parameters into a shape.
Identify whether the hazard is constant, monotone, or potentially nonmonotone before interpreting a fitted or assumed family.

Check the implied survival behavior.
Verify that \(S(t)\) decreases from one and determine whether it tends to zero. In particular, check the Gompertz sign of \(\gamma\).

Keep conditional rates distinct from probabilities.
Use \(S(t)\) for the probability of surviving beyond \(t\); describe \(h(t)\) as an instantaneous rate among those surviving to \(t\).

Model Choice and a Common Pitfall

A parametric family is an assumption about the shape of the event-time distribution, not a discovery guaranteed by the data. Two models can agree at one survival time and disagree elsewhere, as the exponential and Weibull calculation showed. Comparing families therefore requires attention to the full pattern over time and to the scientific meaning of their implied tails and hazard shapes. A hazard that rises and then falls may be plausible in some settings, while a constant hazard may be too restrictive.

A common mistake is to say, “The Weibull shape parameter exceeds one, so the event probability is higher at every later time.” The parameter does imply an increasing hazard, but that sentence confuses a conditional rate with an unconditional probability. Survival is still decreasing, and the probability of an event in an interval depends on the survival at its start as well as the hazard over the interval. A correct statement is: “With \(k>1\), the Weibull hazard increases over time among individuals who remain event-free.”

Another practical check is parameter units. In the Gompertz model, \(\gamma t\) must be dimensionless, so \(\gamma\) has inverse-time units. In the Weibull model, \(t/\eta\) is dimensionless. Such checks can expose an accidental mismatch of time units before it propagates into predicted survival or hazard values.

Key takeaway: Each parametric survival family links a specific survival curve to a specific hazard shape. Report the family, parameterization, time units, and implications together; a parameter value has no useful interpretation without them.

Check Your Understanding

Use the definitions and parameterizations in this tutorial to answer the following.

For a Weibull model with \(k=0.7\), is the hazard increasing, decreasing, or constant? Explain using the hazard formula.
An exponential model has rate \(0.04\) per day. Write its survival probability at 10 days and identify its hazard at that time.
In a Gompertz model, what does a negative value of \(\gamma\) imply for the limiting survival probability?
For a log-normal event time, why is \(\mu\) not generally the mean event time?
A colleague interprets \(h(5)=0.2\) per month as a 20% probability of an event during month 5. Identify the error and state what the hazard means instead.