The Weibull Model and Its Hazard Shape

Why Use a Weibull Model?

The exponential model represents a constant event hazard over time. That can be a useful starting point, but many event processes plausibly become more or less likely as time passes. The Weibull distribution adds one shape parameter that allows the hazard to decrease or increase while retaining a simple survival function. Its shape is therefore substantively important: it describes how the instantaneous event rate changes over elapsed time.

The Weibull model is a parametric model for a positive event time \(T\). We use shape \(k>0\) and scale \(\lambda>0\), with \(\lambda\) measured in the same time units as \(T\). The scale is not generally the mean or median, but it has a direct survival interpretation: survival at time \(\lambda\) is \(e^{-1}\), about 0.368.

Definition: A Weibull event time with shape \(k>0\) and scale \(\lambda>0\) has cumulative hazard \(H(t)=(t/\lambda)^k\) for \(t\geq0\). Consequently, its survival function is \(S(t)=\exp\{-(t/\lambda)^k\}\), its density is \(f(t)=(k/\lambda)(t/\lambda)^{k-1}\exp\{-(t/\lambda)^k\}\), and its hazard is \(h(t)=(k/\lambda)(t/\lambda)^{k-1}\).

Deriving Survival, Hazard, and Cumulative Hazard

For a continuous event time, the survival function is \(S(t)=P(T>t)\), and the hazard is the instantaneous event rate among individuals who have survived to time \(t\). Where the density is positive, \(h(t)=f(t)/S(t)\). The cumulative hazard is the integrated hazard, \(H(t)=\int_0^t h(u)\,du\), and \(S(t)=\exp\{-H(t)\}\).

The Weibull definition specifies a cumulative hazard \(H(t)=(t/\lambda)^k\). Differentiating with respect to time gives the hazard. Applying \(S(t)=\exp\{-H(t)\}\) gives survival, and differentiating \(1-S(t)\) gives the density:

h(t)=H'(t)=\frac{k}{\lambda}\left(\frac{t}{\lambda}\right)^{k-1}, \qquad S(t)=e^{-H(t)}=\exp\left\{-\left(\frac{t}{\lambda}\right)^k\right\}, \qquad f(t)=-S'(t)=h(t)S(t).

These expressions are consistent because \(H(0)=0\), \(S(0)=1\), and \(S(t)\) approaches zero as \(t\) increases without bound. The density is nonnegative and integrates to one. The hazard has units of inverse time: \(k\) is dimensionless, while \(\lambda\) and \(t\) share units.

Worked Example: Deriving Survival at a Specified Time

Suppose a hypothetical event time follows a Weibull model with shape \(k=2\) and scale \(\lambda=5\) years. Find the cumulative hazard and survival probability at 3 years.

Substitute into the cumulative hazard: \(H(3)=(3/5)^2=9/25=0.3600\). Therefore, \(S(3)=\exp(-0.3600)\approx0.6977\). Under this model, about 69.8% of the population survives beyond 3 years. This is a model-based survival probability, not an estimate that accounts for sampling uncertainty.

The corresponding hazard at 3 years is \(h(3)=(2/5)(3/5)^{1}=0.2400\) per year. It is an instantaneous rate at 3 years among those still event-free then, not the probability of an event during the third year.

How Shape Controls Hazard

The shape parameter determines the direction of hazard change because \(h(t)\) is proportional to \(t^{k-1}\). If \(k<1\), the hazard decreases over time; if \(k=1\), it is constant and the Weibull reduces to the exponential model with rate \(1/\lambda\); if \(k>1\), it increases over time. This is a statement about the model’s conditional event rate, not about an individual’s risk being caused by time.

Shape	Hazard pattern	Interpretation
\(0<k<1\)	Decreases with time	The modeled instantaneous event rate is higher earlier and declines among those still at risk.
\(k=1\)	Constant	The model is exponential with hazard \(1/\lambda\).
\(k>1\)	Increases with time	The modeled instantaneous event rate rises among those still at risk.

The cumulative hazard increases in all three cases. A decreasing hazard does not mean cumulative event probability decreases: \(H(t)\) is nondecreasing, and \(S(t)=e^{-H(t)}\) is nonincreasing. The scale controls the time scale of the distribution. In particular, \(S(\lambda)=e^{-1}\), regardless of \(k\).

Worked Example: Comparing Hazard at Two Times

For a hypothetical Weibull event time measured in days, let \(k=0.7\) and \(\lambda=5\) days. Compare the hazard at day 1 with the hazard at day 5.

At day 5, \(h(5)=k/\lambda=0.7/5=0.1400\) per day. At day 1, \(h(1)=0.1400(1/5)^{-0.3}\). Since \((1/5)^{-0.3}=\exp\{0.3\log(5)\}\approx1.6207\), the hazard is approximately \(0.2269\) per day.

The modeled hazard is higher at day 1 than at day 5, consistent with \(k<1\). These values are instantaneous rates for individuals surviving to the respective times; they should not be reported as probabilities of an event on those days.

Estimating Shape and Scale From Complete Event Times

Suppose \(n\) independent event times \(t_1,\ldots,t_n\) are fully observed and each follows the same Weibull distribution. Under these assumptions, the likelihood is the product of their densities. Taking logarithms gives

\ell(k,\lambda) =n\log k-nk\log\lambda+(k-1)\sum_{i=1}^n\log t_i -\sum_{i=1}^n(t_i/\lambda)^k.

For a fixed \(k\), differentiating with respect to \(\lambda\) and setting the score to zero gives \(\widehat{\lambda}^{\,k}=\sum_i t_i^k/n\). Substituting this profile estimate into the shape score yields a one-dimensional equation for \(k\). In general that equation is solved numerically; the estimate of shape does not usually have a closed form.

\frac{1}{k}+\frac{1}{n}\sum_{i=1}^n\log t_i -\frac{\sum_{i=1}^n t_i^k\log t_i}{\sum_{i=1}^n t_i^k}=0, \qquad \widehat{\lambda} =\left(\frac{\sum_{i=1}^n t_i^{\widehat{k}}}{n}\right)^{1/\widehat{k}}.

The shape equation balances the average log event time against a weighted average of log times, with weights proportional to \(t_i^k\). Since the weights themselves depend on \(k\), an iterative root-finding method or numerical likelihood maximization is needed. A fitted root should be checked by substituting it back into the score equation and confirming that the log likelihood is maximized, not merely stationary.

Worked Example: Maximum Likelihood From Four Event Times

Consider four hypothetical, fully observed event times, in years: \(1,2,3,\) and \(4\). Assume independent Weibull event times with common shape and scale, with no censoring. Here \(n=4\).

The mean log time is \((\log 1+\log 2+\log 3+\log 4)/4 =(0+0.6931+1.0986+1.3863)/4 \approx0.7945\). Numerically solving the shape equation gives \(\widehat{k}\approx2.453\). At this value, the powers are approximately \(1^{2.452}=1\), \(2^{2.452}=5.472\), \(3^{2.452}=14.788\), and \(4^{2.452}=29.940\), with sum about \(51.200\).

The corresponding weighted log-time average is \[ \frac{5.472\log2+14.788\log3+29.940\log4}{51.200} \approx\frac{61.544}{51.200} \approx1.2021. \] The shape equation’s left side is therefore approximately \(1/2.452+0.7945-1.2021\approx0.0003\), close to zero given rounding. The scale estimate is \(\widehat{\lambda}=(51.200/4)^{1/2.452}\approx2.828\) years.

Because \(\widehat{k}>1\), this fitted Weibull model has an increasing hazard. The scale estimate is the time at which fitted survival is \(e^{-1}\), not the estimated mean event time.

Right Censoring and Estimation

With right censoring, let \(y_i\) be the observed time (the event time if observed, otherwise the censoring time), and let \(\delta_i=1\) for an observed event and \(0\) for a right-censored observation. If censoring is independent of the event time, the event-time likelihood contribution is \(f(y_i)\) when \(\delta_i=1\), and \(S(y_i)\) when \(\delta_i=0\). Thus the likelihood is

L(k,\lambda)=\prod_{i=1}^n f(y_i)^{\delta_i}S(y_i)^{1-\delta_i}.

This likelihood uses the event status correctly: a censored observation says the event time exceeds the observed time, not that an event occurred at that time. If \(D=\sum_i\delta_i\) events are observed and \(D>0\), then for fixed \(k\), maximizing the likelihood over scale gives \(\widehat{\lambda}^{\,k}=\sum_i y_i^k/D\). The shape must then be estimated using the censored likelihood, rather than the complete-data equation above. Independent censoring is essential to this simple event-time likelihood; informative censoring generally requires additional modeling.

Worked Example: Scale Estimation With Right Censoring

Suppose five hypothetical individuals have observed times \(y=(1,2,3,4,4)\) years and event indicators \(\delta=(1,1,0,1,0)\). Thus events occur at 1, 2, and 4 years; the other two observations are right-censored. Assume independent censoring and, just for this calculation, fix the Weibull shape at \(k=2\).

There are \(D=3\) observed events. The fixed-shape scale estimate is \[ \widehat{\lambda} =\left(\frac{1^2+2^2+3^2+4^2+4^2}{3}\right)^{1/2} =\left(\frac{46}{3}\right)^{1/2} \approx3.916\text{ years}. \] The censored times contribute to the sum because each provides information that the event time exceeded its censoring time. This is a conditional scale estimate for the stipulated shape, not a joint estimate of both parameters.

Assumptions, Diagnostics, and Alternatives

The basic Weibull model assumes positive event times from a common distribution with a single shape and scale. For likelihood estimation, observations are independent, and censoring must be independent of event time (or independent conditional on variables included in the model). A shared distribution is also a substantive assumption: major unmodeled differences between individuals can make a single fitted curve misleading. The calculations above do not account for delayed entry.

Assess fit by comparing the fitted survival curve with a nonparametric estimate such as Kaplan–Meier, especially in the time range with adequate numbers at risk. Large systematic differences suggest that the chosen parametric shape may not describe the data. A useful Weibull-specific diagnostic follows by taking logarithms: \(\log\{-\log S(t)\}=k\log t-k\log\lambda\). Thus a plot of log cumulative hazard against log time should be approximately linear under the model, over a range where the survival estimates are reliable. Curvature can indicate that a single Weibull shape is inadequate; sparse late follow-up can make the plot unstable.

The main alternative in the earlier survival-model material is the Cox proportional hazards model, which leaves the baseline hazard unspecified rather than requiring a Weibull shape. It is more flexible about baseline shape, while the Weibull gives a compact parametric survival curve and can describe a monotone changing hazard. The exponential model is the special Weibull case \(k=1\), and is more restrictive because it requires constant hazard. Choice among them depends on the scientific aim, fit, and extrapolation needs; a plausible fit over observed follow-up does not guarantee reliable extrapolation far beyond it.

A common mistake is to treat a censored time as an event time and include its density contribution. That asserts the event occurred exactly at censoring, which the observation does not establish. The correction is to use \(S(y_i)\) for a right-censored observation, reflecting only that the event time exceeds \(y_i\). Another pitfall is to interpret \(h(t)\) as a probability: it is a rate, with units of inverse time, not a probability without specifying an interval.

Key takeaway: The Weibull shape \(k\) determines whether the modeled hazard decreases, stays constant, or increases; the scale \(\lambda\) sets the time scale. Correct estimation depends on distinguishing observed events from right-censored times and checking whether the assumed hazard shape is credible.

Check Your Understanding

Answer each question, then compare your reasoning with the brief solutions below.

If a Weibull model has \(k=1\), what is its hazard, and which familiar survival model does it represent?
For \(k=1.6\), does the hazard rise or fall with time? State why.
If \(\lambda=8\) months, what is the model survival probability at 8 months?
In a right-censored likelihood, what contribution corresponds to a censoring time \(y\)?

Solutions. (1) The hazard is constant at \(1/\lambda\), so the model is exponential. (2) It rises, because \(h(t)\) is proportional to \(t^{k-1}=t^{0.6}\). (3) \(S(8)=\exp\{-(8/8)^k\}=e^{-1}\approx0.3679\). (4) The contribution is \(S(y)\): the observation establishes that the event time exceeds \(y\), not that an event occurred at \(y\).