Three Shapes of Risk
The survival function tells you what fraction of a population is still running. The hazard rate tells you something more useful: the risk faced right now by the units that are still alive. This page pins one distribution and gives you one knob, so you can see the two curves move together.
New vocabulary is underlined like this — hover or tap any term for a one-line explanation that appears right below the line you are reading.
- Think · 2 minA jet engine has run 200 cycles without failing. A second, identical engine is fresh out of the box. Which one is more likely to fail in the next cycle — and is there any lifetime model in which the honest answer is neither?
I've done the Think step — reveal Pair & Share
- Pair · 3 minCompare with a neighbour. Did either of you argue for the used engine being safer? Under what circumstances would that be right?
- Share · 2 minAs a table, commit to a prediction for what the hazard curve looks like for each of your three answers — rising, flat, or falling — then carry it into the widget below.
Three shapes of risk
Every lifetime distribution can be described three equivalent ways: by its CDF \(F(t)\), by its survival function \(S(t) = 1 - F(t)\), or by its hazard rate. The hazard is the density renormalized by the probability of still being around:
$$h(t) = \frac{f(t)}{S(t)} = \frac{f(t)}{1 - F(t)}$$
Read it as a conditional statement: \(h(t)\,dt\) is roughly the probability of failing in the next instant given survival past \(t\). That conditioning is the whole point — \(f(t)\) describes failures spread across every unit that ever existed, while \(h(t)\) describes the risk to the units you still own.
The Weibull family makes the shape of that risk a single parameter. With the scale fixed at \(\beta = 1\):
$$S(t) = e^{-(\beta t)^k}, \qquad h(t) = k\beta(\beta t)^{k-1}$$
Drag \(k\) and watch both panels. The tint under the hazard curve marks the regime: orange where risk climbs with age, green where it falls.
At \(k = 1\) the hazard is exactly flat — the Weibull collapses to the exponential, and \(h(t) = \beta\) forever. That flatness is memorylessness: a component that has run for 200 cycles is neither safer nor more fragile than a new one. Nothing wears out and nothing settles in.
This is why the exponential is the null model for lifetimes rather than a realistic one. When you plot an empirical hazard and it is not flat, you have learned something real about the mechanism — rising means wear-out, falling means manufacturing defects burning off early.
Set \(k = 0.6\), then \(k = 1.8\). In each case, compare \(h(0.5)\) with \(h(2.0)\) in the readout. Which value of \(k\) describes a population where surviving a long time is good news about the unit you're holding?
Show answer
\(k = 0.6\). The hazard falls from about 0.79 at \(t=0.5\) to about 0.46 at \(t=2\), a ratio of roughly 0.57× — a unit that has lasted to \(t=2\) faces less instantaneous risk than a young one, because the fragile units have already failed and been removed from the at-risk pool. At \(k = 1.8\) the ratio runs the other way (about 3×): every cycle survived is a cycle closer to wearing out.
Note that \(S(t)\) is decreasing in both cases — survival always falls. It is the rate of failing, conditional on still being alive, that can go either way. Confusing those two is the most common error in reading these curves.
Real systems often show a bathtub curve: a falling hazard early (defects), a flat stretch in the middle (random shocks), and a rising hazard late (wear-out). No single Weibull produces all three at once — you need a mixture. The value of the one-parameter version is that it names the three regimes cleanly enough to recognize them in data.