Weibull distribution

You will learn: Connect shape with hazard and compare conditional remaining survival.

Does surviving longer change the chance of failing soon? A Weibull lifetime model can have a decreasing, constant, or increasing hazard. Its shape controls that direction; its scale sets the time units. Choosing a curve is a modeling assumption to check against observed lifetimes.

f(x;\lambda,k)=\frac{k}{\lambda}\!\left(\frac{x}{\lambda}\right)^{k-1}\!e^{-(x/\lambda)^{k}}

PDF (mean 0.90, mode 0.48)

The density starts at 0.0001λ and clips at height 0.85699; for k<1 it diverges at zero. 1 of 1000 sampled times are beyond the horizontal window, with model probability 0.0010000. All samples remain in denominators.

Hazard is a rate, survival is a probability

The separate hazard drawing spans 0.05λ to 3λ without flattening the curve. Its vertical units are inverse time. A hazard above one is possible; it is not a probability.

Time already survived 1.00 Further time 1.00

P(T>1) = 0.367879. Conditional on survival to that time, P(T>2 | T>1) = 0.160666. The fresh-life survival for the same additional duration is 0.367879.

Hazard at the entered age: 1.50000 per time unit. At k=1, rate=1/λ, not λ. The chance of failure in a short interval is approximately hazard × interval length, but the conditional probability above uses the exact model.

Check the lifetime model

Simulated source for the model check

The Weibull source uses the entered shape and scale. The alternative is log-normal with median λ and log-SD 1.

Complete uncensored synthetic lifetimes: empirical quantiles and Weibull transformation
p	Time	log(−log(1−p))	k log(time/λ)
0.1	0.1994	-2.250	-2.419
0.25	0.4294	-1.246	-1.268
0.5	0.7732	-0.367	-0.386
0.75	1.211	0.327	0.287
0.9	1.714	0.834	0.808

Under this Weibull model, the last two columns should approximately agree across probabilities. Sample noise remains; a few close values do not establish a fit. The log-normal alternative can disagree systematically. These quantiles use ordered complete lifetimes, not a method for censored data.

λ (scale) 1.0 k (shape) 1.5 n (samples) 1000

PDF

f(x;\lambda,k)=\frac{k}{\lambda}\!\left(\frac{x}{\lambda}\right)^{k-1}\!\!e^{-(x/\lambda)^{k}}

. Hazard is decreasing for

k<1

(infant mortality), constant at

k=1

(exponential), rising for

k>1

(wear-out).

Three hazard regimes

The defining feature of the Weibull is how its hazard rate — the instantaneous failure rate conditional on survival so far — depends on shape:

h(x)=\frac{k}{\lambda}\!\left(\frac{x}{\lambda}\right)^{k-1}

k < 1: decreasing hazard. The model describes a declining instantaneous rate among survivors; this alone does not identify why the rate declines. Surviving longer is associated with a lower instantaneous failure rate, but the model still assigns eventual failure probability 1.
k = 1: constant hazard. Memoryless. Pure exponential. The component has no memory of how long it’s been running.
k > 1: increasing hazard. The longer it’s been running, the more likely it is to fail. This is compatible with a wear-out pattern, but does not establish a physical mechanism.

The first plot shows a density and sampled lifetimes. The separate ochre hazard plot uses inverse-time units and starts at a positive time because the k<1 hazard diverges at zero.

Why not just exponential?

The exponential distribution is the k=1 case. Here λ is a scale: its exponential rate is 1/λ. Doubling λ doubles each corresponding quantile and halves the hazard at the same scaled age t/λ. It does not change the shape regime.

E[X]=\lambda\,\Gamma\!\left(1+\tfrac{1}{k}\right)

Work a conditional lifetime calculation

For λ=2 and k=2, survival is S(t)=exp(−(t/2)²). At time 2, S(2)=exp(−1)≈0.367879. Among units still running then, the probability of lasting another unit of time is S(3)/S(2)=exp(−1.25)≈0.286505. A fresh unit lasts one time unit with probability S(1)=exp(−0.25)≈0.778801. These are different populations with different denominators.

The hazard at time 2 is 1 per time unit. That does not mean certain failure in the next unit: the exact conditional failure probability is 1−exp(−1.25)≈0.713495. Multiplying a hazard by a duration is only a local approximation for sufficiently short durations.

Make a prediction

At λ=2 and k=1, does surviving two time units change survival over the next one?

Explore the answer

No. Both fresh and conditional survival are exp(−1/2)≈0.606531. The constant hazard is 0.5 per time unit. Change k back to 2 to see why the conditional and fresh probabilities separate.

Use the transformation to question the model

Taking two logarithms of the survival law gives log(−log S(t))=k log(t/λ). The model-check table compares this relationship at five empirical quantiles. Under the chosen Weibull model the transformed columns should approximately agree, with sampling noise. Switch to the log-normal source and inspect systematic differences across the probabilities; one matching median cannot validate the whole tail.

These are complete synthetic lifetimes. Real tests often end before all units fail. Dropping those still running biases the observed lifetime sample; a censored-data method must retain their time at risk. A bathtub-shaped hazard, changing operating conditions, or dependence between units can also defeat a single fixed Weibull model.

References

NIST’s Weibull reference gives survival, hazard, quantiles, and moments. NIST’s cumulative-hazard plotting guide explains the logarithmic model check and treatment of running times. Compare the exact conditional-survival experiment in exponential waiting times.

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