Weibull Distribution

Plain-Language Definition

The Weibull distribution is the working parametric model for the time until an event when the hazard rate is not constant. The hazard rate of a positive random variable is the conditional instantaneous failure rate given survival to time $x$. For an Exponential, the hazard is constant; for many real-world lifetimes, the hazard increases with age (wear-out) or decreases with age (early-life failures dropping off as defective units fail and are replaced).

The Weibull is built around a hazard rate that is a power of $x$. The shape parameter $k$ controls the power, and therefore the trend of the hazard. The scale parameter $\lambda$ rescales the time axis. Together they give a two-parameter family flexible enough for reliability engineering, survival analysis in medicine, and wind-speed modeling, while remaining easy to fit and interpret.

Definition

Definition

Weibull Distribution$X\sim \mathrm{Weibull}(k,\lambda )$

A positive random variable $X$ has a Weibull distribution with shape $k > 0$ and scale $\lambda > 0$ when its density is

$f_X(x) = \frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k - 1} \exp\!\left(-\left(\frac{x}{\lambda}\right)^k\right), \quad x \geq 0,$

equivalently, its survival function is $S_X(x) = \mathbb{P}(X > x) = \exp(-(x/\lambda)^k)$.

The hazard function $h(x) = f_X(x) / S_X(x)$ equals $(k/\lambda)(x/\lambda)^{k - 1}$, a power of $x$ with exponent $k - 1$.

The parameter names vary across references. Some texts use $\beta$ for the shape and $\eta$ for the scale; some use $a$ and $b$. The shape is the parameter that controls the hazard trend, and is what most readers will care about; the scale just sets the time units.

Why This Matters

Three reasons the Weibull is the most-used parametric lifetime model.

1. Hazard flexibility. A constant hazard is a strong assumption that few real-world systems satisfy. Mechanical components have an early-life region where defective units fail (decreasing hazard) and a wear-out region where surviving units accumulate damage (increasing hazard). The Weibull spans the full range of monotone hazards with one parameter.

2. Extreme-value connection. The Weibull is the Type III extreme-value distribution: the limiting law for the minimum of iid samples from a distribution bounded below, after suitable centering and scaling. This is the formal reason wind speeds and failure times often follow Weibulls; both are minima of many local failures and can be characterized via the Fisher-Tippett-Gnedenko theorem in extreme-value theory.

3. Reliability engineering practice. The Weibull is the default fit in reliability software (Minitab, Reliasoft, R's survreg). Engineers can interpret the shape parameter directly as "infant mortality" ($k < 1$), "random failures" ($k = 1$), or "wear-out" ($k > 1$). The Weibull also generalizes the Exponential and the Rayleigh as special cases, so the same workflow covers all three.

Survival, Mean, Variance

Theorem

Weibull Survival Function and Moments

Statement

$\mathbb{P}(X > x) = e^{-(x/\lambda)^k}, \quad \mathbb{P}(X \leq x) = 1 - e^{-(x/\lambda)^k}, \quad x \geq 0.$

All moments exist. The first two are $\mathbb{E}[X] = \lambda\, \Gamma\!\left(1 + \frac{1}{k}\right), \qquad \operatorname{Var}(X) = \lambda^2 \left[\Gamma\!\left(1 + \frac{2}{k}\right) - \Gamma\!\left(1 + \frac{1}{k}\right)^2\right].$

Intuition

The integral that defines the $m$-th moment is $\int_0^\infty x^m \cdot (k/\lambda)(x/\lambda)^{k - 1} e^{-(x/\lambda)^k}\, dx$. The substitution $u = (x/\lambda)^k$ converts it to $\lambda^m \int_0^\infty u^{m/k} e^{-u}\, du = \lambda^m \Gamma(1 + m/k)$. All Weibull moments are finite because the Gamma function is finite at every positive argument and the substitution removes the power-law factor cleanly.

Proof Sketch

For the survival function, integrate the density: $\mathbb{P}(X > x) = \int_x^\infty f_X(t)\, dt$ with substitution $u = (t/\lambda)^k$ giving $du = (k/\lambda)(t/\lambda)^{k-1}dt$, so the integral evaluates to $\int_{(x/\lambda)^k}^\infty e^{-u}\, du = e^{-(x/\lambda)^k}$.

For the $m$-th moment, $\mathbb{E}[X^m] = \int_0^\infty m x^{m-1} \mathbb{P}(X > x)\, dx = \int_0^\infty m x^{m-1} e^{-(x/\lambda)^k}\, dx$. Substitute $u = (x/\lambda)^k$ to obtain $\lambda^m \Gamma(1 + m/k)$.

Why It Matters

Both moments require the Gamma function, which is a closed-form quantity only in the integer-shape case. For $k = 1$ (Exponential), $\Gamma(2) = 1$ and $\mathbb{E}[X] = \lambda$, $\operatorname{Var}(X) = \lambda^2$. For $k = 2$ (Rayleigh), $\mathbb{E}[X] = \lambda \sqrt{\pi}/2$ and $\operatorname{Var}(X) = \lambda^2(1 - \pi/4)$. For other $k$, the moments are computed numerically from a Gamma routine.

Failure Mode

Statistical software is split on the scale parameterization. The Wikipedia / R pweibull convention uses $\lambda$ as defined here. The reliability-engineering convention sometimes uses $\eta = \lambda$ (same thing) but with the survival function written as $e^{-(x/\eta)^k}$ and the density formula modified accordingly. SciPy's weibull_min takes shape and scale and matches the convention used here. Check the docstring before fitting.

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Hazard Function and Its Three Regimes

Theorem

Weibull Hazard Is Monotone, with Shape Setting the Trend

Statement

The hazard function $h(x) = f_X(x) / S_X(x) = (k / \lambda) (x / \lambda)^{k - 1}$ is

• strictly decreasing in $x$ when $0 < k < 1$ ("infant mortality" or early-failure regime);
• constant at $1/\lambda$ when $k = 1$ (the Exponential / memoryless case);
• strictly increasing in $x$ when $k > 1$ ("wear-out" or aging regime).

Intuition

The hazard is the failure rate conditional on survival. The Weibull hazard is a monotone power of time. When $k = 1$, the power is $0$ and the hazard is flat. When $k$ exceeds $1$, the hazard grows; when $k$ is below $1$, it falls. The Weibull is the simplest parametric family that lets you fit a monotone hazard with one parameter.

Proof Sketch

Direct from the formula $h(x) = (k/\lambda)(x/\lambda)^{k - 1}$. The derivative with respect to $x$ is $h'(x) = (k(k - 1)/\lambda^2)(x/\lambda)^{k - 2}$, which is positive when $k > 1$ and negative when $k < 1$. Setting $k = 1$ gives $h(x) = 1/\lambda$, a constant.

Why It Matters

The shape parameter is the most interpretable quantity in a Weibull fit. A reliability engineer who sees $k = 2.3$ on a fitted Weibull immediately knows the failure mode is wear-out, not random. An $k = 0.8$ suggests early-failure issues that burn-in testing can mitigate. The numerical scale $\lambda$ only sets the time axis; the qualitative regime comes from $k$.

Failure Mode

The Weibull cannot fit a hazard that goes up then down (a "bathtub curve" without monotonicity inside each segment) with one set of parameters. Such data needs a mixture of Weibulls or a parametric extension (Generalized Gamma, Burr, or a competing-risks model). Forcing a single Weibull on bathtub data gives a poor fit and an uninterpretable $k$.

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Worked Example: Component Lifetime

A manufacturer fits a Weibull to time-to-failure data on a sealed bearing. The MLE gives $\hat k = 2.5$ and $\hat\lambda = 8000$ hours.

The fit interpretation: $\hat k > 1$ indicates a wear-out regime, consistent with the physical picture that the bearing accumulates damage over time and fails more readily after extended use.

Key quantities:

• $\mathbb{E}[X] = 8000 \cdot \Gamma(1 + 1/2.5) = 8000 \cdot \Gamma(1.4) \approx 8000 \cdot 0.8873 \approx 7099$ hours.
• $\operatorname{Var}(X) = 8000^2 \cdot [\Gamma(1.8) - \Gamma(1.4)^2] \approx 6.4 \times 10^7 \cdot [0.9314 - 0.7873] \approx 9.2 \times 10^6$; $\operatorname{SD}(X) \approx 3036$ hours.
• Median: $q_{0.5} = \lambda \cdot (\ln 2)^{1/k} = 8000 \cdot (0.693)^{0.4} \approx 8000 \cdot 0.866 \approx 6928$ hours.
• 10th percentile (reliability $R(t) = 0.9$): $q_{0.1} = \lambda \cdot (\ln(1 / 0.9))^{1/k} = 8000 \cdot (0.1054)^{0.4} \approx 8000 \cdot 0.408 \approx 3267$ hours.

The 10 percent of bearings fail before about 3267 hours. The median is 6928 hours. The mean is 7099 hours, only slightly above the median because the right skew of a $k = 2.5$ Weibull is modest.

Compare to the Exponential fit (assuming constant hazard, which gives $k = 1$). For the same mean of $7099$ hours, $\lambda_{\text{exp}} = 7099$ and 10 percent fail before $-7099 \cdot \ln 0.9 \approx 748$ hours. The Exponential dramatically overstates early failures and understates the median lifetime. The Weibull fit captures the wear-out behavior that the Exponential cannot.

Common Misconceptions

Watch Out

The Weibull is not memoryless

The memoryless property $\mathbb{P}(X > s + t \mid X > s) = \mathbb{P}(X > t)$ is unique to the Exponential among continuous distributions on the half-line. The Weibull with $k \neq 1$ is explicitly age-dependent: the conditional probability of further survival depends on how long the system has already survived. This is the whole point of using a Weibull instead of an Exponential.

Watch Out

The shape k is not the mean or the median

$k$ is dimensionless and controls the hazard trend. The scale $\lambda$ has units of time (or whatever the random variable measures). $\mathbb{E}[X] = \lambda \Gamma(1 + 1/k)$ depends on both. A larger $k$ at fixed $\lambda$ gives a slightly smaller mean (because $\Gamma(1 + 1/k)$ decreases towards $1$ as $k$ grows past 1). The two parameters are not interchangeable.

Watch Out

Maximum likelihood for k has no closed form

The Weibull log-likelihood is concave in $(k, \lambda)$ but the score equation for $k$ is transcendental. Solving requires a one-dimensional numerical solver (Newton-Raphson is standard). The closed form $\hat \lambda = (1/n \sum X_i^k)^{1/k}$ exists for $\lambda$ once $k$ is fixed, which is why the standard algorithm profiles out $\lambda$ and solves for $k$ alone.

Watch Out

A bathtub curve is not a single Weibull

A "bathtub" hazard rate (high early, low middle, high late) is a common reliability-textbook picture, but no single Weibull produces it. The Weibull hazard is monotone. A bathtub curve is typically modeled as a mixture or a competing-risks decomposition. Fitting a single Weibull to bathtub data and reading off "the $k$" is meaningless: the algorithm will fit the dominant tail and ignore the others.

Comparison: Weibull vs Exponential vs Gamma vs Lognormal

The four are the standard nonnegative lifetime distributions in applied work, and they differ on the hazard shape and the tail.

• Exponential. Constant hazard. Memoryless. Tail decays as $e^{-\lambda x}$. Only one parameter.
• Weibull. Monotone hazard. Tail decays as $e^{-(x/\lambda)^k}$, faster than Exponential if $k > 1$, slower if $k < 1$. Two parameters; $k = 1$ recovers Exponential.
• Gamma. Initially-increasing-then-leveling hazard for shape $> 1$; saturates at $1/\beta$ rather than rising indefinitely. Tail decays as $e^{-\beta x}$ times a polynomial. Two parameters.
• Lognormal. Non-monotone hazard (rises and then falls). Tail decays sub-exponentially, heavier than Gamma or Weibull with $k > 1$.

Diagnostics: a log-log plot of $-\ln S(x)$ against $\ln x$ is linear for a Weibull (the slope is $k$ and the intercept involves $\lambda$). A curved version of the same plot suggests Gamma or Lognormal. A plot of empirical hazard against time shows whether the hazard is monotone (Weibull or Gamma) or non-monotone (Lognormal); whether monotone hazards trend up (wear-out, $k > 1$) or down (early-life, $k < 1$); and whether the hazard ever flattens (Exponential).

For loss-modeling and severity-fitting applications, see ActuaryPath's Weibull page at https://www.actuarypath.com/concepts/weibull-distribution/.

Maximum-Likelihood Estimation

The log-likelihood from an iid sample $X_1, \dots, X_n$ is

$\ell(k, \lambda) = n \ln k - n k \ln \lambda + (k - 1) \sum_i \ln X_i - \sum_i (X_i / \lambda)^k.$

Setting the score $\partial \ell / \partial \lambda = 0$ gives $\hat\lambda(k) = \left[(1/n) \sum_i X_i^k\right]^{1/k}$ as a function of $k$. Substituting into the score for $k$ produces a one-dimensional equation

$\frac{1}{k} + \frac{1}{n}\sum_i \ln X_i - \frac{\sum_i X_i^k \ln X_i}{\sum_i X_i^k} = 0,$

which is solved by Newton-Raphson. The MLE is consistent, asymptotically Normal, and asymptotically efficient under standard regularity. Censored-data extensions (right censoring is the common case in reliability and survival data) require including censored observations through the survival function in the likelihood rather than the density.

Exercises

ExerciseCore

Problem

A battery lifetime is modeled as $X \sim \operatorname{Weibull}(k = 1.5, \lambda = 1000)$ hours. Find the survival function at $x = 500$ and $x = 1500$, the median lifetime, and the mean lifetime.

Hint

Reveal Solution

ExerciseCore

Problem

A patient cohort in a clinical trial has time-to-event modeled as $X \sim \operatorname{Weibull}(k = 0.8, \lambda = 24)$ months. Compute the hazard rate at 6 months and 24 months, and explain what the shape parameter implies clinically.

Hint

Reveal Solution

ExerciseCore

Problem

Wind speeds at a hub-height anemometer are modeled as $X \sim \operatorname{Weibull}(k = 2, \lambda = 8)$ m/s (the $k = 2$ case is the Rayleigh distribution). Find the median, the mean, the 95th percentile, and the probability the wind exceeds 15 m/s.

Hint

Reveal Solution

ExerciseAdvanced

Problem

Show that $Y = (X / \lambda)^k$ is Exponential with rate 1 when $X \sim \operatorname{Weibull}(k, \lambda)$.

Hint

Reveal Solution

ExerciseAdvanced

Problem

For an iid sample $X_1, \dots, X_n$ from $\operatorname{Weibull}(k, \lambda)$ with known shape $k$, show that the MLE of $\lambda$ is $\hat\lambda = \left((1/n) \sum_i X_i^k\right)^{1/k}$ and derive its asymptotic distribution.

Hint

Reveal Solution

ExerciseAdvanced

Problem

A reliability dataset of $n = 100$ failure times has $\sum_i \ln X_i = 750$. Suppose the Weibull MLE for the shape parameter has been computed numerically as $\hat k = 2.1$. Find $\hat\lambda$.

Hint

Reveal Solution

Extensions

The Weibull is the starting point for several richer parametric families used in reliability and survival work.

• Three-parameter Weibull. Adds a location (shift) parameter $\theta$: $X - \theta \sim \operatorname{Weibull}(k, \lambda)$. Used when failure times have a known minimum (a guaranteed warranty period, or a manufacturer-specified minimum life).
• Generalized Gamma. Adds a third parameter to interpolate between Weibull, Gamma, and Lognormal. Used when the shape diagnostic does not clearly favor any one of the three.
• Weibull mixtures and competing risks. Two or more Weibulls combined by a mixing distribution; produces bathtub hazard curves and other non-monotone hazards.

Each extension keeps the Weibull as a special case and adds flexibility when the two-parameter Weibull is rejected by a goodness-of-fit test.

References

• Casella, G., and Berger, R. L. (2002). Statistical Inference, 2nd ed., Duxbury. Section 3.3 includes the Weibull as a transformation of the Exponential and gives the moment formulas via the Gamma function.
• Blitzstein, J. K., and Hwang, J. (2019). Introduction to Probability, 2nd ed., Chapman and Hall / CRC. Chapter 5 covers the Weibull alongside the other lifetime distributions, with the $k = 2$ Rayleigh case worked out.
• For the actuarial / severity-modeling perspective on the Weibull, see ActuaryPath's Weibull page at https://www.actuarypath.com/concepts/weibull-distribution/ and Klugman, Panjer, Willmot (2019), Loss Models, 5th ed., Wiley, Chapter 5.
• For the survival-analysis perspective with censored data, see Klein, J. P., and Moeschberger, M. L. (2003), Survival Analysis: Techniques for Censored and Truncated Data, 2nd ed., Springer. Chapter 12 covers Weibull regression with right censoring.