Weibull Distribution

Modeling time-to-failure, event durations, and reliability.

The "Time-to-Event" Distribution

The Weibull distribution is a highly flexible continuous probability distribution. It's widely used in engineering to model reliability and time-to-failure of components. In finance, it can be applied to model the duration of events, such as the time until a corporate bond defaults or the time a stock price stays above a certain level.

Its flexibility comes from its shape parameter, $k$ . Depending on the value of $k$ , it can mimic the behavior of other distributions like the exponential (when $k=1$ ) or approximate the normal distribution (when $k$ is around 3-4).

Interactive Weibull Distribution

Adjust the shape (k) and scale (λ) parameters to see how the distribution's form changes.

Shape (k): 2.0

Scale (λ): 1.0

Mean (

\mu

): 0.89

Variance (

\sigma^2

): 0.21

Core Concepts

Probability Density Function (PDF)

The PDF is defined by its shape parameter `k` and scale parameter `λ`.

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

$x \ge 0$ is the variable (e.g., time).
$k > 0$ is the shape parameter. If $k < 1$ , the failure rate decreases over time. If $k = 1$ , it's constant (Exponential). If $k > 1$ , the failure rate increases over time (wear-out).
$\lambda > 0$ is the scale parameter, which stretches or contracts the distribution.

Expected Value & Variance

Expected Value (Mean)

E[X] = \lambda \Gamma(1 + 1/k)

Variance

Var(X) = \lambda^2 \left[ \Gamma(1 + 2/k) - (\Gamma(1 + 1/k))^2 \right]

These are expressed using the Gamma function, $\Gamma(z)$ .

Key Derivations

Deriving the Expected Value (Mean)

We derive the mean by calculating the integral of

x \cdot f(x)

over its domain [0, ∞).

Step 1: Set up the Integral for E[X]

The expected value is the integral of $x$ times the PDF.

E[X] = \int_{0}^{\infty} x \cdot \frac{k}{\lambda} \left(\frac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k} dx

Step 2: Simplify the Expression

Combine the $x$ terms and rearrange.

E[X] = \int_{0}^{\infty} \frac{k}{\lambda} \cdot \frac{x^k}{\lambda^{k-1}} \cdot e^{-(x/\lambda)^k} dx = \int_{0}^{\infty} k \frac{x^k}{\lambda^k} e^{-(x/\lambda)^k} dx

Step 3: Apply u-Substitution

This integral becomes much simpler with a substitution. Let $u = (x/\lambda)^k$ . Then:

$x = \lambda u^{1/k}$
Differentiating with respect to u gives: $dx = \lambda \frac{1}{k} u^{1/k - 1} du$

Substituting these back into the integral:

E[X] = \int_{0}^{\infty} k \cdot u \cdot e^{-u} \cdot \left(\lambda \frac{1}{k} u^{1/k - 1}\right) du

Step 4: Simplify and Recognize the Gamma Function

The `k` terms cancel out. We can combine the `u` terms and pull $\lambda$ out of the integral.

E[X] = \lambda \int_{0}^{\infty} u \cdot u^{1/k - 1} e^{-u} du = \lambda \int_{0}^{\infty} u^{1/k} e^{-u} du

The integral $\int_{0}^{\infty} t^{z-1}e^{-t}dt$ is the definition of the Gamma function $\Gamma(z)$ . In our case, $z-1 = 1/k$ , so $z = 1 + 1/k$ .

This gives us the final result:

Final Mean Formula

E[X] = \lambda \Gamma(1 + 1/k)