Weibull Distribution

Study guide: Exam MAS-I

The Weibull Distribution $X \sim \text{Weibull}(\theta, \tau)$ is a continuous distribution on $(0, \infty)$ with scale $\theta > 0$ and shape $\tau > 0$ . It generalises the exponential ( $\tau = 1$ ) and is used for claim severities and for lifetimes whose hazard rate rises or falls with age.

$f(x) = \frac{\tau\,(x/\theta)^{\tau}\,e^{-(x/\theta)^{\tau}}}{x}, \quad x > 0$

$F(x) = 1 - e^{-(x/\theta)^{\tau}}$

$E[X^k] = \theta^k\,\Gamma\!\left(1 + \frac{k}{\tau}\right), \quad k > -\tau$

$\text{Var}(X) = \theta^2\left[\Gamma\!\left(1 + \tfrac{2}{\tau}\right) - \Gamma\!\left(1 + \tfrac{1}{\tau}\right)^{2}\right]$

$h(x) = \frac{\tau\,x^{\tau - 1}}{\theta^{\tau}}$

$\begin{aligned} E[X \wedge x] &= \theta\,\Gamma\!\left(1 + \tfrac{1}{\tau}\right)\Gamma\!\left[1 + \tfrac{1}{\tau};\ \left(\tfrac{x}{\theta}\right)^{\tau}\right] \\ &\quad + x\,e^{-(x/\theta)^{\tau}} \end{aligned}$

This is the parameterization of the CAS MAS-I tables. The mean is $E[X] = \theta\,\Gamma(1 + 1/\tau)$ ; the mode is $\theta\left(\frac{\tau - 1}{\tau}\right)^{1/\tau}$ for $\tau > 1$ , else 0.
Hazard rate: decreasing for $\tau < 1$ , constant $1/\theta$ for $\tau = 1$ (the exponential with mean $\theta$ ), increasing for $\tau > 1$ . This flexibility is why it is a standard lifetime model in reliability. The survival function is $S(x) = e^{-(x/\theta)^{\tau}}$ , so the cumulative hazard is $(x/\theta)^{\tau}$ .
Link to the exponential: $(X/\theta)^{\tau}$ has the exponential distribution with mean 1. Inverting $F$ gives the inversion formula $x = \theta\left[-\ln(1 - u)\right]^{1/\tau}$ .
Limited Expected Value (last block, as the tables give it): $\Gamma(\alpha; y) = \frac{1}{\Gamma(\alpha)}\int_0^y t^{\alpha - 1}e^{-t}\,dt$ is the incomplete gamma function — the Gamma CDF with shape $\alpha$ and scale 1.
Minimum of a sample: if $X_1, \ldots, X_n$ are independent $\text{Weibull}(\theta, \tau)$ , then $P(\min > x) = e^{-n(x/\theta)^{\tau}}$ , so the minimum is $\text{Weibull}(\theta n^{-1/\tau}, \tau)$ — the lifetime of a series system of identical components.
On MAS-I it appears as a claim-severity model (Tse §2.2.3), in hazard-rate and survival calculations, in order statistics and in simulation by inversion.

A Heavy-Tailed Weibull Severity Liability losses follow a Weibull distribution with $\theta = 2{,}000$ and $\tau = 0.5$ .

(a) Find the probability that a loss exceeds 5,000.

(b) Find the mean and standard deviation of a loss.

(c) Find the hazard rate at 500 and at 5,000, and say what they show.
Answer (a) $(5{,}000/2{,}000)^{0.5} = 1.5811$ , so
 $\begin{align*} P(X > 5{,}000) &= e^{-1.5811} \\ &= 0.2057 \end{align*}$ 
(b) With $1/\tau = 2$ , $\Gamma(3) = 2$ and $\Gamma(5) = 24$ :
 $\begin{align*} E[X] &= 2{,}000\,\Gamma(3) \\ &= 4{,}000 \\ \text{Var}(X) &= 2{,}000^2\left[\Gamma(5) - \Gamma(3)^2\right] \\ &= 4{,}000{,}000\,(24 - 4) \\ &= 80{,}000{,}000 \end{align*}$ 
The standard deviation is $\sqrt{80{,}000{,}000} \approx 8{,}944$ , more than twice the mean.

(c) $h(x) = 0.5\,x^{-0.5}/2{,}000^{0.5}$ :
 $\begin{align*} h(500) &= \frac{0.5}{\sqrt{500}\sqrt{2{,}000}} \\ &= 0.000500 \\ h(5{,}000) &= \frac{0.5}{\sqrt{5{,}000}\sqrt{2{,}000}} \\ &= 0.000158 \end{align*}$ 
With $\tau < 1$ the hazard rate falls as the loss grows: the chance that a loss which has reached $x$ ends within the next unit is lower at 5,000 than at 500 — the mark of a heavy tail.

Lifetime of a Series System A sprinkler controller fails if any of its 4 independent sensors fails. Each sensor's lifetime in years is Weibull with $\theta = 10$ and $\tau = 2$ . Find the probability that the controller is still working after 3 years, and its expected lifetime.
Answer The controller's lifetime $T$ is the minimum of the four sensor lifetimes:
 $\begin{align*} P(T > t) &= \left[e^{-(t/10)^2}\right]^4 \\ &= e^{-4(t/10)^2} \\ &= e^{-(t/5)^2} \end{align*}$ 
So $T$ is Weibull with $\theta = 5$ , $\tau = 2$ . Then $P(T > 3) = e^{-0.36} = 0.6977$ , and
 $\begin{align*} E[T] &= 5\,\Gamma(1.5) \\ &= 5(0.8862) \\ &= 4.43 \end{align*}$ 
A single sensor survives 3 years with probability $e^{-0.09} = 0.914$ and lasts $8.86$ years on average; four in series halve the expected lifetime, since $4^{-1/2} = \frac{1}{2}$ .