The Weibull Distribution

Posted by Beetle B. on Tue 06 June 2017

Probability Density Function

Let \(\alpha,\beta>0\)

\begin{equation*} f(x;\alpha,\beta)=\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1}e^{-\left(x/\beta\right)^{\alpha}} \end{equation*}

for \(x\ge0\)

When \(\alpha=1\), you get the exponential distribution.

In the Weibull distribution, \(\beta\) is a scale parameter, but \(\alpha\) is not a location parameter. It is called a shape parameter. Similarly for \(\alpha,\beta\) of the Gamma distribution.

Mean and Variance

\begin{equation*} \mu=\beta\Gamma\left(1+\frac{1}{\alpha}\right) \end{equation*}
\begin{equation*} \sigma^{2}=\beta^{2}\left[\Gamma\left(1+\frac{2}{\alpha}\right)-\left[\Gamma\left(1+\frac{1}{\alpha}\right)\right]^{2}\right] \end{equation*}

Usage

This is often used just for curve fitting purposes.

If the quantity X is a “time-to-failure”, the Weibull distribution gives a distribution for which the failure rate is proportional to a power of time. The shape parameter, \(\alpha\), is that power plus one, and so this parameter can be interpreted directly as follows:

  • A value of \(\alpha<1\) indicates that the failure rate decreases over time. This happens if there is significant “infant mortality”, or defective items failing early and the failure rate decreasing over time as the defective items are weeded out of the population.
  • A value of \(\alpha=1\) indicates that the failure rate is constant over time. This might suggest random external events are causing mortality, or failure.
  • A value of \(\alpha>1\) indicates that the failure rate increases with time. This happens if there is an “aging” process, or parts that are more likely to fail as time goes on.