Intuition behind Weibull distribution?

Question

I don't understand the physical meaning of Weibull distribution's $k$ parameter. Here is a simplified formula of cumulative probability function of Weibull in the simplest form:

$$p(\xi \geq x) = e^{-(\frac{x}{\lambda})^k}$$

What is the physical interpretation of $k$ parameter, what kind of process would exponentiate $x$? $x$ can be interpreted as a stoppage time or a force applied to a chain/fiber, why would it be exponentiated by $k$?

For comparison: I can more or less rationalise the Gumbel distribution:

Suppose we're running $N$ queues of coin tosses of length $n$. Gumbel distribution gives the probability that the longest chain of successes, observed in any queue, is greater than x, because:

$p(\eta \geq x) = 1 - p(\eta < x) = 1 - (1-\frac{ae^{-bx}}{N})^N = e^{-ae^{-bx}}$

Here I used the fact that probability that there were no sequences of successes longer than $x$ in a Markov chain of length $n$ is distributed as $ae^{-bx}$. For example, see this application to bioinformatics.

From this derivation the interpretation of a and b parameters is pretty intuitive (please, forgive me/correct me if this derivation is not perfectly accurate, but at least it conveys some meaning).

Can you provide any rationalisation, an intuitive model that leads to a Weibull distribution, given it is associated with durability/survival?

Answer 1

Since the Weibull distribution is often used in connection with reliability or survival, the hazard rate function is crucial, see Non-monotone hazard functions. Below is a plot of the Weibull hazard rate, for scale 1 and some assorted values of the shape $k$, note that $k=1$ is the exponential distribution:

So this gives one intuition: The Weibull hazard rate is monotone, decreasing for $k<1$ and increasing for $k>1$.

See Wikipedia links to many applications ... the paper by Waloddi Weibull, which gave the distribution its name, can be found here, and is actually quite accessible. He says

The objection has been stated that this distribution function has no theoretical basis. But insofar as the author understands, there are - with very few exceptions - the same against all other df, applied to real populations from natural biological fields, at least insofar as the theoretical has anything to do with the population in question. Furthermore, it is utterly hopeless to expect a theoretical basis for distribution functions of random variables such as strength of materials or of machine parts or particle sizes, the "particles" being fly ash, Cyrtoideae, or even adult males, born in the British Isles

Nevertheless, in the paper he does give a justification,

Assume that we have a chain consisting of several links. If we have found, by testing, the probability of failure $P$ applied to a "single" link, and if we want to find the probability of failure $P_n$ of a chain consisting of $n$ links, we have to base our deductions upon the proposition that the chain as a whole has failed, if anyone of its parts has failed.

If you then start with an exponential distribution for a single link, you will arrive at the Weibull for $n$ links. What is more, if the distribution for a single link is Weibull, the distribution for the chain will also be Weibull. As pointed out in comments by @Scortchi - Reinstate Monica, ultimately this thinking will lead you to the Fisher–Tippett–Gnedenko theorem.

For the record, R code for the plot:

hweibull <- function(x, shape, scale=1) {
    dweibull(x, shape, scale) / pweibull(x, shape, scale, 
                                         lower.tail=FALSE) }

k <- seq(from=0.6, to=1.5, by=0.2)
mypalette <- RColorBrewer::brewer.pal(length(k), "Oranges") 

for (t in seq_along(k)) {
    plot(function(x) hweibull(x, k[t]), from=0,
         to=10, col=mypalette[t], add=if(t==1)FALSE else TRUE,
         main="Weibull hazard", xlab="x", ylab="", lwd=2)
}
legend("topright", paste("k=", round(k, 2)), col=mypalette,
       text.col=mypalette)

Answer 2

As the OP uses the Gumbel distribution (maximum extreme value distribution) as an example having an intuitive explanation, it's worth adding to Kjetil's answer (+1) by pointing out the association of that distribution with the Weibull.

Say that $W$ represents the standard minimum extreme value form of the distribution (replacing $x$ with $-x$, and setting $a=1,b=1$ in the terminology of this question). If survival times $T$ have the following distribution:

$$\log T = \alpha + \sigma W $$

then $T$ follows a Weibull distribution with $\alpha = -log \lambda$ and $k = 1/\sigma$.

Then $W$ represents the random contribution of a standard minimum extreme value distribution to the distribution of $\log T$ values, and one could then interpret $k$ as the "tightness" of that distribution.