3
$\begingroup$

I've recently fallen in love with the Weibull Distribution and have gotten a reason to see if there's a mapping of this distribution to an interval (0,1). After plotting the Beta Distribution against the Weibull it feels like there's a connection:

    n = 100000

    weibull<-function(alpha,beta,n){
                W <- alpha*(-log(runif(n)))^(1/beta)
                return(W)
            }

    par(mfrow=c(2,4),cex=0.4)
    hist(weibull(1,1,n),breaks=100,xlab="W",axes=FALSE) 
    hist(weibull(1,2,n),breaks=100,xlab="W",axes=FALSE)
    hist(weibull(1,3,n),breaks=100,xlab="W",axes=FALSE)
    hist(weibull(1,8,n),breaks=100,xlab="W",axes=FALSE)

    hist(rbeta(n,1,10),breaks=100,xlab="B",axes=FALSE)
    hist(rbeta(n,2,10),breaks=100,xlab="B",axes=FALSE)
    hist(rbeta(n,4,10),breaks=100,xlab="B",axes=FALSE)
    hist(rbeta(n,20,10),breaks=100,xlab="B",axes=FALSE)

Weibull Vs Beta

This is relevant because I'm thinking about waiting times for recurrent events. If we have a timeline of length L where we mark observed events over it, the length of each induced subinterval (waiting time) can empirically be nicely fit with a Weibull-distribution. Here the last observation is censored if we don't stop before an event but this a sidenote.

So is there a relationship with the beta distribution? Or is there some other nice normalization of the Weibull-distribution?

Remember, CDF and PDFs of the Weibull are

$$ F(x) = Pr(T\leq x) = 1-\exp\left[ - {\left(\frac{x}{\alpha} \right)}^\beta \right] $$ $$ f(x) =\frac{\partial}{\partial x} F(x) =\frac{\beta}{\alpha}{\left(\frac{x}{\alpha} \right)}^{\beta-1} \exp\left[ - {\left(\frac{x}{\alpha} \right)}^\beta \right] $$

$\endgroup$
  • 2
    $\begingroup$ Since $F(X)\sim\mathcal{U}(0,1)$ there is always a natural mapping on $(0,1)$. $\endgroup$ – Xi'an Apr 28 '16 at 13:19
  • 1
    $\begingroup$ Good point. I assume that $F(X|\theta)$ is uniform whenever $X$ is drawn from the same distribution with parameter $\theta$. I wonder when this is used for estimation $\endgroup$ – ragulpr Apr 28 '16 at 15:37
  • 2
    $\begingroup$ Other than this generic transform, I do not see much hope in that the Weibull is a power of the exponential distribution, while Beta distributions are connected to ratios of Gamma distributions. The histograms you plot are based on different parameter values and are drawn on different scales. $\endgroup$ – Xi'an Apr 28 '16 at 17:10
  • 2
    $\begingroup$ Your plots show that they can have quite similar looking shapes; that doesn't imply there's a connection. Note that one is on a bounded interval and the other is on the half-line. $\endgroup$ – Glen_b -Reinstate Monica Apr 29 '16 at 6:18
  • 1
    $\begingroup$ Yes, the hiding of scales was to make a point. As Xi'an pointed out ratios of if $X,Y$ are gamma then $X/(X+Y)$ is beta under certain conditions. What's made clear in another question is the difference between Gamma and Weibull. By inspection: $$ f_{\Gamma}(x) \propto x^{k-1} \exp\left(-\frac{x}{\theta}\right) \\ f_{W}(x) \propto x^{k-1} \exp\left(-\left(\frac{x}{\lambda}\right)^k\right)$$ Seems very different but this doesn't necessarily mean there's no transform or connection. $\endgroup$ – ragulpr Apr 29 '16 at 8:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.