From Wikipedia, the Weibull distribution is defined with the exceedance distribution function (aka survival function) $\exp[-(x/\lambda)^k]$.

If I transform the random variable $x$ using $x'=ax^b$ then $x'$ is also Weibull, with new parameters $\lambda'=a \lambda^b$ and $k'=k/b$.

And it seems that any Weibull can be converted to any other Weibull in this way.

From which I conclude that the Weibull has a transitive transformation group, and that the theory related to that applies.

So I’d like to work out the right Haar prior…does anyone know what it is / how to do that?

I'm gradually trying to get to the point where I understand enough of the relevant theory to work it out myself, but haven't quite got there yet.


ps: the only related literature I've come across is a paper by Dongchu Sun from 1997, which gives the reference prior as $\frac{1}{\lambda k}$. I suspect that might be the RHP too, but I'd like to really know.

  • $\begingroup$ what is "harr"? If you are asking about conjugate prior it is going to be Gamma distribution for Weibull. $\endgroup$
    – forecaster
    Jun 26, 2023 at 14:49
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    $\begingroup$ Alfred Haar is the name of a Hungarian mathematician. The right haar prior (as opposed to the left haar prior), if it exists, is a prior that gives predictive probability matching in repeated tests i.e. a probability of 10% means something occurs 10% of the time. I would guess that the conjugate prior doesn't have that property. $\endgroup$ Jun 26, 2023 at 14:54
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    $\begingroup$ @forecaster here's the wiki page on the Haar measure en.wikipedia.org/wiki/Haar_measure $\endgroup$ Jun 26, 2023 at 15:20
  • $\begingroup$ The Haar measure is related with the nature of the parameter, not the distribution family per se. Here $\lambda$ is a scale parameter, so $1/\lambda$ is the solution. $\endgroup$
    – Xi'an
    Jun 27, 2023 at 6:20
  • $\begingroup$ @ Xi'an: hmm. That's not the result I get. If $k$ were constant I would agree, but $k$ is also a parameter. Note that if $\lambda$ were constant and $k$ were the only parameter then the result would be $1/k$. The result I get also depends on $k$. I will check it and post the derivation. $\endgroup$ Jun 27, 2023 at 9:07

1 Answer 1


The stack-exchange effect: you think about something for 3 days, fail to answer it, post a question on stack-exchange, and then come up with an answer yourself 20 minutes later.

A 'simple' way to an answer seems to be:

a) you don't need to think about left and right transformations after all (phew)

b) instead, if you transform the Weibull random variable using $x'=-\log (x)$, and transform the parameters too in the right way, then you get a Gumbel distribution.

c) then you can work out the RHP for the Weibull by transforming the RHP for the Gumbel using the Jacobian.

I will check my working 100x before I actually post the answer, since right now I'm getting something slightly different from the reference prior.


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