# What is the right Haar prior for the Weibull distribution?

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.

thanks

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.

• what is "harr"? If you are asking about conjugate prior it is going to be Gamma distribution for Weibull. Jun 26, 2023 at 14:49
• 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. Jun 26, 2023 at 14:54
• @forecaster here's the wiki page on the Haar measure en.wikipedia.org/wiki/Haar_measure Jun 26, 2023 at 15:20
• 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. Jun 27, 2023 at 6:20
• @ 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. Jun 27, 2023 at 9:07

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.