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This isn't exactly a homework problem but rather a self-selected problem I'm doing to prepare for a midterm.

I can see from Wikipedia that it is an inverse gamma but I am unable to reach the posterior suggested by the parameter updates.

The Weibull as given on Wikipedia is:

$f(x_1,..,x_n|\lambda, k)=\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}\text{exp}\left\{-\left(\frac{x}{\lambda}\right)^k\right\}$

Then

$L(\lambda|x_1,\dots, x_n)\propto \prod_x \lambda^{-k}\text{exp}\left\{-\frac{x^k}{\lambda^{k}}\right\}\\ =\lambda^{-nk}\text{exp}\left\{-\frac{\sum_x {x^k}}{\lambda^{k}}\right\}$

The inverse gamma prior is

$P(\lambda|\alpha,\beta)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{-\alpha-1}\text{exp}\left\{-\frac{\beta}{\lambda}\right\}$

The posterior then is

$P(\lambda|x_1,\dots,x_n)\propto L(\lambda|x_1,\dots, x_n) P(\lambda|\alpha,\beta)\\ =\lambda^{-nk}\text{exp}\left\{-\sum_x {x^k}\lambda^{-k}\right\}\lambda^{-\alpha-1}\text{exp}\left\{-\frac{\beta}{\lambda}\right\}\\ =\lambda^{-nk-\alpha-1}\text{exp}\left\{-\frac{\sum_x {x^k}}{\lambda^{k}}-\frac{\beta}{\lambda}\right\}$

The problem that I am left with is to coerce an update for the $\beta$ hyperparameter that does not contain $\lambda$. The $\alpha$ also looks different from Wikipedia but I am going with explaining that as the difference between the posterior with observations and without, though honestly I'm confused about that as well. Wikipedia simply has the summation, $\sum_x {x^k}$ as the update for $\beta$ and $n$ for the update of $\alpha$ (not $nk$).

All of my efforts so far to coerce that $\beta$ update have failed.

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    $\begingroup$ The inverse gamma conjugate prior is given for $\lambda^k$, not for $\lambda$. This is, you have to use a different parameterisation: $f(x\vert \lambda, k)=\dfrac{k}{\lambda}x^{k-1}\exp\left(-\dfrac{x^k}{\lambda}\right)$. Please, have a look at page 20 of this document by John D. Cook. $\endgroup$
    – user10525
    Oct 2, 2012 at 15:30
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    $\begingroup$ Thanks that explains everything! Midterm went well incidentally. $\endgroup$ Oct 4, 2012 at 2:39
  • $\begingroup$ Maybe is too late, but I have a cuestion about pdf's "A compendium..." by John D. Cook, there says that priori distribution is a normal distribution, but I don't think so... could you please tell me if posterior distribution is a normal distribution or a Inverse gamma distribution? Please! :) And thanks! $\endgroup$
    – user34825
    Nov 14, 2013 at 18:06

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