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Consider the Weibull sampling model for $X_1,\ldots,X_n$ iid, where $$p(x|\lambda,k)=k\lambda^kx^{k-1}e^{-\lambda^kx^k}$$ for $x>0$. Assume $k$ is known and $\lambda$ is unknown. First, if I adopt a $\text{Gamma}(a,b)$ on $\lambda$, this is not a conjuage prior for the Weibull sampling model, since the leading posterior has the term $\lambda^k$ on the exponential, so the posterior is not a Gamma distribution. I'm thinking of if I reparametrize $\theta=\lambda^k$, whether I can find a conjugate model for $\theta$. Then the sampling model will be $p(x|\theta,k)=k\theta x^{k-1}e^{-\theta x^k}$. But I don't know how to proceed. Something that I have in mind currently is that the reparametrized sampling model can be written as an exponential family $h(x)c(\theta)e^{\theta t(x)}$, with $h(x)=kx^{k-1}$, $c(\theta)=\theta$ and $t(x)=-x^k$. I'd like to know how I can derive the conjugate prior for $\theta$. Thank you.

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It is more usual to take a conjugate prior for $\theta= \lambda^{-k}$. So the Weibull likelihood becomes $$\frac{k}\theta x^{k-1}e^{-x^k/\theta}$$ and the prior for $\theta$ is an inverse gamma distribution with density $$\frac{\beta^\alpha}{\Gamma(\alpha)}\frac{1}{\theta^{\alpha+1}} e^{-\beta/\theta}$$ which with observations $x_1,x_2, \ldots, x_n$ gives a posterior density proportional (after dropping multiplicative terms not involving $\theta$) to

$$\frac{1}{\theta^{\alpha+1}} e^{-\beta/\theta} \times \frac1{\theta^n}e^{-\sum x_i^k /\theta} = \frac{1}{\theta^{\alpha+n+1}} e^{-(\beta+\sum x_i^k)/\theta} $$

i.e. with $\alpha$ becoming $\alpha+n$ and with $\beta$ becoming $\beta+\sum\limits_1^n x_i^k$. You can translate this back to $\lambda$ if you must.

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  • $\begingroup$ Hi @Henry, I'd like to know how to derive it. Can you include more details? Does your $\alpha$ and $\beta$ correspond to $a$ and $b$ in my question? $\endgroup$ Commented Oct 30, 2022 at 18:40
  • $\begingroup$ @PseudodifferentialOperators You do not have an $a$ or $b$ in your question. I had some typos in my answer and have expanded it; note your Weibell distribution has a reciprocal $\lambda$ to Wikipedia's - I have stuck with yours $\endgroup$
    – Henry
    Commented Oct 31, 2022 at 2:32
  • $\begingroup$ I don't take the reciprocal of $\lambda$ for each of computation $\endgroup$ Commented Oct 31, 2022 at 3:04

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