Find a conjugate prior for the Weibull distribution under reparametrization 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.

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.
