# 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.
• 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? Commented Oct 30, 2022 at 18:40
• @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 Commented Oct 31, 2022 at 2:32
• I don't take the reciprocal of $\lambda$ for each of computation Commented Oct 31, 2022 at 3:04