# Conjugate prior for the Weibull distribution

On Wikipedia we find a nice overview on conjugate prior distributions. I am interested in the conjugate prior for a random variable $$X$$ with density

$$f(x;\lambda,k) = \begin{cases} \frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}e^{-(x/\lambda)^{k}} & x \geq 0 ,\\ 0 & x<0, \end{cases}$$ the Weibull. With known rate parameter $$k$$ the inverse Gamma distribution with density $$g(\lambda; \alpha, \beta) = \frac{\beta^\alpha}{\Gamma(\alpha)} (1/\lambda)^{\alpha + 1}\exp\left(-\beta/\lambda\right)$$ is a conjugate prior for $$\lambda$$. The posterior distribution of $$\lambda$$ then is apparently $$g$$ with $$\alpha_*=\alpha+n$$ and $$\beta*=\beta+\sum_i x_i^k$$ (with $$i=1,\ldots,n$$).

I cannot seem to be able to show this. Is this result true? And is there a conjugate prior for $$k$$ as well?

• There seems to be some confusion with the notations that led to mathematically erroneous statements. See this post on the statistics site, in particular, the comment by user10525. For the record, the relevant source cited in Wikipedia is page 20 of Compendium Of Conjugate Priors.pdf by Daniel Fink. – Lee David Chung Lin Feb 4 '20 at 4:44
• In particular, let me quote: "The inverse gamma conjugate prior is given for $\lambda^k$, not for $\lambda$". This is a common mishap, since many people would prefer to denote the parameter $\lambda^k$ as its own thing. Namely, $\theta = \lambda^k$ with $\theta$ being the parameter when the shape parameter is considered a constant. You can double check the calculation see in the question statement part of the post linked above. – Lee David Chung Lin Feb 4 '20 at 4:56
• If I'm not mistaken, hope this helps you with the math, as well as saving you some bounty rep. – Lee David Chung Lin Feb 4 '20 at 4:58

I suggest you to use an advise to take $$\theta =\lambda^k$$ (like Lee David Chung Lin said)

but if you assist to use your notation I introduce you one conjugate prior.(you should check identifiability of your model).

$$f(x_1,\cdots,x_n|\lambda)\propto \frac{1}{\lambda^{nk}} e^{-\frac{1}{\lambda^k}\sum x_{i}^{k}}$$

if we choose a prior $$g(\lambda)$$ then

$$g(\lambda|x_1,\cdots ,x_n)\propto \frac{1}{\lambda^k} e^{-\frac{1}{\lambda^k}\sum x_{i}^{k}} g(\lambda)$$

so conjugate prior should be something like

$$g(\lambda)\propto \frac{1}{\lambda^a} e^{-\frac{b}{\lambda^k}} \hspace{.5cm} \lambda>0$$

I found :

$$\int_{0}^{\infty} y^m e^{-b y^k} dy=\frac{\Gamma(\frac{m+1}{k})}{kb^{\frac{m+1}{k}}}$$ (reference: Table of Integrals, Series, and Products, I.S. Gradshteyn and I.M. Ryzhik, page 337)

by choosing $$\lambda=1/y$$ in this integral

$$\int_{0}^{\infty} \frac{1}{\lambda^{m+2}} e^{-b \frac{1}{\lambda^{k}}} d\lambda=\frac{\Gamma(\frac{m+1}{k})}{cb^{\frac{m+1}{k}}}$$ so you can create a distribution

$$g(\lambda)=NEWG(m,b)= \frac{\frac{1}{\lambda^{m+2}} e^{-b \frac{1}{\lambda^{k}}} }{\frac{\Gamma(\frac{m+1}{k})}{kb^{\frac{m+1}{k}}}} \hspace{.5cm} \lambda>0$$

so posterior will be

$$g(\lambda|x_1,\cdots ,x_n) \propto \frac{1}{\lambda^{m+nk+2}} e^{-(b+\sum x_{i}^{k}) \frac{1}{\lambda^{k}}}$$ that is $$NEWG(m+nk,b+\sum x_{i}^{k})$$

another option is reparmetrize $$m_2=mk$$.

If you use standard notation $$\theta$$

$$f(x|\theta)=\frac{kx^{k-1}}{\theta} e^{-\frac{x^k}{\theta}}$$ using inverse gamma for prior $$g(\theta)\propto \frac{1}{\theta^{\alpha+1}} e^{-\frac{\beta}{\theta}}$$

posterior:

$$g(\theta|x_1,\cdots,x_n)\propto f(x_1,\cdots,x_n|\theta) g(\theta) =\frac{k^n (\prod_{i=1}^{n} x_i)^{k-1}}{\theta^n} e^{-\frac{\sum_{i=1}^{n} x_i^k}{\theta}} \frac{1}{\theta^{\alpha+1}} e^{-\frac{\beta}{\theta}}$$

$$\propto \frac{1}{\theta^{n+\alpha+1}} e^{-\frac{\beta+\sum_{i=1}^{n} x_i^k}{\theta}}$$

that is Inverse gamma with $$(n+\alpha,\beta+\sum_{i=1}^{n} x_i^k)$$

• Thanks for your answer. However I'm asking why the inverse gamma is a conjugate prior for the reparameterization. Can you show this? – tomka Feb 13 '20 at 8:45
• for your density(with that notation), inverse gamma is not conjugate! since the posterior is not inverse gamma – Masoud Feb 13 '20 at 11:31
• yes of course. but for your reparametrization it is apparently - you could show this. – tomka Feb 13 '20 at 11:39
• I dont understand. – Masoud Feb 13 '20 at 12:00
• If I use $\theta = \lambda^k$ (Lee David Chung Lin) why is the inverse gamma then a valid prior? In the reference given (compendium of priors) this is not proven. – tomka Feb 13 '20 at 12:02