# Conjugate Prior for Alpha Power Inverse Weibull Distribution

Let $$X$$ has Alpha Power Inverse Weibull (APIW) distribution with pdf

$$f(x) = \frac{\log \alpha}{\alpha - 1} \lambda \beta x^{-(\beta+1)} e^{-\lambda x^{-\beta}} \alpha^{e^{-\lambda x^{-\beta}}}, \; x>0, \alpha >0, \alpha \neq 1, \lambda > 0, \beta > 0$$.

In this case, I want to estimate parameter $$\alpha$$ (by using Bayesian method) given that parameters $$\lambda$$ and $$\beta$$ are both known.

The likelihood function is (assume that $$\lambda$$ and $$\beta$$ are known)

$$L(x_1, ..., x_n) = \frac{(\log \alpha)^n}{(\alpha - 1)^n} \lambda^n \beta^n (\prod x_i^{-(\beta+1)}) e^{-\lambda \sum x_i^{-\beta}} \alpha^{\sum e^{-\lambda x_i^{-\beta}}}$$

My question is: how do we find the conjugate prior for $$\alpha$$? I try for Gamma distribution, but I think it is not. I've shown that Alpha Power Inverse Weibull distribution is a exponential family distribution (where $$\lambda$$ and $$\beta$$ are known), so it must have conjugate prior.

• You can to find a prior for $a$ that is of the form $p(a)\propto \frac{log \ a}{a-1}a^{C}$ where $C$ is constant. But I've never came across of such a distribution Sep 18, 2021 at 14:02

You don't need to have a "named" functional form for your conjugate prior, so @Fodor1234's comment points the way towards an answer. However, the conjugate prior $$p$$ is slightly different than in the comment:
$$p(\alpha|k, C) \propto \left( {\ln \alpha \over \alpha-1}\right)^k \alpha^C$$
The posterior $$p'$$ for $$\alpha$$, assuming (as in the question) that $$\lambda$$ and $$\beta$$ are known, is:
$$p'(\alpha|k, C, x, n) \propto \left( {\ln \alpha \over \alpha-1}\right)^{k+n} \alpha^{C'}$$ $$C' = C + \sum e^{-\lambda x_i^{-\beta}}$$
which clearly has the same functional form as the prior. The prior parameter $$k$$ can be interpreted as the "equivalent sample size" of the prior information, as it adds to $$n$$ when forming the posterior.