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February 4th, 2018, 03:47 PM   #1
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Exclamation Burr Distribution Derivation from Inverse Weibull

A Weibull distribution, with shape parameter alpha and

Am I supposed to find the MGFs of both distributions and then use the iterated rule/smoothing technique/law of total expectation followed by uniqueness theorem to find the PDF of the Burr distribution? Or am I supposed to use the definition of conditional distribution to find the joint distribution and then integrate over the support of $\beta$ in order to get the PDF? I tried the latter, but of course I ended up with an integral that seems impossible to integrate. How would you go about solving this? Here's what I have so far:

$$f_{W^{-1}}(w)=\frac{\lambda w^{-\lambda-1}e^{-(\frac{1}{w\beta})^\lambda}}{\beta^\lambda}$$ on $0<w<\infty$.

This implies that $$f_x(x)=\frac{\lambda \tau x^{-\lambda -1}}{\Gamma(\alpha)}\int_{0}^{\infty}\frac{(\frac{ \beta }{\theta})^\alpha e^{-(\frac{1}{x\beta})^\alpha}e^{(\frac{\beta}{\theta} )^\tau}}{\beta^{\lambda+1}}d\beta$$
But how on Earth do I integrate that? Surely there's a better way, but I have on idea what it is.
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