Burr Distribution Derivation from Conditional Inverse Weibull and Generalized Gamma Distributions

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.

It was a mistake on your Professor's part, the $\lambda$ parameter of the Inverse Weibull should be $\tau$ which would link the Inverse Weibull and the Generalized Gamma which should have $e^{-(\frac{\beta}{\theta})}$. If you have the course textbook 'Loss Models from Data to Decisions' you can find this problem on page 65.  But if you don't, once you make the changes above you will be left with a function of $w,\beta,\tau,\alpha,\text{ and }\theta$; you would then integrate out the $\beta$ using U Substitution leaving you with a function of $w,\tau,\alpha,\text{ and }\theta$ which is the Inverse Burr Distribution PDF $f(x| \tau,\alpha,\beta)$.
• I noticed that too! They're apparently using the reciprocal of one of the parameters $\theta$. There're just so many things wrong with this problem. Feb 6 '18 at 3:34