This is a review-type homework question that I'm very much stuck on.

Basically, these are the assumptions:

  • $\theta|\lambda = N(\mu,\frac{\sigma^2}{\lambda})$
  • $\lambda = \rm{Gamma}(\nu/2,\nu/2)$
  • $\theta=t_{\nu}(\mu,\sigma^2)$

Giving us:

  • $p(\theta|\lambda) = \frac{\sqrt{\lambda}}{\sqrt{2\pi}\sigma} \rm{exp}\{-\frac{\lambda(\theta - \mu)^2}{2\sigma^3}\} $
  • $p(\lambda) = \frac{1}{\Gamma(\nu/2)} \lambda^{\frac{\nu}{2} - 1} \rm{exp}\{-\frac{\nu}{2}\lambda\}(\nu/2)^{\nu/2}$

Then comes the part I can't seem to prove: obtaining the marginal distribution of $\theta$. Here's what I get so far:

  • $p(\theta)=\int_0^\infty p(\theta|\lambda)p(\lambda) \hspace1ex d\lambda$
  • $p(\theta)=\int_0^\infty p(\theta,\lambda) \hspace1ex d\lambda$
  • $p(\theta,\lambda)=p(\theta|\lambda)p(\lambda)$
  • $p(\theta|\lambda)p(\lambda)=\frac{\sqrt{\lambda}}{\sqrt{2\pi}\sigma} \rm{exp}\{-\frac{\lambda(\theta - \mu)^2}{2\sigma^3}\} \frac{1}{\Gamma(\nu/2)} \lambda^{\frac{\nu}{2} - 1} \rm{exp}\{-\frac{\nu}{2}\lambda\}(\nu/2)^{\nu/2}$
  • $p(\theta)=\frac{1}{\Gamma(\nu/2)}(\nu/2)^{\nu/2} \int_0^\infty \frac{\sqrt{\lambda}}{\sqrt{2\pi}\sigma} \lambda^{\frac{\nu}{2} - 1} \rm{exp}\{-\frac{\nu}{2}\lambda\ -\frac{\lambda(\theta - \mu)^2}{2\sigma^3}\} \hspace1ex d\lambda$

And now I'm stuck... The notes say I should be able to obtain the following "after some algebra" but I'm totally clueless:

  • $p(\theta) = \frac{\Gamma(\frac{\nu+1}{2})}{\Gamma(\nu/2)} \frac{1}{(\pi\nu\sigma^2)^{1/2}} \frac{1}{[1+\frac{1}{\nu}(\frac{\sigma-\mu}{\sigma})^2]^\frac{\nu+1}{2}}$

Any help/clues in the right direction would be much appreciated!

  • $\begingroup$ Your notes, as quoted in the last line, make no sense because the right hand side does not depend on $\theta$ at all. $\endgroup$ – whuber Sep 3 '18 at 20:22
  • $\begingroup$ @whuber sorry to clarify, but do you mean that the "after some algebra" bit is incorrect? As in the notes are wrong? Or as in, I messed up in a previous line? (Also thank you for your response!) (Also also, just curious, why was this marked as duplicate?) EDIT: Never mind the link wasn't coming up, sorry! $\endgroup$ – user3684314 Sep 4 '18 at 0:01
  • $\begingroup$ No density function is constant everywhere. The right hand side of your expression for $p(\theta)$ does not depend on $\theta$ and therefore cannot be correct. There is a typographical error. It's unclear whether that is the cause of your difficulty or is just an error in posting your question. $\endgroup$ – whuber Sep 4 '18 at 13:37