Scale mixture of normals; t-distribution; proving marginal distribution equality

Question

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!