How to get the marginal pdf of $p(y)$? Do you just integrate out $p({\sigma}^{2})$?
Say, the following joint distribution for $y \in {{R}^{d}}$ and ${{\sigma }^{2}}\in {{R}^{d}}$
IG: means inverse Gamma
$${{\sigma }^{2}}\sim IG(\alpha ,\beta )\propto {{({{\sigma }^{2}})}^{-(\alpha +1)}}{{e}^{-\beta /{{\sigma }^{2}}}}$$
$$y|{{\sigma }^{2}}\sim N(\mu ,{{\sigma }^{2}}\Sigma )$$
where $a\in R$, $b\in R$,$\mu \in {{R}^{d}}$,$\Sigma \in {{R}^{d\times d}}$ are known parameters.
I know that
$$p(y|\mu,\Sigma)\propto \frac{1}{{{\left| {{\sigma }^{2}}\Sigma \right|}^{\frac{n}{2}}}}\exp \left[ -\frac{1}{2}\sum\limits_{i=1}^{n}{{{\left( {{y}_{i}}-\mu \right)}^{T}}{{({{\sigma }^{2}}\Sigma )}^{-1}}({{y}_{i}}-\mu )} \right]$$
$$\propto \frac{1}{{{\left| {{\sigma }^{2}}\Sigma \right|}^{\frac{n}{2}}}}\exp \left[ -\frac{1}{2}\sum\limits_{i=1}^{n}{{{\left( {{y}_{i}}-\bar{y} \right)}^{T}}{{({{\sigma }^{2}}\Sigma )}^{-1}}({{y}_{i}}-\bar{y})-\frac{n}{2}{{(\bar{y}-\mu )}^{T}}{{({{\sigma }^{2}}\Sigma )}^{-1}}(\bar{y}-\mu )} \right]$$
$$\propto \frac{1}{{{\left| {{\sigma }^{2}}\Sigma \right|}^{\frac{n}{2}}}}\exp \left[ -\frac{n-1}{2}tr\left( {{({{\sigma }^{2}}\Sigma )}^{-1}}S \right)-\frac{n}{2}{{(\bar{y}-\mu )}^{T}}{{({{\sigma }^{2}}\Sigma )}^{-1}}(\bar{y}-\mu ) \right]$$
what I have done is using
$$p{({\sigma}^{2})}{\times}p(y|{{\sigma }^{2}})$$
which gives me,
$$\propto {{\left( {{\sigma }^{2}} \right)}^{-(\alpha +1)}}{{e}^{-\beta /{{\sigma }^{2}}}}\times \frac{1}{{{\left| {{\sigma }^{2}}\Sigma \right|}^{\frac{n}{2}}}}\exp \left[ -\frac{n-1}{2}tr\left( {{({{\sigma }^{2}}\Sigma )}^{-1}}S \right)-\frac{n}{2}{{(\bar{y}-\mu )}^{T}}{{({{\sigma }^{2}}\Sigma )}^{-1}}(\bar{y}-\mu ) \right]$$
This doesn't look like anything to me, am I even on the right track?! Anyways, if any experts knows, please point me out, thanks sooo much!
Best Answer
What you get as your bottom line is of the form $$ (\sigma^2) ^{-\alpha-1-nd/2}\exp\{-A\sigma^{-2}\} $$ so the posterior distribution in $\sigma^{-2}$ is an inverse gamma distribution. (Note that $$ \text{tr}((\sigma^2\Sigma)^{-1}S)=\sigma^{-2}\text{tr}(\Sigma^{-1}S)\,.) $$ From this property, you can derive the normalising constant.