Really struggling with how to approach this question. The lecturer, as per usual, has provided us with the bare minimum in terms of hints on how to approach this. I know how to do it when we want in terms of the conditional pdf for $\mu$ given $\sigma$ and the data and the marginal pdf for $\sigma$ given the data, by expressing and expanding the exponent in the normal pdf via the inclusion of $\bar{y}$, but I have no idea how to manipulate this to get it the way she wants. Anyway, any help or advice you could offer would be greatly appreciated.
$y_i$ are distributed i.i.d normal($\mu, \sigma^2$)
Assuming a non-informative prior
p($\mu,\sigma)=1/\sigma$,
and given
p($\mu,\sigma$|y)$\propto$L($\mu,\sigma$|y) p($\mu,\sigma$),
re-specify the joint pdf, p($\mu,\sigma$|y) as the following decomposition:
p($\mu,\sigma$|y)= p($\sigma|\mu$, y) p($\mu$|y),
providing the complete forms of p($\sigma|\mu$, y) and p($\mu$|y) in the answer.
Thanks guys
We have
$$p(\mu,\sigma)=\frac1\sigma\;,$$
$$L(\mu,\sigma\mid\mathbf y)=\prod_i\frac{\mathrm e^{-\frac{(y_i-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}}=\frac{\mathrm e^{-\frac{\Sigma_i(y_i-\mu)^2}{2\sigma^2}}}{\sigma^n(2\pi)^{\frac n2}}$$
and thus
$$p(\mu,\sigma\mid\mathbf y)=\frac1\sigma\prod_i\frac{\mathrm e^{-\frac{(y_i-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}}=\frac{\mathrm e^{-\frac{\Sigma_i(y_i-\mu)^2}{2\sigma^2}}}{\sigma^{n+1}(2\pi)^{\frac n2}}\;. $$
The marginal distribution $p(\mu\mid\mathbf y)$ is obtained by integrating over $\sigma$. With
$$ \int_0^\infty\frac{\mathrm e^{-a/\sigma^2}}{\sigma^{n+1}}\mathrm d\sigma=\frac12\int_0^\infty\mathrm e^{-ax}x^{\frac n2-1}\mathrm dx=\frac12a^{-\frac n2}\Gamma\left(\frac n2\right)\;, $$
the result is
$$ p(\mu\mid\mathbf y)=\frac{\Gamma\left(\frac n2\right)}{2\left(\pi\Sigma_i(y_i-\mu)^2\right)^{\frac n2}}\;. $$
Then $p(\sigma\mid\mu,\mathbf y)$ follows from
$$ p(\mu,\sigma\mid\mathbf y)=p(\sigma\mid\mu,\mathbf y)p(\mu\mid\mathbf y)\;. $$