Computing the conditional distribution for the mean of a Gaussian I have the following distributional assumptions on some on my RV and model parameters:
$$
y_i \sim N(\beta x_i, w_i^{-1}\Sigma_y)
$$
There is a normal prior on the parameters $\beta$ as well:
$$
\beta \sim N(\beta_0, \Sigma_{\beta})
$$
AFAIK, the conditional distribution for the mean parameter $\beta$, given by $P(\beta|y, w) should also be normally distributed due to conjugate relationship. I have been struggling to derive that.
So, I want to get the expression for the conditional distribution of $\beta$, which in this case is given by $p(y|\beta, w) P(\beta)$ and this is proportional to:
$$
|\Sigma_y|^\frac{-N}{2}\Bigg(\exp \sum_{i=1}^N(y_i - \beta x_i)^T\Sigma_y^{-1}(y-\beta x_i) \Bigg)|\Sigma_{\beta}|^{-0.5} \exp (\beta-\beta_0)^T\Sigma_{\beta}^{-1}(\beta -\beta_0)
$$
Here I am dropping the terms depending on $w$. This becomes
$$
|\Sigma_y|^\frac{-N}{2} |\Sigma_{\beta}|^{-0.5} \exp \big((\beta-\beta_0)^T\Sigma_{\beta}^{-1}(\beta -\beta_0) + \sum_{i=1}^N(y_i - \beta x_i)^T\Sigma_y^{-1}(y-\beta x_i)\big)
$$
However, after this I am quite stuck. I am not sure how I can express this in the form of a normal distribution.
 A: First note that the availability of a closed-form posterior, which essentially amounts to conjugacy as you notice, depends also on $\sigma^2$ and on what prior distribution it is assigned. If  $\sigma^2$ is known, then your prior $\beta \sim N(\beta_0, \Sigma_0)$ is indeed conjugate, otherwise if $\sigma^2$ is random, you need an inverse gamma prior on it, and you also need to change your prior on $\beta$ as  $$\beta\vert\sigma^2 \sim N(\beta_0, \sigma^2\Sigma_0).$$
Showing conjugacy requires some tedious algebra, a crucial point in the algebra is to use the least squares estimate 
$$\hat \beta = (XX^T)^{-1}X^Ty$$
in order to center the posterior distribution of $\beta$. You can show that the posterior is $$\beta\vert \sigma^2, Y, X\sim N(\beta_n, \sigma^2\Sigma_n)$$
where
$$\Sigma_n = (X^TX+\Sigma_0^{-1})^{-1},\quad \beta_n = \Sigma_n((X^TX)\hat\beta+\Sigma_0^{-1}\beta_0).$$
The steps are well detailed on the Wikipedia page of Bayesian linear regression, see here: http://en.wikipedia.org/wiki/Bayesian_linear_regression#Posterior_distribution
PS: mind your expression of the posterior distribution for $\beta$ where you are missing some terms. It should read
$$\Bigg(\prod_{i=1}^N \exp -\frac{w_i}{2 \sigma^2}\big(y_i - \beta^Tx_i\big)^T\big(y_i - \beta^Tx_i\big)\Bigg)\exp -\frac{1}{2 \sigma^2}(\beta-\beta_0)^T\Sigma_0^{-1}(\beta -\beta_0)$$
