The question is to find $p(x|y)$ given that $p(x) \sim \mathcal{N}(\mu, \Sigma)$ and $p(y|x) \sim \mathcal{N}(Ax, \Gamma)$.
I do realize that I may just obtain a posterior through application of the Bayes Theorem, but the integration would be onerous.
$$p(x|y) = \frac{p(y|x)p(x)}{\int{p(y|x)p(x)dx}}$$
I have been made to know a certain trick, where $y$ posed as $y = Ax+ \epsilon$, $\epsilon \sim \mathcal{N}(0, \Gamma)$. Then the $p(y)$ parameters are derived rather easily: $p(y) \sim \mathcal{N}(A\mu, A \Sigma A^T + \Gamma)$. Hence, we have evidence and are relieved from the integration in denominator.
Yet I cannot come up with the next step of solution. Tips and suggestions are the most welcome.
Thanks.
Edit
Let's write down the distributions we have (neglecting the constant term):
$$p(x) = \frac{1}{\sqrt{|\Sigma|}}exp(-\frac 1 2(x-\mu)^T \Sigma^{-1}(x-\mu))$$
$$p(y|x) = \frac{1}{\sqrt{|\Gamma|}}exp(-\frac 1 2(y-Ax)^T \Gamma^{-1}(y-Ax))$$
$$p(y) = \frac{1}{\sqrt{|A \Sigma A^T + \Gamma|}}exp(-\frac 1 2(y-A\mu)^T {[A \Sigma A^T + \Gamma]}^{-1}(y-A\mu))$$
Then, $p(x|y) = f(x,y)exp(g(x,y))$ and f(x,y) might be expressed as:
$$f(x) = \sqrt{ \frac{|A \Sigma A^T + \Gamma|}{{|\Gamma||\Sigma|}}} = \frac{1}{\sqrt{|\Gamma \Sigma [A \Sigma A^T + \Gamma]^{-1}|}}$$.
Then, $Cov(x|y) = \Gamma \Sigma [A \Sigma A^T + \Gamma]^{-1}$.
But I get stuck on rewriting $g(x,y)$ term: $$g(x,y) = -\frac 1 2(x-\mu)^T \Sigma^{-1}(x-\mu) -\frac 1 2(y-Ax)^T \Gamma^{-1}(y-Ax) + \frac 1 2(y-A\mu)^T {[A \Sigma A^T + \Gamma]}^{-1}(y-A\mu)$$