Conditional probability distribution and prior In a linear Gaussian model, when I multiply a prior distribution $p(x)$ with the conditional $p(y|x)$ (here x and y are vectors), which one do I get:


*

*The joint distribution p(z) where z is the concatenation of x and y 

*Or p(y)?


In Bayes scenario, we have posterior p(x|y) is proportional to p(x)*p(y|x). Does this mean that p(x|y) is proportional to p(z)?
 A: We see that
$$P(Y|X) \cdot P(X) = \frac{P(Y,X)}{P(X)}\cdot P(X) = P(Y,X) = P(Z),$$
Where $Z$ is defined as the variable over the cartesian product of the supports of $Y$ and $X$ (as in, a realization $z$ of $Z$ is in $S_Y \times S_X$). Hence, 1. is the more appropriate answer.
A: $\vec z=\vec x\Vert\vec y~$ is the vector formed by concatenating vectors $\vec x$ and $\vec y$.   That is that: $$\vec x=\begin{pmatrix}a_1 \\ a_2\\ \vdots \\ a_n\end{pmatrix}, \vec y=\begin{pmatrix}b_1 \\ b_2\\ \vdots \\ b_m\end{pmatrix}\\ \vec z= \vec x\Vert \vec y = \begin{pmatrix}a_1 \\ a_2\\ \vdots \\ a_n\\b_1 \\\vdots \\ b_m\end{pmatrix}$$
Then the measure of the concatenation equals the joint measure of the vectors.
 $$p_{\vec X,\vec Y}(\vec x,\vec y)=p_{\vec X\Vert \vec Y}(\vec x\Vert \vec y)$$
By the definition of conditional probability (also Bayes' rule)
$$\begin{align}p_\vec X(\vec x)\,p_{\vec Y\mid \vec X}(\vec y\mid \vec x) =&~ p_{\vec X, \vec Y}(\vec x, \vec y) \\[1ex]=&~p_\vec Z(\vec z)\\[1ex]=&~p_{\vec X\mid \vec Y}(\vec x\mid \vec y)\,p_\vec Y(\vec y)\\[3ex]\therefore~p_\vec Y(\vec y)=&~ p_\vec X(\vec x)\,p_{\vec Y\mid \vec X}(\vec y\mid \vec x)\big/p_{\vec X\mid\vec Y}(\vec x\mid \vec y)\end{align}$$
