Equivalence in conditional probability I am wondering the equivalence of the following problem. 
When we computing 
$$ 
P(\mathbf{x}_1 | \mathbf{x}_2, \mathbf{x}_3)
$$
is it equivalent as following, at first define $\mathbf{y} = \mathbf{x}_1 | \mathbf{x}_2$, as new random variable, and 
$$ P(\mathbf{y} | \mathbf{x}_3) $$
In particular when $\mathbf{x}_i \sim N(\mathbf{\mu}_i, \Sigma_{ii}) $, can we say 
$$ P(\mathbf{x}_1 | \mathbf{x}_2,\mathbf{x}_3) = P(\mathbf{y} | \mathbf{x}_3) $$?
 A: Yes, the desired formula holds without the assumption of normal distribution.
Denote a probability space by $(\Omega,\mathcal{F},\mathbb{P})$. Assume that we know the joint probability density function $f_{X,Y,Z}(x,y,z)$. Write
$$ g_{X,Y}^{z}(x,y) := \frac{f_{X,Y,Z}(x,y,z)}{f_{Z}(z)}$$
Note that 
$$g_{X,Y}^{z}(x,y)=f_{X,Y|Z}(x,y|z)$$
where the right-hand side is the conditional probability density function of $(X,Y)$ given $Z=z$. One can interpret the $g_{X,Y}^{z}(x,y)$ as a Radon-Nikodym derivative of a new probability measure $\mathbb{P}_{Z=z}(\cdot)$ on the original $\sigma$-field $\mathcal{F}$:
$$\mathbb{P}_{Z=z}(X\in A, Y\in B):=\int_{A\times B}g_{X,Y}^{z}(x,y)dxdy,\quad \forall A,B\in \mathcal{B}(\mathbb{R})$$
Here $\mathcal{B}(\mathbb{R})$ denotes a Borel $\sigma$-algebra on $\mathbb{R}$. If $X,Y,Z$ are discrete random variables then one would regard $\mathbb{P}_{Z=z}$ as a conditonal probability mass function 
$$\mathbb{P}_{Z=z}(X=x,Y=y)=\mathbb{P}(X=x,Y=y|Z=z)$$
Now consider the conditional probability density (or mass) function $\mathbb{P}_{Z=z}(X=x|Y=y)$ by using the conditioning of $Y=y$ on $g_{X,Y}^{z}(x,y)$ i.e.
$$g_{X|Y}^{z}(x|y):=\frac{g_{X,Y}^{z}(x,y)}{\int_{\mathbb{R}}g_{X,Y}^{z}(x,y)dx}$$
Then observe that
$$\frac{g_{X,Y}^{z}(x,y)}{\int_{\mathbb{R}}g_{X,Y}^{z}(x,y)dx}=\frac{\frac{f_{X,Y,Z}(x,y,z)}{f_{Z}(z)}}{\int_{\mathbb{R}}\frac{f_{X,Y,Z}(x,y,z)}{f_{Z}(z)}dx}=\frac{f_{X,Y,Z}(x,y,z)}{\int_{\mathbb{R}}f_{X,Y,Z}(x,y,z)dx}=\frac{f_{X,Y,Z}(x,y,z)}{f_{Y,Z}(y,z)}$$
Therefore we obtain
$$g_{X|Y}^{z}(x|y) = f_{X|Y,Z}(x|y,z)$$
which implies the following result
\begin{align*}\mathbb{P}_{Z=z}(X=x|Y=y) = \mathbb{P}(X=x|Y=y,Z=z)\tag{1}\end{align*}
Returning to the initial question, we regard $P(\mathbf{y}|\mathbf{x}_{3})$ as $\mathbb{P}_{\mathbf{x}_{3}=x_{3}}(\mathbf{x}_{1}=x_{1}|\mathbf{x}_{2}=x_{2})$. Hence eq. (1) is the desired result.
