Does the formula p(X|(Y,Z)=(y,z)) =p((X|Y=y)|(Z|Y=y)=z) hold? Suppose we have three absolutely continuous random vectors $X, Y, Z$ with joint normal distribution. I want to prove that
$$p(X|(Y,Z)=(y,z)) =p((X|Y=y)|(Z|Y=y)=z)$$
Put differently $p(X|(Y,Z)=(y,z))=p(\hat{X}(y)|\hat{Z}(y))$, where $\hat{X}(y)=(X|Y=y)$ and $\hat{Z}(y)=(Z|Y=y)$.
Example of use
This theorem is usually used to derive Kalman filter without even writing it down.
E.g. Anderson, B. D. O., Moore, J. B.: Optimal filtering. Prentice-Hall 1979.
Update
Sorry for changing the theorem completely, but I have finally figured it out and realized that I was really inaccurate... Steps are to realize that they are both normal and then to compute means and variances and show that they are equal. I would still be grateful if someone could prove/disprove the statement only with assumption that X, Y, Z are normal but not jointly normal.
I have found exactly the same type of derivation that I wanted to make on slides here: http://stanford.edu/class/ee363/lectures/kf.pdf - slides 14, 15 where this problem is stated as obvious.
 A: First, the conditional distribution of $X$ conditionally on $Y$ and the conditional distribution of $Z$ conditionally on $Y$ cannot determine the conditional distribution of $X$ conditionally on $(Y,Z)$.
Counterexample: Assume first that $(X,Y,Z)$ is i.i.d. standard normal, then the conditional distributions of $X$ conditionally on $Y$, of $Z$ conditionally on $Y$ and of $X$ conditionally on $(Y,Z)$ are standard normal. Now, assume that $(X,Y)$ is i.i.d. standard normal and that $X=Z$, then the conditional distributions of $X$ conditionally on $Y$ and of $Z$ conditionally on $Y$ are standard normal yet the conditional distribution of $X$ conditionally on $(Y,Z)$ is the Dirac measure at $Z$.
An exercise which can help understand the situation is to compare this counterexample with your "example of use" and to spot the exact place where the former destroys the latter.
Second, there is no such thing as conditional random variables $\hat X=X\mid Y$ or $\hat Z=Z\mid Y$ hence, for example, the expression
$$
\hat X\mid \hat Z
$$
is simply undefined. All one has to work with are conditional distributions.
A: How about a counter-example?
Suppose we have the 8 states and there probabilities as follow:

In this case, P(X) is just 0.5.  But P(Z|Y) = 0.01
P(X|Y,Z) = 0.5 (probability of state 1 divided by probability of states 1 and 2)
P(X|Y) x P(Z|Y) = 0.5 x 0.01 = 0.05, much less than the 0.5 of P(X|Y,Z)
