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  1. Given random variables $X, Y, Z$, is joint distribution of $X$ and "$Y|Z$" same as distribution of $(X, Y)|Z$?
  2. Given random variables $X_1, X_2, Y_1, Y_2$, is joint distribution of "$X_1 | X_2$" and "$Y_1|Y_2$" same as distribution of $(X_1, Y_1) | (X_2, Y_2)$?
  3. For two independent random vectors $X$ and $Y$, and any two subvectors $X_1$ and $X_2$ of $X$ and any two subvectors $Y_1$ and $Y_2$ of $Y$, will the conditional random vectors "$X_1|X_2$" and "$Y_1|Y_2$" also be independent?

Why? Thanks and regards!

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    @Tim: Indeed, I don't recall seeing a definition for $Y|Z$. I'm not aware of similar concepts. However, you might conclude something by constructing simple concrete examples, say with $\Omega = \{1,2,3 \}$.2011-02-25

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(1) No if one random variable is $X$ and the other is $Y|Z$. Take for example six equally probable cases for $(X,Y,Z)$:

(1, 1, 2), (2, 1, 1), (3, 2, 1), (1, 2, 2), (2, 3, 1), (3, 3, 2)

Then $Y|Z$ is 1, 2 or 3 with equal probability, no matter what $Z$ is: they are pairwise independent. So the joint distribution of $X$ and of $Y|Z$ is the same six pairs as the joint distribution of $X$ and of $Y$. But $(X,Y)$ is not independent of $Z$ since if $Z=1$ then $X$ is not $2$.

(2) No. Not if $X_1|X_2$ is not independent of $Y_2$ , or $Y_1|Y_2$ is not independent of $X_2$

(3) If I understand your use of vectors and subvectors, then yes

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    Henry: Indeed, not a sensible notation--and certainly not a sensible notion... In fact there is no such thing as a *random variable*$U|V$and one very soon encounters some serious pitfalls trying to define it. What is well defined is the *conditional distribution* of U conditionally on V, which is altogether different.2011-05-01