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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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How do you define $Y|Z$? –  Shai Covo Feb 25 '11 at 8:47
    
@Shai: I admit I am not clear either. Do people use Y|Z to represent something? In my opinion, the distribution of Y|Z=z will be a probability measures P(Y|Z=z), and I perhaps should use Y|Z=z for all z, instead of Y|Z in my post? –  Tim Feb 25 '11 at 8:50
    
Maybe try constructing a concrete example. –  Shai Covo Feb 25 '11 at 9:05
    
@Shai: Instead of having a concrete example, I would like to say that Y|Z=z is a particular random variable with distribution as P(Y|Z=z). I use "particular", because if I understand correctly, there can be several r.v.s with a same given distribution, and I also think Y|Z=z might be some specific r.v. with distribution P(Y|Z=z). –  Tim Feb 25 '11 at 9:11
    
But what $Y|Z$ would be? This is quite different from the situation with conditional expectations. –  Shai Covo Feb 25 '11 at 9:19
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1 Answer

up vote 1 down vote accepted

(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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Thanks! (1). I mean "X" and "Y|Z". Why no? (3). I guess we share the same understanding. Why "yes"? –  Tim Feb 25 '11 at 9:01
    
Henry: consider three random variables U, V and W. I would be grateful if you would care to explain what the following assertion means: U|V and W are independent. Your post is based on this notion and I am curious –  Did May 1 '11 at 10:35
    
@Didier Piau: Good question. I think I was trying to fit into Tim's language, so in my counter-example for (1) came up with a particular $Y|Z$. I suppose that in fact $U|V$ is a set of random variables $U|V=v_1$, $U|V=v_2$, etc. and that $W$ is independent of them if for all $i,j,k$: $P\left(W=w_k|(U|V=v_j)=u_i\right) = P(W=w_k)$. But this may not be sensible notation. –  Henry May 1 '11 at 11:03
    
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. –  Did May 1 '11 at 11:10
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