1. ## conditional independence

hello,

I would like to ask the following:

I have 3 random variables X , Y and Z.
If I know that X is independent of Y given Z then does this imply that X is independent of Z?

thanks !

2. ## Re: conditional independence

(X ⊥ Y | Z) => (X|Z)

I don't think so, because I think Z could be a D seperator (screen) between X and Y.

So in causality it could be: X → Z → Y.

Though I could be wrong, since I am just trying to learn these comcepts myself. I think this is in the first chapter of Judea Pearl's, Causality book - which I don't have access to.

3. ## The Following User Says Thank You to hlsmith For This Useful Post:

michinerd (10-30-2014)

4. ## Re: conditional independence

Let X ~ N(0,1), E1 ~ N(0,1), E2 ~ N(0,1) all independent of one another. Now let Z = X + E1, and let Y = Z + E2

X and Y are not independent but conditional on Z they are. X and Z are definitely not independent.

So to answer your question - X being independent of Y given Z does not imply that X is independent of Z.

5. ## The Following User Says Thank You to Dason For This Useful Post:

michinerd (10-30-2014)

6. ## Re: conditional independence

Thank for the confirmation Dason, but I expected something more elegant from you

Z = X + E1
Y = Z + E2

so, Y = (X+ E1) + E2

7. ## The Following User Says Thank You to hlsmith For This Useful Post:

michinerd (10-30-2014)

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