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Belief Networks (Conditional independence in Bayesian Networks)

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• I ve read in the 2nd edition, that a node is conditionally independent of its non-descendants given its parents or given its Markov blanket. The 2nd edition
Message 1 of 2 , May 3, 2007
I've read in the 2nd edition, that a node is conditionally independent
of its non-descendants given its parents or given its Markov blanket.
The 2nd edition makes reference to the 1st edition regarding d-
separation and I have read the 1st edition's information about d-
separation; however, I have come across in some additonal reading
something called Berkson's paradox.

This paradox seems to show circumstances that contradict the
statements made about d-separation and Markov blankets. Am I correct
in thinking that Berkson's paradox is an exception to rules regarding
d-separation and Markov blankets?
• ... The so called Berkson s paradox states: if 0
Message 2 of 2 , May 4, 2007
On Thu, May 03, 2007 07:37:59PM -0000, orondojones wrote:
> I've read in the 2nd edition, that a node is conditionally independent
> of its non-descendants given its parents or given its Markov blanket.
> The 2nd edition makes reference to the 1st edition regarding d-
> separation and I have read the 1st edition's information about d-
> separation; however, I have come across in some additonal reading
> something called Berkson's paradox.
>
> This paradox seems to show circumstances that contradict the
> statements made about d-separation and Markov blankets. Am I correct
> in thinking that Berkson's paradox is an exception to rules regarding
> d-separation and Markov blankets?

The so called Berkson's paradox states:

if 0 < P(A) < 1 and 0 < P(B) < 1,
and P(A|B) = P(A), i.e. they are independent,
then P(A|B,C) < P(A|C) where C = A∪B (i.e. A or B).

Where is the contradiction?

--
Iván F. Villanueva B.
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