0
$\begingroup$

I am trying to prove decomposition property of conditional dependency but I'm kind of stuff in the middle ! This is ow I proceeded!

we want to prove :

$x\bot (y,w)| z $ implies $x\bot y | z$

That means x is independent ($\bot$) of y and w , given(|) z!

$x\bot (y,w)| z = x | (y,w),z= \frac{p(x,y,w,z)}{p(y,w,z)}=\frac{p(x,y,z|w) p(w)}{p(y,z|w)p(w)}$

now here I apply he marginalization on w(This is the part I dont know whether its correct or not!!)

$ \Rightarrow \frac{ \sum_w p(x,y,z|w)p(w)}{\sum_wp(y,z|w)p(w)}=\frac{p(x,y,z)}{p(y,z)}=p(x|y,z)$

so it has proven! But I am not sure about the marginalization part!Thanks in advance

  • 0
    Any proof must follow from definitions, so you'll have to figure those out first... Also, $x \bot(y, w) | z$ is a Boolean statement and $ p(x | z)$ is a number. What do you mean by "$=$" between them?2012-11-10

1 Answers 1

0

This how you should prove it:

  • $p(x,y|z)=p(x|z)p(y|z)$
  • $p(x,y|z)=\sum_w p(x,y,w|z)=\sum_w p(x|z)p(y,w|z)= p(x|z)\sum_w p(y,w|z)=p(x|z)p(y|z) $