# If $x$ is independent of $(y,w)$ given $z$ then $x$ is independent of $y$ given $z$

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

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It might help the questin to explain what the "upside down T" means. Does it mean two things on either side are independent? If so how is this to be used along with the vertical bar? Alternately you could give a link to some page explaining the notation. – coffeemath Nov 4 '12 at 16:53
sorry for that ! That means x is independent (⊥) of y and w , given(|) z!I add it to my question – Moj Nov 4 '12 at 17:12
How do you define $x \bot(y, w) | z$? It seems to me that it would mean $p(x|y, w, z) = p(x|y, z) = p(x|w, z) = p(x | z)$, which would make the result obvious. – Karolis Juodelė Nov 4 '12 at 17:21
yes but it seems vey easy to that! I could say x⊥y|z = p(x|z) and also say that p(x|y,w,z)=p(x|z) , then I'm done.But it doesn't seem to be a good solution! Or is it really this much easy? – Moj Nov 4 '12 at 18:01
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? – Karolis Juodelė Nov 10 '12 at 19:51

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