$$\require{enclose}\begin{array}{c}\enclose{circle}{X_2}&&&&\enclose{circle}{X_1}\\&\searrow&&&&\searrow\\&&\enclose{circle}{X_3}&&&&\enclose{circle} {X_4}\\&&&\searrow&&\swarrow&&\searrow\\&&&&\enclose{circle}{X_5}&&&&\enclose{circle}{X_6}\end{array}$$
we know
$P(X_1,X_2,X_3,X_4,X_5,X_6)=P(X_1)P(X_2)P(X_3|X_2)P(X_4|X_1)P(X_5|X_3,X_4)P(X_6|X_4)$
I'm trying to prove $X_5$ and $X_6$ are conditionally independent
so I want to prove,
$P(X_5,X_6|X_4)=P(X_5|X_4)P(X_6|X_4)$
I've been trying to use my equation of joint distribution and this more general equation:
$P(X_1,X_2,X_3,X_4,X_5,X_6)=P(X_1)P(X_2|X_2)P(X_3|X_1,X_2)P(X_4|X_1,X_2,X_3)P(X_5|X_1,X_2,X_3,X_4)P(X_6|X_1,X_2,X_3,X_4,X_5)$
$X_3$ seems to complicate the calculations.
Also, I've been trying to disprove the independence of E and F, in general.
In other words, disprove
$P(X_5,X_6)=P(X_5)P(X_6)$
I don't know if I need to use Markov's condition or maybe some re-arrangement would help?
$P(X_5|X_3,X_4)P(X_6|X_4)=\frac{P(X_1,X_2,X_3,X_4,X_5,X_6)}{P(X_1)P(X_2)P(X_3|X_2)P(X_4|X_1)}$ ???
$P(X_5|X_1,X_2,X_3,X_4)P(X_6|X_1,X_2,X_3,X_4,X_5)=\frac{P(X_1,X_2,X_3,X_4,X_5,X_6)}{P(X_1)P(X_2|X_2)P(X_3|X_1,X_2)P(X_4|X_1,X_2,X_3)}$ ???
I tried equating the two expressions for $P(X_1,X_2,X_3,X_4,X_5,X_6)$, but that did not seem to help
$P(X_1)P(X_2)P(X_3|X_2)P(X_4|X_1)P(X_5|X_3,X_4)P(X_6|X_4)=P(X_1)P(X_2|X_2)P(X_3|X_1,X_2)P(X_4|X_1,X_2,X_3)P(X_5|X_1,X_2,X_3,X_4)P(X_6|X_1,X_2,X_3,X_4,X_5)$
useful???