Proving Conditional Independence in a Bayesian Network

Question

$$\require{enclose}\begin{array}{c}\enclose{circle}{X_2}&&&&\enclose{circle}{X_1}\\&\searrow&&&&\searrow\\&&\enclose{circle}{X_3}&&&&\enclose{circle} {X_4}\\&&&\searrow&&\swarrow&&\searrow\\&&&&\enclose{circle}{X_5}&&&&\enclose{circle}{X_6}\end{array}$$

bayesian network

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???

Answer 1

Long Story Short

Investigate the subdiagram for those three nodes. $$\require{enclose}\enclose{box}{X_5}\raise{3ex}{\swarrow\raise{3ex}{\enclose{box}{X_4}}\searrow}\enclose{box}{X_6}$$ This encodes the factorisation: $\mathsf P(X_4,X_5,X_6)=\mathsf P(X_4)\mathsf P(X_5\mid X_4)\mathsf P(X_6\mid X_4)$

So applying the definition of conditional probability immediately produces:

$$\mathsf P(X_5,X_6\mid X_4)=\mathsf P(X_5\mid X_4)\mathsf P(X_6\mid X_4)$$

---

Long Story Long

To prove the above is valid, we investigate the entire diagram:

$$\begin{matrix}\enclose{box}{X_2}&&&&\enclose{box}{X_1}\\&\searrow&&&&\searrow\\&&\enclose{box}{X_3}&&&&\enclose{box}{X_4}\\&&&\searrow&&\swarrow&&\searrow\\&&&&\enclose{box}{X_5}&&&&\enclose{box}{X_6}\end{matrix}$$

Thus applying the Law of Total Probability, factorise from the diagram, and distribute the summation, we get:

$$\small\begin{align}\mathsf P(X_4,X_5,X_6)&=\sum_{x\in\{X_1,X_1^\complement\}}\sum_{y\in\{X_2 ,X_2^\complement\}}\sum_{z\in\{X_3,X_3^\complement\}}\mathsf P(x,y,z,X_4,X_5,X_6)\\&=\sum_{x\in\{X_1,X_1^\complement\}}\sum_{y\in\{X_2 ,X_2^\complement\}}\sum_{z\in\{X_3,X_3^\complement\}}\mathsf P(x)\mathsf P(y)\mathsf P(z\mid y)\mathsf P(X_4\mid x)\mathsf P(X_5\mid z,X_4)\mathsf P(X_6\mid X_4)\\&=\mathsf P(X_6\mid X_4)\sum_{z\in\{X_3,X_3^\complement\}}\mathsf P(X_5\mid z,X_4)\sum_{y\in\{X_2 ,X_2^\complement\}}\mathsf P(z\mid y)\sum_{x\in\{X_1,X_1^\complement\}}\mathsf P(X_4\mid x)\mathsf P(x)\end{align}$$

Then you just apply the Law of Total Probability again to produce: $$\mathsf P(X_4,X_5,X_6)=\mathsf P(X_4)\mathsf P(X_5\mid X_4)\mathsf P(X_6\mid X_4)$$