Algebraic proof of conditional independence in a Bayesian network

Question

I have a Bayesian network shown as $a \to b \to c \leftarrow d \leftarrow e$. I want to prove $a \perp e$.

It's easy to show $b \perp d$ since $$P(b,c,d)=P(c|b,d)P(b)P(d)=P(c|b,d)P(b|d)P(d).$$

So $P(b|d)=P(b)$.

I tried the same approach by $P(a,b,c,d,e)=P(c|b,d)P(b|a)P(d|e)P(a)P(e)$. I tried $P(a,b,c,d,e)=P(c|a,b,d,e)P(b|a,d,e)P(d|a,e)P(a|e)P(e)$ but there are too many variables can't be eliminated. How can I relate this to $P(a|e)=P(a)$? Am I on the right approach? Can anyone give a hint?

Answer 1

I want to prove $a\perp e$.

$$\def\P{\operatorname{\sf P}}\begin{align}&\qquad\P(\mathrm {a,e})\\[2ex]&=\sum_b\sum_c\sum_d \P(\mathrm a,b,c,d,\mathrm e)&&\text{Law of Total Probability}\\[2ex]&=\sum_b\sum_c\sum_d \left(\P(\mathrm a)\P(b\mid\mathrm a)\P(c\mid b,d)\P(d\mid\mathrm e)\P(\mathrm e)\right)&&\text{Factorisation}\\[2ex]&={\P(\mathrm a)\P(\mathrm e)\sum_b\left(\P(b\mid \mathrm a)\sum_d\left(\P(d\mid\mathrm e)\sum_c\P(c\mid b,d)\right)\right)}&&\text{Distribution}\\[2ex]&={\P(\mathrm a)\P(\mathrm e)\sum_b\left(\P(b\mid \mathrm a)\sum_d\P(d\mid\mathrm e)\right)}&&\text{LoTP}\\[2ex]&={\P(\mathrm a)\P(\mathrm e)\sum_b\P(b\mid \mathrm a)}&&\text{LoTP} \\[2ex]&=\P(\mathrm a)\P(\mathrm e)&&{\text{Law of Total Probability}}\end{align}$$