Implying independence of variable

Question

Show that $a$ independent of $b$ and $c$ given $d$ implies $a$ independent of $b$ given $d$ or more formally, show that:

$$ a ⊥ b, c | d = a ⊥ b|d $$

I'm unclear how to proceed on this one, I understand that we have the independence rules, where $A$ and $B$ are independent if:

$ P(A|B) = P(A) $ or $ P(B|A) = P(B) $ or $ P(A,B) = P(A) P(B) $

But I don't understand how I should apply them to a question like this. I'm not even sure if I need to make use of them given we are already working under the assumption that $a$ is independent.

If someone could point me in the right direction that would be great!

Answer 1

If a is independent of b and c given d, then this means that

$p(a,b,c \vert d) = p(a \vert d) p(b,c \vert d)$.

Equivalently, it also means that

$p( a \vert b,c,d) = p(a \vert d)$ because

$p( a \vert b,c,d) = p( a,b,c \vert d) / p(b,c \vert d) = p(a \vert d) p(b,c \vert d) / p(b,c \vert d) = p(a \vert d)$

Hopefully this will point you in the right direction!

Answer 2

Since:

P(a,b,c|d)

=P(a|d)*P(b,c|d)

=P(a|d)*P(b|c,d)*P(c|d)

=P(a|d)*P(b|c)*P(c|d)

=P(a|d)*P(b|d)

=P(a,b|d)

Hence, a⊥b,c|d=a⊥b|d

Is it correct?