# Conditional Independence - Bayesian Network

May the probability distribution $P(A,B,C,D)$ given as:

$P(A,B,C,D) = P(A)P(B)P(C|A,B)P(D|C)$

The task is to show that this holds $A \bot B | \emptyset$ and $A\bot D|C$.

First thing I'd like to know is, if this - drawn as a Bayesian Network - looks like this: and the second thing - of course - is how I can show those conditions.

Assuming that the network looks like the graphic shows I can see, that $A \bot B | \emptyset$ and $A\bot D|C$ should apply to this distribution.

Do I have to show that $P(A,B,C,D) = P(A)P(B)$ and $P(A,B,C,D) = P(A)P(C|A)P(D|C)$? And if so: How can I show this?

I can not assume that $P(C|A,B) = \frac{P(A,B,C)}{P(A,B)} = P(C)$ and $P(D|C) = P(D)$ such that $P(A,B,C,D) = P(A,B,C,D) \Leftrightarrow P(A,B) = P(A,B)$ because this would mean, that all events are independet, right?

How can I show that?

$P(A,B) = \int\int P(A,B,C,D)dDdC = \int\int P(A)P(B)P(C|A,B)P(D|C)dDdC = \int P(A)P(B)P(C|A,B)\int P(D|C)dDdC = \int P(A)P(B)P(C|A,B)1dC = P(A)P(B)\int P(C|A,B)dC = P(A)P(B)1 = P(A)P(B)$ from which the definition of independence is evidently true.
$P(A,D|C) = \int P(A,D,B|C)dB = \int P(A)P(B)P(C|A,B)P(D|C)/P(C) dB = \int P(A,B,C)P(D|C)dB/P(C) = \int P(A,C)P(B|A,C)P(D|C)dB/P(C) = P(A,C)P(D|C)\int P(B|A,C)dB / P(C) = P(A,C)P(D|C)/P(C) = P(A|C)P(D|C)$ from which the definition of conditional independence is evident.
The hypothesis is that $P(A,B,C,D) = P(A)P(B)P(C|A,B)P(D|C)$ $(\ast)$, which implies $P(A,B,C,D) = P(A,B,C)P(D|C)$ $(\ast\ast)$.
• Summing $(\ast)$ over every $(C,D)$ yields $P(A,B)=P(A)P(B)$. This proves that $A$ and $B$ are independent.
• Summing $(\ast\ast)$ over every $B$ yields $P(A,C,D)=P(A,C)P(D|C)$. Dividing both sides by $P(C)$ yields $P(A,D|C)=P(A|C)P(D|C)$. This proves that $A$ and $D$ are independent conditionally on $C$.
• This makes sense, but could you explain to me when I can assume that $P(C|A,B) = P(C)$? If I am allowed to do so, I'd end up with $P(A,B,C,D) = P(A)P(B)P(C)P(D)$. I would like to know when I can assume e.g. $P(C|A,B) = P(C)$ – displayname Nov 11 '13 at 14:59
• And we "see" that $P(C)$ is dependent because of what is given, right? So the graphic of the network is also correct? What kind of notation allows one to show that $P(C|A,B) = P(C)$ and P(C|A,B) \new P(C) \$? That means, that I always have to remember which term describes which part of the network such that I don't eliminate factors be accident, doesn't it? – displayname Nov 11 '13 at 15:11