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:

enter image description here

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 $C$.
  • $\begingroup$ 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. $P(C|A,B) = P(C)$ $\endgroup$ – displayname Nov 11 '13 at 14:59
  • $\begingroup$ When C is independent of (A,B) (which is quite untrue here). $\endgroup$ – Did Nov 11 '13 at 15:02
  • $\begingroup$ 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? $\endgroup$ – displayname Nov 11 '13 at 15:11

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