Conditional Independence - Bayesian Network

Question

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?

Answer 1

$$\begin{align}P(A,B) &= \iint P(A,B,C,D)~dD~dC \\[1ex] &= \iint P(A)\,P(B)\,P(C\mid A,B)\,P(D\mid C)~dD~dC\\[1ex] &= P(A)\,P(B)\int P(C\mid A,B)\int P(D\mid C)~dD~dC\\[1ex]& = P(A)\,P(B)\int P(C|A,B)\,1~dC\\[1ex] &=P(A)\,P(B)\,1\\[1ex]& = P(A)\,P(B)\end{align}$$

... by its definition, independence is demonstrated. $$\begin{align}P(A,D\mid C) &= \int P(A,D,B\mid C)~dB \\[1ex]&= \int \dfrac{P(A)\,P(B)\,P(C\mid A,B)\,P(D\mid C)}{P(C)\qquad} ~dB \\[1ex]&= \int P(A\mid C)\,P(B\mid A,C)\,P(D\mid C)~dB\\[1ex]& = P(A\mid C)\,P(D\mid C)\int P(B\mid A,C)~dB \\[1ex]& = P(A\mid C)\,P(D\mid C)\end{align}$$ ... by its definition, conditional independence is demonstrated.

Answer 2

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 $C$.