Computation of Variables in a Bayesian Belief Network

Example 1

Decompose $\text{P}(A,B,C,D)$ step by step according to the chain rule into as many terms as possible and simplify, if $A$ and $B$ are independent of $C$ and $D$.

Given that $A$ and $B$ are independent of $C$ and $D$, we can decompose $P(A,B,C,D)$ using the chain rule and then simplify using this independence property.

Chain rule decomposition

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

Since $A$ and $B$ are independent of $C$ and $D$:

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

Substitute:

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

Since $A$ and $B$ are independent:

• $P(A|B)=P(A)$

Finally

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

Example 2

• Specify the probability $\mathrm{P}(A,B,C,D,E)$ by using the single terms of the Bayesian belief network (hint: First reorder the terms!)

$$P(E,D,A,C,B)=P(E|D)\cdot P(D|A,C)\cdot P(C|B)\cdot P(A)\cdot P(B)$$

• Now let $C$ be an unknown and the probability for $D$ be inquired.Write down the probability $\mathrm{P}(D\mid A,B,E)$ by using the single terms of the Bayesian belief network.

$$P(D|A,B,E)=P(E|D)\sum _C\left[P(D|A,C)\cdot P(C|B)\right]\left(P(A)\cdot P(B)\right)$$

Example 3

Given: Calculate $C$ using the entire network

Joint probability using chain rule:

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

Marginalize over A, B, and D to get P(C):

$P(C)={\sum }_{d\in D}{\sum }_{b\in B}{\sum }_{a\in A}P(D|C,B)\cdot P(C|A)\cdot P(B|A)\cdot P(A)$

1. Since ${\sum }_{d}P(D|C,B)=1$ (sum of probabilities over all possible values of D):

$P(C)={\sum }_{b\in B}{\sum }_{a\in A}P(C|A)\cdot P(B|A)\cdot P(A)$

2. Since ${\sum }_{b}P(B|A)=1$ (sum of probabilities over all possible values of B):

$P(C)={\sum }_{a\in A}P(C|A)\cdot P(A)$

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Slightly wrong Solution: source