Basic Understanding of Bayesian Belief Networks
Bayesian Belief Network is a graphical representation of different probabilistic relationships among random variables in a particular set. It is a classifier with no dependency on attributes i.e it is condition independent. Due to its feature of joint probability, the probability in Bayesian Belief Network is derived, based on a condition β P(attribute/parent) i.e probability of an attribute, true over parent attribute.
(Note: A classifier assigns data in a collection to desired categories.)
- Consider this example:
- In the above figure, we have an alarm βAβ β a node, say installed in a house of a person βgfgβ, which rings upon two probabilities i.e burglary βBβ and fire βFβ, which are β parent nodes of the alarm node. The alarm is the parent node of two probabilities P1 calls βP1β & P2 calls βP2β person nodes.
- Upon the instance of burglary and fire, βP1β and βP2β call person βgfgβ, respectively. But, there are few drawbacks in this case, as sometimes βP1β may forget to call the person βgfgβ, even after hearing the alarm, as he has a tendency to forget things, quick. Similarly, βP2β, sometimes fails to call the person βgfgβ, as he is only able to hear the alarm, from a certain distance.
Q) Find the probability that βP1β is true (P1 has called βgfgβ), βP2β is true (P2 has called βgfgβ) when the alarm βAβ rang, but no burglary βBβ and fire βFβ has occurred.
=> P ( P1, P2, A, ~B, ~F) [ where- P1, P2 & A are βtrueβ events and β~Bβ & β~Fβ are βfalseβ events]
[ Note: The values mentioned below are neither calculated nor computed. They have observed values ]
Burglary βBβ β
- P (B=T) = 0.001 (βBβ is true i.e burglary has occurred)
- P (B=F) = 0.999 (βBβ is false i.e burglary has not occurred)
Fire βFβ β
- P (F=T) = 0.002 (βFβ is true i.e fire has occurred)
- P (F=F) = 0.998 (βFβ is false i.e fire has not occurred)
Alarm βAβ β
B | F | P (A=T) | P (A=F) |
T | T | 0.95 | 0.05 |
T | F | 0.94 | 0.06 |
F | T | 0.29 | 0.71 |
F | F | 0.001 | 0.999 |
- The alarm βAβ node can be βtrueβ or βfalseβ ( i.e may have rung or may not have rung). It has two parent nodes burglary βBβ and fire βFβ which can be βtrueβ or βfalseβ (i.e may have occurred or may not have occurred) depending upon different conditions.
Person βP1β β
A | P (P1=T) | P (P1=F) |
T | 0.95 | 0.05 |
F | 0.05 | 0.95 |
- The person βP1β node can be βtrueβ or βfalseβ (i.e may have called the person βgfgβ or not) . It has a parent node, the alarm βAβ, which can be βtrueβ or βfalseβ (i.e may have rung or may not have rung ,upon burglary βBβ or fire βFβ).
Person βP2β β
A | P (P2=T) | P (P2=F) |
T | 0.80 | 0.20 |
F | 0.01 | 0.99 |
- The person βP2β node can be βtrueβ or falseβ (i.e may have called the person βgfgβ or not). It has a parent node, the alarm βAβ, which can be βtrueβ or βfalseβ (i.e may have rung or may not have rung, upon burglary βBβ or fire βFβ).
Solution: Considering the observed probabilistic scan β
With respect to the question β P ( P1, P2, A, ~B, ~F) , we need to get the probability of βP1β. We find it with regard to its parent node β alarm βAβ. To get the probability of βP2β, we find it with regard to its parent node β alarm βAβ.
We find the probability of alarm βAβ node with regard to β~Bβ & β~Fβ since burglary βBβ and fire βFβ are parent nodes of alarm βAβ.
From the observed probabilistic scan, we can deduce β
P ( P1, P2, A, ~B, ~F)
= P (P1/A) * P (P2/A) * P (A/~B~F) * P (~B) * P (~F)
= 0.95 * 0.80 * 0.001 * 0.999 * 0.998
= 0.00075