![CONDITIONAL PROBABILITY
Events A and B
• P(A|B)- Probability that event A occurs given that event B has already
occurred.
Example:
There are 2 baskets. B1 has 2 red ball and 5 blue ball. B2 has 4 red ball and 3
blue ball. Find probability of picking a red ball from basket 1?
• The question above wants P(red ball | basket 1).
• The answer intuitively wants the probability of red ball from only the sample
space of basket 1.
• So the answer is 2/7
• The equation to solve it is:
P(A|B) = P(A∩B)/P(B) [Product Rule]
P(A,B) = P(A)*P(B) [ If A and B are independent ]
How do you solve P(basket2 | red ball) ???](https://image.slidesharecdn.com/bayesiannetwork-171113204848/85/Bayesian-network-2-320.jpg)
![BAYESIAN THEOREM
• A special case of
Bayesian Theorem:
P(A∩B) = P(B) x P(A|B)
P(B∩A) = P(A) x P(B|A)
Since P(A∩B) = P(B∩A),
P(B) x P(A|B) = P(A) x
P(B|A)
=> P(A|B) = [P(A) x P(B|A)]
/ P(B)
A B
ABPAPABPAP
ABPAP
BP
ABPAP
BAP
||
)|()(
)(
)|()(
)|(
](https://image.slidesharecdn.com/bayesiannetwork-171113204848/85/Bayesian-network-3-320.jpg)
![BAYESIAN THEOREM
Solution to P(basket2 | red ball) ?
P(basket 2| red ball) = [P(b2) x P(r | b2)] / P(r)
= (1/2) x (4/7)] / (6/14)
= 0.66](https://image.slidesharecdn.com/bayesiannetwork-171113204848/85/Bayesian-network-4-320.jpg)
![A Bayesian Network consists of [Jensen, 1996]:
A set of variables and a set of direct edges between
variables.
1. Each variables has a finite set of mutually exclusive
states
2. The variable and direct edge form a DAG (directed
acyclic graph)
3. To each variable A with parents B1, B2 ..Bn there is
attached a conditional probability table P(A| B1, B2 ..
Bn)](https://image.slidesharecdn.com/bayesiannetwork-171113204848/85/Bayesian-network-6-320.jpg)
Bayesian network
- 1. BAYESIAN NETWORKS BY- Ahmad Ali Al Taweel
- 2. CONDITIONAL PROBABILITY Events A and B • P(A|B)- Probability that event A occurs given that event B has already occurred. Example: There are 2 baskets. B1 has 2 red ball and 5 blue ball. B2 has 4 red ball and 3 blue ball. Find probability of picking a red ball from basket 1? • The question above wants P(red ball | basket 1). • The answer intuitively wants the probability of red ball from only the sample space of basket 1. • So the answer is 2/7 • The equation to solve it is: P(A|B) = P(A∩B)/P(B) [Product Rule] P(A,B) = P(A)*P(B) [ If A and B are independent ] How do you solve P(basket2 | red ball) ???
- 3. BAYESIAN THEOREM • A special case of Bayesian Theorem: P(A∩B) = P(B) x P(A|B) P(B∩A) = P(A) x P(B|A) Since P(A∩B) = P(B∩A), P(B) x P(A|B) = P(A) x P(B|A) => P(A|B) = [P(A) x P(B|A)] / P(B) A B ABPAPABPAP ABPAP BP ABPAP BAP || )|()( )( )|()( )|(
- 4. BAYESIAN THEOREM Solution to P(basket2 | red ball) ? P(basket 2| red ball) = [P(b2) x P(r | b2)] / P(r) = (1/2) x (4/7)] / (6/14) = 0.66
- 5. • A Bayesian network is a graph-based model for conditional independence assertions and hence for compact specification of full joint distributions. • A Bayesian network B is defined as a pair B = (G, P), where G = (V (G),A(G)) is an acyclic directed graph with a set of vertices (or nodes) V (G) = {X1,X2, . . . ,Xn} • And a set of arcs A(G) ⊆ V (G) × V (G), and where P is a joint probability distribution defined on the variables corresponding to the vertices V (G). Bayesian Network
- 6. A Bayesian Network consists of [Jensen, 1996]: A set of variables and a set of direct edges between variables. 1. Each variables has a finite set of mutually exclusive states 2. The variable and direct edge form a DAG (directed acyclic graph) 3. To each variable A with parents B1, B2 ..Bn there is attached a conditional probability table P(A| B1, B2 .. Bn)
- 7. p(A,B,C) = p(C|A,B)p(A)p(B) A B C FORMS OF THE BAYESIAN NETWORKS A CB Marginal Independence: p(A,B,C) = p(A) p(B) p(C) A Directed Acyclic Graph
- 8. A CB Conditionally independent effects: p(A,B,C) = p(B|A)p(C|A)p(A) B and C are conditionally independent Given A Each node in these graphs is a random variable Informally, an arrow from node X to node Y means X has a direct influence on Y A CB Markov dependence: p(A,B,C) = p(C|B) p(B|A)p(A)
- 9. A node X is a parent of another node Y if there is an arrow from node X to node Y eg. A is a parent of B & C P(A,B,C,D,E,F) = P(F|C,D,E)P(A,B,C,D,E) = P(F|C,D,E)P(C|A,E)P(D|B)P(E|B)P(B,A) = P(F|C,D,E)P(C|A,E)P(D|B)P(E|B)P(B|A)P(A)
- 10. EXAMPLE
- 11. Learning phase
- 12. Testing phase (inference)
- 14. Weng-Keen Wong, Oregon State University ©2005 14 Using a Bayesian Network Example Using the network in the example, suppose you want to calculate: P(A = true, B = true, C = true, D = true) = P(A = true) * P(B = true | A = true) * P(C = true | B = true) P( D = true | B = true) = (0.4)*(0.3)*(0.1)*(0.95) A B C D
- 15. Weng-Keen Wong, Oregon State University ©2005 15 Using a Bayesian Network Example Using the network in the example, suppose you want to calculate: P(A = true, B = true, C = true, D = true) = P(A = true) * P(B = true | A = true) * P(C = true | B = true) P( D = true | B = true) = (0.4)*(0.3)*(0.1)*(0.95) A B C D This is from the graph structure These numbers are from the conditional probability tables
- 16. Weng-Keen Wong, Oregon State University ©2005 16 Bayesian Networks Two important properties: 1. Encodes the conditional independence relationships between the variables in the graph structure 2. Is a compact representation of the joint probability distribution over the variables
- 17. Why Bayesian Networks? • — Bayesian Probability represents the degree of belief in that event while Classical Probability (or frequents approach) deals with true or physical probability of an event Bayesian Network Handling of Incomplete Data Sets Learning about Causal Networks Facilitating the combination of domain knowledge and data Efficient and principled approach for avoiding the over fitting of data.
- 18. CONCLUSION • Bayesian nets are a network-based framework for representing and analyzing models involving uncertainty • Used for the cross fertilization of ideas between the artificial intelligence, decision analysis, and statistic communities • People are using this nowadays because of the development of propagation algorithms followed by availability of easy to use commercial software. • And growing number of creative applications.