This preview shows pages 1–9. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Stat 302, Introduction to Probability Jiahua Chen JanuaryApril 2011 Jiahua Chen () Lecture 4 JanuaryApril 2011 1 / 22 Independence of two events Given a random experiment and its sample space, there exist many events. We have reasons to believe that some events are independent of each other by instinct. Independence means that the occurrence of one is not an indication that the other one becomes more or less likely to occur. Jiahua Chen () Lecture 4 JanuaryApril 2011 2 / 22 Example Random Experiments: toss two dice. Let E 1 = the first die shows 2. Let E 2 = the second die shows 2. Unless you are psychic, you should not think that these two events will influence each other. In probability theory, we have a rigorous definition. Jiahua Chen () Lecture 4 JanuaryApril 2011 3 / 22 Definition of Independence (2 Events) The events E 1 and E 2 are independent if and only if P ( E 1 E 2 ) = P ( E 1 ) P ( E 2 ) Back to the previous example: E 1 = { ( 2, 1 ) , ( 2, 2 ) , . . . , ( 2, 6 ) } and E 2 = { ( 1, 2 ) , ( 2, 2 ) , . . . , ( 6, 2 ) } It is seen that P ( E 1 ) = P ( E 2 ) = 1 / 6 and P ( E 1 E 2 ) = 1 / 36. Thus, E 1 and E 2 are independent. Jiahua Chen () Lecture 4 JanuaryApril 2011 4 / 22 A less obvious example of independence Consider the same experiment of tossing two dice. Let E 3 = the sum of two readings is 7. We easily find P ( E 3 ) = 6 / 36 = 1 / 6 and that P ( E 1 E 3 ) = P ( { ( 2, 5 ) } ) = 1 / 36. Coupled with P ( E 1 ) = 1 / 6, we find that E 1 and E 3 are independent. Jiahua Chen () Lecture 4 JanuaryApril 2011 5 / 22 A less obvious example of independence Consider the same experiment of tossing two dice. Let E 3 = the sum of two readings is 7. We easily find P ( E 3 ) = 6 / 36 = 1 / 6 and that P ( E 1 E 3 ) = P ( { ( 2, 5 ) } ) = 1 / 36. Coupled with P ( E 1 ) = 1 / 6, we find that E 1 and E 3 are independent. Did you think so before this calculation? Can you convince yourself that the definition is not a problem. Jiahua Chen () Lecture 4 JanuaryApril 2011 5 / 22 Beat this example to death Consider the same experiment of tossing two dice. Let E 4 = E 1 E 2 . We find P ( E 3 ) = 1 / 6; and P ( E 4 ) = 11 / 36. At the same time, P ( E 3 E 4 ) = 2 / 36. Jiahua Chen () Lecture 4 JanuaryApril 2011 6 / 22 Beat this example to death Consider the same experiment of tossing two dice....
View
Full
Document
This note was uploaded on 04/07/2011 for the course STAT 302 taught by Professor Dr.chen during the Spring '11 term at The University of British Columbia.
 Spring '11
 Dr.Chen
 Probability

Click to edit the document details