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 Document
Unformatted text preview: 1 2. Independence and Bernoulli Trials (Euler, Ramanujan and Bernoulli Numbers) Independence : Events A and B are independent if • It is easy to show that A , B independent implies are all independent pairs. For example, and so that or i.e., and B are independent events. ). ( ) ( ) ( B P A P AB P = (21) ; , B A B A B A , ; , B A AB B A A B ∪ = ∪ = ) ( , φ = ∩ B A AB ), ( ) ( ) ( )) ( 1 ( ) ( ) ( ) ( ) ( B P A P B P A P B P A P B P B A P = − = − = ) ( ) ( ) ( ) ( ) ( ) ( ) ( B A P B P A P B A P AB P B A AB P B P + = + = ∪ = A PILLAI 2 As an application, let A p and A q represent the events and Then from (14) Also Hence it follows that A p and A q are independent events! "the prime divides the number " p A p N = "the prime divides the number ". q A q N = 1 1 { } , { } p q P A P A p q = = 1 { } {" divides "} { } { } p q p q P A A P pq N P A P A pq ∩ = = = (22) PILLAI 3 • If P ( A ) = 0, then since the event always, we have and (21) is always satisfied. Thus the event of zero probability is independent of every other event! • Independent events obviously cannot be mutually exclusive, since and A , B independent implies Thus if A and B are independent, the event AB cannot be the null set. • More generally, a family of events are said to be independent, if for every finite sub collection we have A AB ⊂ , ) ( ) ( ) ( = ⇒ = ≤ AB P A P AB P ) ( , ) ( > > B P A P . ) ( > AB P , , , , 2 1 n i i i A A A " ∏ = = = n k i n k i k k A P A P 1 1 ). ( ∩ { } i A (23) PILLAI 4 • Let a union of n independent events. Then by DeMorgan’s law and using their independence Thus for any A as in (24) a useful result. We can use these results to solve an interesting number theory problem. , 3 2 1 n A A A A A ∪ ∪ ∪ ∪ = " (24) n A A A A " 2 1 = . )) ( 1 ( ) ( ) ( ) ( 1 1 2 1 ∏ ∏ = = − = = = n i i n i i n A P A P A A A P A P " (25) , )) ( 1 ( 1 ) ( 1 ) ( 1 ∏ = − − = − = n i i A P A P A P (26) PILLAI 5 Example 2.1 Two integers M and N are chosen at random. What is the probability that they are relatively prime to each other? Solution: Since M and N are chosen at random, whether p divides M or not does not depend on the other number N . Thus we have where we have used (14). Also from (110) Observe that “ M and N are relatively prime” if and only if there exists no prime p that divides both M and N. 2 {" divides both and "} 1 {" divides "} {" divides "} P p M N P p M P p N p = = 2 {" does divede both and "} 1 1 {" divides both and "} 1 P p not M N P p M N p = − = − PILLAI 6 Hence where X p represents the event Hence using (22) and (25) where we have used the Euler’s identity 1 1 See Appendix for a proof of Euler’s identity by Ramanujan....
View
Full
Document
This note was uploaded on 09/01/2011 for the course ISYE 6414 taught by Professor Staff during the Fall '08 term at Georgia Tech.
 Fall '08
 Staff

Click to edit the document details