lect2a - 1 2 Independence and Bernoulli Trials(Euler...

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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 = (2-1) ; , 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 (1-4) 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 ∩ = = = (2-2) PILLAI 3 • If P ( A ) = 0, then since the event always, we have and (2-1) 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 (2-3) PILLAI 4 • Let a union of n independent events. Then by De-Morgan’s law and using their independence Thus for any A as in (2-4) a useful result. We can use these results to solve an interesting number theory problem. , 3 2 1 n A A A A A ∪ ∪ ∪ ∪ = " (2-4) 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 " (2-5) , )) ( 1 ( 1 ) ( 1 ) ( 1 ∏ = − − = − = n i i A P A P A P (2-6) 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 (1-4). Also from (1-10) 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 (2-2) and (2-5) where we have used the Euler’s identity 1 1 See Appendix for a proof of Euler’s identity by Ramanujan....
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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.

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lect2a - 1 2 Independence and Bernoulli Trials(Euler...

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