Dr. Hackney STA Solutions pg 10

# Dr. Hackney STA Solutions pg 10 - hired on the i th trial...

This preview shows page 1. Sign up to view the full content.

Second Edition 1-7 has probability n ! n n . Any other unordered outcome from { x 1 ,...,x n } , distinct from the un- ordered sample { x 1 ,...,x n } , will contain m diﬀerent numbers repeated k 1 ,...,k m times where k 1 + k 2 + ··· + k m = n with at least one of the k i ’s satisfying 2 k i n . The probability of obtaining the corresponding average of such outcome is n ! k 1 ! k 2 ! ··· k m ! n n < n ! n n , since k 1 ! k 2 ! ··· k m ! > 1 . Therefore the outcome with average x 1 + x 2 + ··· + x n n is the most likely. b. Stirling’s approximation is that, as n → ∞ , n ! 2 πn n +(1 / 2) e - n , and thus ± n ! n n ²³ ´ 2 e n ! = n ! e n n n 2 = 2 πn n +(1 / 2) e - n e n n n 2 = 1 . c. Since we are drawing with replacement from the set { x 1 ,...,x n } , the probability of choosing any x i is 1 n . Therefore the probability of obtaining an ordered sample of size n without x i is (1 - 1 n ) n . To prove that lim n →∞ (1 - 1 n ) n = e - 1 , calculate the limit of the log. That is lim n →∞ n log ± 1 - 1 n ² = lim n →∞ log ( 1 - 1 n ) 1 /n . L’Hˆ opital’s rule shows that the limit is - 1, establishing the result. See also Lemma 2.3.14. 1.32 This is most easily seen by doing each possibility. Let P ( i ) = probability that the candidate
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: hired on the i th trial is best. Then P (1) = 1 N , P (2) = 1 N-1 , ... ,P ( i ) = 1 N-i + 1 , ... ,P ( N ) = 1 . 1.33 Using Bayes rule P ( M | CB ) = P ( CB | M ) P ( M ) P ( CB | M ) P ( M ) + P ( CB | F ) P ( F ) = . 05 1 2 . 05 1 2 + . 0025 1 2 = . 9524 . 1.34 a. P (Brown Hair) = P (Brown Hair | Litter 1) P (Litter 1) + P (Brown Hair | Litter 2) P (Litter 2) = 2 3 1 2 + 3 5 1 2 = 19 30 . b. Use Bayes Theorem P (Litter 1 | Brown Hair) = P ( BH | L 1) P ( L 1) P ( BH | L 1) P ( L 1) + P ( BH | L 2) P ( L 2 = ( 2 3 )( 1 2 ) 19 30 = 10 19 . 1.35 Clearly P ( | B ) 0, and P ( S | B ) = 1. If A 1 ,A 2 ,... are disjoint, then P [ i =1 A i B ! = P ( S i =1 A i B ) P ( B ) = P ( S i =1 ( A i B )) P ( B ) = i =1 P ( A i B ) P ( B ) = X i =1 P ( A i | B ) ....
View Full Document

## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

Ask a homework question - tutors are online