This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 55  Homework 7 Solutions Wednesday, July 27, 2011 1. From the book, do problem 6.4.25 Let Y i be the number of tails minus number of heads on just the i th flip. So X n = Y 1 + ... + Y n . Now E ( Y i ) = 1 1 2 + ( 1) 1 2 = 0 and E ( Y 2 i ) = 1 2 1 2 + ( 1) 2 1 2 = 1 Also note that for i 6 = j , Y i and Y j are independent. (a) E ( X n ) = E ( Y 1 + ... + Y n ) = E ( Y 1 ) + ... + E ( Y n ) = 0 + ... + 0 = 0 (b) V ( X ) = E ( ( X E ( X )) 2 ) = E ( X 2 ) = E ( ( Y 1 + ... + Y n ) 2 ) = E n X i =1 Y 2 i + X i 6 = j Y i Y j ! = n X i =1 1 + X i 6 = j E ( Y i ) E ( Y j ) = n + 0 = n 2. From the book, do Chapter 6 Supplemental Excercise 20 (a) The bins are equally likely, so 1 b . (b) Each ball adds 1 b to a bin, so n b . (c) Since each time it lands in a bin with probability 1 n , It should take an expected 1 1 n = n tries. (d) Per the hint, the probability of landing in a new bin after i 1 bins have been hit is n ( i +1) n . So E ( X i ) = n n ( i +1) . So E ( X ) = E ( X 1 + ...X n ) = 1+ n n 1 + ... + n 2 + n = n 1 n + 1 n 1 + ... + 1 2 + 1 ≈ n ln n 3. From the book, do problem 6.2.393....
View
Full Document
 Summer '08
 STRAIN
 Math, Probability, Probability theory, 2 m, 2 m

Click to edit the document details