so7 - Math 55 Homework 7 Solutions Wednesday 1 From the...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the 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

This note was uploaded on 03/14/2012 for the course MATH 55 taught by Professor Strain during the Summer '08 term at Berkeley.

Page1 / 3

so7 - Math 55 Homework 7 Solutions Wednesday 1 From the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online