# hwsol11 - M362K Probability Homework Solutions Homework 11...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: M362K Probability Homework Solutions Homework 11: Due November 24 Chapter 7 theoretical exercises 50, 54 Chapter 8, problems 2, 3, 7, 11, 15, 16, theoretical exercise 9. Theoretical exercise 7.50 Since Ψ( t ) = log( M ( t )), Ψ ′ ( t ) = M ′ ( t ) /M ( t ), and Ψ ′′ ( t ) = M ( t ) M ′′ ( t ) − M ′ ( t ) 2 M ( t ) 2 . Evaluating at t = 0 gives Ψ ′′ (0) = 1 · E ( X 2 ) − ( E ( X )) 2 1 2 = V ar ( X ). This result can also be understood using the Taylor series log(1 + x ) = x- x 2 / 2 + x 3 / 3 + ··· . Since M ( t ) = 1 + tE ( X ) + t 2 E ( X 2 ) / 2+higher powers of t , we have Ψ( t ) = ( tE ( X ) + t 2 E ( X 2 ) / 2)- t 2 E ( X ) 2 / 2+higher powers of t . Thus Ψ ′ (0) = E ( X ) and Ψ ′′ (0) = E ( X 2 )- E ( X ) 2 = V ar ( X ). Theoretical exercise 7.54 Cov ( Z,Z 2 ) = E ( Z 3 )- E ( Z ) E ( Z 2 ) = 0- × 1 = 0. Problem 8.2 (a) Let X be the score of a random student. Since E ( X ) = 75, P ( X > 85) ≤ P ( X ≥ 85) ≤ 75 / 85, by Markov’s inequality. However, we can do much better using Chebyshev’s inequality. P ( X > 85) ≤ P ( | X- 75 | > 10) ≤ V ar ( X ) / (10) 2 = 1 / 4. (b) By the same reasoning, the probability of being between 65 and 85 is at least 3/4. (c) If we average n scores, then ¯ X n has a mean of 75, and a variance of 25 /n . The probability of being within 5 of 75 is then at least 1....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

hwsol11 - M362K Probability Homework Solutions Homework 11...

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

View Full Document
Ask a homework question - tutors are online