{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 2 Solutions

# Homework 2 Solutions - Homework 2 solution X1 X3 = = so its...

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

Homework 2 – solution 1. a) E X 1 + X 3 2 · = μ + μ 2 = μ , so it’s unbiased. b) E X 1 + X 3 + X n 3 · = μ + μ + μ 3 = μ , so it’s unbiased. c) E X 1 + X 3 +3 X 2 - X 4 5 · = μ + μ +3 μ - μ 5 = 4 μ 5 , so it’s biased with bias equal to 4 μ 5 - μ = - 1 5 μ . d) E (3 X 1 - 2 X 3 ) = 3 μ - 2 μ = μ , so it’s unbiased. e) E X 1 - X 2 + X n/ 2 - X n - 1 + X n · = μ - μ + μ - μ + μ = μ , so it’s unbiased. 2. a) V X 1 + X 3 2 · = σ 2 + σ 2 4 = σ 2 2 . Since it’s unbiased, MSE X 1 + X 3 2 · = V X 1 + X 3 2 · = σ 2 2 . b) V X 1 + X 3 + X n 3 · = σ 2 + σ 2 + σ 2 9 = σ 2 3 . Since it’s unbiased, MSE X 1 + X 3 + X n 3 · = V X 1 + X 3 + X n 3 · = σ 2 + σ 2 + σ 2 9 = σ 2 3 . c) V X 1 + X 3 +3 X 2 - X 4 5 · = σ 2 + σ 2 +9 σ 2 + σ 2 25 = 12 σ 2 25 . So the mean square error is MSE ( X 1 + X 3 +3 X 2 - X 4 5 ) = 12 σ 2 25 + μ 2 25 = 13 σ 2 25 . d) V (3 X 1 - 2 X 3 ) = 9 σ 2 +4 σ 2 = 13 σ 2 . Since it’s unbiased, MSE (3 X 1 - 2 X 3 ) = V (3 X 1 - 2 X 3 ) = 13 σ 2 . e) V X 1 - X 2 + X n/ 2 - X n - 1 + X n · = 5 σ 2 . So MSE X 1 - X 2 + X n/ 2 - X n - 1 + X n · = 5 σ 2 as well.

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.

{[ snackBarMessage ]}

### Page1 / 2

Homework 2 Solutions - Homework 2 solution X1 X3 = = so its...

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

View Full Document
Ask a homework question - tutors are online