100BHWS3

100BHWS3 - STAT 100B Solution III Problem 1. Pr(X (t, t +...

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

View Full Document Right Arrow Icon
STAT 100B Solution III Problem 1. Pr( X ( t,t + Δ t )) = (1 - λ Δ t ) t/ Δ t λ Δ t . So Pr( X ( t,t + Δ t )) / Δ t λe λt . E ( X ) = i 0 xλe λx dx = - i 0 xde λx = - xe λx | 0 + i 0 e λx dx = - e λx | 0 = 1 . E ( X 2 ) = i 0 x 2 λe λx dx = - i 0 x 2 de λx = - x 2 e λx | 0 + i 0 e λx dx 2 = 2 i 0 xλe λx dx/λ = 2 2 . Pr( X > t ) = i t λe λx dx = e λt . Problem 2. (1) 1 = ( X 1 + ... + X n ) /n . ˆ λ = n/ n i =1 X i = 1 . 0581. (2) 2 2 = ( X 2 1 + ... + X 2 n ) /n . ˆ λ = r 2 n/ n i =1 X 2 i = 1 . 3000. (3) e λt = n i =1 1 X i >t /n . ˆ λ = - log( n i =1 1 X i >t /n ) /t = 0 . 9163. Problem 3. Estimator (1): 1.0581 1.5190 1.3381 1.0036 1.1160, 5 k =1 ( ˆ λ k - 1) 2 / 5 = 0 . 0801 Estimator (2): 1.3000 1.7728 1.2955 0.9492 1.1376, 5 k =1 ( ˆ λ k - 1) 2 / 5 = 0 . 1592 Estimator (3): 0.9163 1.6094 1.2040 1.6094 1.6094, 5 k =1 ( ˆ λ k - 1) 2 / 5 = 0 . 2326 Problem 4. (1) h ( X ) = X . (2) h ( X ) = X 2 . (3) h ( X ) = 1 X>t . l ( λ ) = n log λ - λ n i =1 X i . l ( λ ) = n/λ - n i =1 X i = 0. ˆ λ = n/ n i =1 X i . Problem 6. (1) l ( λ ) = m log λ - λ ( m i =1 x i ) - λ ( n - m ) T . l ( λ ) = m/λ
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/30/2011 for the course STAT 100B taught by Professor Wu during the Winter '11 term at UCLA.

Ask a homework question - tutors are online