100BHWS2

100BHWS2 - , ] ( x i ) = 1 n I ( x max ) = x max Problem 7...

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

View Full Document Right Arrow Icon
STAT 100B HW2 Solution Problem 1 Bernoulli f ( x ) = p x (1 - p ) 1 - x ,x ∈ { 0 , 1 } L ( p ) = Q n i =1 f ( x i ) = p x i (1 - p ) n - x i l ( p ) = x i ln p + ( n - x i )ln(1 - p ) ∂l ( p ) ∂p = x i p - n - x i 1 - p = 0 x i (1 - p ) = np - p x i ˆ p = ¯ x Problem 2 Normal f ( x ) = 1 2 πσ 2 e - ( x - μ ) 2 2 σ 2 ,x R L ( μ,σ 2 ) = Q n i =1 f ( x i ) = (2 πσ 2 ) - n 2 e - ( x i - μ ) 2 2 σ 2 l ( μ,σ 2 ) = - n 2 ln(2 πσ 2 ) - ( x i - μ ) 2 2 σ 2 ∂l ( μ,σ 2 ) ∂μ = ( x i - μ ) 2 2 σ 2 = 0 ˆ μ = ¯ x ∂l ( μ,σ 2 ) ∂σ 2 = - n 2 σ 2 + ( x i - μ ) 2 2 σ 4 = 0 ˆ σ 2 = ( x i - ˆ μ ) 2 n = ( x i - ¯ x ) 2 n Problem 3 Poisson f ( x ) = e - λ λ x x ! ,x ∈ { 0 , 1 , 2 , ···} L ( λ ) = e - λ x i Q x i ! l ( λ ) = - + x i ln λ - ln( Q x i !) ∂l ( λ ) ∂λ = - n + x i λ = 0 ˆ λ = ¯ x Problem 4 Exponential f ( x ) = λe - λx ,x 0 L ( λ ) = λ n e - λ x i l ( λ ) = n ln λ - λ x i ( λ ) ∂λ = n λ - x i = 0 ˆ λ = 1 ¯ x Problem 5 Geometric f ( x ) = (1 - p ) x - 1 p,x ∈ { 1 , 2 , 3 , ···} L ( p ) = (1 - p ) x i - n p n l ( p ) = ( x i - n )ln(1 - p ) + n ln p ∂l ( p ) ∂p = ( x i - n )( - 1) 1 - p + n p = 0 ˆ p = 1 ¯ x Problem 6 Uniform f ( x ) = 1 θ ,x [0 ] L ( θ ) = 1 θ n Q I [0
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , ] ( x i ) = 1 n I ( x max ) = x max Problem 7 f ( y ) = 1 2 2 e-( y-x ) 2 2 2 L ( , 2 ) = (2 2 )-n 2 e- ( y i-x i ) 2 2 2 l ( , 2 ) =-n 2 ln(2 2 )- ( y i-x i ) 2 2 2 l ( , 2 ) = 2 ( y i-x i ) x i 2 2 = 0 = x i y i x 2 i l ( , 2 ) 2 =-n 2 2 + ( y i-x i ) 2 2 4 = 0 2 = ( y i- x i ) 2 n...
View Full 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