Dr. Hackney STA Solutions pg 83

# Dr. Hackney STA Solutions pg 83 - Second Edition 5-17 b M =...

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

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

Unformatted text preview: Second Edition 5-17 b. M = sup y Γ( a + b ) Γ( a )Γ( b ) y a- 1 (1- y ) b- 1 Γ([ a ]+ b ) Γ([ a ])Γ( b ) y [ a ]- 1 (1- y ) b- 1 < ∞ , since a- [ a ] > 0 and y ∈ (0 , 1). c. M = sup y Γ( a + b ) Γ( a )Γ( b ) y a- 1 (1- y ) b- 1 Γ([ a ]+1+ β ) Γ([ a ]+1)Γ( b ) y [ a ]+1- 1 (1- y ) b- 1 < ∞ , since a- [ a ]- 1 < 0 and y ∈ (0 , 1). b- b > when b = [ b ] and will be equal to zero when b = b , thus it does not affect the result. d. Let f ( y ) = y α (1- y ) β . Then df ( y ) dy = αy α- 1 (1- y ) β- y α β (1- y ) β- 1 = y α- 1 (1- y ) β- 1 [ α (1- y ) + βy ] which is maximize at y = α α + β . Therefore for, α = a- a and β = b- b M = Γ( a + b ) Γ( a )Γ( b ) Γ( a + b ) Γ( a )Γ( b ) a- a a- a + b- b a- a b- b a- a + b- b b- b . We need to minimize M in a and b . First consider a- a a- a + b- b a- a b- b a- a + b- b b- b . Let c = α + β , then this term becomes ( α c ) α ( c- α c ) c- α . This term is maximize at α c = 1 2 , this is at α = 1 2 c...
View Full Document

## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

Ask a homework question - tutors are online