Dr. Hackney STA Solutions pg 83

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

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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...
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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.

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