Dr. Hackney STA Solutions pg 72

Dr. Hackney STA Solutions pg 72 - 5-6 Solutions Manual for...

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Unformatted text preview: 5-6 Solutions Manual for Statistical Inference Note that the pdf factors, showing that X and Y are independent, and we can read off the pdfs of each: X has the F distribution and Y is 2 . If we integrate out y to recover the p+q proper constant, we get the F pdf fX (x) = p 2 p+q 2 q 2 q p p/2 xp/2-1 1+ q px p+q 2 . b. Since Fp,q = 2 /p p 2 /q , q let U 2 , V 2 and U and V are independent. Then we have p q = = E U/p V /q p 1 qE p V = E U p E q V (by independence) (EU = p). EFp,q Then E 1 V = 0 1 v 1 1 q 2 v 2 -1 e- 2 dv = 2q/2 q-2 2 2(q-2)/2 = q v 1 q 2 2q/2 q-2 2 q-2 2 v 0 q-2 2 -1 e- 2 dv = 1 . q-2 v = Hence, EFp,q = q 2 2q/2 q q-2 , 2(q-2)/2 q-2 2 2q/2 p q p q-2 = if q > 2. To calculate the variance, first calculate U 2 q2 p2 V 2 = q2 E(U 2 )E p2 1 V2 . 2 E(Fp,q ) = E Now E(U 2 ) = Var(U ) + (EU )2 = 2p + p2 and E Therefore, 2 EFp,q = 1 V2 = 0 1 1 1 v (q/2)-1 e-v/2 dv = . 2 (q/2) 2q/2 v (q - 2)(q - 4) 1 q2 (p + 2) q2 p(2 + p) = , 2 p (q - 2)(q - 4) p (q - 2)(q - 4) and, hence Var(Fp,q ) = c. Write X = d. Let Y = U/p V /p q 2 (p + 2) q2 =2 - p(q - 2)(q - 4) (q - 2)2 1 X q q-2 2 q+p-2 p(q - 4) , q > 4. then = = V /q U/p Fq,p , since U 2 , V 2 and U and V are independent. p q qY p(1-Y ) (p/q)X 1+(p/q)X pX q+pX , so X = and p 2 dx dy q = p (1 - y)-2 . Thus, Y has pdf p-2 2 fY (y) = B p 2 q+p 2 p q , 2 2 q 2 -1 p q p qy p(1-y) q p+q 2 1+ p qy q p(1-y) q p(1 - y) 2 = y 2 -1 (1 - y) 2 -1 beta p q , . 2 2 ...
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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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