Dr. Hackney STA Solutions pg 56

Dr. Hackney STA Solutions pg 56 - 4-12 Solutions Manual for...

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Unformatted text preview: 4-12 Solutions Manual for Statistical Inference n( + + 1) n - + n2 VarP 2 )( + + 1) 2 )( + + 1) ( + ( + = n - nVarP + n2 VarP + + = nEP (1 - EP ) + n(n - 1)VarP. = 1 b. Var(Y ) = E(Var(Y |)) + Var(E(Y |)) = E + Var() = + 2 since E = = and Var() = 2 = 4.37 a. Let Y = Xi . P (Y = k) ()2 = 2 . The "extra-Poisson" variation is 1 2 . = P (Y = k, = = 0 1 1 1 < c = (1 + p) < 1) 2 2 1 1 (Y = k|c = (1 + p))P (P = p)dp 2 0 1 n 1 1 (a + b) a-1 [ (1 + p)]k [1 - (1 + p)]n-k p (1 - p)b-1 dp k 2 2 (a)(b) n (1 + p)k (1 - p)n-k (a + b) a-1 p (1 - p)b-1 dp k 2k 2n-k (a)(b) k 0 1 = 0 = n (a + b) k 2n (a)(b) n (a + b) k 2n (a)(b) k k j 2n pk+a-1 (1 - p)n-k+b-1 dp j=0 k = j=0 k (k + a)(n - k + b) j (n + a + b) . = j=0 n (a + b) (k + a)(n - k + b) k (a)(b) (n + a + b) b. A mixture of beta-binomial. EY = E(E(Y |c)) = E[nc] = E n Using the results in Exercise 4.35(a), Var(Y ) = nEC(1 - EC) + n(n - 1)VarC. Therefore, Var(Y ) = nE = = 1 1 (1 + P ) + n(n - 1)Var (1 + P ) 2 2 n n(n - 1) (1 + EP )(1 - EP ) + VarP 4 4 2 n a n(n - 1) ab 1- + . 2 (a + b + 1) 4 a+b 4 (a + b) 1-E x 1 (1 + p) 2 = n 2 1+ a a+b . 1 (1 + P ) 2 4.38 a. Make the transformation u = 0 - x , du = -x 2 d, - = x u . Then 1 -x/ 1 r-1 d e (r)(1 - r) (-)r ...
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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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