Dr. Hackney STA Solutions pg 82

Dr. Hackney STA Solutions pg 82 - 5-16 Solutions Manual for...

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Unformatted text preview: 5-16 Solutions Manual for Statistical Inference 5.55 Let X denote the number of comparisons. Then EX = k=0 P (X > k) = 1 + k=1 P (U > Fy (yk-1 )) = 1+ k=1 (1 - Fy (yk-1 )) = 1 + k=0 (1 - Fy (yi )) = 1 + EY 5.57 a. Cov(Y1 , Y2 ) = Cov(X1 + X3 , X2 + X3 ) = Cov(X3 , X3 ) = 3 since X1 , X2 and X3 are independent. b. Zi = 1 0 if Xi = X3 = 0 otherwise pi = P (Zi = 0) = P (Yi = 0) = P (Xi = 0, X3 = 0) = e-(i +3 ) . Therefore Zi are Bernoulli(pi ) with E[Zi ] = pi , Var(Zi ) = pi (1 - pi ) and E[Z1 Z2 ] = P (Z1 = 1, Z2 = 1) = P (Y1 = 0, Y2 = 0) = P (X1 + X3 = 0, X2 + X3 = 0) = P (X1 = 0)P (X2 = 0)P (X3 = 0) = e-1 e-2 e-3 . Therefore, Cov(Z1 , Z2 ) = E[Z1 Z2 ] - E[Z1 ]E[Z2 ] = e-1 e-2 e-3 - e-(i +3 ) e-(2 +3 ) = e-(i +3 ) e-(2 +3 ) (e3 - 1) = p1 p2 (e3 - 1). p1 p2 (e3 -1) Thus Corr(Z1 , Z2 ) = c. E[Z1 Z2 ] pi , therefore p1 (1-p1 ) p2 (1-p2 ) . Cov(Z1 , Z2 ) = E[Z1 Z2 ] - E[Z1 ]E[Z2 ] p1 - p1 p2 = p1 (1 - p2 ), and Cov(Z1 , Z2 ) p2 (1 - p1 ). Therefore, Corr(Z1 , Z2 ) and Corr(Z1 , Z2 ) which implies the result. 5.59 P (Y y) = P (V y|U < = y 0 1 c fY p1 (1 - p2 ) p1 (1 - p1 ) p2 (1 - p2 ) p2 (1 - p1 ) p1 (1 - p1 ) p2 (1 - p2 ) = p1 (1 - p2 ) p2 (1 - p1 ) p2 (1 - p1 ) p1 (1 - p2 ) = 1 fY (V )) = c = 1 c y 0 P (V y, U < 1 fY (V )) c P (U < 1 fY (V )) c fY (v)dv 1 c y (v) 0 1 c dudv = 0 fY (v)dv 5.61 a. M = supy (a+b) y a-1 (1-y)b-1 (a)(b) ([a]+[b]) y [a]-1 (1-y)[b]-1 ([a])([b]) < , since a - [a] > 0 and b - [b] > 0 and y (0, 1). ...
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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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