Dr. Hackney STA Solutions pg 58

# Dr. Hackney STA Solutions pg 58 - 4-14 Solutions Manual for...

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Unformatted text preview: 4-14 Solutions Manual for Statistical Inference ( = pj-1 pj+1 pn )xj-1 ( )xj+1 ( )xn 1 - pj - p i 1 - pj - p i 1 - pj - p i 1- pi 1 - pj m-xi -xj (m - xj )! pi xi ( ) xi !(m - xi - xj )! 1 - pj . pi Thus Xi |Xj = xj binomial(m - xj , 1-pj ). b. f (xi , xj ) = f (xi |xj )f (xj ) = m! x pxi p j (1 - pj - pi )m-xj -xi . xi !xj !(m - xj - xi )! i j Using this result it can be shown that Xi + Xj binomial(m, pi + pj ). Therefore, Var(Xi + Xj ) = m(pi + pj )(1 - pi - pj ). By Theorem 4.5.6 Var(Xi + Xj ) = Var(Xi ) + Var(Xj ) + 2Cov(Xi , Xj ). Therefore, Cov(Xi , Xj ) = 1 1 [m(pi +pj )(1-pi -pj )-mpi (1-pi )-mpi (1-pi )] = (-2mpi pj ) = -mpi pj . 2 2 4.41 Let a be a constant. Cov(a, X) = E(aX) - EaEX = aEX - aEX = 0. 4.42 XY,Y = Cov(XY, Y ) E(XY )-XY Y EXEY 2 -X Y Y = = , XY Y XY Y XY Y 2 where the last step follows from the independence of X and Y. Now compute 2 XY = E(XY )2 - [E(XY )]2 = EX 2 EY 2 - (EX)2 (EY )2 2 2 2 2 2 2 = (X + 2 )(Y + 2 ) - 2 2 = X Y + X 2 + Y 2 . X Y X Y Y X Therefore, XY,Y = 4.43 Cov(X1 + X2 , X2 + X3 ) = E(X1 + X2 )(X2 + X3 ) - E(X1 + X2 )E(X2 + X3 ) = (42 + 2 ) - 42 = 2 2 2 Cov(X 1 +X 2 )(X 1 -X 2 ) = E(X1 + X2 )(X1 - X2 ) = EX1 - X2 = 0. 2 X (Y +2 )-X 2 Y Y 2 2 (X Y 2 +X 2 Y 2 +Y 1/2 2 ) Y X = X Y 2 (2 Y X +2 Y 2 2 2 X +X Y ) 1/2 . 4.44 Let i = E(Xi ). Then n Var i=1 Xi = = = = Var (X1 + X2 + + Xn ) E [(X 1 + X2 + + Xn ) - (1 + 2 + + n )] E [(X 1 -1 ) + (X 2 -2 ) + + (X n -n )] n 2 2 E(Xi - i )2 + 2 i=1 n 1i<jn E(Xi - i )(Xj - j ) = i=1 VarXi + 2 1i<jn Cov(Xi , Xj ). ...
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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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