Dr. Hackney STA Solutions pg 182

# Dr. Hackney STA Solutions pg 182 - 11-10 Solutions Manual...

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Unformatted text preview: 11-10 Solutions Manual for Statistical Inference since the cross terms have expectation zero. Next, expanding the product in the second term again gives all zero cross terms, and we have 1 2 E(Yi Yi ) = ( + i )( + i ) + 2 (rB ), r and 2 Cov(Yi , Yi ) = B /r. Finally, this gives n Var i=1 ai Yi = 1 r2 1 r n 2 a2 (rB + r 2 ) + 2 i i=1 n 2 a2 2 + B ( i i=1 n i=1 n i>i 2 ai ai B /r = = = ai )2 1 2 a2 r i=1 i 1 2 2 ( + B )(1 - ) a2 , i r i=1 i n where, in the third equality we used the fact that 11.25 Differentiation yields a. c RSS d RSS ai = 0. =2 [yi - (c+dxi )] (-1) = 0 nc + d set set xi = x2 i yi = xi yi . = 2 [yi - (ci +dxi )] (-xi ) = 0 c xi + d b. Note that nc + d xi = yi c = y - d. Then x ( - d) y x xi + d x2 = i xi yi and d x2 - n2 = x i xi yi - xi y which simplifies to d = xi (yi - y )/ estimates. c. The second derivatives are 2 RSS = n, c2 (xi - x)2 . Thus c and d are the least squares 2 RSS = cd xi , 2 RSS = d2 x2 . i Thus the Jacobian of the second-order partials is n xi 11.27 For the linear estimator E i i xi x2 i =n x2 - i 2 xi =n (xi - x)2 > 0. ai Yi to be unbiased for we have ai ( + xi ) = i i ai Yi = ai = 1 and i ai xi = 0. Since Var i ai Yi = 2 i a2 , we need to solve: i a2 subject to i i i minimize ai = 1 and i ai xi = 0. ...
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