Dr. Hackney STA Solutions pg 182

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

Ask a homework question - tutors are online