# 251likelihood estimatorsx22 and1 2 yields x x xi

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Unformatted text preview: 1 (Xi − X )2 ¯ 2 Xi ¯¯ ¯ =0 i (Xi − X )2 e (−− −− β1 Xi )(Xi ) Xi 0 X ) Xi X 2.50. k ( = σ ¯ (Yi β ¯ = ∂β1i XiXi −2 X )2 (X − X )2 Xi ¯ ¯¯ ¯ (Xi −i X )(Xi − X ) (Xi − X )X = = n¯ )21= 1 1 ¯ 2 because 2 1 =0 ∂ loge L(Xi − X ¯ ¯ − (Xi − −)X ) −−¯ = − (Xi 2 X )(Xi XYi¯− β0 − β1 Xi ) (Xi(XiX )X )2 + ( because ∂σ 2 = 2 σ 2 σ4 ¯ =0 ¯ −b ¯ ¯ (Xi)− to zero, simplifying, andXi − X )2 ( substituting the maximum 2.51. E {b0 }S= E {Y each1 X } (Xi −equal X )2 X2 ¯ etting derivative = = Prob. 2.51likelihood estimators−−¯X2),2 and1 σ 2 yields: (X X ) (Xi i b0 , ¯1 b ˆ 1 =¯b E {Yi } − XEi{− }¯ )2 =...
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## This note was uploaded on 10/29/2012 for the course STAT W4315 taught by Professor Martinalindquist during the Spring '12 term at Columbia.

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