Dr. Hackney STA Solutions pg 193

Dr. Hackney STA Solutions pg 193 - Second Edition 12-3 Thus...

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Second Edition 12-3 Thus r 2 λ ( β ) 1 - r 2 λ ( β ) = y λ 2 S 2 xy ± β - ˆ β ² 2 ± β + 1 λ ˆ β ² 2 (1 - λβ 2 ) 2 ( S yy S x - S 2 xy ) = ± β - ˆ β ² 2 ˆ σ 2 β ³ 1+ λβ ˆ β 1+ λβ 2 ! 2 (1 + λ ˆ β 2 ) 2 S 2 xy ˆ β 2 h ( S xx - λS yy ) 2 + 4 λS 2 xy i , after substituting ˆ σ 2 β from page 588. Now using the fact that ˆ β and - 1 ˆ β are both roots of the same quadratic equation, we have (1+ λ ˆ β 2 ) 2 ˆ β 2 = ´ 1 ˆ β + λ ˆ β µ 2 = ( S xx - λS yy ) 2 +4 λS 2 xy S 2 xy . Thus the expression in square brackets is equal to 1. 12 . 15 a. π ( - α/β ) = e α + β ( - α/β ) 1 + e α + β ( - α/β ) = e 0 1 + e 0 = 1 2 . b. π (( - α/β ) + c ) = e α + β (( - α/β )+ c ) 1 + e α + β (( - α/β )+ c ) = e βc 1 + e βc , and 1 - π (( - α/β ) - c ) = 1 - e - βc 1 + e - βc = e βc 1 + e βc . c. d dx π ( x ) = β e α + βx [1 + e α + βx ] 2 = βπ ( x )(1 - π ( x )) . d. Because π ( x ) 1 - π ( x ) = e α + βx , the result follows from direct substitution. e. Follows directly from (d). f. Follows directly from ∂α F ( α + βx ) = f ( α + βx ) and ∂β F ( α + βx ) = xf ( α + βx ) . g. For F ( x ) = e x / (1 + e x ), f ( x ) = F ( x )(1 - F ( x )) and the result follows. For
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