Dr. Hackney STA Solutions pg 50

# Dr. Hackney STA Solutions pg 50 - 4-6 Solutions Manual for...

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Unformatted text preview: 4-6 Solutions Manual for Statistical Inference Then, fU (u) = = = = (++) -1 u ()()() 1 v -1 (1 - v)-1 ( u v-u -1 ) dv v y= dv v-u , dy = 1-u 1-u 1 (++) -1 u (1 - u)+-1 y -1 (1 - y)-1 dy ()()() 0 ()() (++) -1 u (1 - u)+-1 ()()() (+) (++) -1 u (1 - u)+-1 , 0 < u < 1. ()(+) Thus, U gamma(, + ). b. Let x = uv, y = u then v J= x u y u x v x v = 1 1/2 -1/2 u 2v 1 -1/2 -1/2 v u 2 1 1/2 -1/2 v 2u - 1 u1/2 v -3/2 2 = 1 . 2v u v -1 fU,V (u, v) = ( + + ) -1 ( uv (1 - uv)-1 ()()() u v +-1 1- 1 . 2v 1 The set {0 < x < 1, 0 < y < 1} is mapped onto the set {0 < u < v < u , 0 < u < 1}. Then, fU (u) 1/u = u fU,V (u, v)dv ( + + ) -1 u (1-u)+-1 ()()() 1/u u = 1 - uv 1-u -1 1 - u/v 1-u -1 ( u/v) dv. 2v(1 - u) Call it A u/v u/v-u To simplify, let z = 1-u . Then v = u z = 1, v = 1/u z = 0 and dz = - 2(1-u)v dv. Thus, fU (u) = A = = z -1 (1 - z)-1 dz ( kernel of beta(, )) (++) -1 ()() u (1 - u)+-1 ()()() (+) (++) -1 u (1 - u)+-1 , 0 < u < 1. ()(+) That is, U beta(, + ), as in a). x 4.24 Let z1 = x + y, z2 = x+y , then x = z1 z2 , y = z1 (1 - z2 ) and |J| = x z1 y z1 x z2 y z2 = z2 1-z 2 z1 -z 1 = z1 . The set {x > 0, y > 0} is mapped onto the set {z1 > 0, 0 < z2 < 1}. fZ1 ,Z2 (z 1 , z2 ) = = 1 (z1 z2 )r-1 e-z1 z2 (r) 1 z r+s-1 e-z1 (r+s) 1 1 (z1 - z1 z2 )s-1 e-z1 +z1 z2 z1 (s) (r+s) r-1 z (1 - z2 )s-1 , 0 < z1 , 0 < z2 < 1. (r)(s) 2 ...
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