Unformatted text preview: 416 Solutions Manual for Statistical Inference as is the variance,
2 2 Var(aX + bY ) = a2 VarX + b2 VarY + 2abCov(X, Y ) = a2 X + b2 Y + 2abX Y . To show that aX + bY is normal we have to do a bivariate transform. One possibility is U = aX + bY , V = Y , then get fU,V (u, v) and show that fU (u) is normal. We will do this in 1 the standard case. Make the indicated transformation and write x = a (u  bv), y = v and obtain 1 1/a b/a J = = . 0 1 a Then fU V (u, v) = 2a 1 12 e

1 2(12 ) 1 [ a (ubv)] 2 2 a (ubv)+v 2 . Now factor the exponent to get a square in u. The result is  b2 + 2ab + a2 1 2) 2(1 a2 b2 u2 2 + 2ab + a2 b + a b2 + 2ab + a2 uv + v 2 . 2 2 Note that this is joint bivariate normal form since U = V = 0, v = 1, u = a2 + b2 + 2ab and Cov(U , V ) E(aXY + bY 2 ) a + b = = = , U V U V a2 + b2 + ab thus (1  2 ) = 1  where a 12 = U fU V (u, v) = a2 2 + ab + b2 (12 )a (1  2 )a = 2 = 2 + b2 + 2ab 2 + 2ab 2 a a +b u 2 2 12 . We can then write 1 2U V 12 exp  1 2 12 u2 uv v2 2 + 2 2 U U V V , which is in the exact form of a bivariate normal distribution. Thus, by part a), U is normal. 4.46 a. EX VarX EY VarY Cov(X, Y ) aX EZ1 + bX EZ2 + EcX = aX 0 + bX 0 + cX = cX a2 VarZ1 + b2 VarZ2 + VarcX = a2 + b2 X X X X aY 0 + bY 0 + cY = cY a2 VarZ1 + b2 VarZ2 + VarcY = a2 + b2 Y Y Y Y EXY  EX EY 2 2 E[(aX aY Z1 + bX bY Z2 + cX cY + aX bY Z1 Z2 + aX cY Z1 + bX aY Z2 Z1 + bX cY Z2 + cX aY Z1 + cX bY Z2 )  cX cY ] = aX aY + bX bY , = = = = = = 2 2 since EZ1 = EZ2 = 1, and expectations of other terms are all zero. b. Simply plug the expressions for aX , bX , etc. into the equalities in a) and simplify. c. Let D = aX bY  aY bX =  12 X Y and solve for Z1 and Z2 , Z1 Z2 = = bY (XcX )  bX (Y cY ) D Y (XX )+X (Y Y ) 2(1)X Y . = Y (XX )+X (Y Y ) 2(1+)X Y ...
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics, Variance

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