By George Casella and Roger Berger

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24. P (W = 0) = P (Y ≤ X) = fXY (x, y) = 1 − 12 [(x−µ)2 +(y−γ)2 ] e 2σ 2πσ 2 (by independence, sofXY = fX fY ) Let u = x + y, v = x − y, then x = 12 (u + v), y = 12 (u − v) and 1/2 1/2 1/2 −1/2 |J| = = 1 . 2 The set {−∞ < x < ∞, −∞ < y < ∞} is mapped onto the set {−∞ < u < ∞, −∞ < v < ∞}. Therefore 2 2 u−v 1 − 2σ12 (( u+v 1 2 )−µ) +(( 2 )−γ ) fU V (u, v) = e · 2πσ 2 2 2 2 2 2 (µ+γ) (µ+γ) 1 − 2σ12 2( u2 ) −u(µ+γ)+ 2 +2( v2 ) −v(µ−γ)+ 2 e 4πσ 2 1 1 − 2(2σ − 1 2 2 2) = g(u) e (u − (µ + γ)) · h(v)e 2(2σ2 ) (v − (µ − γ)) .

EX VarX EY VarY Cov(X, Y ) = = = = = = aX EZ1 + bX EZ2 + EcX = aX 0 + bX 0 + cX = cX a2X VarZ1 + b2X VarZ2 + VarcX = a2X + b2X aY 0 + bY 0 + cY = cY a2Y VarZ1 + b2Y VarZ2 + VarcY = a2Y + b2Y EXY − EX · EY E[(aX aY Z12 + bX bY Z22 + 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 , since EZ12 = EZ22 = 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.

The inverse transformation from B to A1 is x1 = y2 y1 and x2 = y1 −y 1 y22 with Jacobian √ 1 √y2 2 √ y1 2 1−y 2 1 √ y1 2 J1 = y2 √ y1 √ = y1 1−y 22 1 2 1 − y22 . √ The inverse transformation from B to A2 is x1 = y2 y1 and x2 = − y1 −y 1 y22 with J2 = −J1 . 6), fY1 ,Y 2 (y1 , y2 ) is the sum of two terms, both of which are the same in this case. Then fY1 ,Y 2 (y1 , y2 ) = = 1 −y1 /(2σ2 ) 1 e 2πσ 2 2 1−y 22 1 −y1 /(2σ2 ) 1 e , 2πσ 2 1−y 22 2 0 < y1 < ∞, −1 < y2 < 1. b. We see in the above expression that the joint pdf factors into a function of y1 and a function of y2 .