Dr. Hackney STA Solutions pg 61

Dr. Hackney STA Solutions pg 61 - Second Edition 4-17 Then...

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Unformatted text preview: Second Edition 4-17 Then the Jacobian is J= z1 x1 z2 x z1 y z2 y = bY D -aY D -bX D aX D = aY bX 1 1 aX bY - = = , D2 D2 D - 1-2 X Y and we have that fX,Y (x, y) = = 1 -1 2 e 2 (2X Y x - X X (Y (x-X )+X (y-Y ))2 2(1+) 2 2 X Y 1 -1 2 e 2 (Y (x-X )+X (y-Y ))2 2(1-) 2 2 X Y 1 1-2 X Y 1 - 2 )-1 exp - y - Y Y + 1 2(1 - 2 ) 2 x - X X 2 - 2 y - Y Y , - < x < , - < y < , a bivariate normal pdf. d. Another solution is aX aY cX cY = = = = X bX = Y bY = 0 X Y . (1 - 2 )X 2 2 There are an infinite number of solutions. Write bX = X -a2 ,bY = Y -a2 , and X Y substitute bX ,bY into aX aY = X Y . We get 2 aX aY + X -a2 X 2 Y -a2 Y = X Y . Square both sides and simplify to get 2 2 2 2 (1 - 2 )X Y = X a2 - 2X Y aX aY + Y a2 . Y X This is an ellipse for = 1, a line for = 1. In either case there are an infinite number of points satisfying the equations. 4.47 a. By definition of Z, for z < 0, P (Z z) = = = = = = P (X P (X P (X P (X P (X P (X z and XY > 0) + P (-X z and XY < 0) z and Y < 0) + P (X -z and Y < 0) (since z < 0) z)P (Y < 0) + P (X -z)P (Y < 0) (independence) z)P (Y < 0) + P (X z)P (Y > 0) (symmetry of Xand Y ) z)(P (Y < 0) + P (Y > 0)) z). By a similar argument, for z > 0, we get P (Z > z) = P (X > z), and hence, P (Z z) = P (X z). Thus, Z X n(0, 1). b. By definition of Z, Z > 0 either (i)X < 0 and Y > 0 or (ii)X > 0 and Y > 0. So Z and Y always have the same sign, hence they cannot be bivariate normal. ...
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