Exercise426

Exercise426 - Solution to Exercise 4.2.6 The transformation...

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Solution to Exercise 4.2.6 The transformation S : U = X + Y, V = X Y has inverse transformation X = U + V 2 ,Y = U V 2 . The support of ( X,Y ) is R XY = { ( x,y ):0 <x< , 0 <y< } . x y Rxy 0 Figure 1: Under S ( k, 0) ( k,k ) ,k > 0 (0 ) ( k ) k> 0 and thus S maps R into R UV = { ( u,v ): u<v<u, 0 <u< } . 1
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0 v=u v=-u u v Ruv (u,u) (u,-u) (v,v) (-v,v) u v v Figure 2: The Jacobian of the inverse transformation is ( x,y ) ( u,v ) = ¯ ¯ ¯ ¯ 1 2 1 2 1 2 1 2 ¯ ¯ ¯ ¯ = 1 4 1 4 = 1 2 The joint p.d.f. of U and V is given by g ( )= f μ u + v 2 , u v 2 ¶¯ ¯ ¯ ¯ 1 2 ¯ ¯ ¯ ¯ = 1 2 e u for ( ) R UV The marginal p.d.f. of U is given by g 1 ( u Z −∞
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This note was uploaded on 04/27/2010 for the course STAT 330 taught by Professor Paulasmith during the Fall '08 term at Waterloo.

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Exercise426 - Solution to Exercise 4.2.6 The transformation...

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