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Unformatted text preview: Statistics 351 Fall 2006 (Kozdron) Midterm #1 Solutions 1. (a) By definition, f X  Y = y ( x ) is given by f X  Y = y ( x ) = f X,Y ( x,y ) f Y ( y ) . We being by calculating f Y ( y ) = Z 4 y 2 1 dx = 1 p 4 y 2 for 0 < y < 2. Therefore, f X  Y = y ( x ) = 1 1 p 4 y 2 = 1 p 4 y 2 , < x < p 4 y 2 . 1. (b) We find E ( X  Y = y ) = Z  x f X  Y = y ( x ) dx = Z 4 y 2 x 1 p 4 y 2 dx = 1 2 p 4 y 2 2 p 4 y 2 = p 4 y 2 2 . 1. (c) Solving for X and Y we find X = U cos V and Y = U sin V. The Jacobian of this transformation is J = x u x v y u y v = cos v u sin v sin v u cos v = u cos 2 v + u sin 2 v = u. The joint density of ( U,V ) is therefore given by f U,V ( u,v ) = f X,Y ( u cos v,u sin v )  J  = u provided that 0 < u < 2 and 0 < v < 2 . 1. (d) From (c) , we see that U and V are independent since we can write f U,V ( u,v ) = f U ( u ) f V ( v ) where f U ( u ) = u 2 , 0 < u < 2, and f V ( v ) = 2 , 0...
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This note was uploaded on 03/26/2012 for the course STAT 351 taught by Professor Michaelkozdron during the Fall '08 term at University of Regina.
 Fall '08
 MichaelKozdron
 Statistics, Probability

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