STAT330A3 - 1 Solutions of Assignment #3 - STAT 330 Due in...

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Unformatted text preview: 1 Solutions of Assignment #3 - STAT 330 Due in class: Thursday Mar. 25 Important Note: You need to print out this page as the cover page for your assignment. LAST NAME: FIRST NAME: ID. NO.: QUESTION 1. /5 QUESTION 2. /5 QUESTION 3. /6 QUESTION 4. /6 QUESTION 5. /6 QUESTION 6. /7 TOTAL: /35 2 1. Solution: (a). u 2 = (- 2log x )cos 2 (2 y ) v 2 = (- 2log x )sin 2 (2 y ) u 2 + v 2 =- 2log x x = e- 1 2 ( u 2 + v 2 ) , Also, v u = tan(2 y ) y = 1 2 tan- 1 ( v u ). J = x u x v y u y v =- ue- 1 2 ( u 2 + v 2 )- ve- 1 2 ( u 2 + v 2 ) 1 2 - v/u 2 1+( v/u ) 2 1 2 1 /u 1+( v/u ) 2 =- 1 2 e- 1 2 ( u 2 + v 2 ) x 1 , y 1 - < u < ,- < v < i.e. A XY = { ( x,y ) : 0 x 1 , y 1 } B UV = { ( u,v ) :- < u < ,- < v < } , As X and Y are independent, so f X,Y ( x,y ) = f X ( x ) f Y ( y ) = 1 , x 1 , y 1 By the One-to-One Transformation Theorem, f U,V ( u,v ) = f X,Y ( e- 1 2 ( u 2 + v 2 ) , 1 2 tan- 1 ( v u )) | J | = 1 2 e- 1 2 ( u 2 + v 2 ) ,- < u < ,- < v < (b). f U ( u ) = Z - f U,V ( u,v ) dv = Z - 1 2 e- 1 2 ( u 2 + v 2 ) dv,- < u < = 1 2 e- 1 2 u 2 Z - 1 2 e- v 2 / 2 dv = 1 2 e- 1 2 u 2 ,- < u < (By the fact Z N (0 , 1) Z - f ( z ) dz = Z - 1 2 e- z 2 / 2 dz = 1) Similarly, f V ( v ) = 1 2 e- 1 2 v 2 ,- < v < i.e. U N (0 , 1) ,V N (0 , 1) 3 Note: 1. We can further conclude that U and V are independent. 2. This result can be used to generate data from N (0 , 1) distribution. 3. The transformation is called Box-Mueller Transformation. 4 2. Solution: the joint p.d.f of X and Y is f ( x,y ) = f X ( x ) f Y ( y ) = 1 2 e- x 2 2 1 2 e- y 2 2 = 1 2 e- x 2 + y 2 2 ,- < x < ,- < y < the c.d.f of U is: F U ( u ) = P ( U u ) = P ( X + Y u ) = ZZ x + y u f ( x,y ) dxdy = ZZ x + y u 1 2 e- x 2 + y 2 2 dxdy = Z - Z u- x- 1 2 e- x 2 2 e- y 2 2 dydx = Z...
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This note was uploaded on 05/21/2010 for the course STAT stat330 taught by Professor Guo during the Winter '10 term at Waterloo.

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STAT330A3 - 1 Solutions of Assignment #3 - STAT 330 Due in...

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