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STAT330A3

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

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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

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2 1. Solution: (a). u 2 = ( - 2 log x ) cos 2 (2 πy ) v 2 = ( - 2 log x ) sin 2 (2 πy ) u 2 + v 2 = - 2 log x x = e - 1 2 ( u 2 + v 2 ) , Also, v u = tan(2 πy ) y = 1 2 π 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 π - v/u 2 1+( v/u ) 2 1 2 π 1 /u 1+( v/u ) 2 = - 1 2 π e - 1 2 ( u 2 + v 2 ) 0 x 1 , 0 y 1 ⇐⇒ -∞ < u < , -∞ < v < i.e. A XY = { ( x, y ) : 0 x 1 , 0 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 , 0 x 1 , 0 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 π tan - 1 ( v u )) · | J | = 1 2 π e - 1 2 ( u 2 + v 2 ) , -∞ < u < , -∞ < v < (b). f U ( u ) = -∞ f U,V ( u, v ) dv = -∞ 1 2 π e - 1 2 ( u 2 + v 2 ) dv, -∞ < u < = 1 2 π e - 1 2 u 2 -∞ 1 2 π e - v 2 / 2 dv = 1 2 π e - 1 2 u 2 , -∞ < u < (By the fact Z N (0 , 1) -∞ f ( z ) dz = -∞ 1 2 π e - z 2 / 2 dz = 1) Similarly, f V ( v ) = 1 2 π 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.

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4 2. Solution: · the joint p.d.f of X and Y is f ( x, y ) = f X ( x ) f Y ( y ) = 1 2 π e - x 2 2 1 2 π e - y 2 2 = 1 2 π 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 ) = x + y u f ( x, y ) dxdy = x + y u 1 2 π e - x 2 + y 2 2 dxdy = -∞ u - x -∞ 1 2 π e - x 2 2 e - y 2 2 dydx = -∞ 1 2 π e - x 2 2 u - x -∞ e - y 2 2 dy dx Let h ( u, x ) = u - x -∞ e - y 2 2 dy , then F U ( u ) = -∞ 1 2 π e - x 2 2 h ( u, x ) dx
5 · therefore, the p.d.f of U is: f U ( u ) = d du F U ( u ) = d du -∞ 1 2 π e - x 2 2 h ( u, x ) dx = -∞ ∂u 1 2 π e - x 2 2 h ( u, x ) dx = -∞ 1 2 π e -

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

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