Ass2-soln - 1 Solutions of Assignment #2 - STAT 330 Due in...

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1 Solutions of Assignment #2 - STAT 330 Due in class: Thursday Feb. 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. /8 QUESTION 2. /6 QUESTION 3. /4 QUESTION 4. /12 QUESTION 5. /6 QUESTION 6. /4 TOTAL: /40
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2 1. Solution: (1). P ( X 2 <Y )= ±± D f ( x,y ) dxdy = ± 1 0 ± y 0 6 x 2 ydxdy (OR= ± 1 0 ± 1 x 2 6 x 2 ydydx ) = ± 1 0 y · 2 x 3 ² ² ² y 0 dy =2 ± 1 0 y 5 / 2 dy = 4 7 Figure 1: D= { ( ): x 2 <y } (2). P (0 . 5 <X + Y< 1 . 5) = B 1 S B 2 f ( ) dxdy = B 1 f ( ) dxdy + B 2 f ( ) dxdy = ± 0 . 5 0 ± 1 0 . 5 - x 6 x 2 ydydx + ± 1 0 . 5 ± 1 . 5 - x 0 6 x 2 ydydx = ± 0 . 5 0 x 2 · 3 y 2 ² ² ² 1 0 . 5 - x dx + ± 1 0 . 5 x 2 · 3 y 2 ² ² ² 1 . 5 - x 0 dx = ± 0 . 5 0 3 x 2 [1 - (0 . 5 - x ) 2 ] dx + ± 1 0 . 5 3 x 2 (1 . 5 - x ) 2 dx = ± 0 . 5 0 (2 . 25 x 2 +3 x 3 - 3 x 4 ) dx + ± 1 0 . 5 (6 . 75 x 2 - 9 x 3 x 4 ) dx =0 . 5625
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3 (3). P ( Y> 0 . 5 | X> 0 . 5) = P ( 0 . 5 ,Y > 0 . 5) P ( 0 . 5) = ± 1 0 . 5 ± 1 0 . 5 6 x 2 ydydx ± 1 0 . 5 ± 1 0 6 x 2 ydydx = 0 . 75 x 3 ² ² ² 1 0 . 5 x 3 ² ² ² 1 0 . 5 =0 . 65625 / 0 . 875 = 0 . 75 (4). Note: P ( 0 . 6 | X . 6) ± = P ( 0 . 6 ,X . 6) P ( X . 6) P ( 0 . 6 | X . 6) = ³ y> 0 . 6 f Y | X ( y | 0 . 6) dy = ³ 0 . 6 f (0 . 6 ,y ) f X (0 . 6) dy f X ( x )= ³ -∞ f ( x,y ) dy = ³ 1 0 6 x 2 ydy, for 0 < x < 1 =3 x 2 f Y | X ( y | 0 . 6) = 6 × 0 . 6 2 y 3 × 0 . 6 2 =2 y, 0 < y < 1 P ( 0 . 6 | X . 6) = ³ 1 0 . 6 2 ydy = y 2 ² ² ² 1 0 . 5 . 64 Note: Be careful with the diFerence between P ( Y > b | X = a ± b f Y | X ( y | a ) dy and P ( Y > b | X > a P ( Y >b,X>a ) P ( X>a ) where a and b are constants.
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4 2. Solution: (a) Cov ( X, Y )= E ( XY ) - E ( X ) E ( Y ) E ( ± -∞ ± -∞ xyf ( x,y ) dxdy = ± 0 ± y 0 xye - y dxdy = ± 0 ye - y ( ± y 0 xdx ) dy = ± 0 - y ² x 2 2 ³ | y 0 dy = 1 2 ± 0 y 3 e - y dy = 1 2 Γ(4) = 3 f X ( x ± x e - y dy = ´ - e - y µ x = e - x ,x 0 = E ( X ± 0 xe - x dx = Γ(2) = 1 f Y ( y ± y 0 e - y dx = - y ,y 0 = E ( Y ± 0 y 2 e - y dy = Γ(3) = 2 = Cov(X , Y) = E(XY) - E(X)E(Y) = 3 - 2 × 1 = 1 E ( X 2 ± 0 x 2 e - x dx = Γ(3) = 2 V ar ( X E ( X 2 ) - E ( X ) 2 =2 - 1 2 =1 E ( Y 2 ± 0 y 3 e - y dy = Γ(4) = 6 V ar ( X E ( Y 2 ) - E ( Y ) 2 =6 - 2 2 ρ = ( X, Y ) V ar ( X ) V ar ( Y ) = 1 2 × 1 = 1 2
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5 (b) f X | Y ( x | y )= f ( x,y ) f Y ( y ) = e - y ye - y = 1 y , 0 x y f Y | X ( y | x f ( ) f X ( x ) = e - y e - x = e - y + x = e - ( y - x ) , 0 x y E ( X | Y = y ± y 0 x ( 1 y ) dy = 1 y ² x 2 2 ³ y 0 = y 2 ,y 0 E ( Y | X = x ± x - ( y - x ) dy = e x ± x - y dy = e x { ´ - - y µ x + ± x e - y dy } = e x [ xe - x + e - x ]= x +1 ,x
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This note was uploaded on 06/21/2010 for the course STAT 5023 taught by Professor Yi,graceyun during the Spring '10 term at Waterloo.

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Ass2-soln - 1 Solutions of Assignment #2 - STAT 330 Due in...

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