확률_및_랜덤변수_솔&e

확률_및_랜덤변수_솔&e

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem Solutions – Chapter 9 Problem 9.1.1 Solution Under construction. Problem 9.1.2 Solution Under construction. Problem 9.1.3 Solution Under construction. Problem 9.1.4 Solution The joint PDF of X and Y is f X , Y ( x , y ) = ± 6 ( y x ) 0 x y 1 0 otherwise (1) (a) The conditional PDF of X given Y is found by dividing the joint PDF by the marginal with respect to Y . for y < 0or y > 1, f Y ( y ) = 0. For 0 y 1, f Y ( y ) = Z y 0 6 ( y x ) dx = 6 xy 3 x 2 ² ² y 0 = 3 y 2 (2) The complete expression for the marginal PDF of Y is f Y ( y ) = ± 3 y 2 0 y 1 0 otherwise (3) Thus for 0 < y 1, f X | Y ( x | y ) = f X , Y ( x , y ) f Y ( y ) = ( 6 ( y x ) 3 y 2 0 x y 0 otherwise (4) (b) The minimum mean square estimator of X given Y = y is ˆ X M (() y ) = E [ X | Y = y ]= Z −∞ xf X | Y ( x | y ) (5) = Z y 0 6 x ( y x ) 3 y 2 (6) = 3 x 2 y 2 x 3 3 y 2 ² ² ² ² x = y x = 0 (7) = y / 3 (8) 335
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(c) First we must find the marginal PDF for X . For 0 x 1, f X ( x ) = Z −∞ f X , Y ( x , y ) dy = Z 1 x 6 ( y x ) = 3 y 2 6 xy ± ± y = 1 y = x (9) = 3 6 x + 3 x 2 (10) The conditional PDF of Y given X is f Y | X ( y | x ) = f X , Y ( x , y ) f X ( x ) = ² 2 ( y x ) 1 2 x + x 2 x y 1 0 otherwise (11) (d) The minimum mean square estimator of Y given X is ˆ Y M (() x ) = E [ Y | X = x ]= Z −∞ yf Y | X ( y | x ) (12) = Z 1 x 2 y ( y x ) 1 2 x + x 2 (13) = ( 2 / 3 ) y 3 y 2 x 1 2 x + x 2 ± ± ± ± y = 1 y = x (14) = 2 3 x + x 3 3 ( 1 x ) 2 (15) Perhaps surprisingly, this result can be simplified to ˆ Y M (() x ) = x 3 + 2 3 (16) Problem 9.1.5 Solution (a) First we find the marginal PDF f Y ( y ) . For 0 y 2, x y 1 1 x=y x=0 f Y ( y ) = Z −∞ f X , Y ( x , y ) dx = Z y 0 2 = 2 y (1) Hence, for 0 y 2, the conditional PDF of X given Y is f X | Y ( x | y ) = f X , Y ( x , y ) f Y ( y ) = ² 1 / y 0 x y 0 otherwise (2) (b) The optimum mean squared error estimate of X given Y = y is ˆ x M (() y ) = E [ X | Y = y ] = Z −∞ xf X | Y ( x | y ) = Z y 0 x y = y / 2 (3) (c) The MMSE estimator of X given Y is ˆ X M (() Y ) = E [ X | Y Y / 2. The mean squared error is e X , Y = E h ( X ˆ X M (() Y )) 2 i = E ³ ( X Y / 2 ) 2 = E ³ X 2 XY + Y 2 / 4 (4) 336
Background image of page 2
Of course, the integral must be evaluated. e X , Y = Z 1 0 Z y 0 2 ( x 2 xy + y 2 / 4 ) dx dy (5) = Z 1 0 ( 2 x 3 / 3 x 2 y + 2 / 2 ) ± ± x = y x = 0 dy (6) = Z 1 0 y 3 6 = 1 / 24 (7) Another approach to finding the mean square error is to recognize that the MMSE estimator is a linear estimator and thus must be the optimal linear estimator. Hence, the mean squared error of the optimal linear estimator given by Theorem 9.4 must equal e X , Y . That is, e X , Y = Var [ X ] ( 1 ρ 2 X , Y ) . However, calculation of the correlation coefficient ρ X , Y is at least as much work as direct calculation of e X , Y . Problem 9.2.1 Solution (a) The marginal PMFs of X and Y are listed below P X ( x ) = ² 1 / 3 x =− 1 , 0 , 1 0 otherwise P Y ( y ) = ² 1 / 4 y 3 , 1 , 0 , 1 , 3 0 otherwise (1) (b) No, the random variables X and Y are not independent since P X , Y ( 1 , 3 ) = 0 6= P X ( 1 ) P Y ( 3 ) (2) (c) Direct evaluation leads to E [ X ]= 0V a r [ X 2 / 3 E [ Y a r [ Y 5 This implies Cov [ X , Y ] = Cov [ X , Y ] = E [ XY ] E [ X ] E [ Y ] = E [ ] = 7 / 6 (3) (d) From Theorem 9.4, the optimal linear estimate of X given Y is ˆ X L (() Y ) = ρ X , Y σ X σ Y ( Y µ Y ) + µ X = 7 30 Y + 0 (4) Therefore, a = 7 / 30 and b = 0.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/12/2011 for the course EE EE980 taught by Professor Ji during the Spring '11 term at Akademia Rolnicza w Wrocławiu.

Page1 / 16

확률_및_랜덤변수_솔&e

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online