{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

&iacute;™•&euml;&yen;&nbsp;_&euml;&deg;_&euml;žœ&euml;&curren;&euml;&sup3;€&igrave;ˆ˜_&igrave;†”&e

# í™•ë¥ _ë°_ëžœë¤ë³€ìˆ˜_ì†”&e

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

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

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

View Full Document
(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
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.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 16

í™•ë¥ _ë°_ëžœë¤ë³€ìˆ˜_ì†”&e

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

View Full Document
Ask a homework question - tutors are online