This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: braceleftbigg braceleftbigg Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) Problem Set 8 Solutions 1. Let A t (respectively, B t ) be a Bernoulli random variable that is equal to 1 if and only if the t th toss resulted in 1 (respectively, 2). We have E [ A t B t ] = 0 (since A t = 0 implies B t = 0) and 1 1 E [ A t B s ] = E [ A t ] E [ B s ] = k k for s = t. Thus, E [ X 1 X 2 ] = E [( A 1 + + A n )( B 1 + + B n )] 1 1 = n E [ A 1 ( B 1 + + B n )] = n ( n 1) k k and cov( X 1 ,X 2 ) = E [ X 1 X 2 ] E [ X 1 ] E [ X 2 ] n ( n 1) n 2 n = = . k 2 k 2 k 2 2. (a) The minimum mean squared error estimator g ( Y ) is known to be g ( Y ) = E [ X Y ]. Let  us first find f X,Y ( x,y ). Since Y = X + W , we can write 1 x ) = 2 , if x 1 y x + 1; f Y  X ( y  , otherwise and, therefore, 1 if x 1 y x + 1 and 5 x 10; f X,Y ( x,y ) = f Y X ( y  x ) f X ( x ) = 10 ,  , otherwise as shown in the plot below. o o x , y x,y f ( y o 5 10 ) = 1/10 5 10 x o We now compute E [ X Y ] by first determining f X Y ( x y ). This can be done by    looking at the horizontal line crossing the compound PDF. Since f X,Y ( x,y ) is uniformly distributed in the defined region, f X Y ( x y ) is uniformly distributed as well. Therefore,   5+( y +1) , if 4 y < 6; 2 g ( y ) = E [ X Y = y ] = y, if 6 y 9;  10+( y 1) , if 9 < y 11 . 2 The plot of g ( y ) is shown here. Page 1 of 7 Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) 4 5 6 7 8 9 10 11 4 5 6 7 8 9 10 11 o ) g (y y o (b) The linear least squares estimator has the form cov( X,Y ) g L ( Y ) = E [ X ] + ( Y E [ Y ]) , 2 Y where cov( X,Y ) 2 + W 2 25 / 12, W (10 5) 2 / 12 (1 ( 1)) 2 / 12 [( X [ X ])( Y [ Y ])]. We compute [ X ] 7 5, [ Y ] [ X ] + E E E E E E = = = . 2 fact that X and W are independent, Y 2 7 5, . X E [ W ] 4 / 12 and, using the = = = = = 2 X 29 / 12. Furthermore, = = cov( X,Y ) = E [( X E [ X ])( Y E [ Y ])] = E [( X E [ X ])( X E [ X ] + W E [ W ])] 2 X 2 X E [( X E [ X ])( X E [ X ])] + E [( X E [ X ])( W E [ W ])] + E [( X E [ X ])] E [( W E [ W ])] = = 25 / 12 . = = Note that we use the fact that ( X E [ X ]) and ( W E [ W ]) are independent and E [( X E [ X ])] = 0 = E [( W E [ W ])]. Therefore, 25 g L ( Y ) = 7 . 5 + ( Y 7 . 5) . 29 The linear estimator g L ( Y ) is compared with g ( Y ) in the following figure. Note that g ( Y ) is piecewise linear in this problem....
View
Full
Document
 Spring '10
 MuntherDahleh
 Systems Analysis, Bernoulli, Probability

Click to edit the document details