# ps08_sol - Massachusetts Institute of Technology Department...

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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 | 0 , 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 , | · 0 , 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

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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 σ Y where cov( X, Y ) 2 σ + W 2 25 / 12, σ 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, σ Y 2 7 5, σ . X E [ W ] 4 / 12 and, using the = = = = = 2 σ 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 σ X 2 σ 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. 4 5 6 7 8 9 10 11 Linear predictor o L o ) g (y ) , g (y 4 5 6 7 8 9 10 11 y o Page 2 of 7
Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) 3. (a) The Chebyshev inequality yields P ( X 7 3) 9 = 1, which implies the uninfor- 3 2 | | mative/useless bound P (4 < X < 10) 0.

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