examsoln3 - X i-X j ) f x ( j ) ( X j ) = f x ( i-j ) ( X...

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EECS 501 SOLUTIONS TO EXAM #3 Fall 2001 1a. (5 · 2) 3 e - 5 · 2 3! (4 · 2) 2 e - 4 · 2 2! = 16000 3 e - 18 = 0 . 0000812. 1b. Poisson process with λ = 5 + 4 + 3 = 12. f t ( T ) = 12 e - 12 T ,T 0. 1c. Pr [ EE ] = 5 5+4+3 = 5 12 . Pr = ( 5 3 ) ( 5 12 ) 3 ( 7 12 ) 2 = 0 . 246 1d. Pr[0 or 1 arrivals in 5 minutes]= (3 · 5) 0 e - 3 · 5 0! + (3 · 5) 1 e - 3 · 5 1! = 16 e - 15 . 1e. Pr = R 5 3 2 Te - 3 T dT = 16 e - 15 = 0 . 0000049 (as in (d)). 1f. x (10) = 0 5 successes Pr = ( 10 5 ) ( 1 2 ) 10 = 0 . 246 2a. H ( ω ) = F{ 3 e - 2 t 1( t ) } = 3 +2 . S x ( ω ) = 4 | 3 +2 | 2 = 36 ω 2 +4 . 2b. K x ( t - s ) = F - 1 { 36 ω 2 +4 } = 9 e - 2 | t - s | . σ 2 x ( t ) = K x (0) = 9. 2c. x (7) ∼ N (0 , 9) Pr [ x (7) < 6] = Φ(6 / 9) = Φ(2) = 0 . 9776 2d. x ∼ N (0 ,K ) where x = x (3) x (7) x (9) ;0 = 0 0 0 ; K = 9 1 e - 8 e - 12 e - 8 1 e - 4 e - 12 e - 4 1 2e. ˆ x (7) = E [ x (7)] + λ x (7) ,x (5) σ 2 x (5) ( x (5) - E [ x (5)]) = 9 e - 4 9 6 = 6 e - 4 = 0 . 110. 2f. x ( t ) Markov ˆ x (7)[ x (5) ,x (3)] = ˆ x (7)[ x (5)] = 0 (independent of #2e). 3a. ˆ x ( i )[ x ( j ) ,x ( k )] = [ λ x ( i ) ,x ( j ) λ x ( i ) ,x ( k ) ] σ 2 x ( j ) λ x ( j ) ,x ( k ) λ x ( j ) ,x ( k ) σ 2 x ( k ) - 1 x ( j ) x ( k ) = [ j k ] j k k k - 1 x ( j ) x ( k ) = [ j k ] k - k - k j jk - k 2 x ( j ) x ( k ) = x ( j ). 3b. Makes sense: II Markov disregard x ( k ) given x ( j ) to estimate x ( i ). 3c. f x ( i ) ,x ( j ) ( X i ,X j ) = f x ( i ) | x ( j ) ( X i | X j ) f x ( j ) ( X j ) = f x ( i ) - x ( j ) | x ( j ) ( X i - X j | X j ) f x ( j ) ( X j ) = f x ( i ) - x ( j ) (
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Unformatted text preview: X i-X j ) f x ( j ) ( X j ) = f x ( i-j ) ( X i-X j ) f x ( j ) ( X j ) = 2 i e-2 X i ( X i-X j ) i-j-1 X j-1 j ( i-j-1)!( j-1)! ,X i > X j > 0. 3d-f. y ( n ) = 1 with prob. p n ; y ( n ) = 0 with prob. 1-p n . 3d. LIM n →∞ Pr [ | y ( n )-| > ² ] = LIM n →∞ Pr [ y ( n ) = 1] = LIM n →∞ p n = 0. 3e. LIM n →∞ E [( y ( n )-0) 2 ] = LIM n →∞ 1 2 · Pr [ y ( n ) = 1] = LIM n →∞ 1 2 p n = 0. 3f. LIM n →∞ ∑ n i =1 Pr [ | y ( i )-| > ² ] = LIM n →∞ ∑ n i =1 Pr [ y ( n ) = 1] = LIM n →∞ ∑ n i =1 p i = ∑ ∞ i =1 p i = p 1-p < ∞ → converge with prob. 1....
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This note was uploaded on 07/22/2011 for the course EECS 370 taught by Professor Lehman during the Spring '10 term at University of Florida.

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