solution13 - 11/21/03 1:01 PM HW 13page 1 Probability for...

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11/21/03 1:01 PM HW 13page 1 Probability for Engineering Applications SOLUTION: Assignment #13 Two random variables. 1. (2 pts) Let X be the input to a communications channel (cf 4.17). X takes on the values +1 and –1 with equal probability. The output of the channel is Y=X+N, where N is a Laplacian random variable with pdf f N (z) = 0.5 α e - α |z| - < z < 1a. P[Y<y|X=1]P[X=1] = =0.5 P[N + 1 y] ( 29 ( 1) 1 || ( 1) 1 1 4 0.5 1 2 21 4 y y z y ey e dz a a a a - - - -- -∞ == - . P[Y<y|X=-1]P[X=-1] = ( 29 ( 1) ( 1) 1 1 4 1 4 y y a a + -+ ≤- - - 1b. F Y (y) = P[Y y | X = 1] P[X = 1] + P[Y y| X = -1] P[X = -1] = 0.5 P[N + 1 y] + 0.5 P[N - 1 ( ) ( ) 0. 5 ( 1 ) ( 1) Y Y YY d dd f y F y F y Fy d y d y dy  = = - ++   ( 29 0. 5 ( 1 ) ( NN f y fy - ( 29 ( 29 | 1 | | 1 | | 1 | | 1| 0. 5 0. 5 0. 5 0.25 y y yy e e ee a a aa a a - - - + - - = + =+ ( 29 ( 29 ( 29 ( 1 ) ( ( 1 ) ( ( 1 ) ( 0.2 51 0.2 5 11 0.2 fo ry e e fo fo - - -+ + - + = + <- = + -<< = +< L L L
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11/21/03 1:01 PM HW 13page 2 1c. P[X=1|Y>0] [ 0 | [ 1 ] [ 1] [ 0] P Y P X PX PY == = P[X=1] = 0.5, P[Y>0] = 0.5 (by symmetry) P[X=1|Y>0] = P[Y>0|X=1] (a special case! – not usually true) || 1 0.5 y e dy a a - - = ( 29 1 ( 1 ) ( 1) 01 1 11 1 )1 2 22 22 yy e d ye d ee a a a aa
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This note was uploaded on 09/03/2008 for the course ECSE 4510 taught by Professor Woods during the Spring '03 term at Rensselaer Polytechnic Institute.

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solution13 - 11/21/03 1:01 PM HW 13page 1 Probability for...

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