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Unformatted text preview: Boston University Department of Electrical and Computer Engineering EC505 STOCHASTIC PROCESSES Problem Set No. 3 Solutions Fall 2008 Issued: Wednesday, Sept. 17, 2008 Due: Friday, Sept. 26, 2008 Problem 3.1 Let Z be a random variable with the following exponential density and distribution functions: p Z ( z ) = ez , P Z ( z ) = 1ez where > 0 is a parameter. Define the random process: X ( t ) = 1 if 0 t z if t > z (a) Find the equivalent event for Z that corresponds to X ( t ) = 1, assuming a fixed t > 0. Use this to find the firstorder marginal density p X ( t ) ( x ). (b) Assuming 0 < t 1 < t 2 , fill in the tables below with the equivalent events for Z and their corresponding probabilities to specify the second order marginal joint density function of X ( t 1 ) and X ( t 2 ), p X ( t 1 ) ,X ( t 2 ) ( x 1 , x 2 ). Equivalent Events p X ( t 1 ) ,X ( t 2 ) ( x 1 , x 2 ) X ( t 1 ) 1 X ( t 2 ) 1 X ( t 1 ) 1 X ( t 2 ) 1 Solution: (a) For X ( t ) to be one, z must be greater than or equal to t . For t > 0 we thus have: Pr( X ( t ) = 1) = Pr( z t ) = Z t ez dz = (1 / ) ez t =(0et ) = et Thus, Pr( X ( t ) = 0) = 1et , which is all we need to fill in the tables in the next part. Overall we obtain: p X ( t ) ( x ) = et ( x1) + ( 1et ) ( x ) (b) You know t 2 > t 1 so X ( t 1 ) could never be zero when X ( t 2 ) is 1, so this event has probability 0. Ordering also tells you that X ( t 1 ), X ( t 2 ) are both zero only when z < t 1 , which has probability equal to the area under p Z ( z ) from 0 to t 1 . Also, X ( t 1 ), X ( t 2 ) are only both one when z t 2 , which has probability equal to the area under p Z ( z ) from t 2 to . Finally, X ( t 1 ) being 1 and X ( t 2 ) being zero corresponds to t 1 z < t 2 , which has probability equal to the rest of the area under p Z ( z ). Equivalent Events p X ( t 1 ) ,X ( t 2 ) ( x 1 , x 2 ) X ( t 1 ) 1 z < t 1 t 1 z < t 2 X ( t 2 ) 1 Never z t 2 X ( t 1 ) 1 1et 1 et 1et 2 X ( t 2 ) 1 et 2 1 Problem 3.2 (Old Exam Problem) Let and be two statistically independent, identically distributed Gaussian random variables with means E [ ] = E [ ] = 0 and variances 2 = 2 = 1. Define the stochastic process X ( t ) = cos( t ) + sin( t ). (a) Find the mean, m X ( t ), and autocorrelation, R XX ( t 1 , t 2 ), of the process X ( t ). (b) Is the process X ( t ) widesense stationary? Explain. (c) Is the process X ( t ) a Gaussian random process? Explain. Solution: (a) m X ( t ) = E [ X ( t )] = E [ cos( t ) + sin( t )] = E [ ] cos( t ) + E [ ] sin( t ) = 0 cos( t ) + 0 sin( t ) = 0 R XX ( t 1 , t 2 ) = E [ X ( t 1 ) X ( t 2 )] = E [( cos( t 1 ) + sin( t 1 ))( cos( t 2 ) + sin( t 2 ))] = E 2 cos( t 1 ) cos( t 2 ) + cos( t 1 ) sin( t 2 ) + cos( t 2 ) sin( t 1 ) + 2 sin( t 1 ) sin( t 2 ) = E [ 2 ] cos( t 1 ) cos( t 2 ) + E [ ] cos( t 1 ) sin( t 2 ) + E [...
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 Fall '04
 Karl

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