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Unformatted text preview: 6.262 Discrete Stochastic Processes, Spring 2010 Problem Set 7 Solutions due: Friday, April 2, 2010 Problem 1 (Exercise 3.8) a) Additionally conditioning on N ( t ) = n greatly simplifies the problem, so let us first consider the quantity P ( Y ( t ) > x  Z ( t ) = s,N ( t ) = n ). Note that conditioned on { Z ( t ) = s,N ( t ) = n } , the event { Y ( t ) > x } is the same as the event { X n +1 > x + s } . Furthermore, the event { N ( t ) = n } { Z ( t ) = s } is the same as the event { N ( t ) = n } { X n +1 > s } { X 1 + ... + X n = s } . (Why?) Thus, P( Y ( t ) > x  Z ( t ) = s,N ( t ) = n ) = P( X n +1 > x + s  X n +1 > s,X 1 + ... + X n = s,N ( t ) = n ) = P( X n +1 > x + s  X n +1 > s ) = 1 F X ( s + x ) 1 F X ( s ) , where the second equality follows from the fact that the n + 1 st renewal is independent of all the previous renewals. By the total probability theorem, P( Y ( t ) > x  Z ( t ) = s ) = X n =0 P( Y ( t ) > x  Z ( t ) = s,N ( t ) = n )P( N ( t ) = n  Z ( t ) = s ) = 1 F X ( s + x ) 1 F X ( s ) X n =0 P( N ( t ) = n  Z ( t ) = s ) = 1 F X ( s + x ) 1 F X ( s ) . Note: if we assume that t is large, we can instead solve this problem using steadystate distributions of X ( t ) and Z ( t ). Recall that for large t , f Z ( t ) ,X ( t ) ( z,x ) = f X ( x ) E ( X ) for x z , from which we obtain that f Z ( t ) ( z ) = 1 F X ( z ) E( X ) , z and f X ( t ) ( x ) = xf X ( x ) E( X ) , x , where F X ( x ) = R x f X ( ) d . Since { Y ( t ) > x  Z ( t ) = s } = { X ( t ) > s + x  Z ( t ) = s } , we have that P( Y ( t ) > x  Z ( t ) = s ) = P( X ( t ) > s + x  Z ( t ) = s ) = R s + x f Z ( t ) ,X ( t ) ( s, ) d f Z ( t ) ( s ) = R s + x f X ( ) /E ( X ) d (1 F X ( s )) / E( X ) = 1 F X ( s + x ) 1 F X ( s ) . 1 b) Looking at P( Y ( t ) > x  Z ( t + x/ 2) = s ), one needs to consider two cases. If s x/ 2, then t and t + x/ 2 are in the same interrenewal period. Thus, P( Y ( t ) > x  Z ( t + x/ 2) = s ) = P( Y ( t + x/ 2) > x/ 2  Z ( t + x/ 2) = s ) = 1 F X ( s + x/ 2) 1 F X ( s ) , where the second equality follows directly from a). If s < x/ 2, then t and t + x/ 2 are not within the same interrenewal period. Conditioning on Z ( t + x/ 2) = s is then equivalent to conditioning on Y ( t ) x/ 2 s (why?). Thus, P( Y ( t ) > x  Y ( t ) x/ 2 s ) = 0. c) First note that Z ( t + x ) > x implies Y ( t ) x (why?). Thus, if s > x , P( Y ( t ) x  Z ( t + x ) > s ) = 1 regardless of the distribution of the interrenewal periods. If the interrenewal periods have a density (as we are told to assume), then P( Y ( t ) > x  Z ( t + x ) > s ) = 1. Suppose now that s x . For an arbitrary renewal process, conditioning on Z ( t + x ) > s does not provide straightforward information as t + x and t may or may not be in the same inter renewal period. For a Poisson process, on the other hand, the question reduces to finding the probability of no arrivals in [...
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This note was uploaded on 10/21/2010 for the course EE 5581 taught by Professor Moon,j during the Spring '08 term at Minnesota.
 Spring '08
 Moon,J

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