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6.262.PS7.sol

# 6.262.PS7.sol - 6.262 Discrete Stochastic Processes Spring...

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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 steady-state 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 0 and f X ( t ) ( x ) = xf X ( x ) E( X ) , x 0 , where F X ( x ) = R x 0 f X ( ξ ) . 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, ξ ) f Z ( t ) ( s ) = R s + x f X ( ξ ) /E ( X ) (1 - F X ( s )) / E( X ) = 1 - F X ( s + x ) 1 - F X ( s ) . 1

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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 inter-renewal 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 inter-renewal 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 inter-renewal periods. If the inter-renewal 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 [ t, t + x ] conditioned on no arrivals in [ t + x - s, t + x ]. By the stationary and independent increments properties, this in turn reduces to the probability of no arrivals in [ t, t + x - s ], which is given by e - λ ( x - s ) for a process of rate λ . Problem 2 (Exercise 3.23) a) Since N ( t ) + 1 is a stopping time, Wald’s equality yields E( N ( t ) + 1) E ( X ) = E ( S N ( t )+1 ). Thus, E ( N ( t )) = E ( S N ( t )+1 ) E ( X ) - 1 = E ( S N ( t )+1 - t ) E ( X ) + t E ( X ) - 1 Noting that S N ( t )+1 - t = Y ( t ) (i.e. the residual life at t ), assuming a non-arithmetic renewal process we have that lim t →∞ E( Y ( t )) = E ( X 2 ) / (2 E ( X )) (why?). Therefore: lim t →∞ { E ( N ( t )) - t/E ( X ) } = E ( X 2 ) 2( E ( X )) 2 - 1 .
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