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Note for MATH 6397 with Professor Bodmann at UH


Note for MATH 6397 with Professor Bodmann at UH

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This 4 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 18 views.

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Date Created: 02/06/15
Stochastic Processes Spring 2008 Bernhard Bodmann PGH 636 Exercise Sheet 1 Sketch of solutions 1 This is a birth death process with birth parameter An 71A for n E 0K 7 1 An 0 otherwise and death parameters Mn nu As derived in the lectures the expected time to absorption in state 0 given the process starts in state 1 is pi where p f1 AidHm Substituting A M and u m we nd p ifz39 S K and p 0 otherwise Hence the expected time to absorption 1 is K 1 1 A W 7 7 7 l 1 A ltgt Realizing that 231 xi 1 K for z 31 1 and that term by term integration 17m i K71 7 i i i gives 210 14le fox 11f du we have the alternative integral express1on 1 Auli K W17 udu 0 A 17u 2 The components operate independently If they break the repair times are also independent Failure times for operating are exponentially dis tributed with parameter 739 Repair times are also exponentially distributed with parameter p Consider the stochastic process Xi where X counts the number of components undergoing repair Assume n machines are undergoing repair at the beginning of a time interval 0h The probability P of at least one failure occuring in this interval is N7n P 1 7 ll no failure 1 7 H ll j th unit does not fail 1 7 ciwinw 11 Thus the probability P of one more failed machine remaining at the end of the interval differs by 0h which accounts for two or more events occuring such as two failures and one completed repair etc Now taking the derivative at h 0 we get the rate An N 7 n739 Similarly the death77 rate ie of ending up with one machine less un dergoing repair is Mn np Consequently the process can be regarded as a birth death process with parameters An N 7 n and Mn nu 3 Denote the vacant state by 0 the unacceptable molecule attached by state 1 the acceptable absorbing state by 2 The generator of the stochastic process is 7M M176 6 A 739 7739 0 0 0 0 A In order to compute P 6t we diagonalize A A rst left eigenvector of A is 01 001 with corresponding eigenvalue A1 0 which is the stationary distribution in the absorbing state The other two eigenvectors are A 172 A 712 MA 26M 26M with eigenvalue A2 7M 739 A and i MT7A u172 7397A VG 25M 25M 1 with eigenvalue A3 7M 739 7 A where A xp 739 7 46m Thus if we start with an initial distribution 7T C101 62112 03113 7 de ned with appropriate coef cients 0102053 E R then after time t the probability of NOT being in state 2 is ll not absorbed at time t 1 7 wetA3 1 7 cl 7 026A 7 036A Assuming as the book does that we start in state 0 then 01 1 02 M172 T7A i 7M172 TA 2A and 03 7 2A 4 To generate exponentially distributed waiting times with parameter 739 use a rV U with uniform distribution in 07 1 and transform it by T 7 lnU These waiting times depend on the state If there are r rabbits and f foxes7 then 739 on yrf 677 6f At the jump points7 we have transition probabilities for the embedded Markov Chain Let M on rf yrf 6f then r f a r f 1 with rate yrfu r f a hf 71 with rate Sfp r f a 7quot 17f with rate arp and r f a 7quot 717 with rate arp An example plot is given below Rabbits blue7 foxes red Care must be taken to avoid rabbit explosions77 and also the case r 0 or f 0 needs to be handled separately Sometimes the foxes die out and the rabbits are left to multiply The number of rabbits has been capped at 200 5n A nice movie displaying the evolution of the probability distribution in the rabbit fox state space is due to the group of Sheena Branton7 David Johnson and Robert Rosenbaum7 accessible at the website httpwwwmathuhedu NbgbCoursesMath6397rabbitnfoxavi


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