ass5sol - STAT455/855 Fall 2009 Applied Stochastic...

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Unformatted text preview: STAT455/855 Fall 2009 Applied Stochastic Processes Assignment #5, Solutions Total Marks: 32. 1. Ross, Chapter 6 #1. (4 marks) The state space of the process ( N 1 ( t ) , N 2 ( t )) is S = { ( n, m ) : n , m } . If in state i = ( n, m ), we leave state i as soon as there is a pairing. According to the problem, each particular pairing occurs at an exponential rate of . Since there are nm possible pairings when in state i , the rate out of state i is v i = mn . There are only two possible states to enter when the process leaves state ( n, m ). These are states ( n +1 , m ) and ( n, m +1). Since each pairing results in a male or female with equal probability, when we leave state i (a pairing occurs), we go to state ( n + 1 , m ) with probability 1 2 and to state ( n, m + 1) with probability 1 2 . That is, p ij = 1 2 for j = ( n + 1 , m ) or j = ( n, m + 1), and p ij = 0 for all other states j . 2. From Sheet. (5 marks) Suppose Y n = i . To calculate P ( Y n +1 = j Y n = i ) we condition on the time between the n th and the ( n + 1)st transition. This time is exponentially distributed with rate . Let us denote it by T . Conditioning on T = t , we have q ij = P ( Y n +1 = j Y n = i ) = Z P ( Y n +1 = j Y n = i, T = t ) e- t dt = Z p ij ( t ) e- t dt....
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ass5sol - STAT455/855 Fall 2009 Applied Stochastic...

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