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HW6solns

# HW6solns - May 2 2007 HW6 Problem Solutions#8.1.2 Fix t and...

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May 2, 2007 HW6 Problem Solutions #8.1.2. Fix t, and calculate the joint density of ( B, F ) from first principles at ( b, a ) with 0 < b < t, a > 0, by: f B,F ( b, a ) h 2 P ( B ( b, b + h ) , F ( a, a + h )) P ( N ( t - b ) - N ( t - b - h ) = 1 , N ( t + a ) = N ( t - b ) , N ( a + h ) - N ( a ) = 1) = e - 2 λh ( λh ) 2 e - λ ( a + b ) so that the joint density is given by f B,F ( b, a ) = λ 2 e - λ ( a + b ) I [0 <b<t, a> 0] Note that the probability mass does not integrate to 1, because with probability e - λt = P ( N ( t ) = 0) the backward recurrence time has the value t . Thus ( B, F ) is a random variable of mixed discrete-continuous type, but B, F are independent and in the limit of large t are both Expon( λ ) distributed. Also: lim t →∞ E ( B ( t ) + F ( t )) = 2 , while E ( T n - T n - 1 ) = E ( T 1 ) = 1 . This is paradoxical because it says that the inter-event interval surrounding each time t is not typical in the sense of average length. But that is not really so surprising since the fact of containing t makes it more likely to be a large interval than a small one. #8.2.4. In this flip-flop chain, the rate q i of leaving each state i = 0 , 1 is the same value λ , so the transition times can be viewed as the jump times of a Poisson process. We showed in class that the intensity matrix for a Poisson process has - λ on the main diagonal and λ on the above-diagonal and all other entries 0. Since the flip-flop chain moves at each jump to the opposite

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