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AMS 342 ch 5 notes - Stat 620 Homework 2 Jian Kang 2.4 Let...

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Stat 620 Homework 2 Jian Kang September 23, 2008 2.4 Let { N ( t ) , t 0 } be a Poisson process with rate λ . Calculate E [ N ( t ) · N ( t + s )]. Ans: When s > 0, t + s > t , note that N ( t + s ) N ( t ) = [ N ( t + s ) - N ( t )] N ( t ) + N ( t ) 2 . By independent increments, N ( t + s ) - N ( t ) and N ( t ) are independent. Also note that N ( t + s ) - N ( t ) Poisson( λs ) and N ( t ) Poisson( λt ). Thereby, E [ N ( t ) N ( t + s )] = E [ N ( t )( N ( t + s ) - N ( t ))) + E ( N ( t ) 2 ] = E [ N ( t )] E [ N ( t + s ) - N ( t )] + Var[ N ( t )] + [ E ( N ( t ))] 2 = λ 2 ts + λt + ( λt ) 2 = λt [ λ ( s + t ) + 1] When s < 0, let t 0 = t + s and s 0 = - s , so t 0 + s 0 > t 0 , by the above result E [ N ( t ) N ( t + s )] = E [ N ( t 0 ) N ( t 0 + s 0 )] = λt 0 [ λ ( s 0 + t 0 ) + 1] = λ ( t + s )[ λt + 1] When s = 0, E [ N ( t ) N ( t + s )] = E [ N ( t ) 2 ] = λt [ λt + 1]. To sum up, E [ N ( t ) N ( t + s )] = ( λt [ λ ( s + t ) + 1] s 0 λ ( t + s )[ λt + 1] s < 0 2.5 Suppose that { N 1 ( t ) , t 0 } and { N 2 ( t ) , t 0 } are independent Poisson processes with rates λ 1 and λ 2 . (a) Show that { N 1 ( t ) + N 2 ( t ) , t 0 } is a Poisson process with rate λ 1 + λ 2 . (b) Also, show that the probability that the first event of the combined process comes from { N 1 ( t ) , t 0 } is λ 1 / ( λ 1 + λ 2 ), independently of the time of the event. Ans: 1
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(a) By definition 1, we verify the following three condition: i. N 1 (0) + N 2 (0) = 0 ii. For any s, t > 0 and i = 1 , 2, N i ( t ) and N i ( t + s ) - N i ( t ) are independent. Also N 1 ( t ) and N 2 ( t ) are independent. This implies that N 1 ( t ) + N 2 ( t ) and N 1 ( t + s )+ N 2 ( t + s ) - N 1 ( t ) - N 2 ( t ) are independent. So N 1 ( t )+ N 2 ( t ) has independent increments. iii. For s < t and i = 1 , 2, N i ( t ) - N i ( s ) Poisson( λ i ( t - s )), by the properties of Poisson distribution, we have that N 2 ( t ) - N 2 ( s ) + N 1 ( t ) - N 1 ( s ) Poisson[( λ 1 + λ 2 )( t - s )]
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