tu3-Solutions - ACTL2003&5103 Tutorial 3 Solutions 1 Let...

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ACTL2003&5103 Tutorial 3- Solutions 1. Let N 1 ( t ) be the number of payouts with amount equal to 1 or 4, up to time t . Then N 1 ( t ) is a Poisson process with rate (per hour) 5( 1 2 + 1 2 4 ) = 45 16 . Because we need probability over a 10-minute period, the required probability is Pr ( N 1 ( 1 6 ) = 0) = e - 45 16 × 1 6 = 0 . 6257. 2. Customers arrive according to a Poisson process with rate 0 . 1 per minute. Denote this process by N ( t ). Then the required probability equals Pr ( N (10) 2) = 1 - e - 1 - e - 1 = 1 - 2 e - 1 = 0 . 2642 . 3. Ross: Chapter 5, Q36. (a) E [ S ( t ) | N ( t ) = n ] = sE N ( t ) Y i =1 X i | N ( t ) = n = sE " n Y i =1 X i | N ( t ) = n # = sE " n Y i =1 X i # = s ( E [ X ]) n = s 1 μ n E [ S ( t )] = s X n ( 1 μ ) n e - λt ( λt ) n n ! = se - λt X n ( λt μ ) n n ! = se - λt e λt μ (b) E [ S 2 ( t ) | N ( t ) = n ] = s 2 E N ( t ) Y i =1 X 2 i | N ( t ) = n = s 2 E " n Y i =1 X 2 i | N ( t ) = n # = s 2 E " n Y i =1 X 2 i # = s 2 ( E [ X 2 ] ) n = s 2 2 μ 2 n E [ S ( t ) 2 ] = s 2 X n ( 2 μ 2 ) n e - λt ( λt ) n n ! = s 2 e - λt X n ( 2 λt μ 2 ) n n ! = s 2 e - λt e 2 λt μ 2 1
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4. Ross: Chapter 5, Q39. We are interested in the distribution of { S 196 } defined as the waiting time until the 196 th event. As cell division mistakes occur according to a Poisson Process with rate 2.5 per year,
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