This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: STAT455/855 Fall 2009 Applied Stochastic Processes Assignment #4, Solutions Total Marks: 30 for 455 and 35 for 855. 1. Ross, Chapter 5, #44. (6 marks) (a) (2 marks) The probability that the woman’s waiting time is 0 is equal to the probability that when she arrives at the street location, the time until the next event (car passing) in the Poisson process is greater than T . By memorylessness, the time to the next event is distributed as Exponential( λ ), so the probability that the time to the next event is greater than T is given by e- λT . (b) (4 marks) We use the hint and condition on time of the first car. Let W denote the woman’s waiting time. Condition on the time of the first car being t time units. If t > T , then the woman’s waiting time is 0. If t ≤ T , then the woman’s waiting time is t plus an additional waiting time, say W a , where W a has the same distribution as W , since the interarrival times are i.i.d. So by conditioning, and using the law of total expectation, we get the following equation for E [ W ]: E [ W ] = Z ∞ E [ W first car arrives at time t ] λe- λt dt = Z T ( t + E [ W a ]) λe- λt dt + Z ∞ T (0) λe- λt dt = Z T ( t + E [ W ]) λe- λt dt = Z T tλe- λt dt + E [ W ](1- e- λT ) . Then, solving for E [ W ], we get E [ W ] e- λT = Z T tλe- λt dt =- te- λt T + Z T e- λt dt (using integration by parts) =- Te- λT +- e- λt λ T =- Te- λT + 1 λ (1- e- λT ) or E [ W ] = e λT- 1- λT λ . STAT 455/855 -- Assignment 4 Solutions p. 2 2. Ross, Chapter 5, #53. (4 marks) Let N ( t ) be the total number of rainfalls, N 1 ( t ) the number of type 1 rainfalls and N 2 ( t ) the number of type 2 rainfalls by time t , where a type 1 rainfall adds 5000 units of water to the reservoir and a type 2 rainfall adds 8000 units of water to the reservoir. (a) (2 marks) The reservoir will be empty after five days if and only if there is no rainfall after five days. Thus, P ( N (5) = 0) = e- ( . 2)(5) = e- 1 = 0 . 368 . (b) (2 marks) The reservoir will be empty sometime within the next ten days if and only if there is either no rainfall after 5 days or there is exactly one type 1 rainfall within the first 5 days and no other rainfall in the next ten days. Thus,within the first 5 days and no other rainfall in the next ten days....
View Full Document
- Fall '09
- Poisson Distribution, Probability theory, ASCII, ASCII art