Lecture 20. Poisson Processes. Text: A Course in Probability by Weiss STAT 225 Introduction to Probability Models March 26, 2014

Transcription

1 Lecture 20 Text: A Course in Probability by Weiss 12.1 STAT 225 Introduction to Probability Models March 26, 2014 Whitney Huang Purdue University 20.1

2 Agenda

3 For a specified event that occurs randomly in continuous time, an important application of probability theory is on modeling the number of times such an event occurs. The following are several examples of such random phenomenon: The number of patients that arrive at a hospital emergency room 20.3

4 For a specified event that occurs randomly in continuous time, an important application of probability theory is on modeling the number of times such an event occurs. The following are several examples of such random phenomenon: The number of patients that arrive at a hospital emergency room The number of customers that enter a particular bank 20.3

5 For a specified event that occurs randomly in continuous time, an important application of probability theory is on modeling the number of times such an event occurs. The following are several examples of such random phenomenon: The number of patients that arrive at a hospital emergency room The number of customers that enter a particular bank The number of accidents at an intersection 20.3

6 For a specified event that occurs randomly in continuous time, an important application of probability theory is on modeling the number of times such an event occurs. The following are several examples of such random phenomenon: The number of patients that arrive at a hospital emergency room The number of customers that enter a particular bank The number of accidents at an intersection The number of alpha particles emitted by a radioactive substance 20.3

7 Consider an event that occurs randomly and homogeneously in continuous time at an average rate of λ per unit of time Define the occurrence of the event as a success We say such a counting process is a Poisson Process with rate λ if 2 more properties hold. Namely, if: 20.4

8 Consider an event that occurs randomly and homogeneously in continuous time at an average rate of λ per unit of time Define the occurrence of the event as a success Define N(t) as the number of successes by time t. It implies that N(0) = 0 We say such a counting process is a Poisson Process with rate λ if 2 more properties hold. Namely, if: 20.4

9 Consider an event that occurs randomly and homogeneously in continuous time at an average rate of λ per unit of time Define the occurrence of the event as a success Define N(t) as the number of successes by time t. It implies that N(0) = 0 We say such a counting process is a Poisson Process with rate λ if 2 more properties hold. Namely, if: N(t) : t 0 has independent increments (i.e. non-overlapping intervals are independent) 20.4

10 Consider an event that occurs randomly and homogeneously in continuous time at an average rate of λ per unit of time Define the occurrence of the event as a success Define N(t) as the number of successes by time t. It implies that N(0) = 0 We say such a counting process is a Poisson Process with rate λ if 2 more properties hold. Namely, if: N(t) : t 0 has independent increments (i.e. non-overlapping intervals are independent) N(t) N(s), which is the number of successes in the time interval (s, t], is distributed as Poisson(λ(s t)) for 0 < s < t < 20.4

11 cont d The Poisson Process can be used to model arrivals. It is also used for waiting times and inter arrival times For each n {0, 1, 2, 3, }, we let W n denote the time of the occurrence of the n th event The elapsed time between the occurrence of the (n 1) th and n th events is denoted by I n and is called the nth inter arrival time 20.5

12 cont d The Poisson Process can be used to model arrivals. It is also used for waiting times and inter arrival times For each n {0, 1, 2, 3, }, we let W n denote the time of the occurrence of the n th event W 4 = 5.9 means that the fourth success occurred at time 5.9 The elapsed time between the occurrence of the (n 1) th and n th events is denoted by I n and is called the nth inter arrival time 20.5

13 cont d The Poisson Process can be used to model arrivals. It is also used for waiting times and inter arrival times For each n {0, 1, 2, 3, }, we let W n denote the time of the occurrence of the n th event W 4 = 5.9 means that the fourth success occurred at time 5.9 The random variable W n called that n th waiting time The elapsed time between the occurrence of the (n 1) th and n th events is denoted by I n and is called the nth inter arrival time 20.5

14 cont d The Poisson Process can be used to model arrivals. It is also used for waiting times and inter arrival times For each n {0, 1, 2, 3, }, we let W n denote the time of the occurrence of the n th event W 4 = 5.9 means that the fourth success occurred at time 5.9 The random variable W n called that n th waiting time If W n = t then we have n realizations from the Unif (0, t) The elapsed time between the occurrence of the (n 1) th and n th events is denoted by I n and is called the nth inter arrival time 20.5

15 cont d The Poisson Process can be used to model arrivals. It is also used for waiting times and inter arrival times For each n {0, 1, 2, 3, }, we let W n denote the time of the occurrence of the n th event W 4 = 5.9 means that the fourth success occurred at time 5.9 The random variable W n called that n th waiting time If W n = t then we have n realizations from the Unif (0, t) The elapsed time between the occurrence of the (n 1) th and n th events is denoted by I n and is called the nth inter arrival time W n = n i=1 I i 20.5

16 cont d The Poisson Process can be used to model arrivals. It is also used for waiting times and inter arrival times For each n {0, 1, 2, 3, }, we let W n denote the time of the occurrence of the n th event W 4 = 5.9 means that the fourth success occurred at time 5.9 The random variable W n called that n th waiting time If W n = t then we have n realizations from the Unif (0, t) The elapsed time between the occurrence of the (n 1) th and n th events is denoted by I n and is called the nth inter arrival time W n = n i=1 I i I n = W n W n

17 cont d The Poisson Process can be used to model arrivals. It is also used for waiting times and inter arrival times For each n {0, 1, 2, 3, }, we let W n denote the time of the occurrence of the n th event W 4 = 5.9 means that the fourth success occurred at time 5.9 The random variable W n called that n th waiting time If W n = t then we have n realizations from the Unif (0, t) The elapsed time between the occurrence of the (n 1) th and n th events is denoted by I n and is called the nth inter arrival time W n = n i=1 I i I n = W n W n 1 I 1, I 2, i.i.d Exp(λ) 20.5

18 Example 51 Suppose that phone calls arrive at a switchboard according to a Poisson Process at a rate of 2 per minute 1 Let X be the number of calls between 9:30 and 9:45. Find the distribution of X 2 Let T be the time between the 8 th and 9 th calls. What is the distribution (and parameters) of T? 3 What is the probability that exactly 10 calls come in the next 4 minutes? 4 What is the probability that the next call comes in 30 seconds and the second call comes at least 45 seconds after that? 5 Given there are exactly 7 calls in 3 minutes, what is the probability that they all came in the last minute? 20.6

19 Example 51 cont d Solution. 1 X Poisson(λ = 2 15 = 30) 2 T Exp(λ = 2) 3 P(exactly 10 calls come in the next 4 minutes) = e ! = P(I n <.5, I n+1 >.75) = P(I n <.5)P(I n+1 >.75) = (1 e 1 )(e 1.5 ) = = Let Z be the number of calls came in last minute Z Binomial(n = 7, p = 1 3 ) P(Z = 7) =

20 Example 52 At any point during a Stat 225 exam, the next person to drop a calculator will take 5 minutes on average to do so. Let C represent the time until the next person drops their calculator 1 Name the distribution and parameter(s) of C 2 Find the following probabilities: (i) P(C > 5 C < 10) (ii) P(C 8 C < 15) (iii) C is at least 7 given that it is more than

21 Example 52 cont d Solution. 1 C Exp(λ = 1 5 ) 2 (i) P(C > 5 C < 10) = P(5<C<10) P(C<10) (ii) P(C 8 C < 15) = P(8 C<15) P(C<15) = e 1 e 2 = e 2 = e 1.6 e 3 = e 3 (iii) P(C 7 C > 5) = P(C 2) = e.4 =