1. 1 Birth and death process N(t) Depends on how fast arrivals or departures occur Objective N(t) = # of customers at time t. λ arrivals (births) departures (deaths) μ

2. 2 Behavior of the system λ>μ λ<μ Possible evolution of N(t) Time 1 2 3 4 5 6 7 8 9 10 11 1 2 3 busyidle N(t)

3. 3 General arrival and departure rates λ n Depends on the number of customers (n) in the system Example μ n Depends on the number of customers in the system Example

4. 4 Changing the scale of a unit time Number of arrivals/unit time Follows the Poisson distribution with rate λ n Inter-arrival time of successive arrivals is exponentially distributed Average inter-arrival time = 1/ λ n What is the avg. # of customers arriving in dt? Time

5. 5 Probability of one arrival in dt dt so small Number of arrivals in dt, X is a r.v. X=1 with probability p X=0 with probability 1-p Average number of arrivals in dt Prob (having one arrival in dt) = λ n dt dt

6. 6 Probability of having 2 events in dt Departure rate in dt μ n dt Arrival rate in dt λ n dt What is the probability Of having an (arrival+departure), (2 arrivals or departures)

7. 7 Probability distribution of N(t) P n (t) The probability of getting n customers by time t The distribution of the # of customers in system t+dtt ?n n-1: arrival n+1: departure n: none of the above

8. 8 Differential equation monitoring evolution of # customers These are solved Numerically using MATLAB We will explore the cases Of pure death And pure birth

9. 9 Pure birth process In this case μ n =0, n >= 0 λ n = λ, n >= 0 Hence,

10. First order differential equation 10

11. 11 Pure death process In this case λ n =0, n >= 0 μ n = μ

12. 12 Queuing system Transient phase Steady state Behavior is independent of t P n (t) λ μ t transient Steady state

13. 13 Differential equation: steady state analysis Limiting case

14. 14 Solving the equations n=1 n=2 (1) (1) =>

15. 15 PnPn What about P 0

16. 16 Normalization equation

17. 17 Conditional probability and conditional expectation: d.r.v. X and Y are discrete r.v. Conditional probability mass function Of X given that Y=y Conditional expectation of X given that Y=y

18. 18 Conditional probability and expectation: continuous r.v. If X and Y have a joint pdf f X,Y (x,y) Then, the conditional probability density function Of X given that Y=y The conditional expectation Of X given that Y=y

19. 19 Computing expectations by conditioning Denote E[X|Y]: function of the r.v. Y Whose value at Y=y is E[X|Y=y] E[X|Y]: is itself a random variable Property of conditional expectation if Y is a discrete r.v. if Y is continuous with density f Y (y) => (1) (2) (3)

20. 20 Proof of equation when X and Y are discrete

21. 21 Problem 1 Sam will read Either one chapter of his probability book or One chapter of his history book If the number of misprints in a chapter Of his probability book is Poisson distributed with mean 2 Of his history book is Poisson distributed with mean 5 Assuming Sam equally likely to choose either book What is the expected number of misprints he comes across?

22. 22 Solution

23. 23 Problem 2 A miner is trapped in a mine containing three doors First door leads to a tunnel that takes him to safety  After 2 hours of travel Second door leads to a tunnel that returns him to the mine  After 3 hours of travel Third door Leads to a tunnel that returns him to the mine  After 5 hours Assuming he is equally likely to choose any door What is the expected length of time until he reaches safety?

24. 24 Solution

25. 25 Computing probabilities by conditioning Let E denote an arbitrary event X is a random variable defined by It follows from the definition of X

26. 26 Problem 3 Suppose that the number of people Who visit a yoga studio each day is a Poisson random variable with mean λ Suppose further that each person who visit is, independently, female with probability p Or male with probability 1-p Find the joint probability That n women and m men visit the academy today

27. 27 Solution Let N 1 denote the number of women, N 2 the number of men Who visit the academy today N= N 1 +N 2 : total number of people who visit Conditioning on N gives Because P(N1=n,N2=m|N=i)=0 when i != n+m

28. 28 Solution (cont’d) Each of the n+m visit is independently a woman with probability p The conditional probability That n of them are women is  The binomial probability of n successes in n+m trials

29. 29 Solution: analysis When each of a Poisson number of events is independently classified As either being type 1 with probability p Or type 2 with probability (1-p) => the numbers of type 1 and 2 events Are independent Poisson random variables

30. 30 Problem 4 At a party N men take off their hats The hats are then mixed up and Each man randomly selects one A match occurs if a man selects his own hat What is the probability of no matches?

31. 31 Solution E = event that no matches occur P(E) = P n : explicit dependence on n Start by conditioning Whether or not the first man selects his own hat M: if he did, M c : if he didn’t P(E|M c )  Probability no matches when n-1 men select of n-1  That does not contain the hat of one of these men

32. 32 Solution (cont’d) P(E|M c ) Either there are no matches and Extra man does not select the extra hat => P n-1 (as if the extra hat belongs to this man) Or there are no matches Extra man does select the extra hat => (1/n-1)xP n-2

33. 33 Solution (cont’d) P n is the probability of no matches When n men select among their own hats => P 1 =0 and P 2 = ½ =>

34. 34 Problem 5: continuous random variables The probability density function of a non-negative random variable X is given by Compute the constant λ?

35. 35 Problem 6: continuous random variables Buses arrives at a specified stop at 15 min intervals Starting at 7:00 AM They arrive at 7:00, 7:15, 7:30, 7:45 If the passenger arrives at the stop at a time Uniformly distributed between 7:00 and 7:30 Find the probability that he waits less than 5 min? Solution Let X denote the number of minutes past 7 That the passenger arrives at the stop =>X is uniformly distributed over (0, 30)

36. 36 Problem 7: conditional probability Suppose that p(x,y) the joint probability mass function of X and Y is given by P(0,0) =.4, P(0,1) =.2, P(1,0) =.1, P(1,1) =.3 Calculate the conditional probability mass function of X given Y = 1

37. 37 counting process A stochastic process {N(t), t>=0} is said to be a counting process if N(t) represents the total number of events that occur by time t N(t) must satisfy  N(t) >= 0  N(t) is integer valued  If s < t, then N(s) <= N(t)  For s < t, N(s) – N(t) = # events in the interval (s,t] Independent increments # of events in disjoint time intervals are independent

38. 38 Poisson process The counting process {N(t), t>=0} is Said to be a Poisson process having rate λ, if N(0) = 0 The process has independent increments The # of events in any interval of length t is  Poisson distributed with mean λt, that is

39. 39 Properties of the Poisson process Superposition property If k independent Poisson processes A1, A2, …, An Are combined into a single process A => A is still Poisson with rate Equal to the sum of individual λi of Ai

40. 40 Properties of the Poisson process (cont’d) Decomposition property Just the reverse process “A” is a Poisson process split into n processes Using probability Pi The other processes are Poisson With rate Pi.λ