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 Behavior of the system λ>μ λ<μ Possible evolution of N(t) Time busyidle N(t)

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

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

First order differential equation 10

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

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

13 Differential equation: steady state analysis Limiting case

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

15 PnPn What about P 0

16 Normalization equation

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 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 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 Proof of equation when X and Y are discrete

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 Solution

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 Solution

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

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 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 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 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 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 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