Birth and death process

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

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

3. General arrival and departure rates • λn • Depends on the number of customers (n) in the system • Example • μn • 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 λn • Inter-arrival time of successive arrivals • is exponentially distributed • Average inter-arrival time = 1/ λn • 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) = λn dt dt

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

7. Probability distribution of N(t) • Pn (t) • The probability of getting n customers by time t • The distribution of the # of customers in system t t+dt ? 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 • μn =0, n >= 0 • λn = λ, n >= 0 Hence,

10. First order differential equation

11. Pure death process • In this case • λn =0, n >= 0 • μn = μ

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

13. Differential equation: steady state analysis • Limiting case

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

15. Pn • What about P0

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 fX,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 fY (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 • N1 denote the number of women, N2 the number of men • Who visit the academy today • N= N1 +N2 : 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) = Pn : explicit dependence on n • Start by conditioning • Whether or not the first man selects his own hat • M: if he did, Mc : if he didn’t • P(E|Mc) • 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|Mc) • Either there are no matches and • Extra man does not select the extra hat • => Pn-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)xPn-2

33. Solution (cont’d) • Pn is the probability of no matches • When n men select among their own hats • => P1 =0 and P2 = 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