Lecture 12.5 – Additional Issues Concerning Discrete-Time Markov Chains

1. Lecture 12.5 – Additional Issues Concerning Discrete-Time Markov Chains Topics • Review of DTMC • Classification of states • Economic analysis • First-time passage • Absorbing states

2. Discrete-Time Markov Chain A stochastic process {Xn} where nN = {0, 1, 2, . . . } is called a discrete-timeMarkov chain if Pr{Xn+1 = j|X0 = k0, . . . , Xn-1 = kn-1, Xn = i} = Pr{ Xn+1 = j |Xn = i}transition probabilities for every i, j, k0, . . . , kn-1 and for every n. The future behavior of the system depends only on the current state i and not on any of the previous states.

3. The one-step transition matrix for a Markov chain with states S = { 0, 1, 2 } is where pij = Pr{X1 = j| X0 = i } Stationary Transition Probabilities Pr{Xn+1 = j |Xn = i } = Pr{X1 = j |X0 = i } for all n (They don’t change over time) We will only consider stationary Markov chains.

4. Classification of States Accessible: Possible to go from state i to state j (path exists in the network from i to j). Two states communicate if both are accessible from each other. A system is irreducible if all states communicate. State i is recurrent if the system will return to it after leaving some time in the future. If a state is not recurrent, it is transient.

5. Classification of States (continued) A state is periodic if it can only return to itself after a fixed number of transitions greater than 1 (or multiple of a fixed number). A state that is not periodic is aperiodic. a. Each state visited every 3 iterations b. Each state visited in multiples of 3 iterations

6. Classification of States (continued) Anabsorbingstate is one that locks in the system once it enters. This diagram might represent the wealth of a gambler who begins with $2 and makes a series of wagers for $1 each. Let ai be the event of winning in state i and dithe event of losing in state i. There are two absorbing states: 0 and 4.

7. Classification of States (continued) Class: set of states thatcommunicatewith each other. A class is either allrecurrentor alltransientand may be either all periodicor aperiodic. Statesin atransientclass communicate only with each other so no arcs enter any of the corresponding nodes in the network diagram from outside the class. Arcs may leave, though, passing from a node in the class to one outside.

8. Illustration of Concepts Example 1 Every pair of statescommunicates, forming a singlerecurrentclass; however, the states are not periodic. Thus the stochastic process isaperiodic andirreducible.

9. Illustration of Concepts Example 2 States 0 and 1 communicate and for arecurrent class. States 3 and 4 form separatetransient classes. State 2 is an absorbing state and forms arecurrent class.

10. Illustration of Concepts Example 3 Every state communicates with every other state, so we have irreducible stochastic process. Periodic? Yes, so Markov chain is irreducible and periodic.

11. 3 Classification of States Example .6 .7 1 2 .4 4 .5 .4 .5 .5 .3 .8 .1 5 .2

12. A state j is accessible from state i if pij(n) > 0 for some n > 0. In example, state 2 is accessible from state 1 & state 3 is accessible from state 5 but state 3 is not accessible from state 2. States i and jcommunicateif i is accessible from jandj is accessible from i. States 1 & 2 communicate; also states 3, 4 & 5 communicate. States 2 & 4 do not communicate States 1 & 2 form one communicating class. States 3,4 & 5 form a 2nd communicating class.

13. If all states in a Markov chain communicate (i.e.,all states are members of the same communicating class) then the chain is irreducible. The current example is not an irreducible Markov chain. Neither is the Gambler’s Ruin example which has 3 classes: {0}, {1, 2, 3} and {4}. First Passage Times Let fii = probability that the process will return to state i (eventually) given that it starts in state i. If fii = 1 then state i is called recurrent. If fii < 1 then state i is called transient.

14. If pii= 1 then state i is called an absorbing state. Above example has no absorbing states States 0 & 4 are absorbing in Gambler’s Ruin problem. The period of a statei is the smallest k > 1 such that all paths leading back to ihave a length that is a multiple of k; i.e., pii(n) = 0 unless n = k, 2k, 3k, . . . If a process can be in state i at time nor time n + 1 having started at state i then state i is aperiodic. Each of the states in the current example are aperiodic

15. Example of Periodicity - Gambler’s Ruin States 1,2 and 3 each have period 2. 01 2 34 0 1 0 0 0 0 1 1-p 0 p 0 0 2 0 1-p 0 p 0 3 0 0 1-p 0 p 4 0 0 0 0 1 If all states in a Markov chain are recurrent, aperiodic, & the chain is irreducible then it is ergodic.

16. State-transition network For example, 1 2 3 Existence of Steady-State Probabilities A Markov chain is ergodic if it is aperiodic and allows the attainment of any future state from any initial state after one or more transitions. If these conditions hold, then Conclusion: chain is ergodic.

17. The cost (profit) of being in a particular state is represented by the m-dimensional column vector where each component is the cost associated with state i. The cost of a transition is embodied in the mm matrix . where each component specifies the cost of going from state i to state j in a single step. Economic Analysis • Two kinds of economic effects: • those incurred when the system is in a specified state, and • those incurred when the system makes a transition from one state to another.

18. Expected Cost for Markov Chain Expected cost of being in state i: Let C = (c1, . . . cm)T ei = (0, 0, 1, 0, 0) be the ith row of the mm identity matrix, and fn = a random variable representing the economic return associated with the stochastic process at time n. Property 3: Let {Xn: n = 0, 1, . . .} be a Markov chain with finite state space S, state-transition matrix P, and expected state cost (profit) vector C. Assuming that the process starts in state i, the expected cost (profit) at the nth step is given by  E[fn(Xn) |X0 = i] = eiP(n)C.

19. Additional Cost Results What if the initial state is not known? Property 5: Let {Xn: n = 0, 1, . . .} be a Markov chain with finite state space S, state-transition matrix P, initial probability vector q(0),and expected state cost (profit) vector C. The expected economic return at the nth step is given by E[fn(Xn) |q(0)] = q(0)P(n)C. Property 6: Let {Xn: n = 0, 1, . . .} be a Markov chain with finite state space S, state-transition matrix P, steady-state vector π,and expected state cost (profit) vector C. Then the long-run average return per unit time is given by SiSπici = πC.

20. Insurance Company Example An insurance company charges customers annual premiums based on their accident history in the following fashion: • No accident in last 2 years: $250 annual premium •  Accidents in each of last 2 years: $800 annual premium  Accident in only 1 of last 2 years: $400 annual premium • Historical statistics: • If a customer had an accident last year then they have a 10% chance of having one this year; • If they had no accident last year then they have a 3% chance of having one this year.

21. (N, N) (N, Y) (Y, N) (Y, Y) (N, N) 0.97 0.03 0 0 (N, Y) 0 0 0.90 0.10 (Y, N) 0.97 0.03 0 0 (Y, Y) 0 0 0.90 0.10 P = Problem: Find the steady-state probability and the long-run average annual premium paid by the customer. Solution approach: Construct a Markov chain with four states: (N, N), (N, Y), (Y, N), (Y,Y) where these indicate (accident last year, accident this year).

22. N, Y Y, Y Y, N N, N State-Transition Network for Insurance Company .90 .03 .90 .03 .97 .10 .97 .10 • This is an ergodicMarkov chain. • All states communicate (irreducible) • Each state is recurrent (you will return, eventually) • Each state is aperiodic

23. Solving the Steady–State Equations m i=1 (N,N) = 0.97(N,N) + 0.97(Y,N) (N,Y) = 0.03(N,N) + 0.03(Y,N) (Y,N) = 0.9(N,Y) + 0.9(Y,Y) (N,N) + (N,Y)+(Y,N) + (Y,Y) =1 pj= å pipij, j = 0,…,m å pj= 1, pj 0,  j m j =1 Solution: (N,N) = 0.939, (N,Y) = 0.029, (Y,N) = 0.029, (Y,Y) = 0.003 & the long-run average annual premium is 0.939*250 + 0.029*400 + 0.029*400 + 0.003*800 = 260.5

24. Markov Chain Add-in Matrix

25. Economic Data and Solution

26. Transient Analysis for Insurance Company

27. First Passage Times • Let ij = expected number of steps to transition • from state i to state j • If the probability that we will eventually visit state j • given that we start in i is less than 1, then • we will have ij= +. For example, in the Gambler’s Ruin problem, 20 = + because there is a positive probability that we will be absorbed in state 4 given that we start in state 2 (and hence visit state 0).

28. mij= 1 + å pirmrj, for i = 0,1, . . . , m–1 rj Computations for All States Recurrent If the probability of eventually visiting state j given that we start in i is 1 then the expected number of steps until we first visit j is given by We go from i to r in the first step with probability pir and it takes mrj steps from r to j. It will always take at least one step. For j fixed, we have linear system in m equations and m unknowns mij , i = 0,1, . . . , m–1.

29. First-Passage Analysis for Insurance Company Suppose that we start in state (N,N) and want to find the expected number of years until we have accidents in two consecutive years (Y,Y). This transition will occur with probability 1, eventually. • For convenience number the states • 0 1 2 3 • (N,N) (N,Y) (Y,N) (Y,Y) • Then, 03 = 1 + p00 03 + p01 13 + p0223 • 13 = 1 + p10 03 + p11 13 + p1223 • 23 = 1 + p20 03 + p21 13 + p2223

30. Solution: 03 = 343.3, 13 = 310,23 = 343.3 So, on average it takes 343.3 years to transition from (N,N) to (Y,Y). Note, 03 = 23. Why? Note, 13 < 03. First-Passage Computations (N, N) (N, Y) (Y, N) (Y, Y) (N, N) 0.97 0.03 0 0 (N, Y) 0 0 0.90 0.10 (Y, N) 0.97 0.03 0 0 (Y, Y) 0 0 0.90 0.10 0 1 2 3 states Using P = 03 = 1 + 0.9703 + 0.0313 13 = 1 + 0.923 23 = 1 + 0.9703 + 0.0313

31. First Passage Probabilities

32. Game of Craps Probability of win = Pr{ 7 or 11 } = 0.167 + 0.056 = 0.223 Probability of loss = Pr{ 2, 3, 12 } = 0.028 + 0.56 + 0.028 = 0.112

33. First Passage Probabilities for Craps

34. Absorbing States An absorbing state is a state j with pjj = 1. Given that we start in state i, we can calculate the probability of being absorbed in state j. We essentially performed this calculation for the Gambler’s Ruin problem by finding • P(n)= (pij(n) ) for large n. But we can use a more efficient analysis like that used for calculating first passage times.

35. Let 0, 1, . . . , k be transient states and k + 1, . . . , m – 1 be absorbing states. • Let qij = probability of being absorbed in state j given that we start in transient state i. • Then for each j we have the following relationship • qij= pij + pirqrj, i = 0, 1, . . . , k k r = 0 Go directly to j Go to r and then to j For fixed j (absorbing state) we have k + 1 linear equations in k + 1 unknowns, qrj, i = 0, 1, . . . , k.

36. Absorbing States – Gambler’s Ruin Suppose that we start with $2 and want to calculate the probability of going broke, i.e., of being absorbed in state 0. We know p00 = 1 and p40 = 0, thus q20 = p20 + p21q10 + p22q20 + p23q30 (+p24q40) q10 = p10 + p11q10 + p12q20 + p13q30 + 0 q30 = p30 + p31q10 + p32q20 + p33q30 + 0 where P = 0 1 2 3 4 0 1 0 0 0 0 1 1-p 0 p 0 0 2 0 1-p 0 p 0 3 0 0 1-p 0 p 4 0 0 0 0 1

37. Solution to Gambler’s Ruin Example • Now we have three equations with three unknowns. • Using p = 0.75 (probability of winning a single bet) • we have • q20 = 0 + 0.25 q10 + 0.75 q30 • q10 = 0.25 + 0.75 q20 • q30 = 0 + 0.25 q20 • Solving yields q10 = 0.325, q20 = 0.1, q30 = 0.025 • (This is consistent with the values found earlier.)

38. What You Should Know About The Mathematics of DTMCs • How to classify states. • What an ergodic process is. • How to perform economic analysis. • How to compute first-time passages. • How to compute absorbing probabilities.