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- 1. Markov Analysis by Dr.V.V.HaraGopal Professor,Dept of Statistics, Osmania University, Hyderabad-7
- 2. STOCHASTIC PROCESS: Stochastic “denotes the process of selecting from among a group of theoretically possible alternatives those elements or factors whose combination will most closely approximate a desired result” Stochastic models are not always exact Stochastic models are useful shorthand representations of complicated processes. Markov property Only the current value of variable is relevant for future predictions No information from past prices or path
- 3. What's a Markov Process? 3
- 4. • A Markov analysis looks at a sequence of events, and analyzes the tendency of one event to be followed by another. Using this analysis, you can generate a new sequence of random but related events, which will look similar to the original. 4
- 5. • A Markov process is useful for analyzing dependent random events - that is, events whose likelihood depends on what happened last. It would NOT be a good way to model a coin flip, for example, since every time you toss the coin, it has no memory of what happened before. The sequence of heads and tails are not inter- related. They are independent events. 5
- 6. • But many random events are affected by what happened before. For example, yesterday's weather does have an influence on what today's weather is. They are not independent events. 6
- 7. • A Markov model could look at a long sequence of rainy and sunny days, and analyze the likelihood that one kind of weather gets followed by another kind. Let's say it was found that 25% of the time, a rainy day was followed by a sunny day, and 75% of the time, rain was followed by more rain. Let's say we found out additionally, that sunny days were followed 50% of the time by rain, and 50% by sun. Given this analysis, we could generate a new sequence of statistically similar weather by following these steps: 1) Start with today's weather. 2) Given today's weather, choose a random number to pick tomorrow's weather. 3) Make tomorrow's weather "today's weather" and go back to step 2. 7
- 8. • What we'd get is a sequence of days like: Sunny Sunny Rainy Rainy Rainy Rainy Sunny Rainy Rainy Sunny Sunny... • In other words, the "output chain" would reflect statistically the transition probabilities derived from weather we observed. • This stream of events is called a Markov Chain. A Markov Chain, while similar to the source in the small, is often nonsensical in the large. (Which is why it's a lousy way to predict weather.) That is, the overall shape of the generated material will bear little formal resemblance to the overall shape of the source. But taken a few events at a time, things feel familiar. 8
- 9. 9 Markov Analysis In an industry with 3 firms we could look at the market share of each firm at any time and the shares have to add up to 100%. If we had information about how customers might change from one firm to the next then we could predict future market shares. This is just one example of Markov Analysis. In general we use current probabilities and transitional information to figure future probabilities. Here we study an accounts receivable example.
- 10. 10 Say in the accounts receivable department, accounts are in one of 4 states, or categories: state 1 - s1, paid, state 2 – s2, bad debt, here defined as overdue more than three months and company writes off the debt, state 3 – s3, overdue less than one month, state 4 – s4, overdue between one and three months. Note the states are mutually exclusive and collectively exhaustive. At any given time there will be a certain fraction of accounts in each state. Say in the current period we have the % of accounts receivable in each state. In general we have a row vector of probabilities (s1, s2, s3, s4).
- 11. 11 Say now there are 25% of the accounts in each state. We would have (.25, .25, .25, .25). This set of numbers is called the vector of state probabilities. Next the matrix of transition probabilities: 1 0 0 0 0 1 0 0 .6 0 .2 .2 .4 .1 .3 .2 The first row is being in the first state in the current period, the second row is being in the second state in the current period, and so on down the rows.
- 12. 12 Now, in the matrix of transition probabilities let’s think about each column. The first column says an account is in state 1 in the next period. The second column says an account is in state 2 in the next period, and so on. Note the first row has values 1, 0, 0, 0. The values add to one. If an account is all paid this period then it must be all paid next period. So the 1 means there is a 100% chance of being all paid next period and 0 % chance in being in any other category. In the second row we have 0, 1, 0, 0. If an account starts as bad it will always be bad. So it has a zero chance of being paid, less than one period overdue or be between 1 and 3 periods overdue.
- 13. 13 In row three we have .6, 0, .2, .2. If an account is less than 1 month overdue now, next period there is a 60% chance it will be all paid, 0% chance it will be bad because it can not be over 3 months bad, 20% chance it will be less than a month - wait, wait wait. How can an account be bad less than one month now and less than one month next period? Any account can have more than one unpaid bill and we keep track of the oldest unpaid bill for the category. Note that each row has to add up to 1. Now we are ready to ask a question. If each state has 25% of the accounts this period, what percent will be in each state next period? We take the row vector and multiply by the matrix of transition probabilities, as seen on the next screen.
- 14. 14 (t, u, v, w) d e f g h i j k l m n o p q r s We will end up with (a1, a2, a3, a4), where a1 = t(d) + u(h) + v(l) + w(p) a2 = t(e) + u(i) + v(m) + w(q) a3 = t(f) + u(j) + v(n) + w(r) a4 = t(g) + u(k) + v(o) + w(s) Matrix multiplication
- 15. 15 (.25, .25, .25, .25) 1 0 0 0 0 1 0 0 .6 0 .2 .2 .4 .1 .3. .2 We will end up with (a1, a2, a3, a4), where a1 = .25(1) + .25(0) + .25(.6) + .25(.4) = .5 a2 = .25(0) + .25(1) + .25(0) + .25(.1) = .275 a3 = .25(0) + .25(0) + .25(.2) + .25(.3) = .125 a4 = .25(0) + .25(0) + .25(.2) + .25(.2) = .1
- 16. 16 So, if we start with 25% of accounts in state 1, then next period we have 50 % of accounts in state 1, and so on. If you wanted to see what the %’s in each state would be two periods from the start we would do the same calculation, but use the row vector that we ended with in the first period (.5, .275, .125, .1) If I wanted to see the probabilities of being in each state at the end of two months I would put 2 for number of transitions and would get (.615, .285, .055, .045).
- 17. 17 Now, in this particular problem we have what are called absorbing states. Not all problems have absorbing states and if not just do what we have done up to now. An absorbing state is one such that once in it one stays in that state. For instance, once debt is bad it is always bad. Now, in the long run all debt will either be bad or paid. The Markov Analysis problem that has absorbing states, no matter how many transitions you put there is always an output section called matrices and it includes the FA matrix. In our problem we have .9655 .0345 .8621 .1379 The rows represent the non-absorbing states and the columns represent the absorbing states.
- 18. 18 The first row is state 3, debt of less than one month, and row 2 is state 4, debt of 1 to 3 months. Column 1 is paid debt and column 2 is bad debt. So, the first row says 96.55% of less than one month debt will be paid over the long term and only 3.45% of this debt will not be paid. The second row means that 86.21% of 1 to 3 month debt will be paid over the long terms and 13.79% of this debt will go bad. Say that there is $2000 in the less than one month overdue category and $5000 in the 1 to 3 month overdue category. How much can the company expect to collect of this $7000 and how much will it not collect?
- 19. 19 We have to do matrix multiplication, here (2000, 5000) .9655 .0345 .8621 .1379 ([{2000*.9655} + {5000*.8621}], [{2000*.0345} + {5000*.1379}]) or (6241.50, 758.5). So of the $7000 in states 3 and 4, $6241.50 can be expected to be collected and $758.5 would not be collected.
- 20. Markov Processes • Markov process models are useful in studying the evolution of systems over repeated trials or sequential time periods or stages. • Examples: – Brand Loyalty – Equipment performance – Stock performance
- 21. Markov Processes • When utilized, they can state the probability of switching from one state to another at a given period of time • Examples: – The probability that a person buying Colgate this period will purchase Crest next period – The probability that a machine that is working properly this period will break down the next period
- 22. Markov Processes • A Markov system (or Markov process or Markov chain) is a system that can be in one of several (numbered) states, and can pass from one state to another each time step according to fixed probabilities. • If a Markov system is in state i, there is a fixed probability, pij, of it going into state j the next time step, and pij is called a transition probability.
- 23. Markov Processes • A Markov system can be illustrated by means of a state transition diagram, which is a diagram showing all the states and transition probabilities– probabilities of switching from one state to another.
- 24. Transition Diagram 1 2 3 .4 .8 .2 .35 .65 .50 .15 What does the diagram mean?
- 25. Transition Matrix • The matrix P whose ijth entry is pij is called the transition matrix associated with the system. • The entries in each row add up to 1. • Thus, for instance, a 2 2 transition matrix P would be set up as shown at the right. 1 2 1 P11 P12 2 P21 P22 From To
- 26. Diagram & Matrix 1 2 3 .4 .8 .2 .35 .6 .50 .15 1 2 3 1 .2 .8 0 2 .4 0 .6 3 .5 .35 .15 From To
- 27. Vectors & Transition Matrix • A probability vector is a row vector in which the entries are nonnegative and add up to 1. • The entries in a probability vector can represent the probabilities of finding a system in each of the states.
- 28. Probability Vector • Let P = .2 .8 0 .4 0 .6 .5 .35 .15
- 29. State Probabilities • The state probabilities at any stage of the process can be recursively calculated by multiplying the initial state probabilities by the state of the process at stage n.
- 30. State Probabilities Πi (n) Probability that the system is in state i in period n Π(n) = [ Π1 (n) Π2 (n) ] Denotes the vector of state probabilities for the system in period n Π(n+1) = Π(n) P State probabilities for period n+1 can be found by multiplying the known state probabilities for period n by the transition matrix
- 31. State Probabilities • Example: ∀Π(n) = [π1 (n) π2 (n) ] ∀Π(1) = Π(0) P ∀Π(2) = Π(1) P ∀Π(3) = Π(2) P ∀Π(n+1) = Π(n) P
- 32. Steady State Probabilities • The probabilities that we approach after a large number of transitions are referred to as steady state probabilities. • As n gets large, the state probabilities at the (n+1)th period are very close to those at the nth period.
- 33. Steady State Probabilities • Knowing this, we can compute steady state probabilities without having to carry out a large # of calculations Π(n) = [π1 (n) π2 (n) ] [ π1 (n+1) π2 (n+1) ] = p11 p12 [π1 (n) π2 (n)] p21 p22
- 34. Example • Hari, a persistent salesman, calls ABC Hardware Store once a week hoping to speak with the store's buying agent, Shyam. If Shyam does not accept Hari's call this week, the probability he will do the same next week (and not accept his call) is .35. On the other hand, if he accepts Hari's call this week, the probability he will not accept his call next week is .20.
- 35. Example: Transition Matrix Refuses Accepts Refuses .35 .65 Accepts .20 .80 This Week’s Call Next Week’s Call
- 36. Example • How many times per year can Hari expect to talk to Shyam? • Answer: To find the expected number of accepted calls per year, find the long-run proportion (probability) of a call being accepted and multiply it by 52 weeks.
- 37. Example Let π1 = long run proportion of refused calls π2 = long run proportion of accepted calls Then, .35 .65 [π1 π2 ] .20 .80 = [π1 π2 ]
- 38. Example .35π1 + .20π2 = π1 (1) .65π1 + .80π2 = π2 (2) π1 + π2 = 1 (3) Solve for π1 and π2
- 39. • The probability of the system being in a particular state after a large number of stages is called a steady-state probability.
- 40. Example: Machine Adjustment ToTo FromFrom In adj. (1)In adj. (1) Out of adj. (2)Out of adj. (2) InIn adjustmentadjustment (state 1)(state 1) 0.70.7 0.60.6 Out ofOut of adjustmentadjustment (state 2)(state 2) 0.30.3 0.40.4
- 41. Example: Machine Adjustment Day 1 11 22 11 .7.7 .3.3 Day 2 .7.7 .3.3 If the machine is found to be in adjustment on day 1, what is the likelihood it will be in adjustment on day 3? Not in adjustment?
- 42. Example: Machine Adjustment 11 22 11 22 11 22 11 .7.7 .3.3 .7.7 .3.3 .6.6 .4.4 .49.49 .21.21 .18.18 .12.12 Day 1 Day 2 .67.67 .33.33 Day 3
- 43. 11 22 11 22 11 22 11 11 22 11 22 11 22 11 22.7.7 .3.3 .7.7 .3.3 .6.6 .4.4 .7.7 .3.3 .7.7 .3.3 .6.6 .4.4 .6.6 .4.4 Day 4
- 44. 11 22 11 22 11 22 11 11 22 11 22 11 22 11 22.7.7 .3.3 .7.7 .3.3 .6.6 .4.4 .7.7 .3.3 .7.7 .3.3 .6.6 .4.4 .6.6 .4.4 Day 4
- 45. Example: Machine Adjustment • Day 4: P(S1|S1) = .7(.7)(.7) + .7(.3)(.6) +.3(.6)(.7) +.3(.4)(.6) = .667 P(S2|S1) = .7(.7)(.3) + .7(.3)(.4) + .3(.6) (.3) + 3(.4)(.4) = .333
- 46. Day 5 11 22 11 22 11 22 11 11 22 11 22 11 22 11 22 11 22 11 22 11 22 11 22 11 22 11 22 11 22 11 22.7.7 .3.3 .7.7 .3.3 .6.6 .4.4 .7.7 .3.3 .7.7 .3.3 .6.6 .4.4 .6.6 .4.4 .7.7 .3.3 .6.6 .4.4 .7.7 .3.3 .6.6 .4.4 .7.7 .3.3 .6.6 .4.4 .7.7 .3.3 .6.6 .4.4
- 47. Example: Machine Adjustment • Day 5: P(S1|S1) = .7(.7)(.7)(.7)+ .7(.7)(.3)(.6) +.7(.3)(.6)(.7) +.7(.3)(.4)(.6) + .3(.6)(.7)(.7) + .3(.6)(.3)(.6) + .3(.4)(.6)(.7) + .3(.4)(.4)(.6) = .666 P(S2|S1) = .7(.7)(.7)(.3) + .7(.7)(.3)(.4) + .7(.3)(.6)(.3) + .7(.3)(.4)(.4) + .3(.6)(.7)(.3) + .3(.6)(.3)(.4) + .3(.4)(.6)(.3) + .3(.4)(.4)(.4) = .334 Notice anything interesting?
- 48. Steady State Probabilities • These probabilities are called steady state probabilities • The long term probability of being in a particular state no matter which state you begin in – Steady state prob. (state 1)= .667 – Steady state prob. (state 2) = .333
- 49. Example: Machine Adjustment ToTo FromFrom In adj. (1)In adj. (1) Out of adj. (2)Out of adj. (2) InIn adjustmentadjustment (state 1)(state 1) 0.70.7 0.60.6 Out ofOut of adjustmentadjustment (state 2)(state 2) 0.30.3 0.40.4
- 50. Example: Machine Adjustment ToTo FromFrom In adj. (1)In adj. (1) Out of adj.Out of adj. (2)(2) InIn adjustmentadjustment (state 1)(state 1) pp1111 pp2121 Out ofOut of adjustmentadjustment (state 2)(state 2) pp1212 pp2222
- 51. Steady State Probabilities P(S1 Day n+1|S1) =P(S1 Day n+1|S1) = .7 P(S1 Day n|S1) + .6 P(S2 Day n|S1).7 P(S1 Day n|S1) + .6 P(S2 Day n|S1) PP11 PP22 P(S2 Day n+1|S1) =P(S2 Day n+1|S1) = .3 P(S1 Day n|S1) + .4 P(S2 Day n|S1).3 P(S1 Day n|S1) + .4 P(S2 Day n|S1) PP11 PP22
- 52. Steady State Probabilities PP11 == pp1111PP11 ++ pp2121PP22 PP11 = (1 -= (1 - pp1212 )) PP11 ++ pp2121 (1-(1- PP11)) PP11 = P= P11 -- pp1212 PP11 ++ pp2121 -- pp2121 PP11 pp1212 PP11 ++ pp2121 PP11 == pp2121 PP11 == pp2121 pp1212 ++ pp2121
- 53. Steady State Probabilities p21 p12 + p21 P1 = = = = .6 .3+.6 .6 .9 2 3 p12 p12 + p21 P2 = = = = .3 .3+.6 .3 .9 1 3
- 54. Example – Steady State Let p1 = long run proportion of refused calls p2 = long run proportion of accepted calls Then, .70 .30 [p1 p2 ] .60 .40 = [p1 p2 ]
- 55. Example – Steady State .70p1 + .60p2 = p1 (1) .30p1 + .40p2 = p2 (2) p1 + p2 = 1 (3) Solve for p1 and p2 p1 = 1 – p2 p2 = 1 – p1 Can be restated as:
- 56. Example – Steady State Using equations (2) and (3), substitute p1 = 1 – p2 into (2): .30(1 - p2) + .40p2 = p2 This gives p2 = .3333 Substituting back into equation (3) gives p1 = .67 Thus the expected number of accepted calls per year is: (.76471)(52) = 39.76 or about 40

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