1 / 46

CHAPTER 7

CHAPTER 7. Subjective Probability and Bayesian Inference. 7. 1. Subjective Probability. Personal evaluation of probability by individual decision maker Uncertainty exists for decision maker: probability is just a way of measuring it

Download Presentation

CHAPTER 7

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. CHAPTER 7 Subjective Probability and Bayesian Inference

  2. 7. 1. Subjective Probability • Personal evaluation of probability by individual decision maker • Uncertainty exists for decision maker: probability is just a way of measuring it • In dealing with uncertainty, a coherent decision maker effectively uses subjective probability

  3. 7.2. Assessment of Subjective Probabilities Simplest procedure: • Specify the set of all possible events, • ask the decision maker to directly estimate probability of each event • Not a good approach from psychological point of view • Not easy to conceptualize, especially for DM not familiar with probability

  4. Standard Device • Physical instrument or conceptual model • Good tool for obtaining subjective probabilities • Example • A box containing 1000 balls • Balls numbered 1 to 1000 • Balls have 2 colors: red, blue

  5. Standard Device example • To estimate a students’ subjective probability of getting an “A” in SE 447, • we ask him to choose between 2 bets: • Bet X: If he gets an A, he win SR 100 If he doesn’t get A, he wins nothing • Bet Y: If he picks a red ball, he win SR 100 If he picks a blue ball, he wins nothing • We start with proportion of red balls P = 50%, then adjust successively until 2 bets are equal

  6. Other standard devices Pie diagram (spinner) Circle divided into 2 sectors: 1 red 1 blue • Bet Y: If he spins to red section, he win SR 100 If he spins to blue section, he wins nothing • Size of red section is adjusted until 2 bets are equal

  7. Subjective Probability Bias • Standard device must be easy to perceive, to avoid introducing bias 2 kinds of bias: • Task bias: resulting from assessment method (standard device) • Conceptual bias: resulting from mental procedures (heuristics) used by individuals to process information

  8. Mental Heuristics Causing Bias • Representativeness • If x highly represents set A, high probability is given that X  A • Frequency (proportion) ignored • Sample size ignored • Availability • Limits of memory and imagination • Adjustment & anchoring • Starting from obvious reference point, then adjusting for new values. • Anchoring: adjustment is typically not enough • Overconfidence • Underestimating variance

  9. Fractile Probability Assessment • Quartile Assessment: • Determine 3 values • x1, for which p(x > x1) = 0.5 • x2, for which p(x < x2) = p(x2 < x< x1) • x3, for which p(x > x3) = p(x1 < x< x3) • x x2 x1 x3 • F(x) 0.25 0.5 0.75

  10. Fractile Probability Assessment • Quartile Assessment: 4 intervals • Octile Assessment: 8 intervals • Tertile Assessment: 3 intervals. avoids anchoring at the median

  11. Histogram Probability Assessment • Fix the points x1, x2 , …, xm. • Ask the decision maker to assess probabilities p(x1 < x< x2) p(x2 < x< x3) … x1 x2 x3 … x Gives probability distribution (not cumulative p.d. as fractile method)

  12. Assessment Methods & Bias • No evidence to favor either fractile or histogram methods • One factor that reinforces anchoring bias is self-consistency • Bias can be reduced by “pre-assessment conditioning”: training, for/against arguments • The act of probability assessment causes a re-evaluation of uncertainty

  13. 7.3. Impact of New Information(Bayes’ Theorem) • After developing subjective probability distribution • Assume new information becomes available Example: new data is collected • According to coherence principle, DM must take new information in consideration, thus Subjective probability must be revised • How? using Bayes’ theorem

  14. Bayes’ Theorem Example • Suppose your subjective probability distribution for weather tomorrow is: • chances of being sunny P(S) = 0.6 • chances of being not sunny P(N) = 0.4 • If the TV weather forecast predicted a cloudy day tomorrow. How should you change P(S)? • Assume we are dealing with mutually exclusive and collectively exhaustive events such as sunny or not sunny.

  15. Impact of Information • We assume the weather forecaster predicts either • cloudy day C, or • bright day B. • To change P(S), we use the joint probability = conditional probability * marginal probability • P(C,S) = P(C|S)P(S) P(B,S) = P(B|S)P(S) • P(C,N) = P(C|N)P(N) P(B,N)= P(B|N)P(N)

  16. Impact of Information • To obtain j.p.m.f we need the conditional probabilities P(C|S) and P(C|N). • These can be obtained from historical data. How? • In past 100 sunny days, cloudy forecast in 20 days P(C|S) = 0.2 P(B|S) = 0.8 • In past 100 cloudy days, cloudy forecast in 90 days P(C|N) = 0.9 P(B|N) = 0.1

  17. Joint probabilityCalculations • Joint probability P(A,B) = conditional probability (likelihood) P(A|B) * marginal probability P(B) • Cloudy forecast • P(C,S) = P(C|S)P(S) = 0.2(0.6) = 0.12 • P(C,N) = P(C|N)P(N) = 0.9(0.4) = 0.36 • Sunny forecast • P(B,S) = P(B|S)P(S) = 0.8(0.6) = 0.48 • P(B,N) = P(B|N)P(N) = 0.1(0.4) = 0.04

  18. Bayes’ Theorem S N C P(C,S) P(C,N)P(C) B P(B,S)P(B,N) P(B) P(S) P(N) • P(S|C) = P(C,S) / P(C) = P(C|S)P(S ) / P(C) • P(S|C) P(C|S)P(S )

  19. Joint probabilityTable S N C 0.12 0.36 0.48 B 0.48 0.04 0.52 0.6 0.4 • P(S|C) = 0.12/0.48 = 0.25 posterior (conditional) probability • Compare to P(S) = 0.6 prior (marginal) prob. • P(S) decreased because of C forecast

  20. Prior and Posterior Probabilities • Prior means before. Prior probability is the probability P(S) before the information was heard. • Posterior means after. It is probability obtained after incorporating the new forecast information. It is P(S|C). It is obtained using Bayes’ theorem.

  21. Example with 3 states 3 demand possibilities for new product • High P(H) = 0.6 • Medium P(M) = 0.1 • Low P(L) = 0.3 Market research gives Average result: • 30% of time if true demand is High • 50% of time if true demand is Medium • 90% of time if true demand is Low

  22. Example : Probability Table for Average result State Prior Likelihood Joint Posterior S P(S) P(A|S) P(S,A) P(S|A) H 0.6 0.3 0.18 0.36 M 0.1 0.5 0.05 0.10 L 0.3 0.9 0.27 0.54  1.0 0.50 1.00 • If market research gives Average result: P(H), P(L), P(M)

  23. Ex: Sequential Bayesian Analysis • An oil company has 3 drilling sites: X, Y, Z 3 possible reserve states: • No reserves P(N) = 0.5 • Small reserves P(S) = 0.3 • Large reserves P(L) = 0.2 3 possible drilling outcomes: • Dry D • Wet W • Gushing G

  24. Ex: Sequential Bayesian Analysis If reserves are: • None (N) all wells will be dry (D) • Large (L) all wells will be Gushing (G) • Small (S) some Dry (D) and some wet (W) wells P(1D/1) = 0.8 P(2D/2) = 0.2 P(3D/3) = 0 • All sites are equally favorable • Assume order of drilling is: XYZ • Notation: DX = probability of Dry well at site X

  25. Ex: Probability Table for Site X State Prior Conditional Joint K P(K) P(D|K) P(W|K) P(G|K) DX WX GX N 0.5 1 0 0 0.5 0 0 S 0.3 0.8 0.2 0 0.24 0.06 0 L 0.2 0 0 1 0 0 0.2  1.0 0.74 0.06 0.2 • Stop exploratory drilling at X if you get: W: Reserves are S, or G: Reserves are L • If you get D, drill at Y.. (Res. N: P = 0.5/0.74 = 0.68, S: P = 0.24/0.74 = 0.32)

  26. Ex: Probability Table for Site Y State Prior Conditional Joint K P(K) P(D|K) P(W|K) DY WY N 0.68 1 0 0.68 0 S 0.32 0.5 0.5 0.16 0.16 = 0.4/0.8  1.0 0.84 0.16 • Stop exploratory drilling at Y if you get W: Reserves are S • If you get D, drill at Z (Res. N: P = 0.68/0.84 = 0.81, S: P = 0.16/0.84 = 0.19)

  27. Ex: Probability Table for Site Z State Prior Conditional Joint K P(K) P(D|K) P(W|K) DZ WZ N 0.81 1 0 0.81 0 S 0.19 0 1 0 0.19  1.0 0.81 0.19 • If you get W: Reserves are S • If you get D: Reserves are N

  28. Ex: Change in P(N) Prior Probability • Before drilling P(N) = 0.5 Posterior Probabilities • After 1 Dry P(N|Dx) = 0.68 • After 2 Dry P(N|Dx, DY) = 0.81 • After 3 Dry P(N|Dx, DY, DZ) = 1

  29. 7.4. Conditional Independence Two events, A and B, are independent iff: • P(A, B) = P(A)*P(B) Implying • P(A|B) = P(A) • P(B|A) = P(B) • Posterior probability is the same the prior • New information about 1 event does not affect the probability of the other

  30. Conditional Independence Two events, A and B, are conditionally independent iff: • P(A, B)  P(A)*P(B) But their conditional probabilities on a 3rd event, C, are independent • P(A, B|C) = P(A|C)*P(B|C) • Useful property in Bayesian analysis

  31. Ex: Horse Race Probability Horse named WR will race at 3:00 p.m. Probability of WR winning P(WR) depends on track condition • Firm (F) P(F) = 0.3 P(WR|F) = 0.9 • Soft (S) P(S) = 0.7 P(WR|S) = 0.2

  32. Ex: Horse Race Probability • Given the results of 2 previous races • At 1:30, horse named MW won the race P(MW|F) = 0.8 P(MW|S) = 0.4 • At 2:00, horse named AJ won the race P(AJ|F) = 0.9 P(AJ|S) = 0.5 • What the new WR win probability P(WR|MW, AJ)?

  33. Ex: Horse Race Probability • P(WR) wins given MW and AJ have won must sum conditional probabilities of both possible track conditions: F or S • P(WR|MW, AJ) = P(WR|F) * P(F|MW, AJ) + P(WR|S) * P(S|MW, AJ)

  34. Ex: Horse Race Probability • The 2 events MW and AJ are conditionally independent with respect to a 3rd event: track condition F or S • Recall: P(A|B) = P(B|A)P(A)/P(B) P(A|B)  P(B|A)P(A) • P(F|MW, AJ)  P(MW, AJ|F) P(F)  P(MW|F) P(AJ|F) P(F) • P(S|MW, AJ)  P(MW, AJ|S) P(S)  P(MW|S) P(AJ|S) P(S)

  35. Ex: Horse Race Probability • P(F|MW, AJ)  P(MW|F) P(AJ|F) P(F)  0.8 * 0.9 * 0.3 = 0.216 • P(S|MW, AJ)  P(MW|S) P(AJ|S) P(S)  0.4 * 0.5 * 0.7 = 0.14 Normalizing P(F|MW, AJ) = 0.216/(0.216 + 0.14) = 0.61 P(F|MW, AJ) = 0. 14/(0.216 + 0.14) = 0.39

  36. Ex: Horse Race Probability • Substituting into • P(WR|MW, AJ) = P(WR|F) * P(F|MW, AJ) + P(WR|S) * P(S|MW, AJ) = 0.9(0.61) + 0.2(0.39) = 0.63

  37. 7.5. Bayesian Updating with Functional Likelihoods • Posterior probability P(A|B) = P(A,B )/P(B) =P(B|A)P(A)/P(B) • Conditional probability (likelihood) P(B|A) can be described by particular probability distribution: • (1) Binomial • (2) Normal

  38. Binomial likelihood distribution • Dichotomous (2-value) data: defective, not • Sequence of dichotomous outcomes: series of quality tests • In each test, constant probability (p) of one of the 2 outcomes: defective • Outcome of each test is independent of others • Total number (r) of one kind of outcomes (defective) out of (n) tests is • f(r|n, p) = Given in tables

  39. Binomial likelihood example Demand for new product can be: • High (H) P(H) = 0.2 • Medium (M) P(M) = 0.3 • Low (L) P(L) = 0.5 For each case, the probability (p) that an individual customer buys the product is • H: p = 0.25 • M: p = 0.1 • L: p = 0.05 In random sample of 5 customers, 1 will buy

  40. Binomial likelihood example For (n = 5, r = 1), likelihoods are obtained from table, or calculated by: • H: [5!/(4!*1)]0.25(0.75)4 = 0.3955 • M: [5!/(4!*1)]0.1(0.9)4 = 0.3281 • L: [5!/(4!*1)]0.05(0.95)4 = 0.2036 The joint probability table can now be constructed

  41. Binomial Likelihood Example State Prior Likelihood Joint Posterior p P(p) P(1|5,p) P(p,1/5) P(p|1/5) H: 0.25 0.2 0.3955 0.0791 0.2832 M : 0.1 0.3 0.3281 0.0984 0.3523 L: 0.05 0.5 0.2036 0.1018 0.3645  1.0 0.2793 1.00 • If 1 in a sample of 5 customers buys: P(H), P(M), P(L)

  42. Normal likelihood distribution • Most common, symmetric, • Continuous data, can approximate discrete • f(y|, ) = • Given in tables. Formula usually not used. • Two parameters: mean () and standard deviation ().

  43. Normal likelihood Updating • Mean () and standard deviation () Can be updated individually (assuming one is known) or together • P(|y, )  f(y|, ) P(|) • P(|y, )  f(y|, ) P(|) • P(,  |y)  f(y|, ) P(, )

  44. Normal likelihood example Updating Mean () Average weight setting has 2 possibilities: • High (H) P(H) = 0.5  = 8.2,  = 0.1 • Low (L) P(L) = 0.5  = 7.9,  = 0.1 A sample of 1 bottle has weight = 8.0 oz. What is the posterior probability of H and L?

  45. Normal likelihood example • Likelihood values are obtained from table, or calculated by • If  = 8.2 Z = (8.0 – 8.2)/0.1 = – 2 f(8|, ) = 0.054 • If  = 7.9 Z = (8.0 – 7.9)/0.1 = 1 f(8|, ) = 0.242

  46. Normal Likelihood Example State Prior Likelihood Joint Posterior  P() P(8|, ) P(8,,) P( |8) H: 8.2 0.5 0.054 0.027 0.18 L: 7.9 0.5 0.242 0.121 0.82  1.0 0.148 1.00 • After a sample of 1 bottle with weight = 8: P(H), P(L)

More Related