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Prof. Saibal Chattopadhyay IIM Calcutta

IIMC Long Duration Executive Education Executive Programme in Business Management Statistics for Managerial Decisions. Prof. Saibal Chattopadhyay IIM Calcutta. An Outline of the Course. Probability Theory: Basic Concepts Distribution Theory:Random Variables

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Prof. Saibal Chattopadhyay IIM Calcutta

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  1. IIMC Long Duration Executive EducationExecutive Programme in Business ManagementStatistics for Managerial Decisions Prof. Saibal Chattopadhyay IIM Calcutta

  2. An Outline of the Course • Probability Theory: Basic Concepts • Distribution Theory:Random Variables • Utility Theory: Decisions under Uncertainty • Different Probability Distributions and their applications: Modeling real data • Bi-variate Data Analysis: Correlation and regression • Multivariate Data Analysis: Multiple and partial correlations; multiple regression • Sampling Theory: Different Methods • Statistical Inference

  3. Uncertainty and Randomness • Theory of Probability • Random variables & Probability Distributions • Mean & Variance of a distribution Decisions under uncertainty - Utility Theory - Decision Making using expected utility

  4. Theory of Probability Preliminaries • Random Experiment – outcomes bound by chance • Events – outcomes of a random experiment Example 1: A coin is tossed twice A = there is at least one head Event A has the following decompositions: A1: Head in both tosses (HH) A2: Head in 1st toss, Tail in 2nd (HT) A3: Tail in 1st toss, Head in 2nd (TH) No further decomposition possible – these are simple events

  5. Simple events – can’t be decomposed further • Sample Space – Collection of all simple events (Sure event : S) • Impossible event: events impossible to occur (φ ) B = at least 3 heads : B = φ Probability Space : (S, P) S = Sample Space P = Probability function on simple events in S

  6. P(.)  0 for any simple event, and Sum of all probabilities for simple events = 1. Then P(A) = Sum of the probabilities for those simple events which constitute event A. Example 1: S = {HH, HT, TH, TT} P(HH) = P(HT) = P(TH) = P(TT) = ¼ (equally likely outcomes – set-up of classical definition) A = at least one head = {HH, HT, TH} Therefore, P(A) = P(HH) + P(HT) + P(TH) = ¾

  7. If outcomes not equally likely (general case): Suppose the coin is biased, with P(H) = 1/3 & P(T)=2/3 Then P(HH) = 1/9, P(HT) = 2/9, P(TH) = 2/9, P(TT) = 4/9, and P(A) = P(HH) + P(HT) + P(TH) = 5/9 If B = at least 3 heads, then P(B) = 0 In general, P(.) is a number between 0 & 1. P(.) close to 0 : unlikely to occur P(.) close to 1: very likely to occur Probability: An attempt to quantify the degree of uncertainty

  8. Some Set-theoretic Operations with Events • Union: AB is the event which occurs if at least one of A and B occurs; • Intersection: AB is the event which occurs if A and B occur together; • Difference: A – B is the event which occurs if A occurs, but not B; • Complement: Ac is the event which occurs if event A does not occur

  9. Some special types of events • Mutually Exclusive Events: A and B are mutually exclusive (disjoint) if they cannot occur together Notation: AB = φ For three events A, B and C, they are disjoint if no two can occur together, i.e., AB = φ, AC = φ and BC = φ

  10. Exhaustive events: A set of events A1, A2,…,Ak is exhaustive if at least one of them is sure to occur, A1A2…Ak = Sample space = S • Partition of Sample Space: A1, A2, …, Ak form a partition of S if they are mutually exclusive as well as exhaustive. Example: A and Ac are disjoint (by definition) and are exhaustive; so they form a partition

  11. An Example Experiment consists of selecting three items from a manufacturer’s output and observing whether or not each item is defective. Call Defective = D and non-defective = G. S = Sample space = {DDD, DDG, DGD, GDD, DGG, GDG, GGD, GGG} Total no. of elements = 23 = 8 Suppose all are equally probable; then probability of each simple event = 1/8

  12. A = exactly one defective item = {DGG, GDG, GGD}; P(A) = 3/8. B = at most one defective = {GGG, DGG, GDG, GGD}; P(B) = 4/8=1/2 C = items drawn are all of the same type = {GGG, DDD}; P(C) = 2/8. AB = {GGG, DGG, GDG, GGD} = B Here event A is contained in event B: AB. Then AB = A. Ac = {DDD, DDG, DGD, GDD, GGG} = S – A P(Ac) = 5/8 = 1 – P(A).

  13. B C = {GGG}; P(BC) = 1/8. BC = {GGG, DGG, GDG, GGD, DDD}; P(BC) = 5/8. But note that this is SAME as P(B) + P(C) – P(BC) = 4/8 + 2/8 – 1/8. Earlier we also noted that P(Ac) = 1 – P(A). Are these mere coincidences, or are generally true? True, depending on specific conditions.

  14. Some Probability Results • (For disjoint events): If A and B are disjoint, P(AB) = P(A) + P(B) • If they are not disjoint, P(AB) = P(A) + P(B) – P(AB) • P(Ac) = 1 – P(A) • Any event A can occur either with a second event B jointly (i.e. AB occurs) or without the occurrence of the event B (i.e., ABc occurs). Thus P(A) = P(AB) + P(ABc)

  15. More than two events? P(ABC) = P(A) + P(B) + P(C), if disjoint; But P(ABC) = P(A) + P(B) + P(C) - P(AB) – P(AC) – P(BC) + P(ABC). How to tackle probability of intersections? P(AB) = ? P(ABC) = ? Need some further concepts !!

  16. Conditional Probability – Updating prior belief Knowing some event A to have occurred already, what is the chance that another event B will also occur ? P(B | A) = conditional prob. of B, given A = P(AB)/P(A), if P(A) > 0. What happens if P(A) = 0? Then A = φ. P(B | A) is not defined in this case. In general, P(A) = P(A | S) When information given about event A is trivial (A is a sure event), conditional probability of B, given A is same as unconditional probability of B, since no extra information is provided.

  17. Some useful Results Result 1: P(AB) = P(A).P(B | A) = P(B).P(A | B) Result 2: P(A) = P(B).P(A | B) + P(Bc).P(A| Bc) Use: Helps in updating our belief about chance of occurrence of random events, given additional information.

  18. An Application in Medical Science Following information are given: 1. P(a doctor diagnose disease X correctly) = 0.60 2. P(a patient will die by his treatment after correct diagnosis) = 0.40; 3. P(patient will die by his treatment after wrong diagnosis) = 0.70; Question 1: What is P(patient will die) ? Call B = doctor diagnose disease X correctly Then, Bc=doctor diagnose wrongly Given that P(B) = 0.60; we have P(Bc) = 1 – P(B) = 0.40.

  19. Call A = patient will die. To find P(A). Given: P(A | B) = 0.40, and P(A | Bc) = 0.70. Then P(A) = P(B).P(A|B) + P(Bc).P(A| Bc) = (0.60)(0.40) + (0.40).(0.70) = 0.24 + 0.28 = 0.52 How do we update our prior belief? Suppose we now know that the person (with disease X) died; Question 2: What is the probability that his disease was diagnosed correctly?

  20. Without knowing anything extra, this is P(doctor diagnosed correctly) = P(B) = 0.60 But now we know that the person had died (i.e., event A has already occurred). Given this extra information, what is P(B)? [should we expect it to be less now?]  Conditional probability of B, given A = P(B | A) = P(A and B) / P(A) =P(B).P(A | B) / P(A) = (0.60)(0.40)/(0.52) = 0.46 ( Bayes’ Theorem)

  21. Bayes’ Theorem Suppose B1, B2, …, Bk form a partition of S. Then for any event A which is known to have occurred, we have, for any i=1,2,...,k, P(Bi | A) = P(Bi) P(A|Bi) / P(A), where P(A) = P(B1).P(A | B1) + P(B2).P(A | B2) + …. + P(Bk).P(A | Bk). Note: A can occur only if one of B1, B2, …, Bk occurs; knowing A had occurred we now take a fresh look at the initial events B1, B2, …, Bk and examine if we have to update our prior belief about their occurrences.  Posterior probabilities of B1, B2, …, Bk

  22. Statistical Independence of Events • A and B are mutually independent events if (and only if) P(AB) = P(A).P(B) 2a) A, B and C are pair-wise independent events if P(AB) = P(A).P(B); P(AC) = P(A).P(C) and P(BC) = P(B).P(C). 2b) A, B and C are mutually independent if, in addition, we also have P(ABC) = P(A).P(B).P(C) Note: If A and B independent, the so are Ac and Bc. Use: Helps in calculation of probabilities

  23. Another way to quantify uncertainty Random Variable: A real-valued function on S S = {HH, HT, TH, TT} X = Number of heads obtained If {HH} occurs, X = 2; If {HT} occurs, X = 1; If {TH} occurs, X = 1: If {TT} occurs, X = 0. X takes 3 values: 0, 1, and 2 X is a random variable. Are all values of X equally probable? May be not, even if simple events are !

  24. Consider equally likely simple events: P(HH) = P(HT) = P(TH) = P(TT) = ¼. Then P(X=0) = P(TT) = ¼ , P(X=1) = P(HT) + P(TH) = ½ , and P(X=2) = P(HH) = ¼ Total Probability = ¼ + ½ + ¼ = 1 Probability distribution of X:

  25. Mean and Variance of a distribution Mean = E(X) = Sum(value*Probability) = 0. ¼ + 1. ½ + 2. ¼ = 1 Variance = Sum 1 – Square of Mean Sum1 = Sum(value-squared*probability) = 0. ¼ + 1. ½ + 4. ¼ = 1.5 Variance = 1.5 – 1 = 0.5 Standard deviation = SQRT(Variance) = SQRT(0.5) = 0.707

  26. What if simple events are not equally likely? With P(H) = 1/3 and P(T) = 2/3, we get P(HH)=1/9, P(HT) = P(TH) = 2/9, P(TT) = 4/9 Now P(X=0) = P(TT) = 4/9 P(X=1) = P(HT) + P(TH) = 4/9 P(X=2) = 1/9 Prob. Distribution of X(=no. of heads): Mean & Variance: Similar

  27. Expected Utility Theory Some Math preliminaries: 1. Function: y=f(x) is a mapping between two sets of elements ( or numbers) Example: Y=a + bx : linear function Or, Y = a + bx + cx2 : second degree etc. 2. Optimization (maxima & minima) of a function: Differentiable function: f ’(x) = 0: solve for x (say x = x0) f ”(x) at x=x0 is negative: f(x) is maximum at x = x0 f ”(x) at x=x0 is positive: f(x) is minimum at x = x0

  28. In decision making under uncertainty, a decision d may lead to several levels of wealth: w1, w2, …, wk, with corresponding probs. p1, p2, …, pk, total prob.=sum(pi) = 1. Wealth is usually transformed into consumption, and hence utility (for example, in a business decision, it may be profit of the company) Utility function over wealth: u(w) Different levels of utility: u(w1), u(w2),.., u(wk) These are random quantities, with respective probalilities p1, p2, …, pk.

  29. Expected Utility of a decision d E(u(d)) = average of these utilities = u(w1).p1 + u(w2).p2 +… + u(wk).pk An optimal decision d depends on: a) Optimization of expected utility b) Choice of the utility function u(w)

  30. Some Applications Example 1(Dilemma of a Contractor) A contractor has to choose one of the two contracting jobs:both having chances of labour problems (say, strike). Profit possibilities: Job1: 10K if no strike, 2K if strike Job 2: 20K if no strike, 0.5K if strike Chance of strike: P(strike in Job 1) = ¼ ; P(no strike) = ¾ P(strike in Job 2) = ½ ; P(no strike) = ½

  31. Expected Profit from Job 1 = (10000)(3/4) + (2000)(1/4) = 8,000 Expected Profit from Job 2 = (20000)(1/2) + (500)(1/2) = 10,250 Want to maximize your profit ? Choose Job 2 !

  32. Any other consideration for his choice? What if he is a born pessimist? Expects the worst: there will be a strike! Choose Job 1: it maximizes his minimum profit. What if he is an optimist? Expects no strike or neglects the chance of it Choose Job 2: it may give him 20,000 Anything else?

  33. Chance of strike not known ! How does he choose? Go for a randomized decision rule: Choose Job 1 with prob. p and Job 2 with prob. (1-p) such that his profit must be same whether he chooses Job 1 or Job 2 and whether there is a strike or not. His profit is: A = 10000 p + 20000 (1-p) if no strike; B = 2000 p + 500 (1-p) if strike; Find p such that A = B; p > 1(check); Can’t be ! So p = 1. Choose Job 1 !!

  34. Back to Utility Function and expected utility E(u(d)) = expected utility for decision d and utility function u(w) = u(w1).p1 + u(w2).p2 +… + u(wk).pk Different Choices of u(w): • u(w) = √w : risk averse • u(w) = w2 : risk seeker • u(w) = w : risk neutral For a given u(w), choose a decision that maximizes the expected utility

  35. Example 2: An investment decision problem(Ex. 1.22; Aliprantis-Chakrabarti) To invest $10,000 in stocks/bonds Return rate of stock: 2% with prob. 0.37 & 10% with prob. 0.63 Return of Bond: 7% with certainty Individual is Risk-averse: utility function is u(w) = √w How much to invest in stocks? Say a fraction ‘s’ of the total investment.

  36. Investor has a chance 0.37 of getting an amount (10,000 s)(1.02) + 10,000(1-s)(1.07) =10,000(1.07 – 0.05 s) And a chance 0.63 of getting the amount (10,000 s)1.10 + 10,000(1-s)(1.07) =10,000(1.07 + 0.03 s) Investor’s expected utility (risk-averse !) is E(s) = 0.37 √10000(1.07 – 0.05 s) + 0.63 √10000(1.07 + 0.03 s) Choose s such that E(s) is maximum; s=56.9% Invest $5690 in stocks &$4310 in Bonds.

  37. Example 3:Choice between two stocks (Ex.1.23 (Aliprantis & Chakrabarti): $10,000 to invest between two stocks S & M Probability Table for Returns from S & M

  38. Invest proportion ‘s’ in stock S and proportion (1-s) in stock M With 5% return from stock S and 20% from M: Wealth = w1 = 10000 s.1.05 + 10000(1-s).1.20 = 10000(1.20 – 0.15 s) { this event has probability = 0.40} With 5% from S and 5% from M: Wealth = w2 = 10000.1.05 { probability = 0.1} With 20% from S and 20% from M: Wealth = w3 = 10000.1.20 { probability = 0.1} With 20% from S and 5% from M: Wealth = w4 = 10000(1.05 + 0.15 s) { probability= 0.4}

  39. For risk averse, utility function u(w) = √w Expected utility = E(s) = Sum{(√w1)(0.40) + (√w2)(0.10) + (√w3)(0.10) + (√w4)(0.40)} = 40 √(1.2 – 0.15 s) + 40 √(1.05 +0.15 s) + 10 √1.05 + 10 √1.20 Choose ‘s’ so that E(s) is maximum S = 0.5 = 50% Decision for Risk averse: Invest 50% ($5000) in Stock S and 50% ($5000) in Stock M. A Question for you: What happens if he is risk seeker or risk neutral?

  40. Some remarks Are decisions always based on expected utility? Possibly not; consider the following lotteries: L1: Receive $2 million with certainty L2: Receive $10 million with prob. 0.15, $2 m with prob. 0.75 & $ 0 with prob. 0.10 L3: Receive $2 million with prob. 0.25 and $0 with prob. 0.75 L4: Receive $10 million with prob. 0.15 and $0 with prob. 0.85 Choose one between L1 and L2 and one between L3 and L4

  41. If L1 is chosen over L2, then u(2) > 0.15 u(10) + 0.75 u(2) + 0.10 u(0) Add 0.75 u(0) to both side and get 0.25 u(2) + 0.75 u(0) > 0.15 u(10) + 0.85 u(0) i.e., expected utility for L3 > expected utility for L4 So we should choose L3 over L4 Do you agree? I don’t !

  42. Violations of the expected utility theory Lotteries and Gambling If by paying a small amount one has a chance of winning a large amount, individuals often ignore the negative expected payoff, as the loss is small. BUT If potential loss is larger, the same individual may choose very differently  preference reversal in decision making

  43. Suggested Reading • Statistical Methods in Business and Social Sciences: Shenoy, G.V. & Pant, M. (Macmillan India Limited) • Games and Decision Making: Aliprantis, C.D. & Chakrabarti, S.K. (Oxford University Press) • •Complete Business Statistics: Aczel, A.D. & Sounderpandian, J. – Fifth Edition (Tata McGraw-Hill)

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