1. Chapter 15 Risk Analysis

2. Frequency definition of probability zGiven a situation in which a number of possible outcomes might occur, the probability of an outcome is the proportion of times that it occurs if the situation exists repeatedly. zP(A) = r/R is the probability of outcome A, given r A’s in R trials.

3. Subjective definition of probability Given a situation in which a number of possible outcomes might occur, the probability of an outcome reflects the degree of confidence that the decision maker has that this particular outcome will occur.

4. Probability distribution ProfitProbability $1,000,000.6 $ - 600,000.4 Expected value of profit: ($1,000,000).6 + ($-600,000).4 = $360,000

5. Constructing a decision tree zDecision fork: a juncture representing a choice where the decision maker is in control of the outcome zChance fork: a juncture where “chance” controls the outcome

6. Constructing a decision tree.5 $800,000 $100,000 $200,000 Do not increase price Increase price Successful Unsuccessful.5 -$600,000

7. Tomco Oil Corporation - $90,000 $56,000 $0 Do not drill well Drill well No oil =.6 10,000 B =.15 $100,000 20,000 B =.15 $300,000 30,000 B =.10 $500,000

8. Expected utility The sum of the utility if each outcome occurs times the probability of occurrence of the outcome

9. Using a utility function Tomco Oil Corporation Utility Profit 0 -90 10 20 30 40 50 0100300500

10. Expected utility for Tomco Oil E(U) =.6U($-90) +.15U(100) +.15U(300) +.10 U(500) =.6 (0) +.15(20) +.15(40) +.10(50) = 14

11. Attitudes toward risk Utility Profit 0 Risk Seeking Risk Neutral Risk Averse

12. Measures of risk zStandard deviation:  = [  n i=1  i - E (  i )) 2 P i ] 1/2  Coefficient of variation V =  / E (  )

13. Adjusting the valuation model for risk zCertainty equivalent approach zRisk-adjusted discount rates

14. Winner’s curse zIn an auction situation, the highest bidder is likely to pay more for the good than it is worth zThe highest bid, by definition, must be greater than the average bid (unless all bids are equal)

15. Risk vs. uncertainty zRisk occurs when the outcome is not certain, but the probabilities of all outcomes are known or estimable zUncertainty refers to a situation where some or all probabilities are unknown

16. Problem 1 1.a. E( P x ) =.2(20) +.3(8) +.4(10) +.1(3) = $10.7 million. s X = [(20 – 10.7) 2 (.2) + (8 – 10.7) 2 (.3) + (10 – 10.7) 2 (.4) + (3 – 10.7) 2 (.1)] 1/2 = 5.0606. V X = s X /E( P x ) = 5.0606/10.7 =.4729. b. E( P y ) =.1(12) +.3(9) +.1(16) +.5(11) = 11. s Y = [(12 – 11) 2 (.1) + (9 – 11) 2 (.3) + (16 – 11) 2 (.1) + (11 – 11) 2 (.5)] 1/2 = 1.97. V Y = s Y /E( P y ) = 1.97/11 =.18.

17. Problem 1 (cont.) 1.c. V X = V Y  X is riskier. d. Y, since Y is both less risky and has a higher expected value than X, characteristics valued by Martin’s president.

18. Problem 2 2.a. 18 percent – 6 percent = 12 percent. b. 6 percent. c. 18 percent. d. 18 percent.

19. Problem 3 3.a. No, 0 > –$8 million, so the worst outcome is least bad if Zodiac does nothing. b. It assigns 100 percent of the probability to the worst possible case for each option and therefore might lead you to choose an action with a much lower expected value than available alternatives. c. No, it would be 50P if P is the chance of success for the university research, but this is unknown.

20. Problem 4 4. a. E( P ) = $20(100,000) – $1,000,000 = $1,000,000. b. s = 10,000($20) = $200,000. c. V = $200,000/$1,000,000 =.2.

21. Problem 5 5.U(X) = X/100,000 – 1. a. U(400) = 4 – 1 = 3. b. U(40) =.4 – 1 = –.6. c. U(–20) = –.2 – 1 = –1.2.

22. Problem 6 6. b. She is risk neutral since the indifference curve is a straight line.

23. Problem 7 7. b. Only one: To buy or not to buy the firm. c. Only one: The firm is effective or it is not. d. Yes, $50,000 > $0. e.(1)Yes, the “won’t work” branch has another chance fork attached with.2 leading to $100,000 and.8 leading to – $400,000. (2)Yes, (a) the project works, (b) the project is resold, and (c) the project loses money. (3) (a).5, (b).5(.2) =.1, (c).5(.8) =.4. (4) (a) $500,000, (b) $100,000, (c) –$400,000.

24. Problem 7 (cont.) 7. f. E( P ) = 500,000(.5) + 100,000(.1) – 400,000(.4) = $100,000. Buy the firm. g. (1) E( P ) = (500,000 – X) x (.5) – 400,000(.5) = 0 implies X = $100,000. (2) E( P ) = (500,000 – X) x (.5) + 100,000(.1) – 400,000(.4) = 0 implies X = $200,000.

25. Problem 8 8. Given the low risk factor NASA used, the attractiveness of a launch appeared greater than an unbiased appraisal would suggest.

26. Problem 9 9. b. The first fork is a decision fork to publish or not to publish. The second fork is a chance fork, where publishing either succeeds or fails.

27. Problem 10 10.b. No, there are no probabilities given. c. 1/4(800) – 3/4(200 = 50 > 0, so a person who is risk neutral would drill. However, if very risk averse, the person would not want to drill. d. Yes, since the project has both a positive expected value and contains risk, Mr. Lamb will be doubly pleased. e. Yes, Mr. Lamb cares only about expected value, which is positive for this project.