Chapter 15 Risk Analysis. Frequency definition of probability zGiven a situation in which a number...
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Transcript of Chapter 15 Risk Analysis. Frequency definition of probability zGiven a situation in which a number...
![Page 1: Chapter 15 Risk Analysis. Frequency definition of probability zGiven a situation in which a number of possible outcomes might occur, the probability of.](https://reader036.fdocuments.in/reader036/viewer/2022081513/56649ec15503460f94bccc7d/html5/thumbnails/1.jpg)
Chapter 15Risk Analysis
![Page 2: Chapter 15 Risk Analysis. Frequency definition of probability zGiven a situation in which a number of possible outcomes might occur, the probability of.](https://reader036.fdocuments.in/reader036/viewer/2022081513/56649ec15503460f94bccc7d/html5/thumbnails/2.jpg)
Frequency definition of probability
Given 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.
P(A) = r/R is the probability of outcome A, given r A’s in R trials.
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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.
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Probability distribution
Profit Probability
$1,000,000 .6
$ - 600,000 .4
Expected value of profit:
($1,000,000).6 + ($-600,000).4 = $360,000
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Constructing a decision tree
Decision fork: a juncture representing a choice where the decision maker is in control of the outcome
Chance fork: a juncture where “chance” controls the outcome
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Constructing a decision tree
.5$800,000$100,000
$200,000
Do not increase price
Increase price
Successful
Unsuccessful.5
-$600,000
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Tomco Oil Corporation
- $90,000$56,000
$0
Do not drill well
Drill well
No oil = .6
10,000 B = .15$100,00020,000 B = .15
$300,000
30,000 B = .10
$500,000
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Expected utility
The sum of the utility if each outcome occurs times the probability of occurrence of the outcome
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Using a utility functionTomco Oil Corporation
Utility
Profit0
-90
10
20
30
40
50
0 100 300 500
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Expected utility for Tomco Oil
E(U) = .6U($-90) + .15U(100) + .15U(300) + .10 U(500)
=.6 (0) + .15(20) + .15(40) + .10(50)
= 14
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Attitudes toward risk
Utility
Profit0
Risk Seeking
Risk Neutral
Risk Averse
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Measures of risk
Standard deviation:
= [ni=1i - E(i))2 Pi]1/2
Coefficient of variation
V = / E()
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Adjusting the valuation model for risk
Certainty equivalent approachRisk-adjusted discount rates
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Winner’s curse
In an auction situation, the highest bidder is likely to pay more for the good than it is worth
The highest bid, by definition, must be greater than the average bid (unless all bids are equal)
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Risk vs. uncertainty
Risk occurs when the outcome is not certain, but the probabilities of all outcomes are known or estimable
Uncertainty refers to a situation where some or all probabilities are unknown
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Problem 1
1. a. E(Px) = .2(20) + .3(8) + .4(10) + .1(3) = $10.7 million.
sX = [(20 – 10.7)2(.2) + (8 – 10.7)2(.3) + (10 –
10.7)2(.4) + (3 – 10.7)2(.1)]1/2
= 5.0606.
VX = sX /E(Px) = 5.0606/10.7 = .4729.
b. E(Py) = .1(12) + .3(9) + .1(16) + .5(11) = 11.
sY = [(12 – 11)2(.1) + (9 – 11)2(.3) + (16 – 11)2(.1) +
(11 – 11)2(.5)]1/2 = 1.97.
VY = sY /E(Py) = 1.97/11 = .18.
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Problem 1 (cont.)
1. c. VX = VY 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.
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Problem 2
2.a. 18 percent – 6 percent = 12 percent.
b. 6 percent.
c. 18 percent.
d. 18 percent.
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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.
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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.
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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.
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Problem 6
6. b. She is risk neutral since the indifference curve is a straight line.
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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.
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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.
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Problem 8
8. Given the low risk factor NASA used, the attractiveness of a launch appeared greater than an unbiased appraisal would suggest.
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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.
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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.