15-1 COMPLETE BUSINESS STATISTICS by AMIR D. ACZEL & JAYAVEL SOUNDERPANDIAN 6 th edition (SIE)

15-1

COMPLETE COMPLETE BUSINESS BUSINESS

STATISTICSSTATISTICSbyby

AMIR D. ACZELAMIR D. ACZEL

&&

JAYAVEL SOUNDERPANDIANJAYAVEL SOUNDERPANDIAN

66thth edition (SIE) edition (SIE)

15-2

Chapter 15 Chapter 15

Bayesian Statistics Bayesian Statistics and Decision Analysisand Decision Analysis

15-3

• Using Statistics• Bayes’ Theorem and Discrete Probability Models• Bayes’ Theorem and Continuous Probability Distributions• The Evaluation of Subjective Probabilities• Decision Analysis: An Overview• Decision Trees• Handling Additional Information Using Bayes’ Theorem• Utility• The Value of Information• Using the Computer

Bayesian Statistics and Decision Bayesian Statistics and Decision AnalysisAnalysis1515

15-4

• Apply Bayes’ theorem to revise population parameters

• Solve sequential decision problems using decision trees

• Conduct decision analysis for cases without probability data

• Conduct decision analysis for cases with probability data

LEARNING OBJECTIVESLEARNING OBJECTIVES1515

After studying this chapter you should be able to:After studying this chapter you should be able to:

15-5

• Evaluate the expected value of perfect information

• Evaluate the expected value of sample information

• Use utility functions to model the risk attitudes of decision makers

• Solve decision analysis problems using spreadsheet templates

LEARNING OBJECTIVES (2)LEARNING OBJECTIVES (2)1515

After studying this chapter you should be able to:After studying this chapter you should be able to:

15-6

ClassicalInferenceClassicalInference

DataDataStatistical ConclusionStatistical Conclusion

Bayesian InferenceBayesian Inference

DataData

PriorInformation

PriorInformation

Statistical ConclusionStatistical Conclusion

Bayesian statistical analysis incorporates a prior probability distribution and likelihoods of observed data to determine a posterior probability distribution of events.

Bayesian statistical analysis incorporates a prior probability distribution and likelihoods of observed data to determine a posterior probability distribution of events.

Bayesian and Classical StatisticsBayesian and Classical Statistics

15-7

• A medical test for a rare disease (affecting 0.1% of the population [ ]) is imperfect:When administered to an ill person, the test will indicate so

with probability 0.92 [ ]

• The event is a false negativeWhen administered to a person who is not ill, the test will

erroneously give a positive result (false positive) with probability 0.04 [ ]

• The event is a false positive. .

• A medical test for a rare disease (affecting 0.1% of the population [ ]) is imperfect:When administered to an ill person, the test will indicate so

with probability 0.92 [ ]

• The event is a false negativeWhen administered to a person who is not ill, the test will

erroneously give a positive result (false positive) with probability 0.04 [ ]

• The event is a false positive. .

P I( ) .0 001

P Z I P Z I( ) . ( ) . 92 08

( )Z I

( )Z I

P Z I P Z I( ) . ( ) . 0 04 0 96

Bayes’ Theorem: Example 2-10 Bayes’ Theorem: Example 2-10

15-8

P I

P I

P Z I

P Z I

( ) .

( ) .

( ) .

( ) .

0001

0999

092

004

P I ZP I Z

P Z

P I Z

P I Z P I Z

P Z I P I

P Z I P I P Z I P I

( )( )

( )

( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

(. )( . )

(. )( . ) ( . )(. )

.

. .

.

..

92 0001

92 0001 004 999

000092

000092 003996

000092

040880225

Applying Bayes’ TheoremApplying Bayes’ Theorem

15-2 Bayes’ Theorem and Discrete Probability 15-2 Bayes’ Theorem and Discrete Probability Models _ Example 2-10 (Continued)Models _ Example 2-10 (Continued)

15-9

P I( ) .0001

999.0)( IP04.0)( IZP

96.0)( IZP

08.0)( IP Z

P Z I( ) .092 P Z I( ) ( . )( . ) . 0 001 0 92 00092

P Z I( ) ( . )( . ) . 0 001 0 08 00008

03996.)04.0)(999.0()( IZP

95904.)96.0)(999.0()( IZP

Prior ProbabilitiesPrior Probabilities

Conditional ProbabilitiesConditional Probabilities

JointProbabilitiesJointProbabilities

Example 2-10: Decision TreeExample 2-10: Decision Tree

15-10

Bayes’ theorem for a discrete random variable:

where is an unknown population parameter to be estimated from the data. The summation in the denominator is over all possible values of the parameter of interest, i, and x stands for the observed data set.

Bayes’ theorem for a discrete random variable:

where is an unknown population parameter to be estimated from the data. The summation in the denominator is over all possible values of the parameter of interest, i, and x stands for the observed data set.

P xP x P

P x Pii

i

( )( ) ( )

( ) ( )

The likelihood function is the set of conditional probabilities P(x|) for given data x, considering a function of an unknown population parameter, .

The likelihood function is the set of conditional probabilities P(x|) for given data x, considering a function of an unknown population parameter, .

15-2 Bayes’ Theorem and Discrete 15-2 Bayes’ Theorem and Discrete Probability ModelsProbability Models

15-11

Prior Prior DistributionDistributionS P(S)0.1 0.050.2 0.150.3 0.200.4 0.300.5 0.200.6 0.10

1.00

Prior Prior DistributionDistributionS P(S)0.1 0.050.2 0.150.3 0.200.4 0.300.5 0.200.6 0.10

1.00

LikelihoodLikelihoodBinomial with n = 20 and p = 0.100000 x P( X = x) 4.00 0.0898Binomial with n = 20 and p = 0.200000 x P( X = x) 4.00 0.2182Binomial with n = 20 and p = 0.300000 x P( X = x) 4.00 0.1304Binomial with n = 20 and p = 0.400000 x P( X = x) 4.00 0.0350Binomial with n = 20 and p = 0.500000 x P( X = x) 4.00 0.0046Binomial with n = 20 and p = 0.600000 x P( X = x) 4.00 0.0003

LikelihoodLikelihoodBinomial with n = 20 and p = 0.100000 x P( X = x) 4.00 0.0898Binomial with n = 20 and p = 0.200000 x P( X = x) 4.00 0.2182Binomial with n = 20 and p = 0.300000 x P( X = x) 4.00 0.1304Binomial with n = 20 and p = 0.400000 x P( X = x) 4.00 0.0350Binomial with n = 20 and p = 0.500000 x P( X = x) 4.00 0.0046Binomial with n = 20 and p = 0.600000 x P( X = x) 4.00 0.0003

Example 15-1: Prior Distribution and Example 15-1: Prior Distribution and Likelihoods of 4 Successes in 20 TrialsLikelihoods of 4 Successes in 20 Trials

15-12

Prior PosteriorPrior PosteriorDistribution Likelihood DistributionDistribution Likelihood DistributionS P(S) P(x|S) P(S)P(x|S) P(S|x) 0.1 0.05 0.0898 0.00449 0.060070.2 0.15 0.2182 0.03273 0.437860.3 0.20 0.1304 0.02608 0.348900.4 0.30 0.0350 0.01050 0.140470.5 0.20 0.0046 0.00092 0.012300.6 0.10 0.0003 0.00003 0.00040

1.00 0.07475 1.00000

Prior PosteriorPrior PosteriorDistribution Likelihood DistributionDistribution Likelihood DistributionS P(S) P(x|S) P(S)P(x|S) P(S|x) 0.1 0.05 0.0898 0.00449 0.060070.2 0.15 0.2182 0.03273 0.437860.3 0.20 0.1304 0.02608 0.348900.4 0.30 0.0350 0.01050 0.140470.5 0.20 0.0046 0.00092 0.012300.6 0.10 0.0003 0.00003 0.00040

1.00 0.07475 1.00000

93% CredibleSet

Example 15-1: Prior Probabilities, Example 15-1: Prior Probabilities, Likelihoods, and Posterior ProbabilitiesLikelihoods, and Posterior Probabilities

15-13

0.60.50.40.30.20.1

0.5

0.4

0.3

0.2

0.1

0.0

S

P(S

)

P os te rio r D is tributio n o f Marke t S hare

0.60.50.40.30.20.1

0.5

0.4

0.3

0.2

0.1

0.0

SP

(S)

P rio r D is tributio n o f Marke t S hare

Example 15-1: Prior and Posterior Example 15-1: Prior and Posterior DistributionsDistributions

15-14

Prior DistributionPrior Distribution S P(S)0.1 0.060070.2 0.437860.3 0.348900.4 0.140470.5 0.012300.6 0.00040

1.00000

Prior DistributionPrior Distribution S P(S)0.1 0.060070.2 0.437860.3 0.348900.4 0.140470.5 0.012300.6 0.00040

1.00000

LikelihoodLikelihoodBinomial with n = 16 and p = 0.100000 x P( X = x) 3.00 0.1423Binomial with n = 16 and p = 0.200000 x P( X = x) 3.00 0.2463Binomial with n = 16 and p = 0.300000 x P( X = x) 3.00 0.1465Binomial with n = 16 and p = 0.400000 x P( X = x) 3.00 0.0468Binomial with n = 16 and p = 0.500000 x P( X = x) 3.00 0.0085Binomial with n = 16 and p = 0.600000 x P( X = x) 3.00 0.0008

LikelihoodLikelihoodBinomial with n = 16 and p = 0.100000 x P( X = x) 3.00 0.1423Binomial with n = 16 and p = 0.200000 x P( X = x) 3.00 0.2463Binomial with n = 16 and p = 0.300000 x P( X = x) 3.00 0.1465Binomial with n = 16 and p = 0.400000 x P( X = x) 3.00 0.0468Binomial with n = 16 and p = 0.500000 x P( X = x) 3.00 0.0085Binomial with n = 16 and p = 0.600000 x P( X = x) 3.00 0.0008

Example 15-1: A Second Sampling Example 15-1: A Second Sampling with 3 Successes in 16 Trialswith 3 Successes in 16 Trials

15-15

Prior PosteriorPrior PosteriorDistribution Likelihood DistributionDistribution Likelihood Distribution S P(S) P(x|S) P(S)P(x|S) P(S|x) 0.1 0.06007 0.1423 0.0085480 0.0490740.2 0.43786 0.2463 0.1078449 0.6191380.3 0.34890 0.1465 0.0511138 0.2934440.4 0.14047 0.0468 0.0065740 0.0377410.5 0.01230 0.0085 0.0001046 0.0006010.6 0.00040 0.0008 0.0000003 0.000002

1.00000 0.1741856 1.000000

Prior PosteriorPrior PosteriorDistribution Likelihood DistributionDistribution Likelihood Distribution S P(S) P(x|S) P(S)P(x|S) P(S|x) 0.1 0.06007 0.1423 0.0085480 0.0490740.2 0.43786 0.2463 0.1078449 0.6191380.3 0.34890 0.1465 0.0511138 0.2934440.4 0.14047 0.0468 0.0065740 0.0377410.5 0.01230 0.0085 0.0001046 0.0006010.6 0.00040 0.0008 0.0000003 0.000002

1.00000 0.1741856 1.000000

91% Credible Set

Example 15-1: Incorporating a Example 15-1: Incorporating a Second SampleSecond Sample

15-16

Application of Bayes’ Theorem using the Template. The posterior probabilities are calculated using a formula based on Bayes’ Theorem for discrete random variables.

Application of Bayes’ Theorem using the Template. The posterior probabilities are calculated using a formula based on Bayes’ Theorem for discrete random variables.

Example 15-1: Using the TemplateExample 15-1: Using the Template

15-17

Example 15-1: Using the Template Example 15-1: Using the Template (Continued)(Continued)

Display of the Prior and Posterior probabilities.Display of the Prior and Posterior probabilities.

15-18

We define f() as the prior probability density of the parameter . We define f(x|) as the conditional density of the data x, given the value of . This is the likelihood function.

We define f() as the prior probability density of the parameter . We define f(x|) as the conditional density of the data x, given the value of . This is the likelihood function.

Bayes' theorem for continuous distributions:

Total area under f x

f x f

f x f d

f x ff

( )( ) ( )

( ) ( )

( ) ( )( )

x

Bayes' theorem for continuous distributions:

Total area under f x

f x f

f x f d

f x ff

( )( ) ( )

( ) ( )

( ) ( )( )

x

15-3 Bayes’ Theorem and Continuous 15-3 Bayes’ Theorem and Continuous Probability DistributionsProbability Distributions

15-19

• Normal population with unknown mean and known standard deviation

• Population mean is a random variable with normal (prior) distribution and mean M and standard deviation .

• Draw sample of size n:

• Normal population with unknown mean and known standard deviation

• Population mean is a random variable with normal (prior) distribution and mean M and standard deviation .

• Draw sample of size n:

The posterior mean and variance of the normal population ofthe population mean, :

=

1 n

1 n

1 n

2 2

2 2 2 2

MM M

2 1

The Normal Probability ModelThe Normal Probability Model

15-20

M n M s

M

M M

M

15 8 10 1154 684

2 1

815

6841154

8 684

2 1

8 684

=

12

n2

12

n2

12

n2

=

12

102

12

102

12

102

. .

..

. .

MM

= 11.77 Credible Set:

2 2 07795% 196 1177 196 2 077 7 699 15841

.. . ( . ) . [ . , . ]

The Normal Probability Model: The Normal Probability Model: Example 15-2Example 15-2

15-21

LikelihoodLikelihood

11.5411.77

PosteriorDistributionPosteriorDistribution

PriorDistributionPriorDistribution

15

Density

Example 15-2Example 15-2

15-22

Example 15-2 Using the TemplateExample 15-2 Using the Template

15-23

Example 15-2 Using the Template Example 15-2 Using the Template (Continued)(Continued)

15-24

• Based on normal distributionBased on normal distribution95% of normal distribution is within 2 standard deviations of the meanP(-1 < x < 31) = .95= 15, = 8

68% of normal distribution is within 1 standard deviation of the meanP(7 < x < 23) = .68 = 15, = 8

• Based on normal distributionBased on normal distribution95% of normal distribution is within 2 standard deviations of the meanP(-1 < x < 31) = .95= 15, = 8

68% of normal distribution is within 1 standard deviation of the meanP(7 < x < 23) = .68 = 15, = 8

15-4 The Evaluation of Subjective 15-4 The Evaluation of Subjective ProbabilitiesProbabilities

15-25

• Elements of a decision analysisElements of a decision analysis Actions

Anything the decision-maker can do at any time

Chance occurrences Possible outcomes (sample space)

Probabilities associated with chance occurrences Final outcomes

Payoff, reward, or loss associated with action

Additional information Allows decision-maker to reevaluate probabilities and possible rewards

and losses

Decision Course of action to take in each possible situation

• Elements of a decision analysisElements of a decision analysis Actions

Anything the decision-maker can do at any time

Chance occurrences Possible outcomes (sample space)

Probabilities associated with chance occurrences Final outcomes

Payoff, reward, or loss associated with action

Additional information Allows decision-maker to reevaluate probabilities and possible rewards

and losses

Decision Course of action to take in each possible situation

15-5 Decision Analysis15-5 Decision Analysis

15-26

MarketMarket

Do notDo notmarketmarket

ProductProductunsuccessfulunsuccessful(P = 0.25)(P = 0.25)

ProductProductsuccessfulsuccessful(P = 0.75)(P = 0.75)

$100,000$100,000

-$20,000-$20,000

$0$0

DecisionDecisionDecisionDecisionChance Chance OccurrenceOccurrence

Chance Chance OccurrenceOccurrence

Final Final OutcomeOutcome

Final Final OutcomeOutcome

15-6: Decision Tree: New-Product 15-6: Decision Tree: New-Product IntroductionIntroduction

15-27

Product isProduct isActionAction SuccessfulSuccessful Not SuccessfulNot SuccessfulMarket the product $100,000 -$20,000Do not market the product $0 $0

Product isProduct isActionAction SuccessfulSuccessful Not SuccessfulNot SuccessfulMarket the product $100,000 -$20,000Do not market the product $0 $0

The expected value of , denoted ( ):

= 750000 -5000 = 70,000

all x

X E XE X xP x

E Outcome

( ) ( )

( ) (100, )( . ) ( , )( . )

000 0 75 20 000 0 25

The expected value of , denoted ( ):

= 750000 -5000 = 70,000

all x

X E XE X xP x

E Outcome

( ) ( )

( ) (100, )( . ) ( , )( . )

000 0 75 20 000 0 25

15-6: Payoff Table and Expected Values 15-6: Payoff Table and Expected Values of Decisions: New-Product Introductionof Decisions: New-Product Introduction

15-28

Market

Do notmarket

Productunsuccessful(P=0.25)

Productsuccessful(P=0.75)

$100,000

-$20,000

$0

ExpectedPayoff$70,000

ExpectedPayoff$70,000

ExpectedPayoff$0

ExpectedPayoff$0

Nonoptimaldecision branchis clipped

Nonoptimaldecision branchis clipped

Clipping the Nonoptimal Decision BranchesClipping the Nonoptimal Decision Branches

Solution to the New-Product Solution to the New-Product Introduction Decision TreeIntroduction Decision Tree

15-29

OutcomeOutcome PayoffPayoff Probability xP(x)Probability xP(x)Extremely successful $150,000 0.1 15,000Very successful 120.000 0.2 24,000Successful 100,000 0.3 30,000Somewhat successful 80,000 0.1 8,000Barely successful 40,000 0.1 4,000Break even 0 0.1 0Unsuccessful -20,000 0.05 -1000Disastrous -50,000 0.05 -2,500

Expected Payoff: $77,500

OutcomeOutcome PayoffPayoff Probability xP(x)Probability xP(x)Extremely successful $150,000 0.1 15,000Very successful 120.000 0.2 24,000Successful 100,000 0.3 30,000Somewhat successful 80,000 0.1 8,000Barely successful 40,000 0.1 4,000Break even 0 0.1 0Unsuccessful -20,000 0.05 -1000Disastrous -50,000 0.05 -2,500

Expected Payoff: $77,500

New-Product Introduction:New-Product Introduction: Extended-Possibilities Extended-Possibilities

15-30

Market

Do notmarket

$100,000

-$20,000

$0

DecisionDecision Chance OccurrenceChance Occurrence

Payoff

-$50,000

$0

$40,000$80,000

$120,000$150,000

0.2

0.3

0.05

0.1

0.1

0.1

0.1

0.05

ExpectedPayoff$77,500

ExpectedPayoff$77,500

Nonoptimaldecision branchis clipped

Nonoptimaldecision branchis clipped

New-Product Introduction:New-Product Introduction: Extended-Possibilities Decision Tree Extended-Possibilities Decision Tree

15-31

$780,000

$750,000

$700,000

$680,000

$740,000

$800,000

$900,000

$1,000,000

LeaseLease

Not LeaseNot Lease

Pr = 0.9

Pr = 0.1

Pr = 0.05

P r = 0.4

Pr = 0.6

Pr = 0.3

Pr = 0.15

Not PromoteNot Promote

PromotePromotePr = 0.5

Example 15-3: Decision TreeExample 15-3: Decision Tree

15-32

$780,000

$750,000

$700,000

$680,000

$740,000

$800,000

$900,000

$1,000,000

LeaseLease

Not LeaseNot Lease

Pr = 0.9

Pr = 0.1

Pr = 0.05

Pr = 0.4

Pr = 0.6

Pr = 0.3

Pr = 0.15

Not PromoteNot Promote

PromotePromote

Expected payoff: $753,000Expected payoff: $753,000

Expected payoff: $716,000Expected payoff: $716,000Expected payoff:

$425,000Expected payoff: $425,000

Expected payoff: $700,000Expected payoff: $700,000

Pr=0.5Expected payoff: 0.5*425000+0.5*716000=$783,000

Expected payoff: 0.5*425000+0.5*716000=$783,000

Example 15-3: SolutionExample 15-3: Solution

15-33

0

$100,000

$95,000

-$25,000

-$5,000

$95,000

-$25,000

-$5,000

-$20,000

Test

Not test

Test indicatessuccess

Test indicatesfailure

Market

Do not market

Do not market

Do not market

Market

Market

Successful

Failure

Successful

Successful

Failure

Failure

PayoffPayoff

Pr=0.25

Pr=0.75

New-Product DecisionTree with Testing

New-Product DecisionTree with Testing

15-7 Handling Additional Information 15-7 Handling Additional Information Using Bayes’ TheoremUsing Bayes’ Theorem

15-34

P(S)=0.75 P(IS|S)=0.9 P(IF|S)=0.1P(F)=0.75 P(IS|F)=0.15 P(IF|S)=0.85P(IS)=P(IS|S)P(S)+P(IS|F)P(F)=(0.9)(0.75)+(0.15)(0.25)=0.7125P(IF)=P(IF|S)P(S)+P(IF|F)P(F)=(0.1)(0.75)+(0.85)(0.25)=0.2875

P(S| IS) =P(IS|S)P(S)

P(IS|S)P(S) P(IS|F)P(F)

P(F| IS) 1 P(S| IS) 1 0.9474 .0526

P(S| IF) =P(IF|S)P(S)

P(IF|S)P(S) P(IF|F)P(F)

P(F| IF) 1 P(S| IF) 1 0.2609 .7391

( . )( . )

( . )( . ) ( . )( . ).

( . )( . )

( . )( . ) ( . )( . ).

0 9 0 75

0 9 0 75 0 15 0 250 9474

0

0 1 0 75

0 1 0 75 0 85 0 250 2609

0

Applying Bayes’ TheoremApplying Bayes’ Theorem

15-35

0

$100,000

$95,000

-$25,000

-$5,000

$95,000

-$25,000

-$5,000

-$20,000

Test

Not test

P(IS)=0.7125

Market

Do not market

Do not market

Do not market

Market

Market

P(S)=0.75

PayoffPayoff

P(F)=0.25

P(IF)=0.2875

P(S|IF)=0.2609

P(F|IF)=0.7391

P(S|IS)=0.9474

P(F|IS)=0.0526

$86,866$86,866 $86,866$86,866

$6,308$6,308

$70,000$70,000

$6,308$6,308

$70,000$70,000

$66.003$66.003

$70,000$70,000

Expected Payoffs and SolutionExpected Payoffs and Solution

15-36

Prior InformationPrior InformationLevel ofEconomic

Profit Activity Probability$3 million Low 0.20$6 million Medium 0.50$12 million High 0.30

Prior InformationPrior InformationLevel ofEconomic

Profit Activity Probability$3 million Low 0.20$6 million Medium 0.50$12 million High 0.30

Reliability of Consulting FirmReliability of Consulting FirmFutureState of Consultants’ Conclusion Economy High Medium LowLow 0.05 0.05 0.90Medium 0.15 0.80 0.05High 0.85 0.10 0.05

Reliability of Consulting FirmReliability of Consulting FirmFutureState of Consultants’ Conclusion Economy High Medium LowLow 0.05 0.05 0.90Medium 0.15 0.80 0.05High 0.85 0.10 0.05

Consultants say “Low”Consultants say “Low” Event Prior Conditional Joint PosteriorLow 0.20 0.90 0.180 0.818Medium 0.50 0.05 0.025 0.114High 0.30 0.05 0.015 0.068 P(Consultants say “Low”) 0.220 1.000

Consultants say “Low”Consultants say “Low” Event Prior Conditional Joint PosteriorLow 0.20 0.90 0.180 0.818Medium 0.50 0.05 0.025 0.114High 0.30 0.05 0.015 0.068 P(Consultants say “Low”) 0.220 1.000

Example 15-4: Payoffs and Example 15-4: Payoffs and ProbabilitiesProbabilities

15-37

Consultants say “Medium”Consultants say “Medium” Event Prior Conditional Joint PosteriorLow 0.20 0.05 0.010 0.023Medium 0.50 0.80 0.400 0.909High 0.30 0.10 0.030 0.068 P(Consultants say “Medium”) 0.440 1.000

Consultants say “Medium”Consultants say “Medium” Event Prior Conditional Joint PosteriorLow 0.20 0.05 0.010 0.023Medium 0.50 0.80 0.400 0.909High 0.30 0.10 0.030 0.068 P(Consultants say “Medium”) 0.440 1.000

Consultants say “High”Consultants say “High” Event Prior Conditional Joint PosteriorLow 0.20 0.05 0.010 0.029Medium 0.50 0.15 0.075 0.221High 0.30 0.85 0.255 0.750 P(Consultants say “High”) 0.340 1.000

Consultants say “High”Consultants say “High” Event Prior Conditional Joint PosteriorLow 0.20 0.05 0.010 0.029Medium 0.50 0.15 0.075 0.221High 0.30 0.85 0.255 0.750 P(Consultants say “High”) 0.340 1.000

Alternative InvestmentAlternative InvestmentProfit Probability$4 million 0.50$7 million 0.50

Consulting fee: $1 million

Alternative InvestmentAlternative InvestmentProfit Probability$4 million 0.50$7 million 0.50

Consulting fee: $1 million

Example 15-4: Joint and Conditional Example 15-4: Joint and Conditional ProbabilitiesProbabilities

15-38

$3 million

$6 million

$3 million

$11 million

$5 million

$2 million

$6 million

$3 million

$11 million

$5 million

$2 million

$6 million

$7 million

$4 million

$12 million

$6 million

$3 million

$11million

$5 million

$2 million

Hire consultantsHire consultantsDo not hire consultantsDo not hire consultants

L

H L

H

L L L

MM M M

HHH

InvestInvest InvestInvest InvestInvest InvestInvestAlternativeAlternative AlternativeAlternativeAlternativeAlternativeAlternativeAlternative

0.5 0.5 0.50.5 0.5 0.5 0.50.5

0.3 0.2 0.5 0.750 0.221 0.029 0.068 0.909 0.023 0.068 0.114 0.818

5.5 7.2 4.54.54.5 9.413 5.339 2.954

M0.34

0.44

0.22

7.2 9.413 5.339 4.5

6.54

Example 15-4: Decision TreeExample 15-4: Decision Tree

15-39

Dollars

Utility

Additional UtilityAdditional Utility

Additional $1000Additional $1000

Additional Utility

Additional $1000Additional $1000

}}

{

Utility is a measure of the total worth of a particular outcome.It reflects the decision maker’s attitude toward a collection of factors such as profit, loss, and risk.

Utility is a measure of the total worth of a particular outcome.It reflects the decision maker’s attitude toward a collection of factors such as profit, loss, and risk.

15-8 Utility and Marginal Utility15-8 Utility and Marginal Utility

15-40

UtilityUtility

DollarsDollars

Risk AverseRisk Averse

Dollars

UtilityUtility Risk TakerRisk Taker

UtilityUtility

DollarsDollars

Risk NeutralRisk Neutral

Dollars

MixedMixedUtilityUtility

Utility and Attitudes toward RiskUtility and Attitudes toward Risk

15-41

PossiblePossible InitialInitial IndifferenceIndifferenceReturnsReturns UtilityUtility ProbabilitiesProbabilities UtilityUtility$1,500 0 04,300 (1500)(0.8)+(56000)(0.2) 0.2

22,000 (1500)(0.3)+(56000)(0.7) 0.731,000 (1500)(0.2)+(56000)(0.8) 0.856,000 1 1

PossiblePossible InitialInitial IndifferenceIndifferenceReturnsReturns UtilityUtility ProbabilitiesProbabilities UtilityUtility$1,500 0 04,300 (1500)(0.8)+(56000)(0.2) 0.2

22,000 (1500)(0.3)+(56000)(0.7) 0.731,000 (1500)(0.2)+(56000)(0.8) 0.856,000 1 1

6000050000400003000020000100000

1.0

0.5

0.0

Utility

Dollars

Example 15-5: Assessing UtilityExample 15-5: Assessing Utility

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The expected value of perfect information (EVPI): EVPI = The expected monetary value of the decision situation when

perfect information is available minus the expected value of the decision situation when no additional information is available.

The expected value of perfect information (EVPI): EVPI = The expected monetary value of the decision situation when

perfect information is available minus the expected value of the decision situation when no additional information is available.

Expected Net Gain

Sample Size

Max

nmax

Expected Net Gain from SamplingExpected Net Gain from Sampling

15-9 The Value of Information15-9 The Value of Information

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$200Fare

$300Fare

Competitor:$200Pr = 0.6

Competitor:$300Pr = 0.4

Competitor:$300Pr = 0.4

Competitor:$200Pr = 0.6

$8 million

$10 million

$4 million

$9 million

PayoffPayoffCompetitor’sFare

Competitor’sFare

AirlineFare

AirlineFare

8.4

6.4

Example 15-6: The Decision TreeExample 15-6: The Decision Tree

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• If no additional information is available, the best strategy is to set the fare at $200 E(Payoff|200) = (.6)(8)+(.4)(9) = $8.4 million E(Payoff|300) = (.6)(4)+(.4)(10) = $6.4 million

• With further information, the expected payoff could be: E(Payoff|Information) = (.6)(8)+(.4)(10)=$8.8 million

• EVPI=8.8-8.4 = $.4 million

• If no additional information is available, the best strategy is to set the fare at $200 E(Payoff|200) = (.6)(8)+(.4)(9) = $8.4 million E(Payoff|300) = (.6)(4)+(.4)(10) = $6.4 million

• With further information, the expected payoff could be: E(Payoff|Information) = (.6)(8)+(.4)(10)=$8.8 million

• EVPI=8.8-8.4 = $.4 million

Example 15-6: Value of Additional Example 15-6: Value of Additional InformationInformation