Download Bayesian Statistics and Decision Analysis

Bayesian Statistics and
Decision Analysis
Session 10
15-1 Bayesian Statistics and
Decision Analysis
• 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
• Summary and Review of Terms
Bayesian and Classical
Statistics
Data
Data
Classical
Inference
Statistical
Conclusion
Bayesian
Inference
Statistical
Conclusion
Prior
Information
Bayesian statistical analysis incorporates a prior
probability distribution and likelihoods of observed
data to determine a posterior probability distribution
of events.
Bayes’ Theorem:
Example 10.1 (1)
• A medical test for a rare disease (affecting 0.1% of
the population [ P( I )  0.001 ]) is imperfect:
– When administered to an ill person, the test will indicate so
with probability 0.92 [ P ( Z I ) .92  P ( Z I ) .08 ]
•
–
The event ( Z I ) is a false negative
When administered to a person who is not ill, the test will
erroneously give a positive result (false positive) with
probability 0.04 [ P ( Z I )  0.04  P ( Z I )  0.96 ]
•
The event ( Z I ) is a false positive.
.
Example 10.1:
Applying Bayes’ Theorem
P( I )  0.001
P( I )  0.999
P( Z I )  0.92
P( Z I )  0.04
P( 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 )
P( I Z ) 
(.92)( 0.001)
(.92)( 0.001)  ( 0.04)(.999)
0.00092
0.00092


0.00092  0.03996
.04088
.0225

Example 10.1: Decision Tree
Prior
Probabilities
Conditional
Probabilities
P( Z I )  0.92
P( I )  0001
.
P( I )  0999
.
P( Z I )  008
.
P( Z I )  004
.
Joint
Probabilities
P( Z  I )  (0.001)(0.92) .00092
P( Z  I )  (0.001)(0.08) .00008
P( Z  I )  (0.999)(0.04) .03996
P( Z I )  096
.
P( Z  I )  (0.999)(0.96) .95904
10-2 Bayes’ Theorem and
Discrete Probability Models
The likelihood function is the set of conditional probabilities
P(x|) for given data x, considering a function of an unknown
population parameter, .
Bayes’ theorem for a discrete random variable:
P(x  ) P( )
P( x) 
 P(x  i )P( i )
i
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.
Example 10.2: Prior Distribution
and Likelihoods of 4 Successes
in 20 Trials
Prior
Distribution
S
P(S)
0.1
0.05
0.2
0.15
0.3
0.20
0.4
0.30
0.5
0.20
0.6
0.10
1.00
Likelihood
Binomial with n = 20 and p = 0.100000
x
P( X = x)
4.00
0.0898
Binomial with n = 20 and p = 0.200000
x
P( X = x)
4.00
0.2182
Binomial with n = 20 and p = 0.300000
x
P( X = x)
4.00
0.1304
Binomial with n = 20 and p = 0.400000
x
P( X = x)
4.00
0.0350
Binomial with n = 20 and p = 0.500000
x
P( X = x)
4.00
0.0046
Binomial with n = 20 and p = 0.600000
x
P( X = x)
4.00
0.0003
Example 10.2: Prior
Probabilities, Likelihoods, and
Posterior Probabilities
Prior
Distribution
S
P(S)
0.1
0.05
0.2
0.15
0.3
0.20
0.4
0.30
0.5
0.20
0.6
0.10
1.00
Posterior
Likelihood
Distribution
P(x|S)
P(S)P(x|S) P(S|x)
0.0898
0.00449
0.06007
0.2182
0.03273
0.43786
0.1304
0.02608
0.34890
0.0350
0.01050
0.14047
0.0046
0.00092
0.01230
0.0003
0.00003
0.00040
0.07475
1.00000
93%
Credible
Set
Example 10.2: Prior and
Posterior Distributions
P rio r D is trib utio n o f Marke t S hare
0.5
0 .5
0.4
0 .4
0.3
0 .3
P (S)
P (S)
P o s te rio r D is trib utio n o f M arke t S hare
0.2
0 .2
0.1
0 .1
0.0
0 .0
0.1
0.2
0.3
0.4
S
0 .5
0.6
0.1
0.2
0.3
0.4
S
0.5
0.6
Example 10.2: A Second
Sampling with 3 Successes in
16 Trials
Likelihood
Prior Distribution
S
P(S)
0.1
0.06007
0.2
0.43786
0.3
0.34890
0.4
0.14047
0.5
0.01230
0.6
0.00040
1.00000
Binomial with n = 16 and p = 0.100000
x
P( X = x)
3.00
0.1423
Binomial with n = 16 and p = 0.200000
x
P( X = x)
3.00
0.2463
Binomial with n = 16 and p = 0.300000
x
P( X = x)
3.00
0.1465
Binomial with n = 16 and p = 0.400000
x
P( X = x)
3.00
0.0468
Binomial with n = 16 and p = 0.500000
x
P( X = x)
3.00
0.0085
Binomial with n = 16 and p = 0.600000
x
P( X = x)
3.00
0.0008
Example 10.2: Incorporating a
Second Sample
Prior
Distribution Likelihood
S
P(S)
P(x|S)
0.1 0.06007 0.1423
0.2 0.43786 0.2463
0.3 0.34890 0.1465
0.4 0.14047 0.0468
0.5 0.01230 0.0085
0.6 0.00040 0.0008
1.00000
P(S)P(x|S)
0.0085480
0.1078449
0.0511138
0.0065740
0.0001046
0.0000003
0.1741856
Posterior
Distribution
P(S|x)
0.049074
0.619138
0.293444
0.037741
0.000601
0.000002
1.000000
91%
Credible Set
Example 10.2: Using Excel
Application of Bayes’ Theorem using Excel. The spreadsheet uses the BINOMDIST
function in Excel to calculate the likelihood probabilities. The posterior probabilities
are calculated using a formula based on Bayes’ Theorem for discrete random variables.
PRIOR DISTRIBUTION:
S
P(S)
0.1
0.05
0.2
0.15
0.3
0.2
0.4
0.3
0.5
0.02
0.6
0.1
LIKELIHOOD OF 4 OCCURRENCES IN 20 TRIALS GIVEN THE VALUES OF S:
S
0.1
0.2
0.3
0.4
0.5
0.6
P(x|S)
0.089778828 0.218199402 0.130420974 0.03499079 0.004620552 0.000269686
POSTERIOR DISTRIBUDTION AFTER THE FIRST SAMPLE:
S
0.1
0.2
0.3
0.4
0.5
0.6
P(S|x)
0.060051633 0.437850349 0.348946078 0.14042871 0.012362456 0.000360778
LIKEHOOD OF 3 OCCURRENCES IN 16 TRIALS GIVEN THE VALUES OF S:
S
0.1
0.2
0.3
0.4
0.5
0.6
P(x|S)
0.142344486 0.246290605 0.146496184 0.04680953 0.008544922 0.000811749
POSTERIOR DISTRIBUTION AFTER THE SECOND SAMPLE:
S
0.1
0.2
0.3
0.4
P(S|x)
0.049074356 0.619102674 0.293476795 0.03773804
0.5
0.00060646
0.6
1.68E-06
10-3 Bayes’ Theorem and
Continuous Probability
Distributions
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:
f (x  ) f ( )
f (x  ) f ( )
f ( x) 

 f (x  ) f ( )d Total area under f (x )
The Normal Probability Model
•
•
•
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 of
the population mean,  :
 1 
 n
 2 M   2 M
  
 
1
2
M  =
  
 1   n
 1   n
 2  2
 2  2
    
    
The Normal Probability Model:
Example 10.3
M   15
M  =
M  =
  8
n  10
 1 
 n
 2 M   2 M
  
 
 1   n
 2   2
    
 1
 10 
.
 2  15  
 1154
2
8 
 684

.
 1
 2
8 
 10 

2 
 684
.
M  1154
.
 
 
2
2


s  684
.
1
 1   n
 2   2
    
1
 1
 2
8 
 10 

2 
 684
.
2
M  = 11.77
      2.077
95% Credible Set: M   196
.    1177
.  (196
. ) 2.077  [ 7.699 ,15841
. ]
Example 10.3
Density
Posterior
Distribution
Likelihood
Prior
Distribution

11.54
11.77
15
10-4 The Evaluation of
Subjective Probabilities
•
Based on normal distribution
95% of normal distribution is within 2
standard deviations of the mean
P(-1 < x < 31) = .95 = 15,  = 8
68% of normal distribution is within 1
standard deviation of the mean
P(7 < x < 23) = .68  = 15,  = 8
10-5 Decision Analysis
•
Elements 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
Decision Tree: New-Product
Introduction
Decision
Chance
Occurrence
Product
successful
(P=0.75)
Final
Outcome
$100,000
Market
Do not
market
Product
unsuccessful
(P=0.25)
-$20,000
$0
Payoff Table and Expected
Values of Decisions: NewProduct Introduction
Action
Market the product
Do not market the product
Product is
Successful Not Successful
$100,000
-$20,000
$0
$0
The expected value of X , denoted E ( X ):
E ( X )   xP( x )
all x
E ( Outcome)  (100,000)( 0.75)  ( 20,000)( 0.25)
= 750000 -5000 = 70,000
Solution to the New-Product
Introduction Decision Tree
Clipping the Nonoptimal Decision Branches
Expected
Payoff
$70,000
Product
successful
(P=0.75)
$100,000
Market
Product
unsuccessful
(P=0.25)
Nonoptimal
decision branch
is clipped
Do not
market
Expected
Payoff
$0
-$20,000
$0
New-Product Introduction:
Extended-Possibilities
Outcome
Extremely successful
Very successful
Successful
Somewhat successful
Barely successful
Break even
Unsuccessful
Disastrous
Payoff
Probability
xP(x)
0.1
0.2
0.3
0.1
0.1
0.1
0.05
0.05
15,000
24,000
30,000
8,000
4,000
0
-1000
-2,500
Expected Payoff:
$77,500
$150,000
120.000
100,000
80,000
40,000
0
-20,000
-50,000
New-Product Introduction:
Extended-Possibilities
Decision Tree
Chance
Occurrence
Decision
Expected
Payoff
$77,500
Market
Payoff
0.1
$150,000
0.2 $120,000
0.3
$100,000
0.1
$80,000
0.1 $40,000
0.1
$0
0.05
0.05
-$20,000
-$50,000
Nonoptimal
decision branch
is clipped
Do not
market
$0
Example 10.4: Decision Tree
Not Promote
$700,000
Pr=0.4
Pr=0.5
Promote
$680,000
Pr=0.6
$740,000
Lease
Pr=0.3
$800,000
Pr=0.15
Not Lease
Pr=0.05
Pr=0.9
$900,000
$1,000,000
$750,000
Pr=0.1
$780,000
Example 10.4: Solution
Not Promote
Expected payoff:
0.5*425000
+0.5*716000=
$783,000
Lease
Pr=0.5
Expected payoff:
$700,000
Pr=0.4
Promote
Expected payoff:
$425,000
Expected payoff:
$716,000
Pr=0.3
$740,000
$800,000
Pr=0.15
Pr=0.9
Expected payoff:
$753,000
$680,000
Pr=0.6
Pr=0.05
Not Lease
$700,000
$900,000
$1,000,000
$750,000
Pr=0.1
$780,000
10-6 Handling Additional
Information Using Bayes’ Theorem
Payoff
Successful
Market
Test indicates
success
Do not market
$95,000
Failure
-$25,000
-$5,000
Market
Test
Test indicates
failure
Failure
Do not market
Not test
Market
New-Product Decision
Tree with Testing
Successful
Do not market
$95,000
-$25,000
-$5,000
Successful Pr=0.75
$100,000
Pr=0.25
Failure
-$20,000
0
Applying Bayes’ Theorem
P(S)=0.75
P(IS|S)=0.9
P(IF|S)=0.1
P(F)=0.75
P(IS|F)=0.15
P(IF|S)=0.85
P(IS)=P(IS|S)P(S)+P(IS|F)P(F)=(0.9)(0.75)+(0.15)(0.25)=0.7125
P(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)
( 0.9 )( 0.75)
 0.9474
( 0.9 )( 0.75)  ( 0.15)( 0.25)
P(F| IS)  1  P(S| IS)  1  0.9474  0.0526
P(IF|S)P(S)
P(S| IF) =
P(IF|S)P(S)  P(IF| F)P(F)
( 0.1)( 0.75)

 0.2609
( 0.1)( 0.75)  ( 0.85)( 0.25)
P(F| IF)  1  P(S| IF)  1  0.2609  0.7391
Expected Payoffs and Solution
$86,866
Market
$86,866 P(S|IS)=0.9474
$95,000
P(IS)=0.7125
P(F|IS)=0.0526
$6,308
Not test
-$5,000
$6,308
Market
P(IF)=0.2875
$70,000
-$25,000
Do not market
$66.003
Test
Payoff
P(S|IF)=0.2609
P(F|IF)=0.7391
Do not market
$70,000
$70,000
P(S)=0.75
Market
P(F)=0.25
Do not market
$95,000
-$25,000
-$5,000
$100,000
-$20,000
0
Example 10.5: Payoffs and
Probabilities
Prior Information
Level of
Economic
Profit
Activity Probability
$3 million Low
0.20
$6 million Medium 0.50
$12 million High
0.30
Consultants say “Low”
Event Prior Conditional
Low
0.20
0.90
Medium 0.50
0.05
High
0.30
0.05
P(Consultants say “Low”)
Reliability of Consulting Firm
Future
State of
Consultants’ Conclusion
Economy High Medium Low
Low
0.05
0.05
0.90
Medium
0.15
0.80
0.05
High
0.85
0.10
0.05
Joint Posterior
0.180
0.818
0.025
0.114
0.015
0.068
0.220
1.000
Example 10.5: Joint and
Conditional Probabilities
Consultants say “Medium”
Event Prior Conditional
Low
0.20
0.05
Medium 0.50
0.80
High
0.30
0.10
P(Consultants say “Medium”)
Joint Posterior
0.010
0.023
0.400
0.909
0.030
0.068
0.440
1.000
Consultants say “High”
Event Prior Conditional
Low
0.20
0.05
Medium 0.50
0.15
High
0.30
0.85
P(Consultants say “High”)
Joint Posterior
0.010
0.029
0.075
0.221
0.255 0.750
0.340
1.000
Alternative Investment
Profit
Probability
$4 million
0.50
$7 million
0.50
Consulting fee: $1 million
Example 10.5: Decision Tree
Hire consultants
Do not hire consultants
6.54
L
H
0.22
M
0.34
0.44
7.2
9.413
Alternative
Invest
5.5
0.5
H
Alternative
7.2 L
M
0.5
5.339
4.5
0.5
Alternative
4.5
H 9.413 L
M
0.5
0.5
Invest
Alternative
4.5
H 5.339 L
M
0.5
0.5
Invest
H 2.954 L
M
0.5
0.068 0.909 0.023
0.068 0.114 0.818
$2 million
$5 million
$11million
$3 million
$6 million
$2 million
$5 million
$11 million
$3 million
$6 million
$2 million
$5 million
$3 million
$11 million
$6 million
0.750 0.221 0.029
$3 million
0.5
$6 million
0.2
$12 million
$4 million
$7 million
0.3
Invest
4.5
Example 10.5: Using Excel
Application of Bayes’ Theorem to the information in
Example 15-4 using Excel. The conditional probabilities of
the consultants’ conclusion given the true future state and
the prior distribution on the true future state are used to
calculate the joint probabilities for each combination of true
state and the consultants’ conclusion. The joint
probabilities and Bayes’ Theorem are used to calculate the
prior probabilities on the consultants’ conclusions and the
conditional probabilities of the true future state given the
consultants’ conclusion.
SEE NEXT SLIDE FOR EXCEL OUTPUT.
Example 10.5: Using Excel
Distribution of Consultants' Conclusion
Given True Future State:
CONSULTANTS' CONCLUSION
TRUE STATE "High"
"Medium" "Low"
Low
0.05
0.05
0.9
Medium
0.15
0.8
0.05
High
0.85
0.1
0.05
Joint Probability of True State
and Consultants' Conclusion:
Prior
Prob. Of
True State
0.2
0.5
0.3
Prior Probability of Consultants' Conclusion:
Posterior Distribution on True Future State
Given the Consultants' Conclusion:
CONSULTANTS' CONCLUSION
TRUE STATE "High"
"Medium" "Low"
Low
0.01
0.01
0.18
Medium
0.075
0.4
0.025
High
0.255
0.03
0.015
CONSULTANTS' CONCLUSION
"High"
"Medium" "Low"
0.34
0.44
0.22
CONSULTANTS' CONCLUSION
TRUE STATE "High"
"Medium" "Low"
Low
0.029
0.023
0.818
Medium
0.221
0.909
0.114
High
0.75
0.068
0.068
10-7 Utility and Marginal
Utility
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
Additional
Utility
Additional
Utility
{
}
}
Additional $1000
Additional $1000
Dollars
Utility and Attitudes toward
Risk
Utility
Utility
Risk Averse
Risk Taker
Dollars
Utility
Dollars
Utility
Risk Neutral
Dollars
Mixed
Dollars
Assessing Utility
Possible Initial
Indifference
Returns Utility
Probabilities
Utility
$1,500
0
0
4,300
(1500)(0.8)+(56000)(0.2)
0.2
22,000
(1500)(0.3)+(56000)(0.7)
0.7
31,000
(1500)(0.2)+(56000)(0.8)
0.8
56,000
1
1
Utility
1.0
0.5
Dollars
0.0
0
10000
20000
30000
40000
50000
60000
10-8 The Value of Information
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 from Sampling
Expected
Net Gain
Max
Sample Size
nmax
Example 10.6:
The Decision Tree
Competitor’s
Fare
Airline
Fare
$200
Fare
$300
Fare
Payoff
$8 million
Competitor:$200
Pr=0.6
8.4
6.4
Competitor:$300
Pr=0.4
Competitor:$200
Pr=0.6
$9 million
$4 million
Competitor:$300
Pr=0.4
$10 million
Example 10.6: Value of
Additional Information
•
•
•
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.