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# CHAPTER 7 - PowerPoint PPT Presentation

CHAPTER 7. Subjective Probability and Bayesian Inference. 7. 1. Subjective Probability. Personal evaluation of probability by individual decision maker Uncertainty exists for decision maker: probability is just a way of measuring it

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CHAPTER 7

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## CHAPTER 7

Subjective Probability and Bayesian Inference

### 7. 1. Subjective Probability

• Personal evaluation of probability by individual decision maker

• Uncertainty exists for decision maker: probability is just a way of measuring it

• In dealing with uncertainty, a coherent decision maker effectively uses subjective probability

### 7.2. Assessment of Subjective Probabilities

Simplest procedure:

• Specify the set of all possible events,

• ask the decision maker to directly estimate probability of each event

• Not a good approach from psychological point of view

• Not easy to conceptualize, especially for DM not familiar with probability

### Standard Device

• Physical instrument or conceptual model

• Good tool for obtaining subjective probabilities

• Example

• A box containing 1000 balls

• Balls numbered 1 to 1000

• Balls have 2 colors: red, blue

### Standard Device example

• To estimate a students’ subjective probability of getting an “A” in SE 447,

• we ask him to choose between 2 bets:

• Bet X: If he gets an A, he win SR 100

If he doesn’t get A, he wins nothing

• Bet Y:If he picks a red ball, he win SR 100

If he picks a blue ball, he wins nothing

then adjust successively until 2 bets are equal

### Other standard devices

Pie diagram (spinner)

Circle divided into 2 sectors:

1 red

1 blue

• Bet Y: If he spins to red section, he win SR 100

If he spins to blue section, he wins nothing

• Size of red section is adjusted until 2 bets are equal

### Subjective Probability Bias

• Standard device must be easy to perceive, to avoid introducing bias

2 kinds of bias:

• Task bias: resulting from assessment method (standard device)

• Conceptual bias: resulting from mental procedures (heuristics) used by individuals to process information

### Mental Heuristics Causing Bias

• Representativeness

• If x highly represents set A, high probability is given that X  A

• Frequency (proportion) ignored

• Sample size ignored

• Availability

• Limits of memory and imagination

• Starting from obvious reference point, then adjusting for new values.

• Anchoring: adjustment is typically not enough

• Overconfidence

• Underestimating variance

### Fractile Probability Assessment

• Quartile Assessment:

• Determine 3 values

• x1, for which p(x > x1) = 0.5

• x2, for which p(x < x2) = p(x2 < x< x1)

• x3, for which p(x > x3) = p(x1 < x< x3)

• x x2x1x3

• F(x)0.250.50.75

### Fractile Probability Assessment

• Quartile Assessment: 4 intervals

• Octile Assessment:8 intervals

• Tertile Assessment:3 intervals.

avoids anchoring

at the median

### Histogram Probability Assessment

• Fix the points x1, x2, …, xm.

• Ask the decision maker to assess probabilities

p(x1 < x< x2)

p(x2 < x< x3)

x1x2x3…x

Gives probability distribution

(not cumulative p.d. as fractile method)

### Assessment Methods & Bias

• No evidence to favor either fractile or histogram methods

• One factor that reinforces anchoring bias is self-consistency

• Bias can be reduced by “pre-assessment conditioning”: training, for/against arguments

• The act of probability assessment causes a re-evaluation of uncertainty

### 7.3. Impact of New Information(Bayes’ Theorem)

• After developing subjective probability distribution

• Assume new information becomes available

Example: new data is collected

• According to coherence principle, DM must take new information in consideration, thus

Subjective probability must be revised

• How? using Bayes’ theorem

### Bayes’ Theorem Example

• Suppose your subjective probability distribution for weather tomorrow is:

• chances of being sunnyP(S) = 0.6

• chances of being not sunnyP(N) = 0.4

• If the TV weather forecast predicted a cloudy day tomorrow. How should you change P(S)?

• Assume we are dealing with mutually exclusive and collectively exhaustive events such as sunny or not sunny.

### Impact of Information

• We assume the weather forecaster predicts either

• cloudy dayC, or

• bright dayB.

• To change P(S), we use the joint probability = conditional probability * marginal probability

• P(C,S) = P(C|S)P(S)P(B,S) = P(B|S)P(S)

• P(C,N) = P(C|N)P(N) P(B,N)= P(B|N)P(N)

### Impact of Information

• To obtain j.p.m.f we need the conditional probabilities P(C|S) and P(C|N).

• These can be obtained from historical data. How?

• In past 100 sunny days, cloudy forecast in 20 days

P(C|S) = 0.2

P(B|S) = 0.8

• In past 100 cloudy days, cloudy forecast in 90 days

P(C|N) = 0.9

P(B|N) = 0.1

### Joint probabilityCalculations

• Joint probability P(A,B) =

conditional probability (likelihood) P(A|B) * marginal probability P(B)

• Cloudy forecast

• P(C,S) = P(C|S)P(S) = 0.2(0.6)= 0.12

• P(C,N) = P(C|N)P(N) = 0.9(0.4)= 0.36

• Sunny forecast

• P(B,S) = P(B|S)P(S) = 0.8(0.6)= 0.48

• P(B,N) = P(B|N)P(N) = 0.1(0.4)= 0.04

### Bayes’ Theorem

SN

CP(C,S) P(C,N)P(C)

BP(B,S)P(B,N) P(B)

P(S)P(N)

• P(S|C) = P(C,S) / P(C)

= P(C|S)P(S ) / P(C)

• P(S|C) P(C|S)P(S )

### Joint probabilityTable

SN

C0.120.360.48

B0.480.040.52

0.60.4

• P(S|C) = 0.12/0.48 = 0.25 posterior (conditional)

probability

• Compare to P(S) = 0.6 prior (marginal) prob.

• P(S) decreased because of C forecast

### Prior and Posterior Probabilities

• Prior means before.

Prior probability is the probability P(S) before the information was heard.

• Posterior means after.

It is probability obtained after incorporating the new forecast information. It is P(S|C).

It is obtained using Bayes’ theorem.

### Example with 3 states

3 demand possibilities for new product

• HighP(H)= 0.6

• MediumP(M)= 0.1

• LowP(L)= 0.3

Market research gives Average result:

• 30% of timeif true demand is High

• 50% of timeif true demand is Medium

• 90% of timeif true demand is Low

### Example : Probability Table for Average result

StatePriorLikelihoodJointPosterior

SP(S)P(A|S)P(S,A)P(S|A)

H0.60.30.180.36

M0.10.50.050.10

L0.30.90.270.54

1.00.501.00

• If market research gives Average result:

P(H),P(L),P(M)

### Ex: Sequential Bayesian Analysis

• An oil company has 3 drilling sites: X, Y, Z

3 possible reserve states:

• No reservesP(N) = 0.5

• Small reservesP(S) = 0.3

• Large reservesP(L) = 0.2

3 possible drilling outcomes:

• DryD

• WetW

• GushingG

### Ex: Sequential Bayesian Analysis

If reserves are:

• None (N) all wells will be dry (D)

• Large (L) all wells will be Gushing (G)

• Small (S) some Dry (D) and some wet (W) wells P(1D/1) = 0.8

P(2D/2) = 0.2

P(3D/3) = 0

• All sites are equally favorable

• Assume order of drilling is: XYZ

• Notation: DX = probability of Dry well at site X

### Ex: Probability Table for Site X

StatePriorConditionalJoint

KP(K)P(D|K)P(W|K)P(G|K)DXWXGX

N0.51000.500

S0.30.80.200.240.060

L0.2001000.2

1.00.740.060.2

• Stop exploratory drilling at X if you get:

W: Reserves are S, or G: Reserves are L

• If you get D, drill at Y..

(Res. N: P = 0.5/0.74 = 0.68, S: P = 0.24/0.74 = 0.32)

### Ex: Probability Table for Site Y

StatePriorConditionalJoint

KP(K)P(D|K)P(W|K)DYWY

N0.68100.680

S0.320.50.50.160.16= 0.4/0.8

1.00.840.16

• Stop exploratory drilling at Y if you get W:

Reserves are S

• If you get D, drill at Z

(Res. N: P = 0.68/0.84 = 0.81, S: P = 0.16/0.84 = 0.19)

### Ex: Probability Table for Site Z

StatePriorConditionalJoint

KP(K)P(D|K)P(W|K)DZWZ

N0.81100.810

S0.19010 0.19

1.00.810.19

• If you get W:

Reserves are S

• If you get D:

Reserves are N

### Ex: Change in P(N)

Prior Probability

• Before drillingP(N) = 0.5

Posterior Probabilities

• After 1 DryP(N|Dx) = 0.68

• After 2 DryP(N|Dx, DY) = 0.81

• After 3 DryP(N|Dx, DY, DZ) = 1

### 7.4. Conditional Independence

Two events, A and B, are independent iff:

• P(A, B) = P(A)*P(B)

Implying

• P(A|B) = P(A)

• P(B|A) = P(B)

• Posterior probability is the same the prior

• New information about 1 event does not affect the probability of the other

### Conditional Independence

Two events, A and B, are conditionally independent iff:

• P(A, B)  P(A)*P(B)

But their conditional probabilities on a 3rd event, C, are independent

• P(A, B|C) = P(A|C)*P(B|C)

• Useful property in Bayesian analysis

### Ex: Horse Race Probability

Horse named WR will race at 3:00 p.m.

Probability of WR winning P(WR) depends on track condition

• Firm (F)P(F) = 0.3 P(WR|F) = 0.9

• Soft (S)P(S) = 0.7 P(WR|S) = 0.2

### Ex: Horse Race Probability

• Given the results of 2 previous races

• At 1:30, horse named MW won the race

P(MW|F) = 0.8P(MW|S) = 0.4

• At 2:00, horse named AJ won the race

P(AJ|F) = 0.9 P(AJ|S) = 0.5

• What the new WR win probability P(WR|MW, AJ)?

### Ex: Horse Race Probability

• P(WR) wins given MW and AJ have won must sum conditional probabilities of both possible track conditions: F or S

• P(WR|MW, AJ) =

P(WR|F) * P(F|MW, AJ) + P(WR|S) * P(S|MW, AJ)

### Ex: Horse Race Probability

• The 2 events MW and AJ are conditionally independent with respect to a 3rd event: track condition F or S

• Recall:P(A|B) = P(B|A)P(A)/P(B)

P(A|B)  P(B|A)P(A)

• P(F|MW, AJ)P(MW, AJ|F) P(F)

P(MW|F) P(AJ|F) P(F)

• P(S|MW, AJ)P(MW, AJ|S) P(S)

P(MW|S) P(AJ|S) P(S)

### Ex: Horse Race Probability

• P(F|MW, AJ)P(MW|F) P(AJ|F) P(F)

0.8 * 0.9 * 0.3 = 0.216

• P(S|MW, AJ)P(MW|S) P(AJ|S) P(S)

0.4 * 0.5 * 0.7 = 0.14

Normalizing

P(F|MW, AJ) = 0.216/(0.216 + 0.14) = 0.61

P(F|MW, AJ) = 0. 14/(0.216 + 0.14) = 0.39

### Ex: Horse Race Probability

• Substituting into

• P(WR|MW, AJ) =

P(WR|F) * P(F|MW, AJ) + P(WR|S) * P(S|MW, AJ)

= 0.9(0.61) + 0.2(0.39)

= 0.63

### 7.5. Bayesian Updating with Functional Likelihoods

• Posterior probability

P(A|B)= P(A,B )/P(B)

=P(B|A)P(A)/P(B)

• Conditional probability (likelihood) P(B|A) can be described by particular probability distribution:

• (1) Binomial

• (2) Normal

### Binomial likelihood distribution

• Dichotomous (2-value) data: defective, not

• Sequence of dichotomous outcomes: series of quality tests

• In each test, constant probability (p) of one of the 2 outcomes: defective

• Outcome of each test is independent of others

• Total number (r) of one kind of outcomes (defective) out of (n) tests is

• f(r|n, p) =Given in tables

### Binomial likelihood example

Demand for new product can be:

• High (H)P(H) = 0.2

• Medium (M)P(M) = 0.3

• Low (L)P(L) = 0.5

For each case, the probability (p) that an individual customer buys the product is

• H:p = 0.25

• M:p = 0.1

• L:p = 0.05

In random sample of 5 customers, 1 will buy

### Binomial likelihood example

For (n = 5, r = 1), likelihoods are obtained from table, or calculated by:

• H: [5!/(4!*1)]0.25(0.75)4 = 0.3955

• M: [5!/(4!*1)]0.1(0.9)4 = 0.3281

• L: [5!/(4!*1)]0.05(0.95)4 = 0.2036

The joint probability table can now be constructed

### Binomial Likelihood Example

StatePriorLikelihoodJointPosterior

pP(p)P(1|5,p)P(p,1/5)P(p|1/5)

H: 0.250.20.39550.07910.2832

M: 0.10.30.32810.09840.3523

L: 0.050.50.20360.10180.3645

1.00.27931.00

• If 1 in a sample of 5 customers buys:

P(H),P(M),P(L)

### Normal likelihood distribution

• Most common, symmetric,

• Continuous data, can approximate discrete

• f(y|, ) =

• Given in tables. Formula usually not used.

• Two parameters: mean () and standard deviation ().

### Normal likelihood Updating

• Mean () and standard deviation () Can be updated individually (assuming one is known) or together

• P(|y, )  f(y|, ) P(|)

• P(|y, )  f(y|, ) P(|)

• P(,  |y)  f(y|, ) P(, )

### Normal likelihood example

Updating Mean ()

Average weight setting has 2 possibilities:

• High (H)P(H) = 0.5  = 8.2, = 0.1

• Low (L)P(L) = 0.5  = 7.9, = 0.1

A sample of 1 bottle has weight = 8.0 oz.

What is the posterior probability of H and L?

### Normal likelihood example

• Likelihood values are obtained from table, or calculated by

• If  = 8.2

Z = (8.0 – 8.2)/0.1 = – 2

f(8|, ) = 0.054

• If  = 7.9

Z = (8.0 – 7.9)/0.1 = 1

f(8|, ) = 0.242

### Normal Likelihood Example

StatePriorLikelihoodJointPosterior

 P()P(8|, )P(8,,)P( |8)

H: 8.20.50.0540.0270.18

L: 7.90.50.2420.1210.82

1.00.1481.00

• After a sample of 1 bottle with weight = 8:

P(H),P(L)