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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

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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 x2 x1 x3
• F(x) 0.25 0.5 0.75
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)

x1 x2 x3 … 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 sunny P(S) = 0.6
• chances of being not sunny P(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 day C, or
• bright day B.
• 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

S N

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

B P(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

S N

C 0.12 0.36 0.48

B 0.48 0.04 0.52

0.6 0.4

• 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

• High P(H) = 0.6
• Medium P(M) = 0.1
• Low P(L) = 0.3

Market research gives Average result:

• 30% of time if true demand is High
• 50% of time if true demand is Medium
• 90% of time if true demand is Low
Example : Probability Table for Average result

State Prior Likelihood Joint Posterior

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

H 0.6 0.3 0.18 0.36

M 0.1 0.5 0.05 0.10

L 0.3 0.9 0.27 0.54

 1.0 0.50 1.00

• 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 reserves P(N) = 0.5
• Small reserves P(S) = 0.3
• Large reserves P(L) = 0.2

3 possible drilling outcomes:

• Dry D
• Wet W
• Gushing G
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

State Prior Conditional Joint

K P(K) P(D|K) P(W|K) P(G|K) DX WX GX

N 0.5 1 0 0 0.5 0 0

S 0.3 0.8 0.2 0 0.24 0.06 0

L 0.2 0 0 1 0 0 0.2

 1.0 0.74 0.06 0.2

• 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

State Prior Conditional Joint

K P(K) P(D|K) P(W|K) DY WY

N 0.68 1 0 0.68 0

S 0.32 0.5 0.5 0.16 0.16 = 0.4/0.8

 1.0 0.84 0.16

• 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

State Prior Conditional Joint

K P(K) P(D|K) P(W|K) DZ WZ

N 0.81 1 0 0.81 0

S 0.19 0 1 0 0.19

 1.0 0.81 0.19

• If you get W:

Reserves are S

• If you get D:

Reserves are N

Ex: Change in P(N)

Prior Probability

• Before drilling P(N) = 0.5

Posterior Probabilities

• After 1 Dry P(N|Dx) = 0.68
• After 2 Dry P(N|Dx, DY) = 0.81
• After 3 Dry P(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) = 0.8 P(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

State Prior Likelihood Joint Posterior

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

H: 0.25 0.2 0.3955 0.0791 0.2832

M : 0.1 0.3 0.3281 0.0984 0.3523

L: 0.05 0.5 0.2036 0.1018 0.3645

 1.0 0.2793 1.00

• 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

State Prior Likelihood Joint Posterior

 P() P(8|, ) P(8,,) P( |8)

H: 8.2 0.5 0.054 0.027 0.18

L: 7.9 0.5 0.242 0.121 0.82

 1.0 0.148 1.00

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

P(H), P(L)