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CHAPTER 7

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CHAPTER 7

Subjective Probability and Bayesian Inference

- 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

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

- 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

- 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

- We start with proportion of red balls P = 50%,
then adjust successively until 2 bets are equal

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

- 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

- 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

- Adjustment & anchoring
- Starting from obvious reference point, then adjusting for new values.
- Anchoring: adjustment is typically not enough

- Overconfidence
- Underestimating variance

- 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

- Quartile Assessment: 4 intervals
- Octile Assessment:8 intervals
- Tertile Assessment:3 intervals.
avoids anchoring

at the median

- 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)

- 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

- 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

- 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.

- 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)

- 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 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

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 )

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 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.

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

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)

- 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

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

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)

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)

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

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

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

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

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

- 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)?

- 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)

- 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)

- 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

- 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

- 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

- 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

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

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

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)

- Most common, symmetric,
- Continuous data, can approximate discrete
- f(y|, ) =
- Given in tables. Formula usually not used.
- Two parameters: mean () and standard deviation ().

- 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(, )

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?

- 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

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)