Foundations of Decision Analysis by Ronald A. Howard & Ali E. Abbas

Author:Ronald A. Howard & Ali E. Abbas
Language: eng
Format: mobi
Publisher: Pearson Education
Published: 2015-01-05T14:00:00+00:00

T

(a)

1

2

3

4

…

34

35

36

37

38

…

61

62

63

64

65

…

98

99

100

table 20.1

Indication,

402

20.2 • Detector with 100 Indications

403

{ S | T , &}

1.0

0.8

0.6

0.4

0.2

0.0

1

10

19

28

37

46

55

64

73

82

91

100

T

FIgure 20.5 Posterior Probability, 5 S ∣ T, & 6

Column (g) shows the posterior probability of Sun after receiving an indication T, so we

have 5 S T, &6. This is obtained by dividing the corresponding cells in Columns (d) and (f).

5 S, T &6

5 S T, &6 = 5 T&6

Column (h) shows the posterior probability of Rain given an indication T, so we have 5 R T, &6, which is obtained by dividing the corresponding cells in Columns (e) and (f). Thus,

5 R, T &6

5 R T, &6 = 5 T&6

The analysis we have done in tabular form corresponds to the tree reversal we conducted earlier.

The table format is convenient for the large number of indications. Figures 20.5 and 20.6 plot the

posterior probabilities of Sun and Rain for a given indication, T.

Column (g) and Figure 20.5 show that there only three possible values for the probability of Sun after receiving the detector indication: 1, 0.4, and 0. Column (h) and Figure 20.6 show that there are only three possible values for the probability of Rain after receiving the detector

{ R | T , &}

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0 1 10 19 28 37 46 55 64 73 82 91 100

T

FIgure 20.6 Posterior Probability, 5 R ∣ T, & 6

404

Chapter 20 • Detectors with Multiple Indications

indication: 0, 0.6, and 1. Column (i) shows the best decision for Kim after receiving an indica-

tion T. Recall from the sensitivity analysis chapter that if 5 S T, &6 Ú 0.87, she should have her party Outdoors. If 0.47 … 5 S T, &6 6 0.87, she should have her party on the Porch, and if

5 S T, &6 6 0.47, she should have her party Indoors. Therefore, we are not surprised to see that Kim will never have a Porch party after viewing the results of the detector because 5 S T, &6 does not lie between 0.47 and 0.87.

Columns (j) and (k) show the certain equivalent and the e-value of u-values of the party

after receiving the indication. Note that, when the posterior probability of Sun is 1, the u-value of

the Outdoor alternative is 1 (her u-value of Outdoor–Sun prospect). When the posterior probabil-

ity is 0.4, Kim chooses Indoors, and the u-value of this alternative is 0.63 (same as the u-value of the Indoor deal with no detector). When the posterior probability is 0, Kim goes Indoors, and the

u-value of this alternative is 0.67 (her u-value of the Indoor–Rain prospect).

By multiplying the preposterior probability of a detector indication, 5 T &6 by the e-value

of the u-value of the party given this indication, column (k), and summing over all possible values

of T, we get

e@value of the u@values of party deal with free detector = 0.72886

Value of party deal with free detector = +57.07

Since Kim is a deltaperson and values the party at $45.87 with no detector, her value of the 100

degree detector is

Value of the detector = +57.07 - +45.87 = +11.20

Note that this analysis is in principle similar to that of the simple detector analysis.

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