HO12-HW3Solutions

MS&E 252 Handout #12 Decision Analysis I October 26th, 2004

� of 11 HW #3 Solutions

Homework Assignment #3 Solutions

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Distinctions �� An expert:

• has powerful distinctions and knows how they relate to each other, • in some cases, has required physical skills, • knows the history and borders of the field, and • has humility (and knows what he or she does not know)

�� We define clairvoyance as information, which resolves all uncertainty about an uncertain distinction. We call the PIBP of clairvoyance on a distinction the value of clairvoyance on that distinction.

�� We define a distinction as a thought, which separates one large group of things into two or more smaller groups. We call the small groups the degrees of the distinction. We require the degrees to be mutually exclusive (a thing only falls into one degree) and collectively exhaustive (a thing will always fall into one of the degrees). We can create several kinds of distinctions over the same large group; for example, we can create the two distinctions "sex" and "hair length" over the group "humans."

�� We call a distinction clear when it satisfies the clarity test. This means that a clairvoyant could, without exercise of judgment, tell in which degree any particular thing falls.

��We call a distinction observable when we "know it when we see it." In other words, the decision maker could tell in which degree any particular thing falls.

��We call a distinction useful when it • means what we want it to mean, and/or • helps us achieve clarity of action.

��Decision Basis: • Alternatives are what you can do. Need more than one. • Preferences are what you want. You must care about what might happen. • Information is what you know. Information links what you want to what you can do.

��The probability of an event is a number from 0 to 1 describing a person's beliefs about the likelihood of that event occurring. Since a probability describes a person's belief, we say that it is conditioned on that person's background state of information (written &), or the sum of all that person's knowledge and experience, which affects his or her belief.

��An associative logic error occurs when a person confuses conditional probabilities for one another. For example, if a person thinks that since most hemophiliacs are male, most males are hemophiliac. In probabilistic notation, if a person thinks that since {A | B, &} is large, then {B | A, &} must also be large. We use the word associative, since people who make this error think that A and B are associated with each other, and thus tend to occur together. The opposite of associative logic we call distinctive logic.

��We say that two uncertain distinctions are relevant when knowledge of one will affect our beliefs about the other. In probabilistic notation, A is relevant to B given & if and only if {A | B, &} � {A | ¬B, &}.



��We call a possibility a possible event that could occur, usually with a particular probability.

Probabilistic questions 1) Solution: b

If we define events A & B as follows: A: At least one quarter lands “heads” B: Both quarters land “heads”

Then we’re interested in finding: p{B|A&} From the problem’s statement: p{(H,H)|&}= p{(H,T)|&}= p{(T,H)|&}=p{(T,T)|&}=0.25 It follows that: p{B|A&} = p{(H,H)|&}/[p{(H,H)|&}+ p{(H,T)|&}+ p{(T,H)|&}] = 0.25/0.75 = 1/3

2) Solution: a

We’re interested in finding: p{A|B’&} From the problem’s statement: p{B|&}=0.82 and p{A’B’|&}=0.08 Clearly: p{A’B’|&}=p{A’|B’&}* p{B’|&} This can be rewritten as: p{A’B’|&}=[1-p{A|B’&}]*[1- p{B|&}] Solving for p{A|B’&}, we obtain: p{A|B’&}=1- p{A’B’|&}/[1- p{B|&}]=1-0.08/(1-0.82)=1-0.08/0.18=0.1/0.18=0.56

3) Solution: d

It is important for distinctions to pass the clarity test because clear distinctions make for effective communication among all parties in a decision. Answer (a) is incorrect because the value of information does not have anything to do with the clarity of a distinction. Answer (b) is incorrect because there is no such person as a clairvoyant; it is only an imaginary construct to aid in thinking about decisions. Answer (c) is incorrect because we cannot really begin to assign probabilities to outcomes unless those outcomes are clearly defined, let alone the fact that it is never coherent nor necessary to use a range of probabilities to describe uncertainties. 4) Solution: c

Statement I is true: If A and B are mutually exclusive, p{A and B|&}=0. If A and B are irrelevant, p{A and B|&} = p{A|&} p{B|&}=0. But p{A|&}�0 and p{B|&}�0 by assumptions. And so A and B cannot be irrelevant Statement II is true: Let us suppose that A and B are irrelevant. We have p{A and B|&} = p{A |&} p{B|&}. We always have: p{A or B|&} = p{A|&}+ p{B|&} - p{A and B|&}. And so p{A or B|&} = p{A|&}+ p{B|&} - p{A |&} p{B|&} = p{A |&} (1 - p{B|&}) + p{B|&} < 1 * (1 - p{B|&}) + p{B|&} = 1 So p{A or B|&} < 1. A and B cannot be collectively exhaustive. Statement III is true: it is clearly an example of associative logic error



Statement IV is false: Knowing Berkeley’s chances against UCSD and Stanford’s chances against Berkeley do not necessarily say anything about Stanford’s chances against UCSD.

5) Solution: d

b) corresponds to a diagram without any arrow. c) corresponds to a diagram with arrows pointing from C to B, A to B and A to C. This corresponds to a cycle and this cannot be a valid relevance diagram. d) is correct. a) does not correspond to any given diagram and is false. Indeed, we generally do not have {C|A,&}{C|B,&} = {C|A,B,&}

6) Solution: b

Let A, B, and C be outcomes, so that A corresponds to the outcome “A is scheduled to be executed,” and so on. Let “A,” “B,” and “C” denote what the guard says, so that “A” corresponds to “A is pardoned.” We want to compute the following probability: P{A | “B”, &}. The problem statement gives us the values of

P{A | &}, P{B | &}, P{C | &},

P{“B” | A, &}, P{“B” | B, &}, P{“B” | C, &},

P{“C” | A, &}, P{“C” | B, &}, P{“C” | C, &}.

From this information we can construct the following probability tree:

Now to compute P{A | “B”, &}, we flip the tree as follows:

1/2

1/21/6

1/6“B”

“C”

0

11/3

0

1

00

1/3

1/3

1/3

A

B

C 1/3

“B”

“C”“B”

“C”



The answer we want is circled. The guard has not given Prisoner A any useful information, since A’s probability assignment for his own execution has not changed. 7) Solution: b

The original probability tree is:

Flip the tree to make location and industry the first and second distinctions:

0.2

0.8

Small

Large

0.1

0.9

Small

Large

0.7

0.3 0.024

0.056NY

SF

1

0 0

0.32 NY

SF

0.5

0.5 0.03

0.03 NY

SF

0.5

0.5 0.27

0.27 NY

SF

0.4

0.6

I-Banking

Consulting

1/3

00

1/6 A

B

2/3

C1/3

0.5

0.5

“B”

“C”

1/3

2/31/3

A

B

0

C0

1/6



The conditional probability is circled in the previous tree. b) is the correct answer. 8) Solution: d

Flip the tree to make firm size and location the first and second distinctions:

0.614

0.386

NY

SF

0.686

0.314

NY

SF

0.651

0.349 0.03

0.056 I-Banking

Consulting

0.444

0.556 0.03

0.024 I-Banking

Consulting

0.542

0.458 0.27

0.32 I-Banking

Consulting

0

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0 I-Banking

Consulting

0.14

0.86

Small

Large

0.556

0.444

I-Banking

Consulting

0.074

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I-Banking

Consulting

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0.851 0.32

0.056 Small

Large

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0.03 Small

Large

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0 0

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Large

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0.03 Small

Large

0.676

0.324

NY

SF



The conditional probability is circled in the previous tree. d) is the correct answer.

9) Solution: d

Using the given elemental probabilities, we can draw a tree and fill in its branch probabilities:

Statement I is false – a student is as likely to be lazy as to be hard-working given &, since {Y = “Lazy” | &} = {Y = “Hard-Working” | &} = 0.5. Statement II is true – as shown by the probabilities that were circled in the tree above, {X = “Easy” | Y = “Lazy”, &} = 0.6 = {X = “Easy” | Y = “Hard-Working”, &}. Statement III is true – as shown by the probabilities that are surrounded by a box in the tree above, {Z = “A” | Y = “Hard-Working”, X = “Difficult”} = 0.6 > 0.5 = {Z = “A” | Y = “Lazy”, X = “Easy”}. Statement IV is true – we need to flip the tree to see this, by making Z the first distinction in the tree (see flipped tree below). Once this is done, we can see from the conditional probabilities of Y that X is relevant to Y given Z: for example, the two conditional probabilities that were circled on the tree below do not match.

0.6

0.4

Easy

Difficult

0.6

0.4

Easy

Difficult

0.5

0.50.15

0.15 A

B or Less

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0.9 0.18

0.02 A

B or Less

0.9

0.1 0.03

0.27 A

B or Less

0.6

0.4 0.08

0.12 A

B or Less

0.5

0.5

Lazy

Hard-Working



10) Solution: b

Diagram a) implies that Y and X are irrelevant given Z, which we know from the previous question to be false (see statement IV). Diagram d) implies that Z and X are irrelevant given Y, which we know from the first of the two probability trees above to be false. Diagrams b) and c) are both compatible with the fact that Z is relevant to Y given X and with the fact that Z is relevant to X given Y. However, c) also implies a possibility of relevance between X and Y given &, which we know to be false (see statement II). Therefore, although both diagrams are compatible with the probability distribution provided, b) best describes the relationships in this distribution because it reflects this stronger and true statement about the independence of X and Y given &.

Quantitative Problems 1)

Part a) Remember that your PIBP should not depend on the $2000 market buying price, but only on your use of the cheese.

0.75

0.25

Easy

Difficult

0.41

0.59

Easy

Difficult

0.36

0.64 0.27

0.15 Lazy

Hard-Working

0.14

0.86 0.12

0.02 Lazy

Hard-Working

0.83

0.17 0.03

0.15 Lazy

Hard-Working

0.69

0.31 0.08

0.18 Lazy

Hard-Working

0.56

0.44

A

B or Less



Part b) When you determine your bid, you will probably consider the $2000 market buying price. This means that your bid may exceed your PIBP. Part c) The main difference between your PIBP and the amount you would bid are market considerations. In this case, the main effect is the ability to resell the cheese for $2000. 2) Aruna wants to maximize her e-score, what probability, p, should she write down for answer (a)? She should assign 0.7 to the probability of (a) being correct. Aruna’s e-score is:

.7*()4ln(15

)ln(10015

100 p+ ) + .3 *()4ln(15

)3/)1ln((10015

100 p−+ )

If you take the derivative of this expression with respect to p, you will find that the maximum occurs at p=0.7. The best strategy is to assign your true beliefs. It is called a “strictly proper scoring rule.” 3) Weather Part a) Here are some useful clarity test definitions. • R = During the 3-hour party, it rains .05 inches in any 15 minute period, measured in

the center of the outdoor party area. • W = The temperature does not drop below 65°F at all and doesn't drop below 70°F for

any 15 minute period, as measured in the center of the outdoor party area. One may find it difficult to say when the clarity test is satisfied, but may often use the heuristic of asking other people if they understand what is meant by both the distinctions. Think for yourself whether or not the above definitions pass the clarity test. Part b) Here are some definitions of the different probabilistic information. • {¬R|&} = the probability that it does not rain, given only my background state of

information. • {W|¬R, &} = the probability that it will be warm, given only that it doesn't rain and

my background state of information. • {¬R|W, &} = the probability that it doesn't rain, given only that it will be warm and

my background state of information. • {¬R,W|&} = the probability that it doesn't rain and it will be warm, given only my

background state of information.



Part c) Each of you may have different models of the weather to explain the relevance or lack of relevance between R and W. However, the relevance between R and W does not necessarily mean that one causes the other. Relevance does not imply causation. Part d) We wrote this question to give you practice thinking about probabilities and flipping trees. 4) Die Part a) The possibilities are {1, 2, 3, 4, 5, 6}. Here is a tree expressing equal probability for each possibility.

1/6

1/6

3

4

1/65

1/62

1/66

1/61

Part b)

1 2 3 4 5 6

1/6

1 2 3 4 5 6

1/6

2/6

3/6

4/6

5/6

1Cumulative

Excess

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