Department of Computer and Systems Sciences David Sundgren

Written examination in Computer and Systems Sciences Decision Support Methods BSM Friday 8 January to Monday 11 January 2016

The solutions must be complete and easy to follow. Formulations that are verbatim quotations from sources will not contribute to the points given to a solution, whether they are referenced or not, that is, wherever you find a solution, formulate it in your own words and make sure that you understand what you write. For grade A a minimum of 41 points out of 45 possible are required, grade B requires at least 34 points, grade C requires at least 29 points, for grade D at least 25 points are required, for E at least 20 and for Fx at least 18 points are required.

1. What are the problems associated with having (a) too few, (b) too many decision alternatives? (5 p.) 2. Describe three attributes that might be usable when the objective is to improve health in a population and state whether attributes are natural, artificial or proxy attributes. (5 p.) 3. How many (a) scenarios, (b) strategies, are there in the decision situation depicted in the decison tree in Figure 1? (5 p.) 4. Let C = (4, 0.9; 6, 0.1) be a lottery and determine the risk premium and risk attitude of a person whose certainty equivalent is (a) 3.6, (b) 4.2, (c) 4.8. (5 p.) 5. A friend of yours wants to buy a new telephone. There are four attributes that he finds relevant and you want to help him with determining weights. The attributes are price (between e 100 and e 800), quality (2 – 9 points in a consumer test), durability (0 – 6 points in the same consumer test) and battery life (30 – 80 hours).

2

0.1

C A

0.5

0.4 B

0.8 0.2

D E

4 8

H

F

0.6 I

G

0.4

J K

0 5

2

4

Figure 1: Decision tree for Problems 3 and 7 By following a weight determination method you ask your friend about his preferences regarding some artificial alternatives with attribute levels ordered as price, quality, durability, battery life and the result is that (800, 9, 0, 30)  (100, 2, 0, 30)  (800, 2, 0, 80)  (800, 2, 6, 30)  (800, 2, 0, 30) What are the attribute weights? (5 p.) 6. You are thinking of finally buying a blue-eared Norwegian forest frog. The problem is that two percent of all blue-eared Norwegian forest frogs carry a disease that can spread to humans. In the frog the disease causes nothing more than cough but it can be deadly for humans. Looking at the charming little creature in the pet shop you notice that it starts to cough. The salesperson insures that the frog only has a cold. But since you are an amateur expert on blue-eared Norwegian forest frogs you know that over half of them, more precisely 60% of the frogs that carry the disease cough. On the other hand, only 5% of all frogs that do not carry the disease suffer from coughs. What is the probability that the coughing frog in the pet shop carries the disease? Is it worth the risk? (5 p.) 7. Roll back the tree to see what the expected values are of the strategies in the decision tree in Figure 1. (5 p.) 8. In the textbook Rational Decision Making by Eisenf¨ uhr, Weber and Langer stochastic dominance is on page 300 defined as a dominates b stochastically if and only of 1 − Pa (x) ≥ 1 − Pb (x) for all values of X and 1 − Pa (x) > 1 − pb (x) for at least one value of X. But in video number 25 for this course the definition is given as a stochastically dominates b if for every consequence x p (outcome ≤ x|a) ≤ p (outcome ≤ x|b) and for at least one consequence y p (outcome ≤ y|a) < p (outcome ≤ y|b). Show that the two definitions of stochastic dominance are equivalent.

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(5 p.)

9. Does either of the independence axioms of utility theory and subjective expected utility theory imply the other? That is, when the two independence axioms are as follows, is it so that if one holds then the other must also hold? • If a  b holds for two lotteries, then it also has to hold for all lotteries c and all probabilities p, that p · a + (1 − p) · c  p · b + (1 − p) · c. • Let a, b, a0 and b0 be alternatives and let S 0 be a subset of the set of events S with a(s) = a0 (s) as well as b(s) = b0 (s) for s ∈ S 0 and a(s) = b(s) as well as a0 (s) = b0 (s) for s ∈ S \ S 0 , then a  b holds if and only if a0  b0 . (5 p.)

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Answers for examination in Decision Support Methods, Friday 8 January to Monday 11 January 2016.

Solutions for examination in Decision Support Methods, Friday 8 January to Monday 11 January 2016.

Solution to problem 1. See Chapter 4 in general and in particular 4.8.1 for 1b. In short, the over-arching problem with having too few alternatives is the risk of not considering an alternative that is superior to the alternatives that you have. And with too many alternatives you might not have the resources to properly analyse them. Solution to problem 2. Some attributes are • Number of reported sick days (natural or proxy depending on how much you trust that people report sick if and only if they really are sick), • Consumtion of alcohol and tobacco (proxy) • Life expectancy (proxy) Solution to problem 3. There are (a) Nine scenarios, there are three event nodes, one with outcomes C, D, E, one with F, G and one with J, K, but the latter only occurs if F . But C, D; E is independent of F, G. Regardless of whether C, D or E happens we can have F and then J or F and then K, or G. That is, the scenarios can be summarised as • C, F, J • C, F, K • C, G • D, F, J • D, F, K • D, G • E, F, J • E, F, K • E, G (b) Three strategies; A, B(−H), B(−I), that is A is a strategy of its own, B must be complemented with either H or I in case that F happens. Solution to problem 4. The expected value of the lottery is 4 · 0.9 + 6 · 0.1 = 3.6 + 0.6 = 4.2, so the risk premiums are (a) 4.2 − 3.6 = 0.6 (b) 4.2 − 4.2 = 0 (c) 4.2 − 4.8 = −0.6 so the risk attitudes are (a) risk averse since 0.6 > 0 (b) risk neutral since 0 = 0

(c) risk prone since −0.6 < 0. Solution to problem 5. The given information seems geared for the swing method, so let’s assign points that reflect the given preferences. We’ll put 0 points for (800, 2, 0, 30) and 100 points for (800, 9, 0, 30) Then we have an infinite amount of choices but arbitrarily we give 25 points to (800, 2, 6, 30), 50 points to (800, 2, 0, 80) and 75 points to (100, 2, 0, 30); if nothing else equal differnces might make calculations easier. Then 100 points correspond to the second attribute (quality), 75 points to the first (price), 50 to the fourth (battery) and 25 points to the third attribute (durability). Normalisation gives the weigths: 100 100 + 75 + 50 + 25 75 = 100 + 75 + 50 + 25 50 = 100 + 75 + 50 + 25 25 = 100 + 75 + 50 + 25

wquality = wprice wbattery wdurability

100 250 75 = 250 50 = 250 25 = 250 =

= 0.4 = 0.3 = 0.2 = 0.1

Solution to problem 6. Let’s denote cough by c and contagious disease by d. Then we can summarise some of the information given in the problem formulation as p(d) = 0.02, p(c|d) = 0.6 and p(c|d) = 0.05. Hence p(d) = 1 − 0.02 = 0.98. We want to know p(d|c) and by Bayes’ rule p(c|d)p(d) p(c|d)p(d) = = p(c) p(c|d)p(d) + p(c|d)p(d) 0.012 0.6 × 0.02 = ≈ 0.20 0.6 × 0.02 + 0.05 × 0.98 0.012 + 0.049 p(d|c) =

So the probability that the coughing frog is dangerous to you is 20%. Whether it’s worth the risk is subjective but no one should call you a coward for not daring to take the frog home. 20% risk of dying is worse than russian roulette. Solution to problem 7. As we saw in problem 3 there are three stratgies, namely A, B − H and B − I. Starting from the right as one should when rolling back a tree we first compare H with I. The expected utility of H is 0.4 · 0 + 0.6 · 5 = 3. See Figure 2. To find the maxium expected utility we could cancel I but here we want to compute the expected utility fo all three strategies. In Figure 3 we put the expected utility of H as value for F . We look at strategy B − H in Figure 4, the expected utility of B − H is 0.8 · 3 + 0.2 · 4 = 3.2. But to compute the expected utility of B − I we look at Figure 5 where the I alternative is decided for. The expected utility of B − I is 0.8 · 2 + 0.2 · 4 = 2.4, see Figure 6. The expected utility of A is 0.1 · 2 + 0.5 · 4 + 0.4 · 8 = 5.4, see figure 7. So the expected utilities are EU(A) = 5.4, EU(B − H) = 3.2, EU(B − I) = 2.4 Solution to problem 8. p(outcome ≤ z|c) can be written as distribution function Pc (z), so p(outcome ≤ x|a) = Pa (x) and p(outcome ≤ x|b) = Pb (x). And Pa (x) ≤ Pb (x) is equivalent to −Pa (x) ≥ −Pb (x) ⇐⇒ 1 − Pa (x) ≥ 1 − Pb (x), and in the same way Pa (y) < Pb (y) ⇐⇒ 1 − Pa (y) > 1 − Pb (y). Solution to problem 9. If the independence axiom for utility theory holds, then so must the independence axiom for subjective expected utility theory. To prove this, assume the independence axiom for utility theory and assume the conditions for the independence axiom for 6

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0.1

C A

0.5

D

0.4 B

E

4 8

H

F

0.8

I G

0.2

3 2

4

Figure 2: Expected utility of H 2

0.1

C A

0.5

0.4 B

0.8 0.2

D E F G

4 8 3 4

Figure 3: Deciding for H 2

0.1

C A

0.5

0.4 B

D E

4 8

3.2

Figure 4: Expected utility of B − H subjective expected utility theory, i.e. we assume that if a  b holds for two lotteries, then it also has to hold for all lotteries c and all probabilities p, that p·a+(1−p)·c  p·b+(1−b)·c and that a, b, a0 and b0 be alternatives and let S 0 be a subset of the set of events S with a(s) = a0 (s) as well as b(s) = b0 (s) for s ∈ S 0 and a(s) = b(s) as well as a0 (s) = b0 (s) for s ∈ S \ S 0 . Let α = A(s) = A0 (s), s ∈ S 0 , β = B(s) = B 0 (s), s ∈ S 0 and γ = A(s) = B(s), s ∈ S \ S 0 , δ =

7

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C A

0.5

0.4 B

0.8 0.2

D E F G

4 8 2 4

Figure 5: Deciding for I 2

0.1

C A

0.5

0.4 B

D E

4 8

2.4

Figure 6: Expected utility of B − I A

5.4

H B I

3.2 2.4

Figure 7: Expected utility of A A0 (s) = B 0 (s), s ∈ S \ S 0 . Further, let p equal the probability of S 0 . Either α  β or β  α. Assume that α  β. By the independence axiom for utility theory (that we have assumed) both pα + (1 − p)γ  pβ + (1 − p)γ and pα + (1 − p)δ  pβ + (1 − p)δ .

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But this means that both A  B and A0  B 0 , that is, the sure thing principle, the independence axiom for subjective expected utility theory, holds. In the same way we get that B  A and B 0  A0 if β  α. But that the independence axiom for subjective expected utility theory holds is not enough for guaranteeing the independence axiom for utility theory. Even though a  b ⇐⇒ a0  b0 under the conditions of the partitioning of S into S 0 and S \ S 0 there is nothing that excludes the possibility of a lottery C and a probability p such that A  B but pB +(1−p)C  pA+(1−b)C.

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