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PPE 110

PPE 110. Introduction to Decision Theory. 1. Quantifying uncertainty The class will try and translate a lot of the natural language we use to reason about uncertainty to a more formal language.

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PPE 110

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  1. PPE 110 Introduction to Decision Theory

  2. 1. Quantifying uncertainty The class will try and translate a lot of the natural language we use to reason about uncertainty to a more formal language. The cost of this translation will be learning the new language. The benefit will be the ability to gain a more precise way of expressing uncertainty and how to deal with it

  3. The basic building block of this new language will be the use of probability. • Probability is a quantification of uncertainty. When you say something is likely, or moderately likely, or very likely, you only have a small and finite number of phrases to express the uncertainty. • Numbers give you a richer spectrum of options. • Probability is a way of using numbers to capture intuitive notions of uncertainty, richer than what would be otherwise possible with words. It allows for finer gradations.

  4. Let us revisit some basic probability. • Example: 3 flips of a fair coin. What are all the possibilities? Many times, it helps us to draw pictures, to visualize the information. • Here it is cut and dried: there are 8. • Write down each of these 8 possibilities as a point in a rectangle. Let 1=HHH, 2=HHT, 3=HTH,4=THH,5=HTT,6=THT,7=TTH,8=TTT • Each has the same probability, so the probability is 1/8 • 2 3 4 • 5 6 7 8 • These are the elementary atoms or propositions which are the building blocks of our investigation. The set of all propositions we are interested in will form the Universal Set. It is shown above by the numbers enclosed within the rectangle.

  5. A description of the universal set, and the probability of each atom within the universal set is known as a probability distribution. • For example, the probability distribution of the set of rolls of a die is described below:

  6. 2. Formalizing all this. • Let us make all this more formal – which will be useful later. This is the grammar of our new language. • Given a set X, a probability system is a function p which has the following properties • For all x∈X (for all x an element of X, where each element is the same thing as an atom), • p(x) ≥ 0 • ∑p(x)=1 (the sum of all the probabilities of the atoms equals 1) • p(φ)=0 where φ is the null set (what is the null set? It is the set of all impossible events. An example of an impossible event are propositions like: a student in PPE 110 will get both an A and a B for the class).

  7. 3. From atoms to events • When we try to reason about uncertainty, we are often interested in not just the likelihood of the most basic of the propositions, but also about collections of such propositions (George AND Ringo are nice men; The weather in Philadelphia can be very hot OR very cold). • Think of equating an elementary proposition to an atom, and now wanting to be able to represent more complex sentences in the new language. We will use the words atom and propositions interchangeably • The compound sentence (often called ‘event’ in probability theory) will always boil down to a grouping of atoms, where the group is formed by the particular concept we are trying to express. In other words, we want to be able to express our uncertainty not just about individual atoms or fundamental situations, but also about collective, aggregate situations that may involve more than one of our atomistic propositions.

  8. Loosely speaking, compound sentences are just elementary propositions connected together by connectors and qualifiers: and, or, not, etc. Which means that they refer to a one or more atoms, or a collection of atoms. • This collection of atoms (event) will be called a subset of the universal set we encountered earlier. • A is a subset of X implies A is part (possibly all of but not necessarily) of X. It is written as A ⊆X. A ⊂ X means that A is a part of X but not all of it. • The next slide shows an event, and the slides after that uses visual representations of subsets to illustrate a problem.

  9. Depiction of an event A={2,3,6,7}. Refer to slide 4 for the meaning of this event. • 2 3 4 • 5 6 7 8

  10. We continue with trying to think of a problem in visual terms. Here is an example of using events, or a collection of elementary facts. • A town has 100 taxis. 85 green taxis owned by the Green Cab Company and 15blue taxies owned by the Blue Cab Company. • On March 1, 1990, Alice was struck by a speeding cab, and the only witness testified that the cab was blue rather than green. Alice sued the Blue Cab Company. • The judge instructed the jury and the lawyers at the start of the case that the reliability of a witness must be assumed to be 80% in a case of this sort, and that liability requires that the “preponderance of the evidence,’ meaning at least a 50% probability, be on the side of the plaintiff. • Conditional on seeing a blue cab, the witness is likely to have correctly identified it as blue, or incorrectly as blue. The situation is the same for green cabs. • There are 15 blue cabs of which 12 would be correctly identified, and 85 green cabs or which 17 would be incorrectly identified. • Assume that a priori, each cab is equally likely to be the culprit.

  11. We show this situation schematically. We will return to this problem later. For now, just see the visualization. Each atom here is a particular cab (not shown) and each circle is an event, as are the shaded parts of each circle.

  12. 4. From collections of atoms to collections of events • In much the same way we want to talk about events (as an enlargement of our ability to talk about elementary propositions), we also want to talk about collections of events. For this, we need to introduce some new terms. • A∩B (A intersection B) will mean the set of situations where both A and B occur, A U B (A union B) will mean the set of situations where A or B occur, and Ac (A complement) will mean the set of situations incompatible with A. These are to be shown in the class. • Using these constructors, we can talk about various events, such as: the set of blue cabs which are never mistaken for yellow. • Let us check some facts visually: • 1: (AUB)c = Ac∩Bc • 2: (A∩B)c = AcUBc

  13. Here is a visual proof that (A∩B)c = AcUBc • (AUB)c are in black lines, Ac is in green and Bc is in pink. The intersection of the latter two should be the area in black.

  14. 5. Probabilities of events. • An earlier slide talks about the probability of each atom and the fact that such probability must add to 1. • Thus the probability of each atom must be known. If we know that, then we may extend the analysis to find the probability of the larger events: • For instance, for any event A, probability of A or p(A)=sum of all the probabilities of the atoms in A. • Thus, the probability of being hit by a blue cab was 1/100+1/100+..=15/100 • It is possible that whenever one event occurs, the other is impossible. If this is true, then we call the two events mutually exclusive or disjoint.

  15. This means that if we have two sets that do not intersect, say A and B, then P(A U B)=P(A)+P(B). Why? Let Z= A U B. Now, the probability of Z is the sum of the probability of all the atoms in Z. Now atoms in Z can be either in A or in B, which means that the probability of Z is the sum of the probability of all the atoms in A and the atoms in B. • By the way, recall that P(A U B) is the probability of A or B. What is this probability if A and B may or may not intersect?

  16. For any two sets, A and B, We will show that P(AUB)=P(A)+P(B)-P(A ∩ B), by using this information and breaking up AUB into disjoint parts. • P(A)+P(B)-P(A ∩ B)=P(A ∩ Bc)+P(B ∩ Ac)+P(A ∩ B) (draw a diagram and check). • Let us write this expression on the right-hand side as [P(A ∩ Bc)+ P(A ∩ B)] +[P(B ∩ Ac)+P(A ∩ B)] - P(A ∩ B) • But note that • P(A ∩ Bc)+ P(A ∩ B)=P(A) • P(B ∩ Ac)+P(A ∩ B)=P(B) • Hence we are done

  17. Thus, when A and B have no outcomes in common, the probability of AUB is P(A)+P(B) • When A and B are not disjoint, and are just two arbitrary sets, then P(AUB)=P(A)+P(B)-P(A∩B) • Note that the disjoint case is a special version of the arbitrary case (why?)

  18. 6. An example. • Using our new tools, let us attempt problem 5, page 232 of the statistics text. Suppose the percentages refer to a total of 100 men and women. F is the set of freshmen and S is the set of sophomores. The probability of each set is shown in the aggregate. The individual men and women within each set (the individual atoms) are the dots. Not all the dots or atoms are shown, and they will be omitted after this slide. 0.8=prob of men (M) 0.2=prob of women (W) F 0.15 . . . . .. .. ... ... S 0.85 . . . . .. ... ...

  19. (a) The min prob of P(S ∩ W)=0.05 (shown) 0.8=prob of men (M) 0.2=prob of women (W) F 0.15 S 0.85

  20. (b) The max prob of P(S ∩ W)=0.2 0.8=prob of men (M) 0.2=prob of women (W) 0.15 F 0.2 S 0.85

  21. What is the max probability of (M∩S)c? We know this is equal to Mc U Sc=W U F. The max probability of this is when W ∩F= φ and is equal to 0.35 0.8=prob of men (M) 0.2=prob of women (W) F 0.15 0.15 S 0.65

  22. What is the minimum and max probability of Fc U Wc? The former is S and the latter is M. The minimum probability is therefore 0.85 (when P(S ∩ W)=0.05 )and the max is 1 (shown below). 0.8=prob of men (M) 0.2=prob of women (W) F 0.15 S 0.85

  23. 7. Conditional probability. • Now, suppose we want to know if knowing one event has occurred adds information about another event. For instance, if there is a breeze, is there an increased possibility of rain? Given the budget crisis in Philadelphia, will it take longer to rebuild the south street bridge? What is the probability of the 2nd toss of a fair coin landing heads if the first has been already seen to be heads? Always, we are interested in the probability of a 2nd event given a first. • In general, we write P(A|B) to mean the probability of event A given the knowledge that B has occurred. This is also said to be the conditional probability of A given B. • We define P(A|B)=P(A∩B)/P(B). • Let us come back to the cab problem. Let A be the event that Alice was struck by a blue cab. Let B be the event that a witness sees a blue cab. P(B) is the probability that the witness saw a blue cab. • Let us convince ourselves that what we need is P(A|B) • What is P(B)? It is 0.8[15/100]+0.2[85/100]=29/100 • What is P(A∩B)? It is the probability that the Alice was struck by a blue cab and it was correctly identified as blue. It is the part of the blue circle not shaded green. This is 12/100

  24. Thus, the chance that one of those cabs seen to be blue may have caused the accident is 12/29 < 0.5

  25. 8. Independence • When knowing that one event occurred gives us no information about the probability of the other event, we will say that the two events are independent. • Another way of saying this is, given the information of one, or conditional on obtaining the information about one, the assessment of the other likelihood is unchanged. • This is a case of independence. • We are going to make precise the nature of independence between two events by using probabilities, and in particular, conditional probabilities. • For instance, what is the probability of the 2nd toss of a fair coin landing heads if the first has been already seen to be heads? Clearly, the first coin toss does not affect the probabilities of the second, so the probability is 0.5 • In our formula, note that P(A|B)=P(A∩B)/P(B)=P(A) if B gives no information about A. This in turn means that P(A∩B)=P(A) × P(B)

  26. Let us return to the example we have been working. Suppose that P(F ∩ M)=0.12 • This is shown below 0.8=prob of men (M) 0.2=prob of women (W) F 0.03 0.12 S 0.17 0.68

  27. Now, suppose that you know a person you have picked at random is a male. Does this change the probability of this person being a freshman? • The prior probability of encountering a freshman was 0.15. Now, you know this person is a male. Is this useful information in terms of changing our probabilities? • Well, of the 80 men, 12 are freshmen, so in fact, the probability of encountering a freshman =12/80=0.15 remains the same as before. The events “encountering a freshman” and “encountering a male” are independent.

  28. Let us assume that the area of the sets are proportional to their probabilities. This means we cannot leave empty spaces as before and must draw the F and S sets in a different way. This is done below. 0.8=prob of men (M) 0.2=prob of women (W) F 0.12 0.03 0.17 0.68 S

  29. Here is a visual representation of independence. What independence means is that conditional on being in the blue region, the probability of being in the upper left quadrant is equal to the probability of being in the F region. Or, the ratio of the shaded region to the region of M is equal to the ratio of F as a fraction of the whole box. 0.8=prob of men (M) 0.2=prob of women (W) F 0.12 0.03 0.17 0.68 S

  30. Some new but related questions. If M gives no further information about F, does F give any further information about M? • If F is known to have not occurred, then does our probability of M not occurring change? • These questions we can answer generally. In fact, let us prove the following:

  31. For any two sets A and B, is it true that if P(A|B)=P(A), then P(B|A)=P(B)? • Suppose any two sets A and B are independent. Are Ac and Bc independent as well? • The answer to both are yes. The former is very easy. The latter is done in the next slides.

  32. For this statement to be true, we need P(Ac ∩Bc)= P(Ac) ×P(Bc) • First note again that Ac ∩Bc=(AUB)c • Thus, if we can prove that • P(Ac) ×P(Bc)=P(AUB)c, then we are done. • Note that the left hand side • = [1-P(A)][1-P(B)] • =1-P(A)-P(A)+P(A)P(B) • =1-[P(A)+P(B)-P(A ∩B)] • =1-[P(AUB)] • = P(AUB)c (hence proved)

  33. We will do parts of some problems from the book that illustrates how simple probability problems may be conceptualized and solved using sets. • Let us look at problem 3 from page 247 • What is the set of outcomes or the universal set? It is the set of all arrangements of the deck – thus, there are 52! outcomes (52 factorial) in the set. • We will briefly define factorials in class, but since they are coming up already, here is a very simple explanation • :http://www.purplemath.com/modules/factorial.htm

  34. Each of those points have equal probability, since the deck is shuffled. • Thus, we can find the probability of any situation that is described by a collection of outcomes in this universal set. • In general then, if you can write down the set of possibilities and assign a probability to all of its outcomes, a lot of the hard work is done.

  35. (a) Let A = set of all those outcomes which have Jack of clubs at the top place. What is the probability of this set? • There are 51! elements in this set (check), and there are a total of 52! Outcomes. Thus, the likelihood of A happening is 51!/52! which boils down to 1/52 • (c) Let B be the set of outcomes which have the bottom card = J of Diamonds

  36. We need P(AUB) which we know equals P(A)+P(B)-P(A∩B). The first two terms equal 1/52, using logic just encountered. • The last term is that set, call it C, which lies in both A and B. This means that the first and last cards are fixed, and there are 50 cards to be arranged in 50 possible spaces. This can clearly be done in 50! ways. • Which means that the likelihood of C occurring is 50!/52! or 1/(52 ×51). • Thus, the answer is 1/52+ 1/52- 1/(52 ×51). • You are also encouraged to do Problem 4 and 5 (same section)

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