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Uncertainty and probability Using probabilities Using decision trees Probability revision Today’s agenda • Important terms • Simple review (objective, subjective, marginal, joint, and conditional probabilities) • Examples: outcomes, expected values, risk attitudes • Examples: action choices, decision trees Vocabulary • A probability is a number between zero and one representing the likelihood of the occurrence of some event. • Probability – objective vs. subjective – marginal vs. joint – joint vs. conditional – prior vs. posterior – likelihood vs. posterior Vocabulary continued • Outcomes or payoffs (mutually exclusive) • Action choices • States of nature • Decision tree • Expected value • Risk Probability Imagine an urn containing 1500 red, pink, yellow, blue and white marbles. Take one ball from the urn. What is: P(black) = 0 P(~black) = 1 ~ = NOT Probabilities are all greater than or equal to zero and less than or equal to one. Same urn: Suppose the number of balls is as follows: Red 400 Pink 100 Yellow 400 What is: Blue 500 White 100 P(Red) = 400/1500 = .267 Total 1500 P(Pink) = 100/1500 = .067 P(Yellow) = 400/1500 = .267 P(Blue) = 500/1500 = .333 P(White) = 100/1500 = .067 Total = 1 Joint probabilities and independence Define A as the event “draw a red or a pink marble.” We know 500 marbles are either red or pink. What are: P(A) = = .33 P(~A) = (1 - P(A)) = .67 Joint probabilities and independence (we’re getting there) Define B as the event, “draw a pink or white marble.” We know 200 marbles are pink or white. What are: P(B) = .133 P(~B) = .867 Joint probabilities and independence Define A as the event “draw a red or a pink marble.” Define B as the event “draw a pink or white marble.” What is: P(A, B) = P(A B) This is the joint probability of A and B. What color is the marble? Pink P(A, B) = P(pink) = = .0667 Joint probabilities and independence Are A and B independent? Note that P(A, B) P(A) * P(B) = .33 * .13 = .0429 Are A and B mutually exclusive? What is the probability of A or B? P(A or B) = P(A B) = P(A) + P(B) - P(A B) = .40 Joint probabilities and independence Suppose we draw one marble from the urn and replace it. Then, we draw a second marble. What is: P(Red, Red) = = .071 P(Red, Blue) = .088 Are (Red, Red) independent? Are (Red, Blue) independent? Joint and marginal probabilities What are: P(B) = = .133 P(~B) = .867 1500 P(A, B) = P(A or B) = Conditional probabilities The probability that a particular What is event will occur, given we already know that another event has P(A | B) = occurred. We have information to bring P(~A | ~B) = to bear on the base rate 1500 probability of the event P(A | ~B) = P(~B | A) = Definition of independence P(B | A) = Events A and B are independent if P(B | A) = P(B) Here P(B | A) = P(B) P(B) = 1500 Marginal and joint probability table: The joint probabilities are in the box. The marginals are outside. How do you compute conditionals from this? A ~A P(B | A) = B ~B Joint probability tables A ~A What are: B P(A, B) = .0667 ~B P(~A, ~B) = .6 P(~A) = .667 P(~B) = .867 P(~B | A) = .8 Outcomes or payoffs Example: Win $1,000 if you draw a pink marble, win $0 otherwise. Outcomes: $1,000 or $0 Events: A pink marble or a marble of another color Probabilities: P(Pink) = 1/15 P(~Pink) = 14/15 The expected value of this gamble: E(gamble) = (1/15)*($1,000) + (14/15)*($0) = $66.67 Example We expect to sell 10,000 computers if the market is good and to sell 1,000 computers if the market is bad. The marketing department’s best estimate of the likelihood of a good market is .5. Outcomes: 10,000 computers sold or 1,000 computers sold. Probabilities: P(good market) = .5 P(bad market) = .5 Are these objective or subjective probabilities? What are expected computer sales? 5,500 More expected values With discrete outcomes, an expected value is the probability-weighted sum of the outcomes for the decision of interest. Here are some two-outcome lotteries. Compute the expected values. L1 : Win $1,000 with probability .5 or lose $500 with probability .5. E(L1) = $250 L2 : Win $2,000 with probability .5 or lose $1,000 with probability .5. E(L2) = $500 More lotteries Here are some two-outcome lotteries. Compute the expected values. L3 : Win $1,500 with probability 1/3 or lose $750 with probability 2/3. E(L3) = $0 L4 : Win $750 with probability 2/3 or lose $750 with probability 1/3. E(L ) = $250 4 L5 : Win $300 with probability .5 or lose $200 with probability .5. E(L5) = $50 L6 : Win $10,000 with probability 9/10 or lose $85,000 with probability 1/10. E(L6) = $500 L1 : Win $1,000 with probability .5 or lose $500 with probability .5. E(L1) = $250 L2 : Win $2,000 with probability .5 or lose $1,000 with probability .5. E(L2) = $500 L3 : Win $1,500 with probability 1/3 or lose $750 with probability 2/3. E(L3) = $0 L4 : Win $750 with probability 2/3 or lose $750 with probability 1/3. E(L4) = $250 L5 : Win $300 with probability .5 or lose $200 with probability .5. E(L5) = $50 L6 : Win $10,000 with probability 9/10 or lose $85,000 with probability 1/10. E(L6) = $500 Action choices If I build a large hotel (cost = $5,000,000) and tourism is high (P(high) = 2/3), I will make $15,000,000 in revenue, but if it is low, I will make $2,000,000. If I build a small hotel (cost = $2,000,000) and tourism is high, I will make $5,000,000, but if tourism is low, I will make $2,000,000. I can also choose to do nothing. Action choices If I build a large hotel (cost = $5,000,000) and tourism is high (P(high) = 2/3), I will make $15,000,000 in revenue, but if it is low, I will make $2,000,000. If I build a small hotel (cost = $2,000,000) and tourism is high, I will make $5,000,000, but if tourism is low, I will make $2,000,000. I can also choose to do nothing. Action choices: Do nothing, build large, build small Outcomes: $15,000,000; $2,000,000; $5,000,000, $0 States of nature: high tourism, low tourism Probabilities: P(high) = 2/3; P(low) = 1/3 Decision trees Suppose I need to decide whether to invest $10,000 in the market or leave it in the bank to earn interest. If I invest, there is a 50% chance that the market will increase 20% over the coming year and a 50% chance that the market will be stagnant (no change). If I leave the money in the bank, there is an 80% chance that interest rates will increase to 10% and a 20% chance that interest rates will remain at 5%. What should I do? Use a decision tree. A decision by an individual is required Nature makes these decisions $2,000 or Nature decides .5 EV = $1,000 (Good) (Stagnant) I decide .5 Stock $0 or EV = $0 Market $1,000 $1,000 or .8 EV = $800 Bank (Increase) (Same) .2 $500 or $900 EV = $100 Homework assignment Problems 5-15 and 16 Consider two urns: Urn 1 Urn 2 Red balls 7 4 Black balls 3 6 P(R1) = Probability of red on first draw P(R2) = Probability of red on second draw P(B1) = Probability of black on first draw P(B2) = Probability of black on second draw a(1) Take one ball from urn 1, replace it, and take a second ball. What is the probability of two reds being drawn? P(R1, R2) = .7 x .7 = .49 Homework a(2) What is the probability of a red on the second draw if a red is drawn on the first draw? P(R2 | R1) = = a(3) What is the probability of a red on the second draw if a black is drawn on the first draw? P(R2 | B1) = Homework b(1) Take a ball from urn 1; replace it. Take a ball from urn 2 if the first ball was black; otherwise, draw a ball from urn 1. What is the probability of two reds being drawn? P(R1, R2) = .7 x .7 = .49 b(2) What is the probability of a red on the second draw if a red is drawn on the first draw? P(R2 | R1) = Homework b(3) What is the probability of a red on the second draw if a black is drawn on the first draw? P(R2 | B1) = What is the unconditional probability of red on the second draw? P(R2) = P(B1, R2) + P(R1, R2) = (.3 x .4) + (.7 x .7) = .61 Homework 5-16. Draw a tree diagram for Problem 5-15a Red P(R1, R2) = .49 Draw 2 .7 Red .3 Draw 1 .7 Black P(R1, B2) = .21 .3 Red P(B1, R2) = .21 Black .7 .3 Black P(B1, B2) = .09 Homework 5-17. Draw a tree diagram for Problem 5-15b Red P(R1, R2) = .49 Draw 2 .7 Red Draw 1 .3 .7 Black P(R1, B2) = .21 .3 Red P(B1, R2) = .12 .4 Black .6 Black P(B1, B2) = .18 Homework 5-29. This is the survey problem involving home- ownership and income levels. The results can be summarized by the table below Survey A. Suppose a reader of this magazine is selected at random and you are told that the person is a home- owner. What is the probability that the person has income in excess of $25,000? P(>$25,000 | homeowner) = Survey b. Are home ownership and income (measured only as above or below $25,000) independent factors for this group? They are NOT independent. If yes, then P(>$25 | home) = P(>$25) But, P(>$25) = .7 and P(>$25 | home) = .75 Homework 5-38. The president of a large electric utility has to decide whether to purchase one large generator (Big Jim) or four smaller generators (Little Arnies) to attain a given amount of electric generating capacity. On any given summer day, the probability of a generator being in service is 0.95 (the generators are equally reliable). Equivalently, there is a 0.05 probability of a failure. Homework 5-38. a. What is the probability of Big Jim’s being out of service on a given day? Let P(out) = the probability of any generator being out of service = .05 If P(BJout) = the probability of Big Jim’s being out of service. Then P(out) = P(BJout) = .05 5-38. Homework b. What is the probability of either zero or one of the four Arnies being out? (At least three will be running.) Since the probability of a failure (f) for one Arnie is .05, We want P(f 1|n=4, p=.05) P(f = 0 | n=4, p=.05) = P(f = 1 | n=4, p=.05) = P(f 1|n=4, p=.05) = .8145 + .1715 = .9860 Homework 5-38. c. If five Little Arnies are purchased, what is the probability of at least four operating? P(f 1 | n=5, p=.05) = 5-38. Homework d. If six Little Arnies are purchased, what is the probability of at least four operating? P(f 2 | n=6, p=.05) = Homework 5-40. Newspaper articles frequently cite the fact that in any one year, a small percentage (say, 10%) of all drivers are responsible for all automobile accidents. The conclusion is often reached that if only we could single out these accident-prone drivers and either retrain them or remove them from the roads, we could drastically reduce auto accidents. You are told that of 100,000 drivers who were involved in one or more accidents in one year, 11,000 of them were involved in one or more accidents in the next year. A. Given the above information, complete the entries in the joint probability table in Table 5-21. Accidents: Joint probability table A1 = accident in year 1, A2 = accident in year 2 Given: P(A2 | A1) = 11,000/100,000 = .11 P(A2 | A1)= P(A1, A2) = .011 .089 Therefore, .11 P(A1, A2) = P(A2 | A1)x .089 .811 P(A1) = .11 x .10 = .011 Accidents B. Do you think searching for accident-prone drivers is an effective way to reduce auto accidents? Why? If the information in the problem is representative, then searching for accident-prone drivers will not be very helpful, since having had an accident in Year 1 has only a minor effect on the probability of an accident in Year 2. Uncertainty continued . . . Probability revisions Continue decision trees Today’s agenda • Finish the homework problems • Work through a decision tree example that – Uses no information – Uses perfect information – Uses imperfect information • Briefly discuss Freemark Abbey Winery • Group problem solving Homework 5-42. A safety commissioner for a certain city performed a study of the pedestrian fatalities at intersections. He noted that only 6 of the 19 fatalities were pedestrians who were crossing the intersection against the light (i.e., in disregard of the proper signal), whereas the remaining 13 were crossing with the light. He was puzzled because the figures seemed to show that it was roughly twice as safe for a pedestrian to cross against the light as with it. Can you explain this apparent contradiction to the commissioner? 5-42. Homework The commissioner is looking at the wrong conditional frequencies (probabilities). P(?) = 6/19 It’s a conditional probability P(crossing against the light | killed at intersection) = 6/19 It is not the probability of being killed if you cross against the light. Further, 13/19 is not the probability of being killed if you cross with the light. 5-42. Homework The relevant probabilities are: P(killed | crossed with light) and P(killed | crossed against light) The deaths must be considered relative to the number of pedestrians who cross with and against the light. As an extreme possibility, it may be that the only six persons who crossed against the light were killed, a fatality rate of 100%; whereas, 1 million crossed with the light, a fatality rate of .000013. It isn’t likely that this is the case, but the commissioner’s data do not rule it out. 5-43. Homework Probability revision Suppose a new test is available to test for drug addiction. The test is 95 percent accurate “each way”; that is, if the person is an addict, there is a 95 percent chance the test will indicate “yes”; if the person is not an addict, then 95 percent of the time the test will indicate “no.” Suppose it is known that the incidence of drug addiction in urban populations is about 1 out of 1,000. Given a positive (yes) test result, what are the chances that the person being tested is addicted? Homework: Probability revision 5-43. We were given P(“yes” | addicted) = .95 and P(“no” | not addicted) = .95 and the prior P(addicted) = .001 We want P(addicted | “yes”) and P(not addicted | “no”) This is a different conditional probability called a “revised” or “posterior” probability. Test for drug addiction We know: P(addicted | yes) = and P(yes | addicted) = = .95 P(addicted) = .001 We can solve for the joint. Test for drug addiction P(addicted, yes) What is P(yes)? It consists of two joint probabilities. The test can say “yes” and the subject is addicted OR The test can say “yes” and the subject is not addicted P(addicted, yes) + P(not addicted, yes) = P(yes) Test for drug addiction P(not addicted, yes) Therefore, P(yes) = .00095 + .04995 = .0509 Given a positive test result, the probability that a person chosen at random from an urban population is a drug addict is P(addict | yes) = .00095/.0509 = .0187 This is a posterior probability. Drug test: joint probability table .95 .95 Bayes Theorem and the Multiplication Rule P(Ai | B) = and Another revision example Priors: P(disease) = .01 P(~disease) = .99 Test accuracy: P(positive | disease) = .97 P(positive | ~disease) = .05 P(negative | disease) = .03 P(negative | ~disease) = .95 Note that: false positives > false negatives Disease detection continued Priors: P(disease) = .01 P(~disease) = .99 Test accuracy: P(positive | disease) = .97 P(positive | ~disease) = .05 P(negative | disease) = .03 P(negative | ~disease) = .95 What is the probability that an individual chosen at random who tests positive has the disease? P(positive, disease) = .97 * .01 = .0097 P(positive) = (.97 * .01) + (.05 * .99) = .0592 P(disease | positive) = .0097/.0592 = .1639 Disease detection continued Suppose the tested individual was not chosen from the population at random, but instead was selected from a subset of the population with a greater chance of getting the disease? Prior: Suppose P(disease) = .2 Then, P(disease | positive) = 5-45 Revision Homework This is a classical probability problem. Try out your intuition before solving it systematically. Assume there are three boxes and each box has two drawers. There is either a gold or silver coin in each drawer. One box has two gold, one box two silver, and one box one gold and one silver coin. A box is chosen at random and one of the two drawers is opened. A gold coin is observed. What is the probability of opening the second drawer in the same box and observing a gold coin? Coin and box problem Here is a helpful visualization: We know we chose a box with a gold coin. We want P(gold2 | gold1) = Buying information As manager of a post office, you are trying to decide whether to rearrange a production line and facilities in order to save labor and related costs. Assume that the only alternatives are to “do nothing” or “rearrange.” Assume also that the choice criterion is that the expected savings from rearrangement must equal or exceed $11,000. Operating costs if you do nothing will be $200,000 If you rearrange successfully, operating costs will be $100,000. If you rearrange unsuccessfully, operating costs will be $260,000. Buying information Post Office Example Operating costs if you do nothing will be $200,000 If you rearrange successfully (P(success) = .6), operating costs will be $100,000. If you rearrange unsuccessfully (P(fail) = .4), operating costs will be $260,000. What is the expected value of each action choice? Rearrange: .6 x $100,000 + .4 x $260,000 = $164,000 Do nothing: $200,000 What would you choose? Post Office Decision Tree Do nothing $200,000 You decide $100,000 .6 Rearrange Succeed Fail $164,000 .4 $260,000 You can hire a consultant, Joan Zenoff, to study the situation. She would then render a flawless prediction of whether the rearrangement would succeed or fail. Compute the maximum amount you would be willing to pay for the errorless prediction. $164,000 Don’t buy info. rearrange $100,000 $140,000 $100,000 Positive do nothing .6 $200,000 Buy .4 rearrange $260,000 Negative $140,000 do nothing $200,000 $200,000 You can hire a consultant, Joan Zenoff, to study the situation. She would then render a flawless prediction of whether the rearrangement would succeed or fail. Compute the maximum amount you would be willing to pay for the errorless prediction. $164,000 Don’t buy info. rearrange $140,000 $100,000 $100,000 Positive do nothing .6 $200,000 Buy .4 rearrange $260,000 Negative $140,000 do nothing $200,000 $200,000 How much would you pay for Joan’s report? Compute the expected value of perfect information = EVPI EVPI = The expected value of the decision with the report ($140,000) - The expected value of the decision without the report ($164,000) EVPI = $140,000 - $164,000 = -$24,000 The report saves us $24,000 in expected value We would pay up to $24,000 Suppose now that Joan’s reports are not flawless. Suppose you have been provided the following posterior probabilities: This means that: P(success | optimistic) = .818, not 1 P(failure | optimistic) = .182, not 0 What would you now be willing to pay for Joan’s report? EVII = E(decision with imperfect information) - E(decision with no information) Recall that E(decision with no information) = $164,000 Required: 1. Compute the expected cost assuming an optimistic report. success .818 $100,000 We rearrange decide .182 failure optimistic ? $129,120 $260,000 $129,120 do nothing $200,000 2. Compute the expected costs assuming a pessimistic report. success .333 $100,000 rearrange We .667 decide failure $206,720 ? $260,000 pessimistic $200,000 do nothing $200,000 3. We were given the probability of an optimistic report (P(optimistic) = .55) and the probability of a pessimistic report (P(pessimistic) = .45). Compute the expected value of imperfect information. First we the first finish thewe must show on the tree? What is need to decision decision tree. optimistic Buy information .55 $129,120 We decide .45 pessimistic $161,016 $200,000 $161,016 Don’t buy info $164,000 E(decision with imperfect information) = $161,016 E(decision with no information) = $164,000 EVII = $164,000 - $161,016 = $2,984 We were not given likelihoods. We do not know the probability that Joan will render an optimistic report given the rearrangement is a success. What is that probability? P(optimistic | success) = P(success) = .6 P(success | optimistic) = = .818 P(optimistic) = .55 P(optimistic, success) = .55 x .818 = .45 P(optimistic | success) = .45/.6 = .75 Also: P(pessimistic | failure) = (.667 x .45)/.4 = .75 Normally we would be given likelihoods and priors and we would expect to compute: 1. The posterior probability of the outcome given a particular kind of information, and 2. The marginal probability of receiving that particular kind of information Therefore, given the following information about the accuracy of Joan Zenoff’s forecasts, complete a joint probability table and compute the necessary posterior (revised) probabilities. P(optimistic | success) = P(optimistic | success) = .75 P(pessimistic | failure) = .75 P(opt, success) = .75 x .6 = .45 .45 .10 .55 P(pess, failure) = .75 x .4 = .30 .15 .30 .45 P(success | opt) = .45/.55 P(failure | opt) = .1/.55 P(failure | pess) = .3/.45 P(success | pess) = .15/.45 .45 .10 .55 .15 .30 .45 Which ones go on the decision tree?

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