122S:138Bayesian StatisticsCalibration Experiments and Reviewof ProbabilityLecture 2Aug. 23, 2006Kate Cowles374 SH, [email protected] subjective probability aboutevents• We may sometimes need to quantify our sub-jective probability of an event in order tomake a decision or take an action.• Example:– You have been offered a job as a statisti-cian with a marketing firm in Cincinnati.– In order to decide whether to accept thejob and move to Cincinnati, you wish toquantify your subjective probability of theevent that you would like the job and wouldlike Cincinnati– (We will talk about Bayesian decision the-ory later in the semester.)3Calibration experiments• calibration experiment — a scale used toassess a person’s degree of belief that a par-ticular event will occur (or has occurred)• All outcomes of the calibration experimentmust be equally likely in the opinion of theperson whose subjective probability is beingassessed– Example: imagine that I promise to payyou $100 if the roll of a 6-sided die comesup the number you call– If you are indifferent as to which num-ber you call, the 6 possible outcomes areequally likely for you4• Calibration experiments may be useful if– person is not knowedgeable or comfortablewith probability– person is uncertain as to his/her opinionabout the event• principle of using a calibration experiment toassess subjective probability– Person is offered a choice of 2 ways of win-ning a prize:∗ through a realization of the calibrationexperiment with known probability ofsuccess∗ through the occurrence of the event ofinterest.– The calibration experiment is adjusted atsuccessive steps5Example:Using a “chips-in-a-bowl” experiment to assessyour subjective probability regarding event A:that the Department of Physics at Florida StateUniversity has more than 2 female faculty mem-bers.• Experiment is having a blindfolded persondraw one chip at random from a bowl con-taining chips of the same size and shape• Let Ps(A) denote your subjective probabilitythat event A has occurred.6• Step 1: The bowl contains 1 green chip and1 red chip.Imagine that you may choose 1 of two games.1. I will be blindfolded and draw one chip atrandom. I will pay you $100 if the chipdrawn is red. I will pay you nothing if itis green.2. I will pay you $100 if the Physics Dept atFSU has more than 2 female faculty. I willpay you nothing if it has 2 or fewer.If you choose Game 1, then I conclude that0 < Ps(A) < 0.5So I construct Step 2 as follows.7• The bowl contains 3 green chips and 1 redchip. You may choose Game 1 or Game 2.If you choose Game 2, then I conclude thatyour0.25 < Ps(A) < 0.50• Then I may go on to Step 3:Now the bowl contains 5 green chips and 3red chips. You may choose Game 1 or Game2.If you choose Game 1, I conclude that0.25 < P (A) < 0.3758• We continue until it becomes too difficult foryou to choose between the two games.• How the chips are set up in the bowl at eachstep is determined by your answer at the pre-ceding step.• Comments:– The payoffs have to be imaginary, becausewe wish to use this procedure for assessingunverifiable probabilities.– Luckily, a high degree of accuracy in as-sessing subjective probabilties usually isnot needed.9Quick review of probability• event: any outcome or set of outcomes of arandom phenomenon– the basic element to which probability canbe applied– usual notation is a capital letter near be-ginning of alphabet– example: random phenomenon is that weare drawing a patient at random from ahuge database of patients insured by anHMO∗ event A is the event that the patient wedraw is under 6 years of age– we will denote the probability of event Aas P (A)• sample space S: the set of all possible out-comes of a random phenomenon– P (S) = 110• intersection of two events A and B is theevent “both A and B”– notated ATB– example: if event B is the event that thepatient we draw weighs at least 150 pounds,then ATB is the event that the patient wedraw is under 6 years of age and weighs atleast 150 pounds• union of two events is event “either A or Bor both”– notated ASB• A set of events A1, A2, A3, . . . are exhaus-tive ifA1[A2[A3[. . . = S11• complement of an event A is the event“not A”– notated ACor¯A– AS¯A = S• the null event – an event that can neverhappen– notated ∅• events A and B are disjoint or mutuallyexclusive if they cannot occur together– i.e. if ATB = ∅– example: event A is that the patient wedraw is under 6 years of age, and event Bis event that the patient is 6 to 11 years ofage12• additive rule of probability– if two events A and B are mutually exclu-sive, thenP (A[B) = P (A) + P (B)• P (AC) = 1 − P (A)13Conditional Probability• P (B|A) – the probability that event B willoccur given that we already know that eventA has occurred• multiplicative rule of probabilityP (A\B) = P (A)P (B|A)P (A\B) = P (B)P (A|B)• so if P (A) 6= 0P (B|A) =P (ATB)P (A)14Patients in the database example< 150 pounds ≥ 150 pounds Totalunder 6 798 2 800≥ 64702 4498 9200Total 5500 4500 1000015Independence• Two events are independent if the occurrence(or non-occurrence) of one of them does notaffect the probability that the other one oc-curs. Events A and B are independent ifP (A|B) = P (A)P (B|A) = P (B)• multiplicative rule of probability for inde-pendent eventsP (A\B) = P (A)P (B)16Patients in the database example< 150 pounds ≥ 150 pounds Totalgreen eyes 440 360 800not green5060 4140 9200Total 5500 4500 1000017Law of Total Probability• Applies when you wish to know the marginal(unconditional) probability of some event, butyou only know its probability under someconditions.• Example:– I have asked my friend to mail an impor-tant letter.– I want to calculate P (A), the probabil-ity that the letter will reach the addresseewithin the next 3 days.– I believe that P (M), the probability thatmy friend will remember to mail the lettertoday or tomorrow, is .60.18– I believe that if my friend mails the lettertoday or tomorrow, the probability thatthe postal service will deliver it to the ad-dressee within the next 3 days is .95.P (A|M) = .95– I believe there’s only 1 chance in 10000that the letter will get there somehow ifmy friend forgets to mail it.P (A|¯M) = .000119Using the Law of Total Probability tofind P (A)• For any events A and MA = (A\M)[(A\¯M)• Events (ATM) and (AT¯M) are disjoint,
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