EFTA02444385.pdf

Simplifying Bayesian Inference Stefan KrauB, Laura Martignon & Ulrich Hoffrage Max Planck Institute For Human Development Lentzeallee 94, 14195 Berlin-Dahlem Probability theory can be used to model inference under uncertainty. The particular way in which Bayes'formula is stated, which is of only minor importance in standard probability textbooks, becomes central in this context. When events can be interpreted as evidences and hypotheses, Bayes'formula allows one to update one's belief in a hypothesis in light of new data. Is unaided human reasoning Bayesian? Kahneman and "Iversky (1972) affirmed: "In his evaluation of evidence, man is not Bayesian at all." In their book Judgment under uncertainty (1982), they attempted to prove that human judgment is riddled with systematic deviations from the logical and probabilistic norm. In chapter 18 of the same book David M. Eddy stressed that medical doctors do not follow Bayes'formula when solving the following task: Theprobability that a woman at age 40 has breast cancer (B) is I% (P(B) = prevalence = 1%) According to the literature, theprobability that the disease is detected by a mammography (M) is 80%. (P(M+ IB) = sensitivity = 80%) Theprobability that the test misdetects the disease although the patient does not have it is 9.6%. (P(M+ I0B) = 1- specificity = 9.6%) If a woman at age 40 is tested as positive, what is theprobability that she indeed has breast cancer (POW)? Bayes'formula yields the following result: P(M+ BrP(B) 80%?I% P(BI M+) - - 0.078 P(M+ I M?P(B)+ P(M+ - B) 9.11(- B) 80%?1%+ 9.6% ?99% Thus, the probability of breast cancer is only 7.8%, while Eddy reports that 95 out of 100 doctors estimated this probability to be between 70% and 80%. Gigerenzer and Hoffrage (1995) focused on another aspect of the problem: the representation of uncertainty. In Eddy's task, quantitative information was given in probabilities. Gigerenzer and Hoffrage presented Eddy's problem to medical doctors replacing probabilities with a different representation of uncertainty, namely natural frequencies. In their formulation the task was: 100 out ofevery 10000 women at age 40 who participate in routine screening have breast cancer. 80 ofevery 100 women with breast cancer willget a positive mammography. 950 out ofevery 9900 women without breast cancer will also get a positive mammography. Here is a new representative sample of women at ageforty who get a positive mammography in routine screening. How many ofthese women do you expect to actually have breast cancer? Now nearly half (46%) of all doctors gave the Bayesian answer: 80 out of 1030 (7.8%). EFTA_R1_01521127 EFTA02444385  Probabilities Natural Frequencies 10.000 breast N_ no breast cancer cancer p(B) = .01 p(T+ I B) = .80 9.900 p(T+ B) = .096 0 e 8.950 Test Test Test Test posifiv negativ post*/ ) neoafiv (T+) (T-) (T+ (T-) p(B T+) 80T+) • .01 x .80 .01 x .80 + .99 x .096 • p(B T+) 80 + 950 0 0 OO OOO Figure 1 What is the crucial property that helps one to find the Bayesian solution? To answer this question, it is helpful to consider a more general case. In real-life situations, decisions are usually based on several cues. A medical doctor, for instance, seldom diagnoses a disease based on a single test. The usual procedure after a mammography is to perform an ultrasound test (U). For an ultrasound test, sensitivity and specificity are usually given in the instructions: P(U+ IB) = 95% P(U+ I0B) = 4% In an empirical study, we presented this information together with P(B), P(M+ IB) and P(M+ IfaB) to a group of participants. They were asked: What is the probablity that a woman at age 40 has breast cancer, given that she has a positive mammography anda positive ultrasound test? When given this probability format, only 12.2% of our participants reached the correct solution (» 3/3 ). D. Massaro (1998) gave an example describing the same situation with frequencies': EFTA_R1_01521128 EFTA02444386  Wt..* I M./AU- I M•AU. ev1+8.t.l- M-11U+ 11 4/4-8,U• Figure 2 Massaro writes that in the case of two cues „a frequency algorithm will not work" and „it might not be reasonable to assume that people can maintain exemplars of all possible symptom configurations." However, his statements are not based on experimental evidence, and his frequency configuration is not really equivalent to the probability format because he works with combined sensitivity P(M+ & U+ IB) and combined specificity 1-P(M- & U- I0B). One possible frequency format, which does correspond to our probability format, is': re — 1 1100001 women (0000) women d breast cancer/ no breast cancer breast cancer no breast cancer r. /Th 100 99001 IWO) (9900i \_/ \ _/ M+ (80 \ M-(20) 1 950) M+ 8950 M- U U-( 5 e 396)U+ 950:1) U- Figure 3 EFTA_R1_01521129 EFTA02444387 In words: 100 out ofevery /0000 woman at age 40 who participate in routine screening have breast cancer. 80 ofevery 100 women with breast cancer willget a positive mammography. 950 out ofevery 9900 woman without breast cancer will also get a positive mammography. 95 out of100 women with cancer willget a positive ultrasound test 396 out of9900 women, although they do not have cancer, nevertheless obtain a positive ultrasound test. How many ofthe women who get a positive mammography and a positive ultrasound test do you expect to actually have breast cancer? 14.6% of our participants solved this version correctly. Another possibility is to consider the tests sequentially. This is possible because the ultrasound test and the mammography are conditionally independent, i.e. P(U+ IB) = P(U+ IB & M+). Now we have: —• (I,000) women breast cancer no breast cancer ( 100 I N M-1 20 ) 950 M+ 8950 M- fiTh c m % 1Th \-1 76 ( 4 19 ) 1 (38 / U+ U- U+ U- U+ U- U+ U- Figure 4 In words: 100 out ofevery 10000 women at age 40 who participate in routine screening have breast cancer. 80 ofevery 100 women with breast cancer willget a positive mammography. 950 out ofevery 9900 women without breast cancer will also get a positive mammography. 76 out of80 women who had a positive mammography and have cancer also have a positive ultrasound test. 38 out of950 women who had a positive mammography, although they do not have cancer, also have a positive ultrasound test. How many ofthe women who get a positive mammography and a positive ultrasound test do you expect to actually have breast cancer? 53.7% of our participants solved this task correctly. EFTA_R1_01521130 EFTA02444388 Not all frequencies in the tree were actually used. The next step is to eliminate all frequencies irrelevant to the task. Thus we obtain: 11000u women breast cancer no breast cancer 7 11)?) 9900 M+ 80 ) U+ /—• 76 Figure 5 These frequencies, namely those that really foster insight, deserve a special name. We decided to call them Markovfrequencies because of the natural analogy with Markov chains. In fact: 1) Our tree consists of two chains which are joined at the root. 2) Each node corresponds to the reference class that determines the next node. As in a Markov chain, the frequency in each node depends only upon its predecessor, not upon previous nodes. Being able to "think in chains" seems crucial for human insight and fits the modern view that problem solving, unlike perception, is sequential rather than parallel. Markov frequencies are task-oriented, i.e., only information that is relevant for the task appears in the tree. Gigerenzer and Hoffrage (1995) also used a tree (see Figure 1). Their tree contains the information (P(T- IB) and P(T- I0B)), which is not relevant to the question "P(BIT+) =?". In our chains, the odds of the problem can be read directly from the last two nodes. This is because the tree with Markov frequencies corresponds to the well-known likelihood-combination rule (see, for instance, Spies, 1993): prior odds • product of the likelihood ratios = posterior odds 100 The prior odds for breast cancer are 9900 Multiplying this with the likelihood ratio for the mammography', we obtain 985°0 . Again multiplying this with the likelihood ratio of the ultrasound test, we finally . get38 By using Markov frequencies, it is not only clear which information should be given to experts, but also which information should be omitteds. Appropriately deleting useless information is part of the overall computation, as we know from information theory. EFTA_R1_01521131 EFTA02444389 References Eddy, D. M. (1982). Probabilistic reasoning in clinical medicine: Problems and opportunities. In D. Kahneman, P. Slovic & A. Tversky (Eds.), Judgment under uncertainty: Heuristics and biases (pp. 249-267). Cambridge, England: Cambridge University Press. Gigerenzer, G. & Hoffrage, U. (1995). How to improve bayesian reasoning without instruction: Frequency formats. Psychological Review, 102, 684-704. Kahneman, D. & Tversky, A. (1972). Subjective probability: A judgement of representativeness. Cognitive Psychology, 3, 430454. Massaro, D. (1998). Perceiving talkingfaces (pp.174-179). Boston. MIT Press. Spies, M. (1993). Unsicheres Wissen: Wahrscheinlichkeit, Fuzzy-Logik, neuronale Netze and menschliches Denken (pp.51-54). Heidelberg, Berlin, Oxford: Spektrum Akademischer Verlag. Footnotes I) To integrate the research on this topic, we borrowed concepts from various sources and explored them in the breast cancer example. In fact, Gigerenzer and Hoffrage used a sample of 1.000 (not 10.000) women, Massaro speaks of symptoms instead of tests and we tested our subjects with ,tuberculosis tasks" instead of „breast cancer tasks." 2) Gigerenzer and Hoffrage stressed that only frequencies work that can be sampled „naturally". A doctor would get information of this kind when he samples instructions for different tests and translates the information therein into frequencies. 3) A doctor would get information of this kind when he samples patients with respect to their state of illness. (M + I B) 8 4) The likelihood ratio L(B, M+) is defined by , which is 0% » 8.3 P P(M+ B) The likelihood ratio L(B, U+) therefore is 5% = 23.75 4% 5) Because Baycs'formula can be used to model inference under uncertainty, it is also a tool in scientific reasoning. Klaus Hasselmann from the Max Planck Institute for Meteorology in Hamburg is presently applying a Bayesian analysis to hypotheses about changes in climate. The Society for Mathematics and Data Analysis in St. Augustin is investigating various methods for EFTA_R1_01521132 EFTA02444390  estimating credit risks, such as analysis of discriminance, fuzzy-pattern classification, and neural networks with the help ofBayes'theorem. The „Krebsatlas" (almanac of cancer patients) for Germany is being reviewed at the Ludwig Maximilian University in Munich by means of Bayesian methods. The task is to detect and eliminate spurious correlations. Even the Microsoft Office Assistant uses Bayesian procedures. The mathematician Anthony 0' Hagan „elicits" on behalf of the Britsh government hydrological conductivity of the rock at Sellafield from experts. He uses their beliefs to determine a prior distribution, with which the appropriateness of the area as a permanent diposal site for nuclear waste can be estimated (Neue Ziircher Zeitung, May 13, 1998, 5.39.). Even the most expert systems are based on Bayes'formula. A famous example is MUNIN (Muscle and Nerve Inference Network) from Lauritzen and Spiegelhalter (1988), which is used for making diagnoses on the basis of measurements of muscular electrical impulses („electromyography"). Maybe Markov frequencies can also help to facilitate programming those expert systems. Acknowledgments We thank Valerie Chase, Martin Lages, Donna Alexander and Matthias Licha for helpful comments and Ursula Dohme for running the experiments. (Submission to the 1998 Conference on „Model-Based Reasoning in Scientific Discovery") EFTA_R1_01521133 EFTA02444391