Information Theory, Inference, and Learning ... - Inference Group

Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links.60 3 — More about Inference(b) P (p a | s = bbb, F = 3) ∝ (1 − p a ) 3 . The most probable value of p a (i.e.,the value that maximizes the posterior probability density) is 0. Themean value of p a is 1/5.See figure 3.7b.86420-2-4H 0 is truep a = 1/61000/1100/110/11/11/101/1000 50 100 150 20086420-2-4p a = 0.25H 1 is true1000/1100/110/11/11/101/1000 50 100 150 20086420-2-4p a = 0.51000/1100/110/11/11/101/1000 50 100 150 200Solution to exercise 3.7 (p.54). The curves in figure 3.8 were found by findingthe mean and standard deviation of F a , then setting F a to the mean ± twostandard deviations to get a 95% plausible range for F a , and computing thethree corresponding values of the log evidence ratio.Solution to exercise 3.8 (p.57). Let H i denote the hypothesis that the prize isbehind door i. We make the following assumptions: the three hypotheses H 1 ,H 2 and H 3 are equiprobable a priori, i.e.,Figure 3.8. Range of plausiblevalues of the log evidence infavour of H 1 as a function of F .The vertical axis on the left islogP (s | F,H1)P (s | F,H 0); the right-handvertical axis shows the values ofP (s | F,H 1)P (s | F,H 0) .The solid line shows the logevidence if the random variableF a takes on its mean value,F a = p a F . The dotted lines show(approximately) the log evidenceif F a is at its 2.5th or 97.5thpercentile.(See also figure 3.6, p.54.)P (H 1 ) = P (H 2 ) = P (H 3 ) = 1 3 . (3.36)The datum we receive, after choosing door 1, is one of D = 3 and D = 2 (meaningdoor 3 or 2 is opened, respectively). We assume that these two possibleoutcomes have the following probabilities. If the prize is behind door 1 thenthe host has a free choice; in this case we assume that the host selects atrandom between D = 2 and D = 3. Otherwise the choice of the host is forcedand the probabilities are 0 and 1.P (D = 2 | H 1 ) = 1/ 2 P (D = 2 | H 2 ) = 0 P (D = 2 | H 3 ) = 1P (D = 3 | H 1 ) = 1/ 2 P (D = 3 | H 2 ) = 1 P (D = 3 | H 3 ) = 0(3.37)Now, using Bayes’ theorem, we evaluate the posterior probabilities of thehypotheses:P (H i | D = 3) = P (D = 3 | H i)P (H i )(3.38)P (D = 3)P (H 1 | D = 3) = (1/2)(1/3)P (D=3)P (H 2 | D = 3) = (1)(1/3)P (D=3)P (H 3 | D = 3) = (0)(1/3)P (D=3)(3.39)The denominator P (D = 3) is (1/2) because it is the normalizing constant forthis posterior distribution. SoP (H 1 | D = 3) = 1/ 3 P (H 2 | D = 3) = 2/ 3 P (H 3 | D = 3) = 0.(3.40)So the contestant should switch to door 2 in order to have the biggest chanceof getting the prize.Many people find this outcome surprising. There are two ways to make itmore intuitive. One is to play the game thirty times with a friend and keeptrack of the frequency with which switching gets the prize. Alternatively, youcan perform a thought experiment in which the game is played with a milliondoors. The rules are now that the contestant chooses one door, then the game