Information Theory, Inference, and Learning ... - MAELabs UCSD

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Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981 You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links. 448 35 — Random Inference Topics counts, given F = 1000 outcomes, are a = 0 a = 1 b = 0 760 5 765 b = 1 190 45 235 950 50 What are the posterior probabilities of the two hypotheses? (35.4) Hint: it’s a good idea to work this exercise out symbolically in order to spot all the simplifications that emerge. Ψ(x) = d 1 ln Γ(x) ln(x) − dx 2x + O(1/x2 ). (35.5) The topic of inferring causation is a complex one. The fact that Bayesian inference can sensibly be used to infer the directions of arrows in graphs seems to be a neglected view, but it is certainly not the whole story. See Pearl (2000) for discussion of many other aspects of causality. 35.4 Further exercises Exercise 35.6. [3 ] Photons arriving at a photon detector are believed to be emitted as a Poisson process with a time-varying rate, λ(t) = exp(a + b sin(ωt + φ)), (35.6) where the parameters a, b, ω, and φ are known. Data are collected during the time t = 0 . . . T . Given that N photons arrived at times {tn} N n=1 , discuss the inference of a, b, ω, and φ. [Further reading: Gregory and Loredo (1992).] ⊲ Exercise 35.7. [2 ] A data file consisting of two columns of numbers has been printed in such a way that the boundaries between the columns are unclear. Here are the resulting strings. 891.10.0 912.20.0 874.10.0 870.20.0 836.10.0 861.20.0 903.10.0 937.10.0 850.20.0 916.20.0 899.10.0 907.10.0 924.20.0 861.10.0 899.20.0 849.10.0 887.20.0 840.10.0 849.20.0 891.10.0 916.20.0 891.10.0 912.20.0 875.10.0 898.20.0 924.10.0 950.20.0 958.10.0 971.20.0 933.10.0 966.20.0 908.10.0 924.20.0 983.10.0 924.20.0 908.10.0 950.20.0 911.10.0 913.20.0 921.25.0 912.20.0 917.30.0 923.50.0 Discuss how probable it is, given these data, that the correct parsing of each item is: (a) 891.10.0 → 891. 10.0, etc. (b) 891.10.0 → 891.1 0.0, etc. [A parsing of a string is a grammatical interpretation of the string. For example, ‘Punch bores’ could be parsed as ‘Punch (noun) bores (verb)’, or ‘Punch (imperative verb) bores (plural noun)’.] ⊲ Exercise 35.8. [2 ] In an experiment, the measured quantities {xn} come independently from a biexponential distribution with mean µ, P (x | µ) = 1 exp(− |x − µ|) , Z
Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981 You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links. 35.5: Solutions 449 where Z is the normalizing constant, Z = 2. The mean µ is not known. An example of this distribution, with µ = 1, is shown in figure 35.2. Assuming the four datapoints are {xn} = {0, 0.9, 2, 6}, 0 1 2 3 4 5 6 7 8 what do these data tell us about µ? Include detailed sketches in your answer. Give a range of plausible values of µ. 35.5 Solutions Solution to exercise 35.4 (p.446). A population of size N has N opportunities to mutate. The probability of the number of mutations that occurred, r, is roughly Poisson −aN (aN)r P (r | a, N) = e . (35.7) r! (This is slightly inaccurate because the descendants of a mutant cannot themselves undergo the same mutation.) Each mutation gives rise to a number of final mutant cells ni that depends on the generation time of the mutation. If multiplication went like clockwork then the probability of ni being 1 would be 1/2, the probability of 2 would be 1/4, the probability of 4 would be 1/8, and P (ni) = 1/(2n) for all ni that are powers of two. But we don’t expect the mutant progeny to divide in exact synchrony, and we don’t know the precise timing of the end of the experiment compared to the division times. A smoothed version of this distribution that permits all integers to occur is P (ni) = 1 Z 1 , (35.8) where Z = π 2 /6 = 1.645. [This distribution’s moments are all wrong, since ni can never exceed N, but who cares about moments? – only sampling theory statisticians who are barking up the wrong tree, constructing ‘unbiased estimators’ such as â = (¯n/N)/ log N. The error that we introduce in the likelihood function by using the approximation to P (ni) is negligible.] The observed number of mutants n is the sum n = n 2 i r ni. (35.9) i=1 The probability distribution of n given r is the convolution of r identical distributions of the form (35.8). For example, P (n | r = 2) = n−1 n1=1 1 Z 2 1 n 2 1 1 for n ≥ 2. (35.10) (n − n1) 2 The probability distribution of n given a, which is what we need for the Bayesian inference, is given by summing over r. P (n | a) = N P (n | r)P (r | a, N). (35.11) r=0 This quantity can’t be evaluated analytically, but for small a, it’s easy to evaluate to any desired numerical precision by explicitly summing over r from -3 -2 -1 0 1 2 3 Figure 35.2. The biexponential distribution P (x | µ = 1).
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