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

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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.368 29 — Monte Carlo Methods(a)1210864204035302520(b) Metropolis100 iterations0 5 10 15 20400 iterations(c) Independent sampling100 iterations1210864200 5 10 15 20400 iterations4035302520Figure 29.12. Metropolis methodfor a toy problem. (a) The statesequence for t = 1, . . . , 600.Horizontal direction = states from0 to 20; vertical direction = timefrom 1 to 600; the cross bars marktime intervals of duration 50. (b)Histogram of occupancy of thestates after 100, 400, and 1200iterations. (c) For comparison,histograms resulting whensuccessive points are drawnindependently from the targetdistribution.151510105500 5 10 15 201200 iterations00 5 10 15 201200 iterations90807060504030201000 5 10 15 2090807060504030201000 5 10 15 20
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.29.4: The Metropolis–Hastings method 369To illustrate how slowly a random walk explores a state space, figure 29.12shows a simulation of a Metropolis algorithm for generating samples from thedistribution:{ 1/21 x ∈ {0, 1, 2, . . . , 20}P (x) =(29.33)0 otherwise.The proposal distribution isQ(x ′ ; x) ={ 1/2 x ′ = x ± 10 otherwise.(29.34)Because the target distribution P (x) is uniform, rejections occur only whenthe proposal takes the state to x ′ = −1 or x ′ = 21.The simulation was started in the state x 0 = 10 and its evolution is shownin figure 29.12a. How long does it take to reach one of the end states x = 0and x = 20? Since the distance is 10 steps, the rule of thumb (29.32) predictsthat it will typically take a time T ≃ 100 iterations to reach an end state. Thisis confirmed in the present example: the first step into an end state occurs onthe 178th iteration. How long does it take to visit both end states? The ruleof thumb predicts about 400 iterations are required to traverse the whole statespace; and indeed the first encounter with the other end state takes place onthe 540th iteration. Thus effectively-independent samples are generated onlyby simulating for about four hundred iterations per independent sample.This simple example shows that it is important to try to abolish randomwalk behaviour in Monte Carlo methods. A systematic exploration of the toystate space {0, 1, 2, . . . , 20} could get around it, using the same step sizes, inabout twenty steps instead of four hundred. Methods for reducing randomwalk behaviour are discussed in the next chapter.Metropolis method in high dimensionsThe rule of thumb (29.32), which gives a lower bound on the number of iterationsof a random walk Metropolis method, also applies to higher-dimensionalproblems. Consider the simple case of a target distribution that is an N-dimensional Gaussian, and a proposal distribution that is a spherical Gaussianof standard deviation ɛ in each direction. Without loss of generality, we canassume that the target distribution is a separable distribution aligned with theaxes {x n }, and that it has standard deviation σ n in direction n. Let σ max andσ min be the largest and smallest of these standard deviations. Let us assumethat ɛ is adjusted such that the acceptance frequency is close to 1. Under thisassumption, each variable x n evolves independently of all the others, executinga random walk with step size about ɛ. The time taken to generate effectivelyindependent samples from the target distribution will be controlled by thelargest lengthscale σ max . Just as in the previous section, where we needed atleast T ≃ (L/ɛ) 2 iterations to obtain an independent sample, here we needT ≃ (σ max /ɛ) 2 .Now, how big can ɛ be? The bigger it is, the smaller this number T becomes,but if ɛ is too big – bigger than σ min – then the acceptance rate willfall sharply. It seems plausible that the optimal ɛ must be similar to σ min .Strictly, this may not be true; in special cases where the second smallest σ nis significantly greater than σ min , the optimal ɛ may be closer to that secondsmallest σ n . But our rough conclusion is this: where simple spherical proposaldistributions are used, we will need at least T ≃ (σ max /σ min ) 2 iterationsto obtain an independent sample, where σ max and σ min are the longest andshortest lengthscales of the target distribution.
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