problemset3

Z bazant 18366 random walks and diusion problem set 3

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Unformatted text preview: tend to zero. Therefore we have �∞ dl ilζ −l2 /2 φ(ζ, �) → φ(ζ ) = e −∞ 2π 2 = 2 2.1 e−ζ /2 √ . 2π Breakdown of the CLT for decaying walks The PDF of X∞ for a = 0.99 Appendix A shows a simple C++ code to simulate 105 walkers from the decaying PDF. Figure 1 shows the PDF of x component of this distribution, and figure 2 shows a comparison to the M. Z. Bazant – 18.366 Random Walks and Diffusion – Problem Set 3 Solutions 3 0.09 0.08 0.07 P∞ (x) 0.06 0.05 0.04 0.03 0.02 0.01 0 -30 -20 -10 0 10 20 30 x Figure 1: Simulated PDF of X∞ for a = 0.99 based on 109 trials. formulae calculated in question 1. We see that the two curves match to a very high degree of accuracy, particularly in the central region. We know however that this match will not continue � forever: since X∞ is bounded, its PDF will be uniformly zero outside of some region. However a Gaussian features non-zero probabilities everywhere. 2.2 The exact PDF of X∞ for a = 1/2 � � To find the PDF of X∞ for a = 1/2 we first consider the change of variables X = (x, y ) → R = (r, s) given by r= s= 1+x+y 2 1 + x − y . 2 � In the transform variables, the walk starts from R0 = (1/2, 1/2) and the PDF of the nth step is given by 1 pn (r, s) = (δr,−1/2n+1 + δr,1/2n+1 )(δs,−1/2n+1 + δs,1/2n+1 ). 4 From this from, it is clear that the variables r and s are independent. We see that the N th step of the r component can be written as N RN 1 � 2In − 1 =+ 2 2n+1 n=1 M. Z. Bazant – 18.366 Random Walks and Diffusion – Problem Set 3 Solutions 0.1 4 Simulation Gaussian 0.01 P∞ (x) 0.001 0.0001 1e-05 1e-06 1e-07 1e-08 -30 -20 -10 0 10 20 30 x Figure 2: Simulated PDF of X∞ for a = 0.99 based on 109 trials on a log scale, compared with a Gaussian curve with a variance calculated using the formula derived in question 1. where the In ’s are independent random variables which take values 0 and 1 with equal probability. This can be rewritten as RN N N � In 1 � 1 − + 2 2n+1 2n n=1 n=1 � � N � In 1 1 1 − 21 N −2 + 22 2n 1 − 1 2 = = n=1 N � In 1 +...
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This note was uploaded on 01/23/2014 for the course MATH 18.366 taught by Professor Martinbazant during the Fall '06 term at MIT.

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