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# problemset3 - Solutions to Problem Set 3 Chris H Rycroft 1...

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± ± ± ± ± ± ± ± ² Solutions to Problem Set 3 Chris H. Rycroft October 30, 2006 1 Inelastic diﬀusion 1.1 The PDF of P N ( x ) The PDF of the random variable a n Δ X n is p ( xa n ) a n . If Δ X n has characteristic function ˆ p ( k ) then a n Δ X n has characteristic function e ikx p ( xa n ) dx = e ikya n p ( y ) dy a n −∞ −∞ = p ˆ( ka n ) . By the convolution theorem, we know that the characteristic function of X N can be written as P ˆ N ( k ) = N ³ p ka n ) n =1 and hence the PDF is given by e ikx dx P N ( k ) = N ³ 2 π n =1 −∞ p ka n ) N ³ = 2 π n =1 −∞ −∞ 1.2 The cumulants of X N e ikx dx e ikya n p ( y ) dy. The cumulant generating function of the n th step is given by ψ n ( k ) = log ˆ p ( ka n ) and therefore the l th cumulant is given by i l d l ψ n dk l = a ln i l d l dk l ψ n = a ln c l . k =0 k =0 If random variables are added, then their cumulants add. We therefore know that the cumulants of X N are given by N C N,l = a l (1 a lN ) ln c l = 1 a l a c l . n =1 1

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± ² M. Z. Bazant 18.366 Random Walks and Diﬀusion Problem Set 3 Solutions 2 1.3 Limits as N → ∞ By taking N → ∞ in the above expression, we see that the cumulants of X are given by l a C l = c l . 1 a l If a = 1 where > 0 is small, then we see that C 2 m a 2 m ³ 1 a 2 ´ m c 2 m = C 2 2 m a 2 c 2 m 1 a m (2 2 ) m c 2 m = 1 (1 ) 2 m c 2 µ m 2 m m 1 2 m c 2 m = c 2 m 2 m (2 m 1) 2 + . . . m 2 2 = 2 m m 1 1 m + . . . c 2 m 2 m 2 m (2 2 m 1) 2 + . . . c m = O ( m 1 ) . 1.4 A “Central Limit Theorem” The PDF of X can be written as P ( x ) = dk e ikx + ψ ( k ) 2 π −∞ ³ ´ dk C 2 k 2 iC 3 k 3 C 4 k 4 = 2 π exp ikx + 0 2 6 + 24 . −∞ If x = ζC 2 1 / 2 we see that 1 / 2 1 / 2 dl l 2 C 3 l 3 C 4 l 4 φ ( ζ, ) = C 2 P ( 2 ) = 2 π exp ilζ 2 i 6 C 1 / 2 + 24 C 2 2 . −∞ 2 From the previous result, we know that as 0, the terms involving the higher cumulants tend to zero. Therefore we have φ ( ) φ ( ζ ) = dl e ilζ l 2 / 2 2 π −∞ e ζ 2 / 2 = . 2 π 2 Breakdown of the CLT for decaying walks 2.1 The PDF of X for a = 0 . 99 Appendix A shows a simple C++ code to simulate 10 5 walkers from the decaying PDF. Figure 1 shows the PDF of x component of this distribution, and ±gure 2 shows a comparison to the
M. Z. Bazant 18.366 Random Walks and Diﬀusion Problem Set 3 Solutions 3 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 -30 -20 -10 0 10 20 30 P ( x ) x Figure 1: Simulated PDF of X for a = 0 . 99 based on 10 9 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. 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.

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problemset3 - Solutions to Problem Set 3 Chris H Rycroft 1...

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