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# Thus the pdf of x is given by p r s 1 2 0 for x

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Unformatted text preview: . 2N +1 2n = n=1 In the limit, as N → ∞, we obtain R∞ N � In = . 2n n=1 Thus, the binary expansion of R∞ is R∞ = 0.I1 I2 I3 I4 I5 I6 . . . From this form, it is clear that every possible binary expansion between 0 and 1 can be achieved, each with equal probability. Thus R∞ is uniformly distributed on the interval [0, 1]. Since the same is true for S∞ , we know that the joint PDF is given by � 1 for 0 &lt; r &lt; 1, 0 &lt; s &lt; 1 P∞ (r, s) = 0 otherwise. Thus the PDF of X∞ is given by P∞ (r, s) = � 1 2 0 for |x − y | &lt; 1, |x + y | &lt; 1 otherwise. M. Z. Bazant – 18.366 Random Walks and Diﬀusion – Problem Set 3 Solutions 1 1 0.8 C∞ (q ) C∞ (x) 0.8 0.6 0.4 0.6 0.4 0.2 0.2 0 0 -0.4 -0.2 0 0.2 0 0.4 0.2 0.4 0.6 0.8 1 q x Figure 3: The cumulative distribution func� tion of the x component of X∞ for a = 1/3. 2.3 5 Figure 4: The cumulative distribution function of the rotated variable q = 1/2 + x − y for a = 1/3. � The CDF of X∞ for a = 1/3 � To compute the CDF of x component of X∞ for a = 1/3, we ﬁrst consider the possible values that it can take. The maximum that x can increase by at the nth step is 1/3n and thus the maximum value that x∞ can take is ∞ �1 1 1 = 1 = 2. n 3 3(1 − 3 ) n=1 By symmetry, we therefore know that X∞ can take values in the range [−1/2, 1/2]. We also note that the sum of steps starting at n = 2 follows the same distribution as X∞ , but scaled by 1/3. If the ﬁrst step is ΔX1 , then X∞ lies in the interval [ΔX1 − 1/6, ΔX1 + 1/6]. For the three possible choices ΔX1 = −1/3, 0, 1/3 these intervals are mutually exclusive, and this allows us to write the CDF C∞ (x) recursively as ⎧ for x &lt; − 1 ⎨ C∞ (3x + 1)/4 6 1/4 + C∞ (3x)/2 for − 1 &lt; x &lt; 1 C∞ (x) = 6 6 ⎩ 3/4 + C∞ (3x − 1)/4 for 1 &lt; x 6 This can be easily calculated using a recursive function, and a graph is shown in ﬁgure 3. It is also � interesting to consider the CDF of X∞ in a coordinate rotated by 45◦ . Suppose we consider the coordinate transformation p= q= 1 +x+y 2 1 +x−y 2 Then we ﬁnd that the walk from starts fro...
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