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I have a process that generates a series of real numbers.

Specifically, starting from a given arbitary value (Xi-1), the process will generate a new number Xi following the formula: Xi = Xi-1 + ∂, where ∂ is a normally-distributed random variable.


suppose Xi-1 = 100.00000…

assume ∂ ~ N(0,1)

Then I might get a series of numbers such as: 99.83422187603…, 100.70137698…, 102.2048120964…, 103.10328521754…, 103.6376792209…, …


The generated series will exhibit Brownian motion, and follow a normal distribution, with a mean of the arbitrary starting value (in the example, 100).

HOWEVER. My concern is the following. Suppose I pick a arbitrary decimal place to the right of the decimal. So, the thousandths digit, for instance. My contention is that over a sufficient number of iterations of the number-generating process, the distribution of the digits 0 to 9 will be uniform. That is, each of the 10 possible single integer digits 0 through 9 will appear in the (thousandths) place with 10% frequency. So, for instance, if there were 100,000 iterations, the digit 0 would appear in the thousandths place 10,000 times, the digit 1 would would appear in the thousandths place 10,000 times, etc…

Via simulation, I have been able to "prove" this… But I am struggling to do so more rigorously. Does anyone have any thoughts?

Thanks greatly!

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The stationary distribution of the Markov chain on [0,1] with transition x→x+z (mod 1) where z is standard normal, is uniform hence the empirical frequency of every digit at every place converges to 10%. Likewise the empirical frequency of the digits at places from 123,456 to 123,459 being 2014 converges to 0.01%.

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thank you for your response. could i ask, though, if you can be more specific/detailed; maybe provide a step-by-step explanation (and possibly an example) so i can really understand it? i would really appreciate your (or anyone's) assistance. i find this problem interesting but quite difficult to grasp for some reason. Thanks! – JAEmat Feb 22 '14 at 7:44
What is the first place in my answer where you are lost? – Did Feb 22 '14 at 7:46
thanks for responding. your statment, "and hence the empirical freq… is 10%" jumps to the conclusion too quickly; i.e., WHY is it that a markov chain of x-->x+z generates a uniform distribution across every digit. also; i don't follow your statement of "likewise…". thank you – JAEmat Feb 22 '14 at 8:47
0. What is your textbook for Markov chains? 1. Which theorem do you know linking limits of empirical frequencies of Markov chains and their stationary distribution? 2. Did you try to check that the uniform distribution on [0,1] (not on the digits) is stationary for this Markov chain? If you did, what stopped you? 3. Let us save "Likewise" for later on. – Did Feb 22 '14 at 9:21

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