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.
[example:
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!
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%.