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Suppose $X$ is has a uniform distribution on $[0,1]$. Let $X_i$ be the random variable that denotes the $i$th digit in the decimal expansion of $X$, i.e. if $x$ is a realization of $X$ with decimal expansion $0.x_1x_2x_3\ldots$, then $X_i=x_i$.

Intuitively, it should hold that the $X_i $ are i.i.d. $Unif(\{0,\ldots,9\})$.

Is this true? If yes, how would one show this rigorously?

Epiousios
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    I think it's true - someone is likely to provide a proof here. For real data check out Benford's Law if you don't know about it already. https://en.wikipedia.org/wiki/Benford%27s_law – Ethan Bolker Sep 22 '17 at 18:09
  • @EthanBolker Interesting, thanks for the link! – Epiousios Sep 22 '17 at 18:10
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    Issue you need to think about... the "ith digit" of a real number is not always uniquely defined. 0.40000... = 0.39999.... so which is the second digit, 0 or 9? (Then again, this only happens for rational numbers, which are a negligible set - but this must be part of the problem statement, or perhaps of the solution.) –  Sep 22 '17 at 18:13
  • @mathguy - Since the Lesbegue measure of the rational numbers is zero, this shouldn't be an issue. – Nathaniel Bubis Sep 22 '17 at 18:41
  • @nbubis - I already said that myself, and in the same Comment - what is the point of repeating it? –  Sep 22 '17 at 18:42

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You can show that the Lesbegue measure of the set of reals with digit $k$ in the $i^{th}$ place is $\frac 1{10}$ just by considering the length of the intervals. For example, the intervals that have $5$ in the $2^{nd}$ place are $[0.05,0.06), [0.15,0.16) \ldots [0.95,0.96)$. Each of those intervals has length $\frac 1{100}$ and there are $10$ of them.

Ross Millikan
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