I am confused on what a scale invariant distribution is. Can anyone provide a rigorous definition for a scalar invariant distribution? Also, if I consider the set of all powers of 2 and suppose I write each element in scientific notation $x_{m} \times 10^{n_m}$. Would the distribution of the frequencies of $x_m$ be considered scalar invariant? If so, why?
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According to Whittaker, p. 258, a (cumulative) distribution function $G$ is called scale-invariant to base $10^t$ for some number $t>0$ if $$\sum_{n=-\infty}^\infty [G(nt+x+y)-G(nt+y)]$$ does not depend on the value of $y$ for all real numbers $x$ and $y$. Whittaker's Theorem 4 characterizes the scale-invariant distribution functions in terms of their characteristic functions.
As for "the distribution of the frequencies of $x_m$", I think it is impossible to attach any meaning to this phrase. However, the number-theoretic density of the first (say decimal) digit of the powers of $2$ does follow Benford's law, because $2$ is not a rational power of $10$ and hence the sequence $(\log_{10}(2^n)\mod1)$ is equidistributed on $[0,1)$ -- see e.g. Distributions known to obey Benford's law and The First Digit Problem, top paragraph on p. 525.
answered Dec 6, 2020 at 3:31
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$\begingroup$ "Because 2 is not a rational power of 10, the sequence $\{y_{n}\} =\{\log_{10}(2^n) mod1\}$ is uniformly distributed on [0,1)" First of all what does $\log_{10}(2^n) mod1$ mean? Isn't anything mod 1 equal to 0? Second why is this uniform? $\endgroup$– DebbieCommented Dec 6, 2020 at 15:44
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$\begingroup$ Also is this related to my following observation? (if so, how?): Consider a sequence of powers of 2, and write each element in this sequence in scientific notation $x_m \times 10^{n_m}$. Then create a second sequence of first significant digits of $x_m$ (denote as $d_m$). Define $y_m = \log_{10}(x_m)$ and consider it in the interval $x \in [0,1)$. Is this $y_m$ same as the sequence of $\{\log_{10}(2^n) mod1\}$? If I assume its uniform, then the probability distribution follows simply: $\endgroup$– DebbieCommented Dec 6, 2020 at 15:44
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$\begingroup$ $P(d) = Prob(d\le x < d+1) = Prob(\log_{10}(d) \le y < \log_{10}(d+1)) = \int_{\log_{10}(d)}^{\log_{10}(d+1)}1\,dy = \log_{10}(\frac{d+1}{d})$. Hence $P(d) = \log_{10}(\frac{d+1}{d})$ $\endgroup$– DebbieCommented Dec 6, 2020 at 15:44
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$\begingroup$ @user758469 : (i) For a real $x$, the standard notation $x\mod 1$ denotes the fractional part of $x$. (ii) The sequence $(\log_{10}(2^n)\mod1)=(n\log_{10}2\mod1)$ is equidistributed (not uniform!) on $[0,1)$ by the equidistribution theorem en.wikipedia.org/wiki/Equidistribution_theorem, because $\log_{10}2$ is an irrational number. (iii) Use $\{\cdot\}$ for sets and $(\cdot)$ for sequences. (iv) Dealing with the sequences in question, you cannot use probabilities, because such probabilities are undefined -- you need to use number-theoretic densities instead. $\endgroup$ Commented Dec 6, 2020 at 17:15
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