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Let $X_1, \ldots, X_n$ be independent binary uniform random variable taking values in $\{0,1\}$. For $a>1$, we define the random variable $Y_n$ as $$Y_n := \sum_{i=1}^n X_i \frac{a^i}{a^{n+1}-1}. $$ We denote the distribution subject to $Y_n$ by $p_n$. I would like to know the limiting distribution $\lim_{n\to \infty} p_n$.

Indeed, when $a=2$, $(2^{n+1}-1)Y_n$ is subject to the uniform distribution on the set $\{0,1, \ldots, 2^{n+1}-1\}$. Hence, the limiting distribution $\lim_{n\to \infty} p_n$ is the uniform distribution on the interval $[0,1]$. However, it is not so easy to solve this for general $a>1$.
More precisely, I would like to know the probability density function of the limiting distribution when it has the limiting distribution. In particular, I would like to know the answer when $1<a<2$.

Since $a^{n+1}-1$ is close to $a^{n+1}$, we can consider the following variable instead of $Y_n$; $$Z_n := \sum_{i=1}^n X_i a^{i-n-1}=\sum_{i=1}^n X_i a^{-i}. $$

In fact, I would like to know the form of the derivative of the probability density function.

Notice that the support of $Z_n$ is contained in $[0,1/(a-1)]$. Hence, the the support of the limiting distribution is also contained in $[0,1/(a-1)]$. I am interested in the value of the probability density function at the boundary $0$ and $1/(a-1)$.

Masahito Hayashi
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  • Do the $X_i$ have the same distribution? – angryavian Jun 27 '21 at 05:16
  • I forgot to say that $X_i$ is subject to the binary uniform distribution. – Masahito Hayashi Jun 27 '21 at 05:21
  • To Brian Moehring. You are right. Now, I fixed the typo. – Masahito Hayashi Jun 27 '21 at 06:31
  • Random note from my attempt to answer: Your example of $a={\color{red}2}$ is equivalent to the limit $$\prod_{k=0}^{n-1}\cos\left(\frac{{\color{red}2}^kt}{{\color{red}2}^n-1}\right) \to \frac{\sin(t)}{t}$$ (which I'm not sure I've seen before). The general case would effectively require you to find $$\lim_{n\to\infty}\prod_{k=0}^{n-1}\cos\left(\frac{a^kt}{a^n-1}\right).$$ – Brian Moehring Jun 27 '21 at 07:00
  • To Brian Moehring. You are right. Your method was used in https://math.stackexchange.com/questions/1830354/series-of-independent-bernoulli-variables Hence, the case with a=2 is easy, and the case with 1<a<2 is more difficult. – Masahito Hayashi Jun 27 '21 at 07:07
  • Since $a^{n+1}-1$ is close to $a^{n+1}$, we ca replace $a^{n+1}-1$ by $a^{n+1}$. – Masahito Hayashi Jun 27 '21 at 07:09
  • Ahhh, that last comment does change my initial product (with the 2's) into a simpler form, and I do think I remember that one now. I'm certainly not seeing a way to resolve the infinite product for any other value of $a>1$ though. May I ask how you came upon this problem? As in, did it come up within another problem, or are you interested in this for its own sake? – Brian Moehring Jun 27 '21 at 07:19
  • To Brian Moehring. This problem is related to the weight distribution of an error-correcting code. Do you know an error-correcting code, which is mathematically given as a vector space over finite field $F_2$? – Masahito Hayashi Jun 27 '21 at 07:28
  • @NN2 Thank you for your comment. The support of a normal distribution is the set of real numbers, i.e., the non-compact set. However, the support of $p_n$ is contained in $[0,1]$. Hence, the limit is not a normal distribution. – Masahito Hayashi Jun 27 '21 at 11:23
  • @NN2 The support of $p_n$ is contained in $[0,1/(a-1)]$ for any $n$. Hence, the support of the limiting distribution is contained in $[0,1/(a-1)]$. – Masahito Hayashi Jun 27 '21 at 11:58

1 Answers1

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If $a$ is an integer, then I think $Y_n$ is uniform among "$n$-digit decimals" in base $a$ consisting of only $0$ and $1$.

  • When $a=2$, this becomes uniform on $[0,1]$ since all numbers in this interval have a binary representation with only $0$s and $1$s.
  • When $a=3$, numbers that have a $2$ in their ternary representation (like $0.1021$) are excluded. The set of numbers without any $2$s in their ternary representation resembles the Cantor set (the Cantor set consists of numbers whose ternary expansion does not have any $1$s).
  • I think this generalizes further similarly for other integers $a$.
angryavian
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