Suppose that you have a uniform distribution in the Interval $I_0$

where $$ I_0 \in [0,1] $$

With this as a starting interval, now you take another interval $I_1$ which is a subset of $I_0$ but is exactly half the length. You repeat this multiple times.

So if your interval $m = n-1$, then

$$ I_n \subset I_m $$

and the length of interval $n$ is half to that of interval $m$

All the intervals are continuous.

Question -> If $n$ tends to infinity, the interval will converge on a point. Now if you repeat this experiment infinitely many times, you will get infinite such points, which will form a distribution. What is the standard deviation of that distribution?

Note - Since $I_0$ is of length 1, $I_1$ needs to be of length 0.5. So $I_1$ cannot start from (0.5,1], as that would mean that $I_1$ will not be subset of $I_0$. All the possible selections of $I_1$ are equally likely. So, choosing [0.25,0.75], [0.2,0.7],[0,0.5] etc are all equally likely candidates for $I_1$

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    $\begingroup$ Your question is very unclear. How is the interval $I_1$ picked? By ‘continious intervals’ I assume you mean connected? What do you mean with repeat experiment infinity times? Is what you mean that for each $I_n$ we draw an observation $i_n$ from an uniform distribution on $I_n$? If so then like you noted $I_n$ converges to a point and hence limit of the sd is 0. $\endgroup$
    – ScapeProf
    Jul 29, 2023 at 10:55
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    $\begingroup$ @ScapeProf I1 is subset of I0 with half the length. So, one example of I0 can be [0,0.5], another can be [0.2,0.7]. It is a uniform distribution. I am not sure what you mean by connected. No, what I mean is that when n tends to infinity, our interval will become so small that it converges at a point. For one such point the sd is 0 as you correctly pointed out. But if you repeat this experiment infinity times, you will get a distribution. The question is, what is the sd of that distribution $\endgroup$ Jul 29, 2023 at 15:47
  • $\begingroup$ @ScapeProf apologies if question wasn't clear. Have modified it a bit. Hope its clear now :) $\endgroup$ Jul 29, 2023 at 15:48
  • $\begingroup$ @SagarChand, in your context, you can think of connected as having no gaps in your set. For example, $[0,1)\cup(1,2]$ is not connected. However, you do not need to say it is connected because the definition of interval already asks for that. $\endgroup$ Jul 29, 2023 at 15:56
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    $\begingroup$ @P.S.Dester, no 0.9 cannot be starting for I1, as length of I1 needs to be 0.5. So that is not a valid selection. $\endgroup$ Jul 30, 2023 at 2:43

2 Answers 2


Denote $I_n = [L_n, L_n+1/2^n]$ (expressing the interval this way is valid, since $I_n$ should have length $1/2^n$).

If I have understood your construction procedure correctly, the left endpoint of interval $I_{n+1}$ should follow a uniform distribution on $[L_n, L_n + 1/2^{n+1}]$. That is, at each step of the recursion we draw the left endpoint of the interval, which determines entirely the value of the right endpoint. In this case, we can find the variance of the limit.

Let $(U_n)_{n\in\mathbb{N}}$ be a sequence of i.i.d. uniform variables on the interval $[0,1]$. Using the above information, it is clear that the following equality in distribution should hold: $$ L_n \overset{d}{=} \sum_{k = 1}^n \frac{U_k}{2^k}. $$

Denote the limit $L = \lim_n L_n$. Then we can compute the variance of $L$ as $$ V(L) = \sum_{n = 1}^\infty V\left(\frac{U_n}{2^n}\right) = \sum_{n = 1}^\infty \frac{V(U_n)}{4^n} = \frac{1}{12}\sum_{n=1}^\infty \frac{1}{4^n}, $$ where we used the independence of the $U_n$'s and the fact that the variance of a uniformly distributed variable on $[0,1]$ is $1/12$. Computing the geometric series above, we deduce that $$ \text{sd}(L) = \frac{1}{6}. $$ This is in line with my Monte Carlo estimate of $0.1659425$.

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    $\begingroup$ If one wants to be a little more rigorous, the interchange of variance with the infinite sum can be justified by appealing to the monotone convergence theorem. $\endgroup$
    – Bajas
    Jul 29, 2023 at 17:29
  • $\begingroup$ Would you care to explain why you think that? If we draw $L_{n+1}$ as described above we should have $L_{n+1} \in [L_n, L_n + 1/2^{n+1}]$ and hence $L_{n+1} + 1/2^{n+1}\in [L_n+1/2^{n+1}, L_n + 1/2^{n}]$. Thus $I_{n+1} = [L_{n+1}, L_{n+1} + 1/2^{n+1}] \subseteq I_n$ as claimed. $\endgroup$
    – Bajas
    Jul 29, 2023 at 18:00
  • $\begingroup$ Read the interval I am drawing from again. The second draw is incompatible with my procedure. According to me, $L_2$ should be drawn from $[0.05, 0.05 +1/2^2]$, not $[0.05, 0.05 +1/2^1]$ as you did. $\endgroup$
    – Bajas
    Jul 29, 2023 at 18:17

Too long for a comment

The chart below shows a simulation using R of the density of the distribution (even with a million samples, there is a little simulation noise leading to some small bumps in the density, which can be ignored).

It would be tempting, based on the obvious symmetry, a mean of $\frac12$, standard deviation of $\frac16$ i.e. a variance of $\frac1{36}$ (as shown in Bajas's excellent answer) to guess that you might have a $\text{Beta}(4,4)$ distribution since that would give this mean and variance. That density is also illustrated (in light grey) and it is clear that the density we are considering for this question (in black) has a lower peak density and smaller tails and is more platykurtic.

limitpoint <- function(n){ sum(runif(n) / 2^(1:n)) }
sims <- replicate(10^6, limitpoint(64))
# 0.5003589
# 35.99841
curve(dbeta(x,4,4), col="lightgrey")
lines(density(sims), col="black")


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    $\begingroup$ Actually, it is possible to show that the pdf is a bump function that satisfies $f'(t) = 4 f(2t)$, $t\in[0,1/2]$. $\endgroup$ Aug 10, 2023 at 1:10
  • $\begingroup$ Also, the characteristic function is given by this post. $\endgroup$ Aug 10, 2023 at 1:14
  • $\begingroup$ @P.S.Dester - I had not spotted the derivative point, but I suspected $f(x)=2-f(|x-\frac12|)$ which seems to be implied by your derivative equality plus the overall symmetry $\endgroup$
    – Henry
    Aug 10, 2023 at 1:49

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