What is the standard deviation of the distribution

Question

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 $I_1$

Answer 1

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 $0.1659425$.

Answer 2

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.

set.seed(2023)
limitpoint <- function(n){ sum(runif(n) / 2^(1:n)) }
sims <- replicate(10^6, limitpoint(64))
mean(sims)
# 0.5003589
1/var(sims)
# 35.99841
curve(dbeta(x,4,4), col="lightgrey")
lines(density(sims), col="black")