Standard Deviation of Nested Halving Intervals
Start with the interval $I_0 = [0, 1]$. At each step $i$, you pick a new interval $I_i$ uniformly at random inside $I_{i-1}$, where $I_i$ has exactly half the length of $I_{i-1}$ (so $I_i$ has length $2^{-i}$).
The nested sequence $I_0 \supset I_1 \supset I_2 \supset \cdots$ converges to a single point $X \in [0,1]$.
What is the standard deviation of $X$?
Hints
- Try expressing the limit point $X$ as a sum of independent random contributions, one from each step of the nesting process.
- Each step contributes an independent $\text{Uniform}(0,1)$ variable scaled by $2^{-k}$. Since they are independent, the variance of the sum is the sum of the variances.
- You need $\sum_{k=1}^{\infty} \frac{1}{12} \cdot 4^{-k}$. This is a geometric series with ratio $1/4$ -- evaluate it and take the square root.
Worked Solution
How to Think About It: Each step randomly picks where a half-length sub-interval sits inside the current interval. The limit point $X$ accumulates all that randomness. The key structural insight is that $X$ decomposes as an infinite sum of independent uniform contributions, each scaled down by a factor of $2^{-k}$. That makes the variance a geometric series -- quick to compute. Before doing any algebra, notice that $X$ should have mean $1/2$ by symmetry (every step is uniform and centered). The standard deviation should be less than $1/\sqrt{12} \approx 0.289$ (the std dev of a single $U(0,1)$), because the nested structure constrains $X$ more than a single uniform draw would.
Quick Estimate: The first interval $I_1$ has length $1/2$ and is placed uniformly in $[0,1]$, so its midpoint is $U(1/4, 3/4)$ -- that alone gives standard deviation $1/(2\sqrt{12}) \approx 0.144$. Each subsequent step adds less and less variance (geometrically decaying). The total variance is roughly $1/12$ times $\sum 4^{-k}$, which is $1/12 \times 1/3 = 1/36$. So $\sigma \approx \sqrt{1/36} = 1/6 \approx 0.167$. That is slightly more than the first-step-only estimate, which makes sense -- later steps add a small amount of additional spread.
Approach: Represent the limit point $X$ as an infinite sum of independent uniform random variables, then compute the variance term by term.
Formal Solution:
At step $k$, the interval $I_k$ has length $2^{-k}$ and is placed uniformly inside $I_{k-1}$ (which has length $2^{-(k-1)}$). The left endpoint of $I_k$ relative to the left endpoint of $I_{k-1}$ is uniform on $[0, 2^{-(k-1)} - 2^{-k}] = [0, 2^{-k}]$.
Define $U_k \sim \text{Uniform}(0,1)$ independently for each $k \geq 1$. Then the left endpoint of $I_k$ shifts by $U_k \cdot 2^{-k}$ relative to $I_{k-1}$. As the interval lengths shrink to zero, the limit point is:
$$X = \sum_{k=1}^{\infty} U_k \cdot 2^{-k}$$
Since the $U_k$ are independent, the variance of $X$ is:
$$\text{Var}(X) = \sum_{k=1}^{\infty} \text{Var}(U_k) \cdot (2^{-k})^2 = \sum_{k=1}^{\infty} \frac{1}{12} \cdot 4^{-k}$$
This is a geometric series:
$$\text{Var}(X) = \frac{1}{12} \cdot \frac{1/4}{1 - 1/4} = \frac{1}{12} \cdot \frac{1}{3} = \frac{1}{36}$$
The standard deviation is:
$$\sigma = \sqrt{\text{Var}(X)} = \sqrt{\frac{1}{36}} = \frac{1}{6}$$
Answer: The standard deviation of the limit point is $\sigma = 1/6$.
Intuition
The core pattern here is decomposing a complex random object into a sum of independent pieces at different scales. Each nesting step contributes a tiny independent uniform jitter, and because the contributions are independent the variances just add up. The geometric decay ($4^{-k}$ for variance, since length scales as $2^{-k}$ and variance goes as the square of the scale) means most of the randomness comes from the first few steps -- later steps are negligible. This is the same structure you see in random binary expansions, Polya urn limits, and many fractal constructions.
In practice, this kind of hierarchical decomposition shows up whenever you model a quantity that is built from many layers of independent randomness at shrinking scales -- think volatility clustering in markets, or multi-scale noise in signal processing. The lesson: when a process has a natural scale hierarchy, decompose at each scale, compute variance per scale, and sum. The answer is usually dominated by the coarsest scale.