Clusters of uniformly distributed random points This can be viewed as a toy version of this wonderful question. (I'm removing the TSP component, and I'll zoom in on the statistical part.)
Let $X_1,\ldots, X_n$ be iid, with uniform distribution in $[0,1]$. Let $\ell_n$ be the length of the shortest subinterval of $[0,1]$ that contains $n^{\alpha}$ of the points $X_j$, with $0<\alpha<1$. What can we say about the distribution of the normalized length $L_n=n^{1-\alpha}\ell_n$? For example, what are the asymptotics of $EL_n$?
As a warm-up of sorts, we can in fact also consider the shortest interval with $cn$ points and $L_n=\ell_n/c\,$; the answers to the linked question seem to suggest that $EL_n\to 1$ in this case.
(This sounds like it should be well studied, but I couldn't locate anything on the internet.)
 A: As in the $2$-dimensional problem linked, the probability that a particular interval much shorter than $n^{\alpha-1}$ contains $n^\alpha$ points is very small, and we can use the union bound over a small set of such intervals to bound the probability that the shortest interval containing $n^\alpha$ points is small. This implies $EL_n \to 1$ and more precise statements about the difference with $1$.
Suppose we want to bound the probability that $L_n \gt 1-2\epsilon$. Consider intervals of length $(1-\epsilon)n^{\alpha-1}$ starting at $0, \epsilon n^{\alpha-1}, ...$. There are fewer than $n^{1-\alpha}/\epsilon$ such intervals and every interval of length $(1-2\epsilon)n^{\alpha-1}$ fits into one of these. Let us bound the probability that there are $n^\alpha$ points in a particular interval of length $(1-\epsilon)n^{\alpha-1}$ and then use the union bound.
The number of points in an interval of length $(1-\epsilon)n^{\alpha-1}$ is a binomial random variable with mean $(1-\epsilon)n^\alpha$ and standard deviation under $\sqrt{n^\alpha}$. So, having $\epsilon n^\alpha$ points more than average is more than $\epsilon n^{\alpha/2}$ standard deviations above the mean. A normal approximation would suggest that the probability drops rapidly as $n$ increases. To be rigorous we can use a Chernoff bound. For $0 \lt \delta \lt 1,$
$\textrm{Prob}[\textrm{Binom} \ge (1+\delta)\mu] \le \exp(-\mu \delta^2/3).$
We can choose $\delta = \epsilon$, since $(1+\epsilon) \mu = (1+\epsilon)(1-\epsilon)n^\alpha = (1-\epsilon^2)n^\alpha \lt n^\alpha$.
$\textrm{Prob}[\textrm{Binom} \ge n^\alpha]  \le \exp (-c_\epsilon n^\alpha).$
So, the probability that the shortest interval containing $n^\alpha$ points is shorter than $(1-2\epsilon)n^{\alpha-1}$ is at most $\frac{n^{1-\alpha}}{\epsilon}\exp (-c_\epsilon n^\alpha).$ As $n\to\infty$ the exponential term dominates and the probability drops to $0$. As $n\to \infty$, $L_n$ is close to $1$ with high probability.  
A: This 1965 paper of Naus contains formulas for the exact probabilities.  Turning them into the asymptotic results you want might take a little effort but since Naus' paper is cited in at least 255 places there is fair chance someone did the work already.
