Distance From Point to Nearest Value in Series Let's say I have a point, chosen at random from the range [0, 1]. What is the average distance of this point to the nearest point in a set of n points chosen at random from the same range? Intuitively, this distance should decrease as n grows, but I'm not sure by how much.
 A: Let $X_i\sim\mathcal U(0;1)$ iid for $i\in \{0..n\}$
Let $D_{j}$ measure the distance between point $X_0$ and $X_j$.  $D_{j}=\lvert X_j-X_0\rvert$
First we find the p.density function for distance from $X_0$ to any $X_j$. Then use that to find the density function of the minimum distance.  Finally use that to obtain the expected minimum distance.
$$\begin{align}
f_{D_j}(u) & =\frac{\mathrm d}{\mathrm d u} \int_0^1 \int_{\max(0,x-u)}^{\min(1,x+u)} f_X(x)f_X(y)\operatorname dy\operatorname d x\quad \mathbf 1_{u\in (0;1)}
\\[1ex]
& =\frac{\mathrm d}{\mathrm d u}\left( \int_0^u (x+u)\operatorname d x+\int_d^{1-u}2u\operatorname d x + \int_{1-u}^1 (u+1-x)\operatorname d x\right)\quad \mathbf 1_{u\in (0;1)}
\\[1ex]
& = (2-2u)\quad \mathbf 1_{u\in (0;1)}
\\[2ex]
F_{D_j}(u)
& = \int_0^u (2-2v)\operatorname d v\quad \mathbf 1_{u\in (0;1)}
+ \mathbf 1_{u\in [1;\infty)}
\\ 
& = (2u-u^2)\; \mathbf 1_{u\in (0;1)} + \mathbf 1_{u\in [1;\infty)}
\\[2ex]
f_{\min D_j}(u) & = n f_{D_j}(u)(1-F_{D_j}(u))^{n-1}
\\[2ex]
\mathsf E(\min\limits_{j\in \{1;n\}} D_j) & = \int_0^1 n u f_{D_j}(u)(1-F_{D_j}(u))^{n-1}\operatorname d u
\end{align}$$
Can you finish?
