Minimum distance from uniformly distributed points Let $P=\{p_1,\dots,p_n\}$ be $n$ points on the unit square with each of their coordinates drawn from a uniform distribution on $[0,1]$.
Then we add an extra point $q$ by the same distribution and we define
$d_i=\|q-p_i\|$.
Would like to know what is the average and the minimum of the $d_i$'s. I assume that $d_i$'s are independent, so the are identically distributed. Am I right?
For the average distance we have $E[\text{Mean}(d_1,\dots,d_n)] = E[d_1]$ since they are all i.i.d. and the expectation can be found by the integral
$$\int_0^1\int_0^1\int_0^1\int_0^1\sqrt{(x_1-x_2)^2+(x_3-x_4)^2}dx_1dx_2dx_3dx_4$$
which does not have a nice closed form but its value is approximately $0.5214$.
However, I am not sure how to figure out how $Z=E[\text{Min}(d_1,\dots,d_n)]$ depends on $n$.
I would expect that asymptotically it behaves like $c/n$. I say this because the probability that $Z<r$ can be bounded from above (unless I am wrong) by the area of $n$ disks with radius $r$.
When I generated numbers to test this, it seems that the asymptotic behaviour is closer to $1/n^{1/2}$ than $c/n$. Is there any straightforward way to get this exponent?
 A: What follows is really a heuristic rather than rigorous proof, but hopefully provides some intuition.
Consider first the case where $q$ is fixed and is given to be $q_0 = (1/2,1/2)$, so is in the middle of the box. For a fixed radius $r > 0$ let $D_r^{(q_0)} = \left\{x \, \colon \, |x-q_0| < r\right\}$ denote the disk of radius $r$, and define
$$N_r^{(q_0)} = \# \left\{ x_i \, \colon \, x_i \in D_r^{(q_0)}\right\},$$
the number of the (random) points that fall inside the disk. The probability that any given point is inside the disk is equal to 
$$\mathbf P[ x_i \in D_r] = \frac{ \text{Area}( D_r \cap [0,1]^2 )}{\text{Area}([0,1]^2)}$$
for small enough values of $r$, $D_r \subset [0,1]^2$, and the formula above becomes
$$ \mathbf P[x_i \in D_r] = \pi r^2.$$
In particular it follows that for sufficiently small $r$, $N_r^{(q_0)}$ follows a binomial distribution with success probability $\pi r^2$. This is well defined so long as $r < 1/\sqrt{\pi}$.
Now we turn to the random variable you were interested in (with the caveat that we have fixed $q$), which we denote $M_n^{(q)}$
$$M_n^{(q_0)} = \min_{i = 1,\ldots,n} \{|x_i - q|\}$$
We can relate this to the random variable $N_r^{(q)}$ by the formula
$$
\mathbf P[ M_n^{(q_0)} \leq r] = \mathbf P[ N_r^{(q_0)} > 0]$$
That is: there is a point within a distance $r$ of $q_0$ (i.e. $M_n^{(q_0)} \leq r$), if and only if there is at least one point inside the disk $D_r^{(q_0)}$, which is to say $N_r^{(q_0)} > 0$.
So (under the assumption that $r$ is small) we have
$$
\begin{aligned}
\mathbf P[ M_n^{(q_0)} \leq r] &= \mathbf P[\text{Bin}(n, \pi r^2) > 0] \\
& = 1 - (1-\pi r^2)^n
\end{aligned}
$$
That is, we have the CDF for the variable $M_n^{(q_0)}$, from which we can derive the PDF by differentiation
$$
\begin{aligned}
f_n^{(q_0)}(r)& = \frac{d}{dr}\mathbf P[ M_n^{(q_0)} \leq r] \\
&= 2 \pi n r(1 - \pi r^2)^{n-1}
\end{aligned}
$$
which we define on the range $[0, 1/\sqrt{\pi}]$. Then we can calculate the expected value
$$
\mathbf E[M_n^{(q_0)}] = \int_0^{1/\sqrt{\pi}} r f_n^{(q)}(r)dr = \frac{n}{2} \frac{\Gamma(n)}{\Gamma(n+3/2)}$$
(I admit, that I used wolframalpha to derive this. We can now evaluate the Taylor series at infinity to get the asymptotic formula for this, which indeed is $O(n^{-\frac12})$, (again, I had to rely on wolframalpha for this).
So we have a heuristic for why given a fixed $q$ away from the boundary of the box we might expect the minimum distance to be of the order $n^{-\frac12}$.
Extending this to the case where $q$ is not fixed is now quite (heuristicallly!) simple by conditioning on q:
$$ 
\begin{aligned}
\mathbf E[ M_n] & = \int_{[0,1]^2} \mathbf E[ M_n^{(\rho)} ] \mathbf P[q = \rho] d \rho \\
& = \int_{[0,1]^2} \mathbf E[ M_n^{(\rho)} ]d\rho
\end{aligned}
$$
For points $\rho$ away from the boundary, this expectation is independent of $\rho$, and since the vast majority of points are away from the boundary we have
$$
\begin{aligned}
\mathbf E[ M_n] &\sim \int_{[0,1]^2} \mathbf E[ M_n^{(q_0)} ]d\rho \\
& = \mathbf E[ M_n^{(q_0)} ] \\
& = \frac{n}{2} \frac{\Gamma(n)}{\Gamma(n+3/2)}
\end{aligned}
$$
Again, this is all very heuristic, but hopefully gives you some intuition!
A: An exact formula for finite values is probably too difficult.
An asymptotic for large $n$ can be obtained by noticing that it turns equivalent to a 2d Point Process with density $I=n/A=n$. In this case, the amount of points in a (small) circle of radius $r$ follows a Poisson distribution with $\lambda = I \pi r^2 = n \pi r^2$.
Then, for an additional point placed in the square (disregarding border effects, which should be asymptotically negligible), calling $D$ its distance from the nearest point, we have
$$P(D\ge d) = \exp (- n \pi d^2) \tag{1}$$
(the value of the Poisson probability evaluated at zero).
Hence the density is $$f_D(d) =  2  \pi \, n \,  d \, \exp (- n \pi d^2) \tag{2}$$
which is a Rayleigh distribution . And its mean is
$$E[D]=\frac{1}{2\sqrt{n}} \tag{3}$$
Alternatively (perhaps slightly more precise here, given that $n$ is fixed, but also a little more clumsy, and asymptotically equivalent), is to use a Binomial instead of a Poisson, so that $(1)$ turns into $P(D\ge d) = (1-  \pi d^2)^n $ and we get owen88's answer. 
Then, because $\frac{\Gamma(n)}{\Gamma(n+a)} \to n^{-a}$ we get the same asymptotic mean as in $(3)$.
A: See D. Moltchanov Distance distributions in random networks (Ad Hoc Networks 10(6), 2012).
They give the distribution to the $n^{\text{th}}$ nearest neighbour of $q$ as a Gamma distribution.
