Let us have $k$ independent random vectors $x_1, x_2, \dots, x_k$ with uniform distribution over $ \left[0;1 \right]^n $. Then the distance (preferrably Manhattan) between an arbitrary vector $x_a$ and its nearest neighbor among these vectors is a well defined random variable. How can we estimate its momenta? Especially the expected value?

Since I have no idea how to start let us begin with the case of two vectors $a$ and $b$ in two dimensions:

In this case one is the nearest neighbor to the other, so their distance is: $$ d \left(a, b\right) = \left | a_1 - b_1 \right | + \left | a_2 - b_2 \right | $$ Where $a_1, a_2, b_1, b_2 \sim U \left[0;1 \right]$ and are independent.

In this case we can compute the expected value as a double integral: $$ I := \iint_{\left[0;1 \right]^2}^{} \left | x - y \right | + \left | x - y \right | dxdy $$

Since the integrand is symmetric along $x = y$: $$ I = 2 \iint_{\left[0;1 \right]^2}^{} \left | x - y \right | dxdy = 4 \int_{0}^{1} \left ( \int_{0}^{x} x - y dy \right ) dx = 4 \int_{0}^{1} \frac{x^2}{2} dx = \frac{2}{3} $$

It would probably not be so hard to extend this result to $n$ dimensions. The real challenge is to add other vectors to the picture.

If we have three vectors $a,b,c$, then both $b$ and $c$ have equal probability of being the closest individual to $a$. We can then compute the expected distance as the weighted sum of conditional expected values. $$ \frac{1}{2} E \left [ d \left( a, b \right) | b \text{ is closer} \right] + \frac{1}{2} E \left [ d \left( a, c \right) | c \text{ is closer} \right] $$ Both $E \left [ d \left( a, b \right) | b \text{ is closer} \right]$ and $E \left [ d \left( a, c \right) | c \text{ is closer} \right]$ could be computed as integrals in 6 dimensions somehow.

  • $\begingroup$ Do you want the nearest neighbour of (say) vector $x_1$, or the minimal distance among all vectors? $\endgroup$ – leonbloy Aug 9 '13 at 14:26
  • $\begingroup$ @leonboy The first option. The nearest neighbor to $x_1$ $\endgroup$ – Martin Drozdik Aug 9 '13 at 14:33

Consider the variable $D=|A-B|$ where $A$,$B$, and iid uniform on $[0,1]$ Then, $D$ has a triangular density $f_D(d)=2(1-d) \,[0\le d\le1]$ and characteristic function

$$\psi_D(t)=\frac{2}{t^2}(1+i\, t -e^{i t})= 2\left(\frac{1}{2!}+\frac{i\,t}{3!}+\frac{(i \, t)^2}{4!}+\frac{(i \, t)^3}{5!}+\cdots \right)$$

From this we get the first moments of $D$ : $E(D)=1/3$, $E(D^2)=1/6$ or $\sigma_D^2=1/18$, etc

Then, in $n$ dimensions, the Manhattan distance between two points is given by the sum of $n$ variables with that density. The resulting density is obtained by the $n$-th convolution of $f_D(d)$, and the CF as $\psi_D^n(t)$.

Given that, you are insterested in the minimum of $k-1$ realizations of such (independent) variables. That's well known problem.

These steps allow you to obtain the density of the nearest neighbour distance from $x_1$, and hence the moments. Of course, all this is not very straighforward to compute, but I doubt that there is a shorcut (to obtain the first moments only). Perhaps for large $n$ and/or $k$ some asymptotics can be found.

See also this related question

Update: If the number of points $k$ is big (and the dimension $n$ is not, so that that the selected point is, with high probabily, surrounded by other points, and not too near the box edges; specifically, if $k > M^n$ with $M$ , say, greater than 10 or 30) we can get an approximation by computing the volume of the "diamond" shape that corresponds to a bounded Manhattan distance (cross polytope). So, if $X$ is the Manhattan distance of the nearest neighbour, we'd get

$$P(X \le x) \approx 1 - \left(1 - x^n \frac{2^n}{n!}\right)^{k-1}, \;\; 0\le x\le x_M=\frac{(n!)^{1/n}}{2}$$

And so, for example, we could approximate $$E(X) = \int_0^\infty (1-F_X(x))\, dx \approx \int_0^{x_M} \left(1 - x^n \frac{2^n}{n!}\right)^{k-1} dx$$

  • $\begingroup$ Great job! Please, what is $x_M$ in your equations? $\endgroup$ – Martin Drozdik Aug 9 '13 at 21:11
  • 1
    $\begingroup$ Just a shortcut for what's on the right of the equation, so that the CDF ends at 1. $\endgroup$ – leonbloy Aug 9 '13 at 21:18

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