Probability distribution between two unit vectors using the taxicab metric for distance

Question

Suppose we have two positive, real unit vectors $X$ and $Y$ in $\mathbb{R}^n$.

EDIT: As was suggested in a comment, let me describe how $X$ and $Y$ are randomly generated. For both vectors, pick $n$ random values from the uniform distribution $[0,1]$. Then normalize the vector by the taxicab metric (i.e. all the entries sum to $1$).

By the taxicab metric, the distance between these vectors is

$$d(X,Y)=\sum_{i=1}^n |x_i-y_i|.$$

Now, if we perform this operation for thousands of random vectors, we get histograms that look like

$n=2$">

$n=3$">

$n=4$">

which correspond to $n=2,3,4$. First, it is easy to prove that $d(X,Y)\leq 2$. Second, this idea can be extended to other $Lp$ metrics (such as the standard Euclidean metric) by defining

$$d(X,Y)=\sqrt[p]{\sum_{i=1}^n |x_i-y_i|^p}$$

which is interesting but not something I am currently investigating. I am trying to find an analytic form for the histograms above in terms of $n$. Obviously, this function will be $0$ at $0$ and $2$, and the maximum seems to be approaching $2/3$ as $n$ goes to infinity. Unfortunately, I have not had any luck with this problem as it seems to go through several nasty integrals involving absolute values. Any helps, hints, or general terms to look up would be appreciated.

Answer 1

As $n \to \infty$, the histogram will become more and more concentrated at $\frac{2}{3}$. For example, $Pr(|d(X,Y)-\frac{2}{3}| > \epsilon)$ goes to $0$ as $n \to \infty$ for any fixed $\epsilon > 0$. By analyzing the proof below, you could say something stronger (e.g. $Pr(|d(X,Y)-\frac{2}{3}| > \frac{1}{\log N})$ goes to $0$, idk).

For ease of notation, have $x_1,\dots,x_n,y_1,\dots,y_n$ be the points uniformly chosen from $[0,1]$, and then $$d(X,Y) = \sum_{i=1}^n \left|\frac{x_i}{x_1+\dots+x_n}-\frac{y_i}{y_1+\dots+y_n}\right|.$$ The main observation is that $x_1+\dots+x_n$ and $y_1+\dots+y_n$ will be really close to $\frac{1}{2}n$. We have $$\sum_{i=1}^n \left|\frac{x_i}{x_1+\dots+x_n}-\frac{x_i}{\frac{1}{2}n}\right| \le 2\frac{|\frac{1}{2}n-(x_1+\dots+x_n)|}{x_1+\dots+x_n},$$ which is $o(1)$ with probability $1-o(1)$ (this comes from, e.g., $Pr\left(|x_1+\dots+x_n-\frac{1}{2}n| > \frac{n}{\log n}\right) = o(1)$). Similarly with the $y_i$'s. Therefore, $d(X,Y)$ is basically $\frac{2}{n}\sum_{i=1}^n \left|x_i-y_i\right|$. And a concentration inequality again says this is very close to $\frac{2}{3}$ with overwhelming probability, since the expected value of $|x_i-y_i|$ is $\int_0^1\int_0^1 |x-y|dxdy = \frac{1}{3}$.